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Nootan solutions for Mathematics [English] Class 10 ICSE chapter 7 - Ratio and proportion [Latest edition]

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Solutions for Chapter 7: Ratio and proportion

Below listed, you can find solutions for Chapter 7 of CISCE Nootan for Mathematics [English] Class 10 ICSE.


Exercise 7AExercise 7BExercise 7CExercise 7DExercise 7EChapter Test
Exercise 7A [Page 116]

Nootan solutions for Mathematics [English] Class 10 ICSE 7 Ratio and proportion Exercise 7A [Page 116]

1. (i)Page 116

Find the ratio compounded at 2 : 3, 4 : 5, 5 : 6

1. (ii)Page 116

Find the ratio compounded at a : b, b : c, c : d

1. (iii)Page 116

Find the ratio compounded at x + y : x − y, x3 − y3 : x2 − y2

2. (i)Page 116

Find the duplicate ratio of 4 : 7

2. (ii)Page 116

Find the duplicate ratio of m : n

2. (iii)Page 116

Find the duplicate ratio of 5x : 6y

3. (i)Page 116

Find the sub-duplicate ratio of 49 : 81.

3. (ii)Page 116

Find the sub-duplicate ratio of 16x4 : 25y4.

3. (iii)Page 116

Find the sub-duplicate ratio of 169m6 : 196n6.

4. (i)Page 116

Find the triplicate ratio of 3 : 5.

4. (ii)Page 116

Find the triplicate ratio of x : y.

4. (iii)Page 116

Find the triplicate ratio of 2x : 3y.

5. (i)Page 116

Find the sub-triplicate ratio of a9 : b12.

5. (ii)Page 116

Find the sub-triplicate ratio of 64a3 : 343b6.

6. (i)Page 116

Find the reciprocal ratio of 3 : 4.

6. (ii)Page 116

Find the reciprocal ratio of 42 : 52.

6. (iii)Page 116

Find the reciprocal ratio of `1/3 : 4`

7.Page 116

If 4x + 3 : 7x − 1 be the duplicate ratio of 3 : 4, find the value of x.

8.Page 116

If a : b be the triplicate ratio of (a + x) : (b + x), prove that x3 − 3abx − ab(a + b) = 0.

9. (i)Page 116

Find which is greater?

`3/4 or 15/16`

9. (ii)Page 116

Find which is greater?

2 : 3 or 3 : 4

10.Page 116

If x : y = 7 : 12, find the value of 3x + 5y : 6x + 7y.

11.Page 116

If x : y = 3 : 4, find the ratio of 3y − x : 2x + y.

12.Page 116

If 3x + 2y : 5x + 6y = 1 : 2, find the ratio of x : y.

13.Page 161

If 5a + 3b : 3a + 5b = 3 : 4, find a : b.

14.Page 161

What number must be added to each terms of the ratio 14 : 23 so that it may becomes equal to 8 : 11?

15.Page 161

Two numbers are in the ratio of 3 : 4. If 4 be added to each, the ratio becomes 4 : 5, find the numbers.

16.Page 161

If  A : B = 3 : 4, B : C = 2 : 5, find A : B : C.

17.Page 116

If A : B = 5 : 2, B : C = 3 : 4, find A : C.

18.Page 116

If 6A = 3B = 5C, find A : B : C.

19.Page 116

If (4x2+ xy) : (3xy – y2) = 12 : 5, find (2x + y) : (x + 2y).

20.Page 116

If m : n is the duplicate ratio of m + x : n + x; show that x2 = mn.

21.Page 116

If `(5x + 4y)/(8x + 5y) = 2/3`, find the value of 4x + 7y : 9x + 2y.

22.Page 116

Two numbers are in the ratio of 2 : 3, and if 6 subtracted from each, the remainders are in the ratio of 4 : 9. Find them.

23.Page 116

If `x/y = 3(1)/3`, find the value of `(x - 3y)/(2x - 5y)`.

24.Page 116

A’s present age is to B’s present age as 8 : 7. 27 years ago their ages were 5 : 4. Find their present ages.

25.Page 116

If `a/b = 2/3` and `x/y = 3/4`, find the value of `(4ay - 3bx)/(5ax - 2by)`.

26.Page 116

Two numbers are in the ratio of 3 : 5 and the sum of their squares is 1666. Find them.

27.Page 116

The ratio between A and B’s share is 7 : 11. If B’s share is ₹ 22.55, find A’s share.

28.Page 116

A bag contains ₹ 102 in the form of ₹ 1,50 paise and 25 paise coins in the ratio of 16 : 5 : 28. Find the number of coins of each kind.

29.Page 116

In a mixture of 28 litres, the ratio of milk and water is 5 : 2. Another 2 litres of water is added to the mixture. Find the ratio of milk and water in the new mixture.

30.Page 116

The ages of two persons are in the ratio of 5 : 7. Sixteen years ago, they were in the ratio 3 : 5. Find their present age.

Exercise 7B [Pages 125 - 126]

Nootan solutions for Mathematics [English] Class 10 ICSE 7 Ratio and proportion Exercise 7B [Pages 125 - 126]

1. (i)Page 125

State whether the following are in proportion:

6 : 18 :: 5 : 15

1. (ii)Page 125

State whether the following are in proportion:

3 : 7.5 :: 2 : 7

1. (iii)Page 125

State whether the following are in proportion:

10 : 21 :: 4 : 8.4

1. (iv)Page 125

State whether the following are in proportion:

9 : 12 :: 12 : 16

1. (v)Page 125

State whether the following are in proportion:

11 : 13 :: 4 : 5

2. (i)Page 125

Find the fourth proportional to the following:

11, 15, 33

2. (ii)Page 125

Find the fourth proportional to the following:

1.5, 4.5, 3.5

2. (iii)Page 125

Find the fourth proportional to the following:

`1/2, 1/3, 1/4`

2. (iv)Page 125

Find the fourth proportional to the following:

`1/7, 1/5, 1(1)/7`

3. (i)Page 125

Find the value of x in the following:

₹ 10 : ₹ 25 = 14m : x

3. (ii)Page 125

Find the value of x in the following:

75 : 15 = x : 7

3. (iii)Page 125

Find the value of x in the following:

x : 7.5 = 7 : 17.5

4. (i)Page 125

Find the third proportional to the following:

20, 15

4. (ii)Page 125

Find the third proportional to the following:

2.4, 3.6

4. (iii)Page 125

Find the third proportional to the following:

`2(2)/3 m`, 4m

4. (iv)Page 125

Find the third proportional to the following:

3, 12

5. (i)Page 125

Find the mean proportional of the following:

9, 25

5. (ii)Page 125

Find the mean proportional of the following:

`1/16, 4/25`

5. (iii)Page 125

Find the mean proportional of the following:

17.5, 0.007

5. (iv)Page 125

Find the mean proportional of the following:

0.25, 40000

5. (v)Page 125

Find the mean proportional of the following:

₹ 5, ₹ 125

5. (vi)Page 125

Find the mean proportional of the following:

2x, 8x

5. (vii)Page 125

Find the mean proportional of the following:

`(a - b)/(a + b) and (a^2b^2)/(a^2 - b^2)`

6.Page 125

What number must be added to each of the numbers 4, 6, 8, 11 in order to get the four numbers in proportion?

7.Page 125

What number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?

8.Page 125

What least number must be subtracted from each of the numbers 7, 17 and 47 so that the remainders are in continued proportion?

9.Page 125

The following numbers, K + 3, K + 2, 3K – 7 and 2K – 3 are in proportion. Find k. 

10.Page 125

k + 6 is the mean proportion between k − 2 and k2. Find the value of k.

11.Page 125

What number must be added to each of the numbers 6, 14 and 30 so that the resulting numbers may be in continued proportion?

12.Page 125

Find two numbers such that their mean proportional is 28 and third proportional in 224.

13.Page 125

Find two numbers such that their mean proportional is 12 and third proportional in 96.

14. (i)Page 125

If `a/b = c/d` then prove that each of the given ratio is equal to `(3a + 2c)/(3b + 2d)`.

14. (ii)Page 125

If `a/b = c/d` then prove that each of the given ratio is equal to `sqrt((4a^2 - 3c^2)/(4b^2 - 3d^2))`.

14. (iii)Page 125

If `a/b = c/d` then prove that each of the given ratio is equal to `root3 ((a^3 + 4c^3)/(b^3 + 4d^3))`.

15. (i)Page 125

If `a/b = c/d` then prove that `(5a + 3b)/(5c + 3d) = ((4a^3 - 3b^3)/(4c^3 - 3d^3))^(1/3)`

15. (ii)Page 125

If `a/b = c/d` then prove that `(6a + 7b)/(6c + 7d) = sqrt((5a^2 + 3b^2)/(5c^2 + 3d^2))`

16.Page 125

If b is the mean proportional between a and c, prove that a, c, a² + b², and b² + c² are proportional.

17.Page 125

If a + c = mb and `1/b + 1/d = m/c`, prove that a, b, c and d are in proportion.

18. (i)Page 125

If `x/a = y/b = z/c`, prove that: `(x^2a^2 + y^2b^2 + z^2c^2)/(a^3x + b^3y + c^3z) = ((xyz)/(abc))^(1/3)`

18. (ii)Page 125

If `x/a = y/b = z/c`, prove that: `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`

19. (i)Page 126

If `a/b = c/d = e/f`, prove that: (a2 + c2 + e2) (b2 + d2 + f2) = (ab + cd + ef)2

19. (ii)Page 126

If `a/b = c/d = e/f`, prove that: `(a^2 + c^2)^3/(b^2 + d^2)^3 = e^6/f^6`

19. (iii)Page 126

 If `a/b = c/d = e/f`, prove that: `a^2/b^2 + c^2/d^2 + e^2/f^2 = (ac)/(bd) + (ce)/(df) + (ae)/(bf)`

20.Page 126

If b is the mean proportional between a and c, show that `(a^4 + a^2b^2 + b^4)/(b^4 + b^2c^2 + c^4) = a^2/c^2`.

21.Page 126

If ax = by = cz, prove that `(yz)/x^2 + (zx)/y^2 + (xy)/z^2 = a^2/(bc) + b^2/(ca) + c^2/(ab)`

22. (i)Page 126

If a, b, c, d are in proportion, prove that (2a + 3b) (3c + 5d) = (2c + 3d) (3a + 5b)

22. (ii)Page 126

If a, b, c, d are in proportion, prove that (a2 + c2) : (b2 + d2) = ac : bd

22. (iii)Page 126

If a, b, c, d are in proportion, prove that `(a^2 - ab + b^2)/(c^2 - cd + d^2) = (a^2 - b^2)/(c^2 - d^2)`

22. (iv)Page 126

If a, b, c, d are in proportion, prove that `(a + c)^3/(b + d)^3 = (a^2(a - c))/(b^2(b - d))`.

22. (v)Page 126

 If a, b, c, d are in proportion, prove that abcd(a−2 + b−2 + c−2 + d−2) = a2 + b2 + c2 + d2.

23. (i)Page 126

If x, y, z are in continued proportion, prove that: `(x + y)^2/(y + z)^2 = x/z`.

23. (ii)Page 126

If x, y, z are in continued proportion, prove that: `(px^2 + qxy + ry^2)/(py^2 + qyz + rz^2) = x/z`.

23. (iii)Page 126

If x, y and z are in continued proportion, Prove that:

`x/(y^2.z^2) + y/(z^2.x^2) + z/(x^2.y^2) = 1/x^3 + 1/y^3 + 1/z^3`

23. (iv)Page 126

If x, y, z are in continued proportion, prove that: x2y2z2(x−4 + y−4 + z−4) = y−2(x4 + y4 + z4)

23. (v)Page 126

If x, y, z are in continued proportion, prove that: xyz(x + y + z)3 = (xy + yz + zx)3.

24. (i)Page 126

If a, b, c, d are in continued proportion, prove that: `sqrt(ab) + sqrt(bc) - sqrt(cd) = sqrt((a + b - c)(b + c - d))`.

24. (ii)Page 126

If a, b, c, d are in continued proportion, prove that: (a + d)(b + c) – (a + c)(b + d) = (b – c)2 

24. (iii)Page 126

If a, b, c, d are in continued proportion, prove that: (a2 + b2) (c2 + d2) = (b2 + c2)2.

24. (iv)Page 126

If a, b, c, d are in continued proportion, prove that: `(a^3 + b^3 + c^3)/(b^3 + c^3 + d^3) = a/d`

24. (v)Page 126

If a, b, c, d are in continued proportion, prove that: `((a - b)/c + (a - c)/b)^2 - ((d - b)/c + (d - c)/b)^2 = (a - d)^2 (1/c^2 - 1/b^2)`.

Exercise 7C [Pages 138 - 140]

Nootan solutions for Mathematics [English] Class 10 ICSE 7 Ratio and proportion Exercise 7C [Pages 138 - 140]

1.Page 138

If `x/y = p/q`, prove that `(5x + 7y)/(5x - 7y) = (5p + 7q)/(5p - 7q)`.

2.Page 138

If `a/b = c/d`, prove that `(4a + 9b)/(4a - 9b) = (4c + 9d)/(4c - 9d)`.

3.Page 138

If a : b :: c : d, show that (2a + 3b) : (2c + 3d) :: (2a − 3b) : (2c − 3d).

4.Page 138

If `(3x + 4y)/(3a + 4b) = (3x - 4y)/(3a - 4b),` prove that `x/y = a/b`.

5.Page 138

If `(8a - 5b)/(8c - 5d) = (8a + 5b)/(8c + 5d)`, prove that `a/b = c/d.`

6.Page 139

If (6a + 5b) (6c − 5d) = (6a − 5b) (6c + 5d), prove that a, b, c, d are in proportion.

7.Page 139

If (9a2 + 8b2) (9c2 − 8d2) = (9a2 − 8b2) (9c2 + 8d2), prove that a : b :: c : d.

8.Page 139

If (xa + yb) : b :: (xc + yd) : d, prove that a : b :: c : d.

9.Page 139

If x = `(2ab)/(a + b)`, prove that `(x + a)/(x - a) + (x + b)/(x - b)` = 2

10.Page 139

If x = `(8ab)/(a + b)`, prove that `(x + 4a)/(x - 4a) + (x + 4b)/(x - 4b)` = 2

11. (i)Page 139

Using properties of proportion solve for x, given

`(sqrt(5x)+sqrt(2x -6))/(sqrt(5x)- sqrt(2x -6)) = 4`

11. (ii)Page 139

Using properties of proportion, find the value of x from the following:

`(sqrt(1 + x) + sqrt(1 - x))/(sqrt(1 + x) - sqrt(1 - x)) = a/b`

11. (iii)Page 139

Using properties of proportion, find the value of x from the following:

`(a + sqrt(a^2 - 2ax))/(a - sqrt(a^2 - 2ax))` = b

11. (iv)Page 139

Using properties of proportion, find the value of x from the following:

`(sqrt(12x + 1) + sqrt(2x - 3))/(sqrt(12x + 1) - sqrt(2x - 3)) = 3/2`

11. (v)Page 139

Using properties of proportion, solve for x. Given that x is positive:

`(2x + sqrt(4x^2 -1))/(2x - sqrt(4x^2 - 1)) = 4`

11. (vi)Page 139

Using properties of proportion, find the value of x from the following:

`(3x + sqrt(9x^2 - 5))/(3x - sqrt(9x^2 - 5))` = 5

12.Page 139

If x = `(sqrt(2a + 3b) + sqrt(2a - 3b))/(sqrt(2a + 3b) - sqrt(2a - 3b))` then show that 3bx2 − 4ax + 3b = 0.

13.Page 139

If x = `(sqrt(b^2 + ab) + sqrt(b^2 - ab))/(sqrt(b^2 + ab) - sqrt(b^2 - ab))` show that ax2 − 2bx + a = 0.

14.Page 139

Solve for x: `16((a - x)/(a + x))^3 = (a + x)/(a - x)`

15.Page 139

If x = `(sqrt(2a + 1) + sqrt(2a - 1))/(sqrt(2a + 1) - sqrt(2a - 1))` prove that x2 − 4ax + 1 = 0

16.Page 139

Using properties of proportion, find x : y, if `(x^2 + 2x)/(2x + 4) = (y^2 + 3y)/(3y + 9)`

17.Page 139

If `(x^2 + y^2)/(x^2 - y^2) = 17/8`, then find the value of:

  1. x : y
  2. `(x^3 + y^3)/(x^3 - y^3)`
18.Page 139

Solve for x: `(1 + x + x^2)/(1 - x + x^2) = (62(1 + x))/(63(1 - x))`

19.Page 139

Given `x = (sqrt(a^2 + b^2) + sqrt(a^2 - b^2))/(sqrt(a^2 + b^2) - sqrt(a^2 - b^2)`. Use componendo and dividendo to prove that: `b^2 = (2a^2x)/(x^2 + 1)`

20.Page 139

If `(x + y)/(ax + by) = (y + z)/(ay + bz) = (z + x)/(az + bx)`, prove that each of these ratio is equal to `(2)/(a + b)` unless x + y + z = 0.

21.Page 139

If `(b + c - a)/(y + z - x) = (c + a - b)/(z + x - y) = (a + b - c)/(x + y - z)` then prove that each ratio is equal to `a/x = b/y = c/z`

22.Page 139

Using the properties of proportion, solve for x, given. `(x^4 + 1)/(2x^2) = (17)/(8)`.

23.Page 140

If `(a + b)^3/(a - b)^3 = 64/27`

  1. Find `(a + b)/(a - b)`
  2. Hence using properties of proportion, find a : b.
Exercise 7D [Pages 140 - 141]

Nootan solutions for Mathematics [English] Class 10 ICSE 7 Ratio and proportion Exercise 7D [Pages 140 - 141]

Choose the correct answer from the given four options in each of the following questions:

1.Page 140

x, 8, 16 are in continued proportion. The value of x is ______.

  • 2

  • 4

  • 5

  • 6

2.Page 140

The mean proportional of two numbers is 16 and the third proportional is 128. The larger number is ______.

  • 24

  • 48

  • 32

  • 64

3.Page 140

If b is the mean proportional of a and c then the mean proportional of a2 + b2 and b2 + c2 is ______.

  • a + b + c

  • ab + cа

  • ca + cb

  • ab + bc

4.Page 140

If 3x + 4y : 3x − 4y = 17 : 1 then x : y is equal to ______.

  • 2 : 3

  • 3 : 2

  • 2 : 5

  • 5 : 2

5.Page 140

The sub-triplicate ratio of 125 : 27 is ______.

  • 25 : 9

  • 9 : 25

  • 3 : 5

  • 5 : 3

6.Page 140

The quantity added to each term of the ratio 6 : 11 so that it becomes equal to 3 : 4 is ______.

  • 8

  • 9

  • 12

  • 14

7.Page 140

If x, y, z are in continued proportion then `(x^2 + y^2)/(y^2 + z^2)` is equal to ______.

  • `x/z`

  • `x/y`

  • `y/z`

  • `y/x`

8.Page 140

If A : B = 2 : 3, В : C = 4 : 5 then A : B : C is equal to ______.

  • 8 : 12 : 15

  • 6 : 9 : 10

  • 8 : 12 : 25

  • None of these

9.Page 140

If a : b = 3 : 11 then (15a – 3b) : (9a + 5b) is equal to ______.

  • 35 : 6

  • 6 : 35

  • 6 : 41

  • 41 : 6

10.Page 140

If `(1 + x + x^2)/(1 - x + x^2) = (62(1 + x))/(63(1 - x))` then the value of x is ______.

  • `1/5`

  • 5

  • 3

  • `1/3`

11.Page 141

The given table shows the distance covered and the time taken by a train moving at a uniform speed along a straight track:

Distance (in m) 60 90 y
Time (in sec) 2 x 5

The values of x and y are:

  • x = 4, y = 150

  • x = 3, y = 100

  • x = 4, y = 100

  • x = 3, y = 150

Exercise 7E [Page 141]

Nootan solutions for Mathematics [English] Class 10 ICSE 7 Ratio and proportion Exercise 7E [Page 141]

Valid Statements Questions

1.Page 141

In the following question, two statements (i) and (ii) are given. Choose the valid statement.

  1. The duplicate ratio of `sqrt2 : sqrt3` is 2 : 3
  2. 7 : 8 > 4 : 5
  • Only (i)

  • Only (ii)

  • Both (i) and (ii)

  • Neither (i) nor (ii)

2.Page 141

In the following question, two statements (i) and (ii) are given. Choose the valid statement.

  1. If (4a + 3b) : (6a +7b) = 16 : 25 then a : b = 31 : 4.
  2. If (1 + x) : (3x + 1) is the duplicate ratio of 3 : 5 then x = 8.
  • Only (i)

  • Only (ii)

  • Both (i) and (ii)

  • Neither (i) nor (ii)

3.Page 141

In the following question, two statements (i) and (ii) are given. Choose the valid statement.

  1. The third proportional of 5, 10 is 40.
  2. If three quantities are in continued proportion, then first is to the third is the sub-duplicate ratio of first to the second.
  • Only (i)

  • Only (ii)

  • Both (i) and (ii)

  • Neither (i) nor (ii)

4.Page 141

In the following question, two statements (i) and (ii) are given. Choose the valid statement.

  1. If x : y = 2 : 3, y : z = 4 : 5, then x : y : z = 8 : 12 : 15.
  2. lf x, 2, 8, y are in continued proportion, then x = 1.
  • Only (i)

  • Only (ii)

  • Both (i) and (ii)

  • Neither (i) nor (ii)

Chapter Test [Page 143]

Nootan solutions for Mathematics [English] Class 10 ICSE 7 Ratio and proportion Chapter Test [Page 143]

1.Page 143

Using componendo and dividendo solve for x:

`(sqrt(2x + 2) + sqrt(2x - 1))/(sqrt(2x + 2) - sqrt(2x - 1))` = 3

2.Page 143

If `a/c = c/d = e/f` prove that: `(a^3 + c^3)^2/(b^3 + d^3)^2 = e^6/f^6`

3.Page 143

If 3A = 4B = 6C, find: A : B : C.

4.Page 143

 If (4x2 + xy) : (3xy – y2) = 12 : 5, find (x + 2y) : (2x + y).

5.Page 143

If `(by + cz)/(b^2 + c^2) = (cz + ax)/(c^2 + a^2) = (ax + by)/(a^2 + b^2)`, prove that each of these ratio is equal to `x/a = y/b = z/c`

6.Page 143

If x = `(root3 (a + 1) + root3 (a - 1))/(root3 (a + 1) - root(3)(a - 1)`, prove that x³ – 3ax² + 3x – a = 0.

7.Page 143

Find x from the equation: `(a+ x + sqrt(a^2 - x^2))/(a + x - sqrt(a^2 - x^2)) = b/x`

8.Page 143

If x = `"pab"/(a + b)`, prove that `(x + pa)/(x - pa) - (x + pb)/(x - pb) = (2(a^2 - b^2))/(ab)`.

9.Page 143

If x = `(2mab)/(a + b)`, find the value of `(x + ma)/(x - ma) + (x + mb)/(x - mb)`.

10.Page 143

If (3x2 + 2y2) : (3x2 − 2y2) = 11 : 9, find the value of `(3x^4 + 25y^4)/(3x^4 - 25y^4)`.

11. (i)Page 143

If a : b = 9 : 10, find the value of `(5a + 3b)/(5a - 3b)`

11. (ii)Page 143

If a : b = 9 : 10, find the value of `(2a^2 - 3b^2)/(2a^2 + 3b^2)`

12.Page 143

If `x/(b + c - a) = y/(c + a - b) = z/(a + b - c)`  prove that each ratio’s equal to `(x + y + z)/(a + b + c)`.

13.Page 143

If x : a = y : b, prove that `(x^4 + a^4)/(x^3 + a^3) + (y^4 + b^4)/(y^3 + b^3) = ((x + y)^4 + (a + b)^4)/((x+ y)^3 + (a + b)^3`

14.Page 143

If `x/a = y/b = z/c`, prove that `(3x^3 - 5y^3 + 4z^3)/(3a^3 - 5b^3 + 4c^3) = ((3x - 5y + 4z)/(3a - 5b + 4c))^3`.

Solutions for 7: Ratio and proportion

Exercise 7AExercise 7BExercise 7CExercise 7DExercise 7EChapter Test
Nootan solutions for Mathematics [English] Class 10 ICSE chapter 7 - Ratio and proportion - Shaalaa.com

Nootan solutions for Mathematics [English] Class 10 ICSE chapter 7 - Ratio and proportion

Shaalaa.com has the CISCE Mathematics Mathematics [English] Class 10 ICSE CISCE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Nootan solutions for Mathematics Mathematics [English] Class 10 ICSE CISCE 7 (Ratio and proportion) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

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Concepts covered in Mathematics [English] Class 10 ICSE chapter 7 Ratio and proportion are Ratio, Proportion, Direct Applications of Ratio and Proportion, Ratio in Lowest Terms, All About Ratios: Dividing, Comparing, and Modifying Quantities, Commensurable and Incommensurable Quantities, Composition of Ratios, Concept of Continued Proportion, Properties of Proportion.

Using Nootan Mathematics [English] Class 10 ICSE solutions Ratio and proportion exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in Nootan Solutions are essential questions that can be asked in the final exam. Maximum CISCE Mathematics [English] Class 10 ICSE students prefer Nootan Textbook Solutions to score more in exams.

Get the free view of Chapter 7, Ratio and proportion Mathematics [English] Class 10 ICSE additional questions for Mathematics Mathematics [English] Class 10 ICSE CISCE, and you can use Shaalaa.com to keep it handy for your exam preparation.

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