RD Sharma solution class 8 chapter 13 Profit ,Discount and Value Added Tax(VAT) Exercise 13.1

Exercise 13.1

Page-13.11

Question 1:

A student buys a pen for Rs 90 and sells it for Rs 100. Find his gain and gain percent.

Answer 1:

$$\begin{gathered} \mathrm{C}.\mathrm{P} \mathrm{of} \mathrm{pen}=\mathrm{Rs}. 90 \\ \mathrm{S}.\mathrm{P} \mathrm{of} \mathrm{pen} = \mathrm{Rs}. 100 \\ \mathrm{Gain}=\mathrm{SP} -\mathrm{CP} \\ \mathrm{Gain}=100-90 \\ =\mathrm{Rs}. 10 \\ \mathrm{Gain}\%=\left(\frac{\mathrm{gain}\times 100}{\mathrm{C}.\mathrm{P}.}\right) \\ =\left(\frac{10\times 100}{90}\right) \\ =11\frac{1}{9}\% \end{gathered}$$

Question 2:

Rekha bought a saree for Rs 1240 and sold it for Rs 1147. Find her loss and loss percent.

Answer 2:

$$\begin{gathered} \mathrm{C}.\mathrm{P} \mathrm{of} \mathrm{saree}=\mathrm{Rs}. 1240 \\ \mathrm{S}.\mathrm{P} \mathrm{of} \mathrm{saree} = \mathrm{Rs}. 1147 \\ \mathrm{Loss}=\mathrm{CP}-\mathrm{SP} \\ \mathrm{Loss}= \mathrm{Rs}.(1240-1147) \\ =\mathrm{Rs}. 93 \\ \mathrm{loss}\%=\left(\frac{\mathrm{loss}\times 100}{\mathrm{C}.\mathrm{P}.}\right) \\ =\left(\frac{93\times 100}{1240}\right) \\ =7.5\% \end{gathered}$$

Question 3:

A boy buys 9 apples for Rs 9.60 and sells them at 11 for Rs 12. Find his gain or loss percent.

Answer 3:

$$\begin{gathered} \mathrm{C}.\mathrm{P} \mathrm{of} 9 \mathrm{apples}=\mathrm{Rs} 9.60 \\ \therefore \mathrm{C}.\mathrm{P} \mathrm{of} 1 \mathrm{apple}=\mathrm{Rs} \frac{9.60}{9}=\mathrm{Rs}\frac{16}{15} \\ \mathrm{S}.\mathrm{P} \mathrm{of} 11 \mathrm{apples} = \mathrm{Rs} 12 \\ \therefore \mathrm{S}.\mathrm{P} \mathrm{of} 1 \mathrm{apple}=\mathrm{Rs} \frac{12}{11} \\ \mathrm{Clearly}, \mathrm{S}.\mathrm{P} \mathrm{of} 1 \mathrm{apple} >\mathrm{C}.\mathrm{P} \mathrm{of} 1 \mathrm{apple} \\ \mathrm{So}, \mathrm{we} \mathrm{get} \mathrm{profit} \mathrm{on} \mathrm{selling} \mathrm{apples} \\ \mathrm{Gain}\%=\left(\frac{\mathrm{gain}\times 100}{\mathrm{C}.\mathrm{P}.}\right) \\ =\left(\frac{{\displaystyle \frac{12}{11}}-{\displaystyle \frac{16}{15}}}{{\displaystyle \frac{16}{15}}}\right)\times 100 \\ =\left(\frac{12\times 15-16\times 11}{15\times 11}\right)\times \frac{15}{16}\times 100 \\ =\left(\frac{180-176}{11\times 16}\right)\times 100 \\ =\left(\frac{4}{11\times 16}\right)\times 100 \\ =2\frac{3}{11}\% \end{gathered}$$

Question 4:

The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit percent.

Answer 4:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{cost} \mathrm{price} \mathrm{of} \mathrm{one} \mathrm{article} be \mathrm{Rs}. \mathrm{C} \mathrm{and} the \mathrm{selling} \mathrm{price} \mathrm{be} \mathrm{Rs}. \mathrm{S} \\ Therefore, 10\mathrm{C}=9\mathrm{S} \\ \mathrm{C}=\frac{9}{10}\mathrm{S} \\ \mathrm{So}, the \mathrm{cost} \mathrm{price} \mathrm{is} \mathrm{less} \mathrm{than} the \mathrm{selling} \mathrm{price}. \\ \mathrm{S}.\mathrm{P}. = \left(\frac{100+P\mathrm{rofit}\%}{100}\right)\mathrm{C}.\mathrm{P} \\ \mathrm{S}=\left(\frac{100+P\mathrm{rofit}\%}{100}\right)\mathrm{C} \\ \frac{\mathrm{S}}{\mathrm{C}}=\left(\frac{100+P\mathrm{rofit}\%}{100}\right) \\ \frac{10}{9}=\left(\frac{100+P\mathrm{rofit}\%}{100}\right) \\ \frac{1000}{9}=100+P\mathrm{rofit}\% \\ \frac{1000}{9}-100=P\mathrm{rofit}\% \\ P\mathrm{rofit}\%=\frac{1000-900}{9} \\ =11\frac{1}{9} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{required} \mathrm{profit} \mathrm{percent} \mathrm{is} 11\frac{1}{9}\%. \end{gathered}$$

Question 5:

A retailer buys a radio for Rs 225. His overhead expenses are Rs 15. If he sells the radio for Rs 300, determine his profit percent.

Answer 5:

$$\begin{gathered} \mathrm{Radio} \mathrm{costs} =\mathrm{Rs}. 225 \\ \mathrm{Overhead} \mathrm{expenses}=\mathrm{Rs}. 15 \\ \mathrm{C}.\mathrm{P}=\mathrm{Rs}. (225+15)=\mathrm{Rs}. 240 \\ \mathrm{S}.\mathrm{P}=\mathrm{Rs}. 300 \\ \mathrm{Profit}=\mathrm{SP}-\mathrm{CP} \\ = \mathrm{Rs}. (300-240) \\ =\mathrm{Rs}. 60 \\ \mathrm{Profit}\%=\frac{\mathrm{Profit}}{\mathrm{C}.\mathrm{P}}\times 100 \\ =\frac{60}{240}\times 100 \\ =25\% \end{gathered}$$

Question 6:

A retailer buys a cooler for Rs 1200 and overhead expenses on it are Rs 40. If he sells the cooler for Rs 1550, determine his profit percent.

Answer 6:

$$\begin{gathered} \mathrm{Cooler} \mathrm{costs}= \mathrm{Rs}.1200 \\ \mathrm{Overhead} \mathrm{expenses}=\mathrm{Rs}. 40 \\ \mathrm{C}.\mathrm{P}=\mathrm{Rs}. (1200+40)=\mathrm{Rs}. 1240 \\ \mathrm{S}.\mathrm{P}=\mathrm{Rs}. 1550 \\ \mathrm{Profit}=\mathrm{SP}-\mathrm{CP} \\ =\mathrm{Rs}. (1550-1240) \\ =\mathrm{Rs}. 310 \\ \mathrm{Profit}\%=\frac{\mathrm{Profit}}{\mathrm{C}.\mathrm{P}}\times 100 \\ =\frac{310}{1240}\times 100 \\ =25\% \end{gathered}$$

Question 7:

A dealer buys a wristwatch for Rs 225 and spends Rs 15 on its repairs. If he sells the same for Rs 300, find his profit percent.

Answer 7:

$$\begin{gathered} \mathrm{A} \mathrm{dealer} \mathrm{buys} \mathrm{a} \mathrm{wrist} \mathrm{watch} for \mathrm{Rs}. 225 \\ \mathrm{Money} \mathrm{spent} on \mathrm{repair}ing the watch =\mathrm{Rs}. 15 \\ \mathrm{Therefore},\mathrm{C}.\mathrm{P}=\mathrm{Rs}. (225+15)=\mathrm{Rs}. 240 \\ \mathrm{S}.\mathrm{P}=\mathrm{Rs}. 300 \\ \mathrm{Profit}=\mathrm{SP}-\mathrm{CP} \\ =\mathrm{Rs}. (300-240) \\ =\mathrm{Rs}. 60 \\ \mathrm{Profit}\%=\frac{\mathrm{Profit}}{\mathrm{C}.\mathrm{P}}\times 100 \\ =\frac{60}{240}\times 100 \\ =25\% \end{gathered}$$

Question 8:

Ramesh bought two boxes for Rs 1300. He sold one box at a profit of 20% and the other box at a loss of 12%. If the selling price of both boxes is the same, find the cost price of each box.

Answer 8:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{cost} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{box} \mathrm{be} \mathrm{Rs}. x. \\ Therefore, the \cos t of the second box will be \mathrm{Rs}. (1300-x) \\ \mathrm{Profit} \mathrm{on} \mathrm{the} \mathrm{first} \mathrm{box}=20 \% \\ \mathrm{Loss} \mathrm{on} \mathrm{the} \mathrm{second} \mathrm{box}=12\% \\ \mathrm{SP} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{box}=\mathrm{CP}\left(\frac{\mathrm{gain}\%+100}{100}\right) \\ \mathrm{SP}=\mathrm{x}\left(\frac{120}{100}\right) \\ SP \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{box} =\mathrm{Rs}.\frac{120x}{100}=\mathrm{Rs}.\frac{6x}{5} \\ \mathrm{SP} \mathrm{of} \mathrm{the} \mathrm{second} \mathrm{box}=\mathrm{CP}\left(\frac{100-\mathrm{loss}\%}{100}\right) \\ S.\mathrm{P} \mathrm{of} \mathrm{the} \mathrm{second} \mathrm{box}=\frac{88\left(1300-x\right)}{100}=\mathrm{Rs}.\left(\frac{28600-22x}{25}\right) \\ \mathrm{Since} S.\mathrm{P} of both the box \mathrm{are} \mathrm{equal}, \\ \frac{6x}{5}=\left(\frac{28600-22x}{25}\right) \\ 150x=143000-110x \\ 260x=143000 \\ x=\frac{143000}{260} \\ x=550 \\ \mathrm{Therefore}, the \mathrm{cost} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{box} is \mathrm{Rs}. 550. \\ The c\mathrm{ost} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{second} \mathrm{box} will be \mathrm{Rs}.(1300-550)=\mathrm{Rs}. 750 \end{gathered}$$

Question 9:

If the selling price of 10 pens is equal to cost price of 14 pens, find the gain percent.

Answer 9:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{cost} \mathrm{price} \mathrm{of} \mathrm{one} \mathrm{pen} be \mathrm{Rs}. \mathrm{C}, and the s\mathrm{elling} \mathrm{price} be \mathrm{Rs}. \mathrm{S} \\ Therefore, 10\mathrm{S}=14\mathrm{C} \\ \mathrm{C}=\frac{10}{14}\mathrm{S} \\ \mathrm{However}, the \mathrm{cost} \mathrm{price} \mathrm{is} \mathrm{less} \mathrm{than} the \mathrm{selling} \mathrm{price}. \\ \mathrm{S}.\mathrm{P}. = \left(\frac{100+\mathrm{profit}\%}{100}\right)\mathrm{C}.\mathrm{P} \\ \mathrm{S}=\left(\frac{100+\mathrm{profit}\%}{100}\right)\mathrm{C} \\ \frac{\mathrm{S}}{\mathrm{C}}=\left(\frac{100+\mathrm{profit}\%}{100}\right) \\ \frac{14}{10}=\left(\frac{100+\mathrm{profit}\%}{100}\right) \\ \frac{1400}{10}=100+\mathrm{profit}\% \\ 140-100=\mathrm{profit}\% \\ P\mathrm{rofit}\%=40 \\ =40\% \\ \mathrm{Therefore}, \mathrm{the} \mathrm{required} \mathrm{profit} \mathrm{percent} \mathrm{is} 40\%. \end{gathered}$$

Question 10:

If the cost price of 18 chairs be equal to selling price of 16 chairs, find the gain or loss percent.

Answer 10:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{cost} \mathrm{price} \mathrm{of} \mathrm{one} \mathrm{chair} be \mathrm{Rs}. \mathrm{C}, \mathrm{and} \mathrm{selling} \mathrm{price} \mathrm{be} \mathrm{Rs}. \mathrm{S} \\ Therefore, 18\mathrm{C}=16\mathrm{S} \\ \mathrm{C}=\frac{16}{18}\mathrm{S} \\ As \mathrm{cost} \mathrm{price} \mathrm{is} \mathrm{less} \mathrm{than} the \mathrm{selling} \mathrm{price}, \\ \mathrm{S}.\mathrm{P}. = \left(\frac{100+\mathrm{profit}\%}{100}\right)\mathrm{C}.\mathrm{P} \\ \mathrm{S}=\left(\frac{100+\mathrm{profit}\%}{100}\right)\mathrm{C} \\ \frac{\mathrm{S}}{\mathrm{C}}=\left(\frac{100+\mathrm{profit}\%}{100}\right) \\ \frac{18}{16}=\left(\frac{100+\mathrm{profit}\%}{100}\right) \\ \frac{1800}{16}=100+\mathrm{profit}\% \\ \frac{1800}{16}-100=\mathrm{profit}\% \\ P\mathrm{rofit}\%=\frac{1800-1600}{16} \\ =\frac{200}{16} \\ =12.5 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{required} \mathrm{profit} \mathrm{percent} \mathrm{is} 12.5\%. \end{gathered}$$

Question 11:

If the selling price of 18 oranges is equal to the cost price of 16 oranges, find the loss percent.

Answer 11:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{cost} \mathrm{price} \mathrm{of} \mathrm{one} \mathrm{orange} be \mathrm{Rs}. \mathrm{C}, \mathrm{and} \mathrm{its} \mathrm{selling} \mathrm{price} \mathrm{be} \mathrm{Rs}. \mathrm{S} \\ Therefore, 16\mathrm{C}=18\mathrm{S} \\ \mathrm{C}=\frac{18}{16}\mathrm{S} \\ As \mathrm{cost} \mathrm{price} \mathrm{is} \mathrm{more} \mathrm{than} the \mathrm{selling} \mathrm{price}, \\ \mathrm{S}.\mathrm{P}. = \left(\frac{100-\mathrm{loss}\%}{100}\right)\mathrm{C}.\mathrm{P} \\ \mathrm{S}=\left(\frac{100-\mathrm{loss}\%}{100}\right)\mathrm{C} \\ \frac{\mathrm{S}}{\mathrm{C}}=\left(\frac{100-\mathrm{loss}\%}{100}\right) \\ \frac{16}{18}=\left(\frac{100-\mathrm{loss}\%}{100}\right) \\ \frac{1600}{18}=100-\mathrm{loss}\% \\ L\mathrm{oss}\%=100-\frac{1600}{18} \\ L\mathrm{oss}\%=\frac{1800-1600}{18} \\ =\frac{200}{18}=\frac{100}{9} \\ =11\frac{1}{9} \\ \mathrm{Therefore}, \mathrm{the} \mathrm{loss} \mathrm{percent} \mathrm{is} 11\frac{1}{9}\%. \end{gathered}$$

Question 12:

Ravish sold his motorcycle to Vineet at a loss of 28%. Vineet spent Rs 1680 on its repairs and sold the motor cycle to Rahul for Rs 35910, thereby making a profit of 12.5%, find the cost price of the motor cycle for Ravish.

Answer 12:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{cost} \mathrm{price} \mathrm{of} the \mathrm{motorcycle} \mathrm{for} \mathrm{Ravis}h be \mathrm{Rs}. \mathrm{x}. \\ \mathrm{Loss}\%=28\% \\ \mathrm{Therefore}, \mathrm{SP}=\mathrm{CP}\left(\frac{100-\mathrm{Loss}\%}{100}\right) \\ \mathrm{SP}=\mathrm{Rs}. x\left(\frac{72}{100}\right) \\ \mathrm{Selling} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{motorcycle} \mathrm{for} \mathrm{Ravish}= \mathrm{Cost} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{motorcycle} \mathrm{for} \mathrm{Vineet} \\ \mathrm{Money} \mathrm{spent} \mathrm{on} \mathrm{repair}s=\mathrm{Rs}. 1680 \\ Therefore, t\mathrm{otal} \mathrm{cost} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{motorcycle} \mathrm{for} \mathrm{Vineet}=\mathrm{Rs}.( x\left(\frac{72}{100}\right)+1680) \\ \mathrm{Selling} \mathrm{price} \mathrm{of} \mathrm{the} \mathrm{motorcycle} \mathrm{for} \mathrm{Vineet} =\mathrm{Rs}. 35910 \\ \mathrm{Profit}\%=12.5\% \\ \mathrm{SP}=\mathrm{CP}\left(\frac{\mathrm{Profit}\%+100}{100}\right) \\ \Rightarrow 35910=\left(\frac{72x}{100}+1680\right)\left(\frac{12.5+100}{100}\right) \\ \Rightarrow 35910\times 100\times 100=(72x+168000)(112.5) \\ \Rightarrow 359100000=8100x+18900000 \\ \Rightarrow 340200000=8100x \\ \Rightarrow x=\mathrm{Rs}. 42000 \\ \mathrm{Therefore}, \mathrm{Ravis}h bought the motorcycle for \mathrm{Rs}. 42000 \end{gathered}$$

Page-13.12

Question 13:

By selling a book for Rs 258, a bookseller gains 20%. For how much should he sell it to gain 30%?

Answer 13:

$$\begin{gathered} \mathrm{Selling} \mathrm{price} \mathrm{of} the \mathrm{book}=\mathrm{Rs}. 258 \\ \mathrm{Gain}\%=20\% \\ \mathrm{Since}, \mathrm{C}.\mathrm{P} = \left[\frac{100}{\left(100+\mathrm{Gain}\%\right)}\times S.P\right] \\ = \left[\frac{100}{\left(100+20\right)}\times 258\right] \\ =\frac{25800}{120} \\ =\mathrm{Rs}. 215 \\ \mathrm{Let} \mathrm{the} \mathrm{bookseller} sells \mathrm{it} for \mathrm{Rs}. x \\ So, S.P = \left[\frac{\left(100+\mathrm{Gain}\%\right)}{100}\times C.P\right] \\ x=\frac{100+30}{100}\times 215 \\ =\frac{130\times 215}{100} \\ =\mathrm{Rs}. 279.50 \\ \mathrm{Therefore}, the \mathrm{bookseller} \mathrm{must} \mathrm{sell} the book at \mathrm{Rs} 279.50 \mathrm{to} make 30\% profit. \end{gathered}$$

Question 14:

A defective briefcase costing Rs 800 is being sold at a loss of 8%. If the price is further reduced by 5%, find its selling price.

Answer 14:

$$\begin{gathered} \mathrm{C}.\mathrm{P} \mathrm{of} \mathrm{the} \mathrm{briefcase}=\mathrm{Rs}. 800 \\ \mathrm{Loss}=8\% \\ \mathrm{S}.\mathrm{P} =\mathrm{CP} \left(\frac{100-\mathrm{loss}\%}{100}\right) \\ =800\left(\frac{100-8}{100}\right) \\ =800\times 0.92=\mathrm{Rs}. 736 \\ If the p\mathrm{rice} \mathrm{is} \mathrm{further} \mathrm{reduced} \mathrm{by} 5\%, the s\mathrm{elling} \mathrm{price} \mathrm{of} the \mathrm{briefcase} will be \\ = \mathrm{Rs}.\left(736-736\times \frac{5}{100}\right) \\ =736\left(\frac{100-5}{100}\right) \\ =736\times 0.95 \\ =\mathrm{Rs}. 699.20 \\ \mathrm{Thus}, \mathrm{the} \mathrm{selling} \mathrm{price} \mathrm{of} the \mathrm{briefcase} will be \mathrm{Rs} 699.20. \end{gathered}$$

Question 15:

By selling 90 ball pens for Rs 160 a person loses 20%. How many ball pens should be sold for Rs 96 so as to have a profit of 20%?

Answer 15:

$$\begin{gathered} \mathrm{S}.P of 90 ball pens = \mathrm{Rs}. 160 \\ \mathrm{Loss}=20\% \\ \mathrm{Therefore}, \mathrm{C}.\mathrm{P} =\mathrm{SP}\left(\frac{100}{100-l\mathrm{oss}\%}\right) \\ \mathrm{CP}=\frac{100}{100-20}\times 160 \\ =\frac{16000}{80} \\ =\mathrm{Rs}. 200 \\ \mathrm{Now}, \\ \mathrm{S}.\mathrm{P} of 90 ball pens=\mathrm{Rs}. 96 \\ \mathrm{Profit}=20\% \\ \mathrm{C}.\mathrm{P} =\mathrm{SP}\left(\frac{100}{100+\mathrm{Profit}\%}\right) \\ \mathrm{CP}=\frac{100}{100+20}\times 96 \\ =\frac{9600}{120} \\ =\mathrm{Rs}. 80 \\ \mathrm{Rs}. 200 \mathrm{is} \mathrm{the} \mathrm{C}.\mathrm{P} \mathrm{of} 90 \mathrm{ball} \mathrm{pens}. \\ \mathrm{Therefore}, \mathrm{Rs}. 80 \mathrm{is} the \mathrm{C}.\mathrm{P} \mathrm{of} =\frac{90\times 80}{200}=36 ball pens \\ \mathrm{Thus}, 36 \mathrm{ball} \mathrm{pens} \mathrm{should} \mathrm{be} \mathrm{sold} at \mathrm{Rs}. 96 \mathrm{to} earn \mathrm{a} \mathrm{profit} \mathrm{of} 20\%. \end{gathered}$$

Question 16:

A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for Rs 36.75 less, he would have gained 30%. Find the cost price of the article.

Answer 16:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{C}.\mathrm{P} of the article \mathrm{be} \mathrm{Rs}. x. \\ \mathrm{Original} \mathrm{S}.\mathrm{P} = x+\frac{25}{100}x \\ =Rs. \frac{5x}{4} \\ \mathrm{If} \mathrm{he} purchased \mathrm{it} \mathrm{at} 20\% \mathrm{less}, \\ C.\mathrm{P}=x-\frac{20}{100}x \\ =Rs. \frac{4x}{5} \\ He sold the article at \mathrm{Rs} 36.75 \mathrm{less}. \\ So, the selling price =Rs. \frac{5\mathrm{x}}{4}-36.75 \\ \mathrm{Given} \mathrm{that} \mathrm{he} \mathrm{would} \mathrm{have} \mathrm{gained} 30\% selling at that price. \\ Therefore, \mathrm{gain}\%=\frac{\mathrm{S}.\mathrm{P}-\mathrm{C}.\mathrm{P}}{\mathrm{C}.\mathrm{P}}\times 100 \\ \mathrm{S}.\mathrm{P}-\mathrm{C}.\mathrm{P}=\frac{5x}{4}-36.75-\frac{4x}{5} \\ =\frac{5x}{4}-\frac{4x}{5}-36.75 \\ =\frac{25x-16x}{20}-36.75 \\ =\frac{9x}{20}-36.75 \\ \mathrm{So}, \mathrm{gain}\%=\frac{\mathrm{S}.\mathrm{P}-\mathrm{C}.\mathrm{P}}{\mathrm{C}.\mathrm{P}}\times 100 \\ 30=\frac{{\displaystyle \frac{9x}{20}}-36.75}{{\displaystyle \frac{4x}{5}}}\times 100 \\ =\left(\frac{9x}{20}-36.75\right)\times \frac{5}{4x}\times 100 \\ =\frac{9x-735}{16x}\times 100 \\ 30=\frac{9x-735}{16x}\times 100 \\ \frac{225x-18375}{4x}=30 \\ 225x-18375=120x \\ 105x=18375 \\ x=\frac{18375}{105} \\ =175 \\ \mathrm{So}, the \mathrm{cost} \mathrm{price} of the article is \mathrm{Rs}. 175. \end{gathered}$$

Question 17:

A dishonest shopkeeper professes to sell pulses at his cost price but uses a false weight of 950 gm for each kilogram. Find his gain percent.

Answer 17:

$$\begin{gathered} \mathrm{He} \mathrm{sells} 950\mathrm{gm} \mathrm{pulses} \mathrm{and} gets a \mathrm{gain} of 50\mathrm{gm}. \\ \mathrm{If} \mathrm{he} \mathrm{sells} 100\mathrm{gm} \mathrm{of} \mathrm{pulses}, \mathrm{he} \mathrm{will} \mathrm{gain} =\frac{50}{950}\times 100 \\ =\frac{5000}{950} \\ =5\frac{5}{19}\% \end{gathered}$$

Question 18:

A dealer bought two tables for Rs 3120. He sold one of them at a loss of 15% and other at a gain of 36%. Then, he found that each table was sold for the same price. Find the cost price of each table.

Answer 18:

$$\begin{gathered} G\mathrm{iven} \mathrm{that} the \mathrm{selling} \mathrm{price} \mathrm{is} \mathrm{same} \mathrm{for} \mathrm{both} \mathrm{the} \mathrm{tables}. \\ \mathrm{Let} the \mathrm{C}.\mathrm{P} \mathrm{of} 1 \mathrm{table} \mathrm{be} \mathrm{Rs}. x, \mathrm{then} the \mathrm{C}.\mathrm{P} \mathrm{of} the \mathrm{other} will be \mathrm{Rs}.\left(3120-x\right). \\ \mathrm{Los}s on \mathrm{the} \mathrm{first} \mathrm{table}= 15\% \\ \mathrm{Therefore}, S.P=C.P\left(\frac{100-l\mathrm{oss}\%}{100}\right) \\ \mathrm{S}.\mathrm{P}= \frac{85x}{100}=\mathrm{Rs}. 0.85x \\ \mathrm{Gain} on \mathrm{the} \mathrm{second} \mathrm{table} = 36\% \\ \mathrm{Therefore}, \mathrm{S}.\mathrm{P} =C.P\left(\frac{100+g\mathrm{ain}\%}{100}\right) \\ S.P=\mathrm{Rs}. 1.36\left(3120-x\right) \\ \mathrm{Since} \mathrm{both} tables have the same S.P, \\ \mathrm{So}, 0.85\mathrm{x}=1.36\left(3120-x\right) \\ 0.85x=4243.20-1.36x \\ 2.21x=4243.20 \\ x =\frac{4243.20}{2.21} \\ x=\mathrm{Rs}. 1920 \\ \mathrm{So},\mathrm{the} \mathrm{cost} \mathrm{price} \mathrm{of} the \mathrm{first} \mathrm{table} is \mathrm{Rs}. 1920. \\ \mathrm{Cost} \mathrm{price} \mathrm{of} the \mathrm{second} \mathrm{table} = \mathrm{Rs}. ( 3120-1920)=\mathrm{Rs}. 1200 \end{gathered}$$

Question 19:

Mariam bought two fans for Rs 3605. She sold one at a profit of 15% and the other at a loss of 9%. If Mariam obtained the same amount for each fan, find the cost price of each fan.

Answer 19:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} the S.P \mathrm{is} \mathrm{same} \mathrm{for} \mathrm{both} \mathrm{the} \mathrm{fans}. \\ \mathrm{Let} \mathrm{C}.\mathrm{P} \mathrm{of} the \mathrm{first} \mathrm{fa}n be \mathrm{Rs}. \mathrm{x} \\ Therefore, \mathrm{C}.\mathrm{P} \mathrm{of} the \mathrm{second} \mathrm{fan}=\mathrm{Rs}. (3605-\mathrm{x}) \\ \mathrm{Profit} on \mathrm{the} \mathrm{first} \mathrm{fan}=15\% \\ \mathrm{Loss} on \mathrm{the} \mathrm{second} \mathrm{fan}=9\% \\ \mathrm{For} \mathrm{the} \mathrm{first} \mathrm{fan}, \\ \mathrm{S}.\mathrm{P} = \mathrm{C}.\mathrm{P}\left(\frac{100+\mathrm{gain}\%}{100}\right) \\ =x\left(\frac{115}{100}\right) \\ =\frac{23x}{20} \\ \mathrm{For} \mathrm{the} \mathrm{second} \mathrm{fan}, \\ \mathrm{S}.\mathrm{P} = \mathrm{C}.\mathrm{P} \left(\frac{100-\mathrm{loss}\%}{100}\right) \\ =\left(3605-x\right)\left(\frac{91}{100}\right) \\ \mathrm{Since} \mathrm{S}.\mathrm{P} of both the fans is the \mathrm{same}, \\ \frac{23x}{20}=\left(3605-x\right)\left(\frac{91}{100}\right) \\ 2300x=91\left(72100-20\mathrm{x}\right) \\ 2300x=6561100-1820\mathrm{x} \\ 4120x=6561100 \\ x=\mathrm{Rs}.1592.50 \\ \mathrm{Thus}, \mathrm{C}.\mathrm{P} \mathrm{of} the \mathrm{first} \mathrm{fan} is \mathrm{Rs}. 1592.50. \\ \mathrm{C}.\mathrm{P} \mathrm{of} the \mathrm{second} \mathrm{fan} = \mathrm{Rs}. \left(3605-1592.50\right) \\ =\mathrm{Rs}. 2012.50 \end{gathered}$$

Question 20:

Some toffees are bought at the rate of 11 for Rs 10 and the same number at the rate of 9 for Rs 10. If the whole lot is sold at one rupee per toffee, find the gain or loss percent on the whole transaction.

Answer 20:

$$\begin{gathered} \mathrm{Let} \mathrm{the} total \mathrm{number} \mathrm{of} \mathrm{toffees} \mathrm{bought} be x. \\ \mathrm{Let} \frac{x}{2}\mathrm{toffees} \mathrm{at} the \mathrm{rate} \mathrm{of} 11 are bought \mathrm{for} \mathrm{Rs}.10, and \frac{x}{2}\mathrm{toffees} \mathrm{at} \mathrm{the} \mathrm{rate} \mathrm{of} 9 are bought \mathrm{for} \mathrm{Rs}.10 \\ \mathrm{Total} \mathrm{money} \mathrm{spent} \mathrm{on} \mathrm{buying} the \mathrm{toffees}= \left(\frac{x}{2}\right)\left(\frac{10}{11}\right)+\left(\frac{x}{2}\right)\left(\frac{10}{9}\right) \\ =\frac{200x}{198} \\ =\frac{100}{99}x \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} x \mathrm{toffees} \mathrm{are} \mathrm{sold} \mathrm{at} \mathrm{one} \mathrm{rupee} \mathrm{per} \mathrm{toffee}. \\ \mathrm{Therefore}, the \mathrm{selling} \mathrm{price} \mathrm{of}\mathit{ }x\mathit{ }\mathrm{toffees} =\mathrm{Rs}.\mathit{ }x\times 1=\mathrm{Rs}.\mathit{ }x \\ \mathrm{As} C.P \mathrm{is} \mathrm{more} \mathrm{than} S.P, it \mathrm{will} \mathrm{be} a \mathrm{loss}. \\ \mathrm{Loss}=C.P-S.P \\ =\frac{100}{99}x-x \\ =\frac{100x-99x}{99} \\ =\frac{x}{99} \\ \mathrm{Loss}\%=\frac{\mathrm{Loss}}{C.P}\times 100 \\ \frac{{\displaystyle \frac{x}{99}}}{{\displaystyle \frac{100x}{99}}}\times 100=1\% \\ Total \mathrm{loss} \mathrm{on} \mathrm{the} \mathrm{whole} \mathrm{transaction} \mathrm{would} \mathrm{be} 1\%. \end{gathered}$$

Question 21:

A tricycle is sold at a gain of 16%. Had it been sold for Rs 100 more, the gain would have been 20%. Find the C.P. of the tricycle.

Answer 21:

$$\begin{gathered} \mathrm{Let} the \mathrm{S}.\mathrm{P} of the tricycle \mathrm{be} \mathrm{Rs}. x \mathrm{and} \mathrm{C}.\mathrm{P} \mathrm{be} \mathrm{Rs}. y \\ \mathrm{Gain}\%=16\% \\ S.P=C.P\left(\frac{100+\mathrm{gain}\%}{100}\right) \\ \mathrm{Then} \mathrm{we} \mathrm{have}, \\ x=y+\left(\frac{y\times 16}{100}\right) \\ x=y+0.16y \\ x=1.16y \\ \mathrm{When} \mathrm{S}.\mathrm{P} \mathrm{increases} \mathrm{by} \mathrm{Rs}. 100, we get \\ \Rightarrow x+100=y+\left(\frac{y\times 20}{100}\right) \\ \mathrm{Put}ting x=1.6y, we \mathrm{get} \\ \Rightarrow 1.16y+100=y+0.2y \\ 1.16y+100=1.2y \\ 0.04y=100 \\ y=\frac{100}{0.04} \\ =2500 \\ \mathrm{Thus}, \mathrm{C}.\mathrm{P} \mathrm{of} the \mathrm{tricycle} is \mathrm{Rs}. 2500. \end{gathered}$$

Question 22:

Shabana bought 16 dozen ball bens and sold them at a loss equal to S.P. of 8 ball pens. Find
(i) her loss percent
(ii) S.P. of 1 dozen ball pens, if she purchased these 16 dozen ball pens for Rs 576.

Answer 22:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Number} \mathrm{of} \mathrm{pens} \mathrm{bought} =16\times 12=192 \\ \mathrm{Let} \mathrm{the} \mathrm{S}.\mathrm{P} \mathrm{of} \mathrm{one} \mathrm{pen} be \mathrm{Rs}. x. \\ \mathrm{Therefore}, S.P \mathrm{of} 192 \mathrm{pens}=\mathrm{Rs}. 192x \\ \mathrm{S}.\mathrm{P} \mathrm{of} 8 \mathrm{pens}=\mathrm{Rs}. 8x \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} S.P \mathrm{of} 8 \mathrm{pens} \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{loss} \mathrm{on} \mathrm{selling} 192 \mathrm{pens}. \\ \mathrm{Therefore}, \mathrm{loss}= \mathrm{Rs}. 8x \\ \mathrm{C}.\mathrm{P} \mathrm{of} 192 \mathrm{pens} = \mathrm{Rs}. 576 \\ \mathrm{So}, \mathrm{loss}=\mathrm{C}.\mathrm{P}-\mathrm{S}.\mathrm{P} \\ 8x=576-192x \\ 200\mathrm{x}=576 \\ x=\frac{576}{200}=2.88 \\ \mathrm{Therefore}, \mathrm{loss}=8\left(2.88\right)=\mathrm{Rs}. 23.04 \\ \mathrm{Loss}\%=\frac{\mathrm{loss}\times 100}{\mathrm{C}.\mathrm{P}} \\ =\frac{23.04\times 100}{576} \\ =4\% \\ \left(\mathrm{ii}\right) \mathrm{S}.\mathrm{P} \mathrm{of} 1 \mathrm{pen}=\mathrm{Rs}. 2.88 \\ \mathrm{Therefore}, \mathrm{S}.\mathrm{P} \mathrm{of} 1 \mathrm{dozen} \mathrm{pens}=12x=12\times 2.88 \\ =\mathrm{Rs}. 34.56 \end{gathered}$$

Question 23:

The difference between two selling prices of a shirt at profits of 4% and 5% is Rs 6. Find
(i) C.P. of the shirt
(ii) the two selling prices of the shirt

Answer 23:

$$\begin{gathered} L\mathrm{et} the \mathrm{C}.\mathrm{P} \mathrm{of} \mathrm{both} \mathrm{the} \mathrm{shirts} \mathrm{be} \mathrm{Rs}. x. \\ \mathrm{C}.\mathrm{P}=\mathrm{Rs}. x \\ \mathrm{For} \mathrm{shirt} 1: \\ \mathrm{Profit} \mathrm{is} 4\%: \\ \mathrm{Profit}\%=\frac{\mathrm{Profit}}{\mathrm{CP}}\times 100 \\ \mathrm{Profit}=\frac{4}{100}\times \mathrm{C}.\mathrm{P} \\ = \mathrm{Rs}. 0.04x \\ \mathrm{S}.\mathrm{P}=\mathrm{C}.\mathrm{P}+\mathrm{Profit} \\ =x+0.04x \\ = \mathrm{Rs}. 1.04x \\ \mathrm{For} \mathrm{shirt} 2: \\ \mathrm{Profit} = 5\%: \\ \mathrm{C}.\mathrm{P}= \mathrm{Rs}. x \\ \mathrm{Profit}=\frac{5}{100}\times \mathrm{C}.\mathrm{P} \\ = \mathrm{Rs}. 0.05x \\ \mathrm{S}.\mathrm{P}=\mathrm{C}.\mathrm{P}+\mathrm{Profit} \\ =x+0.05x \\ = \mathrm{Rs}. 1.05x \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} the \mathrm{difference} \mathrm{between} \mathrm{the}ir \mathrm{profit}s \mathrm{is} \mathrm{Rs}. 6 \\ \mathrm{So}, 1.05x-1.04x=6 \\ 0.01x=6 \\ x= \mathrm{Rs}. 600 \\ \mathrm{Thus}, \mathrm{C}.\mathrm{P}=\mathrm{Rs}. 600 \\ \mathrm{S}.\mathrm{P} \mathrm{of} \mathrm{shirt} 1=\mathrm{Rs}. 1.04x \\ =\mathrm{Rs}. 1.04\times 600 \\ =\mathrm{Rs}. 624 \\ \mathrm{S}.\mathrm{P} \mathrm{of} \mathrm{shirt} 2= \mathrm{Rs}.1.05x \\ = \mathrm{Rs}.1.05\times 600 \\ =\mathrm{Rs}. 630. \end{gathered}$$

Question 24:

Toshiba bought 100 hens for Rs 8000 and sold 20 of these at a gain of 5%. At what gain percent she must sell the remaining hens so as to gain 20% on the whole?

Answer 24:

$$\begin{gathered} C.P \mathrm{of} 100 \mathrm{hens}=\mathrm{Rs}. 8000 \\ \mathrm{Cost} \mathrm{of} \mathrm{one} \mathrm{hen}=\frac{8000}{100}=\mathrm{Rs}. 80 \\ C.P \mathrm{of} 20 \mathrm{hens} =\mathrm{Rs}.( 80\times 20)=\mathrm{Rs}. 1600 \\ \mathrm{Gain}\%=5\% \\ S.P=C.P\left(\frac{100+g\mathrm{ain}\%}{100}\right) \\ S.P=1600\times \frac{105}{100}=\mathrm{Rs}. 1680 \\ C.P \mathrm{of} 80 \mathrm{hens}=\mathrm{Rs}.(80\times 80)=\mathrm{Rs}.6400 \\ \mathrm{Gain} \mathrm{on} 80 \mathrm{hens}=S.P \mathrm{of} 80 \mathrm{hens}-C.P \mathrm{of} 80 \mathrm{hens} \\ \mathrm{Gain} \mathrm{on} 100 \mathrm{hens} =\mathrm{Gain} \mathrm{on} 80 \mathrm{hens}+\mathrm{Gain} \mathrm{on} 20 \mathrm{hens} \\ \mathrm{Gain} \mathrm{on} 100 \mathrm{hens} =\mathrm{Rs}.(80+S.P \mathrm{of} 80 \mathrm{hens} - 6400) \\ \mathrm{Gain}\% \mathrm{on} 100 \mathrm{hens}=\frac{\mathrm{Gain} \mathrm{on} 100 \mathrm{hens}}{C.P \mathrm{of} 100 \mathrm{hens}} \times 100 \\ 20=\frac{(80+S.P \mathrm{of} 80 \mathrm{hens} - 6400)}{8000}\times 100 \\ 1600=80+S.P \mathrm{of} 80 \mathrm{hens} - 6400S.P \mathrm{of} 80 \mathrm{hens}=\mathrm{Rs}.( 1600+6400-80)S.P \mathrm{of} 80 \mathrm{hens}=\mathrm{Rs}. 7920\mathrm{Gain} \mathrm{on} 80 \mathrm{hens}= S.P \mathrm{of} 80 \mathrm{hens} -C.P \mathrm{of} 80 \mathrm{hens}=\mathrm{Rs}.(7920-6400)=\mathrm{Rs}.1520\mathrm{Gain}\% \mathrm{on} 80 \mathrm{hens} =\frac{\mathrm{Gain} \mathrm{on} 80 \mathrm{hens}}{C.P \mathrm{of} 80 \mathrm{hens}}\times 100=\frac{1520}{6400}\times 100=23.75\%\mathrm{Therefore}, Toshiba \mathrm{gai}ned 23.75\% on 80 hens. \end{gathered}$$