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

Answer: A

They next ring together at 5:00 PM.

The bells coincide after a whole number of each period, that is after the least common multiple of 20, 30 and 50 minutes:

\[ \begin{aligned} 20 &= 2^{2} \times 5,\ 30 = 2 \times 3 \times 5,\ 50 = 2 \times 5^{2} \\ \text{LCM} &= 2^{2} \times 3 \times 5^{2} = 300\ \text{min} \end{aligned} \]

300 minutes is 5 hours, so from 12:00 PM the next joint ring is at 5:00 PM. 5:30, 6:00 and 6:30 PM are 330, 360 and 390 minutes after noon; 330 and 390 are not multiples of 20 or 50, and 360 is not a multiple of 50.

Answer: B

The other number is 99.

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

\[ \begin{aligned} 77 \times n &= 11 \times 693 \\ n &= \frac{7623}{77} = 99 \end{aligned} \]

Check: \(77 = 7 \times 11\) and \(99 = 9 \times 11\), so their HCF is 11 and their LCM is \(7 \times 9 \times 11 = 693\). 33 also shares the factor 11 with 77, but its LCM with 77 is 231, not 693. 96 and 89 are not multiples of 11, so they cannot have 11 as a common factor with 77.

Answer: B

The numbers are 6 and 4.

The product of the LCM and HCF of two numbers equals the product of the numbers themselves, so the two numbers multiply to 24 and differ by 2.

\[ \begin{aligned} n(n + 2) &= 24 \\ n^2 + 2n - 24 &= 0 \\ (n + 6)(n - 4) &= 0 \end{aligned} \]

So n = 4 and the numbers are 4 and 6. Check: their HCF is 2 and their LCM is 12, and \(2 \times 12 = 24\).

The other pairs also differ by 2 but have the wrong product: 2 and 4 give 8, 8 and 6 give 48, and 8 and 10 give 80.

Answer: B

The least square number divisible by 10, 12, 15 and 18 is 900.

Any number divisible by all four is a multiple of their LCM. A square must have every prime factor to an even power.

\[ \begin{aligned} \text{LCM} &= 2^2 \times 3^2 \times 5 = 180 \\ \text{square} &= 180 \times 5 = 900 = 30^2 \end{aligned} \]

Only the 5 has an odd power in 180, so one more 5 makes the smallest square multiple.

400 and 1600 are squares but are not divisible by 3, so they fail 12, 15 and 18. 729 is \(27^2\) but it is odd, so it is not divisible by 10 or 12.

Answer: B

The minimum length is 0.04 feet.

With two sticks you can measure any length of the form \(3.44a + 2.20b\) for whole numbers \(a, b\) (laying sticks end to end, or one against the other). The smallest positive length obtainable that way is their highest common factor.

\[ \begin{aligned} \text{HCF}(344, 220) &= \text{HCF}(220, 124) \\ &= \text{HCF}(124, 96) = \text{HCF}(96, 28) \\ &= \text{HCF}(28, 12) = 4 \end{aligned} \]

So in feet the answer is \(4/100 = 0.04\). The value 0.4 is the same digits out by a factor of ten, and 0.2 or 0.1 cannot be built from 3.44 and 2.20 at all, since neither divides both lengths.

Answer: D

They will next meet on a Thursday.

Reading the visits as every 13, 15, 17 and 18 days, all four are together again after the LCM of those numbers.

\[ \begin{aligned} \text{LCM}(13, 15, 17, 18) &= 13 \times 17 \times 90 \\ &= 19890 \text{ days} \\ 19890 &= 7 \times 2841 + 3 \end{aligned} \]

19890 days is 2841 complete weeks plus 3 days, so the meeting falls 3 days after a Monday, on Thursday. Note that 15 and 18 share a factor 3, so the LCM uses 90, not \(15 \times 18 = 270\). Multiplying all four numbers gives 59670 days, which lands on a Wednesday, not one of the options but a sign of the slip.

Answer: A

Only one pair is correctly matched.

Pair 1 is right. The LCM of numbers is the product of the highest power of each prime that appears in any of them; for 12 and 18, \(2^2 \times 3^2 = 36\).

Pair 2 is wrong. For two numbers, HCF × LCM equals their product a × b, not their sum; for 12 and 18, \(6 \times 36 = 216 = 12 \times 18\).

Pair 3 is wrong. A number with exactly two factors is a prime; a composite number has more than two factors.

Answer: C

The minimum number of rows is 43.

Every row must hold the same number of trees, and each count must divide exactly, so the row size must be a common factor; for the fewest rows it must be the highest common factor:

\[ \begin{aligned} 121 &= 11^{2},\ 154 = 2 \cdot 7 \cdot 11 \\ 198 &= 2 \cdot 3^{2} \cdot 11 \Rightarrow \text{HCF} = 11 \end{aligned} \]

\[ \begin{aligned} &\tfrac{121}{11} + \tfrac{154}{11} + \tfrac{198}{11} \\ &= 11 + 14 + 18 = 43 \end{aligned} \]

11 is the number of trees per row, not the number of rows. 41 and 42 do not match any count.

Answer: D

The pair is 200 and 250.

For any two numbers, HCF \(\times\) LCM = product of the numbers, so the product must be \(50 \times 1000 = 50000\). Of the options, \(200 \times 250 = 50000\), and indeed HCF(200, 250) = 50 and LCM = 1000.

200 and 500 have HCF 100 and LCM 1000; 150 and 250 have HCF 50 but LCM 750; 50 and 100 have HCF 50 but LCM 100. Each fails one of the two conditions.

Answer: B

The sum is 444.

If the H.C.F. is 37, the numbers are \(37a\) and \(37b\), where a and b have no common factor:

\[ \begin{aligned} 37a \times 37b &= 47915 \\ ab &= \frac{47915}{1369} = 35 \end{aligned} \]

Co-prime pairs with product 35 are (1, 35) and (5, 7). The pair (5, 7) gives the numbers 185 and 259:

\[ 185 + 259 = 444 \]

The pair (1, 35) gives 37 and 1295, whose sum, 1332, is not offered. As a check, \(185 \times 259 = 47915\), and both are multiples of 37 with no larger common factor.