Ratio and Proportion Class 10 RS Aggarwal Exe-7C Goyal Brothers ICSE Maths Solutions Ch-7. Step by step solutions as latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-10.
Ratio and Proportion Class 10 RS Aggarwal Exe-7C Goyal Brothers ICSE Maths Solutions Ch-7
| Board | ICSE |
| Subject | Maths |
| Class | 10th |
| Chapter-7 | Ratio and Proportion |
| Writer/ Book | RS Aggarwal |
| Topics | Solution of Exe-7C Questions |
| Academic Session | 2024-2025 |
Useful Result on Ratio and Proportion
In this article you will learn useful result on proportion such as
- Invertendo Property: If a : b :: c : d then b : a :: d : c.
- Alternendo Property: If a : b :: c : d then a : c :: b : d.
- Componendo Property: if a : b = c : d then (a + b) : b :: (c + d) : d.
- Dividendo Property: If a : b :: c : d then (a – b) : b :: (c – d) : d
- Convertendo Property: If a : b :: c : d then a : (a – b) :: c : (c – d)
- Componendo-Dividendo Property: If a : b :: c : d then (a + b) : (a – b) :: (c + d) : (c – d)
- Addendo Property: If a : b = c : d = e : f, value of each ratio is (a + c + e) : (b + d + f)
Solution of Exe-7C Questions
(Ratio and Proportion Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-7)
Que-1: If a:b = c:d, prove that (9a+13b) : (9a-13b) = (9c+13d) : (9c-13d)
Sol: Given a/b = c/d
⇒ 9a/13b = 9c/13d (Multiplying each side by 9/13)
⇒ (9a+13b)/(9a-13b) = (9c+13d)/(9c-13d) (By componendo and divdendo)
= (9a+13b)(9c-13d) = (9c+13d)(9a-13b)
Que-2: If a:b = c:d, prove that (3a+2b) : (3a-2b) = (3c+2d) : (3c-2d)
Sol: a:b = c:d ⇒ a/b = c/d
⇒ 3a/2b = 3c/2d [Multiplying each side by 3/2}
⇒ (3a+2b)/(3a−2b) = (3c+2d)/(3c−2d) [By componendo and dividendo]
⇒ 3a+2b : 3a−2b = 3c+2d : 3c−2d
Que-3: If (3a+5b) : (3a-5b) = (3c+5d) : (3c-5d), prove that a:b =c:d.
Sol: (3a+5b)/(3a-5b) = (3c+5d)/(3c-5d)
{(3a+5b)+(3a-5b)}/{(3a+5b)-(3a-5b)} = {(3c+5d)+(3c-5d)}/{(3c+5d)-(3c-5d)} [By componendo and dividendo]
6a/10b = 6c/10d
a/b = c/d [Divide each side by 6/10]
a:b =c:d.
Que-4: If (4a²+7b²) : (4a²-7b²) = (4c²+7d²) : (4c²-7d²)
Sol: (4a²+7b²)/(4a²-7b²) = (4c²+7d²)/(4c²-7d²)
By using componendo and dividendo we get,
{(4a²+7b²)-(4a²-7b²)}/{(4a²+7b²)+(4a²-7b²)} = {(4c²+7d²)-(4c²-7d²)}/{(4c²+7d²)+(4c²-7d²)}
(14b²/8a²) = (14d²/8c²)
(4a²/7b²) = (4c²/7d²)
a²/b² = c²/d² [Divide each side by 4/7]
a::b = c:d.
Que-5: If (ma+nb) : (mc+nd) = (ma-nb) : (mc-nd), prove that a:b = c:d.
Sol: (ma+nb)/(mc+nd) = (ma-nb)/(mc-nd)
Let’s cross-multiply the given proportion:
(ma⋅mc)−(ma⋅nd)+(nb⋅mc)−(nb⋅nd) = (ma⋅mc)+(ma⋅nd)−(nb⋅mc)−(nb⋅nd)
Simplify and rearrange:
(ma⋅mc)−(nb⋅nd)+(nb⋅mc)−(ma⋅nd) = (ma⋅mc)−(nb⋅nd)−(nb⋅mc)+(ma⋅nd)
(nb⋅mc) − (ma⋅nd) = (−nb⋅mc) + (ma⋅nd)
2(nb⋅mc) = 2(ma⋅nd)
nb⋅mc = ma⋅nd
a/b = c/d
a:b = c:d.
Que-6: If (5x+6y)/(5x-6y) = (5u+6v)/(5u-6v) show that x/y = u/v.
Sol: (5x+6y)/(5x-6y) = (5u+6v)/(5u-6v)
By componendo and dividendo we get,
(5x+6y+5x-6y)/(5x+6y-5x+6y) = (5u+6v+5u-6v)/(5u+6v-5u=6v)
10x/12y = 10u/12v
x/y = u/v
x:y = u:v.
Que-7: If x = 6ab/(a+b), prove that [(x+3a)/(x-3a) + (x+3b)/(x-3b)] = 2
Sol: Given: x = 6ab/(a+b)
x/3a = 2b/(a+b)
Applying componendo and dividendo
(x+3a)/(x-3a) = (2b+a+b)/(2b-a-b)
(x+3a)/(x-3a) = (3b+a)/(b-a) …(1)
Again x = 6ab/(a+b)
⇒ x/3b = 2a/(a+b)
Applying componendo and dividendo
(x+3b)/(x-3b) = (2a+a+b)/(2a-a-b)
(x+3b)/(x-3b) = (3a+b)/(a-b) ….(2)
From (1) and (2)
(x +3a)/(x-3a) + (x+3b)/(x-3b) = (3b+a)/(b-a) + (3a+b)/(a-b)
(x+3a)/(x-3a) + (x+3b)/(x-3b) = {(-3b-a+3a+b)/a-b}
(x+3a)/(x-3a) + (x+3b)/(x-3b) = (2a-2b)/(a-b)
= 2 = R.H.S.
Que-8: If (x³+3x)/(3x²+1) = 341/91, prove that x = 11
Sol: (x³+3x)/(3x²+1) = 341/91
Applying componendo and dividendo
(x³+3x+3x²+1)/(x³+3x-3x²-1) = (341+91)/(341-91)
⇒ (x³+3x²+3x+1)/(x³-3x²+3x-1) = 432/250 = 216/125
⇒ {(x+1)³/(x-1)³} = 216/125 = (6/5)³
∴ (x+1)/(x-1) = 6/5
⇒ 6x – 6 = 5x + 5
⇒ 6x – 5x = 5 + 6
⇒ x = 11 = R.H.S.
Que-9: If [√(x+2)+√(x-3)]/[√(x+2)-√(x-3)] = 5, prove that x = 7.
Sol: [√(x+2)+√(x-3)]/[√(x+2)-√(x-3)] = 5
By using Componendo and Dividendo
[√(x+2)+√(x-3)+√(x+2)-√(x-3)]/[√(x+2)+√(x-3)-√(x+2)+√(x-3)] = 5+1/5-1
√(x+2)/√(x-3) = 6/4
(x+2)/(x-3) = (6/4)² = 36/16
(x+2)/(x-3) = 9/4
4x+8 = 9x-27
9x-4x = 8+27
5x = 35
x = 7 Hence Proved.
Que-10: If [(√x+5)+(√x-16)]/[(√x+5)-(√x-16)] = 7/3, prove that x = 20.
Sol: {√(x+5) + √(x-16)}/{√(x+5) – √(x-16)} = 7/3
Applying componendo and dividendo
{√(x+5) + √(x-16) + √(x+5) – √(x-16)}/{√(x+5) + √(x-16) – √(x+5) + √(x-16)} = (7+3)/(7-3)
{2√(x+5)}/{2√(x-16) = 10/4
√(x+5)/√(x-16) = 5/2
Squaring both sides
(x+5)/(x-16) = 25/4
4x + 20 = 25x – 400
21x = 420
x = 420/21 = 20 Hence Proved.
Que-11: (i) If [√3x+(√2x+1)]/[√3x-(√2x-1)] = 5, prove that x = 3/2
(ii) Using properties of proportion, solve for x. Given that x is positive.
[2x+(√4x²-1)]/[2x-(√4x²-1)] = 4
Sol: (i) √(3x) + √(2x – 1) / √(3x) – √(2x-1) = 5
Using Componendo and Dividendo
√(3x) + √(2x – 1) / √(3x) – √(2x-1) *(√(3x) + √(2x – 1) / √(3x) + √(2x-1) )= 5
(3x + 2x -1 + 2√(3X)(2X-1))/(3X -2X +1) = 5
(5x-1)+2√(3X)(2X-1))/(X +1) = 5
(5x-1)+2√(3X)(2X-1) = 5x +5
2√(3X)(2X-1) = 5x +5 -5x +1
2√(3X)(2X-1) = 6
√(3X)(2X-1) = 3
(3X)(2X-1) = 9
6x² -3x -9 = 0
2x² -x -3 = 0
2x² -3x +2x-3 = 0
(x+1)(2x-3) = 0
2x-3 = 0
x = 3/2
(ii) {2x+√(4x²-1)}/{2x-√(4x²-1)} = 4
By componendo dividendo
⇒ {2x+√(4x²-1)+2x-√(4x²-1)}/{2x+√(4x²-1)-2x+√(4x²-1)} = (4+1)/(4-1)
⇒ 4x/{2√(4x²-1)} = 5/3
⇒ 2x/√(4x²-1) = 5/3
⇒ 4x²/(4x²-1) = 25/9 (squaring both sides)
⇒ (4x²-4x²+1)/(4x²-1) = (25-9)/9 (By dividendo)
⇒ 1/(4x²-1) =16/9
⇒ 9 = 64x²-16
⇒ 64x² = 25
⇒ x² = 25/64
⇒ x = ±5/8
⇒ x = 5/8 (x is positive)
Que-12: Using properties of proportion solve for x, given
[√5x+√(2x-6)]/[√5x-√(2x-1)] = 4
Sol: [√5x+√(2x-6)]/[√5x-√(2x-1)] = 4/1
Using componendo and dividendo on both sides
We know, if a/b = c/d
⇒ (a+b)/(a-b) = (c+d)/(c-d)
∴ {(√5x+√(2x-6))+(√5x-√(2x-6))}/{(√5x+√(2x-6))-(√5x-√(2x-6))} = (4+1)/(4-1)
√5x/√(2x-6) = 5/3
On squaring both sides
5x/(2x-6) = 25/9
45x = 50x-150
150 = 5x
⇒ x = 150/5
⇒ x = 30
Que-13: Using componendo and dividendo solve for x:
[(√2x+2)+(√2x-1)]/[(√2x+2)-(√2x-1)] = 3
Sol: [(√2x+2)+(√2x-1)]/[(√2x+2)-(√2x-1)] = 3/1
Using componendo and dividendo,
{√(2x+2)+√(2x-1)+√(2x+2)-√(2x-1)}/{√(2x+2)+√(2x-1)-√(2x+2)-√(2x-1)} = (3+1)/(3-1)
{2√(2x+2)}/{2√(2x-1)} = 4/2
√(2x+2)/√(2x-1) = 2
On squaring
(2x+2)/(2x-1) = 4
2x+2 = 8x-4
2x–8x = – 2– 4
– 6x = – 6
x = 1
Que-14: If 16[(a-x)/(a+x)]^3 = [(a+x)/(a-x)], prove that x = a/3.
Sol: {(a-x)/(a+x)}³ = (a+x)/(a-x)
⇒ 16 = {(a-x)/(a+x)}^4
⇒ (2)^4 = axax(a-xa+x)^4
⇒ (a-x)/(a+x) = ± 2
⇒ (a-x)/(a+x) = 2/1 or (a-x)/(a+x) = -2/1
Applying componendo and Dividendo, we get
⇒ (a+x+a-x)/(a+x-a+x) = 3/1 or (a+x+a-x)/(a+x-a+x) = -1/-3
⇒ 2a/2x = 3 or 2a/2x = 1/3
⇒ x = a/3 or x = 3a Hence Proved.
Que-15: If x = {(√2a+1)+(√2a-1)}/{(√2a+1)-(√2a-1)}, prove that : x²-4ax+1 = 0.
Sol: x = {√(2a+1)+√(2a−1)}/{√(2a+1)−√(2a−1)}
= {√(2a+1)+√(2a−1)}/{√(2a+1)−√(2a−1)} . {√(2a+1)+√(2a−1)}/{√(2a+1)+√(2a−1)}
= {√(2a+1)+√(2a−1)}²/(2a+1−2a+1)
= [2a+1+2a−1+2√{(2a+1) (2a−1)}]/2
⇒ 2x = 4a+2√{(2a)²−1}
⇒ x = 2a+√{4a²−1}
⇒ x−2a = √{4a²−1}
⇒ (x−2a)² = 4a²−1
⇒ x²+4a²−4ax = 4a²−1
⇒ x²−4ax+1 = 0 Hence Proved.
Que-16: If x = {(√b+3a)+(√b-3a)}/{(√b+3a)-(√b-3a)}, prove that : 3ax²-2bx+3a = 0.
Sol: x = {(√b+3a)+(√b-3a)}/{(√b+3a)-(√b-3a)}
Applying componendo and dividendo we get
(x+1)/(x-1) = {(√b+3a)+(√b-3a)+(√b+3a)-(√b-3a)}/{(√b+3a)+(√b-3a)-(√b+3a)+(√b-3a)}
(x+1)/(x-1) = {2√(b+3a)}/{2√(b-3a)}
Squaring both sides.
(x+1)²/(x-1)² = (b+3a)/(b-3a)
(x²+2x+1)/(x²-2x+1) = (b+3a)/(b-3a)
Again applying componendo and dividendo
(x²+2x+1+x²-2x+1)/(x²+2x+1-x²+2x-1) = (b+3a+b-3a)/(b+3a-b+3a)
(x²+1)/2x = b/3a
3a(x²+1) = 2x.b
3ax²+3a = 2bx
3ax²-2bx+3a = 0.
Hence Proved.
Que-17: If x = {(√2a+3b)+(√2a-3b)}/{(√2a+3b)-(√2a-3b)}, prove that : 3bx²-4ax+3b = 0.
Sol: x = {(√2a+3b)+(√2a-3b)}/{(√2a+3b)-(√2a-3b)}
Using Componendo and Dividendo
{(√2a+3b)+(√2a-3b)+(√2a+3b)-(√2a-3b)}/{(√2a+3b)+(√2a-3b)-(√2a+3b)+(√2a-3b)} = (x+1)/(x-1)
{2√(2a+3b)}/{2√(2a-3b)} = (x+1)/(x-1)
(2a+3b)/(2a-3b)= (x+1)²/(x-1)²
Using Componendo and Dividendo
(2a+3b+2a-3b)/(2a+3b-2a+3b) = {(x+1)²+(x-1)²}/{(x+1)²+(x-1)²}
4a/6b = {2(x²+1)}/4x
2a/3b = (x²+1)/2x
2a . 2x = 3b(x²+1)
4ax = 3bx²+3b
3bx²+3b-4ax = 0. Hence Proved.
Que-18: If x = {(∛m+1)+(∛m-1)}/{(∛m+1)-(∛m-1)}, prove that : x³-3x²m+3x-m = 0.
Sol: x/1 = {(∛m+1)+(∛m-1)}/{(∛m+1)-(∛m-1)}
Applying componendo and dividendo
(x+1)/(x-1) = {∛(m+1)+∛(m-1)+∛(m+1)-∛(m-1)}/{∛(m+1)+∛(m-1)-∛(m+1)+∛(m-1)}
(x+1)/(x-1) = {2∛(m+1)}/{2∛(m-1)}
Cubing both sides
(x+1)³/(x-1)³ = {8(m+1)}/{8(m-1)}
(x³+3x²+3x+1)/(x³-3x²+3x-1) = (m+1)/(m-1)
⇒ (m – 1) (x3 + 3x2 + 3x + 1) = (m + 1 )(x3 – 3x2 + 3x – 1)
⇒ mx3 + 3mx2 + 3mx + m – x3 – 3x2 – 3x – 1 – mx3 – 3mx2 + 3mx – m + x3 – 3x2 + 3x -1
⇒ 6mx2 + 2m – 2x3 – 6x = 0
⇒ 3mx2+ m – x3 – 3x = 0
⇒ x3 – 3mx2 + 3x = m
⇒ x3 – 3mx2 + 3x – m = 0
Hence Proved.
Que-19: What quantity must be added to each term of the ratio a:b to make it (c:d) ?
Sol: Let the required quantity which is to be added be x.
Then, we have:
(a+x)/(b+x) = c/d
By applying componendo and dividendo.
(a+x+b+x)/(a+x-b-x) = (c+d)/(c-d)
(a+b+2a)/(a-b) = (c+d)/(c-d)
(c-d)(a+b+2a) = (c+d)(a-b)
ac+bc+2xc-ad-bd-2xd-ac+ad-ac+2xc-xd = ad+ad-bc-bc
2x(c-d) = 2(ad-bc)
x = (ad-bc)/(c-d)
Que-20: If {a+3b+2c+6d}/{a-3b+2c-6d} = {a+3b-2c-6d}/{a-3b-2c+6d}, prove that a/b = c/d.
Sol: {a+3b+2c+6d}/{a–3b–2c+6d} = {a+3b–2c–6d}/{a–3b+2c–6d}
⇒ {a+3b+2c+6d}/{a+3b–2c–6d} = {a–3b+2c–6d}/{a–3b–2c+6d} …(by altenendo)
Applying componendo and dividendo
{a+3b+2c+6d+a+3b-2c-6d}/{a+3b+2c+6d-a-3b+2c+6d}
= {a-3b+2c-6d+a-3b-2c+6d}/{a–3b+2c-6d-a+3b+2c-6d}
⇒ {2(a+3b)}/{2(2c+6d)} = {2(a-3b)}/{2(2c-6d)}
⇒ a+3b2c+6d=a-3b2c-6d …(Dividing by 2)
⇒ (a+3b)/(a-3b) = (2c+6d)/(2c-6d) …(By alternendo)
Again applying componendo and dividendo
(a+3b+a-3b)/(a+3b-a+3b) = (2c+6d+2c 6d)/(2c+6d-2c+6d)
⇒ 2a/6b = 4c/12d = 2c/6d
⇒ a/b = c/d. Hence Proved. …[Dividing by 2/6]
Que-21: If {2a+2b-3c-3d}/{2a-2b-3c+3d} = {a+b-4c-4d}/{a-b-4c+4d}, prove that a/b = c/d.
Sol: {2a+2b-3c-3d}/{2a-2b-3c+3d} = {a+b-4c-4d}/{a-b-4c+4d}
By applying Componendo and Dividendo
{(2a+2b-3c-3d)+(2a-2b-3c+3d)}/{(2a+2b-3c-3d)-(2a-2b-3c+3d)} = {(a+b-4c-4d)+(a-b-4c+4d)}/{(a+b-4c-4d)-(a-b-4c+4d)}
(4a-6c)/(4b-6d) = (2a-8c)/(2b-8d)
{2(2a-3c)}/{2(2b-3c)} = {2(a-4c)}/{2(b-4d)}
By doing the cross multiplication
(2a-3c)(b-4d)=(a-4c)(2b-3d)
2ab – 8ad – 3bc + 12cd = 2ab – 3ad – 8cd + 12cd
-8ad + 3ad = -8bc + 3bc
-5ad = -5bc
ad = bc
a/b = c/d
Que-22: If (a+b+c+d) : (a+b-c-d) = (a-b+c-d) : (a-b-c+d), prove that a : b = c : d.
Sol: Taking R.H.S We have a/b = c/d
Applying componendo and dividendo
⇒ (a+b)/(a-b) = (c+d)/(c-d)
Applying alternendo
⇒ (a+b)/(c+d) = (a-b)/(c-d)
Again applying componendo and dividendo
(a+b+c+d)/(a+b-c-d) = (a-b+c-d)/(a-b-c+d)
(a+b+c+d) : (a+b-c-d) = (a-b+c-d) : (a-b-c+d)
Que-23: Given : {(x³+12x)/(6x²+8)} = {(y³+27y)/(9y²+27)}. Using componendo, find x : y.
Sol: {(x³+12x)/(6x²+8)} = {(y³+27y)/(9y²+27)}
Using componendo – dividendo
⇒ (x³+12x+6x²+8)/(x³+12x-6x²-8) = (y³+27y+9y²+27)/(y³+27y-9y²-27)
⇒ {(x+2)³}/{(x-2)³} = {(y+3)³}/{(y-3)³}
⇒ {(x+2)/(x-2)}³ = {(y+3)/(y-3)}³
⇒ (x+2)/(x-2) = (y+3)/(y-3)
⇒ 2x/4 = 2y/6 (Using componendo-dividendo)
⇒ x/2 = y/3
x/y = 2/3
x:y = 2:3
Que-24: Using the properties of proportion, find x : y, given (x²+2x)/(2x+4) = (y²+3y)/(3y+9)
Sol: (x²+2x)/(2x+4) = (y²+3y)/(3y+9)
applying componendo and dividendo,
(x²+2x+2x+4)/(x²+2x-2x-4) = (y²+3y+3y+9)/(y²+3y-3y+9)
(x²+4x+4)/(x²-4) = (y²+6y+9)/(y²-9)
(x+2)²/(x-2)² = (y+3)²/(y-3)²
[because when we split (x+2)² and (y+3)² we will get, x²+4x+4 and y²+6y+9]
(x+2)/(x-2) = (y+3)/(y-3)
applying componendo and dividendo again,
(x+2+x-2)/(x+2-x+2) = (y+3+y-3)/(y+3-y+3)
2x/4 = 2y/6
x/2 = y/3
x/y = 2/3
x:y = 2:3
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