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Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
- For any integers a and b with HCF (a, b) = 1, what is HCF (a + b, a - b) equal to?
Options- A. It is always 1
- B. It is always 2
- C. Either 1 or 2
- D. None of the above
- Correct Answer
- Either 1 or 2
ExplanationPutting arbitrary values of a and b.
IIIustration 1:
Let a = 9 and b = 8
HCF (8 + 9, 9 - 8)
HCF (17, 1) = 1
IIIustration 2:
Let a = 23 and b =17
HCF (17 + 23, 23 -17)
HCF ( 40, 6) = 2
Hence, HCF (a + b, a - b) can either be 1 or 2. - 1. For any integer n, What is HCF (22n + 7, 33n + 10) equal to?
Options- A. 0
- B. 1
- C. 11
- D. None of these Discuss
- 2. How many numbers are there between 4000 and 6000 which are exactly divisible by 32, 40, 48 and 60?
Options- A. 2
- B. 3
- C. 4
- D. 5 Discuss
- 3. The HCF and LCM of two natural numbers are 12 and 72, respectively. What is the difference between the two numbers, if one of the number is 24?
Options- A. 12
- B. 18
- C. 21
- D. 24 Discuss
- 4. The sum of two numbers is 1056 and their HCF is 66, Find the Number of such pairs.
Options- A. 6
- B. 2
- C. 4
- D. 8 Discuss
- 5. The ratio of two numbers is 3 : 4 and their HCF is 4. What will be their LCM?
Options- A. 12
- B. 16
- C. 24
- D. 48 Discuss
- 6. 6 bells commence tolling together and toll of intervals are 2, 4, 6, 8, 10 and 12 s, respectively. In 1 h, how many times, do they toll together?
Options- A. 16
- B. 32
- C. 21
- D. 35 Discuss
- 7. The HCF and LCM of two number are 21 and 4641, respectively. If one of the numbers lies between 200 and 300, then find the two numbers.
Options- A. 273, 363
- B. 273, 359
- C. 273, 361
- D. 273, 357 Discuss
- 8. If (x - 6) is the HCF of x 2 - 2x - 24 and x 2- kx - 6, then what is the value of k?
Options- A. 3
- B. 5
- C. 6
- D. 8 Discuss
- 9. If a and b are positive integers, then what is the value of (a / HCF (a,b) , b/HCF (a,b) )?
Options- A. a
- B. b
- C. 1
- D. a/HCF (a,b) Discuss
- 10. What is the HCF of 8 (N 5 - N 3 + N ) and 28 (N 6 + 1)?
Options- A. 4 (N4 - N2 + 1)
- B. N3 - N + 4N2
- C. N3 - N + 3N2
- D. None of these Discuss
Problems on H.C.F and L.C.M
Search Results
Correct Answer: 1
Explanation:
HCF of (22n + 7, 33n + 10) is always 1.
lllustraction
For n = 1, HCF(29, 43) ? HCF = 1
For n = 2, HCF(51, 76) ? HCF = 1
For n = 3, HCF(73, 109) ? HCF = 1
Correct Answer: 4
Explanation:
LCM of 32, 40, 48 and 60 = 480
The number divisible by 480 between 4000 and 6000 are 4320, 4800, 5280 and 5760.
Hence, required number of numbers are 4.
Correct Answer: 12
Explanation:
Second number = (LCM x HCF) / first number
= (72 x 12)/ 24 = 36
Difference between the two numbers
= 36 - 24 = 12
Correct Answer: 4
Explanation:
Let the numbers be 66a and 66b, where a and b are co-primes.
According to the question,
66a + 66b = 1056
? 66(a + b) = 1056
? (a + b) = 1056/66 = 16
? Possible values of a and b are
(a = 1, b = 15), (a = 3, b = 13).
(a = 5, b = 11 ), (a = 7, b = 9)
? Numbers are
(66 x 1, 66 x 15), (66 x 3, 66 x 13),
(66 x 5, 66 x 11), (66 x 7, 66 x 9).
? Possible number of pairs = 4
Correct Answer: 48
Explanation:
According to the question,
1st number = 3M, 2nd number = 4M
where, M = HCF
But given, M = 4
We know that,
LCM = Product of two numbers / HCF
= (3M x 4M) / M
LCM = 12M = 12 x 4 = 48
Correct Answer:
Explanation:
LCM of 2, 4, 6, 8, 10, 12
=(2 x 2 x 3 x 2 x 5) = 120
? After every 2 min, they toll together.
? Number of times they toll together in one hour = (60/2) + 1 times
= 31 times
Correct Answer: 273, 357
Explanation:
Let the numbers be 21a and 21b, where a and b are co-primes.
Then, 21a x 21b = (21 x 4641)
? ab = 221
Two co-primes with product 221 are 13 and 17.
? Required number = (21 x 13, 21 x 17)
= (273, 357)
Correct Answer: 5
Explanation:
Given that, (x - 6) is the HCF of x2 - 2x - 24 and x2 - kx - 6 i.e., (x - 6) is a factor of both expressions.
Let f (x1) = x12 - 2x1 - 24
and f (x2) = x22 - kx2 - 6
Now f(x1) = f (x2) at (x1 = x2 = 6)
? (6)2 - 2(6) - 24 = (6)2 - k(6) - 6 [by condition]
? 0 = 30 - 6k ? 6 k = 30
? k = 5
Correct Answer: 1
Explanation:
HCF (a /HCF (a, b) , b/HCF (a,b) it value is always equal to 1;
IIIustration 1 Let the two positive integers be a = 24 and b = 36.
? HCF (24/HCF (24,36) , 36/HCF (24,36)
? HCF (24/12 , 36/12)
? HCF (2, 3) = 1
IIIustration 2
Let the two positive integers be a = 13 and b = 17
? HCF (13 / HCF (13,17) , 17/HCF (13,17)
? HCF (13/1 , 17/1) = 1
Correct Answer: 4 (N4 - N2 + 1)
Explanation:
Let p(N) = 8 (N5 - N3 + N)
= 4 x 2 x N (N4 - N2 + 1)
and q(N) = 28 (N6 + 1)
= 7 x 4 [( N2)3 + (1)3]
= 4 x 7 (N2 + 1) ( N4 - N2 + 1)
? HCF of p(N) and q (N) = 4 (N4 - N2 + 1)
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