Answerkey for Class 8 Practice Test for the Chapters Profit and Loss and Percentage

 Answer Key Practice  Test -1

Class- 8                                            Subject - Maths                                     Marks : 50 marks  

Chapters - Profit and Loss & Percentage                                    

Section A  

1. Waheeda bought an air cooler for Rs. 3300 including a tax of 10%. The price of the air cooler before VAT was added is:

Explanation:10% VAT on Rs.100 will make it Rs.110

So, for price including VAT Rs.110, the original price is Rs.100

Then, Price including VAT Rs. 3300, the original price = Rs. (100/110) x 3300 = Rs. 3000.

2. A shopkeeper purchased 300 bulbs for Rs 10 each. However 10 bulbs were fused and had to be thrown away. The remaining were sold at Rs 12 each. Find the gain or loss %

Answer:  C 16%

3. A shopkeeper purchased 500 pieces for Rs 20 each. However 50 pieces were spoiled in the way and had to be thrown away. The remaining were sold at Rs 25 each. Find the gain or loss %. 

Answer: 12.5%

4. The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find the population after 3 years. 

Answer: 29484

5. A trader marks his goods at 40% above the cost price and allows a discount of 25%. What is his gain?

Answer: Let the cost price be Rs. 100.

Then, the marked price =100+40100×100 = Rs. 140

Discount = 25% of 140=25100×140=1404 = Rs. 35

Selling price = marked price – discount = 140 –35 = Rs. 105

Profit percentage =Profit /C.P×100=105−100100×100=5%

6.   A dealer purchased a washing for Rs. 7660. He allows a discount of 12% on its marked price and still gains 10%. Find the marked price of the machine.

Answer:  profit = 7660x 10/100=766

7660+766=8426

x= (100 x  8426)/88 = 9575

7. A vendor bought oranges at Rs. 20 for Rs. 56 and sold them at Rs. 35 per dozen. Find his gain or loss percent.

Answer: 

Let the number of oranges bought = LCM of 20 and 12=60. 

∵12C.P. of 20 oranges = Rs.56

∴ C.P. of 1 orange = Rs. 56/20

C.P. of 60 oranges = Rs.(56/20×60)= Rs. 168

S.P. of 12 oranges = Rs. 35

 S.P. of 1 orange = Rs. 35/12

S.P. of 60 oranges = (3512×60)=Rs.175

C.P. = Rs.168 and S.P. = Rs. 175

Gain =(175−168)= Rs. 7

Gain % = (GainC.P.×100)%=(7168×100)%=25/6%

8. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?

Answer: 12

9. A picnic is being planned in a school for CLASS VIII. Girls are 60% of the total number of students and are 18 in number. The ratio of the number of girls to the number of boys in the class 

Answer: 18:12 =3:2

10. Divide 15 sweets between Shiva and Rama so that they get 20 % and 80 % of them respectively.

Answer:  3 sweets and 12 sweets

Section B Profit & Loss

1. 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. 

Solution : 

C.P of two boxes = Rs 1300

 let C.P of one box be Rs x

C.P of other box = Rs 1300 – x

S.P of first box = x + x × ${20}/{100}$  = x + ${x}/{5}$ = Rs 6x/5

S.P of second box = (1300 – x) – (1300 – x) × 12/100

= Rs (28600 – 22x)/25

Equate S.P of first and second box,

6x/5 = (28600 – 22x)/25

150x = 28600 × 5 – 110x

150x + 110x = 28600 × 5

260x = 28600 × 5

x = (28600 × 5)/260 = 550

∴ C.P of first box = Rs. 550

C.P of second box = Rs 1300 – Rs 550 = Rs 750

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

Solution:

S.P of 10 pens = cost price of 14 pens

 let C.P of 1 pen be Rs x

S.P of 10 pens = Rs 14x

S.P of 1 pen =Rs 14x/10

Gain = S.P – C.P

= 14x/10 – x

= 4x/10

Gain % = (gain/cost price) × 100

= (4x/10)/x × 100

= 2/5 × 100

= 40%

3. 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 profit of 20%?

Solution:

S.P for 90 ball pens is = Rs 160

S.P of 1 ball pen = Rs 160/90 = Rs 16/9

The loss percent is = 20% of 100 = 20/100

Let C.P of 1 pen be Rs x

x – x×20/100 = 16/9

x – x/5 = 16/9

(5x-x)/5 = 16/9

4x/5 = 16/9

4x×9 = 16×5

36x = 80

x = 80/36

= Rs 20/9

C.P of 1 ball pen = Rs 20/9

To get a profit of 20%

Let number of pens be ‘x’

S.P of ‘x’ pens is = Rs 96

Selling price of 1 pen is = Rs 96/x

Gain % = (gain/cost price) × 100

20% = [(96/x) – (20/9)] / (20/9) × 100

20/100 = [(96/x) – (20/9)] / (20/9) × 100

(20/100 × 200/9) + 200/90 = 96/x

4/9 + 200/90 = 96/x

(40+200)/90 = 96/x

240/90 = 96/x

24/9 = 96/x

24x = 96×9

x = 864/24 = 36

∴ 36 ball pens can be sold at a price of Rs 96

4. 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.

Solution: 

Solution:

Let us consider the cost price of 1000gm pulses be Rs x

Selling price of 950 gm pulses is also = Rs x

Selling price of 1000 gm pulses = x/950 × 1000

So, Gain = SP – CP

Gain = 1000x/950 – x

= (1000x – 950x)/950

= 50x/950

Gain % = (gain/cost price) × 100

= (50x/950)/x × 100

= 50x/950x × 100

= 5/95 × 100

= 100/19

= 5 5/19 %

∴ The Shopkeeper’s gain percent is

5 5/ 19%

5. 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.

Solution:

(i) Let the CP of shirt be = Rs x

Profit (4%) = 4/100 of CP

= 4/100 × x

= 4x/100

Selling Price = C.P + Profit

= x + 4x/100

= (100x + 4x)/100

= 104x/100

(ii) Let the CP of shirt be = Rs x

Profit (5%) = 5/100 of CP

= 5/100 × x

= 5x/100

Selling Price = C.P + Profit

= x + 5x/100

= (100x + 5x)/100

= 105x/100

difference between the two selling price is Rs 6

 105x/100 – 104x/100 = 6

(105x-104x)/100 = 6

x/100 = 6

x = 600

∴ Now, C.P of the shirt is = Rs 600

Selling Price of one shirt = 104x/100 = (104×600)/100 = Rs 624

Selling Price of other shirt = 105x/100 = (105×600)/100 = Rs 630

 

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.

Solution:

We know the cost price of cooler = Rs 1200

Overhead expenses = Rs 40

Total cost = Rs 1200 + Rs 40 = Rs 1240

Selling price of cooler = Rs 1550

By using the formula,

Gain = selling price – cost price

= Rs 1550 – Rs1240

= Rs 310

By using the formula,

Gain % = (gain/cost price) × 100

= 310/1240 × 100

= 25%

7.  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.

Solution:

Let us consider the cost price of motorcycle for Ravish be Rs x

Loss% for Ravish = 28%

Selling price for Ravish = x – x × 28/100 = (100x – 28x)/100 = 72x/100

= Rs 18x/25

Selling price for Ravish = cost price for Vineet = Rs 18x/25

Repair cost by Vineet = Rs 1680

Total cost price of the motorcycle for Vineet = Rs18x/25 + Rs 1680

Selling price for Vineet = Rs 35910

Profit = 35910 – (18x+42000)/25

= Rs (855750 – 18x)/25

Profit % = 12.5% (Given)

By using the formula,

Gain % = (gain/cost price) × 100

=> [(855750-18x)/25] / [(18x+42000)/25] × 100 = 12.5

=> [(855750-18x)/25] × [25/(18x+42000)] = 125/1000

=> (855750-18x) / (18x+42000) = 1/8

=> By cross multiplying we get

=> 8(855750-18x) = (18x+42000)

=> 6846000 – 144x = 18x + 42000

=> 6846000 – 42000 = 18x + 144x

=> 162x = 6804000

x = 6804000/162

= 42000

∴ Cost price of motorcycle for Ravish = Rs 42000

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

Solution:

Given details are,

Selling price of book is = Rs 258

The man’s gain percent is = 20% of 100 = 20/100

So, let us consider the cost price of book be Rs x

By solving,

x + x×20/100 = 258

x + x/5 = 258

(5x+x)/5 = 258

By cross multiplying

6x= 5×258

x = 1290/6

= 215

Now, the cost price of book is = Rs 215

For a gain of 30% the man should sell the book at = 215 + 215×30/100

= 215 + 64.5

= 279.50

∴ To gain 30% the man should sell the book at Rs 279.50

9. 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.

Solution:

Given, cost price of the defective briefcase is = Rs. 800

The loss percent is = 8% of 100 = 8/100

Selling price of briefcase is = 800 – 800×8/100

= 800 – 64

= Rs 736

When the price is further reduced by 5% (Given) = 5% of 100 = 5/100

New selling price = 736 – 736×5/100

= 736 – 36.8

= Rs 699.2

∴ The selling price of the defective briefcase is Rs 699.2

10. 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.

Solution: Let us consider the cost price of tricycle be = Rs x

Selling price of the tricycle be = Rs x

Given, Gain% = 16% of 100 = 16/100

Selling price of tricycle = x + x×16/100

= (100x+16x)/100

= 116x/100

= 29x/25

Given, if selling price is Rs 100 more

New Selling price = 29x/25 + 100

Then, Gain% = 20%

By using the formula

Gain % = (gain/cost price) × 100 [by using Gain = SP – CP]

20 = [((29x/25)+100) – x] / x × 100

20x/100 = (29x + 2500 – 25x)/25

x/5 = (29x + 2500 – 25x)/25

5x = 4x + 2500

x = 2500

∴ Cost price of tricycle is Rs 2500

Percentage

1. A coolie deposits Rs 150 per month in his post office Savings Bank account. If this is 15% of his monthly income, find his monthly income.

Solution:

Let the monthly income of a coolie be ‘x’

Here, coolie deposits Rs 150 per month which is 15% of his monthly income

From the above we can derive that,

x × (15/100) =150

Which implies, x = (150 × 100) / 15

x = 15000/15

x = 1000

∴ Monthly income is Rs 1000

2. Asha got 86.875% marks in the annual examination. If she got 695 marks, find the total number of marks of the examination.

Solution:

Given, Marks scored by Asha is 695

And Percentage of marks Asha got is 86.875%

Now,

Let the total marks be ‘x’

From the above we can derive that,

x × (86.875/100) =695

Here, 86.875 can be expressed as 86875/1000

Which implies, x = (695 × 100 × 1000) / 86875

x = 6950000/86875 = 800

Total no: of Marks, x = 800

∴ Total number of marks is 800marks

3. Deepti went to school for 216 days in a full year. If her attendance is 90%, find the number of days on which the school was opened.

Solution:

Given, number of days Deepti went to school = 216 days

Deepti Attendance percentage is = 90%

So, let the number of days when school remained opened be x days

Hence,

(x × 90)/100 = 216

By using cross multiplication we get,

x = (216×100)/90

= 240 days

∴ Number of days the school remained opened for 240 days

4. Balanced diet should contain 12% of proteins, 25% of fats and 63% of carbohydrates. If a child needs 2600 calories in this food daily, find in calories the amount of each of these in his daily food intake.

Solution:

The given details are,

Amount of calorie daily needed = 2600 calorie

Amount of protein needed = 12% of 2600

= (12/100) × 2600

= 312 calorie

Amount of fats needed = 25% of 2600

= (25/100) × 2600

= 650 calorie

Amount of carbohydrate needed = 63% of 2600

= (63/100) × 2600

= 1638 calorie

∴ Amount of calories required in protein is 312 calories, fat is 650 calories and carbohydrates is 1638 calories

5. A cricketer hit 120 runs in 150 balls during a test match. 20% of the runs came in 6’s, 30% in 4’s, 25% in 2’s and the rest in 1’s. How many runs did he score in (i) 6’s (ii) 4’s (iii) 2’s (iv) singles What % of his shots were scoring ones?

Solution:

The given details are,

Total number of runs scored by cricketer = 120

(i) 20% of Runs scored in 6’s = (20/100) × 120 = 24 runs

 

(ii) 30% of Runs scored in 4’s = (30/100) × 120 = 36 runs

 

(iii) 25% of Runs scored in 2’s = (25/100) × 120 = 30 runs

 

(iv) Runs scored in singles = 120 – (24+36+30)

= 120 – 90 = 30 runs

Percentage of shots scoring ones = (Runs came in singles/Total runs scored) × 100

= (30/120) × 100

= 25%

6. An alloy contains 32% copper, 40% nickel and rest zinc. Find the mass of the zinc in 1 kg of the alloy.

Solution:

Given details are,

Alloy contains, 32% of copper

40% of nickel

Remaining zinc

Mass of alloy = 1kg = 1000 grams

Mass of copper in alloy = (1000/100) × 32

= 320 grams

Mass of nickel in alloy = (1000/100) × 40

= 400 grams

So, mass of zinc in alloy = 1000 – (320 + 400)

= 1000 – 720

= 280 grams

∴ Mass of zinc in 1kg of alloy is 280 grams

7. Rs 3500 is to be shared among three people so that the first person gets 50% of the second, who in turn gets 50% of the third. How much will each of them get?

Solution:

We know that the total money to be shared is = Rs 3500

Let us consider third person get = Rs x

So, second person gets (50% of third) = 50% of x

= 50/100 × x

= Rs x/2

Now, first person gets (50% of second) = 50% of x/2

= 50/100 × x/2

= Rs x/4

We know that,

x/4 + x/2 + x = 3500

by taking 4 as LCM

(x+2x+4x)/4 = 3500

By cross multiplying

x+2x+4x = 3500 × 4

7x = 14000

x = 14000/7

= 2000

∴ Each of the person gets,

First person (x/4) gets = x/4 = 2000/4 = Rs 500

Second person (x/2) gets = x/2 = 2000/2 = Rs 1000

Third person (x) gets = x = Rs 2000

8. The population of a town increases by 10% annually. If the present population is 60000, what will be its population after 2 years?

Solution:

Given details are,

Present population of town is = 60000

Annually population increases by = 10%

So, Population after 2 years = present population × [(100 + increased %)/100] years

= 60000 × (100+10)/100 × (100+10)/100

= 60000 × 110/100 × 110/100

= 60000 × 11/10 × 11/10

= 72600

∴ After 2 years population will be 72600

9. The value of a machine depreciates every year by 5%. If the present value of the machine be Rs 100000, what will be its value after 2 years?

Solution:

Given details are,

Present value of machine is = Rs 100000

Every year the depreciation in price is = 5%

So, value after two years = 100000 × (100-5)/100 × (100-5)/100

= 100000 × 95/100 × 95/100

= 90250

∴ Value of machine after two years is Rs 90250

10. A motorist travelled 122 kilometers before his first stop. If he had 10%of his journey to complete at this point, how long was the total ride?

Solution:

Given details are,

Motorist total distance travelled before first stop = 122 km

Journey completed at first stop = 10 %

Let us consider total ride to be travelled be ‘x’ km

So, by calculating

(x/100) × 10 = 122

By cross multiplying we get,

x/100 = 122/10

x = (122 × 100)/10

= 1220 km

∴ Motorist total ride is 1220 km