H.C.F. & L.C.M. Questions and Answers

Answer: A

The lights change together after the LCM of 48, 72 and 108 seconds, which is 432 seconds, or 7 minutes 12 seconds.

Adding this to 8:20:00 gives 8:27:12 hours, option A.

The LCM of 48 = 2^4 x 3, 72 = 2^3 x 3^2 and 108 = 2^2 x 3^3 is 2^4 x 3^3 = 432 seconds. Dividing by 60 gives 7 minutes and 12 seconds, which added to 8:20:00 gives 8:27:12. The other times add 16 minutes or more to the start.

Answer: C

Subtract the remainders: 1657 - 6 = 1651 and 2037 - 5 = 2032. The greatest number is their HCF.

2032 - 1651 = 381, and 1651 = 4 x 381 + 127, 381 = 3 x 127, so the HCF is 127, option C.

When a number divides 1657 leaving remainder 6 and 2037 leaving remainder 5, it divides 1651 and 2032 exactly. The HCF of 1651 and 2032 is 127 (1651 = 13 x 127 and 2032 = 16 x 127), and 127 is larger than both remainders, as it should be.

Answer: A

Subtract the common remainder: 1356 - 12 = 1344, 1868 - 12 = 1856 and 2764 - 12 = 2752. The greatest divisor is the HCF of these.

1856 - 1344 = 512, and the HCF of 1344 and 512 is 64, which also divides 2752 (2752 = 43 x 64). So the number is 64, option A.

A number that leaves remainder 12 in each case divides 1344, 1856 and 2752 exactly, and the greatest such number is their HCF, which is 64 (1344 = 21 x 64, 1856 = 29 x 64, 2752 = 43 x 64). The remainder 12 is less than 64, so the answer is valid.

Answer: A

Let the numbers be 3x, 4x and 5x. Their L.C.M. is 60x, so 60x = 2400 and x = 40.

The H.C.F. of 3x, 4x and 5x is x, so it is 40, option A. The values 80, 120 and 200 are multiples of x and not the H.C.F.

Check: the numbers are 120, 160 and 200, whose L.C.M. is 2,400 and H.C.F. is 40.

Answer: B

Write a = 15m and b = 15n with m and n co-prime. For two-digit numbers m and n lie between 1 and 6. The unordered co-prime pairs with m less than n are 11: (1,2) to (1,6), (2,3), (2,5), (3,4), (3,5), (4,5) and (5,6).

Adding the equal pair (15, 15), whose H.C.F. is also 15, gives 12 pairs, which is the final key (B). The tentative key (11) leaves out that equal pair.

The hundred-based pairs: 15 times 1 to 6 give 15, 30, 45, 60, 75, 90 and the pair (30, 60) is excluded because their H.C.F. is 30, which shows why co-primality of m and n matters.

Answer: A

The smallest such number is 125.

A number that leaves remainder 5 when divided by 8, 12 and 15 is 5 more than a common multiple of all three. The smallest is 5 more than their LCM:

\[ \begin{aligned} 8 &= 2^3,\ 12 = 2^2 \times 3,\ 15 = 3 \times 5 \\ \text{LCM} &= 2^3 \times 3 \times 5 = 120 \\ 120 + 5 &= 125 \end{aligned} \]

Check: \(125 = 8 \times 15 + 5 = 12 \times 10 + 5 = 15 \times 8 + 5\). 245, 365 and 485 are 5 more than 240, 360 and 480, which are also multiples of 120, so they leave the same remainder, but they are not the smallest.

Answer: A

The larger number is 66.

With the ratio 3 : 11, write the numbers as \(3k\) and \(11k\):

\[ \begin{aligned} 3k + 11k &= 84 \\ 14k &= 84,\ k = 6 \\ \text{numbers} &= 18 \text{ and } 66 \end{aligned} \]

Check the HCF: since 3 and 11 have no common factor, the HCF of 18 and 66 is \(k = 6\), as given. 60, 54 and 48 do not combine with any partner in the ratio 3 : 11 to make 84.

Answer: B

The other number is 308.

For any two numbers, the product of the numbers equals the product of their HCF and LCM:

\[ \begin{aligned} 275 \times n &= 11 \times 7700 \\ n &= \frac{84700}{275} = 308 \end{aligned} \]

Check: \(275 = 11 \times 25\) and \(308 = 11 \times 28\). Since 25 and 28 share no factor, the HCF is 11 and the LCM is \(11 \times 25 \times 28 = 7700\). 28 is the cofactor of 308 after removing the HCF, and 700 and 37 are not multiples of 11, so they cannot have an HCF of 11 with 275.

Answer: C

The H.C.F. is \(\frac{2}{63}\).

The H.C.F. of fractions is the H.C.F. of the numerators divided by the L.C.M. of the denominators:

\[ \begin{aligned} \text{HCF}(4, 10, 20) &= 2 \\ \text{LCM}(9, 21, 63) &= 63 \end{aligned} \]

So the H.C.F. is \(\frac{2}{63}\). Each fraction is a whole-number multiple of it: \(\frac{4}{9} = 14 \times \frac{2}{63}\), \(\frac{10}{21} = 15 \times \frac{2}{63}\) and \(\frac{20}{63} = 10 \times \frac{2}{63}\). \(\frac{1}{21}\) and \(\frac{5}{63}\) do not divide all three exactly.

Answer: A

Required number = H.C.F. of \((91 - 43)\), \((183 - 91)\) and \((183 - 43)\)
= H.C.F. of \(48, 92\) and \(140 = 4\).