1. Ratio : The ratio of two qualities a and b in the same units, is the fraction a/b and we write it as a:b. In the ratio, a:b, we call ‘a’ as the first term of antecedent and b, the second term consequent.
Ex: The ratio 5:9 represents 5/9 with antecedent=5 ,consequent=9
3. Rule: The multiplication or division of each term of 9 ratio by the same non-zero number does not affect the ratio.
4. Proportion: The equality of two ratios is called proportion. If a:b=c:d, we write a:b::c:d and we say that a,b,c and d are in proportion. Here a and b are called extremes, while b and c are called mean terms.
Product of means=product of extremes
Thus, a:b::c:d => (b*c)=(a*d)
5. Fourth proportional: If a:b::c:d, then d is called the fourth proportional to a,b and c.
6. Third proportional: If a:b::b:c, then c is called third proportional to a and b.
7. Mean proportional: Mean proportional between a and b is SQRT(a*b).
COMPARISION OF RATIOS:
We say that (a:b)>(c:d) => (a/b)>(c/d)
8.Compounded ratio: The compounded ratio of the ratios (a:b), (c:d),(e:f) is (ace:bdf).
9.Duplicate Ratio: If (a:b) is (a^2: b^2 )
10.Sub-duplicate ratio of (a:b) is (SQRT(a):SQRT(b))
11.Triplicate ratio of (a:b) is (a3: b3 )
12.Sub-triplicate ratio of (a:b) is (a^1/3: b^1/3 ).
13.If a/b=c/d, then (a+b)/(a-b)=(c+d)/(c-d) (componend o and dividend o)
VARIATION:
14. We say that x is directly proportional to y, if x=ky for some constant k and we write.
15. We say that x is inversely proportional to y, if xy=k for some constant and we write.
16. X is inversely proportional to y.
If a/b=c/d=e/f=g/h=k then
k=(a+c+e+g)/(b+d+f+h)
If a1/b1,a2/b2, a3/b3..............an/bn are unequal
fractions then the ratio.
Showing posts with label aptitude concepts. Show all posts
Showing posts with label aptitude concepts. Show all posts
Sunday, 6 December 2009
Partnership
Important Facts:
Partnership:When two or more than two persons run a business
jointly, they are called partners and the deal is known as partnership.
Ratio of Division of Gains:
1.When the investments of all the partners are do the same time, the
gain or loss is distributed among the partners in the ratio of their
investments.
Suppose A and B invest Rs x and Rs y respectively for a year in a
business, then at the end of the year:
(A's share of profit):(B's share of profit)=x:y
2.When investments are for different time periods, then equivalent
capitals are calculated for a unit of time by taking (capital*number
of units of time). Now gain or loss is divided in the ratio of these
capitals.
Suppose A invests Rs x for p months and B invests Rs y for q months,
then (A's share of profit):(B's share of profit)=xp:yq
3.Working and sleeping partners:A partner who manages the business is
known as working partner and the one who simply invests the money is
a sleeping partner.
Formulae
1.When investments of A and B are Rs x and Rs y for a year in a
business ,then at the end of the year
(A's share of profit):(B's share of profit)=x:y
2.When A invests Rs x for p months and B invests Rs y for q months,
then A's share profit:B's share of profit=xp:yq
Partnership:When two or more than two persons run a business
jointly, they are called partners and the deal is known as partnership.
Ratio of Division of Gains:
1.When the investments of all the partners are do the same time, the
gain or loss is distributed among the partners in the ratio of their
investments.
Suppose A and B invest Rs x and Rs y respectively for a year in a
business, then at the end of the year:
(A's share of profit):(B's share of profit)=x:y
2.When investments are for different time periods, then equivalent
capitals are calculated for a unit of time by taking (capital*number
of units of time). Now gain or loss is divided in the ratio of these
capitals.
Suppose A invests Rs x for p months and B invests Rs y for q months,
then (A's share of profit):(B's share of profit)=xp:yq
3.Working and sleeping partners:A partner who manages the business is
known as working partner and the one who simply invests the money is
a sleeping partner.
Formulae
1.When investments of A and B are Rs x and Rs y for a year in a
business ,then at the end of the year
(A's share of profit):(B's share of profit)=x:y
2.When A invests Rs x for p months and B invests Rs y for q months,
then A's share profit:B's share of profit=xp:yq
Time and Work Problems
1)If 9 men working 6 hours a day can do a work in 88 days. Then 6 men
working 8 hours a day can do it in how many days?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so (9*6*88/1)=(6*8*d/1)
on solving, d=99 days.
2)If 34 men completed 2/5th of a work in 8 days working 9 hours a day.
How many more man should be engaged to finish the rest of the work in
6 days working 9 hours a day?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so, (34*8*9/(2/5))=(x*6*9/(3/5))
so x=136 men
number of men to be added to finish the work=136-34=102 men
3)If 5 women or 8 girls can do a work in 84 days. In how many days can
10 women and 5 girls can do the same work?
Sol: Given that 5 women is equal to 8 girls to complete a work
so, 10 women=16 girls.
Therefore 10women +5girls=16girls+5girls=21girls.
8 girls can do a work in 84 days
then 21 girls ---------------?
answer= (8*84/21)=32days.
Therefore 10 women and 5 girls can a work in 32days
4)Worker A takes 8 hours to do a job. Worker B takes 10hours to do the
same job. How long it take both A & B, working together but independently,
to do the same job?
Sol: A's one hour work=1/8.
B's one hour work=1/10
(A+B)'s one hour work=1/8+1/10 =9/40
Both A & B can finish the work in 40/9 days
5)A can finish a work in 18 days and B can do the same work in half the
time taken by A. Then, working together, what part of the same work they
can finish in a day?
Sol: Given that B alone can complete the same work in days=half the time
taken by A=9days
A's one day work=1/18
B's one day work=1/9
(A+B)'s one day work=1/18+1/9=1/6
6)A is twice as good a workman as B and together they finish a piece of
work in 18 days.In how many days will A alone finish the work.
Sol: if A takes x days to do a work then
B takes 2x days to do the same work
=>1/x+1/2x=1/18
=>3/2x=1/18
=>x=27 days.
Hence, A alone can finish the work in 27 days.
7)A can do a certain work in 12 days. B is 60% more efficient than A. How
many days does B alone take to do the same job?
Sol: Ratio of time taken by A&B=160:100 =8:5
Suppose B alone takes x days to do the job.
Then, 8:5::12:x
=> 8x=5*12
=> x=15/2 days.
8)A can do a piece of work n 7 days of 9 hours each and B alone can do it
in 6 days of 7 hours each. How long will they take to do it working together
8 2/5 hours a day?
Sol: A can complete the work in (7*9)=63 days
B can complete the work in (6*7)=42 days
=> A's one hour's work=1/63 and
B's one hour work=1/42
(A+B)'s one hour work=1/63+1/42=5/126
Therefore, Both can finish the work in 126/5 hours.
Number of days of 8 2/5 hours each=(126*5/(5*42))=3days
9)A takes twice as much time as B or thrice as much time to finish a piece
of work. Working together they can finish the work in 2 days. B can do the
work alone in ?
Sol: Suppose A,B and C take x,x/2 and x/3 hours respectively finish the
work then 1/x+2/x+3/x=1/2
=> 6/x=1/2
=>x=12
So, B takes 6 hours to finish the work.
10)X can do ¼ of a work in 10 days, Y can do 40% of work in 40 days and Z
can do 1/3 of work in 13 days. Who will complete the work first?
Sol: Whole work will be done by X in 10*4=40 days.
Whole work will be done by Y in (40*100/40)=100 days.
Whole work will be done by Z in (13*3)=39 days
Therefore,Z will complete the work first.
Complex Problems
1)A and B undertake to do a piece of workfor Rs 600.A alone can do it in
6 days while B alone can do it in 8 days. With the help of C, they can finish
it in 3 days, Find the share of each?
Sol: C's one day's work=(1/3)-(1/6+1/8)=1/24
Therefore, A:B:C= Ratio of their one day’s work=1/6:1/8:1/24=4:3:1
A's share=Rs (600*4/8)=300
B's share= Rs (600*3/8)=225
C's share=Rs[600-(300+225)]=Rs 75
2)A can do a piece of work in 80 days. He works at it for 10 days & then B alone
finishes the remaining work in 42 days. In how much time will A and B, working
together, finish the work?
Sol: Work done by A in 10 days=10/80=1/8
Remaining work=(1-(1/8))=7/8
Now, work will be done by B in 42 days.
Whole work will be done by B in (42*8/7)=48 days
Therefore, A's one day's work=1/80
B’s one day's work=1/48
(A+B)'s one day's work=1/80+1/48=8/240=1/30
Hence, both will finish the work in 30 days.
3)P,Q and R are three typists who working simultaneously can type 216 pages
in 4 hours In one hour , R can type as many pages more than Q as Q can type more
than P. During a period of five hours, R can type as many pages as P can
during seven hours. How many pages does each of them type per hour?
Sol:Let the number of pages typed in one hour by P, Q and R be x,y and z
respectively
Then x+y+z=216/4=54 ---------------1
z-y=y-x => 2y=x+z -----------2
5z=7x => x=5x/7 ---------------3
Solving 1,2 and 3 we get x=15,y=18, and z=21
4)Ronald and Elan are working on an assignment. Ronald takes 6 hours to
type 32 pages on a computer, while Elan takes 5 hours to type 40 pages.
How much time will they take, working together on two different computers
to type an assignment of 110 pages?
Sol: Number of pages typed by Ronald in one hour=32/6=16/3
Number of pages typed by Elan in one hour=40/5=8
Number of pages typed by both in one hour=((16/3)+8)=40/3
Time taken by both to type 110 pages=110*3/40=8 hours.
5)Two workers A and B are engaged to do a work. A working alone takes 8 hours
more to complete the job than if both working together. If B worked alone,
he would need 4 1/2 hours more to compete the job than they both working
together. What time would they take to do the work together.
Sol: (1/(x+8))+(1/(x+(9/2)))=1/x
=>(1/(x+8))+(2/(2x+9))=1/x
=> x(4x+25)=(x+8)(2x+9)
=> 2x2 =72
=> x2 = 36
=> x=6
Therefore, A and B together can do the work in 6 days.
6)A and B can do a work in12 days, B and C in 15 days, C and A in 20 days.
If A,B and C work together, they will complete the work in how many days?
Sol: (A+B)'s one day's work=1/12;
(B+C)'s one day's work=1/15;
(A+C)'s one day's work=1/20;
Adding we get 2(A+B+C)'s one day's work=1/12+1/15+1/20=12/60=1/5
(A+B+C)'s one day work=1/10
So, A,B,and C together can complete the work in 10 days.
7)A and B can do a work in 8 days, B and C can do the same wor in 12 days.
A,B and C together can finish it in 6 days. A and C together will do it in
how many days?
Sol: (A+B+C)'s one day's work=1/6;
(A+B)'s one day's work=1/8;
(B+C)'s one day's work=1/12;
(A+C)'s one day's work=2(A+B+C)'s one day's work-((A+B)'s one day
work+(B+C)'s one day work)
= (2/6)-(1/8+1/12)
=(1/3)- (5/24)
=3/24
=1/8
So, A and C together will do the work in 8 days.
8)A can do a certain work in the same time in which B and C together can do it.
If A and B together could do it in 10 days and C alone in 50 days, then B alone
could do it in how many days?
Sol: (A+B)'s one day's work=1/10;
C's one day's work=1/50
(A+B+C)'s one day's work=(1/10+1/50)=6/50=3/25
Also, A's one day's work=(B+C)’s one day's work
From i and ii ,we get :2*(A's one day's work)=3/25
=> A's one day's work=3/50
B's one day’s work=(1/10-3/50)
=2/50
=1/25
B alone could complete the work in 25 days.
9) A is thrice as good a workman as B and therefore is able to finish a job
in 60 days less than B. Working together, they can do it in:
Sol: Ratio of times taken by A and B=1:3.
If difference of time is 2 days , B takes 3 days
If difference of time is 60 days, B takes (3*60/2)=90 days
So, A takes 30 days to do the work=1/90
A's one day's work=1/30;
B's one day's work=1/90;
(A+B)'s one day's work=1/30+1/90=4/90=2/45
Therefore, A&B together can do the work in 45/2days
Top
10) A can do a piece of work in 80 days. He works at it for 10 days and then
B alone finishes the remaining work in 42 days. In how much time will A&B,
working together, finish the work?
Sol: Work Done by A n 10 days =10/80=1/8
Remaining work =1-1/8=7/8
Now 7/8 work is done by B in 42 days
Whole work will be done by B in 42*8/7= 48 days
=> A's one day's work =1/80 and
B's one day's work =1/48
(A+B)'s one day's work = 1/80+1/48 = 8/240 = 1/30
Hence both will finish the work in 30 days.
11) 45 men can complete a work in 16 days. Six days after they started working,
so more men joined them. How many days will they now take to complete the
remaining work?
Sol: M1*D1/W1=M2*D2/W2
=>45*6/(6/16)=75*x/(1-(6/16))
=> x=6 days
12)A is 50% as efficient as B. C does half the work done by A&B together. If
C alone does the work n 40 days, then A,B and C together can do the work in:
Sol: A's one day's work:B's one days work=150:100 =3:2
Let A's &B's one day's work be 3x and 2x days respectively.
Then C's one day's work=5x/2
=> 5x/2=1/40
=> x=((1/40)*(2/5))=1/100
A's one day's work=3/100
B's one day's work=1/50
C's one day's work=1/40
So, A,B and C can do the work in 13 1/3 days.
13)A can finish a work in 18 days and B can do the same work in 15 days. B
worked for 10 days and left the job. In how many days A alone can finish the
remaining work?
Sol: B's 10 day's work=10/15=2/3
Remaining work=(1-(2/3))=1/3
Now, 1/18 work is done by A in 1 day.
Therefore 1/3 work is done by A in 18*(1/3)=6 days.
14)A can finish a work in 24 days, B n 9 days and C in 12 days. B&C start the
work but are forced to leave after 3 days. The remaining work done by A in:
Sol: (B+C)'s one day's work=1/9+1/12=7/36
Work done by B & C in 3 days=3*7/36=7/12
Remaining work=1-(7/12)=5/12
Now , 1/24 work is done by A in 1 day.
So, 5/12 work is done by A in 24*5/12=10 days
15)X and Y can do a piece of work n 20 days and 12 days respectively. X started
the work alone and then after 4 days Y joined him till the completion of work.
How long did the work last?
Sol: work done by X in 4 days =4/20 =1/5
Remaining work= 1-1/5 =4/5
(X+Y)'s one day's work =1/20+1/12 =8/60=2/15
Now, 2/15 work is done by X and Y in one day.
So, 4/5 work will be done by X and Y in 15/2*4/5=6 days
Hence Total time taken =(6+4) days = 10 days
16)A does 4/5 of work in 20 days. He then calls in B and they together finish
the remaining work in 3 days. How long B alone would take to do the whole work?
Sol: Whole work is done by A in 20*5/4=25 days
Now, (1-(4/5)) i.e 1/5 work is done by A& B in days.
Whole work will be done by A& B in 3*5=15 days
=>B's one day's work= 1/15-1/25=4/150=2/75
So, B alone would do the work in 75/2= 37 ½ days.
17) A and B can do a piece of work in 45 days and 40 days respectively. They
began to do the work together but A leaves after some days and then B completed
the remaining work n 23 days. The number of days after which A left the work was
Sol: (A+B)'s one day's work=1/45+1/40=17/360
Work done by B in 23 days=23/40
Remaining work=1-(23/40)=17/40
Now, 17/360 work was done by (A+B) in 1 day.
17/40 work was done by (A+B) in (1*(360/17)*(17/40))= 9 days
So, A left after 9 days.
18)A can do a piece of work in 10 days, B in 15 days. They work for 5 days.
The rest of work finished by C in 2 days. If they get Rs 1500 for the whole
work, the daily wages of B and C are
Sol: Part of work done by A= 5/10=1/2
Part of work done by B=1/3
Part of work done by C=(1-(1/2+1/3))=1/6
A's share: B's share: C's share=1/2:1/3:1/6= 3:2:1
A's share=(3/6)*1500=750
B's share=(2/6)*1500=500
C's share=(1/6)*1500=250
A's daily wages=750/5=150/-
B's daily wages=500/5=100/-
C's daily wages=250/2=125/-
Daily wages of B&C = 100+125=225/-
19)A alone can complete a work in 16 days and B alone can complete the same
in 12 days. Starting with A, they work on alternate days. The total work will
be completed in how many days?
(a) 12 days (b) 13 days (c) 13 5/7 days (d)13 ¾ days
Sol: (A+B)'s 2 days work = 1/16 + 1/12 =7/48
work done in 6 pairs of days =(7/48) * 6 = 7/8
remaining work = 1- 7/8 = 1/8
work done by A on 13th day = 1/16
remaining work = 1/8 – 1/16 = 1/16
on 14th day, it is B’s turn
1/12 work is done by B in 1 day.
1/16 work is done by B in ¾ day.
Total time taken= 13 ¾ days.
So, Answer is: D
20)A,B and C can do a piece of work in 20,30 and 60 days respectively. In how
many days can A do the work if he is assisted by B and C on every third day?
Sol: A's two day's work=2/20=1/10
(A+B+C)'s one day's work=1/20+1/30+1/60=6/60=1/10
Work done in 3 days=(1/10+1/10)=1/5
Now, 1/5 work is done in 3 days
Therefore, Whole work will be done in (3*5)=15 days.
21)Seven men can complete a work in 12 days. They started the work and after
5 days, two men left. In how many days will the work be completed by the
remaining men?
(A) 5 (B) 6 (C ) 7 (D) 8 (E) none
Sol: 7*12 men complete the work in 1 day.
Therefore, 1 man's 1 day's work=1/84
7 men's 5 days work = 5/12
=>remaining work = 1-5/12 = 7/12
5 men's 1 day's work = 5/84
5/84 work is don by them in 1 day
7/12 work is done by them in ((84/5) * (7/12)) = 49/5 days = 9 4/5 days.
Ans: E
22).12 men complete a work in 9 days. After they have worked for 6 days, 6 more
men joined them. How many days will they take to complete the remaining work?
(a) 2 days (b) 3 days (c) 4 days (d) 5days
Sol : 1 man's 1 day work = 1/108
12 men's 6 days work = 6/9 = 2/3
remaining work = 1 – 2/3 = 1/3
18 men's 1 days work = 18/108 = 1/6
1/6 work is done by them in 1 day
therefore, 1/3 work is done by them in 6/3 = 2 days.
Ans : A
23).A man, a woman and a boy can complete a job in 3,4 and 12 days respectively.
How many boys must assist 1 man and 1 woman to complete the job in ¼ of a day?
(a). 1 (b). 4 (c). 19 (d). 41
Sol : (1 man + 1 woman)'s 1 days work = 1/3+1/4=7/12
Work done by 1 man and 1 women n 1/4 day=((7/12)*(1/4))=7/48
Remaining work= 1- 7/48= 41/48
Work done by 1 boy in ¼ day= ((1/12)*(1/4)) =1/48
Therefore, Number of boys required= ((41/48)*48)= 41 days
So,Answer: D
24)12 men can complete a piece of work in 4 days, while 15 women can complete
the same work in 4 days. 6 men start working on the job and after working for
2 days, all of them stopped working. How many women should be put on the job
to complete the remaining work, if it is to be completed in 3 days.
(A) 15 (B) 18 (C) 22 (D) data inadequate
Sol: one man's one day's work= 1/48
one woman's one day's work=1/60
6 men's 2 day's work=((6/48)*2)= ¼
Remaining work=3/4
Now, 1/60 work s done in 1 day by 1 woman.
So, ¾ work will be done in 3 days by (60*(3/4)*(1/3))= 15 woman.
So, Answer: A
25)Twelve children take sixteen days to complete a work which can be completed
by 8 adults in 12 days. Sixteen adults left and four children joined them. How
many days will they take to complete the remaining work?
(A) 3 (B) 4 ( C) 6 (D) 8
Sol: one child's one day work= 1/192;
one adult's one day's work= 1/96;
work done in 3 days=((1/96)*16*3)= 1/2
Remaining work= 1 – ½=1/2
(6 adults+ 4 children)'s 1 day's work= 6/96+4/192= 1/12
1/12 work is done by them in 1 day.
½ work is done by them 12*(1/2)= 6 days
So, Answer= C
26)Sixteen men can complete a work in twelve days. Twenty four children can
complete the same work in 18 days. 12 men and 8 children started working and
after eight days three more children joined them. How many days will they now
take to complete the remaining work?
(A) 2 days (B) 4 days ( C) 6 days (D) 8 days
ol: one man's one day's work= 1/192
one child's one day's work= 1/432
Work done in 8 days=8*(12/192+ 8/432)=8*(1/16+1/54) =35/54
Remaining work= 1 -35/54= 19/54
(12 men+11 children)'s 1 day's work= 12/192 + 11/432 = 19/216
Now, 19/216 work is done by them in 1 day.
Therefore, 19/54 work will be done by them in ((216/19)*(19/54))= 4 days
So,Answer: B
27)Twenty-four men can complete a work in 16 days. Thirty- two women can
complete the same work in twenty-four days. Sixteen men and sixteen women
started working and worked for 12 days. How many more men are to be added to
complete the remaining work in 2 days?
(A) 16 men (B) 24 men ( C) 36 men (D) 48 men
Sol: one man's one day's work= 1/384
one woman's one day's work=1/768
Work done in 12 days= 12*( 16/384 + 16/768) = 12*(3/48)=3/4
Remaining work=1 – ¾=1/4
(16 men+16 women)'s two day's work =12*( 16/384+16/768)=2/16=1/8
Remaining work = 1/4-1/8 =1/8
1/384 work is done n 1 day by 1 man.
Therefore, 1/8 work will be done in 2 days in 384*(1/8)*(1/2)=24men
28)4 men and 6 women can complete a work in 8 days, while 3 men and 7 women
can complete it in 10 days. In how many days will 10 women complete it?
(A) 35 days (B) 40 days ( C) 45 days (D) 50 days
Sol: Let 1 man's 1 day's work =x days and
1 woman's 1 day's work=y
Then, 4x+6y=1/8 and 3x+7y=1/10.
Solving these two equations, we get: x=11/400 and y= 1/400
Therefore, 1 woman's 1 day's work=1/400
=> 10 women will complete the work in 40 days.
Answer: B
29)One man,3 women and 4 boys can do a piece of work in 96hrs, 2 men and 8 boys
can do it in 80 hrs, 2 men & 3 women can do it in 120hr. 5Men & 12 boys can do
it in?
(A) 39 1/11 hrs (B) 42 7/11 hrs ( C) 43 7/11 days (D) 44hrs
Sol: Let 1 man's 1 hour's work=x
1 woman's 1 hour's work=y
1 boy's 1 hour's work=z
Then, x+3y+4z=1/96 -----------(1)
2x+8z= 1/80 ----------(2)
adding (2) & (3) and subtracting (1)
3x+4z=1/96 ---------(4)
From (2) and (4), we get x=1/480
Substituting, we get : y=1/720 and z= 1/960
(5 men+ 12 boy)'s 1 hour's work=5/480+12/960 =1/96 + 1/80=11/480
Therefore, 5 men and 12 boys can do the work in 480/11 or 43 7/11hours.
So,Answer: C
working 8 hours a day can do it in how many days?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so (9*6*88/1)=(6*8*d/1)
on solving, d=99 days.
2)If 34 men completed 2/5th of a work in 8 days working 9 hours a day.
How many more man should be engaged to finish the rest of the work in
6 days working 9 hours a day?
Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2)
so, (34*8*9/(2/5))=(x*6*9/(3/5))
so x=136 men
number of men to be added to finish the work=136-34=102 men
3)If 5 women or 8 girls can do a work in 84 days. In how many days can
10 women and 5 girls can do the same work?
Sol: Given that 5 women is equal to 8 girls to complete a work
so, 10 women=16 girls.
Therefore 10women +5girls=16girls+5girls=21girls.
8 girls can do a work in 84 days
then 21 girls ---------------?
answer= (8*84/21)=32days.
Therefore 10 women and 5 girls can a work in 32days
4)Worker A takes 8 hours to do a job. Worker B takes 10hours to do the
same job. How long it take both A & B, working together but independently,
to do the same job?
Sol: A's one hour work=1/8.
B's one hour work=1/10
(A+B)'s one hour work=1/8+1/10 =9/40
Both A & B can finish the work in 40/9 days
5)A can finish a work in 18 days and B can do the same work in half the
time taken by A. Then, working together, what part of the same work they
can finish in a day?
Sol: Given that B alone can complete the same work in days=half the time
taken by A=9days
A's one day work=1/18
B's one day work=1/9
(A+B)'s one day work=1/18+1/9=1/6
6)A is twice as good a workman as B and together they finish a piece of
work in 18 days.In how many days will A alone finish the work.
Sol: if A takes x days to do a work then
B takes 2x days to do the same work
=>1/x+1/2x=1/18
=>3/2x=1/18
=>x=27 days.
Hence, A alone can finish the work in 27 days.
7)A can do a certain work in 12 days. B is 60% more efficient than A. How
many days does B alone take to do the same job?
Sol: Ratio of time taken by A&B=160:100 =8:5
Suppose B alone takes x days to do the job.
Then, 8:5::12:x
=> 8x=5*12
=> x=15/2 days.
8)A can do a piece of work n 7 days of 9 hours each and B alone can do it
in 6 days of 7 hours each. How long will they take to do it working together
8 2/5 hours a day?
Sol: A can complete the work in (7*9)=63 days
B can complete the work in (6*7)=42 days
=> A's one hour's work=1/63 and
B's one hour work=1/42
(A+B)'s one hour work=1/63+1/42=5/126
Therefore, Both can finish the work in 126/5 hours.
Number of days of 8 2/5 hours each=(126*5/(5*42))=3days
9)A takes twice as much time as B or thrice as much time to finish a piece
of work. Working together they can finish the work in 2 days. B can do the
work alone in ?
Sol: Suppose A,B and C take x,x/2 and x/3 hours respectively finish the
work then 1/x+2/x+3/x=1/2
=> 6/x=1/2
=>x=12
So, B takes 6 hours to finish the work.
10)X can do ¼ of a work in 10 days, Y can do 40% of work in 40 days and Z
can do 1/3 of work in 13 days. Who will complete the work first?
Sol: Whole work will be done by X in 10*4=40 days.
Whole work will be done by Y in (40*100/40)=100 days.
Whole work will be done by Z in (13*3)=39 days
Therefore,Z will complete the work first.
Complex Problems
1)A and B undertake to do a piece of workfor Rs 600.A alone can do it in
6 days while B alone can do it in 8 days. With the help of C, they can finish
it in 3 days, Find the share of each?
Sol: C's one day's work=(1/3)-(1/6+1/8)=1/24
Therefore, A:B:C= Ratio of their one day’s work=1/6:1/8:1/24=4:3:1
A's share=Rs (600*4/8)=300
B's share= Rs (600*3/8)=225
C's share=Rs[600-(300+225)]=Rs 75
2)A can do a piece of work in 80 days. He works at it for 10 days & then B alone
finishes the remaining work in 42 days. In how much time will A and B, working
together, finish the work?
Sol: Work done by A in 10 days=10/80=1/8
Remaining work=(1-(1/8))=7/8
Now, work will be done by B in 42 days.
Whole work will be done by B in (42*8/7)=48 days
Therefore, A's one day's work=1/80
B’s one day's work=1/48
(A+B)'s one day's work=1/80+1/48=8/240=1/30
Hence, both will finish the work in 30 days.
3)P,Q and R are three typists who working simultaneously can type 216 pages
in 4 hours In one hour , R can type as many pages more than Q as Q can type more
than P. During a period of five hours, R can type as many pages as P can
during seven hours. How many pages does each of them type per hour?
Sol:Let the number of pages typed in one hour by P, Q and R be x,y and z
respectively
Then x+y+z=216/4=54 ---------------1
z-y=y-x => 2y=x+z -----------2
5z=7x => x=5x/7 ---------------3
Solving 1,2 and 3 we get x=15,y=18, and z=21
4)Ronald and Elan are working on an assignment. Ronald takes 6 hours to
type 32 pages on a computer, while Elan takes 5 hours to type 40 pages.
How much time will they take, working together on two different computers
to type an assignment of 110 pages?
Sol: Number of pages typed by Ronald in one hour=32/6=16/3
Number of pages typed by Elan in one hour=40/5=8
Number of pages typed by both in one hour=((16/3)+8)=40/3
Time taken by both to type 110 pages=110*3/40=8 hours.
5)Two workers A and B are engaged to do a work. A working alone takes 8 hours
more to complete the job than if both working together. If B worked alone,
he would need 4 1/2 hours more to compete the job than they both working
together. What time would they take to do the work together.
Sol: (1/(x+8))+(1/(x+(9/2)))=1/x
=>(1/(x+8))+(2/(2x+9))=1/x
=> x(4x+25)=(x+8)(2x+9)
=> 2x2 =72
=> x2 = 36
=> x=6
Therefore, A and B together can do the work in 6 days.
6)A and B can do a work in12 days, B and C in 15 days, C and A in 20 days.
If A,B and C work together, they will complete the work in how many days?
Sol: (A+B)'s one day's work=1/12;
(B+C)'s one day's work=1/15;
(A+C)'s one day's work=1/20;
Adding we get 2(A+B+C)'s one day's work=1/12+1/15+1/20=12/60=1/5
(A+B+C)'s one day work=1/10
So, A,B,and C together can complete the work in 10 days.
7)A and B can do a work in 8 days, B and C can do the same wor in 12 days.
A,B and C together can finish it in 6 days. A and C together will do it in
how many days?
Sol: (A+B+C)'s one day's work=1/6;
(A+B)'s one day's work=1/8;
(B+C)'s one day's work=1/12;
(A+C)'s one day's work=2(A+B+C)'s one day's work-((A+B)'s one day
work+(B+C)'s one day work)
= (2/6)-(1/8+1/12)
=(1/3)- (5/24)
=3/24
=1/8
So, A and C together will do the work in 8 days.
8)A can do a certain work in the same time in which B and C together can do it.
If A and B together could do it in 10 days and C alone in 50 days, then B alone
could do it in how many days?
Sol: (A+B)'s one day's work=1/10;
C's one day's work=1/50
(A+B+C)'s one day's work=(1/10+1/50)=6/50=3/25
Also, A's one day's work=(B+C)’s one day's work
From i and ii ,we get :2*(A's one day's work)=3/25
=> A's one day's work=3/50
B's one day’s work=(1/10-3/50)
=2/50
=1/25
B alone could complete the work in 25 days.
9) A is thrice as good a workman as B and therefore is able to finish a job
in 60 days less than B. Working together, they can do it in:
Sol: Ratio of times taken by A and B=1:3.
If difference of time is 2 days , B takes 3 days
If difference of time is 60 days, B takes (3*60/2)=90 days
So, A takes 30 days to do the work=1/90
A's one day's work=1/30;
B's one day's work=1/90;
(A+B)'s one day's work=1/30+1/90=4/90=2/45
Therefore, A&B together can do the work in 45/2days
Top
10) A can do a piece of work in 80 days. He works at it for 10 days and then
B alone finishes the remaining work in 42 days. In how much time will A&B,
working together, finish the work?
Sol: Work Done by A n 10 days =10/80=1/8
Remaining work =1-1/8=7/8
Now 7/8 work is done by B in 42 days
Whole work will be done by B in 42*8/7= 48 days
=> A's one day's work =1/80 and
B's one day's work =1/48
(A+B)'s one day's work = 1/80+1/48 = 8/240 = 1/30
Hence both will finish the work in 30 days.
11) 45 men can complete a work in 16 days. Six days after they started working,
so more men joined them. How many days will they now take to complete the
remaining work?
Sol: M1*D1/W1=M2*D2/W2
=>45*6/(6/16)=75*x/(1-(6/16))
=> x=6 days
12)A is 50% as efficient as B. C does half the work done by A&B together. If
C alone does the work n 40 days, then A,B and C together can do the work in:
Sol: A's one day's work:B's one days work=150:100 =3:2
Let A's &B's one day's work be 3x and 2x days respectively.
Then C's one day's work=5x/2
=> 5x/2=1/40
=> x=((1/40)*(2/5))=1/100
A's one day's work=3/100
B's one day's work=1/50
C's one day's work=1/40
So, A,B and C can do the work in 13 1/3 days.
13)A can finish a work in 18 days and B can do the same work in 15 days. B
worked for 10 days and left the job. In how many days A alone can finish the
remaining work?
Sol: B's 10 day's work=10/15=2/3
Remaining work=(1-(2/3))=1/3
Now, 1/18 work is done by A in 1 day.
Therefore 1/3 work is done by A in 18*(1/3)=6 days.
14)A can finish a work in 24 days, B n 9 days and C in 12 days. B&C start the
work but are forced to leave after 3 days. The remaining work done by A in:
Sol: (B+C)'s one day's work=1/9+1/12=7/36
Work done by B & C in 3 days=3*7/36=7/12
Remaining work=1-(7/12)=5/12
Now , 1/24 work is done by A in 1 day.
So, 5/12 work is done by A in 24*5/12=10 days
15)X and Y can do a piece of work n 20 days and 12 days respectively. X started
the work alone and then after 4 days Y joined him till the completion of work.
How long did the work last?
Sol: work done by X in 4 days =4/20 =1/5
Remaining work= 1-1/5 =4/5
(X+Y)'s one day's work =1/20+1/12 =8/60=2/15
Now, 2/15 work is done by X and Y in one day.
So, 4/5 work will be done by X and Y in 15/2*4/5=6 days
Hence Total time taken =(6+4) days = 10 days
16)A does 4/5 of work in 20 days. He then calls in B and they together finish
the remaining work in 3 days. How long B alone would take to do the whole work?
Sol: Whole work is done by A in 20*5/4=25 days
Now, (1-(4/5)) i.e 1/5 work is done by A& B in days.
Whole work will be done by A& B in 3*5=15 days
=>B's one day's work= 1/15-1/25=4/150=2/75
So, B alone would do the work in 75/2= 37 ½ days.
17) A and B can do a piece of work in 45 days and 40 days respectively. They
began to do the work together but A leaves after some days and then B completed
the remaining work n 23 days. The number of days after which A left the work was
Sol: (A+B)'s one day's work=1/45+1/40=17/360
Work done by B in 23 days=23/40
Remaining work=1-(23/40)=17/40
Now, 17/360 work was done by (A+B) in 1 day.
17/40 work was done by (A+B) in (1*(360/17)*(17/40))= 9 days
So, A left after 9 days.
18)A can do a piece of work in 10 days, B in 15 days. They work for 5 days.
The rest of work finished by C in 2 days. If they get Rs 1500 for the whole
work, the daily wages of B and C are
Sol: Part of work done by A= 5/10=1/2
Part of work done by B=1/3
Part of work done by C=(1-(1/2+1/3))=1/6
A's share: B's share: C's share=1/2:1/3:1/6= 3:2:1
A's share=(3/6)*1500=750
B's share=(2/6)*1500=500
C's share=(1/6)*1500=250
A's daily wages=750/5=150/-
B's daily wages=500/5=100/-
C's daily wages=250/2=125/-
Daily wages of B&C = 100+125=225/-
19)A alone can complete a work in 16 days and B alone can complete the same
in 12 days. Starting with A, they work on alternate days. The total work will
be completed in how many days?
(a) 12 days (b) 13 days (c) 13 5/7 days (d)13 ¾ days
Sol: (A+B)'s 2 days work = 1/16 + 1/12 =7/48
work done in 6 pairs of days =(7/48) * 6 = 7/8
remaining work = 1- 7/8 = 1/8
work done by A on 13th day = 1/16
remaining work = 1/8 – 1/16 = 1/16
on 14th day, it is B’s turn
1/12 work is done by B in 1 day.
1/16 work is done by B in ¾ day.
Total time taken= 13 ¾ days.
So, Answer is: D
20)A,B and C can do a piece of work in 20,30 and 60 days respectively. In how
many days can A do the work if he is assisted by B and C on every third day?
Sol: A's two day's work=2/20=1/10
(A+B+C)'s one day's work=1/20+1/30+1/60=6/60=1/10
Work done in 3 days=(1/10+1/10)=1/5
Now, 1/5 work is done in 3 days
Therefore, Whole work will be done in (3*5)=15 days.
21)Seven men can complete a work in 12 days. They started the work and after
5 days, two men left. In how many days will the work be completed by the
remaining men?
(A) 5 (B) 6 (C ) 7 (D) 8 (E) none
Sol: 7*12 men complete the work in 1 day.
Therefore, 1 man's 1 day's work=1/84
7 men's 5 days work = 5/12
=>remaining work = 1-5/12 = 7/12
5 men's 1 day's work = 5/84
5/84 work is don by them in 1 day
7/12 work is done by them in ((84/5) * (7/12)) = 49/5 days = 9 4/5 days.
Ans: E
22).12 men complete a work in 9 days. After they have worked for 6 days, 6 more
men joined them. How many days will they take to complete the remaining work?
(a) 2 days (b) 3 days (c) 4 days (d) 5days
Sol : 1 man's 1 day work = 1/108
12 men's 6 days work = 6/9 = 2/3
remaining work = 1 – 2/3 = 1/3
18 men's 1 days work = 18/108 = 1/6
1/6 work is done by them in 1 day
therefore, 1/3 work is done by them in 6/3 = 2 days.
Ans : A
23).A man, a woman and a boy can complete a job in 3,4 and 12 days respectively.
How many boys must assist 1 man and 1 woman to complete the job in ¼ of a day?
(a). 1 (b). 4 (c). 19 (d). 41
Sol : (1 man + 1 woman)'s 1 days work = 1/3+1/4=7/12
Work done by 1 man and 1 women n 1/4 day=((7/12)*(1/4))=7/48
Remaining work= 1- 7/48= 41/48
Work done by 1 boy in ¼ day= ((1/12)*(1/4)) =1/48
Therefore, Number of boys required= ((41/48)*48)= 41 days
So,Answer: D
24)12 men can complete a piece of work in 4 days, while 15 women can complete
the same work in 4 days. 6 men start working on the job and after working for
2 days, all of them stopped working. How many women should be put on the job
to complete the remaining work, if it is to be completed in 3 days.
(A) 15 (B) 18 (C) 22 (D) data inadequate
Sol: one man's one day's work= 1/48
one woman's one day's work=1/60
6 men's 2 day's work=((6/48)*2)= ¼
Remaining work=3/4
Now, 1/60 work s done in 1 day by 1 woman.
So, ¾ work will be done in 3 days by (60*(3/4)*(1/3))= 15 woman.
So, Answer: A
25)Twelve children take sixteen days to complete a work which can be completed
by 8 adults in 12 days. Sixteen adults left and four children joined them. How
many days will they take to complete the remaining work?
(A) 3 (B) 4 ( C) 6 (D) 8
Sol: one child's one day work= 1/192;
one adult's one day's work= 1/96;
work done in 3 days=((1/96)*16*3)= 1/2
Remaining work= 1 – ½=1/2
(6 adults+ 4 children)'s 1 day's work= 6/96+4/192= 1/12
1/12 work is done by them in 1 day.
½ work is done by them 12*(1/2)= 6 days
So, Answer= C
26)Sixteen men can complete a work in twelve days. Twenty four children can
complete the same work in 18 days. 12 men and 8 children started working and
after eight days three more children joined them. How many days will they now
take to complete the remaining work?
(A) 2 days (B) 4 days ( C) 6 days (D) 8 days
ol: one man's one day's work= 1/192
one child's one day's work= 1/432
Work done in 8 days=8*(12/192+ 8/432)=8*(1/16+1/54) =35/54
Remaining work= 1 -35/54= 19/54
(12 men+11 children)'s 1 day's work= 12/192 + 11/432 = 19/216
Now, 19/216 work is done by them in 1 day.
Therefore, 19/54 work will be done by them in ((216/19)*(19/54))= 4 days
So,Answer: B
27)Twenty-four men can complete a work in 16 days. Thirty- two women can
complete the same work in twenty-four days. Sixteen men and sixteen women
started working and worked for 12 days. How many more men are to be added to
complete the remaining work in 2 days?
(A) 16 men (B) 24 men ( C) 36 men (D) 48 men
Sol: one man's one day's work= 1/384
one woman's one day's work=1/768
Work done in 12 days= 12*( 16/384 + 16/768) = 12*(3/48)=3/4
Remaining work=1 – ¾=1/4
(16 men+16 women)'s two day's work =12*( 16/384+16/768)=2/16=1/8
Remaining work = 1/4-1/8 =1/8
1/384 work is done n 1 day by 1 man.
Therefore, 1/8 work will be done in 2 days in 384*(1/8)*(1/2)=24men
28)4 men and 6 women can complete a work in 8 days, while 3 men and 7 women
can complete it in 10 days. In how many days will 10 women complete it?
(A) 35 days (B) 40 days ( C) 45 days (D) 50 days
Sol: Let 1 man's 1 day's work =x days and
1 woman's 1 day's work=y
Then, 4x+6y=1/8 and 3x+7y=1/10.
Solving these two equations, we get: x=11/400 and y= 1/400
Therefore, 1 woman's 1 day's work=1/400
=> 10 women will complete the work in 40 days.
Answer: B
29)One man,3 women and 4 boys can do a piece of work in 96hrs, 2 men and 8 boys
can do it in 80 hrs, 2 men & 3 women can do it in 120hr. 5Men & 12 boys can do
it in?
(A) 39 1/11 hrs (B) 42 7/11 hrs ( C) 43 7/11 days (D) 44hrs
Sol: Let 1 man's 1 hour's work=x
1 woman's 1 hour's work=y
1 boy's 1 hour's work=z
Then, x+3y+4z=1/96 -----------(1)
2x+8z= 1/80 ----------(2)
adding (2) & (3) and subtracting (1)
3x+4z=1/96 ---------(4)
From (2) and (4), we get x=1/480
Substituting, we get : y=1/720 and z= 1/960
(5 men+ 12 boy)'s 1 hour's work=5/480+12/960 =1/96 + 1/80=11/480
Therefore, 5 men and 12 boys can do the work in 480/11 or 43 7/11hours.
So,Answer: C
Time and Work
1.If A can do a piece of work in n days, then A's 1 day work=1/n
2.If A's 1 day's work=1/n, then A can finish the work in n days.
Ex: If A can do a piece of work in 4 days,then A's 1 day's work=1/4.
If A's 1 day’s work=1/5, then A can finish the work in 5 days
3.If A is thrice as good workman as B,then: Ratio of work done by
A and B =3:1. Ratio of time taken by A and B to finish a work=1:3
4.Definition of Variation: The change in two different variables
follow some definite rule. It said that the two variables vary
directly or inversely.Its notation is X/Y=k, where k is called
constant. This variation is called direct variation. XY=k. This
variation is called inverse variation.
5.Some Pairs of Variables:
i)Number of workers and their wages. If the number of workers
increases, their total wages increase. If the number of days
reduced, there will be less work. If the number of days is
increased, there will be more work. Therefore, here we have
direct proportion or direct variation.
ii)Number workers and days required to do a certain work is an
example of inverse variation. If more men are employed, they
will require fewer days and if there are less number of workers,
more days are required.
iii)There is an inverse proportion between the daily hours of a
work and the days required. If the number of hours is increased,
less number of days are required and if the number of hours is
reduced, more days are required.
6.Some Important Tips:
More Men -Less Days and Conversely More Day-Less Men.
More Men -More Work and Conversely More Work-More Men.
More Days-More Work and Conversely More Work-More Days.
Number of days required to complete the given work=Total work/One
day’s work.
Since the total work is assumed to be one(unit), the number of days
required to complete the given work would be the reciprocal of one
day’s work.
Sometimes, the problems on time and work can be solved using the
proportional rule ((man*days*hours)/work) in another situation.
7.If men is fixed,work is proportional to time. If work is fixed,
then time is inversely proportional to men therefore,
(M1*T1/W1)=(M2*T2/W2)
2.If A's 1 day's work=1/n, then A can finish the work in n days.
Ex: If A can do a piece of work in 4 days,then A's 1 day's work=1/4.
If A's 1 day’s work=1/5, then A can finish the work in 5 days
3.If A is thrice as good workman as B,then: Ratio of work done by
A and B =3:1. Ratio of time taken by A and B to finish a work=1:3
4.Definition of Variation: The change in two different variables
follow some definite rule. It said that the two variables vary
directly or inversely.Its notation is X/Y=k, where k is called
constant. This variation is called direct variation. XY=k. This
variation is called inverse variation.
5.Some Pairs of Variables:
i)Number of workers and their wages. If the number of workers
increases, their total wages increase. If the number of days
reduced, there will be less work. If the number of days is
increased, there will be more work. Therefore, here we have
direct proportion or direct variation.
ii)Number workers and days required to do a certain work is an
example of inverse variation. If more men are employed, they
will require fewer days and if there are less number of workers,
more days are required.
iii)There is an inverse proportion between the daily hours of a
work and the days required. If the number of hours is increased,
less number of days are required and if the number of hours is
reduced, more days are required.
6.Some Important Tips:
More Men -Less Days and Conversely More Day-Less Men.
More Men -More Work and Conversely More Work-More Men.
More Days-More Work and Conversely More Work-More Days.
Number of days required to complete the given work=Total work/One
day’s work.
Since the total work is assumed to be one(unit), the number of days
required to complete the given work would be the reciprocal of one
day’s work.
Sometimes, the problems on time and work can be solved using the
proportional rule ((man*days*hours)/work) in another situation.
7.If men is fixed,work is proportional to time. If work is fixed,
then time is inversely proportional to men therefore,
(M1*T1/W1)=(M2*T2/W2)
Pipes and Cisterns problem
1)Two pipes A& B can fill a tank in 36 hours and 45 hours respectively.
If both the pipes are opened simultaneously, how much time will be
taken to fill the tank?
Sol: Part filled by A in 1 hour=1/36
Part filled by B in 1 hour= 1/45;
Part filled by (A+B)'s in 1 hour=1/36 +1/45= 9/180 =1/20
Hence, both the pipes together will fill the tank in 20 hours.
2)Two pipes can fill a tank in 10 hours & 12 hours respectively. While
3rd pipe empties the full tank n 20 hours. If all the three pipes
operate simultaneously,in how much time will the tank be filled?
Sol: Net part filled in 1 hour=1/10 +1/12 -1/20
=8/60=2/15
The tank be filled in 15/2hours= 7 hrs 30 min
3)A cistern can be filled by a tap in 4 hours while it can be emptied
by another tap in 9 hours. If both the taps are opened simultaneously,
then after how much time will the cistern get filled?
Sol: Net part filled in 1 hour= 1/4 -1/9= 5/36
Therefore the cistern will be filled in 36/5 hours or 7.2 hours.
4)If two pipes function simultaneously, the reservoir will be filled in
12 days.One pipe fills the reservoir 10 hours faster than the other.
How many hours does it take the second pipe to fill the reservoir.
Sol: Let the reservoir be filled by the 1st pipe in x hours.
The second pipe will fill it in (x+10) hours
1/x + (1/(x+10))= 1/12
=> (2x+10)/((x)*(x+10))= 1/12
=> x=20
So, the second pipe will take 30 hours to fill the reservoir.
5)A cistern has two taps which fill it in 12 min and 15 min respectively.
There is also a waste pipe in the cistern. When all the three are opened,
the empty cistern is full in 20 min. How long will the waste pipe take to
empty the full cistern?
Sol: Work done by a waste pipe in 1 min
=1/20 -(1/12+1/15)= -1/10 (-ve means emptying)
6)A tap can fill a tank in 6 hours. After half the tank is filled, three
more similar taps are opened. What is the total time taken to fill the
tank completely?
Sol: Time taken by one tap to fill the half of the tank =3 hours
Part filled by the four taps in 1 hour=4/6=2/3
Remaining part=1 -1/2=1/2
Therefore, 2/3:1/2::1:x
or x=(1/2)*1*(3/2)=3/4 hours.
i.e 45 min
So, total time taken= 3hrs 45min.
7)A water tank is two-fifth full. Pipe A can fill a tank in 10 min. And B
can empty it in 6 min. If both pipes are open, how long will it take to
empty or fill the tank completely ?
Sol: Clearly, pipe B is faster than A and So, the tank will be emptied.
Part to be emptied=2/5.
Part emptied by (A+B) in 1 min= 1/6 -1/10=1/15
Therefore, 1/15:2/5::1:x or x=((2/5)*1*15)=6 min.
So, the tank be emptied in 6 min.
8)Bucket P has thrice the capacity as Bucket Q. It takes 60 turns for
Bucket P to fill the empty drum. How many turns it will take for both the
buckets P&Q, having each turn together to fill the empty drum?
Sol: Let the capacity of P be x lit.
Then capacity of Q=x/3 lit
Capacity of the drum=60x lit
Required number of turns= 60x/(x+(x/3))= 60x*3/4x=45
Complex Problems
1)Two pipes can fill a cistern in 14 hours and 16 hours respectively. The
pipes are opened simultaneously and it is found that due to leakage in the
bottom it took 32min more to fill the cistern. When the cistern is full,
in what time will the leak empty it?
Sol: Work done by the two pipes in 1 hour= 1/14+1/16=15/112
Time taken by these two pipes to fill the tank=112/15 hrs.
Due to leakage, time taken = 7 hrs 28 min+ 32 min= 8 hours
Therefore, work done by (two pipes + leak) in 1 hr= 1/8
work done by leak n 1 hour=15/112 -1/8=1/112
Leak will empty full cistern n 112 hours.
2)Two pipes A&B can fill a tank in 30 min. First, A&B are opened. After
7 min, C also opened. In how much time, the tank s full.
Sol: Part filled n 7 min = 7*(1/36+1/45)=7/20
Remaining part= 1-7/20=13/20
Net part filled in 1 min when A,B and C are opened=1/36 +1/45- 1/30=1/60
Now, 1/60 part is filled in 1 min.
13/20 part is filled n (60*13/20)=39 min
Total time taken to fill the tank=39+7=46 min
3)Two pipes A&B can fill a tank in 24 min and 32 min respectively. If
both the pipes are opened simultaneously, after how much time B should
be closed so that the tank is full in 18 min.
Sol: Let B be closed after x min, then part filled by (A+B) in x min+
part filled by A in (18-x) min=1
x(1/24+1/32) +(18-x)1/24 =1
=> x=8
Hence B must be closed after 8 min.
4)Two pipes A& B together can fill a cistern in 4 hours. Had they been
opened separately, then B would have taken 6 hours more than A to fill
the cistern. How much time will be taken by A to fill the cistern
separately?
Sol: Let the cistern be filled by pipe A alone in x hours.
Pipe B will fill it in x+6 hours
1/x + 1/x+6=1/4
Solving this we get x=6.
Hence, A takes 6 hours to fill the cistern separately.
5)A tank is filled by 3 pipes with uniform flow. The first two pipes
operating simultaneously fill the tan in the same time during which
the tank is filled by the third pipe alone. The 2nd pipe fills the tank
5 hours faster than first pipe and 4 hours slower than third pipe. The
time required by first pipe is :
Sol: Suppose, first pipe take x hours to fill the tank then
B & C will take (x-5) and (x-9) hours respectively.
Therefore, 1/x +1/(x-5) =1/(x-9)
On solving, x=15
Hence, time required by first pipe is 15 hours.
6)A large tanker can be filled by two pipes A& B in 60min and 40 min
respectively. How many minutes will it take to fill the tanker from
empty state if B is used for half the time & A and B fill it together for
the other half?
Sol: Part filled by (A+B) n 1 min=(1/60 +1/40)=1/24
Suppose the tank is filled in x minutes
Then, x/2(1/24+1/40)=1
=> (x/2)*(1/15)=1
=> x=30 min.
7)Two pipes A and B can fill a tank in 6 hours and 4 hours respectively.
If they are opened on alternate hours and if pipe A s opened first, in
how many hours, the tank shall be full.
Sol: (A+B)'s 2 hours work when opened alternatively =1/6+1/4 =5/12
(A+B)'s 4 hours work when opened alternatively=10/12=5/6
Remaining part=1 -5/6=1/6.
Now, it is A's turn and 1/6 part is filled by A in 1 hour.
So, total time taken to fill the tank=(4+1)= 5 hours.
8)Three taps A,B and C can fill a tank in 12, 15 and 20 hours respectively.
If A is open all the time and B and C are open for one hour each
alternatively, the tank will be full in.
Sol: (A+B)'s 1 hour's work=1/12+1/15=9/60=3/20
(A+C)'s 1 hour's work=1/20+1/12=8/60=2/15
Part filled in 2 hours=3/20+2/15=17/60
Part filled in 2 hours=3/20+2/15= 17/60
Part filled in 6 hours=3*17/60 =17/20
Remaining part=1 -17/20=3/20
Now, it is the turn of A & B and 3/20 part is filled by A& B in 1 hour.
Therefore, total time taken to fill the tank=6+1=7 hours
9)A Booster pump can be used for filling as well as for emptying a tank.
The capacity of the tank is 2400 m3. The emptying capacity of the tank is
10 m3 per minute higher than its filling capacity and the pump needs 8
minutes lesser to empty the tank than it needs to fill it. What is the
filling capacity of the pump?
Sol: Let, the filling capacity of the pump be x m3/min
Then, emptying capacity of the pump=(x+10) m3/min.
So,2400/x – 2400/(x+10) = 8
on solving x=50.
10)A leak in the bottom of a tank can empty the full tan in 8 hr. An inlet
pipe fills water at the rate of 6 lits a minute. When the tank is full,
the inlet is opened and due to the leak, the tank is empty in 12 hrs.
How many liters does the cistern hold?
Sol: Work done by the inlet in 1 hr= 1/8 -1/12=1/24
Work done by the inlet in i min= (1/24)*(1/60)=1/1440
Therefore, Volume of 1/1440 part=6 lit
Volume of whole=(1440*6) lit=8640 lit.
11)Two pipes A and B can fill a cistern in 37 ½ min and 45 minutes
respectively. Both the pipes are opened. The cistern will be filled in
just half an hour, if the pipe B is turned off after:
sol: Let B be turned off after x min. Then,
Part filled by (A+B) in x min+ part filled by A in (30-x)min=1
Therefore, x(2/75+1/45)+(30-x)(2/75)=1
11x/225 + (60-2x)/75=1
11x+ 180-6x=225
x=9.
So, B must be turned off after 9 minutes.
If both the pipes are opened simultaneously, how much time will be
taken to fill the tank?
Sol: Part filled by A in 1 hour=1/36
Part filled by B in 1 hour= 1/45;
Part filled by (A+B)'s in 1 hour=1/36 +1/45= 9/180 =1/20
Hence, both the pipes together will fill the tank in 20 hours.
2)Two pipes can fill a tank in 10 hours & 12 hours respectively. While
3rd pipe empties the full tank n 20 hours. If all the three pipes
operate simultaneously,in how much time will the tank be filled?
Sol: Net part filled in 1 hour=1/10 +1/12 -1/20
=8/60=2/15
The tank be filled in 15/2hours= 7 hrs 30 min
3)A cistern can be filled by a tap in 4 hours while it can be emptied
by another tap in 9 hours. If both the taps are opened simultaneously,
then after how much time will the cistern get filled?
Sol: Net part filled in 1 hour= 1/4 -1/9= 5/36
Therefore the cistern will be filled in 36/5 hours or 7.2 hours.
4)If two pipes function simultaneously, the reservoir will be filled in
12 days.One pipe fills the reservoir 10 hours faster than the other.
How many hours does it take the second pipe to fill the reservoir.
Sol: Let the reservoir be filled by the 1st pipe in x hours.
The second pipe will fill it in (x+10) hours
1/x + (1/(x+10))= 1/12
=> (2x+10)/((x)*(x+10))= 1/12
=> x=20
So, the second pipe will take 30 hours to fill the reservoir.
5)A cistern has two taps which fill it in 12 min and 15 min respectively.
There is also a waste pipe in the cistern. When all the three are opened,
the empty cistern is full in 20 min. How long will the waste pipe take to
empty the full cistern?
Sol: Work done by a waste pipe in 1 min
=1/20 -(1/12+1/15)= -1/10 (-ve means emptying)
6)A tap can fill a tank in 6 hours. After half the tank is filled, three
more similar taps are opened. What is the total time taken to fill the
tank completely?
Sol: Time taken by one tap to fill the half of the tank =3 hours
Part filled by the four taps in 1 hour=4/6=2/3
Remaining part=1 -1/2=1/2
Therefore, 2/3:1/2::1:x
or x=(1/2)*1*(3/2)=3/4 hours.
i.e 45 min
So, total time taken= 3hrs 45min.
7)A water tank is two-fifth full. Pipe A can fill a tank in 10 min. And B
can empty it in 6 min. If both pipes are open, how long will it take to
empty or fill the tank completely ?
Sol: Clearly, pipe B is faster than A and So, the tank will be emptied.
Part to be emptied=2/5.
Part emptied by (A+B) in 1 min= 1/6 -1/10=1/15
Therefore, 1/15:2/5::1:x or x=((2/5)*1*15)=6 min.
So, the tank be emptied in 6 min.
8)Bucket P has thrice the capacity as Bucket Q. It takes 60 turns for
Bucket P to fill the empty drum. How many turns it will take for both the
buckets P&Q, having each turn together to fill the empty drum?
Sol: Let the capacity of P be x lit.
Then capacity of Q=x/3 lit
Capacity of the drum=60x lit
Required number of turns= 60x/(x+(x/3))= 60x*3/4x=45
Complex Problems
1)Two pipes can fill a cistern in 14 hours and 16 hours respectively. The
pipes are opened simultaneously and it is found that due to leakage in the
bottom it took 32min more to fill the cistern. When the cistern is full,
in what time will the leak empty it?
Sol: Work done by the two pipes in 1 hour= 1/14+1/16=15/112
Time taken by these two pipes to fill the tank=112/15 hrs.
Due to leakage, time taken = 7 hrs 28 min+ 32 min= 8 hours
Therefore, work done by (two pipes + leak) in 1 hr= 1/8
work done by leak n 1 hour=15/112 -1/8=1/112
Leak will empty full cistern n 112 hours.
2)Two pipes A&B can fill a tank in 30 min. First, A&B are opened. After
7 min, C also opened. In how much time, the tank s full.
Sol: Part filled n 7 min = 7*(1/36+1/45)=7/20
Remaining part= 1-7/20=13/20
Net part filled in 1 min when A,B and C are opened=1/36 +1/45- 1/30=1/60
Now, 1/60 part is filled in 1 min.
13/20 part is filled n (60*13/20)=39 min
Total time taken to fill the tank=39+7=46 min
3)Two pipes A&B can fill a tank in 24 min and 32 min respectively. If
both the pipes are opened simultaneously, after how much time B should
be closed so that the tank is full in 18 min.
Sol: Let B be closed after x min, then part filled by (A+B) in x min+
part filled by A in (18-x) min=1
x(1/24+1/32) +(18-x)1/24 =1
=> x=8
Hence B must be closed after 8 min.
4)Two pipes A& B together can fill a cistern in 4 hours. Had they been
opened separately, then B would have taken 6 hours more than A to fill
the cistern. How much time will be taken by A to fill the cistern
separately?
Sol: Let the cistern be filled by pipe A alone in x hours.
Pipe B will fill it in x+6 hours
1/x + 1/x+6=1/4
Solving this we get x=6.
Hence, A takes 6 hours to fill the cistern separately.
5)A tank is filled by 3 pipes with uniform flow. The first two pipes
operating simultaneously fill the tan in the same time during which
the tank is filled by the third pipe alone. The 2nd pipe fills the tank
5 hours faster than first pipe and 4 hours slower than third pipe. The
time required by first pipe is :
Sol: Suppose, first pipe take x hours to fill the tank then
B & C will take (x-5) and (x-9) hours respectively.
Therefore, 1/x +1/(x-5) =1/(x-9)
On solving, x=15
Hence, time required by first pipe is 15 hours.
6)A large tanker can be filled by two pipes A& B in 60min and 40 min
respectively. How many minutes will it take to fill the tanker from
empty state if B is used for half the time & A and B fill it together for
the other half?
Sol: Part filled by (A+B) n 1 min=(1/60 +1/40)=1/24
Suppose the tank is filled in x minutes
Then, x/2(1/24+1/40)=1
=> (x/2)*(1/15)=1
=> x=30 min.
7)Two pipes A and B can fill a tank in 6 hours and 4 hours respectively.
If they are opened on alternate hours and if pipe A s opened first, in
how many hours, the tank shall be full.
Sol: (A+B)'s 2 hours work when opened alternatively =1/6+1/4 =5/12
(A+B)'s 4 hours work when opened alternatively=10/12=5/6
Remaining part=1 -5/6=1/6.
Now, it is A's turn and 1/6 part is filled by A in 1 hour.
So, total time taken to fill the tank=(4+1)= 5 hours.
8)Three taps A,B and C can fill a tank in 12, 15 and 20 hours respectively.
If A is open all the time and B and C are open for one hour each
alternatively, the tank will be full in.
Sol: (A+B)'s 1 hour's work=1/12+1/15=9/60=3/20
(A+C)'s 1 hour's work=1/20+1/12=8/60=2/15
Part filled in 2 hours=3/20+2/15=17/60
Part filled in 2 hours=3/20+2/15= 17/60
Part filled in 6 hours=3*17/60 =17/20
Remaining part=1 -17/20=3/20
Now, it is the turn of A & B and 3/20 part is filled by A& B in 1 hour.
Therefore, total time taken to fill the tank=6+1=7 hours
9)A Booster pump can be used for filling as well as for emptying a tank.
The capacity of the tank is 2400 m3. The emptying capacity of the tank is
10 m3 per minute higher than its filling capacity and the pump needs 8
minutes lesser to empty the tank than it needs to fill it. What is the
filling capacity of the pump?
Sol: Let, the filling capacity of the pump be x m3/min
Then, emptying capacity of the pump=(x+10) m3/min.
So,2400/x – 2400/(x+10) = 8
on solving x=50.
10)A leak in the bottom of a tank can empty the full tan in 8 hr. An inlet
pipe fills water at the rate of 6 lits a minute. When the tank is full,
the inlet is opened and due to the leak, the tank is empty in 12 hrs.
How many liters does the cistern hold?
Sol: Work done by the inlet in 1 hr= 1/8 -1/12=1/24
Work done by the inlet in i min= (1/24)*(1/60)=1/1440
Therefore, Volume of 1/1440 part=6 lit
Volume of whole=(1440*6) lit=8640 lit.
11)Two pipes A and B can fill a cistern in 37 ½ min and 45 minutes
respectively. Both the pipes are opened. The cistern will be filled in
just half an hour, if the pipe B is turned off after:
sol: Let B be turned off after x min. Then,
Part filled by (A+B) in x min+ part filled by A in (30-x)min=1
Therefore, x(2/75+1/45)+(30-x)(2/75)=1
11x/225 + (60-2x)/75=1
11x+ 180-6x=225
x=9.
So, B must be turned off after 9 minutes.
Pipes and Cisterns
1.INLET:A pipe connected with a tank or cistern or a reservoir,
that fills it, it is known as Inlet.
OUTLET:A pipe connected with a tank or a cistern or a reservoir,
emptying it, is known as Outlet.
2. i) If a pipe can fill a tank in x hours, then :
part filled in 1 hour=1/x.
ii)If a pipe can empty a tank in y hours, then :
part emptied in 1 hour=1/y.
iii)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where y>x), then on
opening both the pipes, the net part filled in
1 hour=(1/x -1/y).
iv)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where x>y), then on opening
both the pipes, the net part filled in 1 hour=(1/y -1/x).
v) If two pipes A and B can fill a tank in x hours and y hours
respectively. If both the pipes are opened simultaneously, part
filled by A+B in 1 hour= 1/x +1/y.
3. Volume flow rate = Q
Volume = V
Q= V / t
t=time taken to fill or empty the tank.
that fills it, it is known as Inlet.
OUTLET:A pipe connected with a tank or a cistern or a reservoir,
emptying it, is known as Outlet.
2. i) If a pipe can fill a tank in x hours, then :
part filled in 1 hour=1/x.
ii)If a pipe can empty a tank in y hours, then :
part emptied in 1 hour=1/y.
iii)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where y>x), then on
opening both the pipes, the net part filled in
1 hour=(1/x -1/y).
iv)If a pipe can fill a tank in x hours and another pipe can
empty the full tank in y hours( where x>y), then on opening
both the pipes, the net part filled in 1 hour=(1/y -1/x).
v) If two pipes A and B can fill a tank in x hours and y hours
respectively. If both the pipes are opened simultaneously, part
filled by A+B in 1 hour= 1/x +1/y.
3. Volume flow rate = Q
Volume = V
Q= V / t
t=time taken to fill or empty the tank.
Time and Distance questions
Problems
1)A person covers a certain distance at 7kmph .How many meters
does he cover in 2 minutes.
Solution::
speed=72kmph=72*5/18 = 20m/s
distance covered in 2min =20*2*60 = 2400m
2)If a man runs at 3m/s. How many km does he run in 1hr 40min
Solution::
speed of the man = 3*18/5 kmph
= 54/5kmph
Distance covered in 5/3 hrs=54/5*5/3 = 18km
3)Walking at the rate of 4knph a man covers certain distance
in 2hr 45 min. Running at a speed of 16.5 kmph the man will
cover the same distance in.
Solution::
Distance=Speed* time
4*11/4=11km
New speed =16.5kmph
therefore Time=D/S=11/16.5 = 40min
Complex Problems
1)A train covers a distance in 50 min ,if it runs at a speed
of 48kmph on an average.The speed at which the train must run
to reduce the time of journey to 40min will be.
Solution::
Time=50/60 hr=5/6hr
Speed=48mph
distance=S*T=48*5/6=40km
time=40/60hr=2/3hr
New speed = 40* 3/2 kmph= 60kmph
2)Vikas can cover a distance in 1hr 24min by covering 2/3 of
the distance at 4 kmph and the rest at 5kmph.the total
distance is?
Solution::
Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km
3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr.
the usual time is.
Solution::
Usual speed = S
Usual time = T
Distance = D
New Speed is ¾ S
New time is 4/3 T
4/3 T – T = 5/2
T=15/2 = 7 ½
4)A man covers a distance on scooter .had he moved 3kmph
faster he would have taken 40 min less. If he had moved
2kmph slower he would have taken 40min more.the distance is.
Solution::
Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x --------------1
x/y-2 – x/y = 40/60 hr y(y-2) = 3x -----------------2
divide 1 & 2 equations
by solving we get x = 40
5)Excluding stoppages,the speed of the bus is 54kmph and
including stoppages,it is 45kmph.for how many min does the bus
stop per hr.
Solution::
Due to stoppages,it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min
6)Two boys starting from the same place walk at a rate of
5kmph and 5.5kmph respectively.wht time will they take to be
8.5km apart, if they walk in the same direction
Solution::
The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmph
Distance between them is 8.5 km
Time= 8.5km / 0.5 kmph = 17 hrs
7)2 trains starting at the same time from 2 stations 200km
apart and going in opposite direction cross each other ata
distance of 110km from one of the stations.what is the ratio of
their speeds.
Solution::
In same time ,they cover 110km & 90 km respectively
so ratio of their speed =110:90 = 11:9
8)Two trains start from A & B and travel towards each other at
speed of 50kmph and 60kmph resp. At the time of the meeting the
second train has traveled 120km more than the first.the distance
between them.
Solution::
Let the distance traveled by the first train be x km
then distance covered by the second train is x + 120km
x/50 = x+120 / 60
x= 600
so the distance between A & B is x + x + 120 = 1320 km
9)A thief steals a ca r at 2.30pm and drives it at 60kmph.the
theft is discovered at 3pm and the owner sets off in another car
at 75kmph when will he overtake the thief
Solution::
Let the thief is overtaken x hrs after 2.30pm
distance covered by the thief in x hrs = distance covered by
the owner in x-1/2 hr
60x = 75 ( x- ½)
x= 5/2 hr
thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm
10)In covering distance,the speed of A & B are in the ratio
of 3:4.A takes 30min more than B to reach the destion.The time
taken by A to reach the destinstion is.
Solution::
Ratio of speed = 3:4
Ratio of time = 4:3
let A takes 4x hrs,B takes 3x hrs
then 4x-3x = 30/60 hr
x = ½ hr
Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr
11)A motorist covers a distance of 39km in 45min by moving at a
speed of xkmph for the first 15min.then moving at double the
speed for the next 20 min and then again moving at his original
speed for the rest of the journey .then x=?
Solution::
Total distance = 39 km
Total time = 45 min
D = S*T
x * 15/60 + 2x * 20/60 + x * 10/60 = 39 km
x = 36 kmph
12)A & B are two towns.Mr.Fara covers the distance from A t0 B
on cycle at 17kmph and returns to A by a tonga running at a
uniform speed of 8kmph.his average speed during the whole
journey is.
Solution::
When same distance is covered with different speeds,then the
average speed = 2xy / x+y
=10.88kmph
13)A car covers 4 successive 3km stretches at speed of
10kmph,20kmph,30kmph&:60kmph resp. Its average speed is.
Solution::
Average speed = total distance / total time
total distance = 4 * 3 = 12 km
total time = 3/10 + 3/20 + 3/30 + 3/60
= 36/60 hr
speed =12/36 * 60 = 20 kmph
14)A person walks at 5kmph for 6hr and at 4kmph for 12hr.
The average speed is.
Solution::
avg speed = total distance/total time
= 5*6 + 4*12 / 18
=4 1/3 mph
15)A bullock cart has to cover a distance of 80km in 10hrs.
If it covers half of the journeyin 3/5th time.wht should be its
speed to cover the remaining distance in the time left.
Solution::
Time left = 10 - 3/5*10
= 4 hr
speed =40 km /4 hr
=10 kmph
16)The ratio between the speeds of the A& B is 2:3 an
therefore A takes 10 min more than the time taken by B to reach
the destination.If A had walked at double the speed ,he would
have covered the distance in ?
Solution::
Ratio of speed = 2:3
Ratio of time = 3:2
A takes 10 min more
3x-2x = 10 min
A's time=30 min
--->A covers the distance in 30 min ,if its speed is x
-> He will cover the same distance in 15 min,if its speed
doubles (i.e 2x)
17)A is twice as fast as B and B is thrice as fast as C is.
The journey covered by B in?
Solution::
speed's ratio
a : b = 2: 1
b : c = 3:1
Time's ratio
b : c = 1:3
b : c = 18:54
(if c covers in 54 min i..e twice to 18 min )
18)A man performed 3/5 of the total journey by ratio 17/20 by
bus and the remaining 65km on foot.wht is his total journey.
Solution::
Let total distance is x
x-(3/5x + 17/20 x) =6.5
x- 19x/20 = 6.5
x=20 * 6.5
=130 km
19)A train M leaves Meerat at 5 am and reaches Delhi at 9am .
Another train N leaves Delhi at 7am and reaches Meerut at 1030am
At what time do the 2 trains cross one another
Solution::
Let the distance between Meerut & Delhi be x
they meet after y hr after 7am
M covers x in 4hr
N covers x in 3 ½ i.e 7/2 hr
speed of M =x/4
speed of N = 2x/7
Distance covered by M in y+2 hr + Distance covered by N in
y hr is x
x/4 (y+2) +2x/7(y)=x
y=14/15hr or 56 min
20)A man takes 5hr 45min in walking to certain place and riding
back. He would have gained 2hrs by riding both ways.The time he
would take to walk both ways is?
Solution::
Let x be the speed of walked
Let y be the speed of ride
Let D be the distance
Then D/x + D/y = 23/4 hr -------1
D/y + D/y = 23/4 – 2 hr
D/y = 15/8 --------2
substitute 2 in 1
D/x + 15/8 = 23/4
D/x = 23/4 -15/8 =46-15/8 =31/8
Time taken for walk one way is 31/8 hr
time taken to walk to and fro is 2*31/8 = 31/4 hr
=7 hr 45 min
1)A person covers a certain distance at 7kmph .How many meters
does he cover in 2 minutes.
Solution::
speed=72kmph=72*5/18 = 20m/s
distance covered in 2min =20*2*60 = 2400m
2)If a man runs at 3m/s. How many km does he run in 1hr 40min
Solution::
speed of the man = 3*18/5 kmph
= 54/5kmph
Distance covered in 5/3 hrs=54/5*5/3 = 18km
3)Walking at the rate of 4knph a man covers certain distance
in 2hr 45 min. Running at a speed of 16.5 kmph the man will
cover the same distance in.
Solution::
Distance=Speed* time
4*11/4=11km
New speed =16.5kmph
therefore Time=D/S=11/16.5 = 40min
Complex Problems
1)A train covers a distance in 50 min ,if it runs at a speed
of 48kmph on an average.The speed at which the train must run
to reduce the time of journey to 40min will be.
Solution::
Time=50/60 hr=5/6hr
Speed=48mph
distance=S*T=48*5/6=40km
time=40/60hr=2/3hr
New speed = 40* 3/2 kmph= 60kmph
2)Vikas can cover a distance in 1hr 24min by covering 2/3 of
the distance at 4 kmph and the rest at 5kmph.the total
distance is?
Solution::
Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km
3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr.
the usual time is.
Solution::
Usual speed = S
Usual time = T
Distance = D
New Speed is ¾ S
New time is 4/3 T
4/3 T – T = 5/2
T=15/2 = 7 ½
4)A man covers a distance on scooter .had he moved 3kmph
faster he would have taken 40 min less. If he had moved
2kmph slower he would have taken 40min more.the distance is.
Solution::
Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x --------------1
x/y-2 – x/y = 40/60 hr y(y-2) = 3x -----------------2
divide 1 & 2 equations
by solving we get x = 40
5)Excluding stoppages,the speed of the bus is 54kmph and
including stoppages,it is 45kmph.for how many min does the bus
stop per hr.
Solution::
Due to stoppages,it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min
6)Two boys starting from the same place walk at a rate of
5kmph and 5.5kmph respectively.wht time will they take to be
8.5km apart, if they walk in the same direction
Solution::
The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmph
Distance between them is 8.5 km
Time= 8.5km / 0.5 kmph = 17 hrs
7)2 trains starting at the same time from 2 stations 200km
apart and going in opposite direction cross each other ata
distance of 110km from one of the stations.what is the ratio of
their speeds.
Solution::
In same time ,they cover 110km & 90 km respectively
so ratio of their speed =110:90 = 11:9
8)Two trains start from A & B and travel towards each other at
speed of 50kmph and 60kmph resp. At the time of the meeting the
second train has traveled 120km more than the first.the distance
between them.
Solution::
Let the distance traveled by the first train be x km
then distance covered by the second train is x + 120km
x/50 = x+120 / 60
x= 600
so the distance between A & B is x + x + 120 = 1320 km
9)A thief steals a ca r at 2.30pm and drives it at 60kmph.the
theft is discovered at 3pm and the owner sets off in another car
at 75kmph when will he overtake the thief
Solution::
Let the thief is overtaken x hrs after 2.30pm
distance covered by the thief in x hrs = distance covered by
the owner in x-1/2 hr
60x = 75 ( x- ½)
x= 5/2 hr
thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm
10)In covering distance,the speed of A & B are in the ratio
of 3:4.A takes 30min more than B to reach the destion.The time
taken by A to reach the destinstion is.
Solution::
Ratio of speed = 3:4
Ratio of time = 4:3
let A takes 4x hrs,B takes 3x hrs
then 4x-3x = 30/60 hr
x = ½ hr
Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr
11)A motorist covers a distance of 39km in 45min by moving at a
speed of xkmph for the first 15min.then moving at double the
speed for the next 20 min and then again moving at his original
speed for the rest of the journey .then x=?
Solution::
Total distance = 39 km
Total time = 45 min
D = S*T
x * 15/60 + 2x * 20/60 + x * 10/60 = 39 km
x = 36 kmph
12)A & B are two towns.Mr.Fara covers the distance from A t0 B
on cycle at 17kmph and returns to A by a tonga running at a
uniform speed of 8kmph.his average speed during the whole
journey is.
Solution::
When same distance is covered with different speeds,then the
average speed = 2xy / x+y
=10.88kmph
13)A car covers 4 successive 3km stretches at speed of
10kmph,20kmph,30kmph&:60kmph resp. Its average speed is.
Solution::
Average speed = total distance / total time
total distance = 4 * 3 = 12 km
total time = 3/10 + 3/20 + 3/30 + 3/60
= 36/60 hr
speed =12/36 * 60 = 20 kmph
14)A person walks at 5kmph for 6hr and at 4kmph for 12hr.
The average speed is.
Solution::
avg speed = total distance/total time
= 5*6 + 4*12 / 18
=4 1/3 mph
15)A bullock cart has to cover a distance of 80km in 10hrs.
If it covers half of the journeyin 3/5th time.wht should be its
speed to cover the remaining distance in the time left.
Solution::
Time left = 10 - 3/5*10
= 4 hr
speed =40 km /4 hr
=10 kmph
16)The ratio between the speeds of the A& B is 2:3 an
therefore A takes 10 min more than the time taken by B to reach
the destination.If A had walked at double the speed ,he would
have covered the distance in ?
Solution::
Ratio of speed = 2:3
Ratio of time = 3:2
A takes 10 min more
3x-2x = 10 min
A's time=30 min
--->A covers the distance in 30 min ,if its speed is x
-> He will cover the same distance in 15 min,if its speed
doubles (i.e 2x)
17)A is twice as fast as B and B is thrice as fast as C is.
The journey covered by B in?
Solution::
speed's ratio
a : b = 2: 1
b : c = 3:1
Time's ratio
b : c = 1:3
b : c = 18:54
(if c covers in 54 min i..e twice to 18 min )
18)A man performed 3/5 of the total journey by ratio 17/20 by
bus and the remaining 65km on foot.wht is his total journey.
Solution::
Let total distance is x
x-(3/5x + 17/20 x) =6.5
x- 19x/20 = 6.5
x=20 * 6.5
=130 km
19)A train M leaves Meerat at 5 am and reaches Delhi at 9am .
Another train N leaves Delhi at 7am and reaches Meerut at 1030am
At what time do the 2 trains cross one another
Solution::
Let the distance between Meerut & Delhi be x
they meet after y hr after 7am
M covers x in 4hr
N covers x in 3 ½ i.e 7/2 hr
speed of M =x/4
speed of N = 2x/7
Distance covered by M in y+2 hr + Distance covered by N in
y hr is x
x/4 (y+2) +2x/7(y)=x
y=14/15hr or 56 min
20)A man takes 5hr 45min in walking to certain place and riding
back. He would have gained 2hrs by riding both ways.The time he
would take to walk both ways is?
Solution::
Let x be the speed of walked
Let y be the speed of ride
Let D be the distance
Then D/x + D/y = 23/4 hr -------1
D/y + D/y = 23/4 – 2 hr
D/y = 15/8 --------2
substitute 2 in 1
D/x + 15/8 = 23/4
D/x = 23/4 -15/8 =46-15/8 =31/8
Time taken for walk one way is 31/8 hr
time taken to walk to and fro is 2*31/8 = 31/4 hr
=7 hr 45 min
Time and Distance
I)Speed = Distance/Time
II)Time = Distance/speed
III) Distance = speed*time
IV) 1km/hr = 5/18 m/s
V)1 m/s = 18/5 Km/hr
VI)If the ratio of the speed of A and B is a:b,then the ratio of
the time taken by them to cover the same distance is 1/a : 1/b
or b:a
VII) suppose a man covers a distance at x kmph and an equal
distance at y kmph.then the average speed during the whole
journey is (2xy/x+y)kmph
II)Time = Distance/speed
III) Distance = speed*time
IV) 1km/hr = 5/18 m/s
V)1 m/s = 18/5 Km/hr
VI)If the ratio of the speed of A and B is a:b,then the ratio of
the time taken by them to cover the same distance is 1/a : 1/b
or b:a
VII) suppose a man covers a distance at x kmph and an equal
distance at y kmph.then the average speed during the whole
journey is (2xy/x+y)kmph
Trains
(1) Time taken by a train x mt long in passing a signal post
or a pole or a standing man = time taken by the train to cover x mt
(2) Time taken by a train x mt long in passing a stationary
object of length y mt = time taken by the train to cover x+y mt
(3) Suppose two trains or two bodies are moving in the same
direction at u kmph and v kmph such that u > v then their
relative speed is u-v kmph
(4)If two trains of length x km and y km are moving in opposite
diredtions at u kmph and vmph,then time taken by the train to
cross each other = (x+y)/(u+v) hr
(5) Suppose two trains or two bdies are moving in opposite direction
at u kmph and v kmph then,their relative speed = (u+v) kmph
(6)If two train start at the same time from 2 points A & B towards
each other and after crossing they take a & b hours in reaching B & A
respectively then A's speed : B's speed = (b^1/2 : a^1/2 )
or a pole or a standing man = time taken by the train to cover x mt
(2) Time taken by a train x mt long in passing a stationary
object of length y mt = time taken by the train to cover x+y mt
(3) Suppose two trains or two bodies are moving in the same
direction at u kmph and v kmph such that u > v then their
relative speed is u-v kmph
(4)If two trains of length x km and y km are moving in opposite
diredtions at u kmph and vmph,then time taken by the train to
cross each other = (x+y)/(u+v) hr
(5) Suppose two trains or two bdies are moving in opposite direction
at u kmph and v kmph then,their relative speed = (u+v) kmph
(6)If two train start at the same time from 2 points A & B towards
each other and after crossing they take a & b hours in reaching B & A
respectively then A's speed : B's speed = (b^1/2 : a^1/2 )
BOATS AND STREAMS
1)In water, the direction along the stream is called down stream.
2)Direction against the stream is called upstream.
3)The speed of boat in still water is U km/hr and the speed of stream is V
km/hr then
speed down stream =U + V km/hr
speed up stream = U - V km/hr
Formulas:
If the speed down stream is A km/hr and the speed up stream is B km/hr then
speed in still water = 1/2(A+B) km/hr
rate of stream =1/2(A-B) km/hr
2)Direction against the stream is called upstream.
3)The speed of boat in still water is U km/hr and the speed of stream is V
km/hr then
speed down stream =U + V km/hr
speed up stream = U - V km/hr
Formulas:
If the speed down stream is A km/hr and the speed up stream is B km/hr then
speed in still water = 1/2(A+B) km/hr
rate of stream =1/2(A-B) km/hr
Some learning points about prime
Prime Numbers:
A natural number larger than unity is a prime number if it
does not have other divisors except for itself and unity.
Note:-Unity i e,1 is not a prime number.
Properties Of Prime Numbers:
->The lowest prime number is 2.
->2 is also the only even prime number.
->The lowest odd prime number is 3.
->The remainder when a prime number p>=5 is divided by 6 is 1 or 5.However,
if a number on being divided by 6 gives a remainder 1 or 5 need not be
prime....so, p = 6k+1 or 6k+5
->The remainder of division of the square of a prime number p>=5 divide by
24 is 1.....p^2 = 24k+1
->For prime numbers p>3, p²-1 is divided by 24.
->If a and b are any 2 odd primes then a²-b² is composite. Also a²+b²
is composite.
->The remainder of the division of the square of a prime number p>=5
divided by 12 is 1....p^2 = 12k + 1
Process to Check A Number s Prime or not:
Take the square root of the number.
Round of the square root to the next highest integer call this number as Z.
Check for divisibility of the number N by all prime numbers below Z. If
there is no numbers below the value of Z which divides N then the number
will be prime.
Example 239 is prime or not?
√239 lies between 15 or 16.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
239 is not divisible by any of these. Hence we can conclude that 239
is a prime number.
Learn
n-->Number pn--->number of primes less than n
n pn
10 4
50 14
100 24
Sum of 1st 9 primes = 100
Averages
1.Average=Sum of quantities/Number of quantities.
2.Suppose a man covers a certain distance at x kmph
and an equal distance at y kmph ,then the average speed
during the whole journey is (2xy/x+y) kmph.
3. If the elements are in A.P. , than (n=number of elements)mean = (Middle Element) if n is odd = mean of middle 2 elements if n is odd
Examples:
1.Find the average of all these numbers.142,147,153,165,157.
Solution:
142 147 153 165 157
Here consider the least number i.e, 142
comparing with others,
142 147 153 165 157
+5 +11 +23 +15
Now add 5+11+23+15 = 52/5 = 10.8
Now add 10.8 to 142 we get 152.8
(Average of all these numbers).
Answer is 152.8
2.Find the average of all these numbers.4,10,16,22,28
Solution:
4,10,16,22,28
As the difference of number is 6
Then the average of these numbers is central one i.e, 16.
Answer is 16.
3.Find the average of all these numbers.4,10,16,22,28,34.
Solution:
Here also difference is 6.
Then middle numbers 16,22 take average of these
two numbers 16+22/2=19
Therefore the average of these numbers is 19.
Answer is 19.
4.The average marks of a marks of a student in 4 Examination
is 40.If he got 80 marks in 5th Exam then what is
his new average.
Solution:
4*40+80=240
Then average means 240/5=48.
Answer is 48.
5.In a group the average income of 6 men is 500 and that
of 5 women is 280, then what is average income of the group.
Solution:
6*500+5*280=4400
then average is 4400/11=400.
Another Method: here consider for 6 men
6 men – each 500.
so 5th women is 280.
then 500-280=220.
then 220*6/11=120.
therefore 120+280=400.
Answer is 400.
6.The average weight of a class of 30 students is 40 kgs
if the teacher weight is included then average increases by 2 kgs then find the weight of the teacher?
Solution:
30 students average weight is 40 kgs.
So,when teacher weight is added it increases by 2 kgs
so total 31 persons ,therefore 31*2=62.
Now add the average weight of all student to it
we get teachers weight i.e, 62+40=102 kgs.
Answer is 102 kgs.
7.The average age of Mr and Mrs Sharma 4 years ago is 28 years . If the present average age of Mr and Mrs Sharma and their son is 22 years. What is the age of their son.
Solution:
4 years ago their average age is 28 years.
So their present average age is 32 years.
32 years for Mr and Mrs Sharma then 32*2=64 years.
Then present age including their son is 22 years.
So 22*3 =66 years.
Therefore son age will be 66-64 = 2 years.
Answer is 2 years.
8.The average price of 10 books is increased by 17 Rupees when one of them whose value is Rs.400 is replaced by a new book.
What is the price of new book?
Solution:
10 books Average increases by 17 Rupees
so 10*17= 170.
so the new book cost is more and by adding its cost average
increase,therefore the cost of new book is 400+170=570Rs.
Answer is 570 Rs.
9.The average marks of girls in a class is 62.5. The average marks of 4 girls among them is 60.The average marks of remaining girls
is 63,then what is the number of girls in the class?
Solution:
Total number of girls be x+4.
Average marks of 4 girls is 60.
therefore 62.5-60=2.5
then 4*2.5 =10.
the average of remaining girls is 63
here 0.5 difference therefore 0.5*x=10(since we got from 4 girls)
(this is taken becoz both should be equal)
x=10/0.5
x=20.
This clear says that remaining are 20 girls
therefore total is x+4=20+4=24 girls
Answer is 24 girls.
10.Find the average of first 50 natural numbers.
Solution:
Sum of the Natural Numbers is n(n+1)/2
Average = (n+1)/2
=51/2 = 25.5
11.The average of the first nine prime number is?
Solution:
Prime numbers are 2,3,5,7,11,13,17,19,23
therefore 2+3+5+7+11+13+17+19+23=100
then the average 100/9= 11 1/9.
Answer is 11 1/9.
12.The average of 2,7,6 and x is 5 and the average of and the average of 18,1,6,x and y is 10 .what is the value of y?
Solution:
2+7+6+x/4=5
=>15+x=20
=>x=5.
18+1+6+x+y/5=10
=>25+5+y=50
=>y=20.
13.The average of a non-zero number and its square is 5 times the
number.The number is
Solution:
The number be x
then x+x2/2=5x
=>x2-9x=0
=>x(x-9)=0
therefore x=0 or x=9.
The number is 9.
14.Nine persons went to a hotel for taking their meals . Eight of them spent Rs.12 each on their meals and the ninth spent Rs.8 more then the average expenditure of all the nine. What was the total money spent by them?
Solution:
The average expenditure be x.
then 8*12+(x+8)=9x
=>96+x+8=9x.
=>8x=104
=>x=13
Total money spent =9x=>9*13=117
Answer is Rs.117
15.The average weight of A.B.C is 45 Kgs.If the average weight of A and B be 40 Kgs and that of Band C be 43 Kgs. Find the weight of B?
Solution:
The weight of A,B,Care 45*3=135 Kgs.
The weight of A,B are 40*2=80 Kgs.
The weight of B,C are 43*2=86 Kgs.
To get the Weight of B.
(A+B)+(B+C)-(A+B+C)=80+86-135
B=31 kgs.
Answer is 31 Kgs.
16.The sum of three consecutive odd number is 48 more than the average of these number .What is the first of these numbers?
Solution:
let the three consecutive odd numbers are x, x+2, x+4.
By adding them we get x+x+2+x+4=3x+6.
Then 3x+6-(3x+6)/3=38(given)
=>2(3x+6)=38*3.
=>6x+12=114
=>6x=102
=>x=17.
Answer is 17.
17.A family consists of grandparents,parents and three grandchildren.
The average age of the grandparents is 67 years,that of parents is 35 years and that of the grand children is 6 years . What is the average age of the family?
Solution:
grandparents age is 67*2=134
parents age is 35*2=70
grandchildren age is 6*3=18
therefore age of family is 134+70+18=222
average is 222/7=31 5/7 years.
Answer is 31 5/7 years.
18.A library has an average of 510 visitors
on Sundays and 240 on other days .The average number of visitors per day in a month 30 days beginning with a Sunday is?
Solution:
Here specified that month starts with Sunday
so, in a month there are 5 Sundays.
Therefore remaining days will be 25 days.
510*5+240*25=2550+6000
=8550 visitors.
The average visitors are 8550/30=285.
Answer is 285.
19.The average age of a class of 39 students is 15 years .
If the age of the teacher be included ,then average
increases by 3 months. Find the age of the teacher.
Solution: Total age for 39 persons is 39*15=585 years.
Now 40 persons is 40* 61/4=610 years
(since 15 years 3 months=15 3/12=61/4)
Age of the teacher =610-585 years
=>25 years.
Answer is 25 years.
20.The average weight of a 10 oarsmen in a boat is increases by 1.8 Kgs .When one of the crew ,who weighs 53 Kgs is
replaced by new man. Find the weight of the new man.
Solution: Weight of 10 oars men is increases by 1.8 Kgs
so, 10*1.8=18 Kgs
therefore 53+18=71 Kgs will be the weight of the man.
Answer is 71 Kgs.
21.A bats man makes a score of 87 runs in the 17th inning
and thus increases his average by 3. Find the average
after 17th inning.
Solution: Average after 17 th inning =x
then for 16th inning is x-3.
Therefore 16(x-3)+87 =17x
=>x=87-48
=>x=39.
Answer is 39.
22.The average age of a class is 15.8 years .The average age of boys in the class is 16.4 years while that of the girls
is 15.4 years .What is the ratio of boys to girls in the class.
Solution: Ratio be k:1 then
k*16.4 + 1*15.4 = (k+1)*15.8
=>(16.4-15.8)k=15.8-15.4
=>k=0.4/0.6
=>k=2/3
therefore 2/3:1=>2:3
Answer is 2:3
23.In a cricket eleven ,the average of eleven players is 28 years .Out of these ,the average ages of three groups of players each are 25 years,28 years, and 30 years respectively. If in these groups ,the captain and the
youngest player are not included and the captain is
eleven years older than the youngest players ,
what is the age of the captain?
Solution: let the age of youngest player be x
then ,age of the captain =(x+11)
therefore 3*25 + 3*28 + 3*30 + x + x+11=11*28
=>75+84+90+2x+11=308
=>2x=48
=>x=24.
Therefore age of the captain =(x+11)= 24+11= 35 years.
Answer is 35 years.
24.The average age of the boys in the class is twice
the number of girls in the class .If the ratio of
boys and girls in the class of 36 be 5:1, what is
the total of the age (in years) of the boys in the class?
Solution: Number of boys=36*5/6=30
Number of girls =6
Average age of boys =2*6=12 years
Total age of the boys=30*12=360 years
Answer is 360 years.
25.Five years ago, the average age of P and Q was
15 years ,average age of P,Q, and R today is
20 years,how old will R be after 10 years?
Solution: Age of P and Q are 15*2=30 years
Present age of P and Q is 30+5*2=40 years.
Age of P Q and R is 20*3= 60 years.
R ,present age is 60-40=20 years
After 10 years =20+10=30 years.
Answer is 30 years.
26.The average weight of 3 men A,B and C is 84 Kgs.
Another man D joins the group and the average now
becomes 80 Kgs.If another man E whose weight is
3 Kgs more than that of D ,replaces A then the
average weight B,C,D and E becomes 79 Kgs.
The weight of A is.
Solution:Total weight of A, B and C is 84 * 3 =252 Kgs.
Total weight of A,B,C and Dis 80*4=320 Kgs
Therefore D=320-252=68 Kgs.
E weight (68+3)=71 kgs
Total weight of B,C,D and E = 79*4=316 Kgs
(A+B+C+D)-(B+C+D+E)=320-316 =4Kgs
A-E=4Kgs
A-71=4 kgs
A=75 Kgs
Answer is 75 kgs
27.A team of 8 persons joins in a shooting competition.
The best marksman scored 85 points.If he had scored
92 points ,the average score for the team would
have been 84.The team scored was.
Solution: Here consider the total score be x.
therefore x+92-85/8=84
=>x+7=672
=>x=665.
Answer is 665
28.A man whose bowling average is 12.4,takes 5
wickets for 26 runs and there by decrease his average by 0.4. The number of wickets,taken by him before his last match is:
Solution: Number of wickets taken before last match be x.
therefore 12.4x26/x+5=12(since average decrease by 0.4
therefore 12.4-0.4=12)
=>12.4x+2612x+60
=>0.4x=34
=>x=340/4
=>x=85.
Answer is 85.
29.The mean temperature of Monday to Wednesday was 37 degrees and of Tuesday to Thursday was 34 degrees .If the temperature on Thursday was 4/5th that of Monday. The temperature on Thursday was:
Solution:
The total temperature recorded on Monday,Wednesday was 37*3=111.
The total temperature recorded on Tuesday,
Wednesday,Thursday was 34*3=102.
and also given that Th=4/5M
=>M=5/4Th
(M+T+W)-(T+W+Th)=111-102=9
M-Th=9
5/4Th-Th=9
Th(1/4)=9
=>Th=36 degrees.
30. 16 children are to be divided into two groups A and B
of 10 and 6 children. The average percent marks obtained
by the children of group A is 75 and the average percent
marks of all the 16 children is 76. What is the average
percent marks of children of groups B?
Solution: Here given average of group A and whole groups .
So,(76*16)-(75*10)/6
=>1216-750/6
=>466/6=233/3=77 2/3
Answer is 77 2/3.
31.Of the three numbers the first is twice the second and the second is twice the third .The average of the reciprocal of the numbers is 7/72,the number are.
Solution:Let the third number be x
Let the second number be 2x.
Let the first number be 4x.
Therefore average of the reciprocal means
1/x+1/2x+1/4x=(7/72*3)
7/4x=7/24
=>4x=24
x=6.
Therefore
First number is 4*6=24.
Second number is 2*6=12
Third number is 1*6=6
Answer is 24,12,6.
32.The average of 5 numbers is 7.When 3 new numbers
are added the average of the eight numbers is 8.5.
The average of the three new number is:
Solution: Sum of three new numbers=(8*8.5-5*7)=33
Their average =33/3=11.
Answer is 11.
33.The average temperature of the town in the first
four days of a month was 58 degrees. The average
for the second ,third,fourth and fifth days was
60 degree .If the temperature of the first and
fifth days were in the ratio 7:8 then what is
the temperature on the fifth day?
Solution :
Sum of temperature on 1st 2nd 3rd
and 4th days =58*4=232 degrees.
Sum of temperature on 2nd 3rd 4th
and 5th days =60*4=240 degrees
Therefore 5th day temperature is 240-232=8 degrees.
The ratio given for 1st and 5th days be 7x and 8x degrees
then 8x-7x=8
=>x=8.
therefore temperature on the 5th day =8x=8*8=64 degrees.
H.C.F AND L.C.M
Facts And Formulae:
1.Highest Common Factor:(H.C.F) or Greatest Common Meaure(G.C.M)
The H.C.F of two or more than two numbers is the greatest
number that divides each of them exactly.
There are two methods :
1.Factorization method: Express each one of the given numbers as the product of prime factors.
The product of least powers of common prime factors gives HCF.
Example : Find HCF of 26 * 32*5*74 , 22 *35*52 * 76 , 2*52 *72
Sol: The prime numbers given common numbers are 2,5,7
Therefore HCF is 22 * 5 *72 .
2.Division Method : Divide the larger number by
smaller one. Now divide the divisor by
remainder. Repeat the process
of dividing preceding number last obtained till
zero is obtained as
number. The last divisor is HCF.
Example: Find HCF of 513, 1134, 1215
Sol:1134) 1215(1HCF of 81 and 513 is 27.3.Least common multiple[LCM] : The least number which is divisible by each one of given numbers is LCM.There are two methods for this:1.Factorization method : Resolve each one into product of prime factors. Then LCM is product of highest powers of all factors.2.Common division method.Problems:1.The HCF of 2 numbers is 11 and LCM is 693.If one of numbers is 77.find other.Sol: Other number = 11 * 693/77=99.2.Find largest number of 4 digits divisible by 12,15,18,27Sol: The largest number is 9999.LCM of 12,15,18,27 is 540.on dividing 9999 by 540 we get 279 as remainder.Therefore number =9999 – 279 =9720.3.Find least number which when divided by 20,25,35,40 leaves remainders 14,19,29,34.Sol:
1134
----------
81)1134(14
81
-----------
324
324
-----------
0
-----------
HCF of this two numbers is 81.
81)513(6
486
--------
27)81(3
81
-----
0
---20–14=6Therefore number =LCM of (20,25,35,40) - 6=13944.252 can be expressed as prime as :
25-19=6
35-29=6
40-34=6
2 2525.1095/1168 when expressed in simple form is
2 126
3 63
3 21
7
prime factor is 2 *2 * 3 * 3 *71095)1168(1Therefore 1095/1168 = 1095/73/1168/73= 15/166.GCD of 1.08,0.36,0.9 is Sol: HCF of 108,36,90
1095
------
73)1095(15
73
---------
365
365
---------
0
----------
So, HCF is 73
36)90(27.Three numbers are in ratio 1:2:3 and HCF is 12.Find numbers.Sol: Let the numbers be x.Three numbers are x,2x,3xTherefore HCF is
72
----
18)36(2
36
----
0
----
HCF is 18.
HCF of 18 and 108 is 18
18)108(6
108
-------
0
--------
Therefore HCF =0.182x)3x(1Therefore numbers are 12,24,36.8.The sum of two numbers is 216 and HCF is 27.Sol: Let numbers are
2x
-----
x)2x(2
2x
--------
0
-------------
HCF is x so, x is 1227a + 27 b =216Co-primes of 8 are (1,7) and (3,5)
a + b =216/27=8
numbers=(27 * 1 ), (27 * 7)9.LCM of two numbers is 48..The numbers are in ratio 2:3.The sum of numbers isSol: Let the number be x.Numbers are 2x,3xLCM of 2x,3x is 6xTherefore 6x=48x=8.Numbers are 16 and 24Sum=16 +24=40.10.HCF and LCM of two numbers are 84 and 21.If ratio of two numbers is 1:4.Then largest of two numbers is Sol: Let the numbers be x,4xThen x * 4x = 84 * 21 x2 =84 * 21 /4 x = 21 Largest number is 4 * 21.11.HCF of two numbers is 23,and other factors of LCM are 13,14.Largest number isSol: 23 * 14 is Largest number.12.The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets same number of pens and pencils is ?Sol: HCF of 1001 and 910
=27,89910)1001(113.The least number which should be added to 2497 so that sum is divisible by 5,6,4,3 ?Sol: LCM of 5,6,4,3 is 60.On dividing 2497 by 60 we get 37 as remainder.Therefore number to added is 60 – 37 =23.Answer is 23.14.The least number which is a perfect square and is divisible by each of numbers 16,20,24 is ?Sol: LCM of 16,20,24 is 240.2 * 2*2*2*3*5=240To make it a perfect square multiply by 3 * 5Therefore 240 * 3 * 5=3600Answer is 3600.
910
------------
91)910(10
910
--------
0
---------
Therefore HCF=91
Saturday, 5 December 2009
Numbers
Introduction:
Natural Numbers:
All positive integers are natural numbers.
Ex 1,2,3,4,8,......
There are infinite natural numbers and number 1 is the least natural number.
Based on divisibility there would be two types of natural numbers. They are
Prime and composite.
Prime numbers
Composite Numbers:
The numbers which are not prime are known as composite numbers.
Co-Primes:
Two numbers a an b are said to be co-primes,if their H.C.F is 1.
Example (2,3),(4,5),(7,9),(8,11).....
Place value or Local value of a digit in a Number:
place value:
Example 689745132
Place value of 2 is (2*1)=2
Place value of 3 is (3*10)=30 and so on.
Face value:-It is the value of the digit itself at whatever
place it may be.
Example 689745132
Face value of 2 is 2.
Face value of 3 is 3 and so on.
Tests of Divisibility
BASIC FORMULAE:
->(a+b)²=a²+b²+2ab
->(a-b)²=a²+b²-2ab
->(a+b)²-(a-b)²=4ab
->(a+b)²+(a-b)²=2(a²+b²)
->a²-b²=(a+b)(a-b)
->(a+b+c)²=a²+b²+c²+2(ab+b c+ca)
->a³+b³=(a+b)(a²+b²-ab)
->a³-b³=(a-b)(a²+b²+ab)
->a³+b³+c³-3a b c=(a+b+c)(a²+b²+c²-ab-b c-ca)
->If a+b+c=0 then a³+b³+c³=3a b c
DIVISION ALGORITHM
If we divide a number by another number ,then
Dividend = (Divisor * quotient) + Remainder
MULTIPLICATION BY SHORT CUT METHODS
1.Multiplication by distributive law:
a)a*(b+c)=a*b+a*c
b)a*(b-c)=a*b-a*c
Example
a)567958*99999=567958*(100000-1)
567958*100000-567958*1
56795800000-567958
56795232042
b)978*184+978*816=978*(184+816)
978*1000=978000
2.Multiplication of a number by 5n:-Put n zeros to the right of the
multiplicand and divide the number so formed by 2n
Example 975436*625=975436*54=9754360000/16=609647500.
PROGRESSION
Problems
1.Simplify
a.8888+888+88+8
b.11992-7823-456
Solution: a.8888
888
88
8
9872
b.11992-7823-456=11992-(7823+456)
=11992-8279=3713
2.What could be the maximum value of Q in the following equation?
5PQ+3R7+2Q8=1114
Solution: 5 P Q
3 R 7
2 Q 8
11 1 4
2+P+Q+R=11
Maximum value of Q =11-2=9 (P=0,R=0)
3.Simplify: a.5793405*9999 b.839478*625
Solution:
a. 5793405*9999=5793405*(10000-1)
57934050000-5793405=57928256595
b. 839478*625=839478*54=8394780000/16=524673750.
4.Evaluate 313*313+287*287
Solution:
a²+b²=1/2((a+b)²+(a-b)²)
1/2(313+287)² +(313-287)²=1/2(600 ² +26 ² )
½(360000+676)=180338
5.Which of the following is a prime number?
a.241 b.337 c.391
Solution:
a.241
16>√241.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
241 is not divisible by any of these. Hence we can
conclude that 241 is a prime number.
b. 337
19>√337.Hence take the value of Z=19.
Prime numbers less than 16 are 2,3,5,7,11,13 and 17.
337 is not divisible by any of these. Hence we can conclude
that 337 is a prime number.
c. 391
20>√391.Hence take the value of Z=20.
Prime numbers less than 16 are 2,3,5,7,11,13,17 and 19.
391 is divisible by 17. Hence we can conclude
that 391 is not a prime number.
6.Find the unit's digit n the product 2467 153 * 34172?
Solution: Unit's digit in the given product=Unit's digit in 7 153 * 172
Now 7 4 gives unit digit 1
7 152 gives unit digit 1
7 153 gives 1*7=7.Also 172 gives 1
Hence unit's digit in the product =7*1=7.
7.Find the total number of prime factors in 411 *7 5 *112 ?
Solution: 411 7 5 112= (2*2) 11 *7 5 *112
= 222 *7 5 *112
Total number of prime factors=22+5+2=29
8.Which of the following numbers s divisible by 3?
a.541326
b.5967013
Solution: a. Sum of digits in 541326=5+4+1+3+2+6=21 divisible by 3.
b. Sum of digits in 5967013=5+9+6+7+0+1+3=31 not divisible by 3.
9.What least value must be assigned to * so that th number 197*5462 is
divisible by 9?
Solution: Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.
10.What least number must be added to 3000 to obtain a number exactly
divisible by 19?
Solution:On dividing 3000 by 19 we get 17 as remainder
Therefore number to be added = 19-17=2.
11.Find the smallest number of 6 digits which is exactly divisible by 111?
Solution:Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Number to be added =111-100=11.
Hence,required number =10011.
12.On dividing 15968 by a certain number the quotient is 89 and the remainder
is 37.Find the divisor?
Solution:Divisor = (Dividend-Remainder)/Quotient
=(15968-37) / 89
=179.
13.A number when divided by 342 gives a remainder 47.When the same number
is divided by 19 what would be the remainder?
Solution:Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.
The given number when divided by 19 gives 18 K + 2 as quotient and 9 as
remainder.
14.A number being successively divided by 3,5,8 leaves remainders 1,4,7
respectively. Find the respective remainders if the order of
divisors are reversed?
Solution:Let the number be x.
3 x 5 y - 1 8 z - 4 1 - 7 z=8*1+7=15
y=5z+4 = 5*15+4 = 79
x=3y+1 = 3*79+1=238
Now
8 238
5 29 - 6
3 5 - 4
1 - 2
Respective remainders are 6,4,2.
15.Find the remainder when 231 is divided by 5?
Solution:210 =1024.unit digit of 210 * 210 * 210 is 4 as
4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.
16.How many numbers between 11 and 90 are divisible by 7?
Solution:The required numbers are 14,21,28,...........,84
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.
17.Find the sum of all odd numbers up to 100?
Solution:The given numbers are 1,3,5.........99.
This is an A.P with a=1,d=2.
Let it contain n terms 1+(n-1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.
18.How many terms are there in 2,4,6,8..........,1024?
Solution:Clearly 2,4,6........1024 form a G.P with a=2,r=2
Let the number of terms be n
then 2*2 n-1=1024
2n-1 =512=29
n-1=9
n=10.
19.2+22+23+24+25..........+28=?
Solution:Given series is a G.P with a=2,r=2 and n=8.
Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.
=2*255=510.
20.A positive number which when added to 1000 gives a sum ,
which is greater than when it is multiplied by 1000.The positive integer is?
a.1 b.3 c.5 d.7
Solution:1000+N>1000N
clearly N=1.
21.The sum of all possible two digit numbers formed from three
different one digit natural numbers when divided by the sum of the
original three numbers is equal to?
a.18 b.22 c.36 d. none
Solution:Let the one digit numbers x,y,z
Sum of all possible two digit numbers=
=(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y)
= 22(x+y+z)
Therefore sum of all possible two digit numbers when divided by sum of
one digit numbers gives 22.
22.The sum of three prime numbers is 100.If one of them exceeds another by
36 then one of the numbers is?
a.7 b.29 c.41 d67.
Solution:x+(x+36)+y=100
2x+y=64
Therefore y must be even prime which is 2
2x+2=64=>x=31.
Third prime number =x+36=31+36=67.
23.A number when divided by the sum of 555 and 445 gives two times
their difference as quotient and 30 as remainder .The number is?
a.1220 b.1250 c.22030 d.220030.
Solution:Number=(555+445)*(555-445)*2+30
=(555+445)*2*110+30
=220000+30=220030.
24.The difference between two numbers s 1365.When the larger number is
divided by the smaller one the quotient is 6 and the remainder is 15.
The smaller number is?
a.240 b.270 c.295 d.360
Solution:Let the smaller number be x, then larger number =1365+x
Therefore 1365+x=6x+15
5x=1350 => x=270
Required number is 270.
25.In doing a division of a question with zero remainder,a candidate
took 12 as divisor instead of 21.The quotient obtained by him was 35.
The correct quotient is?
a.0 b.12 c.13 d.20
Solution:Dividend=12*35=420.
Now dividend =420 and divisor =21.
Therefore correct quotient =420/21=20.
PROGRESSION
A succession of numbers formed and arranged in a definite order according
to certain definite rule is called a progression.
1.Arithmetic Progression:-If each term of a progression differs from its
preceding term by a constant.
This constant difference is called the common difference of the A.P.
The n th term of this A.P is Tn=a(n-1)+d.
The sum of n terms of A.P Sn=n/2[2a+(n-1)d].
xImportant Results:
a.1+2+3+4+5......................=n(n+1)/2.
b.12+22+32+42+52......................=n(n+1)(2n+1)/6.
c.13+23+33+43+53......................=n2(n+1)2/4
2.Geometric Progression:-A progression of numbers in which every
term bears a constant ratio with ts preceding term.
i.e a,a r,a r2,a r3...............
In G.P Tn=a r n-1
Sum of n terms Sn=a(1-r n)/1-r
Tests of divisibility
A number is divisible by 2 if its last digit is also (i.e. 0,2,4,6 or 8).A number is divisible by 3 if the sum of its digits is also. Example: 534: 5+3+4=12 and 1+2=3 so 534 is divisible by 3.A number is divisible by 5 if the last digit is 5 or 0..
Let L = Last digit, M= Leading truncated number with last digit removedTest for divisibility by 7. Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number....| M - 2L | should be divisible by 7...repeat this till you get some easy number.
Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also.There are similar rules for the remaining primes under 40, i.e. 11,13, 17,19,23,29,31,37,41,43 and 47.Test for divisibility by 11. Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.Example: 19151--> 1915-1 =1914 -->191-4=187 -->18-7=11, so yes, 19151 is divisible by 11.
| M - L | is divisible by 11.Test for divisibility by 13. Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13. | M + 4L | is divisible by 13.Test for divisibility by 17. Subtract five times the last digit from the remaining leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary.Example: 3978-->397-5*8=357-->35-5*7=0. So 3978 is divisible by 17.
| M - 5L | is divisible by 17.Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.EG: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19.
| M + 2L | is divisible by 19.
Test of divisibility learning wayIf you want you can stop here.However, I have extended the list up to 50.Test for divisibility by 23. Add seven times the last digit to the remaining leading truncated number. If the result is divisible by 23, then so was the first number. Apply this rule over and over again as necessary.Example: 17043-->1704+7*3=1725-->172+7*5=207-->20+7*7=69 which is 3*23, so 17043 is also divisible by 23.Test for divisibility by 29. Add three times the last digit to the remaining leading truncated number. If the result is divisible by 29, then so was the first number. Apply this rule over and over again as necessary.Example: 15689-->1568+3*9=1595-->159+3*5=174-->17+3*4=29, so 15689 is also divisible by 29.Test for divisibility by 31. Subtract three times the last digit from the remaining leading truncated number. If the result is divisible by 31, then so was the first number. Apply this rule over and over again as necessary.Example: 7998-->799-3*8=775-->77-3*5=62 which is twice 31, so 7998 is also divisible by 31.Test for divisibility by 37. This is (slightly) more difficult, since it perforce uses a double-digit multiplier, namely eleven. People can usually do single digit multiples of 11, so we can use the same technique still. Subtract eleven times the last digit from the remaining leading truncated number. If the result is divisible by 37, then so was the first number. Apply this rule over and over again as necessary.Example: 23384-->2338-11*4=2294-->229-11*4=185 which is five times 37, so 23384 is also divisible by 37.Test for divisibility by 41. Subtract four times the last digit from the remaining leading truncated number. If the result is divisible by 41, then so was the first number. Apply this rule over and over again as necessary.Example: 30873-->3087-4*3=3075-->307-4*5=287-->28-4*7=0, remainder is zero and so 30873 is also divisible by 41.Test for divisibility by 43. Now it starts to get really difficult for most people, because the multiplier to be used is 13, and most people cannot recognise even single digit multiples of 13 at sight. You may want to make a little list of 13*N first. Nevertheless, for the sake of completeness, we will use the same method. Add thirteen times the last digit to the remaining leading truncated number. If the result is divisible by 43, then so was the first number. Apply this rule over and over again as necessary.Example: 3182-->318+13*2=344-->34+13*4=86 which is recognisably twice 43, and so 3182 is also divisible by 43.We can also do following:, since we are working to modulo43, instead of adding factor 13 times the last digit, we can subtract 30 times it, because 13+30=43.Finally, the Test for divisibility by 47. This too is difficult for most people, because the multiplier to be used is 14, and most people cannot recognise even single digit multiples of 14 at sight. You may want to make a little list of 14*N first. Nevertheless, for the sake of completeness, we will use the same method. Subtract fourteen times the last digit from the remaining leading truncated number. If the result is divisible by 47, then so was the first number. Apply this rule over and over again as necessary.Example: 34827-->3482-14*7=3384-->338-14*4=282-->28-14*2=0 , remainder is zero and so 34827 is divisible by 47.Why d we sometimes say ADD and for other primes say SUBTRACT, and where from the apparently arbitrary factors come from. So let us do some algebra to show the method in my madness.We know that recursive divisibility test of number N as f-M*r where f are the front digits of N, r is the rear digit of N and M is some multiplier. And we want to see if N is divisible by some prime P. We need a method to work out the values of M. What you do is to calculate (mentally) the smallest multiple of P which ends in a 9 or a 1. If it's a 9 we are going to ADD, if it's a 1 we are going to SUBTRACT later. Then we will use the leading digit(s) of the multiple as our multiplier M.Example for P=17 : three times 17 is 51 which is the smallest multiple of 17 that ends in a 1 or 9. Since it's a 1 we are going to SUBTRACT later. The leading digit is a 5, so we are going to SUBTRACT five times the remainder r. The algorithm was stated above. Now let's do the algebraic proof. Writing N=10f+r, we can multiply by -5 (as shown in the example for 17), getting -5N=-50f-5r. Now we add 51f to both sides (because 51 was the smallest multiple of P=17 to end in a 1 or a 9), giving one f (which we want), so 51f-5N=f-5r. Now if N is divisible by P (here P=17), we can substitute to get 51f-5*17*x=f-5r and rearrange the left side as 17*(3f-5x)=f-5r and therefore f-5r is a multiple of P=17 also. Q.E.D.
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