# Quant Boosters - Maneesh - Set 3

• Unit digit of a number and its square will be equal when unit digit is any of 0,1,5,6.
From 0 to 90 : unit digit 0 ->10 numbers
From 0 to 90 : unit digit 1,5,6 -> 9 each
Hence total 37 numbers.

• Q26) 94^3 – 23^3 - 71^3 is at-least divisible by?
a) 71 and 23
b) 23 and 74
c) 71 and 94
d) 23, 71 and 94

• 94^3-23^3 is divisible by 94-23 = 71
94^3-71^3 is divisible by 94-71 =23
23^3+71^3 is divisible by 23+71 = 94
Hence it is divisible by 23, 71 and 94.

• Q27) Find the least number which being divided by 5,6,8,9,12 leaves in each case a remainder 1 but when divided by 13 leaves no remainder?
a) 2987
b) 3601
c) 3600
d) 2986

• LCM [5,6,8,9,12] = 360
360A +1 = 13B
From option the number of the form 360+1 is 3601 only.

• Q28) Find the greatest number consisting of 6 digits which on being divided by 6,7,8,9,10 leaves 4,5,6,7,8 as remainders respectively.
a) 997920
b) 997918
c) 999999
d) 997922

• LCM [6,7,8,9,10] = 2^3 * 3^2 * 7 * 5 = 2520
Greatest 6 digit number which is a multiple of 2520 = 997920
Required number = 997920 – 2 = 997918.
PS : we are subtracting 2 coz 6-4 = 7-5 = 8-6 = 9-7 = 10-8 = 2

• Q29) If X = 13^3 + 14^3 + 15^3 + 16^3 then what is the remainder when x is divided by 58?
a) 0
b) 1
c) 57
d) 29

• A^n + B^n is divisible by A+B when n is odd.
So x is divisible by 29
X = 29(13^2+16^2-13 * 16) + 29(14^2+15^2-14 * 15)
= 29(13^2+14^2+15^2+16^2-13 * 16 - 14 * 15) = 29 * an even number
Hence x is a multiple of 58.

• Q30) A train moving at uniform speed takes 20 sec to pass a cyclist riding at 11 km/hr but only 9 sec to pass a post. Find the length of train.

• Let length of train is is ‘L’ km and speed ‘S’ Km/hr
L/X-11 = 20/3600 = 1/180
L/X = 9/3600 = 1/400
X = 400 L
180 L = X – 11
180L = 400L-11
220L = 11
L = 11/220 = 1/20 KM = 1000/20 =50 meters

46

63

61

54

61

62

61