# Quant Boosters - Gaurav Sharma - Set 4

• 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 non-negative 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 non-negative 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 (x-1)(x-2)(x-3) = 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 answer

Method 2 :
Fibonacci with a gap of 1
1, 1, 2, 3, 4, 6, 9, 13, 19, 28

Method 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

• Q20) Ashu and Manoj start running simultaneously from the ends A and B respectively, of a straight track of length 800 m, with speeds that are in the ratio 5 : 3. Whenever Ashu reaches either of the ends, he turns around and continues running at the same speed. Whenever Manoj meets Ashu, he turns around and continues running at the same speed. When Ashu comes back at A for the first time, how far (in meters) is Manoj from B?
a) 360
b) 435
c) 510
d) None of these

• Q21) The sum of thirty-two consecutive natural numbers is a perfect square.What is the least possible sum of the smallest and largest of thirty -two numbers?

• Q22) A sequence 'S' is formed from the set of first 'N' natural numbers by deleting all the perfect squares and all the perfect cubes. If the elements are arranged in an ascending order then, what is the 240th term of the sequence 'S' ?

• Q23) A packet of toffees is distributed in a class. A child who receives one-eighth of the total number of toffees gets five times the average number of toffees received by the remaining children in the class. What is the strength of the class?
a) 40
b) 36
c) 35
d) Cannot be determined

48

44

34

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63

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63

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