model 2 find lcm of numbers Practice Questions Answers Test with Solutions & More Shortcuts

Question : 11 [SSC CGL Prelim 2004]

If the students of a class can be grouped exactly into 6 or 8 or 10, then the minimum number of students in the class must be

a) 240

b) 180

c) 120

d) 60

Answer: (c)

Required number of students = LCM of 6, 8, 10 = 120

Question : 12 [SSC CGL 2012]

The greatest number of four digits which when divided by 3, 5, 7, 9 leave remainders 1, 3, 5, 7 respectively is :

a) 9765

b) 9766

c) 9764

d) 9763

Answer: (d)

The difference between divisor and the corresponding remainder is equal.

LCM of 3, 5, 7 and 9 = 315 Largest 4-digit number = 9999

31$234/315$ = 9999

∴ Number divisible by 315 = 9999 – 234 = 9765

Required number = 9765 – 2 = 9763

Question : 13 [SSC CGL Prelim 2002]

Find the greatest number of five digits which when divided by 3, 5, 8, 12 have 2 as remainder :

a) 99962

b) 99960

c) 99958

d) 99999

Answer: (a)

Using Rule 4,

i.e. When a number is divided by x, y or z leaving same remainder ‘R’ in each case then that number must be K + R where k is LCM of x, y and z.

The greatest number of five digits is 99999.

LCM of 3, 5, 8 and 12

23,5,8,12
23,5,4,6
33,5,2,3
 1,5,2,1

∴ LCM = 2 × 2 × 3 × 5 × 2 = 120 After dividing 99999 by 120, we get 39 as remainder 99999 – 39 = 99960 = (833 × 120)

99960 is the greatest five digit number divisible by the given divisors.

In order to get 2 as remainder in each case we will simply add 2 to 99960.

∴ Greatest number = 99960 + 2 = 99962

Question : 14 [SSC CGL Tier-II 2017]

Three electronic devices make a beep after every 48 seconds, 72 seconds and 108 seconds respectively. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest is

a) 10 : 07 : 48 hours

b) 10 : 07 : 36 hours

c) 10 : 07 : 24 hours

d) 10 : 07 : 12 hours

Answer: (d)

Required time = LCM of 48, 72 and 108 seconds

248,72,108
224,36,54
212,18,54
36,9,27
32,3,9
 2,1,3

∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds

= 7 minutes 12 second

∴ Required time = 10 : 07 : 12 hours

Question : 15

Which is the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

a) 196

b) 630

c) 1260

d) 2520

Answer: (b)

The LCM of 12, 18, 21, 30

212,18,21,30
36,9,21,15
 2,3,7,5

∴LCM = 2 × 3 × 2 × 3 × 7 × 5 = 1260

∴ The required number

=$1260/2$ = 630

IMPORTANT quantitative aptitude EXERCISES

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