8 Math Comparing Quantities

Finish Line & Beyond

Comparing Quantities 1. Discount is a reduction given on marked price. Discount = Marked Price – Sale Price. 2. Discount can be calculated when discount percentage is given. Discount = Discount % of Marked Price 3. Additional expenses made after buying an article are included in the cost price and are known as overhead expenses. CP = Buying price + Overhead expenses 4. Sales tax is charged on the sale of an item by the government and is added to the Bill Amount. Sales tax = Tax% of Bill Amount 5. Compound interest is the interest calculated on the previous year’s amount (A = P + I), where A is the amount, P is principal and I is interest. 6. (i) Amount when interest is compounded annually A=P(1+

R t ) 100

Where, P is principal, R is rate of interest, t is time period (ii) Amount when interest is compounded half yearly A=P(1+

R t ) 200

Where t is number of half years.

Exercise 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. Answer: Required Ratio =

15 1 = =1:2 30 2

(b) 5 m to 10 km Answer: Required Ratio =

5 1 = =1:2 10 2

(c) 50 paise to Rs 5 Answer: Rs 5 = 500 paise

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Finish Line & Beyond Ratio =

50 1 = =1:10 500 10

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

3 4

On multiplying by 100 you get:

3 × 100 = 75% 4

(b) 2 : 3 Answer: 2:3=

2 3

On multiplying by 100 you get:

2 2 × 100 = 66 % 3 3

3. 72% of 25 students are good in mathematics. How many are not good in mathematics? Answer: Percentage of Students not good in math = 100-72=28 Number of Students not good in Math =

25 ×

28 = 7 100

Or, you can calculate as follows also:

25 ×

72 = 18 Number of Students good in math 100

Number of Students not good in Math = 25-18=7 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? Answer: Let us assume total number of matches = x Then, x × 40% = 10

⇒ x = 10 ×

100 = 25 is the total number of matches played. 40

Alternate method:

 When the team won 40 matches, number of total match is 100 100 ∴ When the team won 1 match, number of total match is 40 100 ∴ When the team won 10 matches, number of total match is × 10 = 25 40

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Finish Line & Beyond 5. If Chameli had Rs 600 left after spending 75% of her money, how much did she have in the beginning? Answer: Money left after spending 75% = 100-75 = 25% Let us assume total money = x Then, x × 25% = 600 ⇒

x = 600 ×

100 = 2400 25

Chameli had Rs. 2,400 in the 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 are 50 lakh, find the exact number who like each type of game. Answer: Cricket Fans=60% Football Fans = 30% So, Other Sports’ Fans = 100-(60+30)=10% people like other games.

60 = 30,00,000 100 30 = 15,00,000 Number of Football Fans = 50,00,000 × 100 10 = 5,00,000 Number of Other Sports Fans = 50,00,000 × 100 Number of Cricket Fans =

50,00,000 ×

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Finish Line & Beyond Exercise 2 1. A man got a 10% increase in his salary. If his new salary is Rs 1,54,000, find his original salary. Answer: Let us assume the initial salary =100 So, salary after hike = 100+100 × 10%=100+10=110

 When increased salary is 110, then initial salary is 100 100 ∴ When increased salary is 1, the initial salary is 110 100 ∴ When increase salary is 154000, the initial salary is × 154000 = 140000 110 Alternate Method: The question can be solved by assuming x for initial salary, as follows:

110 = 154000 100 100 ⇒ x = 154000 × = 140000 110 x×

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? Answer: Actual decrease in number of visits = 845-169=676 Percentage Decrease =

676 × 100 = 80% 845

Alternate method:

 When original number of visitors is 845, the decrease is 676 676 ∴ When original number of visitors is 1, the decrease is 845 676 ∴ When original number of visitors is 100, the decrease is × 100 = 80% 845 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. Answer: Profit = CP × % profit

= 2400 ×

16 = 384 100

So, SP = CP+Profit = 2400+384=2784

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Finish Line & Beyond SP of one article =

2784 = 34.80 80

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. Answer: CP=Cost+Overheads =15500+450=15950 Profit = CP × % profit

= 15950 ×

15 = 2392.50 100

So, SP = CP+Profit =15950+2392.50=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. Answer: Loss on VCR = CP × % Loss

4 = 320 100 Profit on TV = CP × %Profit 8 8000 × = 640 100 = 8000 ×

Overall Profit = 640-320=320 Percentage Profit =

=

Pr ofit × 100 CP

320 × 100 = 2% 16000

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? Answer: Total Price of a pair of jeans and two shirts =1450+2 × 850=3150 Discount = Marked Price × % Discount

= 3150 ×

10 = 315 100

Effective Price After Discount = 3150-315=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. Answer: Suppose the CP of each buffalo = 100 After 5% profit, SP = 105

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Finish Line & Beyond After 10% loss SP = 90 In case of profitable deal, CP = 20000 ×

= 20000 ×

20 21

Net Profit

= 20000 − 20000 ×

=

420000 − 400000 20000 = 21 21

20 21

In case of loss making deal, CP =

= 20000 ×

100 105

20000 ×

100 95

20 19

20 − 20000 19 400000 − 380000 20000 = = 19 19 20000 20000 − Overall Profit/Loss = 21 19 380000 − 420000 400000 = = − 399 399 Net Loss =

= 20000 ×

As 399~400, so the net loss is approx. Rs. 1000 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. Answer: Sales Tax = Price × Tax Rate

= 13000 ×

12 = 1560 100

Final Price = 13000+1560 = 14560 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. Answer: Marked Price =

=

100 × FinalPr ice 100 − Discount

100 100 × 1600 = × 1600 = 2000 100 − 20 80

Alternate Method:

 When final price is 80, then marked price is 100

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Finish Line & Beyond ∴ When final price is 1, then marked price is

100 80

∴ When final price is 1600, then marked price is =

100 × 1600 = 2000 80

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

=

100 × 5400 = 5000 108

100 × Final Pr ice 100 + VAT

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Finish Line & Beyond Exercise 3 1. Calculate the amount and compound interest on (a) Rs 10,800 for 3 years at

1 12 % per annum compounded annually. 2

R t ) 100 1 P = 10800, R= 12 % and t=3 years 2 25 So, A= 10800(1+ )³ 200 1 =10800(1+ )³ 8 9 =10800( )³ 8 9 9 9 = 10800 × × × =15377.34 8 8 8 Answer: A= P(1+

So, Interest = A-P =15377.54-10800=4577.34 (b) Rs 18,000 for

1 2 years at 10% per annum compounded annually. 2

Answer: Amount = 18000(1+

= 18000 ×

10 )² 100

11 11 × 10 10

=21780 Interest after 2 years = 21780-18000=3780 Plus the interest after half year = 21780 ×

10 × 1 )=1089 100 × 2

Total interest after 2.5 years = 3780+1089=4869

(c) Rs 62,500 for 1.5 years at 8% per annum compounded half yearly. Answer: Amount After 1.5 year = 62500(1+

4 )³ 100

Rate has bee halved because interest is compounded half yearly and for the same reason time has been doubled.

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Finish Line & Beyond

= 62500 ×

26 26 26 × × =70304 25 25 25

So, Interest = 70304-62500=7804 (d) Rs 8,000 for 1 year at 9% per annum compounded half yearly. Answer: A= 8000(1+ = 8000 ×

9 )² 200

209 209 × =8736.20 200 200

So, interest =8736.20-8000=736.20 (e) Rs 10,000 for 1 year at 8% per annum compounded half yearly. Answer: A=10000(1+

= 10000 ×

4 )² 100

26 26 × = 10816 25 25

So, interest = 10816-10000 Alternate method: Calculating big multiplication can be tedious. Compound interest can also be calculated by finding yearly amount for each year separately. 1st half year 10000+400=10400 2nd half year 10400+416=10816 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? Answer: Amount after 1st year 26400+3960=30360 Amount after 2nd year 30360+4554=34914 To make things easier interest can be bifurcated into two parts of 10% and 5% as follows: 1st year 26400+2640+1320=30360 2nd year 30360+3036+1518=34914 Conventional Method

15 )² 100 23 23 = 26400 × × = 34914 20 20

A=26400(1+

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Finish Line & Beyond 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?

P× t× r 100

Answer: Interest for Fabina =

=

12500 × 3 × 12 = 4500 100

Amount for Radha = 12500(1+

= 12500 ×

10 )³ 100

11 11 11 × × = 16637.50 10 10 10

Interest for Radha = 16637.50-12500=4137.50 Interset paid by Fabina is Rs. 362.50 more than that paid by radha 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? Answer: The extra amount payable would be interest on the first year’s interest 1st year’s interest = 12000 × 6%=720 Interest on 720 = 720 × 6%=43.20 Extra amount payable = 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? Answer: Amount = 60000+60000 × 6%=60000+3600=63600 (ii) after 1 year? Answer: Amount = 63600+63600 × 6% = 63600+3816=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 1.5 years if the interest is (i) compounded annually. Answer: Amount after 1 year = 80000+80000 × 10%=80000+8000=88000 Interest for the next 6 months = 88000 × 5%=4400 Amount after 1.5 years = 88000+4400 = 92400

(ii) compounded half yearly.

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Finish Line & Beyond Answer: rate of interest will become half and time will be three half years.

5 )³ 100 21 21 21 = 80000 × × × = 92610 20 20 20

Amount = 80000(1+

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. Answer: Amount = 8000(1+

5 )² 100

21 21 × = 8820 20 20

= 8000 ×

Interest after 2 years = 8820-8000=820 (ii) The interest for the 3rd year. Interest for the 3rd year = 8820 × 5%=441 8. Find the amount and the compound interest on Rs 10,000 for 1.5 years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually? Answer: Amount = 10000(1+

= 10000 ×

5 )³ 100

21 21 21 × × = 11576.25 20 20 20

Interest = 11576.25-10000=1576.25 This interest would be more than the interest compounded annually, as interest compounded half yearly is always more than compounded annually at the same rate of interest. 9. Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12.5% per annum, interest being compounded half yearly. Answer: Amount = 4096(1+

= 4096 ×

25 )³ 400

17 17 17 × × = 4913 16 16 16

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Finish Line & Beyond 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. Answer: Let us assume the population to be P in 2001

5 21 21 × )² = P × 100 20 20 20 20 ⇒ P = 54000 × × = 48979.5 which can be said to be 48980 in round figures. 21 21

Then 54000=P(1+

(ii) what would be its population in 2005? Answer: Population in 2005 = 54000(

21 21 × )=59535 20 20

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. Answer: The bacteria’s growth will be compounded after every hour Final Count after 2 hrs = 506000(1+

= 506000 ×

5 )² 200

41 41 × = 531616.25 which is 531616 in round figures 40 40

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. Answer: Value of scooter after 1 year = 42000-42000 × 8%=42000-3360=38640

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