Math 8

Example: 2

A shopkeeper sold a set of cell for Rs. 110 for which Rs.77 was initially paid. Find the
profit percent.

Solution:

Here, The S.P. of the cell = Rs. 110

The C.P. of the cell = Rs. 77

Since, S.P. > C.P., there is profit

Now, by using formula,

Profit = S.P. – C.P.

= Rs. 110 – Rs. 77

= Rs. 33

∴ Profit % = Profit × 100% = 33 × 100%
C.P. 77

= 37 × 100%

= 300 %= 42 6 %
7 7
Example: 3

A man bought 100 apples at Rs. 1,800. 20 of them were rotten and he sold remaining at
the rate of Rs. 25 per piece. Find his profit or loss percent.

Solution:

Here, C.P. of 100 apples = Rs. 1800

Number of rotten apples = 20

Remaining apples = 100 – 20 = 80

S.P. 80 apples = Rs.80 × 25 = Rs. 2,000.

We have,

Profit = S.P. – C.P.

= Rs.2000 – Rs.1,800

= Rs. 200

Profit per cent = Profit × 100%
C.P.

= 1280000 × 100%

= 1900

= 11 1 %.
9

Oasis School Mathematics-8 119955

Example: 4

A fruit -seller buys 7 oranges for Rs. 35 and sells 4 oranges for Rs. 28. Find his profit or
loss percent.

Solution:

Here, C.P. of 7 oranges = Rs. 35

Again, C.P. of 1 orange = Rs. 35 = Rs. 5.
7

S.P. of 4 oranges = Rs. 28

S.P. of 1 orange = Rs. 28/4 = Rs. 7

Now, Profit = SP – CP


= Rs. 7 – Rs. 5 = Rs. 2

We have, Profit % = Profit × 100%
C.P.

= Rs. 2 × 100% = 40%
5

Exercise 15.1

1. Find the profit or loss per cent in each of the following cases.

(a) C.P. = Rs. 900, S.P. = Rs. 1,050 (b) C.P. = Rs. 5,000, S.P. = Rs. 4,500

(c) C.P. = Rs. 1,400, S.P. = Rs. 1,500 (d) C.P. = Rs. 1,200, S.P. = Rs. 1,000

2. Find the profit or loss percent in each of the following cases.

(a) C.P. = Rs. 200, S.P. = Rs. 250 (b) C.P. = Rs. 800, S.P. = Rs. 900.

(c) C.P. = Rs. 750, S.P. = Rs. 600. (d) C.P. = Rs. 1,600, S.P. = Rs. 1,800

3. (a) If C.P. = Rs. 800, profit = Rs. 25, find S.P.

(b) If C.P. = Rs. 550, loss= Rs. 50, find S.P.

(c) If S.P. = Rs. 1,285, profit = Rs. 85, find C.P.

(d) If S.P. = Rs. 1,500, loss = Rs. 150, find C.P.

4. (a) Arun buys a table fan for Rs. 400 and sells it for Rs. 425. Find his profit and profit
percent.

(b) A man buys an article for Rs. 4,000 and sells it for Rs. 3500. Find his

(i) loss, and (ii) loss per cent.

(c) Asmita bought an article at Rs. 2,000 and she sold it at Rs. 1,750. Find her loss
and loss per cent.

196 Oasis School Mathematics-8

5. (a) A man bought a mobile set for Rs. 12,000. What should be its selling price to
gain Rs. 1250?

(b) A man bought an article for Rs. 15,000. What is the selling price of the article, if
the loss is Rs. 2,250.

(c) Sampada sold an article for Rs. 3,000 at a profit of Rs. 750. What is the cost price
of the article?

6. (a) Suman buys an old scooter for Rs. 95,000 and spends Rs. 5,000 on its repair. If
he sells the scooter for Rs. 1,05,000, find his gain or loss per cent.

(b) Salman sold a mobile set at Rs. 12,000 after spending Rs. 1,500 for its repairing.
If he had paid Rs. 10,000 for it, find his profit or loss per cent.

7. (a) A shop-keeper bought 200 eggs at Rs. 5 each. 50 eggs were broken during
transportation and then he sold the remaining eggs at Rs. 8 each. Find, his gain
or loss per cent.

(b) A fruit-seller purchased 1,000 mangoes at Rs. 15 each. 300 of them were rotten
and he sold the rest at Rs. 20 each. Find his profit or loss per cent.

8. (a) A fruit seller purchased apples at 4 for Rs. 2 and sold them at 5 for Rs. 3. Find
his profit per cent.

(b) A fruit seller buys oranges at 5 kg for Rs. 200 and sells them at 2 kg for Rs.90.
Find his profit per cent.

(c) A man buys some pencils at the rate of Rs. 36 per dozen and sells them at the
rate of Rs. 4 per piece. Find his profit percentage.



Answer

1. (a) Profit% = 16 2 % (b) Loss% = 10% (c) Profit% = 7 1 % (d) Loss% = 16 2 %
3 7 3

2. (a) Profit% = 25% (b) Profit% = 12.5% (c) Loss% = 20% (d) Loss% = 12.5%

3. (a) Rs. 825 (b) Rs. 500 (c) Rs. 1200 (d) Rs. 1650

4. (a) Rs. 25, 6 1 % (b) (i) Rs. 500, 12 1 % (c) Rs. 250 , 12 1 %
4 2 2

5. (a) Rs. 13,250 (b) Rs. 12, 750 (c) Rs. 2250 6. (a) 5% (b) 4 8 %
23

7. (a) 20% profit (b) 6 2 % loss 8. (a) 20% (b) 12 1 % (c) 33 1 %
3 2 3

Oasis School Mathematics-8 119977

15.2 Calculation of C.P. or S.P. if Profit and Loss are Given

I. To find out S.P. when C.P. and profit or loss per cent are given

As we know that Profit = P% of C.P.
Then S.P. = C.P. + Profit
Profit = Profit percent of C.P. and Loss = Loss per cent of C.P.
Then, S.P. is calculated as follows

S.P = C.P. + profit % of C.P.………… (i)

or, S.P. = 100 + P% × C.P. ………… (ii)
100

When L% is given, S.P. = C.P. – Loss, S.P. = C.P. – L% of C.P ………… (iii)

or, S.P. = 100 – L% × C.P. ………… (iv)
100

II. To find out C.P. when S.P. and profit or loss per cent are given

(i) When profit P% is given, we can use same relation to obtain C.P.

i.e. S.P. = C.P. + P % of C.P. …………………….. (i)

or, directly we can obtain C.P by using the relation, C.P = S.P. × 100 ….. (ii)
100 + P%

(ii) When loss % is given, we can use the relation

S.P. = C.P. – L % of C.P. ……………………. (iii)

or, directly we can obtain C.P. by using the relation C.P. = S.P. × 100 .....…. (iv)
100 – L%

Remember !

Profit = Profit % of C.P. Loss = Loss % of C.P.

S.P. = C P + P% of C.P. S.P. = C.P.–L% of C.P.

S. P. = 100+P% ×C.P. S.P. = 100–L% ×C.P.
100 100

C. P. = S.P.×100 C.P. = S.P×100
100+P% 100–L%



198 Oasis School Mathematics-8

Worked Out Examples



Example: 1

An article bought for Rs. 1,000 is sold at a profit of 10%. Find the selling price.

Solution: Note: Alternative method-I
Here,
C.P. = Rs. 1000 • In this method, assume C.P. Since there is the profit of 10 %
Profit = 10%, as 100.
S.P. = ? When C.P. is Rs.100, S.P. is Rs.110.
• S.P. as 100 + profit per cent
if profit per cent is given. When, C.P. is Re.1, S.P. is Rs. 110 .
100
• S.P. as 100 – loss percent

if loss per cent is given.

Now, Profit = 10% of C.P. When C.P. is Rs. 1000, S.P. is Rs. 110 × 1000
100
10
= 100 × Rs. 1000 ∴ S.P. = Rs. 110 × 1,000
100
= Rs. 100
S.P. = Rs. 1,100.

∴ S.P. = C .P. + Profit Alternative method-II

= Rs. 1000 + Rs. 100 Here C. P. = Rs. 1000, P% = 10%, S.P. = ?

= Rs. 1100 We have,

S.P. = 100 + P% × C.P.
100

( )=
100 + 10 × 1000
100
Example: 2
= 110 × 1,000 = Rs 1,100.
A man bought a mobile set for Rs. 11000 100
and spent Rs. 2000 for its repairing.
Find the selling price if it was sold at a
profit of 10%.

Solution:

Here, Cost price (C.P.) = Rs. 11,000

Repairing cost = Rs. 2,000

Net C.P. = Rs. (11,000 + 2,000) = Rs. 13,000

Profit% = 10%, S. P = ?

We have, S.P. = C.P. + P% of C.P.

or, S.P. = 13,000 + 10% of 13,000

= 13,000 + 10 × 13,000
100

= 14,300

∴ S.P. = Rs. 14,300

Oasis School Mathematics-8 119999

Example: 3

By selling a radio for Rs. 1,500, a trader gains 20%. Find the cost price of the radio.
Solution:

Here, Alternative method-I
Since, there is the profit of 20%
S.P. = Rs. 1,500 If S.P is Rs. 120, then C.P. = Rs. 100

Profit % = 20%,

C.P. = ?

Let, C.P. = x If S.P. is Re 1, then C.P. = Rs. 100
120

S.P. = C.P.+ P% of C.P. If S.P. is Rs. 1500, then C.P. = Rs. 100 × 1,500
120
or, 1500 = x +20% of x
100
or, 1500 = x + 20 ×x Hence, C.P. = Rs. 120 × 1,500
100
= Rs. 1,250
or, 1500 = x + x
5 Alternative method-II
6x
or, 1500 = 5 Here, S.P.=Rs. 1500, P%=20%, C.P.=?

or, 6 x = 1500 × 5 100 + S.P.
(100 + P%)
or, x = 1500×5 S.P. =
6
= 100 × Rs.1500
= Rs. 1,250 120

Hence, the required C.P. is Rs. 1,250. = Rs 1,250.

Exercise 15.2

1. Find the selling price in each of the following cases.

(a) C.P. = Rs. 1000; profit = 25% (b) C.P. = Rs. 800, gain = 10%

(c) C.P. = Rs. 750, loss = 10% (d) C.P. = Rs. 5000, loss = 20%

2. Find the cost price when

(a) S.P. = Rs. 2200, profit = 10% (b) S.P. = Rs. 1200, profit = 20%

(c) S.P. = Rs. 1920, loss= 25% (d) S.P. = Rs. 2400, loss = 20%

3. (a) A shopkeeper bought a radio for Rs. 1800 and sold it at 20% profit. Find his
amount of profit.

(b) A watch is bought at Rs. 500 and sold at a loss of 15%. Find the amount of loss
.

(c) An article was bought for Rs. 1,500. If sold at a profit of 10%, find the amount
of profit.

200 Oasis School Mathematics-8

4. (a) Aadhya bought a mobile set for Rs. 12000 and sold at a profit of 15%. Find the
selling price.

(b) Lakpa bought a television for a Rs. 20,000 and sold it at the loss of 10%, find the
selling price of TV.

(c) At what price an article should be sold to make a profit of 20%, if it was
initially paid Rs. 4000.

5. (a) A man purchased a second hand bicycle for Rs. 1200 and spent Rs. 300 on its
repairing. If he sold it at 25% profit, find the selling price.

(b) Sundar bought a second hand mobile set for Rs. 7000. He spent Rs. 1000 for its
maintenance. At what price should it be sold to make the profit of 5%.

6. (b) By selling an article for Rs. 900, a man gains 20%. Find his cost price.

(b) By selling a pen for Rs. 40, a man makes a profit of 25%. How much does it cost
him?

(c) A man sold an article at Rs. 1,250 at the gain of 25%. Find its cost price.

Answer
1. (a) Rs. 1250 (b) Rs. 880 (c) Rs. 675 (d) Rs. 4000 2. (a) Rs. 2000 (b) Rs. 1000 (c) Rs. 2560
(d) Rs. 3000 3. (a) Rs. 360 (b) Rs. 75 (c) Rs. 150 4. (a) Rs. 13800 (b) Rs. 18000 (c) Rs. 4800
5. (a) Rs. 1875 (b) Rs. 8400 6. (a) Rs. 750 (b) Rs. 32 (c) Rs. 1000

15.3 Discount & VAT

I. Marked price:

We can observe the price tag which is kept along with the article while purchasing
goods from the shop or department store. The price of the goods so kept using the
tag or the list or mark is called marked price or the quoted price or the labeled price or
the listed price. It is generally fixed to be either greater than or equal to the selling price.

i.e. M.P. ≥ S.P.

II. Discount:

The concession or deduction over the marked price of an article at the time of selling
is called discount. It is normally expressed in terms of percentage of marked price.

Hence, if the discount is d%, that means.
Discount = d% of M.P.

III. Selling price:

The price of the article after allowing discount from its M.P. is called its net price or
the selling price.

Note: Hence relating above three terms we have,
M.P. – Discount = S.P.

Oasis School Mathematics-8 220011

As we know that the discount is always provided on the marked price of an article,
the discount per cent is calculated by the following formula:

Discount percent = Discount × 100%
M.P.

IV. Value Added Tax (VAT):

The value added tax (VAT) is tax charged on the actual selling price of goods and
services. It is an improved form of sales tax and is levied on the value added on
goods and services at each stage in the process of production and distribution. Thus,
VAT is charged as a certain percentage of S.P.

Net S.P. = S.P. with VAT

= S.P. + VAT percentage of S.P.

VAT percent = VSA.PT. × 100%

Remember !

• Discount = Discount % of M.P.

• VAT = VAT% of S.P.

• S.P. = M.P. – Discount
= M.P. – Discount % of M.P.

• S.P. with VAT = S.P. + VAT

= S.P. + VAT % of S.P.

Worked Out Examples



Example: 1

The marked price of an article is Rs. 500. Find the selling price after a discount of Rs. 50.
Solution:

Here, M.P. = Rs. 500
Discount = Rs. 50
∴
S.P. = M.P. – Discount = Rs. 500 – Rs. 50 = Rs. 450.

Example: 2

If a discount of 10% is offered on the goods whose M.P. is Rs.1,000, find its S.P.
Solution:

Here, M.P. = Rs. 1,000

Discount = 10% of M.P. = 10% of Rs. 1,000 = 10 × Rs. 1,000
100

= Rs. 100

∴ S.P. = M.P. – Discount = Rs. 1,000 – Rs. 100 = Rs. 900

202 Oasis School Mathematics-8

Example: 3

Alina sold a book for Rs. 135 after a discount of 10% on M.P.. Find its M.P.

Solution: M.P. be x, then discount = 10% of x = 10 ×x= x
Let, 100 10
Now,
S.P. = Rs. 135

We have, M.P. – Discount = S.P.
or
x – x = Rs. 135
or, 10
or,
9x = Rs. 135
10
∴
∴ 9x = Rs 135 × 10

x = Rs. 135 × 10 = Rs. 150.
9
M.P. = Rs. 150

Example: 4

The marked price of an article is Rs. 260. If the shopkeeper allows some discount and
sells it for Rs. 234, find the discount and discount per cent.

Solution:

Here, M.P. = Rs. 260,

S.P. = Rs. 234

Discount = ?

Discount percent = ?

We have, Discount = M.P. – S.P. = Rs. 260 – Rs. 234 = Rs. 26.
Again,
Hence Discount percent = Discount × 100 % = 26 × 100% = 10%.
M.P. 260

Discount = Rs. 26 and Discount per cent = 10 %

Example: 5

Marked price of an article is Rs. 10,000. Find the cost price if a profit of Rs. 200 is made
after the discount of 10%.

Solution:

Given, M.P. = Rs. 10,000

Discount% = 10%

We have, Discount = Discount % of M.P.

= 10% of 10,000

= 11000 × 10,000

= Rs. 1,000

Oasis School Mathematics-8 220033

Again,

We have, S.P. = M.P. – Discount
= 10,000 – 1,000
= Rs. 9,000
= Rs. 200
= S.P. – Profit

Profit

We have, C.P.

= Rs. (9,000 – 200) = Rs. 8,800

Example: 6

The selling price of an article is Rs. 400 and 13% VAT is levied on it. Find the net selling price.
Solution:

Here, Selling price (S.P.) = Rs. 400

VAT amount = 13% of S.P.

= 13 × 400
100

= Rs. 52

Net S.P. = S.P. + VAT

= Rs. 400 + Rs. 52 = Rs. 452

Example: 7

The marked price of an article is Rs. 3,000. If the shopkeeper allows 10% discount, how
much does a customer pay for it with 13% VAT?

Solution:

Here, Marked price (M.P.) = Rs. 3,000

Discount % = 10%

Now, Discount = 10% of M.P.

= 10 × 3,000 = Rs. 300
100

∴ S.P. = M.P. – Discount

= Rs. 3,000 – Rs. 300 = Rs. 2,700

Again, VAT amount = VAT% of S.P.

= 13% of 2,700

= 13 × 2,700 = Rs. 351
100

∴ Net S.P. = S.P. + VAT

= Rs. 2,700 + Rs. 351 = Rs. 3,051

204 Oasis School Mathematics-8

Exercise 15.3

1. Find discount and discount percent in the following cases.

(a) M. P. = Rs. 600, S.P. = Rs. 480

(b) M. P. = Rs. 5000, S.P. = Rs. 4,000

(c) M. P. = Rs. 5000, S.P = Rs. 4,500

2. Find the selling price in the following cases.

(a) M.P. = Rs. 200, Discount = 20%

(b) M.P. = Rs. 3,600, Discount = 5%

(c) M.P. = Rs. 4,000, Discount = 15%

3. Find the marked price in the following cases.

(a) S.P. = Rs. 360, Discount = 10%

(b) S.P. = Rs. 1,900, Discount = 5%

(c) S.P. = Rs. 950, Discount = 5%

4. (a) The marked price of a mobile is Rs. 15,000 and it is sold for Rs. 12,000. Find the
amount of discount and its percentage.

(b) What should be the discount percent so that a watch marked Rs. 6,000 is sold
for Rs. 4,500?

5. (a) An article marked Rs. 4,500 is sold at a discount of 10%. Find the selling price
of the article.

(b) The marked price of an article is Rs. 700 and it is sold at a cash discount of 10%.
Find its selling price.

6. (a) After a discount of 12%, a radio was sold for Rs.4400. Find its marked price.

(b) After allowing a discount of 8% on marked price, an article is sold for Rs. 2,300.
Find its marked price.

(c) An article is sold at Rs. 2,400 after a discount of 4%. Find its marked price.

7. (a) Marked price of a watch is Rs. 1,000. It is sold allowing a discount of 15%. If the

profit is Rs. 10, find the cost price of the watch.

(b) Ram marked the price of an article as Rs. 400. He sold it allowing a discount of
20% at a loss of Rs. 60. Find the cost price of the article.

(c) Marked price of a mobile set is Rs. 12,000. It is sold at a discount of 10%. If there
is a profit of Rs. 800, find its cost price.

Oasis School Mathematics-8 220055

8. Find the amount of VAT in the following cases:
(a) S.P. = Rs. 2.400, S.P. with VAT = Rs 2.700.
(b) S.P. = Rs. 1.440, S.P. with VAT = RS 1.780.
9. Find the VAT % in the following cases:
(a) VAT = Rs. 180, S.P. = Rs. 1,800
(b) VAT = Rs. 104, S.P = Rs. 800
(c) S.P. = Rs. 2,000, S.P. with VAT = Rs. 2.260
10. Find the net S.P. in the following case.
(a) S.P. = Rs. 1,200, VAT percent = 13%
(b) S.P. = Rs. 1,800, VAT percent = 13%.
(c) S.P. = Rs. 900, VAT percent = 13%.
11. Find S.P. in the following cases.
(a) S.P. with VAT = Rs 2,260, VAT% = 13%
(b) S.P. with VAT = Rs 2,825, VAT% = 13%
(c) S.P. with VAT = Rs. 565, VAT% = 13%

12. (a) The marked price of an article is Rs. 2,500 and the shopkeeper allows 10%
discount. Find;

(i) Selling price of the article.

(ii) Selling price with 13% VAT.

(b) The marked price of an article is Rs. 8,000. What will be the price of the article
if 13% VAT is levied after allowing 20% discount?

(c) The marked price of an article is Rs. 6,000. What will be the price of the article
if 10% VAT is added after allowing a discount of 10%.



Answer

1. (a) Rs. 120, 20% (b) Rs. 1000, 20% (c) Rs. 500, 10% 2. (a) Rs. 160 (b) Rs. 3420
(c) Rs. 3400 3. (a) Rs. 400 (b) Rs. 2000 (c) Rs. 1000 4. (a) Rs. 3000, 20% (b) Rs. 1500, 25%
5. (a) Rs. 4050 (b) Rs. 630 6. (a) Rs. 5000 (b) Rs. 2500 (c) Rs. 2500 7. (a) Rs. 840 (b) Rs. 380
(c) Rs. 10,000 8. (a) Rs. 300, (b) Rs. 340 9. (a) 10% (b) 13% (c) 13% 10. (a) Rs. 1356
(b) Rs. 2034 (c) Rs. 1017 11. (a) Rs. 2000 (b) Rs. 2500, (c) Rs. 500
12. (a) (i) Rs. 2250 (ii) 2542.50 (b) Rs. 7232 (c) Rs. 5940

206 Oasis School Mathematics-8

Unit

16 Simple Interest

16.1 Introduction

Paridhi deposited Rs. 10,000 in a bank. After 1 year, Rs. 600 was added and her
money rose to Rs.10,600. The extra amount of money Rs. 600 paid by the bank is the
interest. Rs. 10,000 deposited is the principal and Rs. 10,600 is the amount.

Hence let's define some basic terms of simple interest.

Principal: The initial amount money borrowed or invested or deposited is called
principal or sum. It is denoted by P.

Interest: The additional money paid by the borrower for having used another
person's money, is called interest. It is denoted by I.

Rate of Interest : The interest on Rs. 100 in 1 year is known as rate percent per
annum. It is denoted by R i.e. if the interest of Rs. 100 in 1 year is Rs. 12 then, Rate
(R) = 12% p.a.

Amount: The total sum of money paid with interest is called amount. It is denoted
by A. Thus, Amount (A) = principal (P) + simple interest (I)

i.e. A = P + I

16.2 Calculation of Simple Interest

Let, principal deposited be Rs. P for T years, rate be R% per year, and the interest be
Rs. I.

By the definition of the rate of interest,

Interest on Rs. 100 for 1 year = Rs. R

Interest on Re 1 for 1 year = Rs. R
100

Interest on Rs. P for 1 year = Rs. R ×P
100

The interest on Rs. P for T years = Rs. R ×P×T
100

= Rs. P ×T× R
100

∴ Simple interest (I) = P×T×R .............. (i)
100

There are four quantities involved in the calculation of simple interest, namely

principal (P), rate per cent (R), time (T) and interest (I). So, if we are given any three,
we can find the fourth.

Oasis School Mathematics-8 220077

To find any one of P, R, T, it is convenient to rearrange the formula as follows.

P = I × 100 ....................... (ii)
R×T ....................... (iii)
....................... (iv) Remember !
I × 100
R = P×T
Again, we have,
I = P×T× R
100
T = I × 100
P×R A= P+I

A = P + I, where A = amount R= 100 × I
P×T

= P + P ×T× R T= 100 × I
100 P×R

= 100P +P× T × R P= 100 × I
100 T×R

A = P(100 + T × R) P = 100 × A
100 100 + T × R

P = A × 100 ....................... (v)
100 + T × R

Worked Out Examples



Example: 1

Find the interest on Rs. 3,600 at the rate of 12% p.a. for 3 years.
Solution:

Her e, Principal (P) = Rs. 3,600

Rate (R) = 12% p.a.

Time (T) = 3 years

We have, Interest (I) = P×T×R
100

= 3,600 × 3 × 12
100

= Rs. 1,296
Example: 2

In what time will Rs. 3000 amount to Rs. 3360 at 6% p.a. ?
Solution: Here,

Principal (P) = Rs. 3,000

Amount (A) = Rs. 3,360

208 Oasis School Mathematics-8

Rate(R) = 6% p.a
= A–P
We have, Interest (I) = Rs. (3,360 – 3,000)
= Rs. 360




Time (T) = ?
We have, T =
I × 100
P×R

= 360 × 100 = 2 years
Hence, the required time is 2 years. 3,000 × 6

Example: 3

Find the rate if the interest per rupee per month is 1 paisa.

Solution:

Here, Principal (P) = Re. 1 Alternative method

Interest (I) = 1 paisa Here, Interest of Re. 1 in 1
month is 1 paisa.
= Re. 1 .
Time (T) 100 Interest of Rs. 100 in 1
1 month is 100 paisa.
= 1 month = 12 year
Interest of Rs. 100 in 12
Rate (R) = ? months is 1200 paisa.

We have, R = 100 × I i.e. Interest of Rs. 100 in 1
P×T year is Rs. 12.

1 ∴ Rate (R) = 12 %
100 × 100

= 1× 1 = 12 %
12
∴ Rate (R) = 12 % p.a.

Example: 4

In how many years will a sum of money double itself at 10% per annum simple interest.
Solution:

Let, Principal (P) = Rs. x.

Then, Amount (A) = Rs. 2x

Rate (R) = 10% p.a.

Time (T) = ?

Now, Interest (I) = A – P

= Rs. 2x – Rs. x = Rs. x

We have, T = 100 × I = 100 × x = 10 years
P×R x × 10

Hence, the required time is 10 years.

Oasis School Mathematics-8 220099

Example: 5

What sum will amount to Rs. 1500 in 5 years at 10% per year simple interest ?
Solution:

Here, Amount (A) = Rs. 1,500

Time (T) = 5 years

Rate (R) = 10% p.a.

Principal (P) = ?

We have, P = 100 × A
100 + T × R

= 100 × 1500
100 + 5 × 10

= 1,50,000 = Rs. 1,000
150

So, the required sum is Rs. 1,000.

Example: 6

A sum of money doubles itself in 20 years at a certain rate. In how many years will it
treble itself at the same rate of interest?

Solution:

Here, in the first case,

Principal (P) = Rs. x (suppose)

Then, Amount (A) = Rs. 2x,

Time (T) = 20 years

Rate (R) = ?

Now, Interest (I) = A – P

= Rs. 2x – Rs. x

= Rs. x

We have, R = 100 × I
P×T

= xx××12000 = 5% p.a.
Again in the second case, Amount = Rs. 3x

Principal (P) = Rs. x

Rate (R) = 5% p.a. (by question)

Time (T) = ?

Now, I = A – P

= Rs. 3x – Rs. x = Rs. 2x

210 Oasis School Mathematics-8

We have, T = 100 × I
P×R

= 2x × 100 = 40 years
x×5

Hence, the sum will treble itself in 40 years.

Example: 7

Aslam deposited Rs. 4500 in a bank for 3 years at the interest rate of 10% p.a. How much
would he get after 3 years if he paid 5% tax on the interest.

Solution:

Here, Principal (P) = Rs. 4,500
Time (T) = 3 years


Rate (R) = 10 %

We have, Interest (I) = P × T × R
100

= 4500 × 3 × 10
100

= Rs.1,350

Tax on interest = 5% of 1,350

= 1500 × 1,350

= Rs. 67.50

Net interest = Rs. 1,350 – Rs. 67.50

= Rs. 1282.50

Amount = P + Net interest

= 4,500 + 1,282.50

= Rs. 5,782.50

Hence, the sum will be Rs. 5,782.50

Exercise 16.1

1. Find the simple interest and amount in the given cases.

(a) P = Rs. 1,600, T = 3 years, R = 5% p.a.

(b) P = Rs. 1,2000, T = 6 months, R = 10% p.a.

(c) P = Rs. 2,400, T = 73 days, R = 15% p.a.

Oasis School Mathematics-8 221111

2. Find the principal and amount in the following cases.

(a) I = Rs. 200, T = 5 years, R = 10% p.a.

(b) I = Rs.1,200, T=2 years, 6 months, R = 20% p.a.

3. Find the time in the following cases.

(a) P = Rs.6,500, I = Rs. 3120, R = 6% p.a.

(b) P = Rs. 2,500, I = Rs. 750, R = 5% p.a.

(c) P = Rs. 8,000, A = Rs. 10,000, R = 6 1 % p.a.
4

4. Find the rate of interest in the following cases:

(a) P = Rs. 6,000, I = Rs. 2,100, T = 5 years

(b) P = Rs. 1,600, A = Rs. 2,832, T = 5 1 years
2

(c) Interest per rupee per month is 1.5 paisa.

5. Find the principal in the following cases.

(a) A = Rs. 1,500, T = 2 years, R = 10% p.a.

(b) A = Rs. 4,000, T = 6 years, R = 10% p.a.

6. (a) Umesh borrowed Rs. 30,000 from a bank. The bank charged him 12% p.a.
simple interest. How much should he pay at the end of 3 years ?

(b) Bikalpa deposits Rs. 4,800 in a bank which pays 6% interest per annum. Find
the amount he would get at the end of 1 year 9 months.

(c) Sankalpa borrows Rs. 25,000 from a bank at 10% per annum. What amount
must he pay to the bank if he repays the loan in 6 months.?

7. (a) In how many years, Rs. 1,250 invested at 6% p.a. will yield Rs. 225 as simple

interest ?

(b) In what time will Rs. 10,000 amount to Rs. 15,250 at 7 1 % p.a. simple interest?
2

(c) In how many years will a sum of money double itself at 5% p.a. simple interest ?

8. (a) At what rate percent will Rs. 1,250 yield an interest of Rs. 100 in 2 years?

(b) At what rate percent p.a. will Rs. 1,600 amount to Rs. 2,000 in 5 years?

(c) At what rate percent p.a. will the simple interest on Rs. 75 for 9 months be Rs.4.50 ?

(d) At what rate percent per annum will a given sum of money treble itself in 25 years ?

9. (a) What sum amounts to Rs. 3,600 in 2 years at the rate of 10% p.a.?

(b) What sum amounts to Rs. 480 in 2 years at 15% per annum?

(c) A received from B a certain sum of money at 10% per annum simple interest.
After 73 days, A returned Rs. 5,100 to B and cleared the debt. Find the sum
borrowed by A.

10. A sum of money doubles itself in 10 years at a certain rate. In how many years will
it treble itself at the same rate of interest?

212 Oasis School Mathematics-8

11. (a) Lakpa deposited Rs. 5,000 in a bank for 3 years at the rate of 10% p.a. How
much will he get after 3 years if he has to pay 5% tax on the interest?

(b) John deposited Rs. 10,000 in a bank for 2 years. He received 6% p.a. interest from
the bank and paid 5% tax on his interest. How much would he get after 2 years?

Answer
1. (a) Rs. 240, Rs. 1840 (b) Rs. 600, Rs. 12,600 (c) Rs. 72, Rs. 2472 2. (a) Rs. 400, Rs. 600

(b) Rs. 2400, Rs. 3600 3. (a) 8 yrs. (b) 6 yrs (c) 4 yrs. 4. (a) 7% (b) 14% (c) 18% 5. (a) Rs. 1250
(b) Rs. 2500 6. (a) Rs. 40800 (b) Rs. 5304 (c) Rs. 26, 250 7. (a) 3 yrs (b) 7 yrs. (c) 20 yrs.
8. (a) 4% (b) 5% (c) 8% (d) 8% 9. (a) Rs. 3000 (b) Rs. 369.23 (c) Rs. 5000 10. 20 yrs.
11. (a) Rs. 6425 (b) Rs. 11,140

Assessment Test Paper Full marks : 30

Attempt all the questions.
Group - A [6 × 1 = 6]

1. (a) Convert 1112 into decimal number.
(b) Simplify: (+30) ÷ (–5)

(c) Find the value of : ( 3 + 1) ( 3 –1)

(d) Write 25,00,000 in scientific notation.

(e) Find the ratio of 25 cm to 2 m.

(f) What percent of 180 is 45?
Group - B [ 6 × 2 = 12]

2. (a) Convert 328 into quinary number.

(b) Simplify: 32 ÷ {16–(2 × 8 ÷ 2)}

(c) Simplify: 2– 3
2+ 3

3. (a) Find the value of x if 2, 4, 8, x are in proportion.

(b) If the cost of 5 articles is Rs. 200, find the cost of 15 articles.

(c) Find the profit percent if S.P. = Rs. 450 and C.P. = Rs. 300.

Group- C [3 × 4 = 12]

4. Simplify: 110112 + 10012 – 11112
5. Simplify: 5[125–{27–3(6–2)}]

6. A sum of money doubles itself in 20 years. In how many years will it be treble at the
same rate of interest?

Oasis School Mathematics-8 221133

Statistics
Statistics

10Estimated Teaching Hours

Contents
• Cumulative Frequency Table
• Line Graph
• Pie-chart
• Arithmetic Mean
• Median
• Quartiles
• Mode
• Range

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:
• Make cumulative frequency table from the given ungrouped data
• Make the line graph from the given data
• Make the pie-chart from the given data
• Calculate the arithmetic mean from the given grouped and ungrouped

data
• Calculate the median and quartiles from the given individual and

discrete data
• Calculate mode from individual and discrete data
• Calculate range from given individual data

Teaching Materials

• Graph sheet, A4 size paper, etc.

214 Oasis School Mathematics-8

Unit

17 Statistics

17.1 Collection of Data

A set of numerical facts collected for the purpose of investigation in order to fulfill
certain aims and targets is called "statistical data". There are mainly two types of
data:

a. Primary data

Data collected by the investigator himself or herself for the first time is called
primary data. The data collection by this method is more relevant and reliable but
more time consuming. Primary data can be collected by the investigator by direct
personal interview, by oral interviews, by correspondence, questionnaires through
the post, etc.

b. Secondary data

The data collected from the secondary sources are called secondary data. For
example, the weather forecast by radio, television or newspaper, the number of
population of different localities by the statistical department, etc, all are secondary
sources of collection of data. Since we have to depend upon other sources, the
secondary data are not reliable and may not be much relevant.

Frequency table

After the collection of data by primary or secondary sources is over, they should be
represented properly for the presentation so that the investigator can work out with
the set of collected data. The data collected at the very first time are called raw data.
In order to represent them systematically, we can take the help of frequency table
where tally marks ( / ) are used to indicate the number of variables of certain type
like age, mark, height. The presentation of data in such a table is called frequency
table. For example, consider the marks of 15 different students that are collected as
below in a unit test of full marks 10.

1, 5, 6, 9, 2, 8, 10, 4, 7, 6, 4, 2, 8, 7, 8,

To represent the above raw data, the following frequency table can be used where
number of repeated items are represented by the number of tally mark.

Oasis School Mathematics-8 221155

If the number of any variable repeats 5 times, we use cross tally over the first
four tally .

Frequency Table

Variable Tally Mark Frequency (f)
Mark (X)
/ 1
1 // 2
2 0
3 // 2
4 / 1
5 // 2
6 // 2
7 /// 3
8 / 1
9 / 1
10

Frequency distribution

When the data are arranged in a table sharing their respective frequencies in the
frequency column, that tabular arrangement of data is called frequency distribution.
Frequency distribution can be prepared in the following three ways:

(i) Individual series or ungrouped data.

(ii) Discrete series or ungrouped frequency distribution.

(iii) Continuous series or grouped frequency distribution.

(a) Individual series

In the individual series, items are listed individually one by one. No frequency of the
variables is mentioned. So, it is just an ungrouped data, e.g. marks of 10 students are
given as.

1, 2, 5, 6, 4, 3, 1, 7, 3 and 8.

(b) Discrete or ungrouped frequency distribution

In this type of distribution, the variable can be only the exact and specific value with
their corresponding frequency. For example, marks of 15 students are tabulated as
below.

Marks 12 3 45

No of Students 34 5 21

(c) Continuous series or grouped frequency distribution

The data which are represented by continuous variables is called continuous series.
For, e.g.,

216 Oasis School Mathematics-8

Class Interval 0-10 10-20 20-30 30-40 40-50

No. of Students 23 5 14

The above table shows the marks of 15 students in terms of different classes. We
can observe that there are two students whose marks lie between the range 0 to
10 (exclusive). Here the lower and upper limits of the first class are 0 and 10. The
difference between them is called class height or width or size or magnitude.

∴ Class size = Upper limit – Lower limit

The mid–value of the class is obtained by the average value of upper and lower limits.

∴ Mid-value = Upper limit + Lower limit
2

In the above case the lower limit of the class is also considered as its value while the
upper limit will be excluded and will be included for the next class. But this is not
applied for the upper most class interval.

17.2 Cumulative Frequency Distribution

(i) Construction of less than cumulative frequency table:

Marks Obtained 0-10 10-20 20-30 30-40 40-50
52
No. of Students 36 8

Solution:
Less than cumulative frequency table

Marks obtained Frequency (f) Cumulative frequency (c.f.)

0-10 3 3
10-20 6 3+6 = 9
20-30 8 9+8=17
30-40 5 17+5 = 22
40–50 2 22+2 = 24
N = 24

Here, 3 students get (0 – 10) marks

6 students get (10 – 20) marks.

Hence, the number of students who get less than 20 marks = 3 + 6 = 9.

Similarly, the number of students who get less than 30 marks = 9 + 8 = 17

Proceeding in this way, we can prepare the cumulative frequency table as given below.

Ascending order of magnitude and frequencies are cumulated from the top to bottom.

Oasis School Mathematics-8 221177

Worked Out Examples



Example: 1

Construct a frequency distribution table from the marks obtained by 20 students:
24, 26, 22, 20, 20, 18, 18, 22, 24, 26,
28, 20. 22, 22, 26, 24, 22, 22, 20, 20.
Solution:
Here, the lowest marks of the class = 18 and the highest marks of the class = 28

Mark (X) Tally Mark Number of Students
18 // 2
20 //// 5
22 6
24 //// / 3
26 /// 3
28 /// 1

/

Example: 2

Prepare a cumulative frequency distribution table for the following ungrouped
frequency distribution.
Wages (Rs.) 40 50 60 70 80 90 100

No. of persons 4 7 9 10 8 5 2

Solution:

Wages (x) Number of persons (f) c. f.

40 44
50
60 7 4 + 7 = 11
70
80 9 11 + 9 = 20
90
100 10 20 + 10 = 30
8 30 + 8 = 38

5 38 + 5 = 43

2 43 + 2 = 45

N = 45

218 Oasis School Mathematics-8

Example: 3

From the following distribution, construct a less than cumulative frequency table and
find the number of students getting below 60 marks.

Marks obtained 0–20 20–40 40–60 60–80 80–100
No. of students 10 15 25 15 10

Solution:
A less than cumulative frequency table

Marks obtained Frequency (f) c.f.
0 - 20 10 10

20 - 40 15 10 + 15 = 25

40 - 60 25 25 + 25 = 50

60 - 80 15 50 + 15 = 65

80 - 100 10 65 + 10 = 75

N = 75
The number of students getting below 60 marks = 50.

Example: 4

Prepare a frequency distribution for the following marks obtained by 30 students in
class intervals of 10.

20, 14, 18, 35, 47, 33, 18, 16, 25, 4, 30, 14, 30, 27, 11,

29, 29, 20, 22, 15, 29, 25, 20, 29, 14, 39, 9, 18, 10, 25.
Also, construct a cumulative frequency table.

Solution:

Marks Tally bars Number of Cumulative
obtained students frequency

0 - 10 // 2 2

10 - 20 //// //// 10 2 + 10 = 12

20 - 30 //// //// // 12 12 + 12 = 24

30 - 40 //// 5 24 + 5 = 29

40 - 50 / 1 29 + 1 = 30

N = 30

Oasis School Mathematics-8 221199

Exercise 17.1

1. The ages of 20 students of class VIII are given below.

12, 14, 13, 12, 14, 14, 15, 14, 13, 12, 13, 14, 15, 14, 15, 13, 14, 15, 14, 13.
Prepare a frequency table of the above data.

2. Construct the cumulative frequency table from each of the following frequency
distributions.

Marks 5 15 25 35 45

No. of Students 12 15 28 25 20

3. The weights (in kilograms) of 25 students are given as follows:

35, 38, 36, 37, 38, 35, 37, 36, 35, 38, 36, 36, 37,

37, 35, 38, 36, 37, 37, 38, 36, 38, 37,35, 36.

Construct a frequency table of the above data and also prepare the cumulative
frequency table.

4. Construct the cumulative frequency table from the following data.

Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60

No. of Students 5 10 15 20 20 5

Also, find the number of students getting marks below 40.

5. From the following data of the weight (in kg) of 39 students, construct a frequency
distribution table taking the first class as 30 - 35.

37, 34, 42, 41, 43, 44, 32, 48, 49, 51, 55, 58, 62, 60, 61, 57, 40, 45, 36, 52,

53, 37, 48, 50, 39, 55, 51, 36, 53, 37, 49, 38, 59, 61, 47, 52, 49, 50, 53.
Also, construct a cumulative frequency table from the above data.

Answer
Consult your teacher.

17.3 Line Graph or Time Series Graph

The graph obtained by plotting the variables along x-axis and corresponding
frequencies along y –axis and by joining all those points on the cardinal plane is
called a line graph. Let's see an example.

Marks 20 25 30 35 40 45
No. of Students 10 3 6 8 15 7

220 Oasis School Mathematics-8

Number of students Y Steps:

50 • Take variable along X-axis and
45 frequency along Y-axis.
40
35 • Plot all the points.
30
25 • Join all the points by a straight
20 line.
15
10 Note: On the questions related to time,
5 take time along x –axis and values
of the variable along y-axis.
X' 20 25 30 35 40 45 50 55 60 65 X

Y' Marks

17.4 Pie–chart or Pie–diagram or Circular Diagram

A pie–chart is a diagrammetical representation of statistical data in circular form
with several subdivisions. It is also known as a circular diagram or pie–diagram or
angular diagram. The radial lines divide the pie–diagram into different sectors in
such a way that the area of the sectors are proportional to the different components
of the data. As there is 360° at the centre, the sum of the angles representing different
items is equal to 360°.

Angle at the centre by an item = Value of an item × 360°
Total value



Example: 1

Represent the following data for the amount spent on the construction of a house by a
pie–chart.

Item Cement Timber Bricks Labour Steel Misc. Total

Expenditure (in 180 90 135 225 135 135 900
Rs. thousands)

Solution:

Here, Total value = 900, total angle = 360°
Value of an item

∴ Angle at the centre by an item = Total value × 360°

The following pie–diagram shows the amount spent in the construction of a house.

Oasis School Mathematics-8 222211

Calculation for construction of a pie diagram.

Items of Expenditure in Angle at centre
Expenditure Rs. thousand
360 × 180 = 720
Cement 180 900
Timber 90
Bricks 135 360 × 90 = 360 54°
900 72°
Labour 225
360 × 135 = 540 54° 54°
Steel 135 900

Misc. 135 360 × 225 = 900
900
36°

360 × 135 = 540 90°
900

360 × 135 = 540
900

Total 900 3600

Example: 2

The total monthly income of a family is Rs. 7,200. The sources of income are shown in
the given pie chart. What is the income of the family from each title?

Solution:

Here, 360° represents the total income Rs. 7,200.

1° represents the total income Rs. 7,200
360
7,200
85° represents the total income = Rs. 360 × 85 = Rs. 1,700

Income from business = Rs 1,700.

222 Oasis School Mathematics-8

Similarly, Income from house rent = Rs. 7,200 × 75
360

= Rs. 1,500

Income from salary = Rs. 7,200 × 125
360

= Rs. 2,500.

Income from agriculture = Rs. 7,200 × 25
360

= Rs. 500

Income from others = Rs. 7,200 × 50
360

= Rs. 1,000.

Remember !

Value of an item = Total value × angle of that item
3600

Exercise 17.2

1. Draw a line graph for the following data:
(a)

Years 2061 2062 2063 2064 2065 2066 2067 2068
200 150 300 400 200 250
No. of students 250 150

(b)

Years 2006 2007 2008 2009 2010 2011

Profits (In lakhs) 12 14 16 14 16 14

(c) The monthly average temperature of a certain town is given in the following
table:

Month Baisakh Jestha Ashadh Shrawan Bhadra Aswin
Temperature
240 280 260 290 320 200

2. (a) The following table gives the details of monthly expenditure of a family. Draw
a pie–chart to present the following information.

Item Food Clothing Rent Eduction Lighting Miscellaneous
Expenditure 6000
2000 2000 2500 2500 3000
(in Rs.)



Oasis School Mathematics-8 222233

(b) The number of students in different levels of school is given in the table. Repre-
sent it in a pie chart.

Level Pre primary Primary Lower Sec. Secondary Higher Sec.

Number of students 240 540 360 600 420

(c) Represent the monthly expenses of a household in a pie-chart.

Heading Food Eduction House Rent Transport Others
Expenditure
6000 2000 2000 2500 2500

3. (a) Circular graph given below shows the number of students enrolled in classes V,
VI, VII and VIII of a school.

Study the graph carefully and answer the following questions. VII, 10%

(i) Which class has the maximum number of students? VIII

(ii) If 1% equals 2 students, how many students are there in VI
class VII ?

(iii) What is the total number of students enrolled?

V

(b) Given pie chart shows the votes obtained by four can- BA

didates A, B, C and D in an election. If the total number of 950 850

vote cast is 5,400. Find the number of votes obtained by each C 750
candidate. 1050 D



Answer

1. Consult your teacher 2. Consult your teacher 3. a. (i) Class VI (ii) 20 (iii) 720
(b) A = 1275 B = 1425 C = 1575 and D = 1125

17.5 Measures of Central Tendency

The single value which represents the given set of variables is called the volume
central tendency. Measure of central tendency is also called average. There are
mainly three types of measurement of central tendency. They are (i) Mean (ii)
Median (ii) Mode.

I. Arithmetic Mean

The most popular and widely used measure for representing the entire data by

single value is arithmetic mean. The arithmetic mean or simply mean or average

is defined as the total sum of observation divided by total number of observations.

That is,

Arithmetic mean = Total sum of observations
Total number of observations

It is denoted by X . It is simply called Mean.

224 Oasis School Mathematics-8

(i) Mean of ungrouped data when none is repeated: Mean of ungrouped data is ob-

tained by adding all the observations and dividing the sum by the total number of

observations.

∴ Arithmetic Mean ( X) = ∑X .
n

Where ∑X = sum of n items, n = number of items.

Note: The Greek letter ∑(sigma) means "Summation of".

Calculation of Arithmetic Mean in repeated data (discrete series)

If the variable x1, x2, x3 ……. xn are repeated f1,f2 f3 …….fn times, respectively, then mean
of this data.

(x) = ∑fx
∑f

where, ∑fx = f1x1 + f2x2 + f3x3 + ……….+ fnxn
and ∑f = f1 +f2 + f3 + ……+ fn
Remember ! N = ∑f

Mean of grouped and continuous data

In the case of grouped and continuous data, we should find the mid -value of each
class interval.

Mid -value of a class (m) = upper limit + lower limit

2
Mid value represents the variable of that class.

∴ In continuous series, = ∑fm
Mean (x) ∑f

Where, m = mid – value.

Mean of the Combined Series

If n1 and n2 be the number of items of first and second series respectively and x1, x2 be their

respective means of two series, then combined mean of the first and the second series is given

by x 12 = n1x1 + n2x2 .
n1+n2

Oasis School Mathematics-8 222255

Worked Out Examples



Example: 1

The heights of 5 students are 145 cm, 150 cm, 149 cm, 150 cm and 151 cm respectively.
Find the mean height per student.

Solution:

Given, height of 5 students

145cm, 150 cm, 149cm, 155cm, 151cm.

Now, ∑ X = (145 + 150 + 149 + 155 + 151)cm = 750 cm

Number of items (n) = 5.

We have, Mean ( X) = ∑X = 7550
n

= 150 cm.

Example: 2

The marks obtained by 10 students of a class are 46, 51, 62, 70, 35, x, 50, 83, 65, 52. If the
average marks is 55, find x.

Solution:

Here, Mean = 55, n = 10,

∑X = (46 + 51 + 62 + 70 + 35 + x + 50 + 83 + 65 + 52)

= 514 + x
Now, mean ( X) =
or, ∑X
or, 55 = n
550 =
514 + x
10

514 + x

∴ x = 550 – 514

E xam ple: 3 = 36

The marks obtained by 50 students in an examination are given below:

Marks 10 15 20 25 30
No. of students 5 10 18 9 8

Find the average marks.

Solution:

Let marks be denoted by X and no. of students by ƒ.

226 Oasis School Mathematics-8

Calculation of average marks

xf fx

10 5 50

15 10 150

20 18 360

25 9 225

30 8 240

N=50 ∑fX= 1025

Here, N = 50, ∑ƒx = 1025, (x) = ?

Now, Mean (x) = ∑Nfx = 1025
50

= 20.5

Hence, their average marks = 20.5

Example: 4

Find the arithmetic mean from the following distribution:

Class-interval 10-20 20-30 30-40 40-50 50-60
Frequency 3 6 8 5 2

Solution:
Let mid -value of classes be denoted by m and frequency by ƒ.
Calculation of mean:

Class-interval f Mid value (m) fm
15 45
10-20 3

20-30 6 25 150

30-40 8 35 280

40-50 5 45 225

50-60 2 55 110
Here, N=24 N = 24, ∑fm = 810

∑ƒm = 810, X = ?

Now, X = ∑fm = 82140
N

= 33.75

Hence, the arithmetic mean is 33.75.

Oasis School Mathematics-8 222277

Example: 5

The average age of 30 boys in a class is 14 years and that of 20 girls is 12 years. Find
average age of the students in the class.

Solution:

Here, n1= 30, x1 = 14 yrs., n2 = 20, x2 = 12 years.

Combined average age of boys and girls (x2) = ?

Now, x 12 = n1x1 + n2x2
n1+n2

30 × 14 + 20 × 12

= 30 + 20

= 420 + 240 = 660 = 13.2 years.
50 50

Hence, the average age of the students in the class is 13.2 years.

Exercise 17.3

1. Find the arithmetic mean for the following data.
(a) Rs. 5, Rs. 10, Rs. 15, Rs. 20, Rs. 25, Rs. 30, Rs. 35
(b) 50 kg, 52 kg, 53 kg, 54 kg, 60 kg, 65 kg, 71 kg, 75 kg
(c) 10 cm, 12 cm, 14 cm, 16 cm, 20 cm, 24 cm.
2. Calculate the average marks of : 12, 15, 18, 17, 16, 20.
3. (a) If the arithmetic mean of 5, 7, 9, x and 15 is 10, find the value of x.
(b) If the mean of 3, 7, 10, 15 and x is 12, find the value of x.
4. (a) From each of the following frequency distribution table, find the mean.

x 10 20 30 40 50
f 5 10 15 10 9

(b) Calculate the mean mark from the following frequency distribution table.
Marks obtained 10 20 30 40 50
No. of students 5 2 3 6 4

(c)

x: 5 10 15 20 25 30
f: 2 5 10 7 4 2

5. From each of the following frequency distribution table, find the mean.
(a)

Class 0-10 10-20 20-30 30-40 40-50

No. of students 5 10 15 10 9

228 Oasis School Mathematics-8

(b)

Marks obtained 10-20 20-30 30-40 40-50 50-60

Frequency 46 8 53

(c)

x: 0-10 10-20 20-30 30-40 40-50 50-60

f: 54 6 2 85

6. Find the combined mean ( x12 ) of the following two series.
(a) n1 = 70, x1 = 75, n2 = 30, x2 = 65
(b) n1 = 40, x1 = 55, n2 = 60, x2 = 65
7. (a) The mean weight of 30 boys in a class is 45 kg. And that of 20 girls is 40 kg.

Find the mean weight of the students in the class.

(b) The mean marks in Maths of 100 students in a class was 72. The mean mark of
70 boys was 75. Find out the mean marks of girls in the class.

Answer
1. (a) 20 (b) 60 (c) 16 2. 16.33 3. (a) 14 (b) 25 4. (a) 31.63 (b) 31 (c) 17
5. (a) 26.63 (b) 33.84 (c) 31.33 6. (a) 72 (b) 61 7. (a) 43 kg (b) 65

II. Median

The data of the variables are generally collected randomly. One must arrange them
systematically in order. Suppose the data are arranged in ascending order of mag-
nitude. The data which lies exactly in the middle is called median. It is denoted by
Md. Hence, there will be almost same number of variables on either side of median.

Calculation of median for individual series

Let there are n observations of variables. Then positional value of median is given

by ( n2+1(th where 'n' is number of observations. From this, we can easily find the

median value.

Calculation of median for discrete distribution

If the variables are given in terms of their respective frequencies, we will find the
cumulative frequency of each variable. The median position will be obtained by using
the formula,

Position of Median = ( N+1 (th item
2

Where, N is the number of observation.

Oasis School Mathematics-8 222299

III. Quartiles

The variate values which divide the given data into 4 equal parts are called quartiles.
There are three points Q1, Q2 and Q3 such that Q1< Q2< Q3.

Calculation of Quartiles

The working principle for calculating quartiles is the same as that for median.

(a) Individual series or an ungrouped data: After data have been arranged in ascending

order, the quartiles are calculated by applying the following formula.

Q1 = value of ( n+1 (th item, Q2 = value of 2 ( n+1 (th item,
4 4

Q3 = value of 3 ( n+1 (th item, where n = number of observations or items
4

(b) Discrete series or ungrouped frequency distribution: After the data have
been arranged in ascending order, quartiles are calculated by following formula.

Q1 = value of ( N+1 (th item, Md = Q2 = value of 2( N+1 (th item = ( N+1 (th
4 4 2

Q3 = value of 3 ( N+1 (th item, where N = total frequency
4

Worked Out Examples



Example: 1

Find the median size of shoes from the given data : 2, 5, 7, 1, 4, 3, 6, 8, 10.

Solution:

Arranging into ascending order of magnitude,

Here, 1, 2, 3, 4, 5, 6, 7, 8, 10

Number of observations (n) = 9

Median = ( n+1 ( th item
2

∴ = ( 9+1 ( th item = 5th item.
2

Median (Md) = 5

Example: 2

Find median of the following data:

x 5 10 15 20 25
ƒ36852

230 Oasis School Mathematics-8

Solution: f c.f.
3 3
x
5

10 6 9

15 8 17

20 5 22

25 2 24
N=24

Now, Md= ( N+1 (th item = ( 24+1 (th = ( 25 (th = 12.5th item.
2 2 2

The c.ƒ. just greater than 12.5 is 17. So corresponding median value is 15.

∴Md = 15

Example: 3

Find the Median from the following distribution:

X: 10 20 30 40 50 60

ƒ: 5 10 13 10 9 8

Solution: c.f.
5
Calculation of median 15
28
xf 38
10 5 47
20 10 55
30 13
40 10
50 9
60 8

N=55

Now, Md = ( N+1 (th item = ( 55+1 (th = 28th item
2 2

∴ Md = 30

Example: 4

Find lower (Q1) and upper quartiles (Q3) from the following data:
15, 6, 17, 25, 30, 9, 19, 35
Solution:
Arranging the given data in ascending order 6, 9, 15, 19, 25, 30, 35

Here, n = 8

Oasis School Mathematics-8 223311

We have, = ( n+1 (th item.
Q1 4

= ( 8+1 (th item
4

= 2.25th item

= 2nd item + 0.25 (3rd item – 2nd item)

= 9 + 0.25 (15 – 9)

= 9 + 1.5

= 10.5

Q3 = 3( n+1 (th
4

= 3(2.25)th item

= 6th item + 0.75 (7th item – 6th item)
= 25th + 0.75 (30 – 25)
= 25 + 0.75 × 5
= 25 + 3.75
= 28.75

Example: 5

Calculate Q1 and Q3 from the following data.

Marks 10 20 30 40 50 60
No. of students 4 16 20 18 11 6

Solution:

Calculation of Q1, Q2 and Q3.

xf c.f.
10 4 4
20 16 20
30 20 40
40 18 58
50 11 69
60 6 75

N = 75

For Q1, ( N+1 (th item = ( 75+1 (th item = (746 (th item = 19th item.
4 4

The c.ƒ. just greater than 19 is 20.

So corresponding lower quartile value is 20.

∴ Q1 = 20

232 Oasis School Mathematics-8

For Q3, 3 ( N+1 (th item = 3 ( 75+1 (th item = 57th item.
4 4

The c.ƒ. just greater than 57 is 58. So corresponding 3rd quartile value is 40.

∴ Q3 = 40.

Exercise 17.4

1. Find the median for each of the following data.
(a) 4, 3, 8, 6, 1, 12, 15
(b) 3, 7, 11, 12, 10, 8, 6, 5, 9, 4
(c) 3 kg, 5 kg, 10 kg, 7 kg, 2 kg, 15 kg, 2 kg, 10 kg
2. In an examination, out of 10 marks 9 students got the following marks:
9, 5, 8, 4, 2, 3, 1, 6, 7.
Find the median mark.
3. Find the median of the following frequency distribution.

(a) x 5 10 15 20 25 30 35
ƒ 4 5 7 10 8 6 5

(b) x 2 4 6 8 10 12 14 16
ƒ 3 5 9 12 15 10 8 2

(c) Size 2 34 5 6 7

No. of households 2 3 9 21 11 5

4. Find Q1 and Q3 from the following ungrouped data:
(a) 5, 10, 3, 12, 8, 15, 2

(b) 10, 5, 20, 15, 12, 30, 10, 25

(c) Rs. 5, Rs. 3, Rs. 7, Rs. 2, Rs. 12, Rs. 15, Rs. 10

5. Find lower and upper quartiles from the following data.

(a) x 3 6 9 12 15 18 21
ƒ 5 6 8 10 7 5 2

(b) Size 1 3 5 7 9 11

Frequency 4 5 7 3 2 2

Oasis School Mathematics-8 223333

(c) Marks 30 40 50 60 70 80 90

Number of Students 10 15 20 25 10 5 2

Answer
1. (a) 6 (b) 7.5 (c) 6 2. 5 3. (a) 20 (b) 10 (c) 5 4. (a) Q1 = 3, Q3 = 12
(b) Q1 = 10, Q3 = 23.75 (c) Q1=3, Q3 = 12 5. (a) Q1= 6, Q3= 15 (b) Q1= 3, Q3= 7 (c) Q1= 40, Q3= 60

IV. Mode

The mode or the modal value is that value in a distribution which occurs most
frequently. It is the most common of value found in a distribution. Thus, the most
typical, repeated value of a distribution is called the mode. It is denoted by M0.

Calculation of Mode for ungrouped data

a. Individual series:

In case of individual series, mode can often be found by inspection. To find the
mode, count the number of times various values repeat themselves. Then, the value
occurring maximum number of times is the modal value. For example, the modal
value of the given series of individual observation is as follows.

Calculation of Mode: 5, 6, 2, 0, 2, 1, 1, 4, 2, 2, 5, 4

Variable No. of occurrence

0 1
1 2
2 4
5 2
6 1

Here, a variate value 2 repeats maximum number of times 4 times.

∴ Mode = 2

b. Discrete series (or ungrouped frequency distribution)

In case of discrete series also, mode can be obtained by inspection only. Mode occurs
with maximum frequency.

Variable values having greatest frequency is the mode.

Example: 1

Find the modal value for the following data:

Sizes of shoes 5 6 7 8 9 10 11

Number of persons 15 20 25 35 27 18 10

Solution:
Here, the maximum frequency is 35. So corresponding modal size is 8.
∴ Mode = 8

234 Oasis School Mathematics-8

Exercise 17.5

1. Find the mode for each of the following data:
(i ) 65, 6, 1, 0, 2, 1, 2, 5, 2, 6, 4, 4, 2, 4
(ii) Rs. 3, Rs. 4, Rs. 2, Rs. 5, Rs. 3, Rs. 4, Rs. 5, Rs. 5, Rs. 4,
Rs. 6, Rs. 5, Rs. 6, Rs. 6, Rs. 3, Rs. 7, Rs. 5, Rs. 7

2. A boy scored the following marks in various class tests during term exams, each
test being marked out of 20.

15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16.
What are his modal marks ?
3. Find the mode for the following data.

x 10 15 20 25 30 35 40
ƒ 3 6 10 12 7 5 2

4. Calculate the mode for the following data.

Age in years 10 11 12 13 14 15 16
18 12 7
No. of students 10 15 20 25

Answer
1. (i) 2 (ii) 5 2. 16 3. 25 4. 13



V. Range

Marks obtained by 20 students of class VIII are given below:
72, 78, 42, 90, 63, 48, 30, 25, 82, 92, 46, 89, 97, 29, 53, 76, 58, 43, 39, 75
If we arrange the given data in ascending order,
25, 29, 30, 39, 42, 43, 46, 48, 53, 58, 63, 72, 75, 76, 78, 82, 89, 90, 92, 97
It can be observed that,
The lowest marks of the class = 25
The highest marks of the class = 97
The marks of other students are spread between 25 and 97.
Difference between the highest marks and the lowest marks = 97 – 25 = 72
This value is called the range.

Oasis School Mathematics-8 223355

Hence the difference between the highest and the lowest values of the observation
is called the range.

Range = highest value of the observation – lowest value of the observation .

i.e. Range = Largest item – Smallest item

= L – S

Worked Out Examples



Example: 1

Find the range of the following data: 20, 25, 31, 32, 18, 15

Solution:

Given data: 15, 18, 20, 25, 31, 32

Largest item (L) = 32

Smallest item (S) = 15

We have, Range = L – S = 32 – 15 = 17

Exercise 17.6

1. Find the range of the following data:

(a) 3, 7, 9, 12, 6, 15

(b) 19, 12, 8, 17, 23, 25, 30

(c) 65, 60, 52, 35, 80, 68, 92, 31, 56.

2. The height (in cm) of 20 students in a class is given below:
140, 141, 151, 152, 162, 157, 136, 145, 151, 150, 165, 135,
142, 146, 148, 140, 141, 152, 156, 160
(i) What is the height of the tallest student?
(ii) What is the height of the shortest student?
(iii) Find the range.


Answer

1. (a) 12 (b) 22 (c) 61 2. (i) 165 cm (ii) 135cm (iii) 30

236 Oasis School Mathematics-8

Assessment Test Paper

Attempt all the questions

Group 'A' [4 × 1 = 4]

1. (a) If ∑x = 54, n = 9, find X .
(b) Find the median from the given data: 15, 17, 19, 23, 27.
2. (a) Find the first quartile from the given data: 7, 12, 13, 17, 19, 24, 27.
(b) Find the mode from the given data :

x 12 15 17 18 20
f2 3 8 6 3


Group 'B' [4 × 2 = 8]

3. (a) Construct the frequency distribution table from the given marks.
25 24 30 32 38 42 47 52 48 36
42 31 49 52 54 41 23 32 42 52

(b) If x1 = 12, x2 = 15, n1 = 20, n2 = 16, find the combined mean x12.

4. (a) Find the range from the given data:
15, 21, 27, 42, 84, 75, 63, 58, 22, 91, 48

(b) Find Q3 from the given data:
28, 37, 44, 40, 38

Group 'C' [2 × 4 = 8]

5. (a) The number of students indifferent levels of school is given below. Represent
this data in a pie-chart.

Level Pre-primary Primary Lower Secondary Higher
secondary 600 secondary
Number of 240 360
students 300 300

6. Calculate the mean from the given data:

Value 10 20 30 40 50 60 70
Frequency 8 497 9 48

Oasis School Mathematics-8 223377

Algebra

40Estimated Teaching Hours

Contents
• Some Special Product Formulae
• Factorisation
• HCF and LCM
• Rational Expressions
• Indices
• Equation, Inequality and Graph

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:
• State and use the product formulae (a ± b)2 and (a ± b)3
• Factorise simple algebraic expression (binomial and trinomial)
• Find the HCF and LCM of algebraic expressions
• Simplify the simple rational expressions
• State and use the properties of indices to solve the simple problems

related to indices
• Solve the linear equation of one variable
• Solve two simultaneous equations graphically
• Find the slope and intercepts of given line
• Solve the simple problems of inequality

Teaching Materials

• Flash cards, chart paper, A4 size papers, etc.

238 Oasis School Mathematics-8

Unit Some Special Product
Formulae
18

18.1 Some Special Products and Formula

Formula 1

(a + b)2 = a² + 2ab + b²

Proof: We have,

(a + b)2 = (a + b) (a + b)

= a(a + b) + b(a + b)

= a² + ab + ba + b² [∵ ba = ab]

= a² + ab + ab + b²

= a² + 2ab + b²

∴ (a + b)2 = a² + 2ab + b²

∴ (a + b)² = a² + 2ab + b²

Geometrical Interpretation A a Gb B

In the figure, the length of each side of square ABCD is a² ab a
(a + b) unit which divides each side into two parts and a
square ABCD into four parts. OF
ab b² b
Here, AB = AG + GB = a + b E a Hb C
BC = BF + FC = a + b as shown in figure, b
Area of square AEOG = a2
D

Area of square HOFC = b2

Area of rectangle GOFB = ab

Area of rectangle DEOH = ab.

Here, area of square ABCD = area of small square AEOG + Area of small square
HOFC + Area of rectangle GOFB + Area of rectangle DEOH.

or (side)2 = a² + b² + ab + ab [∵ Area of square ABCD = (side)2]

or (AB)2 = a² + b² + 2ab

∴ (a + b)² = a² + 2ab + b²

Oasis School Mathematics-8 223399

Formula 2

(a – b)2 = a² – 2ab + b² Remember

Proof: We have a² + b² = (a + b)² – 2ab

(a – b)2 = (a – b) (a – b) a² + b² = (a – b)² + 2ab
= a(a – b) – b (a – b) (a + b)² = (a – b)² + 4ab
(a – b)² = (a + b)² – 4ab

= a² – ab – ab + b²

∴ (a – b)² = a² – 2ab + b²

Geometrical Interpretation

In the figure, the sides of square ABCD is a and each
sides are decreased by b as shown.

Here, area of square AEOG = (a–b)2
Area of square HOFC = b2
Area of rectangle GOFB = b((a–b)

Area of rectangle EDHO = b(a–b)

Area of square ABCD = Area of square AEOG

+ Area of rectangle GOFB + Area of rectangle EDHO

+ Area of square HOFC

or, (side)2 = (a – b)2 + b(a – b) + b(a – b) + b²

or, (AD)2 = (a – b)2 + ba –b² + ba – b² + b²

or, (a)2 = (a – b)2 + ab + ab – b²

[∴ba = ab and side of square ABCD = AD = a]

or, a² = (a – b)2 + 2ab – b²

or a² – 2ab + b² = (a – b)2

∴ (a – b)2 = a² – 2ab + b²

Formula 3

a² – b² = (a+b) (a–b)

Proof: We have, (a + b) (a – b) = (a + b) (a – b)

= a(a – b) + b(a – b) = a² – ab + ab – b² = a² – b²

∴ a² – b² = (a + b) (a – b)

In words, the product of the sum and difference of any two quantities is equal
to the difference of their squares.

240 Oasis School Mathematics-8

Worked Out Examples

Example: 1

Write down the square of: 2x +y12
Solution:

The required square of (2x + y12) = (2x+ y12)2

= (2x)2 + 2 × 2x × y12 + (y12)2
4x 1
= 4x2+ y2 + y4

Example: 2

If a + b = 8, ab = 5, find the value of a² + b².

Solution: Alternate method

Here, a + b = 8 Now, we know that

Now, squaring on both sides, we get a2 + b2 = (a + b)2 – 2ab

(a + b)2 = (8)2 = (8)2 – 2 × 5 [∵ a + b =8, ab = 5]

or a² + 2ab + b² = 64 = 64 – 10 = 54
or a² + 2× 5 + b² = 64 ∴ a2+b2 = 54

or a² + 10 + b² = 64

or a² + b² = 64 – 10

∴ a² + b² = 54

Example: 3 Alternate method
We know that
If x – 1 = 8, find the value of x2 + 1 .
x x2

Solution: a2 + b2 = (a – b)2 + 2ab

Here, x – 1 = 8 ∴ x2 + 1 = (x)2 + ( 1 )2
x x2 x

Now, squaring both sides, we get = (x– 1 )2 + 2 × x × 1
xx
( )x – 1 2 = (8)2 = (8)2 + 2 = 64 + 2 = 66
x

or x2 – 2 ×x × 1 + 1 = 64 ∴ = x2 + 1 = 66
x x2 x2

or x2 – 2 + 1 = 64
x­2

∴ x2 + 1 = 64 + 2 = 66
x2

Oasis School Mathematics-8 224411

Example: 4

If (a+b)=10, ab=21, find the value of (a–b).

Solution: (a + b) = 10, ab = 21
Here, (a – b)2 = (a+b)2 – 4ab
We have, (a – b)2 = (10)2 – 4 × 21
= 100 – 84
(a – b)2 = 16
(a – b)2 = (4)2
a – b = 4.
∴

Example: 5

Express as a perfect square of: 16a2 + 40ab + 25b2

Solution: 16a2 + 40ab + 25b2 = (4a)2 + 2.4a.5b + (5b)2
Here,

= (4a + 5b)2

Example: 6

Find the value of: 3.5 × 3.5 + 2 × 3.5 × 1.5 + 1.5 × 1.5

Solution:

Here, 3.5 × 3.5 + 2 × 3.5 × 1.5 + 1.5 × 1.5
= (3.5)2 + 2 × 3.5 × 1.5 + (1.5)2

= (3.5 + 1.5)2

= (5.0)2 = (5)2 = 25

Example: 7

Find the product of: (x+y+z) (x+y-z)

Solution:

(x+y+z) (x+y-z) = (x+y)2 – z2
= x2+2xy+y2-z2


Example: 8

Without actual multiplication, find the value of 103 × 97.

Solution:

Here, 103 × 97 = (100 + 3) (100 – 3)
= (100)2 – (3)2 [∵ (a + b) (a – b) = a² – b²]
= 10000 – 9 = 9991

242 Oasis School Mathematics-8

Example: 9

Simplify : 1210 × 1210 – 1110 × 1110
1210 – 1110
Solution:

Here, (1210)2 – (1110)2 = (1210 + 1110) (1210 – 110)
(1210 – 110) 1210 – 1110

= 1210 + 1110 = 2320

Example: 10

Simplify: (x + y)2 – (x – y)2 Alternate method
Solution:
(x + y)2 – (x – y)2 = {(x + y) + (x – y)} {(x + y) – (x – y)} Here,
= (x + y + x – y) (x + y – x + y) (x + y)2 – (x–y)2
= (x + x + y – y) (x – x + y + y) = (x2 + 2xy + y2) (x2–2xy + y2)
= 2x × 2y = 4xy. = x2 + 2xy + y2–x2 + 2xy – y2
= 4xy.

Exercise 18.1

1. Write down the square of
(a) (x + 3) (b) (x – 4) (c) x + 8y (d) p–5q

2. Find the value of a² + b² when

(a) a + b = 5, ab = 9. (b) a + b = 3, ab = 2.

(c) a – b = 3, ab = 10 (d) a – b = 5, ab = 24

3. (a) If a + 1 = 6, find the value of a² + 1 .
a a2

(b) If x + 1 = 7, find the value of x2 + 1 .
x x2

(c) If x– 1 = 2, find the value of x2 + x12.
x

(d) If x – 1 = 9, find the value of x2 + 1 .
x x2

4. (a) If (a – b) = 5, ab = 24, find (a+b).

(b) If (a–b) = 6, ab = 16, find (a+b).

(c) If (a+b) =10, ab = 24, find (a–b).

(d) If (a+b) = 13, ab = 40, find (a–b).

5. Express as perfect square

(a) 9x2 + 12x + 4 (b) 25a² + 60ab + 36b²
(c) 36x2 – 84xy + 49y2
(d) a²b² – 2 + 1
a2b2

Oasis School Mathematics-8 224433

6. Find the value of
(a) 1.25 × 1.25 + 2 × 1.25 × 1.75 + 1.75 × 1.75
(b) 2.75 × 2.75 – 2 × 2.75 × 1.25 + 1.25 × 1.25

7. Find the following products

(a) (2a + 3b) (2a – 3b)

(b) (2xy + 3yz) (2xy – 3yz)

(c) (2a – 3b + 7c) (2a + 3b + 7c)

(d) (p + q) (p – q) (p2 + q2)

(e) (a² – b²) (a² + b²) (a4 + b4)

8. Without actual multiplication, find the value of:
(a) 39 × 41 (b) 104 × 96 (c) 298 × 302

9. Simplify (without actual calculation)

(a) 2.5 × 2.5 – 1.5 × 1.5
2.5 – 1.5

(b) 3.75 × 3.75 – 2.25 × 2.25
3.75 + 2.25
(c) 0.5 × 0.5 – 0.4 × 0.4

(0.5)2 – 2 × 0.5 × 0.4 + (0.4)2


2.1 × 2.1 – 1.5 × 1.5
(d) (2.1)2 + 2 × 2.1 × 1.5 + (1.5)2

10. Simplify: (b) (8p + 9q)2 – (8p – 9q)2
(a) (a + b)2 – (a – b)2 (d) (3x – 2y)2 – (3x + 2y)2

(c) (6x – 8y)2 – (6x + 8y)2



Answer
1. (a) x2 + 6x + 9 (b) x2 – 8x + 16 (c) x2 + 16xy + 64y2 (d) p2 – 10p + 25p2
2. (a) 7 (b) 5 (c) 29 (d) 73 3. (a) 34 (b) 47 (c) 6 (d) 83
4. (a) 11 (b) 10 (c) 2 (d) 3 5. (a) (3x+2)2 (b) (5a+6b)2 (c) (6x–7y)2 (d) (ab – a1b)2
6. (a) 9 (b) 2.25 7. (a) 4a2 – 9b2 (b) 4x2y2 – 9y2z2 (c) 4a2 + 28ca + 49c2 – 9b2 (d) (p4 – q4)

(e) a8–b8 8. (a) 1599 (b) 9984 (c) 89996
9. (a) 4 (b) 1.5 (c) 9 (d) 1/6 10. (a) 4ab (b) 288pq (c) –192xy (d) –24xy

244 Oasis School Mathematics-8