# Excercise 8.3 Comparing-Quantities- NCERT Solutions Class 7

Exercise 8.3

## Question 1

Tell what the profit or loss in the following transactions is. Also find profit per cent or loss per cent in each case.

a) Gardening shears bought for \(\rm Rs\, 250\) and sold for \(\rm Rs \,325.\)

b) A refrigerator bought for \(\rm Rs \,12,000\) and sold at \(\rm Rs\, 13,500.\)

c) A cupboard bought for \(\rm Rs \,2,500\) and sold at \(\rm Rs \,3,000.\)

d) A skirt bought for \(\rm Rs \,250\) and sold at \(\rm Rs \,150.\)

### Solution

**Video Solution**

**What is Known?**

Cost price and selling price of different items.

**What is Unknown?**

Profit and Loss percent in each case.

**Reasoning:**

Cost price is the price on which the product is bought. Selling price is the price at which the product is sold. If selling price is greater than the cost price, then there is profit and if cost price is greater than selling price then there is loss. This can be represented as

Profit \(=\) Selling price \(–\) Cost price

And Loss \(=\) Cost price \(–\) Selling price

Also, to find percentage profit/loss, we will divide the profit or loss with cost and then multiply it by \(100.\)

Profit\(/\)Loss \(\%\)

\[=\frac{{{\text{Profit}}\;{\text{(or}}\;{\text{Loss)}}}}{{{\text{Cost}}\;{\text{Price}}}}\times{\text{100}}\]

**Steps:**

(a) In this question gardening shears are bought for \(\rm Rs \,250\) and sold for \(\rm Rs \,325.\) Since sale price is greater than cost price, so there is a profit

Profit \(=\) Selling price \(-\) Cost price

\[\begin{align} &= 325 -250\\\rm &= \rm{ Rs}\, 75\end{align}\]

And, Profit \(\%\)

\[\begin{align}&= \frac{\text{Profit}}{\text{Cost Price}}\times 100\\ &= \frac{75}{250} \times 100\\ &= \frac{750}{25}\\ &= 30\% \end{align}\]

(b) In this question refrigerator bought for \(\rm Rs \,12,000\) and sold at \(\rm Rs \,13,500.\) Since sale price is greater than cost price so there is a profit.

Profit \(=\) Selling price \(-\) Cost price

\[\begin{align}&= 13500\, - 12000\,\\&= \rm{Rs} 1500\end{align}\]

And, Profit\(\%\)

\[\begin{align} &= \frac{\text{Profit}} {\text{Cost Price}} \times 100\\&= \frac{1500}{12000} \times 100\\ &= \frac{150000}{12000}\\ &= 12.5\% \end{align}\]

(c) In this question sale price of cupboard is \(\rm Rs \,2,500\) and cost price is \(\rm Rs \,3,000.\) Since sale price is greater than cost price so there is profit

Profit \(=\) Selling price \(-\) Cost price

\[\begin{align}&= 3000 - 2500 \\&= \rm{Rs} 500\end{align}\]

And,Profit\(\%\)

\[\begin{align}\rm &= \frac{{{\rm{Profit}}\;}}{{{\rm{Cost}}\;{\rm{Price}}}} \times {\rm{100}}\\&= \frac{{500}}{{2500}} \times 100\\&= 20\% \end{align}\]

(d) Cost price of skirt is \(\rm Rs \,250\) and sale price is \(\rm Rs \,150.\) Since, cost price is greater than sale price so there is Loss

Loss \(=\) Cost price \(-\) Sale price

\[\begin{align}&= 250 - 150\\&= \text{Rs}100\end{align}\]

And Loss\( \%\)

\[\begin{align}&= \,\frac{{\;{\rm{Loss}}}}{{{\rm{Cost}}\;{\rm{Price}}}} \times {\rm{100}}\\&= \frac{{100}}{{250}} \times 100\\&= 40\% \end{align}\]

## Question 2

Convert each part of the ratio to percentage:

(a) \(\begin{align}{ 3:1 }\end{align}\)

(b) \(\begin{align}\text{ 2:3:5 }\end{align}\)

(c) \(\begin{align}\rm{1:4}\end{align}\)

(d) \(\begin{align} \rm{ 1:2:5} \end{align}\)

### Solution

**Video Solution**

**What is Known?**

Different ratios are given

**What is Unknown?**

Percentage form of the given ratios.

**Reasoning:**

To convert each part of ratio into percentage we will first divide each part with total of all parts and then multiply it with \(100\) to convert each part into percentage.

**Steps:**

(a) In the ration \(3:1,\) total number of parts

\[\begin{align}&= 3 + 1{\text{ }}\\ &= {\text{ }}4\end{align}\]

Percentage of first part

\[\begin{align}&= \frac{{\rm{3}}}{{\rm{4}}} \times {\rm{100}}\\&= 75\%\end{align}\]

Percentage of second part

\[\begin{align}&=\frac{{\rm{1}}}{{\rm{4}}} \times {\rm{100}}\\&= 25\% \end{align}\]

(b) In \(2:3:5,\) total number of parts

\[\begin{align} &= 2 + 3 + 5 \\&= 10\end{align}\]

Percentge of first part

\[\begin{align} &=\frac{{\text{2}}}{{{\text{10}}}} \times {\text{100}}\\ &= \,{\text{20% }}\end{align}\]

Percentge of second part

\[\begin{align}&= \frac{{\text{3}}}{{{\text{10}}}} \times {\text{100}}\\&= 30\%\end{align}\]

Percentge of third part

\[\begin{align}&=\frac{{\text{5}}}{{{\text{10}}}} \times {\text{100}}\,\\&= 50\% \end{align}\]

(c) In ratio \(1:4,\) total number of parts

\[\begin{align}&= 1 + 4 \\ &= 5\end{align}\]

Percentge of first part

\[\begin{align}&= \frac{{\text{1}}}{{\text{5}}} \times {\text{100}}\\ &= 20\% \end{align}\]

Percentge of second part

\[\begin{align} &= \,\frac{{\text{4}}}{{\text{5}}}{\text{ x 100}}\\\,&= 80\% \end{align}\]

(d) In ratio \(1:2:5,\) total number of parts

\[\begin{align} &= 1 +2 + 5 \\&= 8\end{align}\]

Percentge of first part

\[\begin{align}&= \frac{{\text{1}}}{{\text{8}}} \times {\text{100}}\,\\&= 12{\text{.5% }}\end{align}\]

Percentge of second part

\[\begin{align} &= \,\frac{2}{8}{\rm{ \times 100}}\\&= 25\% \end{align}\]

Percentge of third part

\[\begin{align}&= \frac{{\text{5}}}{{\text{8}}} \times {\text{100}}\\&= \,{\text{62}}{\text{.5% }}\end{align}\]

## Question 3

The population of a city decreased from \(25,000\) to \(24,500.\) Find the percentage decrease.

### Solution

**Video Solution**

**What is Known?**

The change in the population of a city (from \(25,000\) to \(24,500\))

**What is Unknown?**

The percentage decrease

**Reasoning:**

The decrease in population can be calculated by subtracting earlier (initial) population from present (final) population and percentage decrease in population can be obtained using the formula

Percentage decrease

\[\begin{align}=\frac{{{\text{Change}}\;{\text{in}}\;\;{\text{quantity}}}}{{{\text{Initial}}\;{\text{quantity}}\;}}{\rm{ \times 100}}\end{align}\]

**Steps:**

Decrase in city population

\[\begin{align}&= 25000 - 24500\\&= 500\end{align}\]

Percentage decrease

\[\begin{align}&= \,\frac{{{\text{Change}}\;{\text{in}}\;\;{\text{population}}}}{{{\text{Initial}}\;{\text{population}}\;}}{\rm{ \times 100}}\\&= \,\frac{{{\text{500}}}}{{{\text{25000}}}}{\rm{ \times 100}}\\&= \,{\text{2% }}\end{align}\]

## Question 4

Arun bought a car for \(\rm Rs\, 3,50,000.\) The next year, the price went upto \(\rm Rs\, 3,70,000.\) What was the percentage of price increase?

### Solution

**Video Solution**

**What is Known?**

Price at which Arun bought a car \((\rm Rs\, 3,50,000)\) and the price of the car next year \(\rm(Rs\, 3,70,000).\)

**What is Unknown?**

The percentage of price increase

**Reasoning:**

The increase in price can be calculated by subtracting earlier (initial) price from the next year (final) price and percentage increase in price can be obtained using the formula

Percentage increase

\[\begin{align}{\text{ = }}\frac{{{\text{Change}}\;{\text{in}}\;\;{\text{quantity}}}}{{{\text{Initial}}\;{\text{quantity}}\;}} \times {\text{100}}\end{align}\]

**Steps:**

Increase in price

\[\begin{align} &= 3,70,000 - 3,50,000\,\\&= 20000\end{align}\]

Percentage Increase

\[\begin{align}&= \,\,\frac{{{\text{Change}}\;{\text{in}}\;\;{\text{quantity}}}}{{{\text{Initial}}\;{\text{quantity}}\;}} \times {\text{100}}\,\\&=\frac{{{\text{20000}}}}{{{\text{350000}}}} \times {\text{100}}\\&= \,{\text{ }}\frac{{{\text{200}}}}{{{\text{35}}}}\\&= 5\frac{{\text{5}}}{{\text{7}}}{\text{% }}\end{align}\]

## Question 5

I buy a T.V. for \(\rm Rs\, 10,000\) and sell it at a profit of \(20\%.\) How much money do I get for it?

### Solution

**Video Solution**

**What is Known?**

Cost price \((\rm Rs\, 10,000)\) of T.V. and profit on selling it \((20\%).\)

**What is Unknown?**

The price at which TV was sold (selling price).

**Reasoning:**

In this question profit percent is given. Total profit can be calculated by multiplying cost price by profit percent and dividing by \(100\). Selling price can be obtained by adding total profit to the cost price.

**Steps:**

Cost price of TV \(= 10000\) and Profit percent \(= 20\%\)

Total profit

\[\begin{align}&= \frac{{{\rm{20}}}}{{{\rm{100}}}}{\rm{ \times 10000}}\\ &= \rm{Rs}{\rm{.2000}}\end{align}\]

So, selling price will be \(=\) Cost price \(+\) profit

Thus,

Selling price

\(=10000+2000=\text{Rs}\text{12000}\)

## Question 6

Juhi sells a washing machine for \(\rm Rs\, 13,500.\) She loses \(20\%\) in the bargain. What was the price at which she bought it?

### Solution

**Video Solution**

**What is Known?**

Price at which Juhi sold the washing machine \(\rm (Rs \,13,500)\) and loss she made \((20\%).\)

**What is Unknown?**

The price at which Juhi bought the washing machine.

**Reasoning:**

This question can be solved by assuming the cost price of the washing machine as \(\rm Rs\, 100.\) Since loss is \(20\%,\) the selling price will be \(\rm Rs \,80.\) Now a simple logic can be used. If selling price is \(\rm Rs\, 80\) then cost price is \(\rm Rs\, 100.\) What will be the cost price if seeling price is \(\rm Rs \,13500.\)

**Steps:**

Let us assume cost of washing machine \(= \rm Rs\, 100\)

Loss \(= 20\%\) of cost price

So, the selling price will be

\[\begin{align}&= \text{Cost price - loss}\\&= 100\,{\rm{- 20}}\\&= \rm{Rs}\,{\rm{80}}\end{align}\]

If selling price is \(\rm Rs \,80,\) the cost price \(= \rm Rs \,100\)

If selling price is Rs \(13,500\) the cost price

\[\begin{align}&= \frac{{{\rm{100}}}}{{{\rm{80}}}} \times{\rm{13500}}\\ &= \,{\rm{Rs}}\,{\rm{16875}}\end{align}\]

## Question 7

(i) Chalk contains calcium, carbon and oxygen in the ratio \(10:3:12.\) Find the percentage of carbon in chalk.

(ii) If in a stick of chalk, carbon is \(3\rm g,\) what is the weight of the chalk stick?

### Solution

**Video Solution**

**What is Known?**

(i) Ratio in which chalk contains calcium, carbon and oxygen \((10:3:12). \)

(ii) Weight of carbon in stick of chalk \((3\rm g).\)

**What is Unknown?**

(i) The percentage of carbon in chalk.

(ii) what is the weight of the chalk stick if it contains \(3\rm g\) carbon.

**Reasoning:**

(i) To find the percentage of carbon, use the following formula

Percentage of carbon

\[\begin{align}=\frac{\text{Parts of carbon}}{\text{Sum of all parts}} \times 100\end{align} \]

(ii) the concept of percentage can be used find the total quantity.

**Steps:**

Parts of carbon \(= 3\) and sum of all parts \(= 10 + 3 + 12 = 25\)

Percentage of carbon

\[\begin{align}&= \frac{{{\rm{Parts}}\;{\rm{of}}\;{\rm{carbon}}}}{{{\rm{Sum}}\;{\rm{of}}\;{\rm{all}}\;{\rm{parts}}}} \times {\rm{100}}\\&=\frac{{\rm{3}}}{{{\rm{25}}}} \times {\rm{100}}\\&= 12\% \end{align}\]

(ii) Carbon in the stick of chalk \(= 3\rm g,\) and carbon percentage in chalk is \(12\%.\)

This means \(12\%\) of carbon \(= 3\rm g\)

Let total quantity be \(Q\)

\[{\rm{I}}{\rm{.E}}{\rm{.}}\frac{{{\rm{12}}}}{{{\rm{100}}\;}} \times {\rm{Q}}\,{\rm{ = }}\,{\rm{3}}\]\[{\rm{Total}}\,{\rm{Quantity}}\,{\rm{ = }}\,{\rm{3}} \times \frac{{{\rm{100}}}}{{{\rm{12}}}}\;\,{\rm{ = }}\,{\rm{25}}\,{\rm{g}}\]

## Question 8

Amina buys a book for \(\rm Rs \,275\) and sells it at a loss of \(15\%.\) How much does she sell it for?

### Solution

**Video Solution**

**What is Known?**

The price at which Amina buys a book \(\rm(Rs\, 275)\) and the loss percent \((15\%)\) she makes after selling it.

**What is Unknown?**

The price at which she sells the book.

**Reasoning:**

Loss \(=\) \(\begin{align}\frac{{\text{Loss}\% }}{{100}} \times \end{align}\) Cost Price

Selling Price \(=\) Cost price \(-\) Loss

**Steps:**

Cost of book \(= 275\) and loss \(= 15\%\) of Cost price

Total loss

\[= \frac{{15}}{{100\;}} \times 275 = {\text{Rs }}41.25\]

So,

Selling Price \(=\) Cost price \(-\) loss

\[\begin{align} &=\text{ }275-41.25 \\& =\text{Rs }233.75\end{align}\]

## Question 9

Find the amount to be paid at the end of \(3\) years in each case:

(a) Principal \(= \rm Rs\, 1,200\) at \(12\%\) p.a.

(b) Principal \(= \rm Rs\, 7,500\) at \(5\%\) p.a.

### Solution

**Video Solution**

**What is Known?**

Principal, Time and the rate of interest.

**What is Unknown?**

Amount to be paid in three years.

**Reasoning:**

Simple interest can be calculated using the formula

Simple Interest

\[\begin{align} & =\frac{\left[ \begin{align} & \text{Principal} \times \text{Rate of interest} \\ & \times \text{Time (in years)} \\ \end{align} \right]}{\text{100}\ } \\ \end{align}\]

And amount is obtained by adding Principal to the interest

**Steps:**

(a) Principal = \(\rm Rs\, 1,200\) at \(12\%\) p.a.

Simple Interest

\[\begin{align} & =\frac{\left[ \begin{align} & \text{Princial} \times \text{Rate of interest} \\ & \times \text{Time (in years)} \\ \end{align} \right]}{100} \\ & =\frac{1200 \times 12 \times 3}{100} \\ & =\text{Rs }432 \\ \end{align}\]

\[\begin{align}\text{Amount} &=\text{Principal + Interest }\\&= \text{Rs }1200 + \text{Rs }432 \\ &= \text{Rs }1632\end{align}\]

So, the amount to be paid after \(3\) years will be \(\rm Rs. \,1632\)

(b) Principal \(= \rm Rs \,7,500\) at \(5\%\) p.a.

Simple Interest

\[\begin{align}& =\frac{\left[ \begin{align} & \text{Princial} \times \text{Rate of interest} \\& \times \text{Time (in years)} \\ \end{align} \right]}{100} \\ & =\frac{7500 \times 5 \times 3}{100} \\ & =1125 \end{align}\]

\[\begin{align}\text{Amount} &= \text{Principal + Interest}\\&= 7500 +1125 \\&= 8625\end{align}\]

So, the amount to be paid after \(3\) years will be \(\rm Rs.\,8625\)

## Question 10

What rate gives \(\rm Rs \,280\) as interest on a sum of \(\rm Rs\, 56,000\) in \(2\) years?

### Solution

**Video Solution**

**What is Known?**

Principal, interest and number of years.

**What is Unknown?**

Rate of interest.

**Reasoning:**

Rate of Interest can be calculated using the formula

Simple Interest

\[\begin{align} & =\frac{\left[ \begin{align} & \text{Principal} \times \text{Rate of interest} \\ & \times \text{Time (in years)} \\ \end{align} \right]}{100} \\ \end{align}\]

**Steps:**

Let us assume that rate of interest to be \(\rm{R.}\)

So, Simple Interest

\[\begin{align}{\rm{ = }}\,\frac{{{\rm{Principal}}\,{\rm{ \times Rate}}\,{\rm{ \times }}\,{\rm{Time}}}}{{{\rm{100}}}}\end{align}\]

\[\begin{align}{\rm{I}}{\rm{.E}\,}{\rm{.\,280}}\, &=\frac{{{\rm{56000}} \times {\rm{R}}\, \times {\rm{2}}}}{{{\rm{100}}}}\\{\rm{R}} &= \frac{{280 \times 100}}{{56000 \times 2}}\\ &= 0.25\%\end{align}\]

So, the rate of interest is \(0.25\%\)

## Question 11

If Meena gives an interest of \(\rm Rs\, 45\) for one year at \(9\%\) rate p.a.. What is the sum she has borrowed?

### Solution

**Video Solution**

**What is Known?**

If Meena gives an interest of \(\rm Rs\, 45\) for one year at \(9\%\) rate p.a.

**What is Unknown?**

What is the sum she has borrowed?

**Reasoning:**

This question can be solved by using the formula of simple interest

So Simple Interest

\[\begin{align}{\rm{ = }}\frac{{{\rm{Principal}} \times {\rm{Rate}} \times {\rm{Time}}}}{{{\rm{100}}}}\end{align}\]

**Steps:**

Let us assume principal to be \(P\)

So Simple Interest

\[\begin{align}{\rm{ = }}\frac{{{\rm{Principal}}{\rm{ \times }}{\rm{Rate}}{\rm{ \times }}{\rm{Time}}}}{{{\rm{100}}}}\end{align}\]

\[\begin{align}{\rm{I}}{\rm{.E}}{\rm{.}}{\rm{45}}&= \frac{{{\rm{P}} \times {\rm{9}} \times {\rm{1}}}}{{{\rm{100}}}}\\{\rm{P}} &=\frac{{{\rm{45}}{\rm{ \times }}{\rm{1}}{\rm{00}}}}{{{\rm{9}}{\rm{ \times }}{\rm{1}}}}\\&= 500\end{align}\]

So, Meena borrowed \(\rm Rs\, 500\)

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