Problems on HCM and LCM

Problems on HCM and LCM

Learn and practice Problems on HCM and LCM with easy explaination and shortcut tricks. All questions and answers on HCM and LCM covered for various Competitive Exams

Solve Problems:

1) What is the HCF of 1095 and 1168?

  1. 37
  2. 73
  3. 43
  4. 83
View Answer

Answer: B

Explanation:

By Division Method

Therefore the HCF is 73

2) Find the HCF of 210, 385, and 735.

  1. 7
  2. 14
  3. 21
  4. 35
View Answer

Answer: D

Explanation:

210 = 2 x 5 x 3 x 7 x 1
385 = 5 x 7 x 11 x 1
735 = 5 x 7 x 3 x 7 x 1
HCF = 5 x 7 = 35

3)

View Answer

Answer: B

Explanation:

Formula:

HCF:

2= 2 x 1
8= 2 x 2 x 2
64= 2 x 2 x 2 x 2 x 2 x 2
10= 2 x 5
HCF =2

LCM:

LCM = 3 x 3 x 3 x 3 = 81

4) What will be the HCF of 608, 544; 638, 783; and 425, 476 respectively?

  1. 32, 29, 17
  2. 17, 32, 29
  3. 29, 32, 17
  4. 32, 17, 29
View Answer

Answer: A

Explanation:

First of all we have to write the numbers as product of the prime factor.
Prime factor of 608 = 25 * 19
i.e.,

And the prime factor of 544 = 25 * 17
i.e.,

So, the HCF of 608 and 544 = common factor in both = 25 = 32
Similarly, The prime factor of 638 = 2 * 11 * 29
The prime factor of 783 = 3* 3 * 3 * 29
So, the HCF of 638 and 783 = 29

Now, the prime factor of 425 = 5 * 5 * 17
The prime factor of 476 = 2*2*7*17
So, the HCF of 425 and 476 = 17

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5)

View Answer

Answer: C

Explanation:
Formula:

LCM:

LCM = 2 x 2 x 5 = 20

HCF:

3=3 x 1
6=3 x 2 x 1
9=3 x 3 x 1
27=3 x 3 x 3 x 1
HCF =3
Formula:

6) If the HCF of two numbers is 27, and their sum is 216, find these numbers.

  1. 27, 189
  2. 154, 162
  3. 108, 108
  4. 81, 189
View Answer

Answer: A

Explanation:

Let the numbers be 27x and 27y.

Then, 27x + 27y = 216
27 (x+ y) = 216
(x + y) = 216/27
X + y = 8

co-prime numbers with sum 8 are (1, 7) and (3, 5).

Therefore we can say that the required numbers are (27 x1, 27 x 7) and (27 x 3, 27x 5), i.e. (27, 189) and (81, 135).
Out of the above two answers, the given one in the answer is the pair (27, 189).

7) Two numbers are in the ratio of 15:11. If the HCF of numbers is 13, find the numbers.

  1. 75, 55
  2. 105, 77
  3. 15, 11
  4. 195, 143
View Answer

Answer: D

Explanation:

Let the two numbers are 15x and 11x
HCF = x
x = 13
Therefore, the numbers are (15 x 13 and 11 x 13), i.e. 195 and 143.

8) Find the greatest integer that divides 358, 376, and 334 and leaves the same remainder in each case.

  1. 6
  2. 7
  3. 8
  4. 9
View Answer

Answer: A

Explanation:

Take the difference between two numbers

376 – 358 = 18;
376 – 334 = 42;
358 – 334 = 24;

By taking the HCF of 18, 42, 24.

18 = 2 x3 x 3
42 = 2 x 3 x 7
24 = 2 x 2 x 2 x 3
HCF = 2 x 3 = 6

Hence the required number is 6.

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9) Three bells toll at intervals of 36 sec, 40 sec, and 48 sec respectively. They start singing together at a particular time. When will they toll next together?

  1. 6 minutes
  2. 12 minutes
  3. 18 minutes
  4. 24 minutes
View Answer

Answer: B

Explanation:

Firstly find the LCM of 36, 40, and 48

LCM = 2 x 2 x 2 x 2 x 3 x 3 x 5 = 720 seconds

Hence, after 720 seconds or 12 minutes three bells will toll together.

10) The LCM of two numbers is 7700, and their HCF is 11. If one of these numbers is 275, what is the other number?

  1. 279
  2. 283
  3. 308
  4. 318
View Answer

Answer: C

Explanation:

Apply Formula: HCF x LCM = first number x Second number 11 x 7700 = 275 x second number

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