Answers
1) b
2) c
3) b
4) a
5) d
6) d
Solutions:
1)
Number of ordered pairs = 1;
because D=5, G=1
Step 1.
a + b + c + d = d + e + f + g = g + h + i = 17
Means, ( a + b + c + d )+ (d +e + f + g )+ (g +h + i ) = 17 +17+17 = 17 x 3 = 51
a + b + c + d + e + f + g + h + i +( d + g ) = 51
Step 2.
But 'a' to ' i ' takes values only from 1 to 9
So a + b + c + d + e + f + g + h + i = sum of the numbers from 1 to 9 = 45
Step 3.
From step 1 and step 2
d + g = 51 - 45 = 6
possible values of d & g are (1, 5), ( 5,1 ), (2,4) & (4, 2 )
( 2, 4 ) & ( 4 , 2 ) are not possible as "a = 4 "
We have to try with other values (1, 5 ) & ( 5 , 1 )
Step 4.
Case1: When d = 1 and g = 5
1. a + b + c + d = 4 + b +c + 1 = 17
b + c = 12 ; only possible values of b & c are ( 9, 3 ) or ( 3, 9 )
2. d + e + f + g = 1 + e + f + 5 = 17
e + f = 17 - 6 = 11
possible combinations for 11 are ( 2,9 ) ; ( 3,8 ) ; (4,7 ) & ( 5,6 )
Out of the above four combinations nothing is possible beacuse
b & c takes values 3 & 9 g = 5 (assumed) and a = 4 (given).
So, d = 1 & g = 5 is not the solution.
Case 2. When d = 5 and g = 1
1. a + b + c + d = 4 + b + c + 5 = 17
b + c = 17 - 9 = 8
only possible combination is 2 & 6
a + b + c + d = 4 + ( 2 + 6 ) + 5 = 17
2. d + e + f + g = 5 + e + f + 1 = 17
e + f = 17 - 6 = 11
The only possible value is 3 & 8
d + e + f + g = 5 + ( 3 + 8 ) + 1 = 17
g + h + i = 1 + h + i = 17
h + i = 17 - 1 = 16
The only possible value for h & i are 7 & 9
g + ( h+ i ) = 1 + ( 7 + 9 ) + 17
So the values of d = 5 & g = 1
2) Let us write number in base of 3.
T1=1;T2=(10)3;T3=(11)3;T4=(100)3
T1=1;T2=(10)3;T3=(11)3;T4=(100)3
hence T50= (110010)3=327
(10)3 >= 10 in base 3;(10)3= (3+0)10 ie 3 in base 10
(11)3=(3+1)10=(4)10
(100)3=(9)10
(10)3 >= 10 in base 3;(10)3= (3+0)10 ie 3 in base 10
(11)3=(3+1)10=(4)10
(100)3=(9)10
now (10)2=2,(11)2=3,(100)2=4 are the respective term numbers
so the 50th term would be (110010)3 as (110010)2=50
so the 50th term would be (110010)3 as (110010)2=50
3)g(h(g(x)))=2g(x)^2 + 3g(x)
to calculate g(-4) we have to find out x such that h(g(x))=-4
i.e. x^2 + 4x=0 i.e. x=-4
i.e. x^2 + 4x=0 i.e. x=-4
hence we get g(-4)= 2g(-4)^2+ 3g(-4)
hence g(-4)=-1
hence g(-4)=-1
as the problem says ‘includes all numbers that are a sum of one or more distinct powers of 3′ so
4) 2x-1=ax^3 + bxy^2 – 1; 2y-1= ayx^2 + by^3 – 1
(2x-1)(2y-1)=(ax^3 + byx^2 – 1)(ayx^2 + by^3 – 1)
after multiplying ,rearranging the terms and using the other conditions given in the problem we can easily get the ans as 4
5)If k1 is the first then k2 will be the (n+2)th toffee k3 will be (2n+3)th toffee. let us assume that in one complete circle there are k consecutive toffees, as there should be no overlap hence the (k+1)toffee shouldnot overlap onto the 1st toffee.
as there are 30 toffees each toffee will be placed at 12degrees
so (k+1)(n+1)12 shouldnt be equal to 360
ie (k+1)(n+1) =! 30
as there are 30 toffees each toffee will be placed at 12degrees
so (k+1)(n+1)12 shouldnt be equal to 360
ie (k+1)(n+1) =! 30
out of the options only for n=12 we cannot have any value of k satisfying (k+1)(n+1)=30
hence there should be 12 toffees in between!!!
