Aptitude - H.C.F and L.C.M
11.
The least number which when divided by 5, 6 , 7 and 8 leaves remainder 3, but when divided by 9 leaves no remainder, is :
Answer
L.C.M of 5, 6, 7, 8 = 840.
Let the required number be 840k + 3, which is a multiple of 9.
Smallest value of k is 2.
Required number = (840 x 2 + 3) = 1683
Answer :
Option BExplanation :
L.C.M of 5, 6, 7, 8 = 840.
Let the required number be 840k + 3, which is a multiple of 9.
Smallest value of k is 2.
Required number = (840 x 2 + 3) = 1683
12.
What is the least number which when diminished by 5, is divisible by each one of 21, 28, 36, 45 ?
Answer
Required number = L.C.M. of 21, 28, 36, 45 + 5
= 1260 + 5 = 1265
Answer :
Option DExplanation :
Required number = L.C.M. of 21, 28, 36, 45 + 5
= 1260 + 5 = 1265
13.
The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is :
Answer
Required length = H.C.F of 700 cm, 385 cm and 1295 cm = 35 cm.
Answer :
Option CExplanation :
Required length = H.C.F of 700 cm, 385 cm and 1295 cm = 35 cm.
14.
The sum of numbers is 45. Their difference is one-ninth of their sum. Their L.C.M is :
Answer
Let the numbers be x and y . Then,
x + y = 45
x − y = 1 9 x 45 = 5
⇒ x = 25, y = 20.
L.C.M. of 25 and 20 = 5 x 5 x 4 = 100.
Answer :
Option AExplanation :
Let the numbers be x and y . Then,
x + y = 45
x − y = 1 9 x 45 = 5
⇒ x = 25, y = 20.
L.C.M. of 25 and 20 = 5 x 5 x 4 = 100.
15.
Two 3 digit numbers have their H.C.F 29 and L.C.M. 4147. The sum of the numbers is :
Answer
Let the numbers be 29a and 29b , where a and b are co-primes.
Then, 29a x 29b = 29 x 4147 ⇒ ab = 29 x 4147 29 x 29 = 143
Co-primes with product 143 are 11 and 13.
∴ Numbers are (29 x 11, 29 x 13) i.e. 319, 377
Their sum = (319 + 377) = 696.
Answer :
Option CExplanation :
Let the numbers be 29a and 29b , where a and b are co-primes.
Then, 29a x 29b = 29 x 4147 ⇒ ab = 29 x 4147 29 x 29 = 143
Co-primes with product 143 are 11 and 13.
∴ Numbers are (29 x 11, 29 x 13) i.e. 319, 377
Their sum = (319 + 377) = 696.
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