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Home ‣ Aptitude ‣ Problems on H.C.F and L.C.M Comments
- Question
- 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
- Correct Answer
- 4
ExplanationGiven 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. - 1. 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
- 2. 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 Discuss
- 3. 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
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. 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
- 9. In a school 391 boys and 323 girls have been divided into the largest possible equal classes. So that there are equal number of boys and girls in each class. What is the number of classes?
Options- A. 23 girls classes, 19 boys classes
- B. 23 boys classes, 19 girls classes
- C. 17 boys classes, 23 girls classes
- D. 23 boys classes, 17 girls classes Discuss
- 10. The product of two number is 7168 and their H.C.F. is 16. find the sum of all possible numbers?
Options- A. 640
- B. 860
- C. 460
- D. Data inadequate Discuss
Problems on H.C.F and L.C.M
Search Results
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: 124
Explanation:
Let 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
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: 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.
Correct Answer: 23 boys classes, 19 girls classes
Explanation:
The largest possible number of persons in a class is given by the H.C.F. of 391 and 323 i,e. 17
? No of classes of boys = 391/17 = 23 and
No. of classes of girls = 323 / 17 = 19.
Correct Answer: 640
Explanation:
Let the number be 16a and 16b, where a and be are two number prime to each other.
? 16a x 16b = 7168
? ab = 28
Now, the pairs of number whose products is 28, are (28, 1) : (14, 2), (7,4)
14 and 2 which are not prime to each other should be rejected.
Hence, the required answer
= (448 + 16) + (112 + 64) = 640
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