Discussion
Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
- What is the greatest number that divides 263, 935 and 1383 leaving a remainder of 7 in each case?
Options- A. 30
- B. 31
- C. 32
- D. 35
- Correct Answer
- 32
ExplanationThe required greatest number is the HCF of (263 - 7, 935 - 7, 1383 - 7)
i.e., HCF of 256, 928 and 1376 = 32
- 1. What are the values of c when the HCF of x 3 + c x 2 - x + 2c and x 2 + cx - 2 over the rational is a linear polynomial?
Options- A. +1
- B. +2
- C. +1
- D. +4 Discuss
- 2. The LCM and HCF of two number are 240 and 16, respectively,If the two numbers are in the ratio 3 : 5, the numbers are?
Options- A. 24, 40
- B. 21, 35
- C. 36, 60
- D. 48, 80 Discuss
- 3. If the HCF of m and n is 1, then what are the HCF of m + n, m and HCF of m - n, n, respectively (where, m > n)?
Options- A. 2 and 1
- B. 1 and 1
- C. 1 and 2
- D. Cannot be determined Discuss
- 4. If the HCF of two positive integers is 26. Then their LCM cannot be?
Options- A. 78
- B. 104
- C. 144
- D. 234 Discuss
- 5. Six bell commence tolling together and toll at intervals of 5, 10, 15, 20, and 30 s, respectively. In 60 min, how many times do they toll together?
Options- A. 20
- B. 15
- C. 13
- D. 11 Discuss
- 6. What is the greatest number which divides 390, 480 and 620 so as to leave the same remainder in each case?
Options- A. 15
- B. 10
- C. 20
- D. 19 Discuss
- 7. Consider the number between 300 and 400 which when divided by 12, 18 and 36 leaves 4 as remainder in each case. what is the sum of the number?
Options- A. 840
- B. 692
- C. 324
- D. 1024 Discuss
- 8. Find the HCF of 132, 204 and 228.
Options- A. 12
- B. 18
- C. 6
- D. 21 Discuss
- 9. Find the HCF of 4/5 and 7/15?
Options- A. 1/13
- B. 1/5
- C. 1/15
- D. 1/25 Discuss
- 10. Find the HCF of 1/2, 3/4, and 4/5?
Options- A. 1/20
- B. 1/40
- C. 20
- D. 15 Discuss
Problems on H.C.F and L.C.M
Search Results
Correct Answer: +1
Explanation:
Find HCF of x3 +c x2 -x + 2c and x2 + cx - 2 by long division method and get remainder
? Remainder = 2c - 2/c
Since,remainder should be zero
? 2c2 - 2 = 0
? 2(c2 -1) = 0 ? c= ± 1
Correct Answer: 48, 80
Explanation:
Suppose two numbers are 3N and 5N.
Then, 3N x 5N = HCF x LCM
? 15N2 = 16 x 240
? N2 = 256
? N = 16
So, the number are 3N = 3 x 16 = 48 and 5N = 5 x 16 = 80.
Correct Answer: 1 and 1
Explanation:
The HCF of m and n is 1, so m and n are prime number.
Let m = 7 and n = 5 ? m + n = 12
HCF of(m + n)and m = HCF of 12 and 7 = 1
Similarly, HCF of (m - n) and n = 1
Correct Answer: 144
Explanation:
LCM of the number is always divisible by the HCF of the same numbers.
So, 78, 104 and 234 are all divisible by 26. Whereas, 144 is not divisible by 26.
Thus, 144 cannot be the LCM of the number whose HCF is 26.
Correct Answer: 13
Explanation:
LCM of 5, 10, 15, 20, 25 and 30 is 300. So, the bell will toll together after every 300s (5min).
So, the number of times they toll together = 60/5 + 1=13
Correct Answer: 10
Explanation:
The HCF of [(480 - 390), (620 - 480), (620 - 390)]
= HCF of 90, 140, 230 = 10
Correct Answer: 692
Explanation:
LCM of (12,18, 36) = 36
? A number greater than 300 on dividing by 36 leaves a remainder 4 = (36 x 9) + 4 = 328
Next numbe r = 328 + 36 = 364
? Sum of number = 328 + 364 = 692
Correct Answer: 12
Explanation:
132 = 2 x 2 x 3 x 11
240 = 2 x 2 x 2 x 2 x 3 x 5
228 = 2 x 2 x 3 x 19
Hence, HCF of 132, 204 and 228 = 2 x 2 x 3 = 12
Correct Answer: 1/15
Explanation:
We know that,
HCF of fractions = (HCF of numerators) / (LCM of denominators)
? Required HCF = (HCF of 4 and 7) / (LCM of 5 and 15)
=1/15
Correct Answer: 1/20
Explanation:
Required HCF = (HCF of numerators) / (LCM of denominators)
= (HCF of 1, 3 and 4) / (LCM of 2, 4, and 5)
= 1/20
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.