8 . Comparing Quantities

Chapter 8 . Comparing Quantities

8. Comparing Quantities

Exercise 8.1

1. Find the ratio of the following: (a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour. (b) 5 m to 10 km (c) 50 paise to Rs 5 .

Solution: (a) We have,

The ratio =

(b) We have,

The ratio =

[Note : 1km = 1000 m ]

The ratio =

[Note : Rs 1 = 100 Paise ]

2. Convert the following ratios to percentages: (a) 3 : 4 (b) 2 : 3

Solution: (a) We have,

(b) We have,

3. 72% of 25 students are interested in mathematics. How many are not interested in mathematics?

Solution: The students are interested in mathematics

The students are not interested in mathematics

Therefore, the percentage of the students

4. A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all ?

Solution: Total number of matches = 100 .

Therefore, the number of matches

5. If Chameli had Rs 600 left after spending 75% of her money, how much did she have in the beginning ?

Solution: Suppose , the money of Chameli is Rs 100 .

The money spent by Chameli

The money left after spent by chameli = Rs 100 – Rs 75 = Rs 25

Therefore, the money of Chemali in beginning

6. If 60% people in a city like cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.

Solution: Total number of people = 50,00,000 .

The percentage of other game = (100 – 60 – 30)% = (100 – 90) % = 10 %

The number people who like cricket

The number people who like football

The number people who like other game

Exercise 8.2

1. A man got a 10% increase in his salary. If his new salary is Rs 1,54,000, find his original salary.

Solution: Let be the original salary .

Given, new salary (A) = Rs 154000 and increase in salary (R) = 10 %

We know that,

Therefore, the original salary is Rs 1,40,000 .

2. On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the Zoo on Monday?

Solution: Given , the number of visitor on Sunday = 845 and on Monday = 169 .

The decrease the number of visitor = 845 – 169 = 676

The per cent decrease in the people visiting the Zoo on Monday

3. A shopkeeper buys 80 articles for Rs 2,400 and sells them for a profit of 16%. Find the selling price of one article.

Solution: Price of each articles

Now , Profit

Thus , the selling price of one article = Rs (30+4.8) = Rs 34.8

4. The cost of an article was Rs 15,500. Rs 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.

Solution: The total cost of an article = The cost of an article + The cost of repairs

= Rs (15500 +450) = Rs 15950

Profit

Thus , the selling price of the article = Rs (15950 + 2392.50) = Rs 18342.50

5. A VCR and TV were bought for Rs 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.

Solution: Given, Price of a VCR = Rs 8000 and loss 4 % (This means cost price is Rs 100 , then selling price is Rs (100 – 4) =Rs 96).

Selling price of a VCR

Again , price of a TV = Rs 8000 and profit 8% (This means cost price is Rs 100 , then selling price is Rs (100 + 8) =Rs 108).

Selling price of a VCR

Total cost price = Rs 8000 +Rs 8000 = Rs 16000

And total selling price = Rs 7680 + Rs 8640 = Rs 16320

Since , total SP > total CP

Profit = Rs (16320 - 16000) = Rs 320

Therefore , the gain percent on the whole transaction

6. During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs 1450 and two shirts marked at Rs 850 each?

Solution: Marked price = Cost of a jeans + Cost of two shirts

= Rs 1450 +Rs 2×850

=Rs (1450 + 1700) = Rs 3150

and the discount percentage = 10 %

Discount

The sale price = Marked price – discount = Rs (3150 – 315) = Rs 2835

Therefore , the customer has to pay Rs 2835 .

7. A milkman sold two of his buffaloes for Rs 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint: Find CP of each)

Solution: The cost price of each buffaloes = Rs 20,000 and gain 5% .

This means if CP is Rs 100 then SP is Rs (100 + 5 )= Rs 105 .

Selling price

Again , the cost price of each buffaloes = Rs 20,000 and loss 10 % .

This means if CP is Rs 100 then SP is Rs (100 – 10 )= Rs 90 .

The cost price

Total selling price = Rs (20000 + 20000) = Rs 40000

And total cost price = Rs (19047.62 + 22222.22) = Rs 41269.84

Since total cost price > selling price .

Loss = Rs (41269.84 – 40000) = Rs 1269.84

8. The price of a TV is Rs 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.

Solution: The price of a TV = Rs 13000 and rate 12% .

On Rs 13000 , the tax paid would be

Therefore, the amount that Vinod will have to pay = The Price of a TV + Sales tax

= Rs (13000 + 1560) = Rs14560

9. Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is Rs 1,600, find the marked price.

Solution: let the marked price of the skates is Rs 100 .

Then selling price = Rs (100 – 20) = Rs 80

The marked price

10. I purchased a hair-dryer for Rs 5,400 including 8% VAT. Find the price before VAT was added.

Solution: Let the original price of the article be Rs 100 and VAT = 8 %

Price after VAT in including = Rs 108 .

When the selling price is Rs 118 then original price = Rs 100 .

When the selling price is Rs 5400 then original price

Therefore, the price before VAT was Rs 5000 .

11. An article was purchased for Rs 1239 including GST of 18%. Find the price of the article before GST was added ?

Solution: Let the original price of the article be Rs 100 and GST = 18 %

Price after GST in included = Rs (100+18) = Rs 118

When the selling price is Rs 118 then original price = Rs 100 .

When the selling price is Rs 1239 then original price

Exercise 8.3

1. Calculate the amount and compound interest on
(a) Rs10,800 for 3 years at

% per annum compounded annually.
(b) Rs 18,000 for

years at 10% per annum compounded annually.
(c) Rs 62,500 for

years at 8% per annum compounded half yearly.
(d) Rs 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).
(e) Rs 10,000 for 1 year at 8% per annum compounded half yearly.

(a) Rs10,800 for 3 years at % per annum compounded annually.

Solution: Here , , ,

We have ,

Amount = Rs 15377.34

Compound interest = A – P

= Rs(15377.34 – 10800)

= Rs 4577.34

(b) Rs 18,000 for years at 10% per annum compounded annually.

Solution: Here, , , ,

We have ,

Simple interest for 2 years = Rs (21780 – 18000)

= Rs 3780

For next half year ,

Simple Interest

Compound interest = Rs (3780 + 1089) = Rs 4869

Amount = P + C.I = Rs (18000 + 4869) = Rs 22869

Solution: Here, , ,

Year

We know that,

C.I = A – P = Rs (70304 – 62500) = Rs 7804
(d) Rs 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify).

Solution: Here , ,

and Year

We know that,

Compound interest = A – P = Rs (8736.20 – 8000)

= Rs 736.20

(e) Rs 10,000 for 1 year at 8% per annum compounded half yearly.

Solution: Here, , ,

Year, ,

We know that ,

Compound interest = Rs (10816 – 10000)

= Rs 816 .

2. Kamala borrowed Rs 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for

years).

Solution: Here , P = Rs 26,400 , R = 15 % ,and 2 years 4 months

Firstly , for 2 years in CI ,

We have ,

S.I for 2 years = Rs (34914 – 26400) = Rs 8514

Next , for 1/3 year in SI .

Simple interest

C.I. = Rs (8514 + 1745.70) = Rs 10259.70

Amount = P + CI = Rs (26400 + 10259.70)

= Rs 36659.70

3. Fabina borrows Rs 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Solution: Here , P = Rs 12500 , R = 12 % ,n = 3

For Simple Interest ,

S.I

For CI : Here, P = Rs 12500 , R = 10 % ,n = 3

Amount

Compound interest = A – P

= Rs (16637.50 – 12500)

= Rs 4137.50

Fabina pay more interest than Radha .

Fabina pay = Rs (4500 – 4137.50 ) = Rs 362.5

4. I borrowed Rs 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

Solution: Here , P = Rs 12000 , R = 6 % ,n = 2

For Simple Interest ,

Simple interest

For CI : Here, P = Rs 12000 , R = 6 % ,n = 2

Amount

Compound interest = A – P

= Rs (13483.20 – 12000)

= Rs 1483.20

Pay extra amount = Rs (1483.20 – 1440 )

= Rs 43.20

5. Vasudevan invested Rs 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get: (i) after 6 months? (ii) after 1 year?

Solution: (i) Here , P = Rs 60000 , R = 12% ,

n = 6 month

Simple interest

Amount = P + S.I = Rs (60000 + 3600) = Rs 63600

(ii) Here, P = Rs 60000 , R =12 % and n = 1year

Therefore, the amount is Rs 67416 .

6. Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after

years if the interest is
(i) compounded annually. (ii) compounded half yearly.

Solution: (i) Here , P = Rs 80,000 , R = 10%

and and half years

For 1 year ,

We know that ,

Interest of 1 year = Rs(88000 – 80000) = Rs 8000

For half year ,

Simple interest

Total Compound interest = Rs (8000 + 4400)

= Rs 12400
Amount = P + C.I. = Rs (80000+12400) = Rs 92400

(ii) Here , P = Rs 80000 , R=10%

and year

We know that ,

7. Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find :
(i) The amount credited against her name at the end of the second year. (ii) The interest for the 3rd year.

Solution: Her, P = Rs 8000 , R = 5 %

and n = 2 years

We know that,

(ii) Here, P = Rs 8000 , R = 5 % and n = 3 years

We know that,

Compound interest = A – P = Rs (9261 – 8820)

= Rs 441

8. Find the amount and the compound interest on Rs 10,000 for

years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Solution: Here , P = Rs 10000 , R = 10%

and

We know that,

Compound interest = Rs (11576.25 – 10000)

= Rs 1576.25

For 1 year (i.e., annually) , then n = 1 year

and R = 10 %

Simple interest

Here , P = Rs 10000 + Rs 1000 = Rs 11000

Simple interest for :

Total interest = Rs (1000+550) = Rs 1550

The difference interest = Rs 1576.25 – Rs 1550

= Rs 26.25

Yes , this interest is more than the interest he would get if it were compounded annually .

9. Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at

% per annum, interest being compounded half yearly.

Solution: Here, P = Rs 4096 ,

and

We know that,

10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001. (ii) what would be its population in 2005?

Solution: let be the population in 2001 .

Given , the population in 2003 is 54,000

Here , A = 54000 , R = 5 %

and n = 2003 – 2001 = 2 years

We know that,

Therefore, the population in 2001 is 48,980 .

(ii) let be the population in 2005 .

Given , the population in 2003 is 54,000 .

Here , P = 54000 , R = 5 % ,

n = 2005 – 2003 = 2 years

We know that,

Therefore, the population in 2005 is 59,535 .

11. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5, 06,000.

Solution: Here , P = 506000 , R = 2.5%

and n = 2 hours

We know that ,

Therefore, the bacteria at the end of 2 hours is 531616 (approx)

12. A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Solution: Here, P = Rs 42000 , R = 8%

and n = 1 year

The value of depreciation

Therefore, the value after one year

= Rs (42000 – 3360) = Rs 38,640