Question 1

Product of two coprime numbers is 117. Then, their LCM is

Accepted Answer

LCM of two coprimes is equal to their product.

Question 2

The LCM of two numbers is 2376 while their HCF is 33. If one of the number is 297, then the other number is

Accepted Answer

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

Question 3

The HCF and LCM of two numbers are 13 and 1989, respectively. If one of the numbers is 117, then determine the other

Accepted Answer

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

Question 4

Three numbers are in the ratio of 3 : 4 : 5 and their LCM is 1200. Find the HCF of the numbers?

Accepted Answer

Let the numbers are 3N, 4N and 5N. Where, N = HCF Then, LCM = 60N According to the question, 60N = 1200 ∴ N = 20

Question 5

Four numbers are in the ratio of 10 : 12 : 15 : 18. If their HCF is 3, then find their LCM.

Accepted Answer

Let numbers be 10N, 12N, 15N and 18N. Then, LCM = 180N and HCF = N Hence, required LCM = 180 x 3 = 540

Question 6

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?

Accepted Answer

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

Question 7

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.

Accepted Answer

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

Question 8

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?

Accepted Answer

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

Question 9

Find the greatest number that divides 130, 305 and 245 leaving remainders 6, 9 and 17, respectively?

Accepted Answer

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.

Question 10

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.

Accepted Answer

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

Question 11

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?

Accepted Answer

Time at which they meet at starting point = LCM of 2, 4 and 5.5 = 44 h

Question 12

Find the two numbers whose L.C.M. is 1188 and H.C.F. is 9?

Accepted Answer

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.

Question 13

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?

Accepted Answer

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.

Question 14

The product of two number is 7168 and their H.C.F. is 16. find the sum of all possible numbers?

Accepted Answer

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

Question 15

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?

Accepted Answer

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 .

Question 16

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?

Accepted Answer

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

Question 17

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?

Accepted Answer

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

Question 18

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?

Accepted Answer

Biggest measure = H.C.F. of 403, 434 and 465 = 31 kg

Question 19

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?

Accepted Answer

Greatest possible length of each plank = H.C.F of 42, 49, 63 = 7 m

Question 20

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?

Accepted Answer

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.

Problems on H.C.F and L.C.M Questions