WBBSE Solutions For Class 10 Maths Chapter 5 Ration And Proportion Exercise 5.3
Question 1. If a: b = c:d, let us show that:
1. (a² + b²): (a²-b²) = (ac + bd): (ac-bd)
Solution: \(\left(a^2+b^2\right):\left(a^2-b^2\right)=(a c+b d):(a c-b c)\)
Let \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{d}}=\mathrm{k}\) [k ≠ 0]
∴ a = bk, c = dk
L.H.S. = \(\frac{a^2+b^2}{a^2-b^2}=\frac{(b k)^2+b^2}{(b k)^2-b^2}=\frac{b^2 k^2+b^2}{b^2 k^2-b^2}=\frac{b^2\left(k^2+1\right)}{b^2\left(k^2-1\right)}=\frac{k^2+1}{k^2-1}\)
R.H.S. = \(\frac{a c+b d}{a c-b d}=\frac{b k \cdot d k+b d}{b k \cdot d k-b d}=\frac{b d k^2+b d}{b d k^2-b d}=\frac{b d\left(k^2+1\right)}{b d\left(k^2-1\right)}=\frac{k^2+1}{k^2-1}\)
∴ L.H.S. = R.H.S.
2. √a²+ b²: √b² + d² = (pa + qc): (pb + qd)
Solution: \(\sqrt{a^2+b^2}: \sqrt{b^2+d^2}=(p a+q c):(p b+q d)\)
L.H.S. = \(\frac{\sqrt{\mathrm{a}^2+\mathrm{c}^2}}{\sqrt{\mathrm{b}^2+\mathrm{d}^2}}=\frac{\sqrt{\mathrm{b}^2 \mathrm{k}^2+\mathrm{d}^2 \mathrm{k}^2}}{\sqrt{\mathrm{b}^2+\mathrm{d}^2}}=\frac{\mathrm{k}\left(\sqrt{\mathrm{b}^2+\mathrm{d}^2}\right)}{\left(\sqrt{\mathrm{b}^2+\mathrm{d}^2}\right)}=\mathrm{k}\)
R.H.S. = \(\frac{p a+q c}{p b-q d}=\frac{p d k+q d k}{p b-q d}=\frac{k(p d+q d)}{p b-q d}=k\)
∴ L.H.S. = R.H.S
Question 2. (a²+ b² + c²) (x² + y²+z²) = (ax + by + cz)²
Solution:
To prove, (a² + b² + c²) (x² + y²+z²) = (ax + by + cz)²
L.H.S= (a²+ b² + c²) (x² + y² + z²)
= (a²+ b² + c²) (a²k² + b²k² + c²k²) = k²(a² + b² + c²) (a² + b² + c²) = k²(a² + b² + c²)²
= k²(a² + b² + c²)²
R.H.S. = (ax+by+ cz)²
(a.ak + b.bk+c.ck)² = {k(a² + b² + C²)}²
(a² + b²+ c²) (x² + y²+z²) = (ax + by + cz)² Proved.
L.H.S = R.H.S
Question 3. If a: b = c : d = e: f, let us prove that,.
1. Each ratio = \(\frac{5 a-7 c-13 e}{5 b-7 d-13 f}\)
Solution: Each ratio = \(\frac{5 a-7 c-13 e}{5 b-7 d-13 f}\)
Let, \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{e}}{\mathrm{f}}=\mathrm{k}\) (where k ≠ 0)
∴ a = bk; c = dk; e = fk
\(\frac{5 a-7 c-13 e}{5 b-7 d-13 f}\)= \(\frac{5 b k-7 d k-13 f k}{5 b-7 d-13 f}\)
= \(\frac{k(5 b-7 d-13 f)}{(5 b-7 d-13 f)}=k\)
∴ \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{d}}=\frac{\mathrm{e}}{\mathrm{f}}=\mathrm{k}\)
= \(\frac{5 a-7 c-13 e}{5 b-7 d-13 f}\)
2. (a² + c² + e²) (b² + c² + e²) (b² + d² + f²) = (ab + cd + ef)²
Solution: To prove \(\left(a^2+c^2+e^2\right)\left(b^2+c^2+e^2\right)\left(b^2+d^2+f^2\right)=(a b+c d+e f)^2\)
L.H.S. = \(\left(a^2+c^2+e^2\right)\left(b^2+d^2+f^2\right)\)
= \(\left(b^2 k^2+d^2 k^2+d^2 k^2\right)\left(b^2+d^2+f^2\right)\)
= \(k^2\left(b^2+d^2+f^2\right)\left(b^2+d^2+f^2\right)=k^2\left(b^2+d^2+f^2\right)^2\)
R.H.S. = \((a b+c d+e f) 2\)
= \((\mathrm{bk} \cdot \mathrm{b}+\mathrm{dk} \cdot \mathrm{d}+\mathrm{fk} \cdot \mathrm{f}) 2\)
= \(\left\{\mathrm{k}\left(\mathrm{b}^2+\mathrm{d}^2+\mathrm{f}^2\right)\right\}^2=\mathrm{k}^2\left(\mathrm{~b}^2+\mathrm{d}^2+\mathrm{f}^2\right)^2\)
L.H.S. = R.H.S.
\(\left(a^2+c^2+e^2\right)\left(b^2+c^2+e^2\right)\left(b^2+d^2+f^2\right)=(a b+c d+e f)^2\) Proved.
Question 4. If a, b, c, and d are in continued proportion, let us prove that
1. \(\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)=(a b+b c+c d)^2\)
Solution: \(\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)=(a b+b c+c d)^2\) as a, b, c, d are continued proportional.
Let \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{b}}{\mathrm{c}}=\frac{\mathrm{c}}{\mathrm{d}}=\mathrm{k}\) (where k ≠ 0).
⇒ \(\mathrm{c}=\mathrm{dk} ; \mathrm{b}=\mathrm{ck}=\mathrm{dk} \cdot \mathrm{k}=\mathrm{dk}^2, \mathrm{a}=\mathrm{bk}=\mathrm{dk}^2 \cdot \mathrm{k}=\mathrm{dk}^2\)
L.H.S. = \(\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)\)
= \(\left\{\left(\mathrm{dk}^3\right)^2+\left(\mathrm{dk}^2\right)^2+(\mathrm{dk})^2\right\}\left\{\left(\mathrm{dk}^2\right)^2+(\mathrm{dk})^2+(\mathrm{d})^2\right\}\)
= \(\left(d^2 k^6+d^2 k^4+d^2 k^2\right)\left(d^2 k^4+d^2 k^2+d^2\right)\)
= \(d^2 k^2\left(k^4+k^2+1\right) d^2\left(k^4+k^2+1\right)\)
= \(d^4 k^2\left(k^4+k^2+1\right)^2\)
R.H.S = \((a b+b c+c d)^2\)
= \(\left(\mathrm{dk}^3 \cdot \mathrm{dk}^2+\mathrm{dk}^2 \cdot \mathrm{dk}+\mathrm{dk} \cdot \mathrm{d}\right)^2\)
= \(\left(d^2 k^5+d^2 k^3+d^2 k\right)^2\)
= \(\left\{\mathrm{d}^2 \mathrm{k}\left(\mathrm{k}^4+\mathrm{k}^2+1\right)\right\}^2\)
= \(d^4 k^2\left(k^4+k^2+1\right)^2\)
L.H.S. = R.H.S. Proved.
2. (b – c)²+(c – a)² + (b – d)² = (a – d)²
Solution: To prove, \((b-c)^2+(c-a)^2+(b-d)^2=(a-d)^2\)
L.H.S. = \((b-c)^2+(c-a)^2+(b-d)^2\)
= \(\{\mathrm{dk}(\mathrm{k}-1)\}^2+\left\{\mathrm{dk}\left(1-\mathrm{k}^2\right)\right\}^2+\left\{\mathrm{d}\left(\mathrm{k}^2-1\right)\right\}^2\)
= \(d^2 k^2\left(k^2-2 k+1\right)+d^2 k^2\left(1-2 k^2+k^4\right)+d^2\left(k^4-2 k^3+1\right)\)
= \(d^2\left(k^4-2 k^3+k^2+k^2-2 k^4+k^6+k^4-2 k^2+1\right)\)
= \(d^2\left(k^6-2 k^3+1\right)=d^2\left(k^3-1\right)^2\)
R.H.S. = \((a-d)^2=\left(d k^2-d\right)^2=d^2\left(k^3-1\right)^2\)
∴ L.H.S. = R.H.S. Proved.
Question 5:
1. If m/a = n/b, let us show that (m²+n²) (a²+b²) = (am + bn)².
Solution:
Given
m/a = n/b
If m/a = n/b, prove that (m² + n²) (a² + b²) = (am + bn)².
Let, m/a = n/b = k(where k ≠ 0)
m = ak; n=bk
L.H.S = (m²+n²) (a²+b²)
= (a²k² + b²k²) (a²+ b²) = k²(a²+ b²) (a²+ b²) = {k(a²+ b²)}²
R.H.S. = (am bm)²
= (a.ak + b.bk)² = (a²k + b²k)² = {k(a² + b²)}²
∴ L.H.S. R.H.S. Proved.
2. If a/b = x/y, let us show that (a + b) (a² + b²) x³ = (x + y)(x² + y²) a³.
Solution: If a/b = x/y, prove that (a + b) (a² + b²) x³ = (x + y)(x² + y²) a³.
Let a/b = x/y = k(where k ≠ 0)
∴ a=bk & x=yk
L.H.S = (a + b) (a² + b²) x³
(bk + b) (b²k² + b²) (yk)³ = b(k + 1) b²(k² + 1) y³k³
= b3k3y (k + 1) (k² + 1)
R.H.S.= (x + y) (x²+ y²) a³
=(yk + y) (y²k² + y²) (b³k³) = y(k + 1) y²(k² + 1)
= b³k³ b³k³y³(k+1) (k² + 1)
∴ L.H.S. = R.H.S. Proved.
3. If, \(\frac{x}{\mid m-n^2}=\frac{y}{m n-\left.\right|^2}=\frac{z}{n \mid-m^2}\) , let us show that lx + my + ny = 0.
Solution: If \(\frac{x}{\mid m-n^2}=\frac{y}{m n-\left.\right|^2}=\frac{z}{n \mid-m^2}\)
Prove that, lx + my + nz = 0.
Let \(\frac{x}{\mid m-n^2}=\frac{y}{m n-1^2}=\frac{z}{n \mid-m^2}=k\) (where k ≠ 0)
∴ \(\left.x=k\left(\mid m-n^2\right), y=k(m n-l) ; x=k(n)-m^2\right)\)
Now, lx + my + nz
= \(\left.\left.\mathrm{k}(\mathrm{k}) \mathrm{m}-\mathrm{n}^2\right)+\mathrm{mk}(\mathrm{mn}-\mathrm{P})+\mathrm{rk}(\mathrm{n})-\mathrm{m}^2\right)\)
= \(\mathrm{k}\left[1^2 \mathrm{~m}-\ln ^2+\mathrm{m}^2 \mathrm{n}-1 P \mathrm{~m}+\ln ^2-\mathrm{m}^2 \mathrm{n}\right]\)
= k x 0 = 0 Proved.
4. If \(\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}\) let us show that (b-c) x + (c-a)y + (a-b) x=0.
Solution: \(\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}\)
Prove that (b – c)x + (c – a)y + (a – b)z = 0
Let \(\frac{x}{b+c-a}=\frac{y}{c+a-b}=\frac{z}{a+b-c}=k\) (where k ≠ a)
∴ x = k(b + c – a)
y = k(c + a – b)
z = k(a + b – c)
L.H.S. = (b – c)x + (c – a)y + (a – b)z
= (b – c)(b + a – a)k + (c – a)(c + a – b)k + (a – b)(a + b – c)k
= \(k\left[b^2-c^2-a b+a c+c^2-a^2-b c+a b+\left(a^2-b^2\right)-a c+b c\right)\)
= k x 0 = 0 = R.H.S. Proved.
5. If \(\frac{x}{y}=\frac{a+2}{a-2}\), let us show that \(\frac{x^2-y^2}{x+y^2}=\frac{4 a}{a^2+4}\).
Solution: If \(\frac{x}{y}=\frac{a+2}{a-2}\)
prove that \(\frac{x^2-y^2}{x+y^2}=\frac{4 a}{a^2+4}\)
squaring both sides.
\(\frac{x^2}{y^2}=\frac{(a+2)^2}{(a-2)^2}\)or, \(\frac{x^2-y^2}{x^2+y^2}=\frac{(a+2)^2-(a-2)^2}{(a+2)^2+(a-2)^2}\)
or, \(\frac{x^2-y^2}{x^2+y^2}=\frac{\left(a^2+4 a+4\right)-\left(a^2-4 a+4\right)}{\left(a^2+4 a+4\right)+\left(a^2-4 a+4\right)}\)
or, \(\frac{x^2-y^2}{x^2+y^2}\)
= \(\frac{a^2+4 a+4-a^2+4 a+4}{a^2+4 a+4+a^2-4 a+4}\)
= \(\frac{x-4 a}{x\left(a^2+4\right)}\)
∴ \(\frac{x^2-y^2}{x^2+y^2}=\frac{4 a}{a^2+4}\) Proved.
6. If \(x = \frac{8 a b}{a+b}\) let us write by calculating the value of \(\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}\)
Solution: \(x = \frac{8 a b}{a+b}\), find the value of \(\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}\).
Given, \(\frac{x}{1}=\frac{8 a b}{a+b}\)
\(\frac{x}{4 a}=\frac{2 b}{a+b}\) \(\frac{x+4 a}{x-4 a}=\frac{2 b+a+b}{2 b-a-b}\) \(\frac{x+4 a}{x-4 a}=\frac{3 b+a}{b-a}\)Again, \(\frac{x}{1}=\frac{8 a b}{a+b}\)
or, \(\frac{x}{4 b}=\frac{2 a}{a+b}\)
or, \(\frac{x+4 b}{x-4 a}=\frac{2 a+a+b}{2 a-a-b}=\frac{3 a+b}{a-b}\)
∴ \(\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}=\frac{3 b+a}{b-a}=\frac{3 a+b}{a-b}\)
or, \(=\frac{(3 b+a)}{b-a}-\frac{(3 a+b)}{b-a}\)
= \(\frac{3 b+a-3 a-b}{h-a}\)
= \(\frac{2 b-2 a}{b-a}\)
= \(\frac{2(b-2 a)}{(b-a)}\)
∴ \(\frac{x+4 a}{x-4 a}+\frac{x+4 b}{x-4 b}=2\)
Question 6. If \(\frac{x+y}{3 a-b}=\frac{y+z}{3 b-c}=\frac{z+x}{3 c-a}\), let us show that \(\frac{x+y+z}{a+b+c}=\frac{a x+b y+c z}{a^2+b^2+c^2}\).
Solution. If \(\frac{x+y}{3 a-b}=\frac{y+z}{3 b-c}=\frac{z+x}{3 c-a}\), prove that \(\frac{x+y+z}{a+b+c}=\frac{a x+b y+c z}{a^2+b^2+c^2}\)
Let \(\frac{x+y}{3 a-b}=\frac{y+z}{3 b-c}=\frac{z+x}{3 c-a}=k\) (where k ≠ 0).
∴ x + y = k(3a – b) …(1)
x + z = k(3b – c) …(2)
z + x = k(3c – a)
Adding, 2(x + y + z) = k.2(a + b + c)
∴ x + y + z = k(a + b + c) …(4)
Subtracting (2) from (4), we get
x = k(a – 2b + 2c)
Similarly, y = k(b – 2c + 2a) and z = (c – 2a + 2b)
∴ \(\frac{x+y+z}{a+b+c}=\frac{k(a+b+c)}{(a+b+c)}=k\)
Again, \(\frac{a x+b y+c z}{a^2+b^2+c^2}=\frac{a k(a-2 b+2 c)+b k(b-2 c+2 a)+c k(c-2 a+2 b)}{\left(a^2+b^2+c^2\right)}\)
= \(\frac{k\left(a^2-2 a+b+2 a c+b^2-2 b c+2 a b+c^2-2 a c+2 b c\right)}{\left(a^2+b^2+c^2\right)}\)
= \(\frac{k\left(a^2+b^2+c^2\right)}{\left(a^2+b^2+c^2\right)}=k\)
∴ \(\frac{x+y+z}{a+b+c}=\frac{a x+b y+c z}{a^2+b^2+c^2}\) Proved.
Question 7. If \(\frac{\mathrm{x}}{\mathrm{a}}=\frac{\mathrm{y}}{\mathrm{b}}=\frac{\mathrm{z}}{\mathrm{c}}\) , let us show that \(\frac{x^2-y z}{a^2-b c}=\frac{y^2-z x}{b^2-c a}=\frac{z^2-x y}{c^2-a b}\).
Solution: If \(\frac{\mathrm{x}}{\mathrm{a}}=\frac{\mathrm{y}}{\mathrm{b}}=\frac{\mathrm{z}}{\mathrm{c}}\)
prove that, \(\frac{x^2-y z}{a^2-b c}=\frac{y^2-z x}{b^2-c a}=\frac{z^2-x y}{c^2-a b}\)
Let \(\frac{\mathrm{x}}{\mathrm{a}}=\frac{\mathrm{y}}{\mathrm{b}}=\frac{\mathrm{z}}{\mathrm{c}}=\mathrm{k}\) (where k ≠ 0).
∴ x = ak, y = bk, & z = ck
∴ \(\frac{x^2-y z}{a^2-b c}=\frac{a^2 k^2-b k \cdot c k}{a^2-b c}=\frac{k^2\left(a^2-b c\right)}{a^2-b c}=k^2\)
\(\frac{\mathrm{y}^2-\mathrm{zx}}{\mathrm{b}^2-\mathrm{ca}}=\frac{\mathrm{b}^2 \mathrm{k}^2-\mathrm{ck} \cdot \mathrm{ak}}{\mathrm{b}^2-\mathrm{ca}}=\frac{\mathrm{k}^2\left(\mathrm{~b}^2-\mathrm{ca}\right)}{\left(\mathrm{b}^2-\mathrm{ca}\right)}=\mathrm{k}^2\) \(\frac{z^2-x y}{c^2-a b}=\frac{c^2 k^2-a k \cdot b k}{c^2-a b}=\frac{k^2\left(c^2-a b\right)}{\left(c^2-a b\right)}=k^2\)∴ \(\frac{x^2-y z}{a^2-b c}=\frac{y^2-z x}{b^2-c a}=\frac{z^2-x y}{c^2-a b}\) Proved.
Question 8.
1. If \(\frac{3 x+4 y}{3 u+4 v}=\frac{3 x+4 y}{3 u-4 v}\) , let us show that \(\frac{x}{y}=\frac{u}{v}\).
Solution: If \(\frac{3 x+4 y}{3 u+4 v}=\frac{3 x+4 y}{3 u-4 v}\) prove that \(\frac{x}{y}=\frac{u}{v}\)
or, \(\frac{3 x+4 y}{3 x-4 y}=\frac{3 u+4 v}{3 u-4 v}\)
or, \(\frac{3 x+4 y+3 x-4 y}{3 x+4 y-3 x+4 y}=\frac{3 u+4 v+3 u-4 v}{3 u+4 v-3 u+4 v}\)
or, \(\frac{6 x}{8 y}=\frac{6 u}{8 v}\)
∴ \(\frac{x}{y}=\frac{u}{v}\) Proved.
2. If (a+b+c+d) : (a+b-c-d) = (a-b+c-d) : (a-b-c+d),let us prove that a:b = c:d.
Solution: If (a + b + c + d):(a + b – c – d) = (a – b + c – d):(a – b – c + d),
Prove that a:b = c:d
\(\frac{a+b+c+d}{a+b-c-d}=\frac{a-b+c-d}{a-b-c+d}\)or, \(\frac{a+b+c+d+a+b-c-d}{a+b+c+d-a-b+c+d}=\frac{a-b+c-d+a-b-c+d}{a-b+c-d-a+b+c-d}\)
or, \(\frac{2(a+b)}{2(c+d)}=\frac{2 a-2 b}{2 c-2 d}\)
or, \(\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
or, ac – ad + bc – bd = ac + ad – bc – bd
or, 2bc = 2ad
or, ad = bc
∴ \(\frac{a}{b}=\frac{c}{d}\)
i.e., a:b = c:d Proved.
Question 9.
1. If \(\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1\), let us show that \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1\).
Solution: If \(\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1\)
prove that \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1\)
\(\frac{a^2}{b+c}=\frac{b^2}{c+a}=\frac{c^2}{a+b}=1\)∴ \(a^2=b+c ; b^2=c+a ; c^2=a+b\)
\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\)= \(\frac{a}{a+a^2}+\frac{b}{b+b^2}+\frac{1}{c+c^2}\)
= \(\frac{a}{a+b+c}+\frac{b}{b+c+a}+\frac{c}{c+a+b}\)
= \(\frac{a+b+c}{a+b+c}=1\) Proved.
2. If x² : (by+cz)=y² :(cz + ax) = z² : (ax+by)=1, let us prove that \(\frac{a}{a+x}+\frac{b}{b+y}+c / c+z=1\).
Solution: If \(x^2:(b y+c z)=y^2:(c z+a x)=z^2:(a x+b y)=1\)
Prove that \(\frac{a}{a+x}+\frac{b}{b+y}+c / c+z=1\)
\(\frac{x^2}{b y+c z}=\frac{y^2}{c z+a x}=\frac{z^2}{a x+b y}=1\)∴ \(x^2=b y+c z ; y^2=c z+a x ; z^2=a x+b y\)
L.H.S. = \(\frac{a}{a+x}+\frac{b}{b+y}+\frac{c}{c+z}\)
= \(\frac{a x}{a x+x^2}+\frac{\text { by }}{b y+y^2}+\frac{c z}{c z+z^2}\)
= \(\frac{\mathrm{ax}}{\mathrm{ax}+\mathrm{by}+\mathrm{cz}}+\frac{\mathrm{by}}{\mathrm{by}+\mathrm{cz}+\mathrm{ax}}+\frac{\mathrm{by}}{\mathrm{by}+\mathrm{cz}+\mathrm{ax}}\)
= \(\frac{a x+b y+c z}{a x+b y+c z}=1\) Proved.
Question 10.
1. If \(\frac{x}{x a+y b+z c}=\frac{y}{y a+z b+x c}=\frac{z}{z a+x b+y c}\) and x+y+z ≠ 0, let us show that each ratio is equal to \(\frac{1}{a+b+c}\).
Solution: Given, \(\frac{x}{x a+y b+z c}=\frac{y}{y a+z b+x c}=\frac{z}{z a+x b+y c}\)
& x + y + z ≠ 0.
Prove that each ratio = \(\frac{1}{a+b+c}\)
= \(\frac{x}{x a+y b+z c}=\frac{y}{y a+z b+x c}=\frac{z}{z a+x b+y c}=\frac{x+y+z}{x a+y b+z c+y a+z b+x c+z a+x b+y c}\)
= \(\frac{(x+y+z)}{a(x+y+z)+b(x+y+z)+c(x+y+z)}=\frac{(x+y+z)}{(x+y+z)+(a+b+c)}\)
= \(\frac{1}{a+b+c}\) Proved.
2. If \(\frac{x^2-y z}{a}=\frac{y^2-z x}{b}=\frac{z^2-x y}{c}\), let us prove that (a+b+c) (x+y+z) = ax +by+cz.
Solution: \(\frac{x^2-y z}{a}=\frac{y^2-z x}{b}=\frac{z^2-x y}{c}\)
Prove that (a + b + c)(x + y + z) = ax + by + cz
Let \(\frac{x^2-y z}{a}=\frac{y^2-z x}{b}=\frac{z^2-x y}{c}=\frac{1}{k}\) (where ≠ 0).
∴ \(a=k\left(x^2-y z\right) ; b=k\left(y^2-z x\right) ; c=k\left(z^2-x y\right)\)
L.H.S. = (a + b + c)(x + y + z)
= \(\left(k\left(x^2-y z+y^2-z x+z^2-x y 1\right)(x+y+z)\right.\)
= \(k(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)=k\left(x^2+y^2+z^2-3 x y z\right)\)
R.H.S. = \(\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{k}\left(\mathrm{x}^2-\mathrm{yz}\right) \mathrm{x}+\mathrm{k}\left(\mathrm{y}^2-\mathrm{zx}\right) \mathrm{y}+\mathrm{k}\left(\mathrm{z}^2-\mathrm{xy}\right) \mathrm{z}\)
= \(k\left(x^2-x y z+y^2-x y z+z^2-x y z\right)\)
= \(\mathrm{k}\left(\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2-\mathrm{xy} 2\right)\)
∴ L.H.S. = R.H.S. Proved.
3. If \(\frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}\), let us prove that \(\frac{a(b-c)}{y^2-z^2}=\frac{b(c-a)}{z^2-x^2}=\frac{c(a-b)}{x^2-y^2}\).
Solution: If \(\frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}\)
Prove that \(\frac{a(b-c)}{y^2-z^2}=\frac{b(c-a)}{z^2-x^2}=\frac{c(a-b)}{x^2-y^2}\)
Let, \(\frac{a}{y+z}=\frac{b}{z+x}=\frac{c}{x+y}=k\) (where k ≠ 0)
∴ a = k(y + z); b = k(z + x); c = k(x + y)
\(\frac{a(b-c)}{y^2-z^2}=\frac{k(y+z) k(z+x-x-y)}{y^2-z^2}=\frac{k^2(z+y)(z-y)}{-(y+z)(z-y)}=-k^2\) \(\frac{b(c-a)}{z^2-x^2}=\frac{k(z+x) k(x+y-y-z)}{z^2-x^2}=\frac{k^2(z+x)(z-y)}{-(z+x)(x-y)}=-k^2\) \(\frac{c(a-b)}{x^2-y^2}=\frac{k(x+y) k(y+z-z-x)}{x^2-y^2}=\frac{k^2(x+y)(x-y)}{-(x+y)(x-y)}=-k^2\)∴ \(\frac{a(b-c)}{y^2-z^2}=\frac{b(c-a)}{z^2-x^2}=\frac{c(a-b)}{x^2-y^2}\) Proved.
Chapter 5 Ration And Proportion Exercise 5.3 Multiple Choice Questions
1. The fourth proportion of 3, 4, and 6 are
1. 8
2. 10
3. 12
4. 24
Answer. The 4th proportional of 3,4 & 6 =4×6/3 = 8———-(1)
2. The 3rd proportion of 8 and 12 is
1. 12
2. 16
3. 18
4. 20
Answer. The 3rd proportional of 8 &12 = 12 x 12/8 = 18 ———-(3)
3. The mean proportion of 16 and 25 is
1. 400
2. 100
3. 20
4. 40
Answer. The mean proportional of 16 & 25 = √16×25 = 4 x 5=20———-(3)
4. a is a positive number and if a: 27/64 = 3/4: a, then the value of a is
1. 81/256
2. 9
3. 9/16
4. 16/9
Answer: If a: 27/64 = 3/4: a
∴ a= √27/64 x 3/4 = √81/256 = 9/16—————–(3)
5. If 2a = 3b = 4c, then a:b:c is
1. 3:4:6
2. 4:3:6
3. 3:6:4
4. 6:4:3
Answer: If 2a=3b=4c, find a:b:c
or, 2a/12 = 3b/12 = 4c/12
or a/6 = b/4 = c/3
∴ a:b:c = 6:4:3——–(4)
Chapter 5 Ration And Proportion Exercise 5.3 True or False
1. Compound ratio of ab:c², bc:a² and ca:b² is 1:1
Answer: ab/c² x bc/a² x ca/b² = 1:1
True
2. x³y, x²y² and xy³ are continued proportional.
Answer: If x³y/x²y² = x²y²/xy³
or, x/y = x/y
Chapter 5 Ration And Proportion Exercise 5.3 True Or False
1. If the product of three positive consecutive numbers is 64, then their mean proportion is 4
2. If a: 2 = b: 5 = c: 8, then 50% of a = 20% of b = 12.5% of c.
Chapter 5 Ration And Proportion Exercise 5.3 Short Answers
Question 1. If a/2 = b/3 = c/a = 2a-3b+4c/p , let us find the value of p.
Solution: \(\frac{\mathrm{a}}{2}=\frac{\mathrm{b}}{3}=\frac{\mathrm{c}}{4}=\frac{2 \mathrm{a}-3 \mathrm{~b}+4 \mathrm{c}}{\mathrm{p}}=\mathrm{k} \text { (let) }\)
∴ a = 2k, b = 3k & c = 4k
∴ 2a – 3b + 4c = pk
or, 2.2k – 3.3k + 4.4k = pk
11k = p.k
∴ p = 11.
Question 2. If \(\frac{3 x-5 y}{3 x+5 y}=\frac{1}{2}\), let us find the value of \(\frac{3 x^2-5 y^2}{3 x^2+5 y^2}\)
Solution: If \(\frac{3 x-5 y}{3 x+5 y}=\frac{1}{2}\) or, 6x – 10y = 3x + 5y
or, 6x – 3x = 10y + 5y
or, 3x + 5y
∴ x = 5y
\(\frac{3 x^2-y^2}{3 x^2+5 y^2}=\frac{3(5 y)^2-5 y^2}{3(5 y)^2+5 y^2}\)= \(\frac{75 y^2-5 y^2}{75 y^2+5 y^2}\)
= \(\frac{70 y^2}{80 y^2}=\frac{7}{8}\)