Quant Boosters  Gaurav Sharma  Set 4

7^12 + 7^13 + 7^14 + 7^15 = 7^12 (1 + 7 + 49 + 343)
= 7^12 x 400
= 7^12 x 2^4 x 5^2Number of factors:
(12 + 1) (4 + 1) (2 + 1) = 195

Q9) There are 10 numbers and none of them are divisible by 3. What will be the remainder when sum of their squares is divided by 3?
a) 0
b) 1
c) 2
d) 1 or 2

The Square of a natural number other than one is either a multiple of 3 or exceeds a multiple of 3 by 1. In other words, a perfect square leaves remainder 0 or 1 on division by 3.
Here it is said that none of them is divisible by 3. Hence each square term will give a reminder of 1
So for 10 terms, 1 + 1 + 1 … + 1 (10 times) = 10
10 mod 3 = 1

Q10) Find the value of [root (1)] x [root (2)] x [root (3)] x … x [root (100)]
a) 2^87 x 3^59 x 5^13 x 7^14
b) 2^88 x 3^58 x 5^12 x 7^15
c) 2^81 x 3^59 x 5^12 x 7^17
d) 2^79 x 3^61 x 5^21 x 7^12

To simplify, find the number of 7’s in the product [root (1)] x [root (2)] x [root (3)] x … x [root (100)]
7s will be from [root (49)] to [root (63)]
i.e 63 – 48 = 15
From the given options only b satisfies the condition

Q11) Find the number of nonnegative integral solutions of the equation 3a + 4b + 12c = 1200

a has to be multiple of 4 ( coefficient in all other terms r multiple of 4)
similarly b has to be multiple of 3.
so we can say A + B + C = 100
non negative integer solutions = 102C2

Q12) Find A and B if A^3  B^3 = 56765

a  b = 5, a^2 + b^2 + ab = 11353
So ab = 3776
A + B = 123
(A, B) = (64, 59)

Q13) The number of nonnegative real roots of 2^x – x – 1 = 0
a) 0
b) 1
c) 2
d) 3
e) More than 3

Draw a graph ..
y = 2^x (similar to e^x ) and y = x + 1
both intersect at x = 1, 0

Q14) How many integer solutions exist for the equation (x – 1)(x – 2)(x – 3) = 6^y

Product of thee consecutive nos can be never = 1
also it will include only one 6
so (x1)(x2)(x3) = 6
only one possible value of x = 4

Q15) There is a frog who could climb either 1 stair or 3 stairs in one shot. In how many ways he could reach at 10th stair ?

Method 1 :
1 step in 1 way
2 steps in 1 way
3 steps in 2 ways
4 steps in 3 ways
5 steps in 4 ways
6 steps in 6 ways
7 steps in 9 ways
8 steps in 13 ways
9 steps in 19 ways
10 steps in 28 ways
So 28 should be the answerMethod 2 :
Fibonacci with a gap of 1
1, 1, 2, 3, 4, 6, 9, 13, 19, 28Method 3 :
x + 3y = 10
(1 , 3 ) > 4
(4 , 2 ) > 6!/4!2! = 15
(7 , 1) > 8!/7! = 8
(10 , 0) > 1
TOTAL = 15 + 4 + 8 + 1 = 28

Q16) In how many ways you can climb up 8 steps if minimum and maximum numbers of steps you can take at a time are 1 and 6 respectively?

Minimum steps required is 8, which can be written as a + b + c + d + e + f + g + h = 8, only 1 way.
If we complete in 7 steps, then positive integral solution of a + b + c + d + e + f + g = 8 i.e. 7C6 ways
If we complete in 6 steps, then positive integral solution of a + b + c + d + e + f = 8 i.e 7C5 ways
and so on upto 7C1
But these solutions contains two cases where a or b is more than 6 so eliminate those cases
Final answer is
( 7C7 + 7C6 + 7C5 + 7C4 + 7C3 + 7C2 + 7C1 )  2 = 2^7  7C0  2 = 125

Q17) Rs. 4,500 was distributed among Aman, Baman and Chaman. From the amount that they received Aman, Baman and Chaman spent Rs.110, Rs.120 and Rs.140 respectively. The amounts then left with Aman and Baman were in the ratio 3 : 4 and with Baman and Chaman were in the ratio 5 : 6. What amount (in Rs.) did Baman receive?
a) 1520
b) 1400
c) 1600
d) 1420

Q18) If the nth day of August lies on the same day as the 2nd day of October, then how many values of n are possible?
a) 4
b) 5
c) 2
d) 0

Q19) A two digit number is divided by the sum of its digits. What is the maximum possible remainder?
a) 13
b) 14
c) 15
d) 16