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

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

Questions : What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7 ?

(a) 30

(b) 39

(c) 36

(d) 37

The correct answers to the above question in:

Answer: (b)

LCM of 9, 10 and 15 = 90

⇒ The multiple of 90 are also divisible by 9, 10 or 15.

∴ 21 × 90 = 1890 will be divisible by them.

∴ Now, 1897 will be the number that will give remainder 7.

1936 – 1897

Required number = 1936 – 1897 = 39

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

Question : 1

The least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is :

a) 56

b) 50

c) 48

d) 46

Answer: (b)

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.

L.C.M. of 4, 6, 8, 12 and 16 = 48

∴ Required number = 48 + 2 = 50

Question : 2

The smallest number, which when divided by 12 and 16 leaves remainder 5 and 9 respectively, is :

a) 29

b) 39

c) 41

d) 55

Answer: (c)

Using Rule 5,

When a number is divided by a, b or c, leaving remainders p, q or r respectively such that the difference between divisor and remainder in each case is the same

i.e., (a – P) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t),

Where k is LCM of a, b and c

Here, 12 – 5 = 7, 16 – 9 = 7

∴ Required number = (L.C.M. of 12 and 16) – 7

= 48 – 7 = 41

Question : 3

From a point on a circular track 5 km long A, B and C started running in the same direction at the same time with speed of 2$1/2$ km per hour, 3 km per hour and 2 km per hour respectively. Then on the starting point all three will meet again after

a) 15 hours

b) 10 hours

c) 6 hours

d) 30 hours

Answer: (b)

A makes one complete round of the circular track in $5/{5/2}$ = 2 hours,

B in $5/3$ hours and C in $5/2$ hours.

That is after 2 hours A is at the starting point, B after $5/3$ hours and C after $5/2$ hours.

Hence the required time = LCM of 2, $5/3$ and $5/2$ hours

= ${\text"LCM of " 2, 5, 5}/ {\text"HCF of " 3, 2}$

= $10/1$ = 10 hours.

Question : 4

Three men step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. The minimum distance each should cover so that all can cover the distance in complete steps is

a) 6950 cm

b) 6930 cm

c) 9360 cm

d) 9630 cm

Answer: (b)

Required distance = LCM of 63, 70 and 77 cm.

= 6930 cm.

Illustration :

763,70,77
 9,10,11

∴ LCM = 7 × 9 × 10 × 11

= 6930

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