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 number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, and when divided by 8 leaves a remainder of 7, is :

(a) 1359

(b) 359

(c) 539

(d) 1539

The correct answers to the above question in:

Answer: (b)

Using Rule 5,

Here, Divisor – remainder = 1

e.g., 10 – 9 = 1, 9 – 8 = 1, 8 – 7 = 1

∴ Required number = (L.C.M. of 10, 9, 8) –1

= 360 – 1 = 359

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

Question : 1

The least multiple of 7, which leaves the remainder 4, when divided by any of 6, 9, 15 and 18, is

a) 364

b) 184

c) 94

d) 76

Answer: (a)

LCM of 6, 9, 15 and 18

26,9,15,18
33,9,15,9
31,3,5,3
 1,1,5,1

∴ LCM = 2 × 3 × 3 × 5 = 90

∴ Required number = 90k + 4,

which must be a multiple of 7 for some value of k.

For k = 4,

Number = 90 × 4 + 4 = 364,

which is exactly divisible by 7.

Question : 2

4 bells ring at intervals of 30 minutes, 1 hour, 1$1/2$ hour and 1 hour 45 minutes respectively. All the bells ring simultaneously at 12 noon. They will again ring simultaneously at :

a) 9 a.m.

b) 6 a.m.

c) 3 a.m.

d) 12 mid night

Answer: (a)

1$1/2$ hours = 90 minutes

1 hour and 45 minutes = 105 minutes

1 hour = 60 minutes

∴ LCM of 30 minutes, 60 minutes, 90 minutes and 105 minutes

330,60,90,105
510,20,30,35
22,4,6,7
 1,2,3,7

∴ LCM = 3 × 5 × 2 × 2 × 3 × 7 = 1260 minutes

1260 minutes =$ 1260/60$ = 21 hours

∴ The bell will again ring simultaneously after 21 hours.

∴ Time will be

= 12 noon + 21 hours

= 9 a.m

Question : 3

Four bells ring at the intervals of 5, 6, 8 and 9 seconds. All the bells ring simultaneously at some time. They will again ring simultaneously after

a) 24 minutes

b) 18 minutes

c) 12 minutes

d) 6 minutes

Answer: (d)

The LCM of 5, 6, 8 and 9 = 360 seconds = 6 minutes

Question : 4

The number nearest to 10000, which is exactly divisible by each of 3, 4, 5, 6, 7 and 8, is :

a) 10000

b) 9996

c) 10080

d) 9240

Answer: (c)

LCM of 3, 4, 5, 6, 7, 8 = 840

11$760/840$ = 10000

Since, the remainder 760 is more than half of the divisor 840.

∴ The nearest number

= 10000 + (840 – 760) = 10080

Question : 5

Three bells ring simultaneously at 11a.m. They ring at regular intervals of 20 minutes, 30 minutes, 40 minutes respectively. The time when all the three ring together next is

a) 1.30 p.m.

b) 1.15 p.m.

c) 1 p.m.

d) 2 p.m.

Answer: (c)

LCM of 20, 30 and 40 minutes = 120 minutes

Hence, the bells will toll together again after 2 hours i.e. at 1 p.m.

Question : 6

The least number which when divided by 16, 18, 20 and 25 leaves 4 as remainder in each case but when divided by 7 leaves no remainder is

a) 18004

b) 18002

c) 18000

d) 17004

Answer: (a)

LCM of 16, 18, 20 and 25 = 3600

∴ Required number = 3600K + 4 which is exactly divisible by 7 for certain value of K.

When K = 5,

number = 3600 × 5 + 4

= 18004 which is exactly divisible by 7.

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