Quant Boosters  Maneesh  Set 1

36 = 4 * 9
123456….40 mod 4 => 40 mod 4 = 0
123456…mod 9
= sum of digits mod 9
= 45 * 3 + 10 * (1+2+3) + 4 mod 9
= 135 + 60 + 4 mod 9 = 199 mod 9 = 1
Hence 4a = 9b+1
Remainder > 28 (a = 7 b = 3)

Q22) Find the remainder when 1 * 2 + 2 * 3 + 3 * 4 +… 9 * 100 is divided by 101

1 * 2 + 2 * 3 + 3 * 4 + ... 99 * 100 = Sigma [n * (n+1)] From 1 to 99
sigma (n(n+1))
= sigma(n^2+n)
= n(n+1)(2n+1)/6 + n(n+1)/2
= n(n+1)(2n+1)+3n(n+1)/6
= (n+1)/6 * (2n^2+4n)
= 2n(n+1)(n+2)/6
= 198 * 100 * 101/6
198 * 100 * 101/6 mod 101 = 0

Q23) Find the smallest number with 15 divisors

15 = 3 * 5 = (2+1)(4+1)
so number of the form a^2 * b^4
Smallest number means a = 3 b= 2
3^2 * 2^4 = 9 * 16 = 144

Q24) How many divisors of 21600 are perfect squares?

21600 = 2^5 * 3^3 * 5^2
The even powers of 2 can be selected in 3 ways
Even powers of 3 can be selected in 2 ways
Even powers of 5 can be selected in 2 waysTotal > 3 * 2 * 2 = 12 divisors are perfect squares

Q25) How many of the first 1200 natural numbers are not divisible by any of 2, 3 and 5?

1200 is a multiple of 2, 3 and 5
Hence we have to find out the number of numbers which are less than and coprime to 1200
1200 * (11/2) * (11/5) * (11/3) [As per Euler]
> 1200 * 1 * 2 * 4/2 * 3 * 5 = 320

Q26) N is the smallest number such that N/2 is a perfect square and N/3 is a perfect cube. Then the number of divisors of N?

N/2 is a perfect square > power of 2 in N can be 1,3,5….
Power of 3 in N can be 0,2,4…….
N/3 is a perfect cube > power of 2 in N can be 3,6,9….
Power of 3 in N can be 1,4,7…..
Smallest N = 2^3 * 3^ 4
Number of divisors = 4 * 5 = 20

Q27) Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first met after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters ?

Let length is x.
At first meeting point, Brenda runs 100 meters and sally runs (x/2) – 100
At second meeting point, Sally runs 150 meters and Brenda runs x150 meters
100/x150 = (x/2) 100/150
15000 = x^2/2100x 75x+15000
x^2/2 – 175x = 0
x^2350x = 0
x = 350 meters.

Q28) A and B start from same point on the circular track in same direction. A completes a track in 7 minutes and B takes 3 minutes. When they started running it was 12.00 PM by clock. When they meet 4th time what is time by clock?

If A and B takes m and n minutes respectively to complete one circular track then A and B would meet for the first time after
T = (m * n) /mn
They will keep meeting at the interval T.
m = 7 n= 3
T = 21/4 minutes
4th time it will take 4T = > 4 * 21/4 = 21 minutes
So time 12:21

Q29) Vijay left P for Q at 10:00 AM. At the same time Ajay left Q for P. After their meeting on a point on the way Vijay took 24 minute to reach Q and Ajay took 54 minute to reach P. At what time they met?

Speed of Vijay/Speed of Ajay = root (54/24) = root (9/4) = 3/2
Distance = 3x*24/60 + 2x * 54/60 = 180x/60 = 3x
Time taken to meet = 3x/5x = 3/5 = 36/60
= > after 36 minutes …that is 10:36

Q30) A fires two bullets from same place at an interval of 6 minutes but B sitting in a car approaching the place of firing hears the second fire 5 minutes 32 seconds after the first firing. What is the speed of the car if speed of sound is 332m/s?

It took 28 seconds for him to travel from the place where he heard sound first to current place.
Distance (sound) = 28 *332 m
Same distance he travelled in 5 minutes 32 seconds.
Speed of car = 28 * 332/332 = 28 m/s

LCM (9,18,24) = 72
9  5 = 18  14 = 24  20 = 4
Since difference is same , answer will be LCM – common difference = 724 = 68