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Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
- The LCM of two numbers is 48. The numbers are in the ratio of 2 : 3. Find the sum of the numbers.
Options- A. 28
- B. 32
- C. 40
- D. 64
- Correct Answer
- 40
ExplanationLet numbers are 2N and 3N.
According to the question, 6N = 48
? N = 8 ( ? LCM = 6N )
? Required sum = ( 2N + 3N ) = 5N
= 5 x 8 = 40 - 1. If a number is exactly divisible by 11 and 13, which of the following types the number must be?
Options- A. Divisible by (11 + 13)
- B. Divisible by (13 - 11)
- C. Divisible by (11 x 13)
- D. Divisible by (13 ÷ 11) Discuss
- 2. The ratio of two numbers is 5 : 6 and their LCM is 480, then their HCF is
Options- A. 20
- B. 16
- C. 6
- D. 5 Discuss
- 3. The HCF and LCM of two numbers m and n are respectively 6 and 210. If m + n = 72, then 1/m + 1/n is equal to
Options- A. 1/35
- B. 3/35
- C. 5/37
- D. 2/35 Discuss
- 4. The HCF of three numbers is 23. If they are in the ratio of 1 : 2 : 3, then find the numbers?
Options- A. 69, 15, 22
- B. 23, 46, 69
- C. 25, 31, 41
- D. 23, 21, 35 Discuss
- 5. The HCF of two numbers is 15 and their LCM is 225. If one of the numbers is 75, then find the another number.
Options- A. 105
- B. 90
- C. 60
- D. 45 Discuss
- 6. The Product of two whole numbers is 1500 and their HCF is 10, Find the LCM.
Options- A. 15000
- B. 150
- C. 1500
- D. 15 Discuss
- 7. The least number which should be added to 2497, so that the sum is exactly divisible by 5, 6, 4 and 3, is
Options- A. 3
- B. 13
- C. 23
- D. 33 Discuss
- 8. The least number which when divided by 12,16 and 18 leaves 5 as remainder in each case. Find the number.
Options- A. 139
- B. 144
- C. 149
- D. 154 Discuss
- 9. What least number must be subtracted from 1294 so that the remainder when divided by 9, 11,13 will leave in each case the same remainder 6?
Options- A. 0
- B. 1
- C. 2
- D. 3 Discuss
- 10. Find the number lying between 900 and 1000 which when divided by 38 and 57, leaves in each case a remainder 23?
Options- A. 935
- B. 945
- C. 925
- D. 955 Discuss
Problems on H.C.F and L.C.M
Search Results
Correct Answer: Divisible by (11 x 13)
Explanation:
LCM of 11 and 13 will be (11 x 13). Hence, if a number is exactly divisible by 11 x 13, then the same number must be exactly divisible by their LCM or by (11 x 13).
Correct Answer: 16
Explanation:
Let numbers are 5N and 6N.
Now, HCF of these two numbers is N.
We know that,
LCM x HCF = Product of two numbers
? 480 x N = 5N x 6N
? 480N = 30N2
? N =16
Correct Answer: 2/35
Explanation:
We have, m x n = 6 x 210 = 1260
? 1/m + 1/n = (m + n)/mn = 72/1260 = 4/70 = 2/35
Correct Answer: 23, 46, 69
Explanation:
Let the numbers are N, 2N and 3N.
Where, N = HCF
Given that, N = 23
? The numbers are 23, 46 and 69.
Correct Answer: 45
Explanation:
Let the other number = N
Product of LCM and HCF = Product of two numbers
Then, N x 75 = 15 x 225
? N = (15 x 225)/75 = 45
Correct Answer: 150
Explanation:
Given that, product of two numbers =1500
HCF = 10
According to the formula,
Product of two numbers = HCF x LCM
? 1500 = 10 x LCM
? LCM = 1500/10 =150
Correct Answer: 23
Explanation:
LCM of 5, 6, 4 and 3 = 60
On dividing 2497 by 60, the remainder is 37.
? Number to be added = 60 - 37 = 23
Correct Answer: 149
Explanation:
Given, a = 12, b =16 , c =18, k = 5
According to the formula,
Required number =(LCM of a, b and c) + k
= (LCM Of 12, 16, 18) + 5
LCM of 12, 16, 18 is
? LCM = 2 x 2 x 3 x 4 x 3 = 144
? Required number =144 + 5 = 149
Correct Answer: 1
Explanation:
The number when divided by 9, 11, 13 leaving remainder 6 = (i.c.m. of 9, 11, 13) + 6 = 1293
Required least number = 1294 - 1293 = 1.
Correct Answer: 935
Explanation:
The least common multiple of 38 and 57 is 114 and the multiple which is between 900 and 1000 is 912.
Now, 912 + 23 i,e 935 lies between 900 and 1000 and when divided by 38 and 57 leaves in each case 23 as the remainder.
Therefore, 935 is number required.
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