Practice Find lcm using formula - quantitative aptitude Online Quiz (set-1) For All Competitive Exams

Q-1)   Let the least number of six digits which when divided by 4, 6, 10, 15 leaves in each case same remainder 2 be N. The sum of digits in N is :

(a)

(b)

(c)

(d)

Explanation:

LCM of 4, 6, 10, 15 = 60

Least number of 6 digits = 100000

The least number of 6 digits which is exactly divisible by 60 = 100000 + (60 – 40) = 100020

∴ Required number (N) = 100020 + 2 = 100022

Hence, the sum of digits = 1 + 0 + 0 + 0 + 2 + 2 = 5


Q-2)   The LCM of two prime numbers x and y, (x > y) is 161. The value of (3y – x) :

(a)

(b)

(c)

(d)

Explanation:

LCM of x and y = 161

∴ xy = 23 × 7

∴ x = 23; y = 7

∴ 3y – x = 3 × 7 – 23

= 21 – 23 = – 2


Q-3)   The smallest number, which, when divided by 12 or 10 or 8, leaves remainder 6 in each case, is

(a)

(b)

(c)

(d)

Explanation:

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 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

The smallest number divisible by 12 or 10 or 8

= LCM of 12, 10 and 8 = 120

⇒ Required number =120 + 6 = 126


Q-4)   The least number which when divided by 4, 6, 8 and 9 leave zero remainder in each case and when divided by 13 leaves a remainder of 7 is :

(a)

(b)

(c)

(d)

Explanation:

LCM of 4, 6, 8, 9

24,6,8,9
22,3,4,9
31,1,2,3
 1,1,2,3

∴ LCM = 2 × 2 × 3 × 2 × 3 = 72

∴ Required number = 72, because it is exactly divisible by 4, 6, 8 and 9 and it leaves remainder 7 when divided by 13.


Q-5)   The smallest perfect square divisible by each of 6, 12 and 18 is

(a)

(b)

(c)

(d)

Explanation:

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


Q-6)   The LCM of four consecutive numbers is 60. The sum of the first two numbers is equal to the fourth number. What is the sum of four numbers?

(a)

(b)

(c)

(d)

Explanation:

260
230
315,
 5

∴ 60 = 2 × 2 × 3 × 5

i.e., Numbers = 2, 3, 4 and 5

∴ Required sum = 2 + 3 + 4 + 5 = 14


Q-7)   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)

(b)

(c)

(d)

Explanation:

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


Q-8)   The least number, which when divided by 18, 27 and 36 separately leaves remainders 5,14, and 23 respectively, is

(a)

(b)

(c)

(d)

Explanation:

The difference between the divisor and the corresponding remainder is same in each case ie. 18 – 5 = 13, 27 – 14 = 13, 36 – 23 = 13

∴ Required number = (LCM of 18, 27, and 36 ) – 13

= 108 – 13 = 95


Q-9)   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)

(b)

(c)

(d)

Explanation:

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


Q-10)   The least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is :

(a)

(b)

(c)

(d)

Explanation:

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