S.chand books class 8 maths solution chapter 8 Profit,Loss and Discount Exercise 8A

EXERCISE 8 (A)

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Q1 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 1

A peon purchased a chair for ₹ 700, spent ₹ 170 on its repair and ₹ 30 on the cartage.If he sold the chair for ₹ 1080, what is his gain percent?

Sol :

Cost price of chair = 700 + 170 + 30= 900

Selling price of chair = 1080

Gain = Selling price - Cost price

= 1080 - 900

= 180

Gain Percent $=\dfrac{\text{gain}}{\text{cost price}}\times 100$

$=\dfrac{180}{900}\times 100$

$=\dfrac{180}{9}$

= 20 %

 

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 Q2 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 2

Ramesh bought 10 cycles for ₹ 500 each.He spent ₹ 2000 on the repair of all the cycles.He sold five of them for ₹ 750 each and the remaining for ₹ 550each.What is total gain or loss %?

Sol :

Cost price of 1 cycle = 500 ₹

Cost price of 10 cycle = 500 × 10

= 5000 ₹

Total spending on repair = 2000 ₹

Total cost price = 5000 ₹ + 2000 ₹

= 7000 ₹

Selling price of 5 cycles for ₹ 750 = 5 × 750

= 3750 ₹

Selling price of 5 cycles for ₹ 550 = 5 × 550

= 2750 ₹

Selling price of 10 cycles = S.P of 5 cycles for 750 + S.P of 5 cycles for 550

= 3750 + 2750

= 6500 ₹

Loss = Cost price - Selling price

= 7000 - 6500

= 500 ₹

Loss percentage $=\dfrac{\text{loss}}{\text{Total cost price}}$

$=\dfrac{500}{7000}\times 100$

$=\dfrac{50}{7}$

$=7\dfrac{1}{7}$

 

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Q3 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 3

By selling an article for ₹ 960 a man incurrs a loss of 4%.What was the cost price of the article?

Sol :

Cost price of article $=\dfrac{100}{100-\text{loss percent}}\times\text{Selling Price}$

$=\dfrac{100}{100-4}\times 960$

= 1000

 

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Q4 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 4

If the selling price of 20 articles is equal to the cost price of 15articles, then what is the loss percentage ?

Sol :

Quantity=15

Let the CP of 15 articles=100

SP of 20 articles=CP of 15 articles=100

SP of 20 articles=100

20×(SP of 1 article)=100

(SP of 1 article)$=\frac{100}{20}$

∴SP of 15 articles=15×(SP of 1 article)

$=15 \times \frac{100}{20}=15\times \frac{10}{2}$

=75

As SP<CP . So, its

Loss=CP-SP

=100-75=25

Loss %$=\frac{\text{Loss}}{\text{CP}}\times 100$

$=\frac{25}{100}\times 100$

=25%

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Q5 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 5

A hawker gains the selling price of 4 ball-point pens on selling 1 dozen pens.What is his gain percentage?

Sol :

As we know 1 dozen = 12 pens

Let the selling price of of one ball point pen be x then the profit will be equal to 4x

And the selling price of 12 pens = 12 x

Then cost price of 12 pens is equals to (C.P = S.P - profit)

Cost price of 12 pens = Selling price - Profit

C.P = 12x - 4x

C.P = 8x

Then , $\text{profit }\% = \dfrac{\text{profit}}{C.P}\times 100$

$=\dfrac{4x}{8x}\times 100$

$=\dfrac{1}{2}\times 100$

= 50 %

 

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Q6 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 6

A man bought pencils at the rate of 6 for ₹ 4 and sold them at the rate of 4 for ₹ 6. What is his gain percentage in transaction?

Sol :

6 pencils are bought for ₹ 4

So, 1 is bought for $=\dfrac{4}{6}$

4 are sold for ₹ 6

So, 1 is sold for $=\dfrac{6}{4}$

Here, S.P > C.P , so , it is a gain

gain = S.P - C.P 

$=\frac{6}{4}-\frac{4}{6}$

LCM of 4 and 6 is 12

$=\frac{6\times 3-4\times 2}{12}$

$=\frac{18-8}{12}=\frac{10}{12}=\frac{5}{6}$

∴ gain percent $=\dfrac{\text{gain}}{\text{cost price}}\times 100$$=\dfrac{\frac{5}{6}}{\frac{4}{6}}\times 100$

$=\frac{5}{4}\times 100$

= 125%

 

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Q7 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 7

By selling a basket of mangoes for ₹ 105, a fruit seller loses 30%. For how much should he sell it to gain 40%?

Sol :

As we know

$\text{S.P}=\left(\dfrac{100-\text{loss} \%}{100}\right)\times\text{C.P}$  ....(i)

And here we know S.P = 105 ,

Loss % = 30 % and putting these in (i)

$\text{105}=\left(\dfrac{100-30}{100}\right)\times \text{C.P}$

$\text{105}=\left(\dfrac{70}{100}\right)\times \text{C.P}$

$\text{105}=\dfrac{7}{10}\times \text{C.P}$

$\text{C.P}=105 \times \dfrac{10}{7}$

C.P = 150

Gain % = 40 and putting these two in (ii) , we get

$\text{S.P}=\left(\dfrac{100+\text{Gain}\%}{100}\right)\times \text{C.P}$.....(ii)

$\text{S.P}=\left(\dfrac{100+40}{100}\right)\times 150$

$\text{S.P}=\dfrac{140}{100}\times 150$

S.P = 14 × 15

S.P = 210

 

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Q8 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 8

A man sold a book at a loss of 20%.If he had sold the books for ₹ 12 more,he would have gained 10% .Find the cost price of the book.

Sol :

We know that

Loss % = 20 % and in this case let the Selling Price bex

Putting these in (i)

$\text{S.P}=\left(\dfrac{100-\text{loss} \%}{100}\right)\times\text{C.P}$  .....(i)

$x=\dfrac{100-20}{100}\times \text{C.P}$

$x=\dfrac{80}{100}\times \text{C.P}$

$x=\dfrac{8}{10}\times \text{C.P}$

$\text{C.P}=\dfrac{10x}{8}$ ...(ii)

 

$\text{S.P}=\left(\dfrac{100+\text{Gain}\%}{100}\right)\times \text{C.P}$...(iii)

And here we know gain % = 10 % , in this case we have Selling Price = x + 12

Putting these two in (iii) we get

$x + 12 = \dfrac{100+10}{100}\times \text{C.P}$

$x + 12 = \dfrac{110}{100}\times \text{C.P}$

$x + 12 = \dfrac{11}{10}\times \text{C.P}$

$\text{C.P}=\dfrac{10}{11}\times (x+12)$ ...(iv)

From (ii) and (iv) , we get

$\text{C.P}=\dfrac{10x}{8}$

$\text{C.P}=\dfrac{10}{11}\times (x+12)$

Which means

$\dfrac{10x}{8}=\dfrac{10}{11}\times (x+12)$

$\dfrac{x}{8}=\dfrac{1}{11}\times (x+12)$

$\dfrac{x}{8}=\dfrac{x}{11}+\dfrac{12}{11}$

$\dfrac{x}{11}-\dfrac{x}{8}=-\dfrac{12}{11}$

$\dfrac{8x-11x}{88}=-\dfrac{12}{11}$

$-\dfrac{3x}{88}=-\dfrac{12}{11}$

$\dfrac{3x}{8}=12$

3x = 12 × 8

$x=\dfrac{96}{3}$

x = 32 ...(v)

putting (v) in (ii)

$\text{C.P}=\dfrac{10\times 32}{8}$

$\text{C.P}=\dfrac{320}{8}$

C.P = 40

 

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Q9 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 9

A trader sells two cycles at ₹ 1188 each, gaining 10% on the first cycle and losing 10% on the second cycle.What is the profit or loss percentage in the whole transaction ?

Sol :

Case 1 (First cycle)

$\text{C.P}=\dfrac{100}{100+\text{gain} \%}\times \text{S.P}$

$\text{C.P}=\dfrac{100}{100+10}\times 1188$

$\text{C.P}=\dfrac{100}{110}\times 1188$

$\text{C.P}=\dfrac{11880}{11}$

C.P = 1080

Case 2 (Second cycle)

$\text{C.P}=\dfrac{100}{100-\text{loss} \%}\times \text{S.P}$

$\text{C.P}=\dfrac{100}{100-10}\times 1188$

$\text{C.P}=\dfrac{100}{90}\times 1188$

$\text{C.P}=\dfrac{11880}{9}$

C.P = 1320

 

S.P = 1188 + 1188

S.P = 2376

 

C.P  = 1080 + 1320

C.P = 2400

Here , we come to know that S.P < C.P . So , it is a loss

Loss = C.P - S.P

Loss = 2400 - 2376

Loss = 24

$\text{Loss} \% = \dfrac{\text{loss}}{C.P}\times 100$

$\text{Loss }\% = \dfrac{24}{2400}\times 100$

Loss % = 1 %

 

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Q10 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 10

A merchant bought two calculator which together cost him ₹ 4800.He sold one of them at a loss 15% and the other at a gain of 19%.If the selling price of both the calculator is equal, find the cost price of the lower priced calculator.

Sol :

Let the  S.P. be x

We know that $\text{C.P}=\left(\dfrac{100}{100-loss\%}\right)\times\text{S.P}$

also $\text{C.P}=\left(\dfrac{100}{100+gain\%}\right)\times \text{S.P}$

C.P. of 1st calculator $=\left(\dfrac{100}{100-15}\right)\times x$

$=\dfrac{20x}{17}$

C.P. of 2nd calculator $=\left(\dfrac{100}{100+19}\right)\times x$

$=\dfrac{100x}{119}$

Now , C.P. of 1st calculator + C.P. of 2nd calculator = Total cost

$=\dfrac{20x}{17} + \dfrac{100x}{119}=4800$

$=\dfrac{240x}{119}=4800$

x = 2380

C.P. Of 1st calculator $=\dfrac{20}{17}\times 2380$= 2800

C.P. of 2nd calculator $=\dfrac{100}{119}\times 2380$ = 2000

C.P. of lower priced calculator = Rs. 2000

 

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Q11 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 11

A manufacturer sells an article to a wholesale dealer at a profit of10%.The wholesale dealer sells it to a shopkeeper at 20% profit.The shopkeeper sells it to the customer for ₹ 56,100 at a loss of 15%.What is the cost price of the article to the manufacturer?

Sol :

Given : Selling price of article = 56,100 and at loss of 15 %

We know that $\text{C.P}=\left(\dfrac{100}{100-loss\%}\right)\times\text{S.P}$

$\text{C.P}=\left(\dfrac{100}{100-15}\right)\times 56100$

$\text{C.P}=\left(\dfrac{100}{85}\right)\times 56100$

Cost price of article for shopkeeper =  66,000 (Shopkeeper's cost price)

 

Now , lets find C.P of article for the wholesale dealer . In this case S.P =66000 and at a gain of 20 %

$\text{C.P}=\left(\dfrac{100}{100+gain\%}\right)\times \text{S.P}$

$\text{C.P}=\left(\dfrac{100}{100+20\%}\right)\times \text{66000}$

= 55000

 

Now , lets find C.P of article for manufacturer . In this case S.P = 55000 andat a gain of 10 %

$\text{C.P}=\left(\dfrac{100}{100+ gain\%}\right)\times \text{S.P}$

$\text{C.P}=\left(\dfrac{100}{100+10\%}\right)\times \text{55000}$

= 50000

Cost price of the article to the manufacturer = 50,000

 

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Q12 | Ex-8A | Class 8 | S.Chand | Composite maths | Profit,Loss and Discount |Ch-8 | myhelper

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QUESTION 12

A man buys a plot of agricultural land for ₹ 3,60,000. He sells one third of it at a loss of 25% and two fifth at a gain of 20%.At what price must he sell the remaining field so as to make an overall profit of 10% ?

Sol :

C.P = 360000

To gain 10% on whole land ,

S.P = 360000 + 10% of 360000

= 396000

$\dfrac{1}{3}$ of the land sold on 20% loss

S.P of $\dfrac{1}{3}$ land

$=\left(\dfrac{360000}{3}\right)-20\% \text{ of }\left(\dfrac{360000}{3}\right)$

= 96000

S.P of $\dfrac{2}{5}$ of the land

$=\dfrac{360000\times 2}{5}+25\% \text{ of } \dfrac{360000\times 2}{5}$

= 180000

Thus , S.P of the remaining land

= 39600 - 96000 - 180000

= 120000

ALTERNATE METHOD

S.P of total agricultural field at a profit of 10% $=\dfrac{360000\times110}{100}$

= 396000

So , S.P of $\dfrac{1}{3}$ of field

$=\left(\dfrac{360000}{3}\right) \times \left(\dfrac{80}{100}\right)$

= 96000

S.P of $\dfrac{2}{5}$ of the field

$=\dfrac{2\times360000\times 125}{5\times 100}$

= 180000

Hence ,

S.P of the remaining field

= 396000-96000-180000

= 120000

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