chapter 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 a scooter 30 km per hour.

(b) 5 m to 10 km

(c) 50 paise to ₹ 5

Explanation:

a) Speed of a cycle is given as 15km per hour and

Speed of a scooter is given as 30km per hour

Therefore, Speed of a cycle : Speed of a scooter = 15:50 or 1:2

Therefore, the required ratio is 1:2


b) We know, 1 km = 1000 m

So, 10km= 10 x1,000m= 10,000m

5m/10000m= 1/ 2,000

Therefore, the required ratio is 1:2,000


c) We know, ₹1 = 100paise

So, ₹5= 5 × 100paise = 500paise

50 paise/₹5= 50/500 = 1/10 = 1:10

Therefore, the ratio = 1:10


2. Convert the following ratio to percentages:

a) 3:4

b) 2:3

Explanation:

a) 3:4 

= ¾

 = 3/4 × 100%

= 0.75 × 100%

= 75%

b) 2:3 

= 2/3

= 2/3  × 100% 

= 0.66 x 100%

 = 66.666 % 


3.72% of 25 students are good in mathematics. How many are not good in mathematics?

Explanation:

Given that 72% are good in mathematics out of a total of 25 students

Thus, number of students who are good in Mathematics= 72/100 × 25

                                                                                         =  18

Thus, the number of students who are not good in mathematics can be calculated as

= Total number of students – Number of students good in mathematics

= 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?

Explanation:

Let the total number of matches played by team be x.

Matches won by the team is 10 and win % is 40

⇒ 40/100 × x = 10

4/10 ×  x = 10

4x = 100

x = 100 / 4

= 25

Hence, the total number of matches played by the team is 25. 


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

Explanation:

In the beginning let Chameli have x amount of money

Given, she was left with 600 after spending 75% of x

Percentage of money left after spending 600= 100% - 75% = 25%

So, 25% of x = ₹600

25/100 × x = ₹600

25x = 60000

x = ₹60000/25

= ₹2400

Hence, amount of money Chameli had in the beginning was 2400

6.If 60% of people in the 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 lakhs, find the exact number who like each type of game.

Explanation:

Given, % of people liking cricket= 60%

% of people liking football= 30%

% of people liking other games = (100 – 60 – 30)%

                                                                      = 10%

Total number of people = 50 lakhs

So,

Number of people who like cricket = 60/100 x 50 = 30 lakhs

Number of people who like football = 30/100  x 50 = 15 lakhs

Number of people who like other games = 10/100  x 50 = 5 lakhs


Exercise: 8.2

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

Explanation:

Let the original salary be x

Given,

New salary = ₹ 1,54,000

Original salary + Increase in salary = New salary

Increase in salary is 10% of original salary, i.e., x

So, (x + 10/100 × x) = 154000

x + x/10 = 154000

11x/10 = 154000

x = 154000 × 10/11

= 140000

Hence, the original salary =₹1,40,000.


8.On Sunday, 845 people went to the zoo. On Monday, only 169 people went. What is the percent decrease in the number of people visiting the zoo on Monday?

Explanation:

Given, 

No. of visitors in a zoo on Sunday= 845

No. of visitors in a zoo on Monday= 169

Decline in the number of visitors = 845 – 169 = 676

Thus,

Percentage decrease can be calculated as,

(Decrease in the number of visitors/Number of visitors in the zoo on Sunday) x 100%

= (676 / 845 × 100)%

= 80%


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

Explanation:

Given, Cost of 80 articles for the seller= 2400

Cost of one article = 2400/80 = ₹ 30

Rate of profit = 16%

Profit %  = Profit/Cost Price. x 100

16 = Profit / 30 x 100

Profit/100 = (16 x 30)

Profit= 480/100

= ₹ 4.8

Therefore, the selling price of one article = Cost price + Profit

= ₹ (30 + 4.80)

= ₹ 34.80


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

Explanation:

Given, C.P. of article = 15,500

Repair cost= 450

Aggregate cost of an article = Cost + Repairs

                                           = ₹15500 + ₹450

                                           = ₹15950

Rate of profit= 15%

Profit % = Profit/C.P. x 100

15 = Profit/15950 x 100

Profit/100 = (15 x 15950)

Profit== 239250/100

= 2392.50

Therefore, the S.P. of the article = C.P. + Profit

= ₹(15950 + 2392.50)

= ₹18342.50


11.A VCR and TV were bought for ₹ 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 per cent on the whole transaction.

Explanation:

Cost of a VCR = ₹ 8000

Loss on VCR= 4%

This means that if C.P. is ₹ 100, then S.P. is ₹ 96.

When C.P. is ₹ 8000,

S.P. = (96/100 x 8000)

 = ₹ 7680

Cost. of a TV = ₹ 8000

Profit on TV= 8%

This means that if cost price is ₹ 100, then sale price is ₹ 108.

When Cost price is ₹ 8000,

Selling price = (108/100 x 8000) 

= ₹ 8640

Therefore, total Sales price = ₹ 7680 + ₹ 8640 

                                            = ₹ 16320

Total cost price = ₹ 8000 + ₹ 8000 = ₹ 16000

Since, sales is more than the cost it implies gain.

Profit = ₹ (16320 – 16000) 

= ₹ 320

Rate of profit in total= Profit / Total CP × 100

= 320 / 16000 ×  100

= 2%

Thus, on this transaction the shopkeeper bags a profit of 2%


12.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 ₹ 1450 and two shirts marked at ₹ 850 each?

Explanation:

Total marked price of items= 1450+850+850

                                             = 3150

The discount rate is given as 10%

So, Discount = ₹ (10/100 x 3150) = ₹ 315

Therefore, S.P. = ₹3150 −₹ 315

                                    = ₹ 2835

Hence, the amount customer will have to pay ₹ 2,835.



13.A milkman sold two of his buffaloes for ₹ 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 the C.P. of each)

Explanation:

One buffalo was sold for 20,000

Therefore if 2 buffalos were sold, then total sale was 40,000

Gain on sale of one buffalo= 5%

If cost is 100, then sale price is 105

Cost of one buffalo = 100/105 × 20000

                               = ₹ 19,047.62

Now, loss on sale of second buffalo is 10%

If cost is Rs.100, then sale price is Rs.90

Therefore, Cost of other buffalo = 100 / 90 × 20000

                                                      = ₹ 22222.22

Total Cost = ₹ 19047.62 + ₹ 22222.22 

                 = ₹ 41269.84

Total Sale price = ₹ (20000 + 20000) = ₹ 40000

Since, sale < cost, the seller made loss.

Loss = ₹ 41269.84  − ₹ 40000 

         = ₹ 1269.84

Thus, the total loss of milkman = ₹ 1,269.84


14. The price of a TV is ₹ 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.

Explanation:

Given, Price of the TV = 13,000

Sales tax rate= 12%

Tax to be paid= 12 / 100 x 13000

                       = 1560

Total amount that Vinod has to pay = Price + Sales Tax

                                                        = ₹ (13000 + 1560)

                                                        = ₹ 14560

Hence, amount Vinod has to pay for TV = ₹ 14,560 


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

Explanation:

Let the marked price be x

Discount given= 20% of x

Amount paid by Arun + Discount= Marked Price

=>1600+ 20/100 × x = x

=>1600 + 2x/10 = x

=>16000 + 2x = 10x

=>8x = 16000

=>x= 16000/8

=>x= 2000

Therefore, the marked price = ₹ 2000.


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

Explanation:

Since, the price of dryer includes VAT of 8%, 5400 includes VAT

Let price before VAT be 100

Then, Price including VAT= 108

Similarly, when price including VAT is ₹ 5400, original price = ₹ (100 / 108 × 5400)

                                                                                                   = ₹ 5000

Therefore, the price of the hair dryer before charging VAT was ₹ 5,000.                            


 Exercise: 8.3

Calculate the amount and compound interest on ( from question 17 to 21)

17. ₹ 10,800 for 3 years at 12½% per annum compounded annually.

Explanation:

 Principal = ₹ 10,800

Rate = 12½ % = 25/2 % p.a.

Time = 3yrs

Amount (A) = P(1 + R/100)n

                     = 10800 (1 + 25 / 200)3

                     = 10800 (225 / 200)3

                     = 15377.34375

                     = ₹ 15377.34

C.I. = A – P = ₹ (15377.34 – 10800)

                    = ₹ 4,577.34


18.₹ 18000 for 2½ years at 10% per annum compounded annually.

Explanation:

Principal = ₹ 18,000

Rate = 10% p.a.

Number of years  = 2½yrs

The interest amount for 2 years and 6 months can be calculated by calculating the amount for 2 years first using the compound interest formula, then calculating 6 months of simple interest on the amount earned after 2 years.

First, the amount for 2 years has to be calculated as

 A = P(1 + R/100)n

= 18000(1 + 1/10)2

= 18000(11/10)2

= ₹ 21780

By taking ₹ 21,780 as the principal amount, the Simple Interest for the next ½ year will be calculated

S.I. = (21780 x ½ x 10)/100

= ₹ 1089

Hence, the interest for the first 2 years = ₹ (21780 – 18000) = ₹ 3780

And, interest for the next ½ year = ₹ 1089

Total C.I. = ₹ 3780 + ₹ 1089

= ₹ 4,869

Therefore,

Amount, A = P + C.I.

= ₹ 18000 + ₹ 4869

= ₹ 22,869


19.₹ 62500 for 1½ years at 8% per annum compounded half yearly.

Explanation:

 (P) = ₹ 62,500

R = 8% per annum or 4% per half-year

n = 1½

There will be 3 half-years in 1½ years

Amount, A = P(1 + R/100)n

= 62500(1 + 4/100)3

= 62500(104/100)3

= 62500(26/25)3

= ₹ 70304

C.I. = A – P = ₹ 70304 – ₹ 62500 = ₹ 7,804


20. ₹ 8000 for 1 year at 9% per annum compound half yearly.

(You can use the year-by-year calculation using S.I. formula to verify)


Explanation:

 (P) = ₹ 8000

R = 9% per annum or 9/2% per half-year

n= 1 year

There will be 2 half-years in 1 year

A = P (1 + R/100)n

= 8000 (1 + 9/200)2

= 8000 (209/200)2

= 8736.20

C.I. = A – P = ₹ 8736.20 – ₹ 8000 = ₹ 736.20


21.₹ 10000 for 1 year at 8% per annum compounded half yearly.

Explanation:

P = ₹ 10,000

R = 8% per annum or 4% per half-year

n= 1 year

There are 2 half-years in 1 year, so n=2

Amount, A = P(1 + R/100)n

= 10000 (1 + 4/100)2

= 10000 (1 + 1/25)2

= 10000 (26/25)2

= ₹ 10816

C.I. = A – P = ₹ 10816 – ₹ 10000 = ₹ 816


22.Kamala borrowed ₹ 26400 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 compounded yearly and then find S.I. on the 2nd year amount for 4/12 years.)

Explanation:

Principal (P) = ₹ 26,400

Rate (R) = 15% per annum

Number of years (n) = 2 4/12

First we calculate amount for 2 years using compound interest formula

A = P(1 + R/100)n

= 26400(1 + 15/100)2

= 26400(1 + 3/20)2

= 26400(23/20)2

= ₹ 34914

Now, interest for the remaining 4 months has to be calculated which can be done with the help of simple interest formula by considering compounded amount of 2 years.

By taking ₹ 34,914 as principal, the S.I. for the next 4 months will be calculated as,

S.I. = (34914 × 1/3 x 15)/100

       = ₹ 1745.70

Interest for the first two years = ₹ (34914 – 26400) 

                                                 = ₹ 8,514

And interest for the next 1/3 year = ₹ 1,745.70

Total C.I. = ₹ (8514 + ₹ 1745.70) 

                 = ₹ 10,259.70

Amount = P + C.I. = ₹ 26400 + ₹ 10259.70 

                               = ₹ 36,659.70


23.Fabina borrows ₹ 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?

Explanation:

For Fabina,

P= 12500

R= 12%

n= 3 years

Interest paid by Fabina at simple interest = (P x R x T)/100

                                                                  = (12500 x 12 x 3)/100

                                                                  = 4500

For Radha,

P= 12500

R= 10%

n= 3 yrs

Amount paid by Radha at the end of 3 years ,

A = 12500(1 + 10/100)3

= 12500(110/100)3

= ₹ 16637.50

Interest paid by Radha at compound interest = A – P = ₹ 16637.50 – ₹ 12500 = ₹ 4,137.50

The interest paid by Fabina is ₹ 4,500 and by Radha is ₹ 4,137.50

Thus, Fabina pays more interest

₹ 4500 − ₹ 4137.50 = ₹ 362.50

Hence, Fabina will have to pay ₹ 362.50 more.


24.I borrowed ₹ 12000 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?

Explanation:

Case 1- Borrowed at simple interest

P = ₹ 12000

R = 6% per annum

T = 2 years

S.I. = (P × R × T)/100

       = (12000 × 6 × 2)/100

       = ₹ 1440

Case 2- Borrowed at Compound interest

To find the compound interest, the amount (A) has to be calculated

A = P(1 + R/100)n

= 12000(1 + 6/100)2

= 12000(106/100)2

= 12000(53/50)2

= ₹ 13483.20

∴ C.I. = A − P

= ₹ 13483.20 − ₹ 12000

= ₹ 1,483.20

Therefore, extra amount I have to pay

C.I. − S.I. = ₹ 1,483.20 − ₹ 1,440

= ₹ 43.20

Therefore, the extra amount to be paid is ₹ 43.20.


25.Vasudevan invested ₹ 60000 at an interest rate of 12% per annum compounded half yearly. What amount would he get

(i) after 6 months?

(ii) after 1 year?

Explanation:

(i) P = ₹ 60,000

R= 12% per annum = 6% per half-year

n = 6 months = 1 half-year

Amount, A = P(1 + R/100)n

= 60000(1 + 6/100)1

= 60000(106/100)

= 60000(53/50)

= ₹ 63600

(ii) Since, there are 2 half-years in 1 year

So, n = 2

Amount, A = P(1 + R/100)n

= 60000(1 + 6/100)2

= 60000(106/100)2

= 60000(53/50)2

= ₹ 67416


26.Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1½ years if the interest is

(i) Compounded annually

(ii) Compounded half yearly

Explanation:

(i) P = ₹ 80,000

R = 10% per annum

n = 1½ years

First, we calculate interest amount for 1 year using compound interest formula as

Amount, A = P(1 + R/100)n

= 80000(1 + 10/100)1

= 80000 x 11/100

= ₹ 88000

Now, interest for the remaining 6 months has to be calculated which can be done with the help of simple interest formula by considering compounded amount of 1 year as the principal.

S.I. = (P × R ×  T)/100

= (88000 × 10 × ½)/100

= ₹ 4400

So, Interest for the 1st year = ₹ 88000 – ₹ 80000 = ₹ 8000

And interest for the next ½ year = ₹ 4,400

Total compound interest = ₹ 8,000 + ₹ 4,400 = ₹ 12,400

A = P + C.I.= ₹ (80000 + 12400)

= ₹ 92,400

Thus, if compounded annually Arif has to pay 92,400 after 1 ½ years

(ii) half yearly

Rate = 10% p.a. and 5% per half-year

In 1½ years there will be three half-years so n=3

Amount, A = P(1 + R/100)n

= 80000 × (1 + 5/100)3

= 80000 × (105/100)3

= ₹ 92610

Thus, if compounded half yearly Arif has to pay 92610 after 1 ½ years

Hence, the difference between the annual and half yearly amount = ₹ 92,610 – ₹ 92,400 

= ₹ 210


27.Maria invested ₹ 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

Explanation:

(i) P = ₹ 8,000

R = 5% per annum

T = 2 years

A = P(1 + R/100)n

= 8000 × (1 + 5/100)2

= 8000(105/100)2

= ₹ 8820

(ii) The interest for the third year can be calculated by taking the compounded amount i.e, ₹ 8,820 as principal, the S.I. for the next year will be calculated as

S.I. = (P × R × T)/100

= (8820 × 5 × 1)/100

= ₹ 441


28. Find the amount and the compound interest on ₹ 10,000 for 1½ years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Explanation:

P = ₹ 10,000

Rate = 10% per annum = 5% per half-year

T = 1½ years

In 1½ years there will 3 half-years so n=3

Amount, A = P(1 + R/100)n

= 10000 × (1 + 5/100)3

= 10000 × (105/100)3

= ₹ 11576.25

C.I. = A − P

= ₹ 11576.25 − ₹ 10000

= ₹ 1,576.25

First we calculate amount for 1 year using compound interest formula

A = P(1 + R/100)n

= 10000(1 + 10/100)1

= 10000(110/100)

= ₹ 11000

By taking ₹ 11,000 as the principal, the S.I. for the next ½ year will be calculated as

S.I. = (P x R x T)/100

= (11000 x 10 x ½)/100

= ₹ 550

So, the amount of interest for 1st year = ₹ 11000 − ₹ 10000 = ₹ 1,000

Hence, total compound interest = ₹ 1000 + ₹ 550 = ₹ 1,550

So the difference between two interests = 1576.25 – 1550 = 26.25

The interest compounded half yearly would be 26.25 more than the amount of interest when compounded annually.


29.Find the amount which Ram will get on ₹ 4,096, if he gave it for 18 months at 12½ per annum, interest being compounded half-yearly.

Explanation:

Principle = ₹ 4,096

Rate of interest = 12½ p.a. = 25/2 p.a. = 25/4 half yearly

Time = 18 months (18 months = 3 half years; n=3)

Thus, A = P(1 + R/100)n

= 4096 × (1 + 25/(4 x 100))3

= 4096 ×  (1 + 1/16)3

= 4096 ×  (17/16)3

= ₹ 4913

Hence, the amount ram will get is ₹ 4,913.


30.The population of a place increased to 54000 in 2003 at a rate of 5% per annum

(i) find the population in 2001

(ii) what would be its population in 2005?

Explanation:

(i) Population in 2003 = 54,000

Rate= 5%      n= 2yrs

Population in 2003= Population in 2001 x ( 1 + r/100)n

54,000 = (Population in 2001)x (1 + 5/100)2

54,000 = (Population in 2001)x (105 / 100)2

Population in 2001 = 54000 x (100/105)2

                               = 48979.59

Hence, the population in 2001 was approximately 48,980

(ii) Similarly, Population in the year 2005 = 54000 × ( 1 + 5/100 )2

                                                      = 54000×  (105/100)2

                                                      = 54000 × (21/20)2

                                                      = 59535

Therefore, the population in 2005 would be 59,535.


31.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.

Explanation:

Given, the initial count of bacteria 5,06,000

Rate of increase= 2.5%

Bacteria after two hrs = 506000 × (1 + 2.5/100)2

                                                = 506000 × (1 + 1 / 40)2

                                                = 506000× ( 41 / 40 )2

                                                = 531616.25

Hence, the count of bacteria after 2 hrs will be 5,31,616 (approximate)


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

Explanation:

C.P. of the scooter = ₹ 42,000

Depreciation = 8% × ₹ 42,000 per year

                      = 3360

After 1 year, the value can be calculated as,

Cost of scooter – Depreciation for the year

Therefore,  ₹ 42000 − ₹ 3360 

= ₹ 38,640.