Ratio and Proportion Class 10 RS Aggarwal Exe-7B 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-7B 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-7B Questions |
| Academic Session | 2024-2025 |
Solution of Exe-7B Questions
(Ratio and Proportion Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-7)
In this article you will learn what are continue proportion, Mean and third proportion in detail with practice questions and their solutions.
Continue Proportion:
Any three quantities are said to be in continued proportion if the ratio between the first and the second is equal to the ratio between the second and the third
Third Proportional
The third proportional of a continue proportion is the last term For example, if we have a:b = b:c, then the term ‘c’ is the third proportional
Mean Proportional
Multiply the two given numbers together, then calculate the square root out of their product, and it will be the mean proportional such that mean proportional of “a” and “b” will be √(ab)
Que-1: Find x, when :
(i) 3:4 :: 2.4:x (ii) 1:3 :: x:7 (iii) x:1.5 :: 3:5
Sol: (i) 3/4 = 2.4/x
x = (4 x 2.4)/3
x = 3.2
(ii) 1/3 = x/7
x = (1×7)/3
x = 7/3 = 2*(1/3)
(iii) x/1.5 = 3/5
x = (1.5×3)/5
x = 0.9.
Que-2: Find the fourth proportional to :
(i) 3,8 and 21 (ii) 1.4, 3.2 and 7 (iii) 1.5, 4.5 and 3.6 (iv) a²,ab and b² (v) (a²-ab+b²), (a³+b³) and (a-b)
Sol: (i) Let the fourth proportional be x
So, the ratio will be 3:8 :: 21:x
3/8 = 21/x
x = (21×8)/3
x = 56
(ii) Let the fourth proportional be x
So, the ratio will be 1.4:3.2 :: 7:x
1.4/3.2 = 7/x
x = (3.2×7)/1.4
x = 16
(iii) Let the fourth proportional be x
So, the ratio will be 1.5:4.5 :: 3.6:x
1.5/4.5 = 3.6/x
x = (4.5×3.6)/1.5
x = 10.8
(iv) Let the fourth proportional be x
So, the ratio will be a²:ab :: b²:x
a²/ab = b²/x
x = (b²xab)/a²
= b³/a
(v) Let the fourth proportional be x
So, the ratio will be (a²-ab+b²):(a³+b³) :: (a-b):x
If we expand (a³+b³) = (a² -ab +b²)(a+b)
(a²-ab+b²)/(a² -ab +b²)(a+b) = (a-b)/x
x = (a+b)(a-b)
x = (a²-b²)
Que-3: Find the third proportional between :
(i) 9 and 6 (ii) 2*(2/3) and 4 (iii) 1.6 and 2.4 (iv) (2+√3) and (5+4√3) (v) (a/b+b/a) and √(a²+b²)
Sol: (i) 9:6 :: 6:x
as we know –
product of mean = product of extreme
so…..
9×x = 6×6
9x = 36
x = 36÷9 = 4
so x = 4
(ii) 8/3:4 :: 4:x
as we know –
product of mean = product of extreme
so…..
8/3×x = 4×4
x = (4×4×3)/8
x = 6.
(iii) 1.6:2.4 :: 2.4:x
as we know –
product of mean = product of extreme
so…..
1.6 × x = 2.4×2.4
x = (2.4×2.4)/1.6
x = 3.6
(iv) (2+√3):(5+4√3) :: (5+4√3):x
as we know –
product of mean = product of extreme
so…..
(2+√3)×x = (5+4√3)×(5+4√3)
x = (5+4√3)²/(2+√3)
x = {25+48+40√3}/(2+√3)
x = (73+40√3)/(2+√3)
x = (73+40√3)(2-√3)/(2+√3)(2-√3)
x = {146+80√3-73√3-120}/(4-3)
x = 26+7√3
(v) Let x be the third proportional then
{(a/b)+(b/c)}, {(√a²+b²)} = a²+b²:x
⇒ (a²+b²)/ab : (√a²+b²) = (a²+b²) : x
⇒ {(a²+b²)/ab(a²+b²)} = (a²+b²)/x
⇒ x = {ab(a²+b²)}/(a²+b²)
⇒ x = ab.
Que-4: Find the mean proportion between :
(i) 28 and 63 (ii) 2.5 and 0.9 (iii) 6.25 and 1.6 (iv) (√26-√17) and (√26+√17) (v) (6+3√3) and (8-4√3)
Sol: (i) Given = x : 28 : : 63 : x
(x/28) = (63/x)
x² = 28 * 63
x² = 1764
x² = 42 * 42
x = 42.
(ii) Given = x : 2.5 : : 0.9 : x
(x/2.5) = (0.9/x)
x² = 2.5 * 0.9
x² = 2.25
x² = 1.5 * 1.5
x = 1.5
(iii) Given = x : 6.25 : : 1.6 : x
(x/6.25) = (1.6/x)
x² = 6.25 * 1.6
x² = 10
x² = √10 * √10
x = √10.
(iv) Given two numbers are (√26 – √17) and (√26 + √17)
Here
a = √26 – √17
b = √26 + √17
G = √ab
G² = ab
G² = (√26 – √17)(√26 + √17)
G² = (√26)² – (√17)²
G² = 26 – 17
G² = 9
G = √9
G = 3
(v) Let the mean proportional between 6 + 3√3 and 8 – 4√3 be x.
⇒ 6 + 3√3, x and 8 – 4√3 are in continued proportion.
⇒ 6 + 3√3 : x = x : 8 – 4√3
⇒ x × x = (6 + 3√3) (8 – 4√3)
⇒ x² = 48 + 24√3- 24√3 – 36
⇒ x² = 12
⇒ x = 2√3
Que-5: 6 is the mean proportion between two numbers x and y and 48 is the third proportion of x and y. Find the numbers.
Sol: 6 is the mean proportion between two numbers x and y,
i.e 6 = √xy
So, 36 = xy …(1)
It is given that 48 is the third proportional to x and y
So, y2 = 48x … (2)
From (1) and (2), we get
y² = 48(36/y)
⇒ y³ = 1728
Hence, y = 12
∴ x = 36/y = 36/12 = 3
Thus, the required numbers are 3 and 12.
Que-6: What least number must be added to each of the numbers, 5,11,19 and 37, so that the resulting numbers are proportional.
Sol: Let x be added to 5, 11, 19 and 37 to make them in proportion.
5 + x : 11 + x : : 19 + x : 37 + x
⇒ (5 + x) (37 + x) = (11 + x) (19 + x)
⇒ 185 + 5x + 37x + x2 = 209 + 11x + 19x + x2
⇒ 185 + 42x + x2 = 209 + 30x + x2
⇒ 42x – 30x + x2 – x2 = 209 – 185
⇒ 12x = 24
⇒ x = 2
∴ Least number to be added = 2.
Que-7: What number must be added to each of the numbers 4,6,8,11 in order to get the four numbers in proportion ?
Sol: Let x be added to make them in proportion.
∴ (4 + x) : (6 + x) :: (8 + x) : (11 + x)
⇒ (4 + x)(11 + x) = (6 + x)(8 + x)
⇒ 44 + 4x + 11x + x2 = 48 + 6x + 8x + x2
⇒ 44 + 15x = 48 + 14x
⇒ 15x – 14x = 48 – 44
⇒ x = 4
Que-8: What least number be subtracted from each of the numbers 23,30,57 and 78, so that the remainder are in proportion ?
Sol: Let, m be the number subtracted.
According to the question,
(23-m)/(30-m) = (57-m)/(78-m)
⇒ (23 – m)(78 – m) = (30 – m)(57 – m)
⇒ 1794 – 23m – 78m – m2 = 1710 – 30m – 57m – m2
⇒ 84 = 14m
⇒ m = 6
Que-9: If (x-2), (x+2), (2x+1) and (2x+19) are in proportion, find the value of x.
Sol: (X-2),(X+2),(2X+1),(2X+19) are in proportion
(X-2)(2X+19) = (X+2)(2X+1)
2X²+19X-4X-38 = 2X²+X+4X+2
15X-38 = 5X+2
10X = 40
X = 40/10
X = 4.
Que-10: The following numbers, K+3, K+2, 3K-7 and 2K-3 are in proportion. Find K.
Sol: K+3, K+2, 3K – 7 and 2K – 3 are in proportional then
⇒ (K+3)/(K+2) = (3K-7)/(2K-3)
⇒ ( K +3) (2K -3) = (3k -7) (k+2)
⇒ 2K²-3K+6K-9 = 3K²+6K-7K-14
⇒ 2K²+3K-9 = 3K²-K-14
⇒ K²-4K-5 = 0
⇒ K²-5K+K-5 = 0
⇒K(K-5)+1( K-5) = 0
⇒ K = 5 or K = -1
Que-11: If (x+5) is the geometric mean between (x+2) and (x+9), find the value of x.
Sol: Given, x + 5 is the mean proportional between x + 2 and x + 9.
⟹ (x + 2), (x + 5) and (x + 9) are in continued proportion.
⟹ (x + 2) : (x + 5) = (x + 5) : (x + 9)
⟹ (x+5)² = (x+2)(x+9)
⟹ x2+25+10x = x²+2x+9x+18
⟹ 25 − 18 = 11x − 10x
⟹ x = 7
Que-12: Find two numbers whose mean proportion is 36 and the third proportion is 288.
Sol: Let the required number be a and b.
36 is the mean proportion between a and b
a/36 = 36/b
ab = 1296
a =1296/b
288 is the third proportion
a/b = b/288
b² = 288a
b² = 288 x (1296/b)
b³ = 373248
b = ∛373248
b = 72
a = 1296/b
a = 1296/72
a = 18.
Que-13: If a:b :: c:d, prove that :
(i) (a²+ab):(c²+cd) :: (b²-2ab):(d²-2cd)
(ii) (a²+b²):(c²+d²) :: (ab+ad-bc):(cd-ad+bc)
(iii) (a²+ac+c²):(a²-ac+c²) :: (b²+bd+d²):(b²-bd+d²)
Sol: (i) ∵ a, b, c, d are in proportion
a/b = c/d = k(say)
a = bk, c = dk.
L.H.S. = (a²+ab)/(c²+cd)
= (k²b²+bk.b)/(k²d²+dk.d)
= kb²(k+1)/d²k(k+1)
= b²/d²
R.H.S. = (b²-2ab)/(d²-2cd)
= (b²-2.bkb)/(d²-2dkd)
= b²(1-2k)/d²(1-2k)
= b²d²
∴ L.H.S. = R.H.S.
(ii) ∵ a, b, c, d are in proportion
a/b = c/d = k(say)
a = bk, c = dk.
L.H.S. = (a²+b²/(c²+d²)
= (b²k²+b²)/(d²k²+d²)
= b²(k²+1)/d²(k²+1)
= b²/d²
R.H.S. = (ab+ad-bc)/(bc+cd-ad)
= (bk.b+bk.d-b.dk)/(b.kd+dk.d-bk.d)
= k(b²+bd-bd)/k(bd+d²-bd)
= b²/d²
∴ L.H.S. = R.H.S.
(iii) ∵ a, b, c, d are in proportion
a/b = c/d = k(say)
a = bk, c = dk.
L.H.S = (a²+ac+c²)/(a²-ac+c²)
= {(bk)²+(bk)(dk)+(dk)²}/{(bk)²-(bk)(dk)+(dk)²}
= (b²k²+bdk²+d²k²)/(b²k²-bdk²+d²k²)
= {k²(b²+bd+d²)}/{k²(b²-bd+d²)}
= (b²+bd+d²)/(b²-bd+d²) = R.H.S.
Que-14: If a:b::c:d, show that :
(i) (a+b)/(c+d) = {√(2a²+7b²)}/{√(2c²+7d²)}
(ii) (ma²+nc²)/(mb²+nd²) = {√(a^4 + c^4)}/{√(b^4 + d^4)}
(iii) (a²+ab+b²)/(a²-ab+b²) = (c²+cd+d²)/(c²-cd+d²)
(iv) (a+c)³/(b+d)³ = {a(a-c)²}/{b(b-d)²}
Sol: (i) since, a, b, c, d are in proportion , so,
→ a/b = c/d = Let k .
→ a = bk
→ c = dk
then, LHS :-
→ (a + b)/(c + d)
→ (bk + b) / (dk + d)
→ b(k + 1)/d(k + 1)
→ b/d
R.H.S :-
→ √(2a² + 7b²) / √(2c² + 7d²)
→ √(2b²k² + 7b²) / √(2d²k² + 7d²)
→ √b²(2k² + 7) / √(d²(2k² + 7)
→ √b² * √(2k² + 7) / √d² * √(2k² + 7)
→ √b²/√d²
→ b/d .
therefore,
→ LHS = RHS (Proved.)
(ii) a/b = b/c = c/d = k
b = a/k, c = b/k, c = {(a/k)/k} = a/k²
c/d = k, d = c/k = {(a/k²)/k} = a/k³
R.H.S = {√(a^4 + c^4)}/{√(b^4 + d^4)}
= √{a^4+(a^4/k^8)}/{√(a^4/k^4)+(a^4/k^12)}
= √{1+(1/k^8)}/1/k²√{1+(1/k^8)}
= k²
L.H.S = (ma²+nc²)/(mb²+nd²)
= {ma²+n(a/k²)²}/{m(a/k)²+n(a/k³)²}
= {a²(m+(n/k^4))}/{a²(m/k²)+(n/k^6)}
= {m+(b/k^4)}/[(1/k²){m+(b/k^4)}]
= k²
L.H.S = R.H.S (Proved)
(iii) ∵ a, b, c, d are in proportion
a/b = c/d = k(say)
a = bk, c = dk.
L.H.S. = (a²+ab+b²)/(a²-ab+b²)
= (b²k²+bk.b+b²)/(b²k²-bk.b+b²)
= {b²(k2+k+1)}/{b²(k2-k+1)}
= (k²+k+1)/(k²-k+1)
R.H.S. = (c²+cd+d²)/(c²-cd+d²)
= (d²k²+dkd+d²)/(d²k²-dk.d+d²)
= {d²(k²+k+1)}/{d²(k²-k+1)}
= (k²+k+1)/(k²-k+1)
∴ L.H.S. = R.H.S.
(iv) ∵ a, b, c, d are in proportion
a/b = c/d = k(say)
a = bk, c = dk.
L.H.S. = (a+c)³/(b+d)³
= (bk+dk)³/(b+d)³
= {k³(b+d)³}/(b+d)³
= k³
R.H.S. = {a(a-c)²}/{b(b-d)²}
= {bk(bk-dk)²}/{b(b-d)²}
= {bk.k²(b-d)²}/{b(b-d)²}
= k3
∴ L.H.S. = R.H.S.
Que-15: If x/a = y/b = z/c, prove that :
(i) (x²+y²+z²)/(a²+b²+c²) = {(px+qy+rz)/(pa+qb+rc)}²
(ii) (x³/a³)+(y³/b³)+(z³/c³) = 3xyz/abc
(iii) (x³/a²)+(y³/b²)+(z³/c²) = (x+y+z)³/(a+b+c)²
(iv) ax-by/{(a+b)(x-y)} + by-cz/{(b+c)(y-z)} + cz-ax/{(c+a)(z-x)} = 3
Sol: (i) We know that x/a = y/b = z/c = k.
Multiplying both sides by a², b², and c², respectively, we get:
x² = a²k², y² = b²k², and z² = c²k².
Substituting these values into the numerator, we have:
x² + y² + z² = a²k² + b²k² + c²k² = (a² + b² + c²)k².
[(a² + b² + c²)k²]/(a² + b² + c²)
= k²
R.H.S = Simplifying the denominator (px + qy + rz / pa + qb + rc):
x = ak, y = bk, and z = ck.
Substituting these values into the denominator, we get:
px + qy + rz = p(ak) + q(bk) + r(ck) = k(pa + qb + rc).
{k²(pa + qb + rc)} = (pa + qb + rc)
= k².
L.H.S = R.H.S (Proved)
(ii) Let x/a = y/b = z/c = k
= x = ak, y = bk, z = ck
L.H.S = (x³/a³) + (y³/b³) + (z³/c³)
= {(ak)³/a³} + {(bk)³/b³} + {(ck)³/c³}
= {(a³k³)/a³} + {(b³k³)/b³} + {(c³k³)/c³}
= k³+k³+K³
= 3k³
R.H.S = 3xyz/abc
= {3(ak)(bk)(ck)}/abc
= 3k³
= L.H.S
= L.H.S = R.H.S
(iii) Let x/a = y/b = z/c = k …[By k method]
x = ak, y = bk and z = ck.
L.H.S.
= (a³k³)/a² + (b³k³)/b² + (c³k³)/c²
⇒ k³ [a + b + c]
R.H.S.
= [ak+bk+ck]³/[a+b+c]²
⇒ {k³[a+b+c]³}/[a+b+c]²
= k³(a + b + c)
L.H.S. = R.H.S.
Hence proved.
(iv) x/a = y/b = z/c = k
∴ x = ak, y = bk, z = ck
L.H.S. = ax-by/{(a+b)(x-y)} + by-cz/{(b+c)(y-z)} + cz-ax/{(c+a)(z-x)}
= (a.ak-b.bk)/{(a+b)(ak-bk)} + (b.bk-c.ck)/{(b+c)(bk-ck)} + (c.ck-a.ak)/{(c+a)(ck.ak)}
= (a²k-b²k)/{(a+b)k(a-b)} + (b²k-c²k)/{(b+c)k(b-c)} + (c²k-a²k)/{(c+a)k(c-a)}
= {k(a²-b²)}/{k(a²-b²)} + {k(b²-c²)}/{k(b²-c²)} + {k(c²-a²)}/{k(c²-a²)}
= 1 + 1 + 1
= 3
= R.H.S.
Que-16: If a/b = c/d = e/f prove that :
(i) (b²+d²+f²)(a²+c²+e²) = (ab+cd+ef)²
(ii) (a³+c³+e³)/(b³+d³+f³) = ace/bdf
(iii) {(a²/b²)+(c²/d²)+(e²/f²)} = {(ac/bd)+(ce/df)+(ae/bf)}
(iv) (bdf).{(a+b)/b + (c+d)/d + (e+f)/f}³ = 27(a+b)(c+d)(e+f)
Sol: a/b = c/d = e/f = k
∴ a = bk, c = dk, e = fk
L.H.S. = (b2 + d2 + f2) (a2 + c2 + e2)
= (b2 + d2 + f2) (b2 k2 + d2 k2 + f2 k2)
= k2 (b2 + d2 + f2) k2 (b2 + d2 + f2)
= k2 (b2 + d2 + f2)2
R.H.S. = (ab + cd + ef)2
= (b. kb + dk. d + fk. f)2
= (kb2 + kd2 + kf2)
= k2 (b2 + d2 + f2)2
∴ L.H.S. = R.H.S.
(ii) Given:- a/b = c/d = e/f
To proof:- (a³+c³+e³)/(b³+d³+f³) = ace/bdf
Proof:-
Let a/b = c/d = e/f = k
⇒ a=bk, c=dk, e=fk
Now,
(a³+c³+e³)/(b³+d³+f³) = ace/bdf
{(bk)³+(dk)³+(fk)³}/(b³+d³+f³) = {(bk)(dk)(fk)}/bdf
{(b³k³)+(d³k³)+(f³k³)}/(b³+d³+f³) = (bdf⋅k³)/bdf
{k³(b³+d³+f³)}/(b³+d³+f³) = {(bdf).k³}/bdf
⇒ k³ = k³
L.H.S. = R.H.S.
Hence proved.
(iii) a/c = c/d = c/f = k(say)
∴ a = bk, c = dk, e = fk
L.H.S. = (a²/b²)+(c²/d²)+(e²/f²)
= {(b²k²)/b²} + {(d²k²)/d²} + {(f²k²)/f²}
= k²+k²+k²
= 3k²
R.H.S. = (ac/bd) + (ce/df) + (ae/bf)
= {(bk.dk)/b.d} + {(dk.fk)/d.f} + {(bk.fk)/b.f}
= k²+k²+k²
= 3k²
∴ L.H.S. = R.H.S.
(iv) a/b = c/d = e/f = k
∴ a = bk, c = dk, e = fk
L.H.S. = (bdf).{(a+b)/b + (c+d)/d + (e+f)/f}³
= bdf.[{(bk+b)/b} + {(dk+d)/d} + {(fk+f)/k}]³
= bdf.[{b(k+1)/b} + {d(k+1)/d} + {f(k+1)/f}]³
= bdf(k + 1 + k + 1 + k + 1)3
= bdf(3k + 3)3 = 27bdf(k + 1)3
R.H.S. = 27(a + b)(c + d)(e + f)
= 27(bk + b)(dk + d)(fk + f)
= 27b(k + 1)d(k + 1)f(k + 1)
= 27bdf(k + 1)3
∴ L.H.S. = R.H.S.
Que-17: If a,b,c are are in continued proportion, prove that :
(i) (a+b)/(b+c) = {a²(b-c)}/{b²(a-b)} (ii) (a+b+c)/(a-b+c) = (a+b+c)²/(a²+b²+c²) (iii) (a²+ab+b²)/(b²+bc+c²) = a/c (iv) (a+b+c)(a-b+c) = (a²+b²+c²) (v) a²b²c²(a¯³+b¯³+c¯³) = (a³+b³+c³)
Sol: (i) Since, a, b, c are in continued proportion, a/b = b/c
Let, a/b = b/c = k
Then, a = bk and b = ck
Hence, a = bk = (ck). k = ck²
LHS = (a+b)/(b+c)
= (ck²+ck)/(ck+c)
= {ck(k+1)}/{c(k+1)}
= k
R.H.S = {a²(b-c)}/{b²(a-b)}
= {(ck²)²(ck-c)}/{(ck²)(ck²-ck)}
= {c²k^4(ck-c)}/{c²k²(ck²-ck)}
= {c³k^4(k-1)}/{c³k³(k-1)}
= k
L.H.S = R.H.S (Proved)
(ii) Given, a, b and c are in continued proportion.
Therefore,
a/b = b/c
ac = b2
LHS = (a+b+c)²/a²+b²+c²
= (a+b+c)²/{a²+b²+c²+2ab+2bc+2ac-(2ab+2bc+2ac)}
= (a+b+c)²/(a+b+c)²-2(ab+bc+b²)
= (a+b+c)²/(a+b+c)²-2b(a+c+b)
= (a+b+c)²/{(a+b+c)(a+b+c-2b)}
= (a+b+c)/(a-b+c).
Hence proved.
(iii) Given, a, b and c are in continued proportion.
Therefore
a/b = b/c
b² = ac
Here,
(a²+ab+b²)/(b²+bc+c²)
= (a²+ab+bc)/(ac+bc+c²)
= {a(a+b+c)}/{c(a+b+c)}
= a/c = R.H.S.
(iv) As a, b, c, are in continued proportion
Let a/b = b/c = k
L.H.S. = (a + b + c) (a – b + c)
= (ck2 + ck + c)(ck2 – ck + c)
= c(k2 + k + 1)c(k2 – k + 1)
= c2(k2 + k + 1)(k2 – k + 1)
= c2(k4 + k2 + 1)
R.H.S. = a2 + b2 + c2
= (ck2)2 + (ck)2 + (c)2
= c2k4 + c2k2 + c2
= c2(k4 + k2 + 1)
∴ L.H.S. = R.H.S.
(v)
Que-18: If a,b,c,d are in continued proportion, prove that :
(i) (b+c)(b+d) = (c+a)(c+d)
(ii) (a+b)(b+c) – (a+c)(b+d) = (b-c)²
(iii) (a²-b²)(c²-d²) = (b²-c²)²
(iv) [{(a-b)/c}+{(a-c)/b}]² – [{(d-b)/c}+{(d-c)/b}]² = (a-d)²{(1/c²)+(1/b²)}
Sol: (i)
(ii) Given a,b,c,d are in continued proportion
⟹ a/b = b/c = c/d = k(say)
⟹ c =dk, b = ck = k²d, a = bk = k³d
LHS = (a+d)(b+c)−(a+c)(b+d)
= (k³d+d)(k²d+kd) − (k³d+kd)(k²d+d)
= d²(k^5+k²+k^4+k−k^5−2k³−k)
= k²d²(k−1)² = (k²d−kd)² = (b−c)² = RHS
Hence Proved.
(iii) a, b, c, d are in continued proportion
a/b = b/c = c/d = k
∴ c = dk, b = ck = dk. k = dk2,
a = bk = dk2. k = dk3
L.H.S. = (a2 – b2) (c2 – d2)
= [(dk3)2 – (dk2)2] [(dk)2 – d2]
= (d2k6 – d2k4) (d2k2 – d2)
= d2k4 (k2 – 1) d2(k2 – 1)
= d4k4 (k2 – 1)2
R.H.S. = (b2 – c2)2
= [(dk2)2 – (dk)2]2
= [d2k2 – d2k2]2
= [d2k2 (k2 – 1)]2
= d4k4(k2 – 1)2
∴ L.H.S. = R.H.S.
(iv) a, b, c, d are in continued proportion
a/b = b/c = c/d = k
∴ c = dk, b = ck = dk. k = dk2,
a = bk = dk2. k = dk3
L.H.S. = [{(a-b)/c}+{(a-c)/b}]² – [{(d-b)/c}+{(d-c)/b}]²
= [{(dk³-dk²)/dk} + {(dk²-dk)/dk²}]² – [{(d-dk²)/dk} + {(d-dk)/dk²}]²
= [{dk²(dk-1)/dk} + {dk(k²-1)/dk²}]² – [{d(1-k²)/dk} + {d(1-k²)/dk²}]
= [{k(k-1)} + {(k²-1)/k}]² – [{(1-k²)/k} + {(1-k)/k²}]²
= [{k²(k-1)+(k²-1)}/k]² – [{k(1-k²)+1-k}/k²]²
= (k³-1)²/k² – (-k³+1)²/k^4
= (k³-1)²/k² – (1-k³)/k^4
= {(k³-1)/k²}² – {1-(1/k²)}
= {(k³-1)²(k²-1)}/k^4
R.H.S. = (a-d)²[(1/c²)-(1/b²)]
= (dk³-d)² [(1/d²k²) – (1/d²k^4)]
= d²(k³-1) [(k²-1)/d²k^4]
= {(k³-1)²(k²-1)}/k^4
L.H.S = R.H.S .
Que-19: If ax = by = cz, prove that (x²/yz)+(y²/zx)+(z²/xy) = (bc/a²)+(ca/b²)+(ab/c²)
Sol: Let ax = by = cz = k,
then x = k/a, y = k/b, z = k/c
L.H.S = (x²/yz)+(y²/zx)+(z²/xy)
= [k²/{a²x(k/b)x(k/c)}] + [k²/{b²x(k/c)x(k/a)}] + [k²/{c²x(k/a)x(k/b)}]
= bc/a² + ca/b² + ab/c²
= R.H.S.
Hence Proved.
Que-20: If {x/(b+c-a)} = {y/(c+a-b)} = {z/(a+b-c)}, prove that each ratio is equal to {(x+y+z)/(a+b+c)}. Also, show that (b-c)x + (c-a)y + (a-b)z = 0.
Sol: {x/(b+c-a)} = {y/(c+a-b)} = {z/(a+b-c)} = k
x = k(b + c- a)
y = k(c+a-b)
z = k(a+b-c)
Now ,
(x+y+z)/(a+b+c)
k(b+c-a)+k(c+a-b)+k(a+b-c)/(a + b + c)
{k(b + c – a + c + a – b + a + b – c)}/(a + b + c)
{k(a + b + c)}/(a + b + c) = k
Hence ,
{x/(b+c-a)} = {y/(c+a-b)} = {z/(a+b-c)} = (x+y+z)/(a+b+c)
Proved.
let {x/(b+c-a)} = {y/(c+a-b)} = {z/(a+b-c)} =k
x/(b+c-a) = k
x = k(b+c-a)
y/(c+a-b) = k
y = k(c+a-b)
z/(a+b-c) = k
z = k(a+b-c)
(b-c)x + (c-a)y + (a-b)z = (b-c)k(b+c-a) + (c-a)k(c+a-b) + (a-b)k(a+b-c)
= k(b²+bc-ba-bc-c²+ca+c²+ca-bc-ac-a²+ab+a²+ab-ac-ba-b²+bc)
= k(0)
= 0 Hence Proved.
Que-21: If b is the mean proportion between a and c, show that :
{(a^4 + a²b² + b^4)/(b^4 + b²c² + c^4)} = a²/c².
Sol: Given, b is the mean proportion between a and c.
ab = b/c = k
⇒ a = bk, b = ck
⇒ a = (ck)k = ck² , b = ck
L.H.S = {(a^4 + a²b² + b^4)/(b^4 + b²c² + c^4)}
= {(ck²)^4 + (ck²)²(ck)² + (ck)^4}/{(ck)^4 + (ck)²c² + c^4}
= {c^4 k^8 + (c²k^4)(c²k²) + c^4 k^4}/{c^4 k^4 + (c²k²)c² + c^4}
= {c^4 k^8 + c^4 k^6 + c^4 k^4}/{c^4 k^4 + c^4 k² + c^4}
= {c^4 k^4 (k^4 + k² + 1)}/{c^4 (k^4 + k² + 1)}
= k^4.
R.H.S = a²/c²
= (ck²)²/c²
= c² k^4/c²
= k^4.
Hence, L.H.S = R.H.S
— : End of Ratio and Proportion Class 10 RS Aggarwal Exe-7B Goyal Brothers ICSE Maths Solutions Ch-7 :-
Return to: ICSE Class 10 Maths RS Aggarwal Solutions
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