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 : Find the least number which when divided separately by 15, 20, 36 and 48 leaves 3 as remainder in each case.

(a) 723

(b) 483

(c) 243

(d) 183

The correct answers to the above question in:

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.

Required number = (LCM of 15, 20, 36 and 48) + 3

215,20,36,48
215,10,18,24
315,5,9,12
55,5,3,4
 1,1,3,4

∴ LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720

∴ Required number = 720 + 3 = 723

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

Question : 1

What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 13 leaves no remainder ?

a) 1586

b) 1562

c) 962

d) 312

Answer: (c)

LCM of 3, 5, 6, 8, 10 and 12 = 120

∴ Required number = 120x + 2, which is exactly divisible by 13.

120x + 2 = 13 × 9x + 3x + 2

Clearly 3x + 2 should be divisible by 13.

For x=8,3x + 2 is divisible by 13.

∴ Required number = 120x + 2 = 120 × 8 + 2

= 960 + 2 = 962

Question : 2

The smallest perfect square divisible by each of 6, 12 and 18 is

a) 36

b) 108

c) 144

d) 196

Answer: (a)

The LCM of 6, 12 and 18 = 36 = $6^2$

Question : 3

The least perfect square, which is divisible by each of 21, 36 and 66 is

a) 231444

b) 213444

c) 214434

d) 214344

Answer: (b)

LCM of 21, 36 and 66

∴ LCM = 3 × 2 × 7 × 6 × 11

= 3 × 3 × 2 × 2 × 7 × 11

∴ Required number

= $3^2 × 2^2 × 7^2 × 11^2$

= 213444

Question : 4

Four bells ring at intervals of 4, 6, 8 and 14 seconds. They start ringing simultaneously at 12.00 O’clock. At what time will they again ring simultaneously ?

a) 12 hrs. 3 min. 44 sec.

b) 12 hrs. 3 min. 20 sec.

c) 12 hrs. 3 min.

d) 12 hrs. 2 min. 48 sec.

Answer: (d)

LCM of 4, 6, 8, 14

= 168 seconds

= 2 minutes 48 seconds

They ring again at 12 + 2 min. 48 sec.

= 12 hrs. 2 min. 48 sec.

Question : 5

The LCM of two prime numbers x and y, (x > y) is 161. The value of (3y – x) :

a) 2

b) 1

c) –1

d) –2

Answer: (d)

LCM of x and y = 161

∴ xy = 23 × 7

∴ x = 23; y = 7

∴ 3y – x = 3 × 7 – 23

= 21 – 23 = – 2

Question : 6

Five bells begin to toll together and toll respectively at intervals of 6, 7, 8, 9 and 12 seconds. After how many seconds will they toll together again ?

a) 318 Sec.

b) 504 Sec.

c) 612 Sec.

d) 72 Sec.

Answer: (b)

Required time = LCM of 6, 7, 8, 9 and 12 seconds

= 504 seconds

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