Problems on H.C.F and L.C.M
- 1. 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
Correct Answer: 44 h
Explanation:
Time at which they meet at starting point
= LCM of 2, 4 and 5.5
= 44 h
- 2. 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
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.
- 3. 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
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.
- 4. 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
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
- 5. The Sum of two number is 1215 and their H.C.F is 81, how many pairs of such number can be formed? Find them?
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
Correct Answer: 4
Explanation:
Let the number be 81a and 81b where a and b are two numbers prime to each other.
? 81a + 81b = 1215
? a + b = 1215/81 = 15
Now, find two numbers, whose sum is 15, the possible pairs are (14, 1) , (13, 2), (12, 3), (11, 4), (10, 5), (9, 6 ), (8, 7) of these the only pairs of numbers that are prime to each other are (14, 1) , (13, 2), (11, 4), and (8, 7).
Hence, the required number are
(14 x 81, 1 x 81), (13 x 81, 2 x 81), (11 x 81, 4 x 81), (8 x 81, 7 x 81)
or (1134, 81), (1053, 162), (891, 324), (648, 567)
So, there are four such pairs .
- 6. 21 mangoes trees, 42 apples trees and 56 oranges trees have to be planted in rows such that each row contains the same number of trees one variety only. Minimum number of rows in which the above trees may be planted is?
Options- A. 15
- B. 17
- C. 3
- D. 20 Discuss
Correct Answer: 17
Explanation:
H.C.F. of 21, 42, 56 = 7
Number of rows of mango trees, apple trees and oranges trees are 21/7 = 3, 42/7 = 6 and 56/7 = 8
? Required number of rows = (3 + 6 + 8 ) = 17
- 7. Five bells begin to toll together and toll respectively at intervals of 6, 7, 8, 9 and 12 seconds. How many times they will toll together in one hour , excluding the one at the start?
Options- A. 3
- B. 5
- C. 7
- D. 9 Discuss
Correct Answer: 7
Explanation:
L.C.M of 6, 7, 8, 9, 12 is 504
So, the bells will toll together after 504 sec.
In hour, they will toll together
= (60 x 60) / 504 times
= 7 times
- 8. Three different containers contain different qualities of mixtures of milk and water, whose measurement are 403 kg, 434 kg and 465 kg. What biggest measure must be there to measure all the different qualities exactly?
Options- A. 1 kg
- B. 7 kg
- C. 31 kg
- D. 41 kg Discuss
Correct Answer: 1 kg
Explanation:
Biggest measure = H.C.F. of 403, 434 and 465
= 31 kg
- 9. Three pieces of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length, What is the greatest possible length of each plank?
Options- A. 7 m
- B. 14 m
- C. 42 m
- D. 63 m Discuss
Correct Answer: 7 m
Explanation:
Greatest possible length of each plank = H.C.F of 42, 49, 63 = 7 m
- 10. Six bells commence tolling together and toll intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
Options- A. 4
- B. 10
- C. 15
- D. 16 Discuss
Correct Answer: 16
Explanation:
L . C . M of 2, 4, 6, 8, 10 and 12 is 120
So the bells will toll together after 120 second i,e 2 minutes. In 30 min. they will together in (30/2) + 1 times, i.e 16 times.
More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.