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Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
- The sum of HCF and LCM of two numbers is 403 and their LCM is 12 times their HCF. If one number is 93, then find the another number.
Options- A. 115
- B. 122
- C. 124
- D. 138
- Correct Answer
- 124
ExplanationLet LCM = m, HCF = n,
According to the question,
m = 12n, .......(i)
and m + n = 403.......(ii)
? 12n + n = 403 [from Eq.(i)]
? 13n = 403
? n = 403/13 = 31
? m = 12 x 31 = 372
Let the another number = k
? 93 x k = 372 x 31
[as product of two numbers HCF x LCM]
? k = (372 x 31)/93 = 124 - 1. If the HCF of a and b are 12 and a, b are positive integers and a > b > 12, then what will be the values of a and b?
Options- A. 12, 24
- B. 24, 12
- C. 24, 36
- D. 36, 24 Discuss
- 2. Four numbers are in the ratio of 10 : 12 : 15 : 18. If their HCF is 3, then find their LCM.
Options- A. 420
- B. 540
- C. 620
- D. 680 Discuss
- 3. Three numbers are in the ratio of 3 : 4 : 5 and their LCM is 1200. Find the HCF of the numbers?
Options- A. 40
- B. 30
- C. 80
- D. 20 Discuss
- 4. The HCF and LCM of two numbers are 13 and 1989, respectively. If one of the numbers is 117, then determine the other
Options- A. 121
- B. 131
- C. 221
- D. 231 Discuss
- 5. The LCM of two numbers is 2376 while their HCF is 33. If one of the number is 297, then the other number is
Options- A. 216
- B. 264
- C. 642
- D. 792 Discuss
- 6. The LCM of two numbers is 20 times of their HCF and (LCM + HCF) = 2520. If one number is 480, what will be the triple of another number?
Options- A. 1200
- B. 1500
- C. 2100
- D. 1800 Discuss
- 7. Find the greatest number that divides 130, 305 and 245 leaving remainders 6, 9 and 17, respectively?
Options- A. 4
- B. 5
- C. 14
- D. 24 Discuss
- 8. Find the least number which when divided by 16, 18 and 20 leaves a remainder 4 in each case, but is completely divisible by 7.
Options- A. 2884
- B. 2256
- C. 865
- D. 3332 Discuss
- 9. Three runners running around a circular track can complete one revolution in 2, 4 and 5.5 respectively. When will they meet at starting point?
Options- A. 40 h
- B. 44 h
- C. 20 h
- D. 22 h Discuss
- 10. Find the two numbers whose L.C.M. is 1188 and H.C.F. is 9?
Options- A. 27, 396
- B. 9, 27
- C. 36,99
- D. Data inadequate Discuss
Problems on H.C.F and L.C.M
Search Results
Correct Answer: 36, 24
Explanation:
By Hit and Trial
From option (d), we can say that the HCF of 36 and 24 is 12 and it is also satisfies the given condition
a > b > 12
? a = 36 and b = 24
Correct Answer: 540
Explanation:
Let numbers be 10N, 12N, 15N and 18N.
Then, LCM = 180N and HCF = N
Hence, required LCM = 180 x 3 = 540
Correct Answer: 20
Explanation:
Let the numbers are 3N, 4N and 5N.
Where, N = HCF
Then, LCM = 60N
According to the question,
60N = 1200
? N = 20
Correct Answer: 221
Explanation:
Given,
LCM = 1989, HCF = 13
1st number = 117 and
2nd number = ?
According to the formula,
Product of LCM and HCF = Product of two numbers
? 1989 x 13 = 117 x ?
? = (1989 x 13)/117
? = 221
Correct Answer: 264
Explanation:
Given, LCM of two number = 2376
HCF of two number = 33
One of the number = 297
? (HCF of two numbers) x (LCM of two numbers) = (First number) x (Second number)
? Second number = (33 x 2376)/297
= 264
Correct Answer: 1800
Explanation:
Let HCF = N
According to the question, LCM = 20N
Given that, HCF + LCM = 2520
? N + 20N = 2520
? N = 2520 / 21 = 120
Now, LCM = 20N = 20 x 120 = 2400
We know that,
1st number x 2nd number = HCF x LCM
? 2nd number = (LCM x HCF) / (1st number)
= (120 x 2400) / 480 = 600
? Required answer = 600 x 3 = 1800
Correct Answer: 4
Explanation:
Given that,
x = 130, y = 305, z = 245
a = 6, b = 9, c = 17
According to the formula,
Required number = HCF of [(130 - 6), (305 - 9), (245 - 17)]
= HCF of 124, 296, 228 = 4.
Correct Answer: 2884
Explanation:
LCM of 16, 18 and 20 = 720
? Required number = 720k + 4
Where, k is a natural number to be divisible by 7, (720 k + 4) will be a multiple of 7.
Smallest value of k = 4
? Required number = 720 x 4 + 4 = 2884
Correct Answer: 44 h
Explanation:
Time at which they meet at starting point
= LCM of 2, 4 and 5.5
= 44 h
Correct Answer: 27, 396
Explanation:
Let the two number be 9a and 9b where a and b are two numbers prime to each other. The L.C.M. of 9a and 9b is 9ab.
? 9ab = 1188
? ab = 132
Now, the possible pairs of factors of 132 are 1 x 132, 2 x 66, 3 x 44, 6 x 22, 11 x 12 of these pairs (2, 66) and (6, 22) are not prime to each other and, therefore, not admissible.
Hence the admissible pairs are
(1, 132), (3, 44), (4, 33), (11, 12)
? a = 1, b = 132; or
a=3, b=44 or
a=4, b=33 or
a=11, b = 12
Hence, the possible number are 9, 9 x 13, 9 x 3, 9 x 44, 9 x 4, 9 x 33, 9 x 11, 9 x 12
So option A is correct 27, 396.
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