Problems on H.C.F and L.C.M
- 1. If the HCF of two positive integers is 26. Then their LCM cannot be?
Options- A. 78
- B. 104
- C. 144
- D. 234 Discuss
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.
- 2. 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
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
- 3. 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
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.
- 4. 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
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
- 5. 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 Discuss
Correct Answer: 32
Explanation:
The required greatest number is the HCF of (263 - 7, 935 - 7, 1383 - 7)
i.e., HCF of 256, 928 and 1376 = 32
- 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
Correct Answer: 10
Explanation:
The HCF of [(480 - 390), (620 - 480), (620 - 390)]
= HCF of 90, 140, 230 = 10
- 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
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
- 8. Find the HCF of 132, 204 and 228.
Options- A. 12
- B. 18
- C. 6
- D. 21 Discuss
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
- 9. Find the HCF of 4/5 and 7/15?
Options- A. 1/13
- B. 1/5
- C. 1/15
- D. 1/25 Discuss
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
- 10. Find the HCF of 1/2, 3/4, and 4/5?
Options- A. 1/20
- B. 1/40
- C. 20
- D. 15 Discuss
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
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