# Quant Boosters - Maneesh - Set 3

• Total Amount = 30000+4347 = 34347
34347 = 30000 * (1+7/100) ^N
(107/100)^N = 34347/30000 = 11449/10000
(107/100)^N = 11449/10000
N = 2 Years

• Q11) Ram’s speed is with the current is 10 m/s and speed of current is 2 m/s .Then What’s Ram’s speed against current?

• Let speed of current y = 2 m/s
Let Ram’s speed in still water is x m/s
x+2 = 10 m/s
x = 8 m/s

• Q12) A boat can travel with a speed of 15 m/s in still water. If the speed of the stream is 5 m/s, find the time taken by the boat to go 100 m downstream?

• x = 15 m/s
y= 5 m/s
downstream speed ,d = 15+5 = 20 m/s
distance = 100 m
time taken = 100/20 = 5 seconds

• Q13) A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours and 30 minutes. Find the speed of the stream?

• x = 15 km/hr
t = 4 1/2 h = 9/2 hours
9/2 = 30/15+y + 30/15 - y where y is speed of stream
9/2 = 30(30)/225-y^2
225-y^2 = 900 * 2/9 = 200
y^2 = 25
y = 5 km/hr

• Q14) A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively?

• D/x-y = 8 4/5 = 44/5
D/x+y = 4
x+y/x-y = 44/20
20x+20y = 44x- 44y
24x = 64y
x/y = 64/24 = 8/3

• Q15) A boat takes 90 minutes less to travel 36 km downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 kmph, find the speed of the stream?

• 3/2 = 36/10-x -36/10+x = 36 * 2x/100-x^2
300-3x^2 = 144x
3x^2+144x-300 = 0
x^2+48x-100 = 0
(x+50)(x-2) = 0
Since x can only be positive value, speed of stream = 2 kmph

• Q16) A positive whole number less than 100 is represented in base 2, base 3 and base 5 notations. It is found that exactly in two cases the last digit is 1 and in exactly two cases first digit is 1.What is the number in decimal representation?
a) 31
b) 61
c) 71
d) 81

• In exactly two cases last digit is 1.This means that when the number is divided by two of this three numbers (2, 3, 5), the remainder should be 1.
Since the LCM of (2, 3, 5) is 30, numbers 31 and 61 give 1 remainder in all 3 cases.
Hence 31 and 61 can be eliminated.
71 in base 10 -> 1000111 in base 2 -> 2122 in base 3 -> 241 in base 5
81 in base 10 -> 1010001in base 2 -> 10000 in base 3 -> 311 in base 5
Hence only 81 have the first digit is 1 in two cases.

• Q17) Sachin writes all numbers from 1 to 1000 on a paper in order. Find the 2883 rd digit written by him?

• One digit numbers digits -> 9
Two digit numbers digits -> 180
Three digit numbers digits -> 2700
Total -> 2889
Last 3 digits will be 999…..then 998….hence 2883 rd digit will be 3.

• Q18) If N = 72 then find the number of pairs (a,b) such that a and b are the factors of N and a * b = N?

• N = 72
Factors are 1,2,3,4,6,8,9,12,18,24,36,72
If we form pairs (1,72) (2,36) ( 3,24) (4,18) (6,12) ( 8,9)
Total 6 pairs.

• Q19) If N = 72 then find the number of pairs (a,b) for N such that a and b are the factors of N and are co prime to each other?

• 72 = 3^2 * 2^3
Total number of factors = 3 * 4 = 12
Factor 1 is co prime to all other factors. -> 11 pairs
Powers of 2 will be co prime with powers of 3
Powers of 2 -> 2,4,8 (3 values)
Powers of 3 -> 3,9 (2 values)
Hence 3 * 2 = 6 pairs possible.
So total 11+6 = 17 pairs possible.

• Q20) The LCM of first 100 natural numbers is K .Find the LCM of first 105 natural numbers?

64

65

61

61

62

46

51

61