# Quant Boosters - Sagar Gupta - Set 2

• HCF (2^a -1, 2^b -1) = 2^(HCF a, b) - 1
= 2^20 - 1

• Q5) Find last two digits of 15 * 37 * 63 * 51 * 97 * 17

• 15 * 37 * 63 * 51 * 97 * 17 mod 100
15 * 37 * (-37) * 51 * (-3) * 17 mod 100
45 * 37^2 * 17^2 * 3 mod 100
45 * 89 * 03 * 69
35

• Q6) A five digit number divisible by 3 is to be formed using numbers 0, 1,2,3,4,5 without repetition . The total number of ways in which this can be done is....
a. 216
b. 240
c. 600
d. 3125

• 1 + 2 + 3 + 4 + 5 = 15
-> 5! ways
0 + 1 + 2 + 4 + 5 =12
->5! - [ 5! /5 ] =120 - 24 =96
total 120 +96 =216

• Q7) The total number of ways of selecting two numbers from the set {1,2,3,4,......,30}, so that their sum is divisible by 3, is
a. 95
b. 145
c. 190
d. None of the above.

• 3k, 3k + 1, 3k + 2
divisible by 3:
case 1 : 3k + 3k
10 * 9 = 90 ways
case 2 : (3k+2) + (3k+1)
10 * 10 = 100 ways
total = 190 ways

• Q8) Train X started from point A at 9:00 am with a speed 72km/hr towards station Y. 2 hrs after train Y starts from point B towards X at a speed of 90 km/hr. They cross each other at 1:30 pm. But owing to signal problems at 12:00 noon the speeds of both the trains is reduced by same quantity such that they now cross each other at 4:30 pm. Calculate the new speed of train that started from point A.

• Train X starts at 9:00 am @72 km/hr
at 11:00 am, it is 144 kms from A
A...144 kms...A'......d...........B
Now Y starts from B at 11:00am @90 km/hr
A' and B is the position at 11:00 am
:
They should have met at 1:30pm
72 * 2.5 + 90 * 2.5 = d =405 kms
:
But they meet at 4:30 pm
from 11:00 to 12:00 , they travel at normal speed
Distance reduced = 72+90=162
Distance between them at 12:00 = 243 kms
Now they reduce their speed by k
(72-k) * 4.5 + (90-k) * 4.5 =243
k=54
72-54 =18km/hr

• Q9) abcd is a 4 digit perfect square, then if each digit is increased by 3 the number is still a perfect square. Find abcd.

• abcd = x^2
1000a +100b +10c +d = x^2
1000(a+3) +100(b+3) +10(c+3) +d+3 = y^2
y^2-x^2=3000 +300 + 30 + 3
y^2 - x^2 = 3 [ 1000 +100 +10 +1 ]
y^2 - x^2 =3333
3333=101*33
y+x =101
y-x =33
y = 67
x = 34
x^2 =1156

• Q10) What is the remainder when 72 to the power 7202 is divided by 625?

• E[625]=500
7202 mod 500 = 2
72^2 mod 625 = 5184 mod 625 = 184

• Q11) If a^3 + b^3 = 10 and a^2 + b^2 = 60 then find a + b?

• (a+b)^3 = 10 + 3ab (a+b)
(a+b)^2 = 60 + 2ab
a+b = p
(p^3 - 10 ) / 3p = ab
p^2 = 60 + 2 [ (p^3 - 10 ) / 3p ]
3p^3=180p + 2p^3 - 20
p^3 +20 = 180p
p=13.36

• Q12) A and B can build a wall in 4 days and 6 days respectively working alone, while C and D can destroy the wall in 8 days and 12 days respectively working alone. If three of them start working together and the wall was built in 24 days. Which of the four persons did not work?

• Total work =36 units
A -> +9 units/day
B -> +6 units/day
C- > -4.5 units/day
D- > -3 units/day
24 (x+y+z) = 36
x+y+z =1.5
9-4.5-3 =1.5
Hence B did not work

• Q13) Three persons A, B and C rent the grazing of a park for Rs. 570. A put 126 oxen in the park for 3 months, B puts in 162 oxen for 5 months and C puts in 216 oxen for 4 months. What part of the rent should each person pay?
a. 105, 220, 245
b. 125, 205, 245
c. 105, 225, 240
d. None of these

• 126 * x * 3 + 162 * 5 * x + 216 * 4 * x = 570
x = 5/18
A= 126 * 3 * 5/18 =105
B=162 * 5 * 5/18 =225
C=216 * 4 * 5/18=240
Hence option C

• Q14) a,b and c are three positive real numbers. The minimum value that the expression [(a+b)/2 * (b+c)/2 * (c+a)/2] can take when the product of the three numbers is 3/2 is?

61

43

67

48

61

61

61

107