Tuesday, December 19, 2023

Class – 7 CH-9 RATIONAL NUMBERS MATHS NCERT SOLUTIONS

Class – 7 CH-9 RATIONAL NUMBERS

MATHS NCERT SOLUTIONS

Exercise 9.1

Question 1:

(i) 1 and 0 (ii) 2 and 1

(iii) 4 2

and (iv) 1 2

and

List five rational numbers between:

5 3 2 3

SOLUTION 1:

(i) 1 and 0

Let us write 1 and 0 as rational numbers with denominator 6.

  1 6 and 0 = 0

6 6

6 5 4 3 2 1

       0

6 6 6 6 6 6

  1 5  2  1 1 1 0

6 3 2 3 6

Therefore, five rational numbers between 1 and 0 would be

    5 2 1 1 1

, , , ,

6 3 2 3 6

(ii) 2 and 1

Let us write 2 and 1 as rational numbers with denominator 6.

  2 12 and  1 6

6 6

12 11 10    9 8 7 6

       6 6 6 6 6 6 6

  2 11 5  3  4  7 1

6 3 2 3 6

Therefore, five rational numbers between 2 and 1 would be

    11 5 3 4 7

, , , ,

6 3 2 3 6

(iii)

(iv)

4 2

and

5 3

4 2

Let us write and as rational numbers with the same denominators.

5 3

4 36 2 30   and 

5 45 3 45

36 35 34 33 32 31 30

      

45 45 45 45 45 45 45

4 7 34 11 32 31 2

      

5 9 45 15 45 45 3

4 2

Therefore, five rational numbers between and would be

5 3

 7 34 11 32 31 2

, , , , ,

9 45 15 45 45 3

1 2

and

2 3

1 2

Let us write and as rational numbers with the same denominators.

2 3

1 3 2 4

  and 

2 6 3 6

3 2 1 1 2 3 4        0

6 6 6 6 6 6 6

  1 1 1 1 1 1 2        0

2 3 6 6 3 2 3

1 2

Therefore, five rational numbers between and would be

2 3

 1 1 1 1

, ,0, , .

3 6 6 3

Question 2:

Write four more rational numbers in each of the following patterns:

   3 6 9 12

(i) , , , ,.........

1 2 3

(ii) , , ,..........

1 2 3 4

(iii) , , , ,.........

6   12 18 24

2 2 4 6

(iv) , , , ,..........

3   3 6 9

SOLUTION 2:

   3 6 9 12

(i) , , , ,.........

3 1 3 2 3 3 3 4

 , , , ,.........

Therefore, the next four rational numbers of this pattern would be

 3 5,  3 6,  3 7,  3 8 = 15, 18, 21, 24

5 5 5 6 5 7 5 8 25 30 35 40

  1 2 3

(ii) , , ,..........

1 1 1 2 1 3

 , , ,..........

Therefore, the next four rational numbers of this pattern would be

 1 4,  1 5,  1 6,  1 7 =    4, 5, 6, 7

4 4 4 5 4 6 4 7 16 20 24 28

1 2 3 4

(iii) , , , ,.........

6 12 18 24

 1 1 1 2 1 3 1 4

 , , , ,.........

6 1  6 2    6 3 6 4

Therefore, the next four rational numbers of this pattern would be

1 5 1 6 1 7 1 8 5 6 7 8

, , , = , , ,

 6 5  6 6  6 7  6 8 30 36 42 48

2 2 4 6

(iv) , , , ,..........

3   3 6 9

 2 1 2 1 2 2 2 3

 , , , ,..........

3 1    3 1 3 2  3 3

Therefore, the next four rational numbers of this pattern would be

2 4 2 5 2 6 2 7 8 10 12 14

, , , = , , ,

 3 4  3 5  3 6  3 7 12 15 18 21

Question 3:

Give four rational numbers equivalent to:

2 5 4

(i) (ii) (iii)

7 3 9

SOLUTION 3:

2

(i)

 2 2 4  2 3 6  2 4 8  2 5 10

 ,  ,  , 

7 2 14 7 3 21 7 4 28 7 5 35

   4 6 8 10

Therefore, four equivalent rational numbers are , , , .

14 21 28 35

(ii)

3

5 2 10 5 3 15 5 4 20 5 5 25

 ,  ,  , 

 3 2 6  3 3 9  3 4 12  3 5 15

10 15 20 25

Therefore, four equivalent rational numbers are , , , .

  6 9 12 15

(iii)

8 12 16 20

Therefore, four equivalent rational numbers are , , , .

18 27 36 45

Question 4:

Draw the number line and represent the following rational numbers on it:

3 5 7 7

(i) (ii) (iii) (iv)

4 8 4 8

SOLUTION 4:

(i)

(iv)

Question 5:

The points P, Q, R, S, T, U, A and B on the number line are such that, TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S.

Each part which is between the two numbers is divided into 3 parts.

Therefore, A = , P = , Q = and B =

Similarly T = 3, R = 4, S = 5 and U = 6

3 3 3 3

Thus, the rational numbers represented P, Q, R and S are 7 8, , 4 and 5

3 3 3 3 respectively.

Question 6:

Which of the following pairs represent the same rational numbers:

7 3

(i) and

21 9

16 20

(ii) and

20 25

2 2

(iii) and

3 3

3 12

(iv) and

5 20

8 24

(v) and

5 15

1 1

(vi) and



(vii) and

9 9

SOLUTION 6

(i)





(ii)





(iii)





(iv)



7 3

and

21 9

7 1 3 1

= and =

21 3 9 3

1 1



3 3

7 3



21 9

16 20

and

20 25

16 4 20 4 4

= and = 

20 5 25 5 5

4 4

5 5

16 20

= 20 25

2 2

and

3 3

2 2 2 2

= and =

3 3 3 3

2 2

3 3

3 12

and

5 20

3 3 12 3  = and =

5 5 20 5

3 3

5 5

3 12

5 20

[Converting into lowest term]

8 24

(v) and

5 15

8 8 24 8

 = and = [Converting into lowest term]

5 5 15 5

8 8

= 5 5

8 24

 =

5 15

1 1

(vi) and

3 9

1 1 1 1

 = and = [Converting into lowest term]

3 3 9 9

1 1 

3 9

1 1  

3 9

5 5

(vii) and

9 9

5 5 5 5

 = and = [Converting into lowest term]

9 9 9 9

5 5 

9 9

5 5  

9 9

Question 7:

Rewrite the following rational numbers in the simplest form:

8 25 44 8

(i) (ii) (iii) (iv)

6 45 72 10

SOLUTION 7:

8  8 2 4

(i) = = [H.C.F. of 8 and 6 is 2]

6 62 3

(ii) = [H.C.F. of 25 and 45 is 5]

44  44 4 11

72 724 18

(iv) 8  8 2 4

= = [H.C.F. of 8 and 10 is 2]

(iii) = = [H.C.F. of 44 and 72 is 4]

10 102 5

Question 8:

Fill in the boxes with the correct symbol out of <, > and =:

52 45 714 87

(i) (ii) (iii) (iv)

73 57 816 54

11 557

(v) (vi) (vii) 0

34 11116

SOLUTION 8:

5 2

(i) < Since, the positive number if greater than negative number.

7 3

    

5 7 7 5 35 35 5 7

7 2 14 1 14 14 7 14

(ii) 4 75 5  28 < 25  4 < 5

     

82   16  1 16 16 8 16

8 4 7 5 32 35 8 7

(iii)  =   =

    

(iv)  >   > 

5 44 5 20 20 5 4

11 1 1

(v)  <

34 3 4

55 5 5

11 11

7 11 11

(vi)  =



(vii) 0 > Since, 0 is greater than every negative number. 6

Question 9:

Which is greater in each of the following:

2 5  5 4 3 2 1 1

(i) , (ii) , (iii) , (iv) ,

3 2 6 3 4 3 4 4

2 4

(v) 3 , 3

7 5

SOLUTION 9:

(i) and

4 15 2 5

Since < Therefore 

6 6 3 2

  5 1 5  4 2 8

(ii)  and  6 1 6 3 2 6

Since 5 > 8 Therefore 5 > 4

6 6 6 3

 3 3 9 2  4 8

(iii)  and 

4 3 12   3  4 12

Since 9 < 8 Therefore 3  2

12 12 4 3

1 1

(iv) < Since positive number is always greater than negative

4 4 number.

(v) 32  23   23 5  115 and 34  19   19 7  133

7 7 7 5 35 5 5 5 7 35

Since 115 > 133 Therefore 3 2 > 3 4

35 35 7 5

Question 10:

Write the following rational numbers in ascending order:

(i)   3, 2, 1

1 2 4

(ii)

(iii)   3, 3, 3

7 2 4

SOLUTION 10:

  3 2 1

(i) , ,

5 5 5

3 2 1

  

5 5 5

1  2 4 3  2 12

(ii) , ,  , , [Converting into same denominator]

3 9 3 9 9 9

12 2 3 4 2 1 Now     

9 9 9 3 9 3

  3 3 3

(iii) , ,

7 2 4

3 3 3

  

2 4 7

Exercise 9.2

Question 1:

Find the sum:

5 11 5 3

(i) 4 4  (ii) 3  5



9 22 3 5

(iii)  (iv) 

10 15 11 9

8 2 2

(v)  (vi) 0

19 57 3

1 3

(vii) 2 4

3 5

SOLUTION 1:

5 11 5 11 6 3

(i)  4  = 4 = 4  2 4 

5 3 25 9

(ii)  =  [L.C.M. of 3 and 5 is 15]

3 5 15 15

259 34 4

=   2

15 15 15

9 22  9 3 22 2 27 44

(iii)  =  =  [L.C.M. of 10 and 15 is 30]

10 15 10 3 15 2 30 30

 27 44 17

= 

30 30

3 5  3 9 5 11 27 55

(iv)  =  =  [L.C.M. of 11 and 9 is 99]

11 9  11 9 9 11 99 99

2755 82

= 

99 99

8 2  8 3  2 1 24 2

(v)  =  =  [L.C.M. of 19 and 57 is 57]

19 57 19 3 57 1  57 57

 24 2 26

= =

57 57

2 2

(vi)  0

3 3

(vii) 21 43 = 7  23 =  7 5  23 3 = 35  69 [L.C.M. of 3 and 5 is 15]

3 5 3 5 3 5 5 3 15 15

34 4

= =  2

15 15

Question 2:

(i) 7 17 

24 36 (ii) 5 6

 21

63 

(iii) 6 7

15

13  (iv) 3 7 

8 11

Find:

(v)  2 6

SOLUTION 2:

7 1721 34

(i)  = =  [L.C.M. of 24 and 36 is 72]

24 3672 72

21 34 13

= =

72 72

5 6 5 1  6 3 5 18

(ii) 63 21 = 63 1  21 3  = 63  63 [L.C.M. of 63 and 21 is 63]

5  18 5 18 23

= = 

63 63 63

6 7  6 15  7 13 90 91

(iii) 13 15 = 13 15 15 13  = 195 195 [L.C.M. of 13 and 15 is 195]

  90  91  90 91 1

195 195 195

3 7  3 11 7 8 33 56 (iv)  =  = 

8 11 8 11 11 8 88 88

 33 56 89 1

= = 1

88 88 88 [L.C.M. of 8 and 11 is 88]

(v) 2 6 = 19  6 =  19 1 6 9

9 9 1 9 1 1 9

19 54  19 54 73 1 [L.C.M. of 9 and 1 is 9]

= = 

=  = = 8

9 9 9 9 9

Question 3:

Find the product:

9 7

(i)  4 

2 

6 9

(iii) 

5 11

3 2

(v) 

11 5

SOLUTION 3:

9 7 9  7 63 7

(i) 2 4  = 2 4 = 8 7 8

3 3  9

(ii)    9 272 7

10 10 10 10

6 9  6 9 54

(iii)   

5 11 5 11 55

3 2 3  2 6 (iv)  5  7 5  35

3 2 3 2 6 (v)   

11 5 11 5 55

3 5 3  5

(vi)   5 3 1

5  3   

Question 4:

Find the value of:

(i)  4

4

(iii)   3

2 1

(v) 

13 7

3 4

(vii)  

13 65 

(ii)

(iv)

(vi)

(ii)

(iv)

(vi)   9

3 2

 5 

7 

3 5

 5 3

3

2

1 3



8 4

7  2 

13

SOLUTION 4:

(i)  4 =      4  2 3 6

3 3 1  3 1

(ii) 2 =   3

5 5 2 5 2 10

4 4 1  4 1 4 (iii)   3 =    5 5 3 5   3 15

1 3 1 4  1 1 1 (iv)  =  = 

8 4 8 3 2 3 6

2 1 2 7  2 7 14 1

(v)  =    1

13 7 13 1 13 1 13 13

7 2 7 13  7 13 91 19

(vi) 13 = 12 2 = 12  2  24 324

12 

3 4 3 65 3  5 15 3

(vii) 65 = 134  1 4  4 34 13 

Class – 7 CH-10 PRACTICAL GEOMETRY MATHS NCERT SOLUTIONS

Class – 7 CH-10 PRACTICAL GEOMETRY

MATHS NCERT SOLUTIONS

Exercise 10.1

Question 1:

Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.

SOLUTION 1:

To construct: A line, parallel to given line by using ruler and compasses.

Steps of construction:

(a) Draw a line-segment AB and take a point C outside AB.

(b) Take any point D on AB and join C to D.

(d) With C as centre and same radius as in step 3, draw an arc GH cutting CD at I.

(e) With the same arc EF, draw the equal arc cutting GH at J.

(f) Join JC to draw a line l.

This the required line 𝐴𝐵 ∥ 𝑙.

Question 2:

Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.

SOLUTION 2:

To construct: A line parallel to given line when perpendicular line is also given.

Steps of construction:

(a) Draw a line l and take a point P on it.

(b) At point P, draw a perpendicular line n.

(d) At point X, again draw a perpendicular line m. It is the required construction.

Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose? SOLUTION 3:

To construct: A pair of parallel lines intersecting other part of parallel lines.

Steps of construction:

(a) Draw a line l and take a point P outside of l .

(b) Take point Q on line l and join PQ.

(d) Extend line at P to get line m.

(e) Similarly, take a point R online m, at point R, draw angles such that  P =  R.

(f) Extended line at R which intersects at S online l. Draw line RS. Thus, we get parallelogram PQRS.

10.2

Question 1:

Construct XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.

SOLUTION 1:

To construct: XYZ, where XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.

Steps of construction:

(a) Draw a line segment YZ = 5 cm.

(b) Taking Z as centre and radius 6 cm, draw an arc.

(d) Join XY and XZ.

It is the required XYZ.

Question 2:

Construct an equilateral triangle of side 5.5 cm.

SOLUTION 2:

To construct: A ABC where AB = BC = CA = 5.5 cm Steps of construction:

(a) Draw a line segment BC = 5.5 cm

(b) Taking points B and C as centers and radius 5.5 cm, draw arcs which intersect at point A.

It is the required ABC.

Draw PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this? SOLUTION 3:

To construction: PQR, in which PQ = 4 cm, QR = 3.5 cm and PR = 4 cm.

Steps of construction:

(a) Draw a line segment QR = 3.5 cm.

(b) Taking Q as centre and radius 4 cm, draw an arc.

(d) Join PQ and PR.

It is the required isosceles PQR.

Question 4: Construct ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure  B. SOLUTION 4:

To construct: ABC in which AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm.

Steps of construction:

(a) Draw a line segment BC = 6 cm.

(b) Taking B as centre and radius 2.5 cm, draw an arc.

(d) Join AB and AC.

(e) Measure angle B with the help of protractor. It is the required ABC where  B = 80 .

10.3

Question 1: Construct DEF such that DE = 5 cm, DF = 3 cm and mEDF = 90 .

SOLUTION 1:

To construct: DEF where DE = 5 cm, DF = 3 cm and mEDF = 90 .

Steps of construction:

(a) Draw a line segment DF = 3 cm.

(b) At point D, draw an angle of 90 with the help of compass i.e.,  XDF = 90.

It is the required right angled triangle DEF.

Question 2:

Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110 . SOLUTION 2:

To construct: An isosceles triangle PQR where PQ = RQ = 6.5 cm and  Q = 110 .

Steps of construction:

(a) Draw a line segment QR = 6.5 cm.

(b) At point Q, draw an angle of 110 with the help of protractor, i.e.,  YQR = 110 .

6.5 cm, which cuts QY at point P.

(d) Join PR

It is the required isosceles triangle PQR.

Q R X

Construct ABC with BC = 7.5 cm, AC = 5 cm and mC = 60 .

SOLUTION 3:

To construct: ABC where BC = 7.5 cm, AC = 5 cm and mC = 60 .

Steps of construction:

(a) Draw a line segment BC = 7.5 cm.

(b) At point C, draw an angle of 60 with the help of protractor, i.e.,  XCB = 60 .

It is the required triangle ABC.

10.4

Question 1: Construct ABC, given mA = 60, mB = 30 and AB = 5.8 cm.

SOLUTION 1:

To construct: ABC where mA = 60, mB = 30 and AB = 5.8 cm.

Steps of construction:

(a) Draw a line segment AB = 5.8 cm.

(b) At point A, draw an angle  YAB = 60 with the help of compass.

(d) AY and BX intersect at the point C. It is the required triangle ABC.

Question 2: Construct PQR if PQ = 5 cm, mPQR = 105 and mQRP = 40 . SOLUTION 2:

Given: mPQR = 105 and mQRP = 40

We know that sum of angles of a triangle is 180 .

 mPQR + mQRP + mQPR = 180

 105  40 m QPR = 180

 145 + mQPR = 180

 mQPR = 180 – 145

 mQPR = 35

To construct: PQR where mP = 35, mQ = 105 and PQ = 5 cm. Steps of construction:

(a) Draw a line segment PQ = 5 cm.

(b) At point P, draw  XPQ = 35 with the help of protractor.

(d) XP and YQ intersect at point R.

It is the required triangle PQR.

Question 3:

Examine whether you can construct DEF such that EF = 7.2 cm, mE = 110 and mF = 80 . Justify your SOLUTION.

SOLUTION 3:

Given: In DEF, mE = 110 and mF = 80 .

Using angle sum property of triangle

∠𝐷 + ∠𝐸 + ∠𝐹 = 180°

⟹ ∠𝐷 + 110° + 80° = 180°

⟹ ∠𝐷 + 190° = 180°

⟹ ∠𝐷 = 180° − 190° = −10°

Which is not possible.

10.5

Question 1: Construct the right angled PQR, where mQ = 90, QR = 8 cm and PR = 10 cm.

SOLUTION 1:

To construct:

A right angled triangle PQR where mQ = 90, QR = 8 cm and PQ = 10 cm.

Steps of construction:

(a) Draw a line segment QR = 8 cm.

(b) At point Q, draw QX  QR.

(d) This arc cuts QX at point P.

(e) Join PQ.

It is the required right angled triangle PQR.

Question 2:

Construct a right angled triangle whose hypotenuse is 6 cm long and one the legs is 4 cm long.

SOLUTION 2:

To construct:

A right angled triangle DEF where DF = 6 cm and EF = 4 cm Steps of construction:

(a) Draw a line segment EF = 4 cm.

(b) At point Q, draw EX  EF.

(e) Join DF.

It is the required right angled triangle DEF.

Question 3: Construct an isosceles right angled triangle ABC, where mACB = 90 and AC = 6 cm.

SOLUTION 3:

To construct:

An isosceles right angled triangle ABC where mC = 90, AC = BC = 6 cm.

Steps of construction:

(a) Draw a line segment AC = 6 cm.

(b) At point C, draw XC  CA.

(d) This arc cuts CX at point B.

(e) Join BA.

It is the required isosceles right angled triangle ABC.

Miscellaneous Questions

Questions:

Below are given the measures of certain sides and angles of triangles. Identify those which cannot be constructed and say why you cannot construct them. Construct rest of the triangle.

Triangle Given measurements

1. ABC mA = 85 ; mB = 115 ; AB = 5 cm

2. PQR mQ = 30 ; mR = 60 ; QR = 4.7 cm

3. ABC mA = 70 ; mB = 50 ; AC = 3 cm

4. LMN mL = 60 ; mN = 120 ; LM = 5 cm

5. ABC BC = 2 cm; AB = 4 cm; AC = 2 cm

6. PQR PQ = 3.5 cm; QR = 4 cm; PR = 3.5 cm

7. XYZ XY = 3 cm; YZ = 4 cm; XZ = 5 cm

8. DEF DE = 4.5 cm; EF = 5.5 cm; DF = 4 cm

SOLUTION 1:

In ABC, mA = 85 ,m B = 115, AB = 5 cm

Construction of ABC is not possible because mA = 85 m B = 200, and we know that the sum of angles of a triangle should be 180 .

SOLUTION 2:

To construct: PQR where mQ = 30, mR = 60 and QR = 4.7 cm.

Steps of construction:

(a) Draw a line segment QR = 4.7 cm.

(b) At point Q, draw  XQR = 30 with the help of compass.

(d) QX and RY intersect at point P. It is the required triangle PQR.

We know that the sum of angles of a triangle is 180 .

 mA + mB + mC = 180

 70  50 m C = 180

 120 + mC = 180

 mC = 180 – 120

 mC = 60

To construct: ABC where mA = 70, mC = 60 and AC = 3 cm.

Steps of construction:

(a) Draw a line segment AC = 3 cm. (b) At point C, draw  YCA = 60 .

SOLUTION 4:

In LMN , mL = 60, mN = 120, LM = 5 cm

This LMN is not possible to construct because mL + mN = 60120180 which forms a linear pair.

SOLUTION 5:

ABC, BC = 2 cm, AB = 4 cm and AC = 2 cm

This ABC is not possible to construct because the condition is

Sum of lengths of two sides of a triangle should be greater than the third side. AB < BC + AC

 4 < 2 + 2

 4 = 4,

To construct: PQR where PQ = 3.5 cm, QR = 4 cm and PR = 3.5 cm Steps of construction:

(a) Draw a line segment QR = 4 cm.

(b) Taking Q as centre and radius 3.5 cm, draw an arc.

(c) Similarly, taking R as centre and radius 3.5 cm, draw an another arc which intersects the first arc at point P. It is the required triangle PQR.

SOLUTION 7:

To construct: A triangle whose sides are XY = 3 cm, YZ = 4 cm and XZ = 5 cm.

Steps of construction:

(a) Draw a line segment ZY = 4 cm.

(b) Taking Z as centre and radius 5 cm, draw an arc.

(d) Both arcs intersect at point X. It is the required triangle XYZ.

To construct:

A triangle DEF whose sides are DE = 4.5 cm, EF = 5.5 cm and DF = 4 cm. Steps of construction:

(a) Draw a line segment EF = 5.5 cm.

(b) Taking E as centre and radius 4.5 cm, draw an arc.

It is the required triangle DEF.

Class – 7 CH-8 COMPARING QUANTITIES MATHS NCERT SOLUTIONS

Class – 7 CH-8 COMPARING QUANTITIES

MATHS NCERT SOLUTIONS

Exercise 8.1

Question 1:

Find the ratio of:

(a) ₹5 to 50 paise (b) 15 kg to 210 g

SOLUTION 1:

To find ratios, both quantities should be in same unit.

(a) ₹5 to 50 paise

 5 x 100 paise to 50 paise [ ₹ 1 = 100 paise]

 500 paise to 50 paise

500 10

Thus, the ratio is =  = 10 : 1

50 1

(b) 15 kg to 210 g

 15 x 1000 g to 210 g [ 1 kg = 1000 g]

 15000 g to 210 g

15000 500

Thus, the ratio is =  = 500 : 7

210 7

 9 x 100 cm to 27 cm [ 1 m = 100 cm]

 900 cm to 27 cm

900 100

Thus, the ratio is =  = 100 : 3

27 3

(d) 30 days to 36 hours

 30 x 24 hours to 36 hours [ 1 day = 24 hours]

 720 hours to 36 hours

720 20

Thus, the ratio is = = 20 : 1

36 1

In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?

SOLUTION 2:

6 students need = 3 computers

 1 student needs = computers

 24 students need = 24 = 12 computers

Thus, 12 computers will be needed for 24 students.

Question 3:

Population of Rajasthan = 570 lakhs and population of U.P. = 1660 lakhs. Area of Rajasthan = 3 lakh km2 and area of U.P. = 2 lakh km2.

(i) How many people are there per km2 in both states? (ii) Which state is less populated?

SOLUTION 3:

2 = Population (i) People present per km

Area

570 lakhs 2

In Rajasthan = 3 lakhs per km2 = 190 people km

1660 lakhs 2

In U.P. = 2 lakh per km2 = 830 people per km

(ii) Rajasthan is less populated.

Exercise 8.2

Question 1:

Convert the given fractional numbers to percent:

(a) (b) (c)

SOLUTION 1:

1 25

(a) = 100% % = 12.5%

8 2

(b) = 100% 5 x 25% = 125%

3 3 15

40 2 2

2 200 4

(d) = 100% % 28 %

7 7 7

Question 2:

Convert the given decimal fractions to per cents:

(a) 0.65 (b) 2.1 (c) 0.02

SOLUTION 2:

(a) 0.65 = 100% = 65%

(b) 2.1 = × 100% = 210%

(b) 12.35 = × 100% = 1235%

(d)

(d) 12.35

Estimate what part of the figures is coloured and hence find the percent which is coloured.

SOLUTION 3:

(i) Coloured part =

 Percent of coloured part = 100% = 25% (ii) Coloured part =

 Percent of coloured part = 100% = 60%

(iii) Coloured part =

 Percent of coloured part = 100% = 25%

= 37.5%

Question 4:

Find:

(a) 15% of 250 (b) 1% of 1 hour (c) 20% of ₹2500 (d) 75% of 1 kg

SOLUTION 4:

(a) 15% of 250 = 250 = 15 x 2.5 = 37.5

(b) 1% of 1 hours = 1% of 60 minutes = 1% of (60 x 60) seconds

=  60 60 = 6 x 6 = 36 seconds

(d) 75% of 1 kg = 75% of 1000 g = 1000 = 750 g = 0.750 kg

Question 5:

Find the whole quantity if:

(a) 5% of it is 600 (b) 12% of it is ₹1080

(e) 8% of it is 40 litres

SOLUTION 5:

Let the whole quantity be x in given questions:

(a) 5% of x = 600

  x 600

 x = 12,000

(b) 12% of x = ₹1080

  x 1080

 x = ₹ 9,000

  x 500

 x = 1,250 km

(d) 70% of x = 14 minutes

  x 14

 x = 20 minutes

(e) 8% of x = 40 litres

  x 40

 x = 500 litres

Convert given per cents to decimal fractions and also to fractions in simplest forms:

(a) 25% (b) 150% (c) 20% (d) 5% SOLUTION 6:

S. No. Per cents Fractions Simplest form Decimal form

(a) 25% 0.25

(b) 150% 1.5

(d) 5% 0.05

Question 7:

In a city, 30% are females, 40% are males and remaining are children. What percent are children?

SOLUTION 7:

Given: Percentage of females = 30% Percentage of males = 40%

Total percentage of females and males = 30 + 40 = 70%

Percentage of children = Total percentage – Percentage of males and females

= 100% – 70%

= 30%

Hence, 30% are children.

Question 8:

Out of 15,000 voters in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?

SOLUTION 8:

Total voters = 15,000

Percentage of voted candidates = 60%

Percentage of not voted candidates = 100 – 60 = 40% Actual candidates, who did not vote = 40% of 15000

= 15000 = 6,000

Hence, 6,000 candidates did not vote.

Question 9: Meeta saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?

SOLUTION 9:

Let Meera’s salary be ₹x.

Now, 10% of salary = ₹ 400

 10% of x = ₹ 400

  x 400

 x

 x 4,000

Hence, Meera’s salary is ₹ 4,000.

Question 10:

A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?

SOLUTION 10:

Number of matches played by cricket team = 20

Percentage of won matches = 25%

Total matches won by them = 25% of 20

= 20

= 5

Hence, they won 5 matches.

Exercise 8.3

Question 1:

Tell what is the profit or loss in the following transactions. Also find profit percent or loss percent in each case.

(a) Gardening shears bought for ₹ 250 and sold for ₹ 325.

(b) A refrigerator bought ₹12,000 and sold at ₹ 13,500.

(a) Cost price of gardening shears = ₹ 250

Selling price of gardening shears = ₹ 325

Since, S.P. > C.P., therefore here is profit.

 Profit = S.P. – C.P. = ₹325 – ₹250 = ₹ 75

Profit

Now Profit% = 100

C.P.

= 100 = 30%

Therefore, Profit = ₹75 and Profit% = 30%

(b) Cost price of refrigerator = ₹ 12,000

Selling price of refrigerator = ₹13,500

Since, S.P. > C.P., therefore here is profit.  Profit = S.P. – C.P. = ₹13500 – ₹12000 = ₹1,500

Profit

Now Profit% = 100

C.P.

= 100 = 12.5%

Therefore, Profit = ₹1,500 and Profit% = 12.5%

Selling price of cupboard = ₹ 3,000

Since, S.P. > C.P., therefore here is profit.

 Profit = S.P. – C.P. = ₹3,000 – ₹2,500 = ₹ 500

Profit

Now Profit% = 100

C.P.

= 100 = 20%

Therefore, Profit = ₹ 500 and Profit% = 20%

(d) Cost price of skirt = ₹ 250

Selling price of skirt = ₹ 150

Since, C.P. > S.P., therefore here is loss.

 Loss = C.P. – S.P. =₹250 – ₹150 = ₹100

Loss Now Loss% = 100

C.P.

= 100 = 40%

Therefore, Profit = ₹ 100 and Profit% = 40%

Question 2:

Convert each part of the ratio to percentage:

(a) 3 : 1 (b) 2 : 3 : 5 (c) 1 : 4

SOLUTION 2:

(a) 3 : 1

Total part = 3 + 1 = 4

3 1

Therefore, Fractional part = :

4 4

3 1

 Percentage of parts = 100: 100

4 4

 Percentage of parts = 75% : 25%

(b) 2 : 3 : 5

Total part = 2 + 3 + 5 = 10

2 3 5

Therefore, Fractional part = : :

10 10 10

2 3 5

 Percentage of parts = 100: 100: 100

10 10 10

 Percentage of parts = 20% : 30% : 50%

Total part = 1 + 4 = 5

1 4

Therefore, Fractional part = :

5 5

1 4

 Percentage of parts = 100: 100

5 5

 Percentage of parts = 20% : 80%

(d) 1 : 2 : 5

Total part = 1 + 2 + 5 = 8

1 2 5

Therefore, Fractional part = : :

8 8 8

1 2 5

 Percentage of parts = 100: 100: 100

8 8 8

 Percentage of parts = 12.5% : 25% : 62.5%

Question 3: The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.

SOLUTION 3:

The decreased population of a city from 25,000 to 24,500.

Population decreased = 25,000 – 24,500 = 500

Population decreased

Decreased Percentage = 100

Original population

= 100 = 2%

Hence, the percentage decreased is 2%.

Question 4:

Arun bought a car for ₹3,50,000. The next year, the price went up to ₹3,70,000. What was the percentage of price increase?

SOLUTION 4:

Increased in price of a car from ₹ 3,50,000 to ₹ 3,70,000.

Amount change = ₹ 3,70,000 – ₹ 3,50,000 = ₹ 20,000.

Amount of change

Therefore, Increased percentage = 100

Original amount

= 100 = 5 %

Hence, the percentage of price increased is 5 %.

Question 5: I buy a T.V. for ₹10,000 and sell it at a profit of 20%. How much money do I get for it?

SOLUTION 5:

The cost price of T.V. = ₹ 10,000

Profit percent = 20%

Now, Profit = Profit% of C.P.

= 10000

= ₹ 2,000

Selling price = C.P. + Profit = ₹10,000 + ₹2,000 = ₹ 12,000 Hence, he gets ₹12,000 on selling his T.V.

Question 6:

Juhi sells a washing machine for ₹13,500. She loses 20% in the bargain. What was the price at which she bought it?

SOLUTION 6:

Selling price of washing machine = ₹13,500

Loss percent = 20%

Let the cost price of washing machine be ₹x. Since, Loss = Loss% of C.P.

20 x

 Loss = 20% of ₹ x =  x

100 5

Therefore, S.P. = C.P. – Loss x

 13500 = x

 13500 =

 x = ₹16,875

Hence, the cost price of washing machine is ₹16,875.

(i) Chalk contains Calcium, Carbon and Oxygen in the ratio 10:3:12. Find the percentage of Carbon in chalk.

(ii) If in a stick of chalk, Carbon is 3 g, what is the weight of the chalk stick?

SOLUTION 7:

(i) Given ratio = 10 : 3 : 12

Total part = 10 + 3 + 12 = 25

Part of Carbon =

Percentage of Carbon part in chalk = 100 = 12%

(ii) Quantity of Carbon in chalk stick = 3 g Let the weight of chalk be x g. Then, 12% of x = 3

  x 3

 x = 25 g

Hence, the weight of chalk stick is 25 g.

Question 8: Amina buys a book for ₹275 and sells it at a loss of 15%. How much does she sell it for?

SOLUTION 8:

The cost of a book = ₹275

Loss percent = 15%

Loss = Loss% of C.P. = 15% of ₹275

= 275 = ₹ 41.25

Therefore, S.P. = C.P. – Loss = ₹275 – ₹41.25 = ₹233.75 Hence, Amina sells a book for ₹233.75.

Find the amount to be paid at the end of 3 years in each case:

(a) Principal = ₹1,200 at 12% p.a. (b) Principal = ₹ 7,500 at 5% p.a.

SOLUTION 9:

(a) Here, Principal (P) = ₹1,200, Rate (R) = 12% p.a., Time (T) = 3 years

P   R T 1200 12 3 

Simple Interest = =

100 100

= ₹ 432

Now, Amount = Principal + Simple Interest

= ₹1200 + ₹432

= ₹1,632

(b) Here, Principal (P) = ₹7,500, Rate (R) = 5% p.a., Time (T) = 3 years

P   R T 7500 5 3 

Simple Interest = =

100 100

= ₹1,125

Now, Amount = Principal + Simple Interest

= ₹7,500 + ₹1,125

= ₹ 8,625

Question 10: What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years?

SOLUTION 10:

Here, Principal (P) = ₹56,000, Simple Interest (S.I.) = ₹280, Time (T) = 2 years

P   R T

Simple Interest =

100

56000 R 2

 280 =

 R =

 R = 0.25%

Hence, the rate of interest on sum is 0.25%.

11:

If Meena gives an interest of ₹45 for one year at 9% rate p.a. What is the sum she has borrowed?

SOLUTION 11:

Simple Interest = ₹45, Rate (R) = 9% p.a., Time (T) = 1 years

P   R T

Simple Interest =

100 P 9 1 

 45 =

100

 P =

 P = ₹ 500

Hence, she borrowed ₹ 500.

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Tuesday, December 19, 2023