LCM and HCF Questions with Solutions

  1. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

A) 1
B) 2
C) 3
D) 4
Answer: B) 2
Explanation:
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
=>ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.

  1. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

A) 123
B) 127
C) 235
D) 305
Answer: B) 127
Explanation:
Required number = H.C.F. of (1657 – 6) and (2037 – 5)
= H.C.F. of 1651 and 2032 = 127.

  1. Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

A) 40
B) 80
C) 120
D) 200
Answer: A) 40
Explanation:
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.

  1. The sum of two numbers is 528 and their H.C.F is 33. The number of pairs of numbers satisfying the above condition is

A) 4
B) 6
C) 8
D) 12
Answer: A) 4
Explanation:Let the required numbers be 33a and 33b.
Then 33a +33b= 528   =>   a+b = 16.
Now, co-primes with sum 16 are (1,15) , (3,13) , (5,11) and (7,9).herefore, Required numbers are  ( 33 x 1, 33 x 15), (33 x 3, 33 x 13), (33 x 5, 33 x 11), (33 x 7, 33 x 9)The number of such pairs is 4

  1. The L.C.M of two numbers is 495 and their H.C.F is 5. If the sum of the numbers is 100, then their difference is

A) 10
B) 46
C) 70
D) 90
Answer: A) 10
Explanation:
Let the numbers be x and (100-x).
Then,x(100−x)=5*495×100-x=5*495
=>  x2−100x+2475=0x2-100x+2475=0
=>  (x-55) (x-45) = 0
=>  x = 55 or x = 45
The numbers are 45 and 55
Required difference = (55-45) = 10

  1. The H.C.F and L.C.M of two numbers are  84 and 21 respectively.  If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is

A) 12
B) 48
C) 84
D) 108
Answer: C) 84
Explanation:
Let the numbers be x and 4x. Then,  x×4x=84×21 ⇔ x2=84×214⇔ x=21x×4x=84×21 ⇔ x2=84×214⇔ x=21
Hence Larger Number = 4x = 84

  1. The H.C.F of two numbers is 11 and their L.C.M is 7700. If one of the numbers is 275 , then the other is

A) 279
B) 283
C) 308
D) 318
Answer: C) 308
Explanation:
Other number = (11×7700275)11×7700275= 308.

  1. Two number , both greater than 27,have hcf 27 and lcm 162 The sum of the number is
    a)1

b)162
c)135
d)81
Answer:C)135
Explanation:
H.C.F*l.C.M= product of two no.
162*27=4374
now find out the factors of 4374
ANd find out the suitavle factors.I.e 54*81=4374 which have h.cf=27 and L.c.m-162.
therefore 54+81=135

  1. Find the greatest number, which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively.

A.127
b. 132
c. 114
d. 108
Answer:A
Explanation: Hence, make the dividend completely divisible by the divisor. This is possible, if we subtract remainder from the dividend.
Therefore,
1657 – 6 = 1651
2037 – 5 = 2032
H.C.F. of 1651 and 2032 is 127. 127 is the common factor.
127 × 13 = 1651
Thus by adding 6, we get 1651 + 6 = 1657
127 is the correct answer.

  1. Find the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder.

A.1963
b. 2523
c. 1683
d. 1536
Correct Option: (c)
This is truly a logical question and can just be solved using divisibility test.
Given:
The least number when divided by 9 leaves no remainder.
Divide all the given options by 9 and check whether any number leaves no remainder.

Option a) 1963 = 218.11
9
Option b) 2523 = 280.33
9
Option c) 1683 = 187
9
Option d) 1536 = 170.66
9

Option c) 1683 is completely divisible by 9. Hence, the least number is 1683

  1. The traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 5 : 20 : 00 hours, then find the time at which they will change simultaneously.

A. 5 : 28 : 00 hrs
b. 5 : 30 : 00 hrs
c. 5 : 38 : 00 hrs
d. 5 : 40 : 00 hrs
Correct Option: (b)

Explanation:
Traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively.
Therefore, find the L.C.M. of 40, 72 and 108.
L.C.M. of 40, 72 and 108 = 1080
The traffic lights will change again after 1080 seconds = 18 min
The next simultaneous change takes place at 5 : 38 : 00 hrs.

  1. A rectangular courtyard 4.55 meters long and 5.25 meters wide is paved exactly with square tiles of same size. Find the largest size of the tile used for this purpose?

A. 25 cm
b. 45 cm
c. 21 cm
d. 35 cm
Correct Option: (d)
Explanation:
Here, we are asked to find the largest size of tile. Therefore, calculate H.C.F.
Step 1: Covert numbers without decimal places i.e 455 cm and 525 cm
Step 2: Find the H.C.F. of 455 and 525
H.C.F. of 455 and 525 = 35 cm
Hence, the largest size of the tile is 35 cm.

  1. John, Smith and Kate start at same time, same point and in same direction to run around a circular ground. John completes a round in 250 seconds, Smith in 300 seconds and Kate in 150 seconds. Find after what time will they meet again at the starting point?

A. 30 min
b. 25 min
c. 20 min
d. 15 min
Correct Option: (b)
Explanation:
L.C.M. of 250, 300 and 150 = 1500 sec
Dividing 1500 by 60 we get 25, which mean 25 minutes.
John, Smith and Kate meet after 25 minutes.

  1. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A) 4
B) 10
C) 15
D) 16
Answer: D) 16
Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes,they will together (30/2)+1=16 times

  1. The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is

A) 1677
B) 1683
C) 2523
D) 3363
Answer: B) 1683
Explanation:
L.C.M of 5, 6, 7, 8 = 840
Therefore, Required Number is of the form 840k+3.
Least value of k for which (840k+3) is divisible by 9 is k = 2
Therefore, Required  Number = (840 x 2+3)=1683

  1. A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in Width. Find the least number of square tiles of equal size required to cover the entire floor of the room.

A) 107
B) 117
C) 127
D) 137
Answer: B) 117
Explanation:
Let us calculate both the length and width of the room in centimeters.
Length = 6 meters and 24 centimeters = 624 cm
width = 4 meters and 32 centimeters = 432 cm
As we want the least number of square tiles required, it means the length of each square tile should be as large as possible.Further,the length of each square tile should be a factor of both the length and width of the room.
Hence, the length of each square tile will be equal to the HCF of the length and width of the room = HCF of 624 and 432 = 48
Thus, the number of square tiles required = (624 x 432 ) / (48 x 48) = 13 x 9 = 117

  1. Find the largest number of four digits exactly divisible by 12,15,18 and 27.

A)9720
B)6580
C)4578
D)1023
Answer:A
Explanation:
The Largest number of four digits is 9999.
Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.
On dividing 9999 by 540,we get 279 as remainder .
Required number = (9999-279) = 9720.

  1. Find the least number which when divided by 5,6,7, and 8 leaves a remainder 3, but when divided by 9 leaves no remainder .

A)1683
B)1982
C)2001
D)none
Ans:A
Sol. L.C.M. of 5,6,7,8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 X 2 + 3)=1683

  1. L.C.M of two prime numbers x and y (x>y) is 161. The value of 3y-x is :

A) -2
B) -1
C)1
D)2
Answer:C
Solution:
Let’s say the 7 numbers are a, b, c, d, e, f and g.
Given that:
(a + b + c + d + e + f + g) / 7 = 60
(a + b + c) / 3 = 50
(e + f + g) / 3 = 70
To find: The remaining number i.e. ’d’
Solving the averages we get:
(a + b + c + d + e + f + g) / 7 = 60
a + b + c + d + e + f + g = 420
(a + b + c) / 3 = 50
a + b + c = 150
(e + f + g) / 3 = 70
e + f + g = 210
Inserting the values obtained above in the equation given below:
a + b + c + d + e + f + g = 420
150 + d + 210 = 420
360 + d = 420
d = 60