Quant Boosters  Hemant Malhotra  Set 12

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

@hemant_malhotra ans C?

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 => bc=1 => b=5, c=4.
=> 1542or a=2, d=1 => b+c= 9 and 2+b=c+1 => cb=1 => c=5, b=4
=> 2451if a+d=4 =>a=1 d=3 or a=3, d=1
this gives b+c=12 and bc=2 or cb=2 => b,c= (7,5) or (5,7)
=> 1753 or 3571if a+d=5 => a=1, d=4 or a=4, d=1
b+c=15 and bc=3 or cb=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.

while buying.....5 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

Q25) A problem is given to two students  Tom and Jerry. The probability that Tom can solve the problem correctly is 0.55 and that Jerry can solve it correctly is 0.8. If the two of them try to solve the problem independently, then what is the probability that the problem would be solved correctly ?
(a) 0.25
(b) 0.44
(c) 0.86
(d) 0.91

The probability that Tom cannot solve the question correctly = 1  0.55 = 0.45
The probability that Jerry cannot solve the question correctly = 1  0.8 = 0.2
The probability that none of the two can answer the question correctly = 0.45 * 0.2 = 0.09
the probability that the problem would be solved correctly = 1  0.09 = 0.91

Q26) The sum of the roots of the quadratic equation ax^2 + bx+ c = 0 is equal to the sum of the squares of their reciprocals. If a, b and c are real numbers, and a ≠ 0, then bc^2 , ca^2 and ab^2 are in
(a) G.P
(b) A.P.
(c) H.P
(d) None of these

Let the roots x and y
x + y = b/a
xy = c/a
Now x + y = (1/x^2 + 1/y^2) [ Given]
Solve this
x + y = (x^2 + y^2)/(x^2 y^2)
x+y = (x + y)^2  2xy/(xy)^2
Now equate their values
b/a = (b^2/a^2)  2(c/a)/(c/a)^2
2ca^2 = ab^2 + bc^2
So bc^2, ca^2 and ab^2 are in AP

Q27) Aman and eight of his friends took a test of 100 marks. Each of them got a different integer score and the average of their scores was 86. The score of Aman was 90 and it was more than that of exactly three of his friends. What could have been the maximum possible absolute difference between the scores of two of his friends?
(a) 83
(b) 73
(c) 54
(d) 44