model 2 find lcm of numbers Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 5 EXERCISES

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The following question based on LCM & HCF topic of quantitative aptitude

Questions : A, B, C start running at the same time and at the same point in the same direction in a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds. After what time will they meet again at the starting point ?

(a) 46 minutes 12 seconds

(b) 45 minutes

(c) 42 minutes 36 seconds

(d) 26 minutes 18 seconds

The correct answers to the above question in:

Answer: (a)

Required time = LCM of 252, 308 and 198 seconds

2252,308,198
2126,154,99
763,77,99
99,11,99,
111,11,11
 1,1,1

∴ LCM = 2 × 2 × 7 × 9 × 11 = 2772 seconds

= 46 minutes 12 seconds

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Read more find lcm using formula Based Quantitative Aptitude Questions and Answers

Question : 1

Three bells ring at intervals of 36 seconds, 40 seconds and 48 seconds respectively. They start ringing together at a particular time. They will ring together after every

a) 24 minutes

b) 18 minutes

c) 12 minutes

d) 6 minutes

Answer: (c)

Required answer = LCM of 36, 40 and 48 seconds

= 720 seconds

= $(720/60)$minutes = 12 minutes

Illustration :

236,40,48
218,20,24
29,10,12
39,5,6
 3,5,2

∴ LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720

Question : 2

The greatest number of four digits which when divided by 12, 16 and 24 leave remainders 2, 6 and 14 respectively is

a) 9998

b) 9807

c) 9970

d) 9974

Answer: (d)

Using Rule 5,

When a number is divided by P, Q or R leaving remainders X, Y or Z respectively such that the difference between divisor and remainder in each case is the same i

.e., (P – X) = (Q – Y) = (R – Z) = T

(say) then that (least) number must be in the form of (K – T),

Where K is LCM of P, Q and R

Here,12 – 2 = 10; 16 – 6 = 10; 24 – 14 = 10

Now, LCM of 12, 16 and 24 = 48

∴ The greatest 4–digit number exactly divisible by 48 = 9984

∴ Required number = 9984 – 10 = 9974

Question : 3

The least number, which when divided by 12, 15, 20 or 54 leaves a remainder of 4 in each case, is :

a) 544

b) 540

c) 454

d) 450

Answer: (a)

Using Rule 4,

When a number is divided by a, b or c leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b and c.

LCM of 15, 12, 20, 54 = 540

Then number = 540 + 4 = 544 [4 being remainder]

Question : 4

The largest 4-digit number exactly divisible by each of 12, 15, 18 and 27 is

a) 9960

b) 9930

c) 9720

d) 9690

Answer: (c)

Using Rule 8,

Greatest n digit number which when divided by three numbers A, B, C leaves no remainder will be

Required Number = (n – digit greatest number) – R

R is the remainder obtained on dividing greatest n digit number by L.C.M of A, B, C.

The largest number of 4-digits is 9999. L.C.M. of divisors

212,15,18,27
36,15,9,27
32,5,3,9
 2,5,1,3

LCM = 2 × 2 × 3 × 3 × 3 × 5 = 540

Divide 9999 by 540, now we get 279 as remainder.

9999 – 279 = 9720

Hence, 9720 is the largest 4-digit number exactly divisible by each of 12, 15, 18 and 27.

Question : 5

The least multiple of 13, which on dividing by 4, 5, 6, 7 and 8 leaves remainder 2 in each case is:

a) 840

b) 2522

c) 842

d) 2520

Answer: (b)

Using Rule 4,

LCM of 4, 5, 6, 7 and 8

=

24,5,6,7,8
22,5,3,7,4
 1,5,3,7,2

= 2 × 2 × 2 × 3 × 5 × 7 = 840.

let required number be 840 K + 2 which is multiple of 13.

Least value of K for which (840 K + 2) is divisible by 13 is K = 3

∴ Required number

= 840 × 3 + 2

= 2520 + 2 = 2522

Question : 6

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

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