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Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
- 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
- Correct Answer
- 23
ExplanationLCM of 5, 6, 4 and 3 = 60
On dividing 2497 by 60, the remainder is 37.
? Number to be added = 60 - 37 = 23 - 1. 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
- 2. 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 Discuss
- 3. 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
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. 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
- 9. Find the sum of three numbers which are prime to one another such that the product of the first two is 437 and that of the last two is 551?
Options- A. 91
- B. 81
- C. 71
- D. 70 Discuss
- 10. What least number must be subtracted from 1936, so that the remainder when divided by 9, 10, 15 will leave in each case the same remainder 7?
Options- A. 46
- B. 53
- C. 39
- D. 44 Discuss
Problems on H.C.F and L.C.M
Search Results
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: 40
Explanation:
Let numbers are 2N and 3N.
According to the question, 6N = 48
? N = 8 ( ? LCM = 6N )
? Required sum = ( 2N + 3N ) = 5N
= 5 x 8 = 40
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: 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.
Correct Answer: 71
Explanation:
From the question we see that the second number is a common factor of the two products and since the numbers are prime to one another. It is their . H. C. F . and is, therefore, 19.
? The first number = 437 / 19 = 23
and the third number = 551 / 19 = 29
Hence, the number are 23, 19 and 29
? Sum = 23 + 19 + 29
= 71.
Correct Answer: 39
Explanation:
The L.C.M of 9, 10, 15 = 90
On dividing 1936 by 90, the remainder = 46
But a part of this remainder = 7
Hence, the two numbers = 46 - 7 = 39.
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