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

Question : 41 [SSC Section Officer 2007]

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 : 42 [SSC CGL Prelim 2002]

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 : 43

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 : 44 [SSC CGL 2013]

L.C.M. of $2/3, 4/9, 5/6$ is

a) $20/27$

b) $10/3$

c) $20/3$

d) $8/27$

Answer: (c)

Using Rule 2,

$\text"L.C.M of fractions" = \text"L.C.M.of numerators"/\text"H.C.F.of denominators"$

LCM = ${\text"LCM of " 2, 4, 5}/{\text"HCF of " 3, 9, 6}$= $20/3$

Question : 45

The smallest number, which when divided by 5, 10, 12 and 15, leaves remainder 2 in each case; but when divided by 7 leaves no remainder, is

a) 91

b) 175

c) 182

d) 189

Answer: (c)

LCM of 5, 10, 12, 15

25,10,12,15
35,5,6,15
55,5,2,5
 1,1,2,1

∴ LCM = 2 × 3 × 5 × 2 = 60

∴ Number = 60k + 2

Now, the required number should be divisible by 7.

Now, 60k + 2 = 7 × 8k + 4k + 2 If we put k = 3, (4k + 2) is equal to 14 which is exactly divisible by 7.

∴ Required number = 60 × 3 + 2 = 182

IMPORTANT quantitative aptitude EXERCISES

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