Class 8: Ratio and Proportion

$\displaystyle \textbf{Question 1: }\text{Express the following ratios in simplest form:}$
$\displaystyle \text{(i) }18:30\qquad\text{(ii) }7.5:9\qquad \text{(iii) }6\frac{2}{3}:7\frac{1}{2}$
$\displaystyle \text{(iv) }\frac{1}{6}:\frac{1}{9}:\frac{1}{12}\qquad \text{(v) }7:5:\frac{9}{2}$
$\displaystyle \text{(vi) }3\frac{1}{5}:5\frac{1}{3}:6\frac{2}{3}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }18:30=\frac{18}{30}=\frac{3}{5}=3:5.$

$\displaystyle \text{(ii) }7.5:9=\frac{7.5}{9}=\frac{75}{90}=\frac{5}{6}=5:6.$

$\displaystyle \text{(iii) }6\frac{2}{3}:7\frac{1}{2} =\frac{20}{3}:\frac{15}{2}$
$\displaystyle =\frac{20}{3}\times\frac{2}{15} =\frac{8}{9}=8:9.$

$\displaystyle \text{(iv) }\frac{1}{6}:\frac{1}{9}:\frac{1}{12}$
$\displaystyle =\left(\frac{1}{6}\times36\right): \left(\frac{1}{9}\times36\right):\left(\frac{1}{12}\times36\right)$
$\displaystyle =6:4:3.$

$\displaystyle \text{(v) }7:5:\frac{9}{2} =(7\times2):(5\times2):\left(\frac{9}{2}\times2\right)$
$\displaystyle =14:10:9.$

$\displaystyle \text{(vi) }3\frac{1}{5}:5\frac{1}{3}:6\frac{2}{3} =\frac{16}{5}:\frac{16}{3}:\frac{20}{3}$
$\displaystyle =\left(\frac{16}{5}\times15\right): \left(\frac{16}{3}\times15\right):\left(\frac{20}{3}\times15\right)$
$\displaystyle =48:80:100=12:20:25.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Express the following ratios in simplest form:}$
$\displaystyle \text{(i) }75\text{ paisa}:4\text{ rupees}$
$\displaystyle \text{(ii) }1\text{ m }8\text{ cm}:72\text{ cm}$
$\displaystyle \text{(iii) }1\text{ hour }15\text{ minutes}:45\text{ minutes}$
$\displaystyle \text{(iv) }2\text{ kg }750\text{ g}:3\text{ kg}$
$\displaystyle \text{(v) }1\text{ year }9\text{ months}:2\text{ years }4\text{ months}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }75\text{ paisa}:4\text{ rupees} =75\text{ paisa}:400\text{ paisa}$
$\displaystyle =75:400=3:16.$

$\displaystyle \text{(ii) }1\text{ m }8\text{ cm}:72\text{ cm} =108\text{ cm}:72\text{ cm}$
$\displaystyle =108:72=3:2.$

$\displaystyle \text{(iii) }1\text{ hour }15\text{ minutes}:45\text{ minutes} =75\text{ minutes}:45\text{ minutes}$
$\displaystyle =75:45=5:3.$

$\displaystyle \text{(iv) }2\text{ kg }750\text{ g}:3\text{ kg} =2750\text{ g}:3000\text{ g}$
$\displaystyle =2750:3000=11:12.$

$\displaystyle \text{(v) }1\text{ year }9\text{ months}:2\text{ years }4\text{ months}$
$\displaystyle =21\text{ months}:28\text{ months}=21:28=3:4.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Which ratio is greater:}$
$\displaystyle \text{(i) }(4:9)\text{ or }(3:7)$
$\displaystyle \text{(ii) }\left(2\frac{1}{3}:3\frac{1}{3}\right) \text{ or }(3.6:4.8)$
$\displaystyle \text{(iii) }\left(\frac{1}{2}:\frac{1}{3}\right) \text{ or }\left(\frac{1}{6}:\frac{1}{4}\right)$
$\displaystyle \text{(iv) }\left(3\frac{1}{3}:4\frac{1}{6}\right) \text{ or }(0.9:1)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\frac{4}{9}\text{ or }\frac{3}{7} =\frac{28}{63}\text{ or }\frac{27}{63}.$
$\displaystyle \text{Since }\frac{28}{63}>\frac{27}{63}, \text{ therefore }(4:9)>(3:7).$

$\displaystyle \text{(ii) }2\frac{1}{3}:3\frac{1}{3} =\frac{7}{3}:\frac{10}{3} =\frac{7}{3}\times\frac{3}{10}=\frac{7}{10}.$
$\displaystyle 3.6:4.8=\frac{3.6}{4.8}=\frac{3}{4}.$
$\displaystyle \frac{7}{10}=\frac{14}{20}\quad\text{and}\quad \frac{3}{4}=\frac{15}{20}.$
$\displaystyle \text{Since }\frac{15}{20}>\frac{14}{20}, \text{ therefore }(3.6:4.8)>\left(2\frac{1}{3}:3\frac{1}{3}\right).$

$\displaystyle \text{(iii) }\frac{1}{2}:\frac{1}{3} =\frac{1}{2}\times3=\frac{3}{2}.$
$\displaystyle \frac{1}{6}:\frac{1}{4} =\frac{1}{6}\times4=\frac{2}{3}.$
$\displaystyle \text{Since }\frac{3}{2}>\frac{2}{3}, \text{ therefore }\left(\frac{1}{2}:\frac{1}{3}\right)> \left(\frac{1}{6}:\frac{1}{4}\right).$

$\displaystyle \text{(iv) }3\frac{1}{3}:4\frac{1}{6} =\frac{10}{3}:\frac{25}{6} =\frac{10}{3}\times\frac{6}{25}=\frac{4}{5}.$
$\displaystyle 0.9:1=\frac{0.9}{1}=\frac{9}{10}.$
$\displaystyle \frac{4}{5}=\frac{8}{10}<\frac{9}{10}.$
$\displaystyle \therefore (0.9:1)> \left(3\frac{1}{3}:4\frac{1}{6}\right).$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Arrange the following ratios in ascending order of}$
$\displaystyle \text{magnitude:}$
$\displaystyle \text{(i) }(2:3),\ (5:9),\ (11:15)$
$\displaystyle \text{(ii) }(5:7),\ (9:14),\ (20:21)\text{ and }(3:8)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(2:3),\ (5:9),\ (11:15)$
$\displaystyle \frac{2}{3}=\frac{30}{45},\qquad \frac{5}{9}=\frac{25}{45},\qquad\frac{11}{15}=\frac{33}{45}.$
$\displaystyle \text{Since }25<30<33,\text{ therefore}$
$\displaystyle \frac{5}{9}<\frac{2}{3}<\frac{11}{15}.$
$\displaystyle \therefore (5:9)<(2:3)<(11:15).$

$\displaystyle \text{(ii) }(5:7),\ (9:14),\ (20:21),\ (3:8)$
$\displaystyle \frac{5}{7}=\frac{120}{168},\qquad \frac{9}{14}=\frac{108}{168},$
$\displaystyle \frac{20}{21}=\frac{160}{168},\qquad \frac{3}{8}=\frac{63}{168}.$
$\displaystyle \text{Since }63<108<120<160,\text{ therefore}$
$\displaystyle \frac{3}{8}<\frac{9}{14}<\frac{5}{7}<\frac{20}{21}.$
$\displaystyle \therefore (3:8)<(9:14)<(5:7)<(20:21).$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Divide Rs. }142.20\text{ between }A\text{ and }B \text{ in the ratio }\frac{1}{4}:\frac{1}{5}.$
$\displaystyle \text{Answer:}$
$\displaystyle A:B=\frac{1}{4}:\frac{1}{5} =\left(\frac{1}{4}\times20\right):\left(\frac{1}{5}\times20\right)=5:4.$
$\displaystyle \text{Sum of the terms of the ratio}=5+4=9.$
$\displaystyle \text{A's share}=\text{Rs. }142.20\times\frac{5}{9} =\text{Rs. }79.$
$\displaystyle \text{B's share}=\text{Rs. }142.20\times\frac{4}{9} =\text{Rs. }63.20.$
$\displaystyle \therefore \text{A's share is Rs. }79\text{ and B's share is Rs. }63.20.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Divide Rs. }3726\text{ among }A,B\text{ and }C \text{ in the ratio }\frac{1}{3}:\frac{1}{4}:\frac{1}{6}.$
$\displaystyle \text{Answer:}$
$\displaystyle A:B:C=\frac{1}{3}:\frac{1}{4}:\frac{1}{6} =\left(\frac{1}{3}\times12\right):\left(\frac{1}{4}\times12\right): \left(\frac{1}{6}\times12\right)$
$\displaystyle =4:3:2.$
$\displaystyle \text{Sum of the terms of the ratio}=4+3+2=9.$
$\displaystyle \text{A's share}=\text{Rs. }3726\times\frac{4}{9} =\text{Rs. }1656.$
$\displaystyle \text{B's share}=\text{Rs. }3726\times\frac{3}{9} =\text{Rs. }1242.$
$\displaystyle \text{C's share}=\text{Rs. }3726\times\frac{2}{9} =\text{Rs. }828.$
$\displaystyle \therefore \text{A, B and C receive Rs. }1656,\text{ Rs. }1242 \text{ and Rs. }828\text{ respectively}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Divide Rs. }810\text{ among }A,B\text{ and }C \text{ in the ratio }\frac{1}{4}:\frac{2}{5}:1\frac{3}{8}.$
$\displaystyle \text{Answer:}$
$\displaystyle A:B:C=\frac{1}{4}:\frac{2}{5}:\frac{11}{8}.$
$\displaystyle =\left(\frac{1}{4}\times40\right): \left(\frac{2}{5}\times40\right):\left(\frac{11}{8}\times40\right)$
$\displaystyle =10:16:55.$
$\displaystyle \text{Sum of the terms of the ratio}=10+16+55=81.$
$\displaystyle \text{A's share}=\text{Rs. }810\times\frac{10}{81} =\text{Rs. }100.$
$\displaystyle \text{B's share}=\text{Rs. }810\times\frac{16}{81} =\text{Rs. }160.$
$\displaystyle \text{C's share}=\text{Rs. }810\times\frac{55}{81} =\text{Rs. }550.$
$\displaystyle \therefore \text{A, B and C receive Rs. }100,\text{ Rs. }160 \text{ and Rs. }550\text{ respectively}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Divide Rs. }1050\text{ between }A\text{ and }B \text{ in the ratio }2\frac{2}{3}:6\frac{2}{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle A:B=\frac{8}{3}:\frac{20}{3} =\left(\frac{8}{3}\times3\right):\left(\frac{20}{3}\times3\right)$
$\displaystyle =8:20=2:5.$
$\displaystyle \text{Sum of the terms of the ratio}=2+5=7.$
$\displaystyle \text{A's share}=\text{Rs. }1050\times\frac{2}{7} =\text{Rs. }300.$
$\displaystyle \text{B's share}=\text{Rs. }1050\times\frac{5}{7} =\text{Rs. }750.$
$\displaystyle \therefore \text{A's share is Rs. }300\text{ and B's share is Rs. }750.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Divide Rs. }747\text{ among }A,B\text{ and }C \text{ such that }4A=5B=7C.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }4A=5B=7C=k.$
$\displaystyle \therefore A=\frac{k}{4},\qquad B=\frac{k}{5},\qquad C=\frac{k}{7}.$
$\displaystyle \therefore A:B:C=\frac{1}{4}:\frac{1}{5}:\frac{1}{7} =35:28:20.$
$\displaystyle \text{Sum of the terms of the ratio}=35+28+20=83.$
$\displaystyle \text{A's share}=\text{Rs. }747\times\frac{35}{83} =\text{Rs. }315.$
$\displaystyle \text{B's share}=\text{Rs. }747\times\frac{28}{83} =\text{Rs. }252.$
$\displaystyle \text{C's share}=\text{Rs. }747\times\frac{20}{83} =\text{Rs. }180.$
$\displaystyle \therefore \text{A, B and C receive Rs. }315,\text{ Rs. }252 \text{ and Rs. }180\text{ respectively}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{A bag contains one-rupee, }50\text{-paisa and}$
$\displaystyle 25\text{-paisa coins in the ratio }5:6:8. \text{ Their total value is Rs. }210.$
$\displaystyle \text{Find the number of coins of each type.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numbers of one-rupee, }50\text{-paisa and }25\text{-paisa coins}$
$\displaystyle \text{be }5x,\ 6x\text{ and }8x\text{ respectively}.$
$\displaystyle \text{Their values in paisa are }500x,\ 300x\text{ and }200x \text{ respectively}.$
$\displaystyle \text{Since Rs. }210=21000\text{ paisa,}$
$\displaystyle 500x+300x+200x=21000.$
$\displaystyle 1000x=21000.$
$\displaystyle \therefore x=21.$
$\displaystyle \text{Number of one-rupee coins}=5x=5\times21=105.$
$\displaystyle \text{Number of }50\text{-paisa coins}=6x=6\times21=126.$
$\displaystyle \text{Number of }25\text{-paisa coins}=8x=8\times21=168.$
$\displaystyle \therefore \text{the numbers of coins are }105,\ 126\text{ and }168 \text{ respectively}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Find }A:B:C\text{ when:}$
$\displaystyle \text{(i) }A:B=2:5\text{ and }B:C=7:9$
$\displaystyle \text{(ii) }A:B=3:4\text{ and }B:C=6:11$
$\displaystyle \text{(iii) }A:B=\frac{1}{2}:\frac{1}{3} \text{ and }B:C=\frac{1}{4}:\frac{1}{5}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }A:B=2:5\text{ and }B:C=7:9.$
$\displaystyle \text{The L.C.M. of }5\text{ and }7\text{ is }35.$
$\displaystyle A:B=(2\times7):(5\times7)=14:35.$
$\displaystyle B:C=(7\times5):(9\times5)=35:45.$
$\displaystyle \therefore A:B:C=14:35:45.$

$\displaystyle \text{(ii) }A:B=3:4\text{ and }B:C=6:11.$
$\displaystyle \text{The L.C.M. of }4\text{ and }6\text{ is }12.$
$\displaystyle A:B=(3\times3):(4\times3)=9:12.$
$\displaystyle B:C=(6\times2):(11\times2)=12:22.$
$\displaystyle \therefore A:B:C=9:12:22.$

$\displaystyle \text{(iii) }A:B=\frac{1}{2}:\frac{1}{3} \text{ and }B:C=\frac{1}{4}:\frac{1}{5}.$
$\displaystyle A:B=3:2\qquad\text{and}\qquad B:C=5:4.$
$\displaystyle \text{The L.C.M. of }2\text{ and }5\text{ is }10.$
$\displaystyle A:B=(3\times5):(2\times5)=15:10.$
$\displaystyle B:C=(5\times2):(4\times2)=10:8.$
$\displaystyle \therefore A:B:C=15:10:8.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{If }A:B=4:9\text{ and }A:C=2:3, \text{ find }B:C\text{ and }A:B:C.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{A}{B}=\frac{4}{9}\qquad\text{and}\qquad \frac{A}{C}=\frac{2}{3}.$
$\displaystyle \frac{B}{C}=\frac{A}{C}\times\frac{B}{A} =\frac{2}{3}\times\frac{9}{4}=\frac{3}{2}.$
$\displaystyle \therefore B:C=3:2.$
$\displaystyle \text{Now, }A:B=4:9\text{ and }B:C=3:2.$
$\displaystyle \text{The L.C.M. of }9\text{ and }3\text{ is }9.$
$\displaystyle A:B=4:9.$
$\displaystyle B:C=(3\times3):(2\times3)=9:6.$
$\displaystyle \therefore A:B:C=4:9:6.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{If }A:C=5:8\text{ and }B:C=5:6, \text{ find }A:B\text{ and }A:B:C.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{A}{C}=\frac{5}{8}\qquad\text{and}\qquad \frac{B}{C}=\frac{5}{6}.$
$\displaystyle \frac{A}{B}=\frac{A}{C}\times\frac{C}{B} =\frac{5}{8}\times\frac{6}{5}=\frac{3}{4}.$
$\displaystyle \therefore A:B=3:4.$
$\displaystyle \text{Now, }A:C=5:8\text{ and }A:B=3:4.$
$\displaystyle \text{The L.C.M. of }5\text{ and }3\text{ is }15.$
$\displaystyle A:C=(5\times3):(8\times3)=15:24.$
$\displaystyle A:B=(3\times5):(4\times5)=15:20.$
$\displaystyle \therefore A:B:C=15:20:24.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Two numbers are in the ratio }6:11. \text{ On adding }2\text{ to the first}$
$\displaystyle \text{and }7\text{ to the second, their ratio becomes }8:15. \text{ Find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two numbers be }6x\text{ and }11x.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{6x+2}{11x+7}=\frac{8}{15}.$
$\displaystyle 15(6x+2)=8(11x+7).$
$\displaystyle 90x+30=88x+56.$
$\displaystyle 90x-88x=56-30.$
$\displaystyle 2x=26.$
$\displaystyle \therefore x=13.$
$\displaystyle \text{First number}=6x=6\times13=78.$
$\displaystyle \text{Second number}=11x=11\times13=143.$
$\displaystyle \therefore \text{the two numbers are }78\text{ and }143.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{A sum of money is divided between }A\text{ and }B \text{ in the ratio }5:7.$
$\displaystyle \text{If B's share is Rs. }665,\text{ find the total amount.}$
$\displaystyle \text{Answer:}$
$\displaystyle A:B=5:7.$
$\displaystyle \text{Sum of the terms of the ratio}=5+7=12.$
$\displaystyle \text{Therefore, B's share}=\frac{7}{12}\text{ of the total amount}.$
$\displaystyle \text{Let the total amount be Rs. }x.$
$\displaystyle \frac{7}{12}x=665.$
$\displaystyle x=665\times\frac{12}{7}=1140.$
$\displaystyle \therefore \text{the total amount is Rs. }1140.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{Two numbers are in the ratio }7:4. \text{ If their difference is }72,$
$\displaystyle \text{find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two numbers be }7x\text{ and }4x.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 7x-4x=72.$
$\displaystyle 3x=72.$
$\displaystyle \therefore x=24.$
$\displaystyle \text{First number}=7x=7\times24=168.$
$\displaystyle \text{Second number}=4x=4\times24=96.$
$\displaystyle \therefore \text{the two numbers are }168\text{ and }96.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{A certain sum of money is divided among }A,B \text{ and }C\text{ in the ratio }5:6:7.$
$\displaystyle \text{If A's share is Rs. }175,\text{ find the total amount and the shares of }B \text{ and }C.$
$\displaystyle \text{Answer:}$
$\displaystyle A:B:C=5:6:7.$
$\displaystyle \text{Sum of the terms of the ratio}=5+6+7=18.$
$\displaystyle \text{Let the total amount be Rs. }x.$
$\displaystyle \text{Since A's share is }\frac{5}{18}\text{ of the total amount,}$
$\displaystyle \frac{5}{18}x=175.$
$\displaystyle x=175\times\frac{18}{5}=630.$
$\displaystyle \therefore \text{the total amount is Rs. }630.$
$\displaystyle \text{B's share}=\text{Rs. }630\times\frac{6}{18} =\text{Rs. }210.$
$\displaystyle \text{C's share}=\text{Rs. }630\times\frac{7}{18} =\text{Rs. }245.$
$\displaystyle \therefore \text{B's share is Rs. }210\text{ and C's share is Rs. }245.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{The ratio of the number of boys to the number of girls in a school of }1440$
$\displaystyle \text{students is }7:5.\text{ If }40\text{ new boys are admitted, find how many new girls must be admitted}$
$\displaystyle \text{to make the ratio }4:3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Number of boys}=\frac{7}{12}\times1440=840.$
$\displaystyle \text{Number of girls}=\frac{5}{12}\times1440=600.$
$\displaystyle \text{After }40\text{ new boys are admitted, the number of boys}=840+40=880.$
$\displaystyle \text{Let the number of new girls admitted be }x.$
$\displaystyle \text{Then the number of girls}=600+x.$
$\displaystyle \frac{880}{600+x}=\frac{4}{3}.$
$\displaystyle 3\times880=4(600+x).$
$\displaystyle 2640=2400+4x.$
$\displaystyle 4x=240.$
$\displaystyle \therefore x=60.$
$\displaystyle \therefore 60\text{ new girls must be admitted to make the ratio }4:3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{The sum of three numbers is }212. \text{ If the ratio of the first to the}$
$\displaystyle \text{second is }13:16\text{ and that of the second to the third is }2:3, \text{ find the numbers.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three numbers be }A,B\text{ and }C.$
$\displaystyle A:B=13:16\qquad\text{and}\qquad B:C=2:3.$
$\displaystyle \text{The L.C.M. of }16\text{ and }2\text{ is }16.$
$\displaystyle B:C=(2\times8):(3\times8)=16:24.$
$\displaystyle \therefore A:B:C=13:16:24.$
$\displaystyle \text{Sum of the terms of the ratio}=13+16+24=53.$
$\displaystyle A=\frac{13}{53}\times212=52.$
$\displaystyle B=\frac{16}{53}\times212=64.$
$\displaystyle C=\frac{24}{53}\times212=96.$
$\displaystyle \therefore \text{the three numbers are }52,\ 64\text{ and }96.$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{Find the number which, when added to each term of the}$
$\displaystyle \text{ratio }27:35,\text{ changes the ratio to }4:5.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x\text{ be the number added to each term of the ratio}.$
$\displaystyle \frac{27+x}{35+x}=\frac{4}{5}.$
$\displaystyle 5(27+x)=4(35+x).$
$\displaystyle 135+5x=140+4x.$
$\displaystyle \therefore x=5.$
$\displaystyle \therefore \text{the required number is }5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{The present ages of Mr. X and his son are in the ratio}$
$\displaystyle 17:9.\text{ If the ratio of their ages }9\text{ years ago was }7:3, \text{ find their present ages.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the present ages of Mr. X and his son be }17x\text{ and }9x \text{ years respectively}.$
$\displaystyle \text{Nine years ago, their ages were }(17x-9)\text{ and }(9x-9) \text{ years respectively}.$
$\displaystyle \frac{17x-9}{9x-9}=\frac{7}{3}.$
$\displaystyle 3(17x-9)=7(9x-9).$
$\displaystyle 51x-27=63x-63.$
$\displaystyle 36=12x.$
$\displaystyle \therefore x=3.$
$\displaystyle \text{Present age of Mr. X}=17\times3=51\text{ years}.$
$\displaystyle \text{Present age of his son}=9\times3=27\text{ years}.$
$\displaystyle \therefore \text{their present ages are }51\text{ years and }27\text{ years respectively}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{The salaries of }A,B\text{ and }C\text{ are in the ratio }2:3:5.$
$\displaystyle \text{If increments of }15\%,10\%\text{ and }20\%\text{ are given respectively,}$
$\displaystyle \text{find the new ratio of their salaries.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the salaries of }A,B\text{ and }C\text{ be Rs. }200,\text{ Rs. }300 \text{ and Rs. }500\text{ respectively}.$
$\displaystyle \text{After a }15\%\text{ increment, A's salary} =200\times\frac{115}{100}=230.$
$\displaystyle \text{After a }10\%\text{ increment, B's salary} =300\times\frac{110}{100}=330.$
$\displaystyle \text{After a }20\%\text{ increment, C's salary} =500\times\frac{120}{100}=600.$
$\displaystyle \therefore \text{the new ratio of their salaries} =230:330:600=23:33:60.$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{A got }58\text{ marks out of }75\text{ in Physics and }97 \text{ marks out of }120$
$\displaystyle \text{in Mathematics. In which of the two subjects did he perform better?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Fraction of marks obtained in Physics}=\frac{58}{75}.$
$\displaystyle \text{Fraction of marks obtained in Mathematics}=\frac{97}{120}.$
$\displaystyle \text{The L.C.M. of }75\text{ and }120\text{ is }600.$
$\displaystyle \frac{58}{75}=\frac{58\times8}{75\times8}=\frac{464}{600}.$
$\displaystyle \frac{97}{120}=\frac{97\times5}{120\times5}=\frac{485}{600}.$
$\displaystyle \text{Since }\frac{485}{600}>\frac{464}{600}, \text{ A performed better in Mathematics}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{A mixture consists of only two components }A\text{ and }B. \text{ In }60\text{ litres}$
$\displaystyle \text{of the mixture, }A\text{ and }B\text{ are in the ratio }2:1. \text{ What quantity of component }B$
$\displaystyle \text{must be added so that the new ratio of }A\text{ to }B\text{ becomes }1:2?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Quantity of A}=60\times\frac{2}{3}=40\text{ litres}.$
$\displaystyle \text{Quantity of B}=60\times\frac{1}{3}=20\text{ litres}.$
$\displaystyle \text{Let the quantity of B added be }x\text{ litres}.$
$\displaystyle \text{After adding }x\text{ litres of B, its quantity becomes }(20+x)\text{ litres}.$
$\displaystyle \frac{40}{20+x}=\frac{1}{2}.$
$\displaystyle 80=20+x.$
$\displaystyle \therefore x=60.$
$\displaystyle \therefore 60\text{ litres of component B must be added}.$
$\displaystyle \\$