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Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
- 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
- Correct Answer
- 4
ExplanationLCM 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. - 1. 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
- 2. 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
- 3. 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
- 4. What is the sum of digits of the least number,which when divided by 52 leaves 33 as remainder, when divided by 78 leaves 59 as remainder and when divided by 117 leaves 98 as remainder?
Options- A. 21
- B. 27
- C. 19
- D. 36 Discuss
- 5. What is the um of the digits of the least number which when divided by 54 leaves 35 as remainder, when divided by 80 leaves 61 as remainder and when divided by 119 leaves 100 leaves 100 as remainder?
Options- A. 21
- B. 19
- C. 23
- D. 28 Discuss
- 6. 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
- 7. 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 Discuss
- 8. 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
- 9. 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
- 10. 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
Problems on H.C.F and L.C.M
Search Results
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: 19
Explanation:
(52 - 33) = 19, (78 - 59) = 19 and (117 - 98) = 19
LCM of 52, 78 and 117 is 468.
Required least number = 468 + 19 = 487
Sum of digits of 487 = 4 + 8 + 7 = 19
Correct Answer: 28
Explanation:
(54 - 35) = 19, (80 - 61) = 19 and(119 - 100) = 19
LCM of 54, 80 and 119 is 257040.
Required least number = 257040 + 19 = 257059
Sum of the digits of 257059 = 2 + 5 + 7 + 0 + 5 + 9 = 28
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: Either 1 or 2
Explanation:
Putting 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.
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
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