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

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

Answer: B

We want 4 digit number, so option A is not the answer

Now we want greatest numnber, So out of remaining options, 9600 is greatest

9600 is divisible by 25, 75, 40, and 15

So answer is 9600

Answer: A

Using factorization method,
\(18 = 2 \times 3^2\)
\(288 = 2^5 \times 3^2\)
\(360 = 2^3 \times 3^2 \times 5\)
So H.C.F. will be minimum term present in all three,
i.e. \(2 \times 3^2 = 18\)