17 September Solutions

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

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

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

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

hence we get g(-4)= 2g(-4)^2+ 3g(-4)
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

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!!!