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

Answer: A

The bells next toll together after 12 minutes.

They ring together again after the least common multiple of their intervals:

\[ \begin{aligned} 36 &= 2^{2} \times 3^{2} \\ 40 &= 2^{3} \times 5 \\ 48 &= 2^{4} \times 3 \\ \text{LCM} &= 2^{4} \times 3^{2} \times 5 = 720 \end{aligned} \]

720 seconds is 12 minutes.

6 minutes (360 s) is a multiple of 36 and 40 but not of 48. 24 minutes is the second time they toll together, not the next, and 18 minutes (1080 s) is not a multiple of 48.

Answer: A

The number is 147.

A number that leaves remainder 3 on division by each of 12, 16, 24 and 36 is 3 more than a common multiple of them. The least common multiple is

\[ \begin{aligned} 12 &= 2^{2}\cdot 3,\ 16 = 2^{4},\ 24 = 2^{3}\cdot 3,\ 36 = 2^{2}\cdot 3^{2} \\ \text{LCM} &= 2^{4}\cdot 3^{2} = 144 \end{aligned} \]

so the least such number is 144 + 3 = 147. 144 itself leaves remainder 0, 141 is 3 less than the LCM, and 145 leaves remainder 1.

Answer: B

The numbers are 21 and 35.

Let the numbers be 3k and 5k. Since 3 and 5 have no common factor, their LCM is 15k:

\[ \begin{aligned} 15k &= 105 \\ k &= 7 \end{aligned} \]

So the numbers are 21 and 35. Check: \(21 = 3 \times 7\) and \(35 = 5 \times 7\), so their LCM is \(3 \times 5 \times 7 = 105\), and their HCF is the common factor 7. The pairs 15, 25 (LCM 75), 27, 45 (LCM 135) and 9, 15 (LCM 45) are also in the ratio 3 : 5 but have the wrong LCM.

Answer: C

The second number is (2079 × 27) ÷ 189.

For two numbers, the product equals HCF × LCM:

\[ \begin{aligned} 189 \times b &= 27 \times 2079 \\ b &= \frac{2079 \times 27}{189} = 297 \end{aligned} \]

Check: 189 = 27 × 7 and 297 = 27 × 11, so the HCF is 27 and the LCM is 27 × 7 × 11 = 2079. The other expressions misplace the factors.

Answer: C

The least capacity is 150 ml.

A quantity that each jar can measure exactly, in whole fillings, must be a multiple of 15, 50 and 30, so the least such quantity is their least common multiple:

\[ \begin{aligned} 15 &= 3 \times 5, \quad 30 = 2 \times 3 \times 5, \quad 50 = 2 \times 5^{2} \\ \text{LCM} &= 2 \times 3 \times 5^{2} = 150 \end{aligned} \]

150 ml is 10 fillings of the 15 ml jar, 5 of the 30 ml and 3 of the 50 ml. 100 is not a multiple of 15 or 30, and 75 is not a multiple of 50 or 30. 450 is a common multiple but not the least.

Answer: B

The greatest common divisor is 20.

Split 200 in the ratio 7 : 3. The ratio has 10 parts, so each part is 20:

\[ \begin{aligned} 7 \times 20 &= 140 \\ 3 \times 20 &= 60 \end{aligned} \]

Since 7 and 3 have no common factor, the greatest common divisor of 140 and 60 is the size of one part, 20. Check: 140 = 20 × 7 and 60 = 20 × 3, and 7 and 3 share no further factor.

10 divides both numbers but is not the greatest common divisor. 30 and 40 do not divide both: 30 does not divide 140, and 40 does not divide 60.

Answer: C

The clocks beep together soonest when T is 28 minutes.

They beep together again after the least common multiple of the three intervals. For 20 and 21 alone that is 420 minutes. A value of T that already divides 420 adds nothing, so the clocks meet again at 420 minutes, 7 P.M.

\[ \begin{aligned} 28 = 4 \times 7, \ 420 &= 28 \times 15 \\ \text{LCM}(20, 21, 28) &= 420 \end{aligned} \]

The other choices push the meeting later: T = 8 gives 840 minutes, T = 18 gives 1,260 and T = 38 gives 7,980.

Answer: A

The difference is 1416.

For two numbers, product = LCM \(\times\) GCD. With LCM = 60 \(\times\) GCD = 60\(g\):

\[ \begin{aligned} 60g \times g &= 34560 \\ g^{2} &= 576 \Rightarrow g = 24 \\ \text{LCM} &= 60 \times 24 = 1440 \end{aligned} \]

So LCM - GCD = 1440 - 24 = 1416. 1464 is the sum 1440 + 24, a common slip, and 1424 and 1460 do not come from these values.

Answer: D

The HCF of numbers always divides their LCM, so any HCF of three numbers whose LCM is 120 must be a factor of 120. 8, 12 and 24 are factors of 120, but 35 is not (120 / 35 is not a whole number).

So 35 cannot be their HCF, option D.

The same test applies to LCM and HCF whatever the numbers: the HCF is a divisor of every number and the LCM is a multiple of every number, so the HCF must divide the LCM. The divisors of 120 include 8, 12 and 24, but 35 contains the factor 7, which 120 does not have.

Answer: A

With a = 2^2 x 3^3, b = 3^2 x 5^3 and c = 2^3 x 5^2: a^2 = 2^4 x 3^6 and b^3 = 3^6 x 5^9, so their HCF is 3^6. Next b^2 = 3^4 x 5^6 and c^3 = 2^9 x 5^6, so the HCF is 5^6. Last c^2 = 2^6 x 5^4 and a^3 = 2^6 x 3^9, so the HCF is 2^6.

The sum is 3^6 + 2^6 + 5^6, option A.

The HCF of two numbers takes the smallest power of each prime factor. In the first pair the only common prime is 3, with the smaller power 6, in the second pair only 5 with power 6, and in the third pair only 2 with power 6. The sum of these three common factors is the expression of option A.