# Quant Boosters - Hemant Malhotra - Set 12

• Q13) For how many values of y, y(y + 4)(y + 6)(y + 8) < 300 is satisfied

• Q14) What is the minimum value of (p + q)(q + r)(r + p) if p, q and r are positive numbers

• Q15) Find the sum of the first 10 terms of the series 1, 2, 5, 12, 27 ...

• The given term can be written as (2-1), (4-2), (8 - 3), (16 - 4) and so on
so sum of first 10 terms will be = (2 + 4 + 8 + 16 + - - - - - ) - (1 + 2 +3 +4 +5 +6 +7 - - -)
2046 - 55 = 1991

• Q16) If 2x + 9y ≤ 90, x ≥ 0, y ≥ 0 and x & y are integers, find the total number of solutions

• Q17) x, y and z are real numbers such that -6 ≤ x ≤ -2, -4 ≤ y ≤ 4 and 3 ≤ z ≤ 7. If w = y/xz then which of the below options is necessarily true ?
a) 0 ≤ w ≤ 2
b) -1/2 ≤ w ≤ 1/2
c) -2/3 ≤ w ≤ 1/3
d) -2/3 ≤ w ≤ 2/3

• Q18) What is the maximum value of p for which the below equations have a unique value of x
(2x - p)^2 + y^2 + 32 = 0
3x^2 + y^2 - 16 = 0

• Q19) Sum of two natural numbers is 600 and their hcf is 15. How many such pairs are possible?

• Let numbers be 15a and 15b
so 15a + 15b = 600
so a + b = 40
so E(40)/2 will be number of pairs

Why E(40)/2 ?
E(40) means all numbers below 40 which are coprime to 40

Example- E(10) = 10 * 1/2 * 4/5 = 4
1, 3, 7, 9 - there are 4 numbers below 10 which are coprime to 10
2 pairs 1,9 and 3,7 whose sum is 10

• Q20) Which one of the following is a prime number?
(a) 999,991
(b) 999,973
(c) 999,983
(d) 1,000,001
(e) 7,999,973

• a) (10^3)^2 - 3^2 = (1003)(997)
b) (10^2)^3 - 3^3 divisible by (10^2 -3)= 97
d) (10^2)^3 + 1^3 divisible by 101
e) (2*10^2)^3 - 3^3 divisible by 200 -3 =197
so c) is prime

• Q21) N = 373839404142 ... 9192 (numbers from 37 to 92 written in order to form a number). Find the maximum value of k if the number N is perfectly divisible by 3^k

• N is not divisible by 9 but divisible by 3. As it is not divisible by 9, no need to check for higher powers
so OA = 1

• Q22) In a 4 digit no. having non zero and distinct digits, the sums of the digits at the unit place and the tens place is equal to the sum of other 2 digits. The sum of the digits at the tens and the hundreds places is three times the sum of the remaining 2 digits. If the sum of the digits is atmost 20, then how many such 4 digit numbers are possible ?

• abcd
a+b=c+d --(1)
b+c= 3 (a+d) => 4(a+d) a+d a=1, d=2 => b+c= 9
to fulfill (1), 1+b= c+2 => b-c=1 => b=5, c=4.
=> 1542

or a=2, d=1 => b+c= 9 and 2+b=c+1 => c-b=1 => c=5, b=4
=> 2451

if a+d=4 =>a=1 d=3 or a=3, d=1
this gives b+c=12 and b-c=2 or c-b=2 => b,c= (7,5) or (5,7)
=> 1753 or 3571

if a+d=5 => a=1, d=4 or a=4, d=1
b+c=15 and b-c=3 or c-b=3 => (b,c)= (9,6) or (6,9)
=> 1964 or 4691
similarly 3782 and 2873 for a=2, d=3 and a=3, d=2

=> 8 cases

(bruce wayne)

• Q23) A dishonest ration guy cheats 20% while buying and 25% while selling. If he buys ration for 176 Rs/kg and manages to earn a profit of 10% even after making an offer of buy 2 get 3 free. Find the Marked price per unit kg.

he actually cheats and buys 6 kg so total cost = 880
while selling he actually sells 3.75 coz he cheats 25% so calculate the actual cost of buying 3.75 it will be (880/6)*(15/4)=550.
he makes a profit of 10% so the cost is 605
he claims to have sold 2
so actual cost per unit is 302.5

• Q24) What is the number of common roots of the two equations given below?
x^3 - 3x^2 + 2x + 5 = 0
x^3 + x^2 + 7x + 6 = 0
(a) 0
(b) 1
(c) 2
(d) 3

• To find the common roots
x^3 - 3x^2 + 2x + 5 = x^3 + x^2 + 7x + 6
4x^2 + 5x + 1 = 0
x = -1 or -1/4
Now check the value of x in both the equations
They won't satisfy so there are no common roots

33

58

62

44

64

61

63

54