for class 7th

1

Mathematics Notes

For Class 7th

Syllabus of 2nd Term

Unit # 3: Decimal Numbers

Unit # 4: Squares and Square Roots

Unit # 7: Financial Arithmetic

Unit # 14: Circles

2

Unit #3

“Decimal Numbers” Decimals:

Decimals are fractions which

have denominators with powers of 10.

Terminating decimals:

Decimals that have a finite

number of digits after the decimal point

are known as terminating decimals.

Example:

1 0.52=

Non-terminating decimals:

Decimal numbers with an infinite

number of digits after the decimal point

are known as non-terminating decimals.

Example:

1 0.3333.........3=

Recurring decimal:

If a set of digits is repeated again

and again in a non-terminating decimal,

it is called a recurring decimal.

Example:

2 0.181818........11=

Exercise 3

1. Fill in the blanks.

(Do it in book pg#43)

i. 0.5

ii. 0.23

iii. 40%

iv. 219

100

v. Recurring

2. State whether the following are

true or false? (D0 it in book pg#43)

1. True 3. True 5. True

2. False 4. False

3. Match each rational number in

the left hand column with its correct

decimal representation in the right

hand column: (book pg#44)

(i) 23

(a) 0.655

(ii) 131200

(b) 0.285,714

(iii) 16

(c) 0.666......

(iv) 27

(d) 0.166.......

4. Fill in the blanks.(Do it in book)

Approximate values of the numbers.

Numbe

rs

Corre

ct to

the

neare

st

whol

e #

Corre

ct to

one

decim

al

place

Corre

ct to

two

decim

al

places

Corre

ct to

three

decim

al

places

.

(i)

3.1276

3 3.1 3.13 3.128

(ii)

312.76

313 312.8 312.7

6

312.7

60

(iii)

1.001,

25

1 1.0 1.00 1.001

(iv)

0.012,

5

0 0 0.01 0.013

(v)

0.001,

25

0 0 0 0.001

5. Round off to the decimal place

indicated by the encircled digit.

i) 2 9 .06

Solution:

29.06= (Whole number)

3

29= 0<5; drop 0 29.06 29 =

ii) 37.678

Solution: (Whole number)

37.678=

38= (6<5; Add 1to7). 37.678 38 =

ix) 3.5 45,45

Solution:

3.5 45,45=

3.5= As 4<5; drop 4

3.545,45 3.5 =

6. Write the fractions in decimal

form by changing the denominators

to a power of 10?

i. 1

10

Solution: 1

10

= 1 x 10-1

= 0.1Ans

viii. 2

5

Solution: 2

5

= 2×2

5×2

= 4

10

= 4 x 10-1Ans

= 0.4Ans

iv. 17

5

Solution: 17

5

=17×2

5×2

= 34

10

= 34x10-1

= 3.4Ans

v. 1

25

Solution: 1

25

= 1×4

25×4

= 4

100

=4

10×10

2

4

10=

= 4 x 10-2

= 0.04Ans

7. Express the following rational

numbers as decimals by division.

i. 2

5

Solution:

4

2

5

0.4

5 20

20

x

−

20.4

5= Ans.

ii. 18

125−

Solution:

7

18

125

125 1 8

−

500

500

x

0.144

00

1750

180.144

125Ans

−

−

−= −

8. Which of the following rational

numbers can be represented as

terminating decimals?

i. 13

1

Solution:

10

8

20

20

x

134

3.254 13

12

133.25

4Ans

−

−

−

=

(Terminating decimal)

v. 2

3−

Solution:

20

18

20

18

2

2

30.666........

3 20

18

20.666......

3Ans

−

−

−

−

−=

(Non-terminating decimal)

9. Which of the following rational

numbers cannot be represented as

terminating decimals?

iii. 17

3

−

Solution: 17

20

18

20

18

2

35.66......

3 17

15

175.66......

3Ans

−

−

−

−

−= −

(Non-Terminating decimal)

5

viii. 7

11

Solution:

40

33

70

66

4

7

110.636......

11 70

66

70.636.....

11Ans

−

−

=


ii. 3

7

Solution:

20

14

60

56

40

50

49

10

7

30

28

2

3

70.4285714....

7 30

28

35

30.4285714......

7Ans

−

−

−

−

−

−

−

=


10. Express the following decimals

as fractions?

ii. 0.007

Solution:

0.007

0 .=

007

1000

7.

1000Ans=

iv. 0.009,5

Solution: 0.009,5

0 .=

009,5

10000

95= (÷by5)

10000

19

2000=

Ans

v. 4.33

0.05

Solution:

4.33

0.05

4.33 0.05

4 .

=

=33 0 .

100

05

100

433 5

100 100

433

100

=

=100

5

433

5Ans=

6

11. Express the following as

recurring decimals?

i. 2

3

Solution:

20

18

20

18

20

2

2

30.6666.......

3 20

18

18

20.6666...... 0.66

3Ans

−

−

−

−

= =

v. 4

333

Solution:

4

670

666

400

333

670

4

40.012012........ 0.012

333

3330.012012....

300 400

333

666

Ans

−

−

= =

−

−

13. Henry had a piece of cloth

measuring 674.95 cm. He cut off a

217.43 cm long piece from it. What

length of cloth remains? Round off

the answer to one decimal place.

Given:

Length of a piece of cloth = 674.95cm.

Cut off piece of cloth = 217.43cm.

Required:

Length of cloth remains?

Solution:

Length of cloth remains = Total length

– cut off piece

= 674.95cm-217.43cm.

=457.52cm.

ii) Answer round off to one decimal

place

=457.5cm.

Multiple choice questions:

(Do it in book pg #46)

1. (B) 36

2. (C) 5

10

3. (A) 2.474, 747…….

4. (B) 71.986

5. (D) to convert fractions to decimal

numbers.

6. (D) Powers of 10

7. (B) 1.875

8. (A) 8

10

9. (C) 29.841

10. (A) 13.188

7

Unit # 4

“Squares & Square

Roots” 1. Find by factors, the positive

square roots of?

i. 529

Solution:

529

23 529

23 23

1

2529 23 23 (23)= =

By taking square root on both sides.

2529 (23)=

529 23=

v. 11,025

Solution: 11,025

3 11,025

3 3675

5 1225

5 245

7 49

7 7

1

2 2 2

11,025 3 5 7 7

3 5 7

=

=

By taking square root on both sides. 2 2 2

2

11,025 3 5 7

(3 5 7)

=

=

11,025 3 5 7

11,025 15

11,025 105

=

=

=

vii. 8464

Solution:

8464

2 8464

2 4232

2 2116

2 1058

23 529

23 23

1

2 2 2

8464 = 2×2×2×2×23×23

= 2 ×2 ×23

By taking square root on both sides. 2 2 2

2

8464 = 2 ×2 ×23

= (2×2×23)

8464 = 2×2×23

8464 = 4×23

8464 = 92

2. Find the smallest whole

number by which 512 must be

multiplied in order to give a perfect

square.

Solution:

2 512

2 256

2 128

2 64

2 32

2 16

2 8

2 4

2 2

1

512 = 2×2×2×2×2×2×2×2×2

So, 2 is the smallest whole number by

which 512 must be multiplied to give a

perfect square.

8

4. The product of two positive

numbers is 15,876 and the larger one

is 9 times the other. Find the

numbers.

Solution:

Let the first number = x

Then second number = 9x

According to the given condition.

2

(x)(9x) = 15876

9x = 15876

By dividing 9 on both sides.

9 2x

92

15876= (÷by9)

9

x =1764

By taking square root on both sides. 2x

2

= 1764

x = 2×2×3×3×7×7

x = 22

×32

×72

= (2×3×7)

x = 2×3×7

x = 6×7

x = 42

R.Work

2 1764

2 882

3 441

3 147

7 49

7 7

1

First number 42=

Second number 9 42=

378=

So, the product of 42 and 378 is 15876.

And 378 is 9 times greater than 42.

6. The sum of the squares of two

positive whole numbers is 794. If one

of the numbers is 13, find the other.

Solution:

The 1st number 13=

Let the 2nd number be “x”

According to the given condition.

2 2

2

(13) + x = 794

169 + x = 794

Subtracting 169 from both sides.

169 - 169 2

2

+ x = 794-169

x 625=

By taking square root on both sides.

2

2

2 2

x = 625

x = 5×5×5×5

= 5 ×5 = (5×5)

= 5×5

x = 25

R.Work

625

5 125

5 25

5 5

1 So, the sum of the squares of 13 and

25 is 794.

Exercise 4b 1. Fill in the blanks. (Book pg#59)

i. 0.25

ii. 10

iii. 5

iv. Exponent

v. 16

9


true or false? (Book pg#59)

1. False 2. False 3. True

4. True 5. False

3. Derive the positive square roots

of:

9

i. 3969

Solution:

3969

63

6 3969

-36

123 369

-369

X

3969 = 63 Answer

vi. 151,321

Solution:

151,321

389

3 151,321

-9

68 613

-544

769 6921

-6921

X

151,321 = 389 Answer

3. 982,081

Solution:

982,081

991

9 982,081

-81

189 1720

-1701

1981 1981

-1981

X

982,081 = 991 Answer

4. Find the positive number which

when multiplied by itself gives

678,976.

Solution:

First we find the square root of 678,976.

824

8 678,976

-64

162 389

-324

1644 6576

-6576

X 824 is the positive number, when

multiplied by itself gives 678,976.

5. Mr. Ali wants to buy a carpet

for her square drawing room. If the

carpet measures 324 square meters,

what is the length of each side of the

drawing room?

Solution:

Area of carpet 2= 324m

Length of each side of the drawing

room = 324 .

18

1 324

-1

28 224

-224

X

So, the length of each side of the

drawing room is 18m.

6. What is the smallest number

that will divide into 216 to get a

perfect square?

Solution:

2 216

10

2 108

2 54

3 27

3 9

3 3

1

2 2

216 = 2× 2× 2×3×3×3

= 2× 2×3×3× 2×3

= 2 ×3 ×6

So, 6 is the smallest number that will

divide into 213 to get a perfect square.

12. A gardener plants trees in a

square formation. If he plants 25 trees

in a row, how many trees are there?

Solution:

Number of trees in each row 25=

So total number of trees 2(25)=

25 25=

625= Ans.

Unit #7

“Financial Arithmetic” Cost price (CP):

The price at which an object is

purchased is known as its cost price.

Selling price (SP):

The price at which an object is

sold is called its selling price (SP).

Profit:

A profit is earned when the selling

price is greater than the cost price

(SP>CP)

Therefore, Profit = selling price - cost price

Profit = Sp-CP.

Profit

x100 %CP

Profit % =

Loss:

If the cost price is greater than the

selling price (CP>SP), there will be a

loss.

Therefore, Loss = cost price - selling price

lossx100 %

CPloss % =

Exercise 7a 1. A man purchased a dozen pens

for 25 each and sold them at Rs 28

each. Find the total profit as well as

the profit per cent on the transaction.

Given:

Cost price of 1 pen = Rs. 25

Selling price of 1 pen = Rs. 28

Required:

Profit=? Profit % =?

Solution:

As CP of 1 pen = Rs. 25

Cost price (CP) of dozen pens = 25×12 = Rs. 300

Selling price (SP) of dozen pens= 28×12 = Rs. 336

Profit=SP-CP

By putting values.

Profit = Rs. (336-300)

Profit = Rs. 36 Profit

Profit % = x100 %CP

By putting values.

36Profit %=

300+100 %

36%

3

=

Profit %=12%

3. Mr Arsalan bought a plot of

land for Rs.108,000 and sold it for a

profit of 20%. At what price did he

11

sell it?

Given:

Cost price (CP) = Rs 108,000

Profit % = 20%

Required:

Selling price (SP) =?

Solution:

ProfitProfit%= x100 %

CP

By putting values.

20 %Profit

= x100 %108,000

Profit20=

1080

By multiplying 1080 from both sides. Profit

20 1080=1080

21600 = Profit Profit =Rs21600

Profit =SP-CP

By putting values.

21600 = SP – 108,000

By adding 108,000 on both sides. 21600 +108,000 = SP - 108,000 +108,000

1, 29, 600=SP

SP = Rs. 129600

6. If 13% is lost by selling goods

for Rs 1475, what would be the loss or

gain if the goods are sold for Rs 1615?

Given:

Loss % = 13%

Selling price (SP) = Rs. 1475

Selling price = Rs. 1615

Required:

Loss =?

Solution:

Let the cost price be Rs x

LossLoss%= ×100 %

CP

We know that

Loss = CP-SP

So,

CP-SPLoss%= x100 %

CP

By putting values.

13 % %x-s.p

= 100x

By multiplying x on both sides.

x -147513x =

x×100 x

13x =100x-147500

By subtracting 100x from both sides.

13x -100x = 100x - 100x -147500

-87x = -147500

By dividing -87 on both sides.

-87

-87

x -147500=

-87

x = Rs.1695.40

CP = Rs.1695.40

Loss = CP-SP

By putting values. Loss = Rs. (1695.40-1615)

Loss = Rs80.40

Discount:

The reduction made on the marked price

of an article is called the discount.

Discount = marked price-selling price

Discount percentage: Discount

Discount% = ×100%Marked price

12

Marked price:

The price printed on the tag of an article

is called the marked price (MP).

Exercise 7b 1. Fill in the blank spaces?

S.No Cost

price

Selling

price

Profit Profit

percent

i Rs.520 Rs.650

ii Rs.48 15%

iii Rs.242 10%

iv Rs.256 Rs.56

i. Given:

Cost price (CP) = Rs.520

Selling price (SP) = Rs.650

Required:

Profit =?

Profit %=?

Solution:

We know that

Profit = SP-CP

Profit = Rs. (650-520)

Profit = Rs.130

ProfitProfit% = ×100 %

CP

13 0Profit% =

52 0

1

100 %

13

=4 52

100 %

1

4

= 25

%

Profit % = 25%

ii. Given:

Cost price (CP) = 48

Profit % = 15%

Required:

Selling price (SP) =?

Profit =?

Solution: Profit

Profit% = ×100 %CP

15 %Pr

100 %48

ofit

=

By multiplying 48 on both sides.

Profit15×48=

48×100 48

720 = Prfoit×100

By dividing 100 on both sides. 72 0

10 0

×100=Profit

100

72=Profit

10

7.2 =Profit or

Profit =Rs.7.2

Profit = SP-CP

Profit +CP = SP

Rs(7.2+ 48) = SP

Rs.55.2 = SP

Sellingprice,SP =Rs.55.2

iii. Given:

Selling price, SP=RS.242

Profit % =10%

Required:

Cost price CP=?

Profit =?

Solution:

( )

ProfitProfit %= ×100 %

CP

SP -CPProfit %= ×100 %

CP

Profit = SP -CP

By putting values.

Let the cost price be Rs. x

10%242-x

= ×100 %x


13

242- x10×x =

xx×100 ×

10x = 242- x×100

10x = 24200-100x

By adding 100x on both sides.

10x +100x =24200- 100x + 100x

110x = 24200

By dividing 110 on both sides.

110x

110

24200=

110

2420x =

11x = Rs.220

Cost price CP= Rs.220

Profit =SP -CP

Profit =Rs.(242-220)

Profit =Rs.22

iv. Given:

Selling price SP=Rs. 256

Profit = Rs.56

Required:

Cost price CP=?

Profit %=?

Solution: Profit = SP -CP

CP = SP -Profit

By putting values. CP = Rs.(256-56)

CP = Rs.200

ProfitProfit %= ×100 %

CP

56Profit %=

2 00

×100 %

56= %

2Profit %= 28%

2. If the marked price of an

article is Rs. 400 and it is sold at a

discount of 10%, find the net selling

price of the article.

Given:

Marked price of article = Rs.400

Rate of discount = 10%

Required:

%

DiscountDiscount % = ×100%

M.P

1000

Discount=

400×1 %

Discount10 =

4

By multiplying 4 on both sides.

Discount10×4=

4× 4

40 = Discount or

Discount = Rs.40

Discount =MP-SP

SP = MP-Discount SP = Rs(400-40)

SP =Rs.360

So, the net selling price of the

article is Rs.360.

4. If an article is sold for Rs 450 after

14

allowing a discount of 10%, find the

list price of the article.

Given:

Selling price of article =Rs.450

Rate of discount = 10%

Required:

List price of the article (MP)=?

Solution:

DiscountDiscount%= ×100%

MP

( )

MP-SPDiscount% = ×100%

MP

Discount = MP-SP

Let the marked price of the article is Rs x.

10 %( )x -450

= ×100 %x


( )x -45010×x =

x×100× x

10x =100x -45000

By subtracting 100x from both sides. 10x -100x = 100x - 100x -4500

-90x = -45000

By dividing -90 on both sides.

- 90x

- 90

-=

4500 0

- 9 0

4500x =

9x = 500

x = Rs.500

So the list price of the article is Rs.

500.

Exercise 7c 1. Haider owns a factory that has

a value of Rs. 200,000,000. If the

property tax rate is 2%, how much

property tax should Haider pay?

Given:

Value of factory = Rs. 200,000,000

Property tax rate =2%

Required:

Property tax=?

Solution: Property tax 200,000,000 2%

200,000,0 00

=

=2

100

2,000,000=

Property tax = Rs. 4,000,000

4. Find the sales tax and total cost

of a radio set whose list price is Rs.

1200. The rate of sales tax is 7%.

Given:

List price of radio set = Rs.1200

Rate of sales tax = 7%

Required:

Sales tax=?

Total cost of a radio tax=?

Solution:

Sales tax = Rs.1200×7%

=12 007

×100

=12×7

Sales tax = Rs.84

Total cost of radio set = Rs (1200+84)

Total cost of radio set = Rs 1284

Exercise 7d Do the definitions and formulas from

book at Pg# 100-103.

1. Fill in the blanks.(Book pg#104)

i. Marked

ii. Overhead

iii. Interest

iv. Agricultural assets

v. Successive


true or false?

15

1. False 2. False 3. True

4. True 5. True

3. Find the interest in each of the

following cases:

i. On Rs.400 at 5% per annum

for 4 years.

Given:

Principal amount, P=Rs400

Rate of interest per annum, R=5%

Time, T=4years.

Required:

Interest, I =?

Solution: (P×R×T)

Interest, I =100

4 00I=

×5×4

100I =4×20

Interest, I=Rs80

iii. On Rs 450 at 4% per annum

for two years and 9 months.

Given:



Time, T=2years 9months 9 3 11

=2 = 2 = = 2.75years12 4 4

Required:

Interest, I=?

Solution: (P×R×T)

I =100

450×4×2.75I=

1001800

I =×2.75

100I =18×2.75

Interest,i = Rs49.5

4. Find the rate % per annum by

applying the formula.

i. Rs 700 gives Rs 210 simple

interest in 3years.

Given:

Principal amount, P=Rs.700

Interest, I =Rs.210

Time, T=3years

Required:

Rate % per annum=?

Solution: 100×I

R =T×P

100×210R =

3×70021,0 00

R =2100

210R =

21R =10

Rate%Per annum,R =10%

4. Find the time by applying the

formula.

ii. Rs 1500 will give Rs 450 simple

interest at 6% per annum.

Given:


Interest, I = Rs450


Required:

Time, T=?

Solution: 100×I

T =P×R

100T =

×450

15 00 ×6450

T =90

Time,T = 5years.

12. How much zakat should be

paid on yearly savings of Rs 938000?

16

Given data:

Yearly savings =Rs 93800

Required:

Zakat=?

Solution:

On Rs. 100 rate of Zakat is Rs. 2.50

On Rs. 1 rate of Zakat is 2.

Rs.50

100

On Rs. 93800 amount of Zakat =

250Rs.

100×93800 = Rs.2345

So, Rs. 2,345 Zakat should be paid on

yearly savings of Rs.93800.

Multiple choice questions Select the correct answer from the

given options. (Do it in book pg #105)

1. (A) Wealth

2. (B) Rs 3000

3. (C) Rs 909

4. (B) Rs 550

5. (A) Ushr

6. (C) 16,500

7. (C) 10%

8. (D) 8.04% pa

9. (C) 5yaers

Unit #14

“Circles” Do the definitions of circle and

elements of circle from book Pg# 172-

177.

Exercise 14a

1. Fill in the blanks.(book pg#179)

i. Diameter

ii. Circumference

iii. Segment

iv. Interior

v. Chord


true or false?

1. True 2. False 3. False

4. True 5. True

3. Draw a circle of radius 2.8cm.

Mark its centre as O. Draw two of its

chords not passing through the

centre.

Steps of construction:

i. Draw a circle of

radius 2.8cm.

ii. O is the centre of

the circle.

iii. With the help of

ruler, Draw two chords AB and CD

which do not pass through the centre of

the circle.

8. Draw a circle of radius 3.5 cm.

Shade the minor segment formed

with PQ as a chord of length 2.5cm.

Steps of construction:

i. Draw a circle of radius 3.5cm.

A B

O 2.8cm

C D

17

A B

O

C A =8cmB

A =4cmC

ii. With the help of ruler, draw PQ

as a chord of

length 2.5 cm.

iii. Shade the

smaller segment

formed byPQ .

iv. The shaded

segment is minor

segment.

9. State which of the points shown

in the figures are interior or exterior

points of the circle or point on the

circle.

Ans:Interior points of the circle:

Points A, C, G, D are in the interior

region of the circle.

• Exterior points

of the circle:

Points F and B

are in the exterior

region of the

circle.

• Points on the circle:

Points E and H lie on the circle.

Exercise 14b

1. In the given figure, R is the

mid-point of the chord PQ of the

circle with the centre O. If

om ROP = 42 , find the size of OPR.

Solution:

As, R is the midpoint

of chord PQ so OR is

perpendicular bisector

of PQ .

om ROP = 90 , om ROP = 42 , (given)

The sum of angles in a triangle is

180o. o

o o o

o o

o o

o

+

+

+ +

m ORP m ROP + m OPR =180

90 42 + m OPR =180

132 m OPR 180

m OPR =180 132

m OPR = 48

−

2. In the given figure. AB is a

chord of the circle with centre O.

⊥OC AB . If mAC= 4cm,findmAB.

Solution:

As AB is a chord of circle with centre O

and OC AB⊥ therefore, C is the

midpoint of AB

Now, AC =CB

AB = 2AC

AB=2×4cm

AC=8cm

Multiple choice questions. Select the correct answer from the

given options. (Book pg#184)

1.(A)2r

2.(B) its diameter

3.(C)9cm

4.(B)900

5.(A) Same centre

P

Q

O 3.5cm

OR

Q

P

for class 7th

Documents