ML Aggarwal Solution Class 10 Chapter 7 Ratio and Proportion Exercise 7.3
Exercise 7.3
Question 1
If a : b : : c : d, prove that
(i) 2a+5b2a−5b=2c+5d2c−5d
(ii) 5a+11b5c+11d=5a−11b5c−11d
Applying componendo and dividendo,
2a+5b2a−5b=2c+5d2c−5d
(ii) ∵ a: b:: c: d
∴ab=cd⇒5a11b=5c11d( Multiply by 511)
Applying componendo and dividendo,
5a+11b5a−11b=5c+11d5c−11d
⇒5a+11b5c+11d=5a−11b5c−11d (Applying alternendo)
(iii) ∵ a: b:: c: d
Applying componendo and dividendo,
2a+3b2a−3b=2c+3d2c−3d
⇒(2 a+3 b)(2 c-3 d)
=(2 a-3 b)(2 c+3 d) (By cross multiplication)
(iv) ∵ a: b:: c: d
⇒lamb=lcmd( Multiply by lm)
Applying componendo and dividendo,
la+mbla−mb=lc+mdlc−md
⇒la+mblc+md=la−mblc−md (By alternendo)
⇒(l a+m b):(l c+m d)::(l a-m b) :(l c-m d)
Question 2
(i) If 5x+7y5u+7y=5x−7y5u−7y, Show that xy=uv
(ii) 8a−5b8c−5d=8a+5b8c+5d, prove that ab=cd
Sol :
(i) 5x+7y5x+7y=5x−7y5x−7y
Applying alternendo 5x+7y52+7y=5x−7y54−7
Applying componendo and dividendo
5x+7y+5x−7y5x+7y−5x+7y=5u+7v+5u−7v5u+7u−5u+7v
⇒10x14y=10u14v⇒xy=uv
Hence proved. ( Dividing by =1014)
(ii) 8a−5b8c−5d=8a+5b8c+5d
⇒8a+5b8a−5b=8c+5d8c−5d (using alternendo)
Applying compounde and dividendo,
8a+5b+8a−5b8a+5b−8a+5b=8c+5d+8c−5c8c+5d−8c+5d
∴16a10b=16c10d
⇒ab=cd
(Dividing by 1610)
Hence proved.
Question 3
If (4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d), prove that a, b, c, d are in proporton.
Sol :
(4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d)
Applying componendo and dividendo
Hence, a,b,c,d are in proportion
Question 4
If (pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd) prove that a : b : : c : d
Sol :
(pa + qb) : (pc + qd) :: (pa – qb) : (pc – qd)
Applying componendo and dividendo
⇒2pa2qb=2pc2qd
⇒ab=cd ( Dividing by 2p2q)
hence a : b:: c : d
Question 5
If (ma + nb): b :: (mc + nd) : d, prove that a, b, c, d are in proportion.
Sol :
(ma + nb): b :: (mc + nd) : d
⇒ mad + nbd = mbc + nbd
⇒ mad = mbc
⇒ ad = bc
Hence a : b :: c : d.
Question 6
If (11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²), prove that a : b :: c : d.
Sol :
(11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²)
Applying componendo and dividendo
⇒22a226b2=22c226d2
⇒a2b2=c2d2 ( Dividing by 2226)
⇒ab=cd
hence a:b::c:d
Question 7
If (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d), prove that a : b :: c : d.
a+3b+2c+6da−3b+2c−6d=a+3b−2c−6da−3b−2c+6d
⇒a+3b+2c+6da+3b−2c−6d=a−3b+2c−6da−3b−2c+6d (by altenendo)
Applying componendo and dividendo
=a−3b+2c−6d+a−3b−2c+6da−3b+2c−6d−a+3b+2c−6d
⇒2(a+3b)2(2c+6d)=2(a−3b)2(2c−6d) (Dividing by 2)
⇒a+3b2c+6d=a−3b2c−6d (by alternendo)
Again applying componendo and dividendo)
a+3b+a−3ba+3b−a+3b=2c+6d+2c−6d2c+6d−2c+6d
⇒2a6b=4c12d=2c6d
⇒ab=cd [ Dividing by 26]
Question 8
If x=2aba+b find the value of x+ax−a+x+bx−b
Sol :
x=2aba+b
⇒xa=2ba+b
Applying componendo and dividendo,
Applying componendo and dividendo,
x+bx−b=2a+a+b2a−a−b=3a+ba−b..(ii)
Adding (i) and (ii)
x+ax−a+x+bx−b=3b+ab−a+3a+ba−b
=−a+3ba−b+3a+ba−b
=−a−3b+3a+ba−b
=2a−2ba−b=2(a−b)a−b=2
Question 9
If x=8aba+b find the value of x+4ax−4a+x+4bx−4b
Sol :
x=8aba+b
⇒x4a=2ba+b
Applying componendo and dividendo,
Againx4b=2aa+b
Applying compondndo and dividendo,
x+4bx−4b=2a+a+b2a−a−b=3a+ba−b..(ii)
Adding (i) and (ii)
x+4ax−4a+x+4bx−4b=3b+ab−a+3a+ba−b
=−a+3ba−b+3a+ba−b
=−a−3b+3a+ba−b=2a−2ba−b
=2(a−b)a−b=2
Question 10
If x=4√6√2+√3 find the value of x+2√2x−2√2+x+2√3x−2√3
Sol :
x=4√6√2+√3
⇒4√2×√3√2+√3
x2√2=2√3√2+√3
Applying componendo and dividendo,
x+2√2x−2√2=2√3+√2+√32√3−√2−√3
=3√3+√2√3−√2..(i)
Again x2√3=2√2√2+√3
Applying componendo and dividendo,
x+2√3x−2√3=2√2+√2+√32√2−√2−√3
=3√2+√3√2−√3..(ii)
Adding (i) and (ii)
x+2√2x−2√2+x+2√3x−2√3
=3√3+√2√3−√2+3√2+√3√2−√3
=3√3+√2√3−√2−3√2+√3√3−√2
=3√3+√2−3√2−√3√3−√2
=2√3−2√2√3−√2=2(√3−√2)√3−√2=2
Question 11
Solve x:√36x+1+6√x√36x+1−6√x=9
Sol :
√36x+1+6√x√36x+1−6√x=91
Applying componendo and dividendo,
⇒900x=576x+16
⇒900x−576x=16
⇒324x=16
∴x=16324=481
Question 12
Find x from the following equations :
(ii) √x+4+√x−10√x+4−√x−10=52
Applying componendo and dividendo,
√x+4+√x−10+√x+4−√x−10√x+4+√x−10−√x+4+√x−10=5+25−2
⇒2√x+42√x−10=73⇒√x+4√x−10=73
Squaring both sides,
x+4x−10=499
⇒49x−490=9x+36
⇒49x−9x=36+490⇒40x=526
∴x=52640=26320
(iii) √1+x+√1−x√1+x−√1−x=ab
Applying componendo and dividendo,
√1+x+√1−x+√1+x−√1−x√1+x+√1−x−√1+x+√1−x=a+ba−b
⇒2√1+x2√1−x=a+ba−b⇒√1+x√1−x=a+ba−b
Squaring both sides, 1+x1−x=(a+b)2(a−b)2
Again applying componendo and dividendo,
1+x+1−x1+x−1+x=(a+b)2+(a−b)2(a+b)2−(a−b)2
⇒22x=2(a2+b2)4ab
⇒1x=a2+b22ab
∴x=2aba2+b2
(iv) √12x+1+√2x−3√12x+1−√2x−3=32
Applying componendo and dividendo,
√12x+1+√2x−3+√12x+1−√2x−3√12x+1+√2x−3−√12x+1+√2x−3
=3+23−2
⇒2√12x+12√2x−3=51
⇒√12x+1√2x−3=51
Squaring both sides, 12x+12x−3=251
⇒50x−75=12x+1
⇒50x−12x=1+75
⇒38x=76⇒x=7638=2
∴ x=2
(v) 3x+√9x2−53x−√9x2−5=51
Applying componendo and dividendo.
3x+√9x2−5+3x−√9x2−53x+√9x2−5−3x+√9x2−5=5+15−1
6x2√9x2−5=64⇒3x√9x2−5=32
Squaring both sides
9x29x2−5=94⇒81x2−45=36x2
⇒81x2−36x2=45⇒45x2=45
⇒x2=1⇒x=±1
∴x=1,−1
CHECK
(i) When x=1, then in the given equation
3×1+√9×1−53×1−√9×1−5=3+√43−√4=3+23−2=51
which is given
∴x=1
(ii) When x=-1, then
3(−1)+√9(−1)2−5
3(−1)−√9(−1)2−5
=−3+√9−5−3−√9−5=−3+√4−3−√4
=−3+2−3−2=−1−5=15≠51
∴x=−1 is not its solution.
Hence x=1
(vi) √a+x+√a−x√a+x−√a−x=cd
Applying componendo and dividendo,
√a+x+√a−x+√a+x−√a−x√a+x+√a−x−√a+x+√a−x=c+dc−d
⇒2√a+x2√a−x=c+dc−d⇒√a+x√a−x=c+dc−d
Squaring both sides a+xa−x=(c+d)2(c−d)2
Again applying componendo and dividendo
a+x+a−xa+x−a+x=(c+d)2+(c−d)2(c+d)2−(c−d)2
⇒2a2x=2(c2+d2)4cd⇒ax=c2+d22cd
⇒x(c2+d2)=2acd
⇒x=2acdc2+d2
Question 13
Solve 1+x+x21−x+x2=62(1+x)63(1−x)
Sol :
1+x+x21−x+x2=62(1+x)63(1−x)
⇒(1−x)(1+x+x2)(1+x)(1−x+x2)=6263
⇒(1+x)(1−x+x2)(1−x)(1+x+x2)=6362
⇒1+x31−x3=6362
Applying componendo and dividendo,
1+x3+1−x31+x3−1+x3=63+6263−62
⇒22x3=1251
⇒1x3=1251
⇒x3=1125=(15)3
∴x=15 Ans.
Question 14
Solve for x:16(a−xa+x)3=a+xa−x
Sol :
x:16(a−xa+x)3=a+xa−x
⇒(a+xa−x)×(a+xa−x)3=16
⇒(a+xa−x)4=16=(±2)4
⇒a+xa−x=±2
When a+xa−x=21
Applying componendo and dividendo,
a+x+a−xa+x−a+x=−2+1−2−1
⇒=2a2x=−1−3
⇒ax=13
⇒x=3a
Hence x=a3,3a
Question 15
If x=√a+x+√a−1√a+1−√a−1 , using properties of proportion , show that x² – 2ax + 1 = 0
Sol :
We have x=√a+x+√a−1√a+1−√a−1
(Applying componendo and dividendo)
⇒(x+1)2(x−1)2=a+1a−1
⇒(x+1)2+(x−1)2(x+1)2−(x−1)2=2a2
(Again applying componendo and dividendo)
⇒x2+1+2x+x2+1−2xx2+1+2x−x2−1+2x=a
⇒2x2+24x=a⇒2(x2+1)4x=a
⇒(x2+1)2x=a⇒2ax=x2+1
⇒x2−2ax+1=0
Question 16
Given : x=√α2+b2+√α2−b2√α2+b2−√α2−b2 Use componendo and dividendo to prove that
Sol :
If x1=√α2+b2+√α2−b2√α2+b2−√α2−b2
Applying componendo and dividendo both sides
Squaring, both sides we have
⇒(x+1)2(x−1)2=a2+b2a2−b2
⇒x2+1+2xx2+1−2x=a2+b2a2−b2
Applying componendo and dividendo both sides
⇒x2+1+2x+x2+1−2xx2+1+2x−x2−1+2x=a2+b2+a2−b2a2+b2−a2+b2
2x2+24x=2a22b2
⇒x2+12x=a2b2⇒b2=2a2xx2+1
Question 17
Given that a3+3ab2b3+3a2b=6362. Using componendo and dividendo find a: b. (2009)
⇒ab=64
⇒ab=32
a : b = 3 : 2
Question 18
Give x3+12x6x2+8=y3+27y9y2+27 Using componendo and dividendo find x : y.
Using componendo-dividendo, we have
⇒(x+2)3(x−2)3=(y+3)3(9y−3)3
⇒(x+2x−2)3=(y+3y−3)3
⇒x+2x−2=y+3y−3
Again using componendo-dividendo, we get
x+2+x−2x+2−x+2=y+3+y−3y+3−y+3
⇒2x4=2y3
⇒x2=y3
⇒xy=23
Thus the required ratio is x: y=2: 3
Question 19
Using the properties of proportion, solve the following equation for x; given
Applying componendo and dividendo
⇒x3+3x2+3x+1x3−3x2+3x−1=432250=216125
⇒(x+1)3(x−1)3=216125=(65)3
∴x+1x−1=65
⇒6 x-6=5 x+5
x=11
Question 20
If x+yax+by=y+zay+bz=z+xaz+bx , prove that each of these ratio is equal to 2a+b unless x + y + z = 0
x+yax+by=y+zay+bz=z+xaz+bx
=x+y+y+z+z+xax+by+ay+bz+αz+bx
=2(x+y+z)x(a+b)+y(a+b)+z(a+b)
=2(x+y+z)(a+b)(x+y+z)=2a+b if x+y+z≠0
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