Class 8: Factors and Multiples – Exercise 4C – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1. Find the L.C.M. of the following numbers using common division method:

i) $24, 36, 40$ ii) $15, 24, 30, 40$ iii) $9, 12, 15, 18, 24, 56$

iv) $22,54,108,135,198$ v) $576, 672, 720$ vi) $1620, 1728, 1890$

Answer:

i) $24, 36, 40$

$\begin{array}{r | r | r | r } 2 & 24 & 36 & 40 \\ \hline 2 & 12 & 18 & 20 \\ \hline 3 & 6 & 9 & 10 \\ \hline 2 & 2 & 3 & 10 \\ \hline 3 & 1 & 3 & 5 \\ \hline 5 & 1 & 1 & 5 \\ \hline {\ } & 1 & 1 & 1 \end{array}$

LCM $= 2 \times 2 \times 3 \times 2 \times 3 \times 5 = 360$

ii) $15, 24, 30, 40$

$\begin{array}{r | r | r | r | r } 3 & 15 & 24 & 30 & 40 \\ \hline 5 & 5 & 8 & 10 & 40 \\ \hline 2 & 1 & 8 & 2 & 8 \\ \hline 2 & 1 & 4 & 1 & 4 \\ \hline 2 & 1 & 2 & 1 & 2 \\ \hline {\ } & 1 & 1 & 1 & 1 \end{array}$

LCM $= 3 \times 5 \times 2 \times 2 \times 2 = 120$

iii) $9, 12, 15, 18, 24, 56$

$\begin{array}{r | r | r | r | r | r | r } 3 & 9 & 12 & 15 & 18 & 24 & 56 \\ \hline 3 & 3 & 4 & 5 & 6 & 8 & 56 \\ \hline 2 & 1 & 4 & 5 & 2 & 8 & 56 \\ \hline 2 & 1 & 2 & 5 & 1 & 4 & 28 \\ \hline 2 & 1 & 1 & 5 & 1 & 2 & 14 \\ \hline 7 & 1 & 1 & 5 & 1 & 1 & 7 \\ \hline 5 & 1 & 1 & 5 & 1 & 1 & 1 \\ \hline {\ } & 1 & 1 & 1 & 1 & 1 & 1 \end{array}$

LMC $= 3 \times 3 \times 2 \times 2 \times 7 \times 5 = 2520$

iv) $22,54,108,135,198$

$\begin{array}{ r | r | r | r | r | r } 2 & 22 & 54 & 108 & 135 & 198 \\ \hline 3 & 11 & 27 & 54 & 135 & 99 \\ \hline 3 & 11 & 9 & 18 & 45 & 33 \\ \hline 11 & 11 & 3 & 6 & 15 & 11 \\ \hline 3 & 1 & 3 & 6 & 15 & 1 \\ \hline 2 & 1 & 1 & 2 & 5 & 1 \\ \hline 5 & 1 & 1 & 1 & 5 & 1 \\ \hline {\ } & 1 & 1 & 1 & 1 & 1 \end{array}$

LCM $= 2 \times 3 \times 3 \times 11 \times 3 \times 2 \times 5 = 5940$

v) $576, 672, 720$

$\begin{array}{r | r | r | r } 2 & 576 & 672 & 570 \\ \hline 2 & 288 & 336 & 360 \\ \hline 2 & 144 & 168 & 180 \\ \hline 2 & 72 & 84 & 90 \\ \hline 3 & 36 & 42 & 45 \\ \hline 3 & 12 & 14 & 15 \\ \hline 2 & 4 & 14 & 5 \\ \hline 2 & 2 & 7 & 5 \\ \hline 5 & 1 & 7 & 5 \\ \hline 7 & 1 & 7 & 1 \\ \hline {\ } & 1 & 1 & 1 \end{array}$

LCM $= 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 2 \times 2 \times 5 \times 7 = 20160$

vi) $1620, 1728, 1890$

$\begin{array}{r | r | r | r } 2 & 1620 & 1728 & 1890 \\ \hline 5 & 810 & 864 & 945 \\ \hline 2 & 162 & 864 & 189 \\ \hline 3 & 81 & 432 & 189 \\ \hline 3 & 27 & 144 & 63 \\ \hline 3 & 9 & 48 & 21 \\ \hline 3 & 3 & 16 & 7 \\ \hline 7 & 1 & 16 & 7 \\ \hline 2 & 1 & 16 & 1 \\ \hline 2 & 1 & 8 & 1 \\ \hline 2 & 1 & 4 & 1 \\ \hline 2 & 1 & 2 & 1 \\ \hline {\ } & 1 & 1 & 1 \end{array}$

LCM $= 2 \times 5 \times 2 \times 3 \times 3 \times 3 \times 3 \times 7 \times 2 \times 2 \times 2 \times 2 = 181440$

$\\$

Question 2: Use the prime factorization method to find the L.C.M. of the following numbers:

i) $72, 192, 240$ ii) $56, 72, 84$ iii) $48, 72, 108, 144$

iv) $168, 180, 330$ v) $180, 384, 432$ vi) $288, 432, 486$

Answer:

i) $72, 192, 240$

$\begin{array}{r | r } 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r } 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r } 2 & 240 \\ \hline 2 & 120 \\ \hline2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

$72 = 2^3 \times 3^2$

$192 = 2^6 \times 3$

$240 = 2^4 \times 3 \times 5$

Hence LCM $= 2^6 \times 3^2 \times 5 = 2880$

ii) $56, 72, 84$

$\begin{array}{r | r } 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r } 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r } 2 & 84 \\ \hline 2 & 42 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$

$56 = 2^3 \times 7$

$72 = 2^3 \times 3^2$

$84 = 2^2 \times 3 \times 7$

Hence LCM $= 2^3 \times 3^2 \times 7 = 504$

iii) $48, 72, 108, 144$

$\begin{array}{r | r } 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r } 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r } 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r } 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$48 = 2^4 \times 3$

$72 = 2^3 \times 3^2$

$108 = 2^2 \times 3^3$

$144 = 2^4 \times 3^2$

Hence LCM $= 2^4 \times 3^3 = 432$

iv) $168, 180, 330$

$\begin{array}{r | r } 2 & 168 \\ \hline 2 & 84 \\ \hline 2 & 42 \\ \hline 3 & 21 \\ \hline 7 & 7 \\ \hline & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r } 2 & 180 \\ \hline 2 & 90 \\ \hline 5 & 45 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r } 2 & 330 \\ \hline 5 & 165 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

$168 = 2^3 \times 3 \times 7$

$180 = 2^2 \times 5 \times 3^2$

$330 = 2 \times 3 \times 5 \times 11$

Hence LCM $= 2^3 \times 3^2 \times 5 \times 7 \times 11 = 27720$

v) $180, 384, 432$

$\begin{array}{r | r } 2 & 180 \\ \hline 2 & 90 \\ \hline 5 & 45 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r } 2 & 384 \\ \hline 2 & 192 \\ \hline 2 & 96 \\ \hline 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \hspace{1.5cm} \begin{array}{r | r } 2 & 432 \\ \hline 2 & 216 \\ \hline 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

$180 = 2^2 \times 5 \times 3^2$

$384 = 2^7 \times 3$

$432 = 2^4 \times 3^3$

Hence LCM $= 2^7 \times 3^3 \times 5 = 17280$

vi) $288, 432, 486$

$180 = 2^2 \times 5 \times 3^2$

$384 = 2^7 \times 3$

$432 = 2^4 \times 3^3$

Hence LCM $= 2^7 \times 5 \times 3^3 =17280$

$\\$

Question 3. Find the H.C.F and L.C.M of the following:

i) $23 \times 32 \times 5, 22 \times 33 \times 5$ and $24 \times 3 \times 53 \times 7$

ii) $22 \times 3 \times 52, 23 \times 32 \times 7$ and $2 \times 33 \times 5 \times 7$

iii) $52 \times 7 \times 2, 5 \times 7 \times 32$ and $22 \times 3 \times 11$

iv) $22 \times 33 \times 54, 23 \times 32 \times 5$ and $2 \times 3 \times 7 \times 52$

Answer:

i) $23 \times 32 \times 5, 22 \times 33 \times 5$ and $24 \times 3 \times 53 \times 7$

HCF $= 2^2 \times 3 \time 5 = 60$

LCM $= 2^4 \times 3^2 \times 5^3 \times 7 = 126000$

ii) $22 \times 3 \times 52, 23 \times 32 \times 7$ and $2 \times 33 \times 5 \times 7$

HCF $= 2 \times 3 = 6$

LCM $= 2^3 \times 3^3 \times 5^2 \times 7 = 37800$

iii) $52 \times 7 \times 2, 5 \times 7 \times 32$ and $22 \times 3 \times 11$

HCF $= 1$

LCM $= 2^2 \times 3^2 \times 5^2 \times 7 \times 11 = 69300$

iv) $22 \times 33 \times 54, 23 \times 32 \times 5$ and $2 \times 3 \times 7 \times 52$

HCF $= 2 \times 3 \times 5 = 30$

LCM $= 2^2 \times 3^3 \times 5^4 \times 7 = 945000$

$\\$

Question 4. Find the H.C.F. and L.C.M of the following fractions:

i) $\frac{5}{6}$ $,$ $\frac{5}{9}$ $,$ $\frac{10}{27}$ ii) $\frac{2}{3}$ $,$ $\frac{8}{15}$ $,$ $\frac{20}{27}$ $,$ $\frac{40}{81}$ iii) $\frac{3}{7}$ $,$ $\frac{9}{14}$ $,$ $\frac{18}{35}$ $,$ $\frac{15}{28}$

Note: HCF of a given fraction $=$ $\frac{HCF of Numerators}{LCM of Denominators}$

LCM of a given fraction $=$ $\frac{LCM of Numerators}{HCF of Denominators}$

Answer:

i) $\frac{5}{6}$ $,$ $\frac{5}{9}$ $,$ $\frac{10}{27}$

HCF of numerators $=$ HCF of $5, 5, 10 = 5$

HCF of denominators $=$ HCF of $6, 9, 27 = 3$

LCM of numerators $=$ LCM of $5, 5, 10 = 10$

LCM of denominators $=$ LCM of $6, 9, 27 = 54$

HCF of a given fraction $=$ $\frac{5}{54}$

LCM of a given fraction $=$ $\frac{10}{3}$

ii) $\frac{2}{3}$ $,$ $\frac{8}{15}$ $,$ $\frac{20}{27}$ $,$ $\frac{40}{81}$

HCF of numerators $=$ HCF of $2, 8, 20, 40 = 10$

HCF of denominators $=$ HCF of $3, 15, 27, 81 = 3$

LCM of numerators $=$ LCM of $2, 8, 20, 40 = 40$

LCM of denominators $=$ LCM of $3, 15, 27, 81 = 405$

HCF of a given fraction $=$ $\frac{2}{405}$

LCM of a given fraction $=$ $\frac{40}{3}$

iii) $\frac{3}{7}$ $,$ $\frac{9}{14}$ $,$ $\frac{18}{35}$ $,$ $\frac{15}{28}$

HCF of numerators $=$ HCF of $3, 9, 18, 15 = 3$

HCF of denominators $=$ HCF of $7, 14, 35, 28 = 7$

LCM of numerators $=$ LCM of $3, 9, 18, 15 = 90$

LCM of denominators $=$ LCM of $7, 14, 35, 28 = 140$

HCF of a given fraction $=$ $\frac{3}{140}$

LCM of a given fraction $=$ $\frac{90}{7}$

$\\$

Question 5. Find the smallest number exactly divisible by $28, 36, 42$ and $54$

Answer:

The required number is the LCM of $28, 36, 42$ and $54$

$\begin{array}{r | r | r | r | r } 2 & 28 & 36 & 42 & 54 \\ \hline 2 & 14 & 18 & 21 & 27 \\ \hline 7 & 7 & 9 & 21 & 27 \\ \hline 3 & 1 & 9 & 3 & 27 \\ \hline 3 & 1 & 3 & 1 & 9 \\ \hline 3 & 1 & 1 & 1 & 3 \\ \hline {\ } & 1 & 1 & 1 & 1 \end{array}$

LCM $= 2 \times 2 \times 7 \times 3 \times 3 \times 3 = 756$

$\\$

Question 6. Find the least number which when divided by $12, 16, 18, 21$ and $28$ leaves the same remainder $7$ in each case.

Answer:

Required number is the LCM of $12, 16, 18, 21$ and $28$ plus $7$ added to it.

$\begin{array}{r | r | r | r | r } 2 & 12 & 16 & 18 & 21 \\ \hline 3 & 6 & 8 & 9 & 21 \\ \hline 2 & 2 & 8 & 3 & 7 \\ \hline 2 & 1 & 4 & 3 & 7 \\ \hline 2 & 1 & 2 & 3 & 7 \\ \hline 3 & 1 & 1 & 1 & 7 \\ \hline {7} & 1 & 1 & 1 & 1 \end{array}$

LCM $= 2 \times 3 \times 2 \times 2 \times 2 \times 3 \times 7 = 1008$

Hence the number is $1008 + 7 = 1015$

$\\$

Question 7. Find the least number which when diminished by $8$ , is divisible by each of the numbers $12, 15, 20$ and $54$ .

Answer:

Required number is the LCM of $12, 15, 20$ and $54$ plus $8$ added to it.

$\begin{array}{r | r | r | r | r } 3 & 12 & 15 & 20 & 54 \\ \hline 5 & 4 & 5 & 20 & 18 \\ \hline 2 & 4 & 1 & 4 & 18 \\ \hline 2 & 2 & 1 & 2 & 9 \\ \hline 3 & 1 & 1 & 1 & 9 \\ \hline 3 & 1 & 1 & 1 & 3 \\ \hline {\ } & 1 & 1 & 1 & 1 \end{array}$

LCM $= 3 \times 5 \times 2 \times 2 \times 2 \times 3 = 540$

Hence the number is $540 + 8 = 548$

$\\$

Question 8. Find the smallest number which when increased by $5$ is divisible by each of the numbers $48, 60, 72$ and $108$ .

Answer:

Required number is the LCM of $48, 60, 72$ and $108$ minus $5$ added to it.

$\begin{array}{r | r | r | r | r } 3 & 48 & 60 & 72 & 108 \\ \hline 2 & 16 & 20 & 24 & 36 \\ \hline 2 & 8 & 10 & 12 & 18 \\ \hline 3 & 4 & 5 & 6 & 9 \\ \hline 2 & 4 & 5 & 2 & 3 \\ \hline 5 & 1 & 5 & 1 & 3 \\ \hline 3 & 1 & 1 & 1 & 3 \\ \hline {\ } & 1 & 1 & 1 & 1 \end{array}$

LCM $= 3 \times 2 \times 2 \times 3 \times 2 \times 2 \times 5 \times 3 = 2160$

Hence the number is $2160 - 5 = 2155$

$\\$

Question 9. Find the greatest number of four digits which is exactly divisible by each one of the numbers $12, 18, 21$ and $28$ .

Answer:

The required number must be divisible by LCM of $12, 18, 21$ and $28$

The LCM $= 252$

Now the largest four digit number is $9999$

On dividing the $9999$ by $252$ , we get $9999 \div 252 = 39$ and remainder of $71$

Therefore the number is $( 9999 - 71) = 9828$

$\\$

Question 10. Five bells begin to toll together and toll respectively at intervals of $6,7,8,9$ and $12$ seconds. After how much time would they toll together again?

Answer:

Required number is the LCM of $6,7,8,9$ and $12$

$\begin{array}{r | r | r | r | r | r} 2 & 6 & 7 & 8 & 9 & 12 \\ \hline 3 & 3 & 7 & 4 & 9 & 6 \\ \hline 2 & 1 & 7 & 4 & 3 & 2 \\ \hline 2 & 1 & 7 & 2 & 3 & 1 \\ \hline 3 & 1 & 7 & 1 & 3 & 1 \\ \hline 7 & 1 & 7 & 1 & 1 & 1 \\ \hline {\ } & 1 & 1 & 1 & 1 & 1 \end{array}$

LCM $= 2 \times 3 \times 2 \times 2 \times 3 \times 7 = 504$

Hence all the bells will toll together again in $504$ seconds or $8$ minutes and $24$ seconds.

$\\$

Question 11. Find the least perfect square number which is divisible by $3, 4, 5, 6$ and $8$ .

Answer:

Required number is the square of LCM of $3,4,5,6$ and $8$

$\begin{array}{r | r | r | r | r | r} 3 & 3 & 4 & 5 & 6 & 8 \\ \hline 2 & 1 & 4 & 5 & 2 & 8 \\ \hline 2 & 1 & 2 & 5 & 1 & 4 \\ \hline 2 & 1 & 1 & 5 & 1 & 2 \\ \hline 5 & 1 & 1 & 5 & 1 & 1 \\ \hline {\ } & 1 & 1 & 1 & 1 & 1 \end{array}$

LCM $= 3 \times 2 \times 2 \times 2 \times 5 = 120$

The required perfect square is $120 \times 3 \times 2 \times 5 = 3600$

$\\$

Question 12. Find the least number that should be added to $2500$ so that the sum is divisible by each one of the numbers $3, 4, 5$ and $6$ .

Answer:

LCM of $3, 4, 5$ , and $6 = 60$

$2500 \div 60 = 41$ and remainder of $40$ .

Required number $= 60-40 = 20$ so that the sum is divisible by each of the number $3, 4, 5$ , and $6$ .

The number is $2500 + 20 = 2520$

$\\$

Question 13. Find the least number of five digits which is exactly divisible by each one of the numbers $12, 15$ and $20$ .

Answer:

LCM of $12, 15$ , and $20 = 60$

The smallest $5$ digit number is $10000$

$10000 \div 60 = 166$ and remainder of $40$ .

Required number $= 10000 + ( 60-40) = 10020$ so that the sum is divisible by each of the number $3, 4, 5$ , and $6$ .

$\\$

Question 14. Three boys start cycling around a circular park from the same point at the same time and in the same direction. If these boys, each cycling at a constant speed, complete a round in $24$ min, $36$ min and $54$ min respectively, then after what time would they meet again.

Answer:

The boys complete the circle in $24, 36$ and $54$ minutes respectively.

LCM of $24, 36$ and $54 = 216$ minutes or $3$ hours $36$ minutes.

$\\$

Question 15. The product of two numbers is $16184$ and their L.C.M. is $952$ . Find their H.C.F.

Answer:

Note: The product of two given numbers $=$ Product of HCF and LCM

Therefore $16184 =$ HCF $\times 952 \Rightarrow$ HCF $= 17$

$\\$

Question 16. The product of two numbers is $591223$ and their H.C.F. is $952$ . Find their L.C.M.

Answer:

Note: The product of two given numbers $=$ Product of HCF and LCM

Therefore $591223 = 952 \times$ LCM $\Rightarrow$ LCM $= 20387$

$\\$

Question 17. The H.C.F of two numbers is $14$ and their L.C.M is $11592$ . If one number is $504$ , find the other.

Answer:

Note: The product of two given numbers $=$ Product of HCF and LCM

Therefore Number $\times 504 = 14 \times 11592 \Rightarrow$ Number $= 322$

$\\$

Question 18. The sum of the H.C.F and L.C.M of two numbers is $1150$ . If the L.C.M is $45$ times the H.C.F. and one of the numbers is $125$ , then find the other number.

Answer:

Note: The product of two given numbers $=$ Product of HCF and LCM

$HCF + LCM = 1150$

$LCM = 45 \times HCF$

$\therefore 46 HCF = 1150$

$\Rightarrow HCF = 25$ and $LCM = 45 \times 125 = 1125$

Now $Number \times 125 = 25 \times 1125 \Rightarrow Number = 225$

$\\$

Question 19. The difference between the L.C.M. and H.C.F of two numbers is $294$ . If the L.C.M is $15$ times the H.C.F and one of the numbers is $105$ , find the other.

Answer:

Note: The product of two given numbers $=$ Product of HCF and LCM

$LCM - HCF = 294$

$LCM = 15 \times HCF$

$\therefore 14 HCF = 294$

$\Rightarrow HCF = 21$ and $LCM = 15 \times 21 = 315$

Now $Number \times 105 = 21 \times 315 \Rightarrow Number = 63$

$\\$

Question 20. The product of two co-prime numbers is $609$ . What would be the L.C.M of these numbers?

Answer:

Co-prime are those numbers where HCF is $1$ . So if the product of two co-primes is $609$ then the numbers are $1$ and $609$ . Therefore LCM is $609$ .

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 8: Factors and Multiples – Exercise 4C

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