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+5b2a5b=2c+5d2c5d

(ii) 5a+11b5c+11d=5a11b5c11d

(iii) (2a + 3b)(2c – 3d) = (2a – 3b)(2c + 3d)

(iv) (la + mb) : (lc + mb) :: (la – mb) : (lc – mb)
Sol :
(i) a : b : : c : d
then ab=cd
2a5b=2c5d (multiply by 25)

Applying componendo and dividendo,

2a+5b2a5b=2c+5d2c5d


(ii) ∵ a: b:: c: d

ab=cd5a11b=5c11d( Multiply by 511)

Applying componendo and dividendo,

5a+11b5a11b=5c+11d5c11d

5a+11b5c+11d=5a11b5c11d (Applying alternendo)


(iii) ∵ a: b:: c: d

ab=cd2a3b=2c3d( Multiply by 23)

Applying componendo and dividendo,

2a+3b2a3b=2c+3d2c3d

⇒(2 a+3 b)(2 c-3 d)

=(2 a-3 b)(2 c+3 d) (By cross multiplication)


(iv) ∵ a: b:: c: d

ab=cd

lamb=lcmd( Multiply by lm)

Applying componendo and dividendo,

la+mblamb=lc+mdlcmd

la+mblc+md=lamblcmd (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=5x7y5u7y, Show that xy=uv

(ii) 8a5b8c5d=8a+5b8c+5d, prove that ab=cd

Sol :

(i) 5x+7y5x+7y=5x7y5x7y

Applying alternendo 5x+7y52+7y=5x7y547

Applying componendo and dividendo

5x+7y+5x7y5x+7y5x+7y=5u+7v+5u7v5u+7u5u+7v

10x14y=10u14vxy=uv

Hence proved. ( Dividing by =1014)


(ii) 8a5b8c5d=8a+5b8c+5d

8a+5b8a5b=8c+5d8c5d (using alternendo)

Applying compounde and dividendo,

8a+5b+8a5b8a+5b8a+5b=8c+5d+8c5c8c+5d8c+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)

4a+5b4a5b=4c+5d4c5d

Applying componendo and dividendo

4a+5b+4a5b4a+5b4a+5b=4c+5d+4c5d4c+5d4c+5d
8a10b=8c10dab=cd

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)

pα+qbpc+qd=pqqbpcqd
pa+qbpcqd=pq+qbpcqd

Applying componendo and dividendo

pa+qb+pαqbpa+qbpα+qb=pc+qd+pcqdpcqdpc+qd

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

ma+nbb=mc+ndd

⇒ mad + nbd = mbc + nbd

⇒ mad = mbc

⇒ ad = bc

ab=cd

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)

11a+13b211a213b2=11c2+13d211c213d2

Applying componendo and dividendo

11a2+13b2+11a213b211a2+13b211a2+13b2=11c2+13d2+11c213d211c2+13d211c2+13d2

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.

Sol :

a+3b+2c+6da3b+2c6d=a+3b2c6da3b2c+6d

a+3b+2c+6da+3b2c6d=a3b+2c6da3b2c+6d (by altenendo)

Applying componendo and dividendo

a+3b+2c+6d+a+3b2c6da+3b+2c+6da3b+2c+6d

=a3b+2c6d+a3b2c+6da3b+2c6da+3b+2c6d

2(a+3b)2(2c+6d)=2(a3b)2(2c6d) (Dividing by 2)

a+3b2c+6d=a3b2c6d (by alternendo)

Again applying componendo and dividendo)

a+3b+a3ba+3ba+3b=2c+6d+2c6d2c+6d2c+6d

2a6b=4c12d=2c6d

ab=cd  [ Dividing by 26]


Question 8

If x=2aba+b find the value of x+axa+x+bxb

Sol :

x=2aba+b

xa=2ba+b

Applying componendo and dividendo,

x+axa=2b+a+b2bab=3b+aba...(i)

Again xb=2aa+b

Applying componendo and dividendo,

x+bxb=2a+a+b2aab=3a+bab..(ii)

Adding (i) and (ii)

x+axa+x+bxb=3b+aba+3a+bab

=a+3bab+3a+bab

=a3b+3a+bab

=2a2bab=2(ab)ab=2


Question 9

If x=8aba+b find the value of x+4ax4a+x+4bx4b

Sol :

x=8aba+b

x4a=2ba+b

Applying componendo and dividendo,

x+4ax4a=2b+a+b2bab=3b+aba(i)

Againx4b=2aa+b

Applying compondndo and dividendo,

x+4bx4b=2a+a+b2aab=3a+bab..(ii)

Adding (i) and (ii)

x+4ax4a+x+4bx4b=3b+aba+3a+bab

=a+3bab+3a+bab

=a3b+3a+bab=2a2bab

=2(ab)ab=2


Question 10

If x=462+3 find the value of x+22x22+x+23x23

Sol :

x=462+3

42×32+3

x22=232+3

Applying componendo and dividendo,

x+22x22=23+2+32323

=33+232..(i)

Again x23=222+3

Applying componendo and dividendo,

x+23x23=22+2+32223

=32+323..(ii)

Adding (i) and (ii)

x+22x22+x+23x23

=33+232+32+323

=33+23232+332

=33+232332

=232232=2(32)32=2


Question 11

Solve x:36x+1+6x36x+16x=9

Sol :

36x+1+6x36x+16x=91

Applying componendo and dividendo,

36x+1+6x+36x+16x36x+1+6x36x1+6x
=9+191
236x+112x=108
36x+16x=54 (Squaring both sides)
36x+136x=2516
36x×25=16(36x+1)

900x=576x+16

900x576x=16

324x=16

x=16324=481


Question 12

Find x from the following equations :

(i) 2x+2+x2x2+x=3
(ii) x+4+x10x+4x10=52
(iii) 1+x+1x1+x1x=ab
(iv) 12x+1+2x312x+12x3=32
(v) 3x+9x253x9x25=5
(vi) a+x+axa+xax=cd
Sol :
(i) 2x+2+x2x2+x=3

Applying componendo and dividendo,

2x+2+x+2x2+x2x+2+x2x+2+x=3+131

22x22+x=422x2+x=21

Squaring both sides

2x2+x=418+4x=2x

4x+x=285x=6

x=65 Ans.

(ii) x+4+x10x+4x10=52

Applying componendo and dividendo,

x+4+x10+x+4x10x+4+x10x+4+x10=5+252

2x+42x10=73x+4x10=73

Squaring both sides,

x+4x10=499

49x490=9x+36

49x9x=36+49040x=526

x=52640=26320


(iii) 1+x+1x1+x1x=ab

Applying componendo and dividendo,

1+x+1x+1+x1x1+x+1x1+x+1x=a+bab

21+x21x=a+bab1+x1x=a+bab

Squaring both sides, 1+x1x=(a+b)2(ab)2

Again applying componendo and dividendo,

1+x+1x1+x1+x=(a+b)2+(ab)2(a+b)2(ab)2

22x=2(a2+b2)4ab

1x=a2+b22ab

x=2aba2+b2


(iv) 12x+1+2x312x+12x3=32

Applying componendo and dividendo,

12x+1+2x3+12x+12x312x+1+2x312x+1+2x3

=3+232

212x+122x3=51

12x+12x3=51

Squaring both sides, 12x+12x3=251

50x75=12x+1

50x12x=1+75

38x=76x=7638=2

∴ x=2


(v) 3x+9x253x9x25=51

Applying componendo and dividendo.

3x+9x25+3x9x253x+9x253x+9x25=5+151

6x29x25=643x9x25=32

Squaring both sides

9x29x25=9481x245=36x2

81x236x2=4545x2=45

x2=1x=±1

x=1,1

CHECK 

(i) When x=1, then in the given equation

3×1+9×153×19×15=3+434=3+232=51

which is given

∴x=1

(ii) When x=-1, then

3(1)+9(1)25

3(1)9(1)25

=3+95395=3+434

=3+232=15=1551

x=1 is not its solution.

Hence x=1


(vi) a+x+axa+xax=cd

Applying componendo and dividendo,

a+x+ax+a+xaxa+x+axa+x+ax=c+dcd

2a+x2ax=c+dcda+xax=c+dcd

Squaring both sides a+xax=(c+d)2(cd)2

Again applying componendo and dividendo

a+x+axa+xa+x=(c+d)2+(cd)2(c+d)2(cd)2

2a2x=2(c2+d2)4cdax=c2+d22cd

x(c2+d2)=2acd

x=2acdc2+d2


Question 13

Solve 1+x+x21x+x2=62(1+x)63(1x)

Sol :

1+x+x21x+x2=62(1+x)63(1x)

(1x)(1+x+x2)(1+x)(1x+x2)=6263

(1+x)(1x+x2)(1x)(1+x+x2)=6362

1+x31x3=6362

Applying componendo and dividendo,

1+x3+1x31+x31+x3=63+626362

22x3=1251

1x3=1251

x3=1125=(15)3

x=15 Ans.


Question 14

Solve for x:16(axa+x)3=a+xax

Sol :

x:16(axa+x)3=a+xax

(a+xax)×(a+xax)3=16

(a+xax)4=16=(±2)4

a+xax=±2

When a+xax=21

Applying componendo and dividendo,

a+x+axa+xa+x=2+121

⇒=2a2x=13

ax=13

x=3a

Hence x=a3,3a 


Question 15

If x=a+x+a1a+1a1 , using properties of proportion , show that x² – 2ax + 1 = 0

Sol :

We have x=a+x+a1a+1a1

x+1x1=2a+12a1

(Applying componendo and dividendo)

(x+1)2(x1)2=a+1a1

(x+1)2+(x1)2(x+1)2(x1)2=2a2

(Again applying componendo and dividendo)

x2+1+2x+x2+12xx2+1+2xx21+2x=a

2x2+24x=a2(x2+1)4x=a

(x2+1)2x=a2ax=x2+1

x22ax+1=0


Question 16

Given : x=α2+b2+α2b2α2+b2α2b2  Use componendo and dividendo to prove that 

Sol :

If x1=α2+b2+α2b2α2+b2α2b2

Applying componendo and dividendo both sides

x+1x1=a2+b2+a2b2+a2+b2a2b2a2+b2+a2b2a2+b2+a2b2

x+1x1=2a2+b22a2+b2

x+1x1=a2+b2a2+b2

Squaring, both sides we have

(x+1)2(x1)2=a2+b2a2b2

x2+1+2xx2+12x=a2+b2a2b2

Applying componendo and dividendo both sides

x2+1+2x+x2+12xx2+1+2xx21+2x=a2+b2+a2b2a2+b2a2+b2

2x2+24x=2a22b2

x2+12x=a2b2b2=2a2xx2+1


Question 17

Given that a3+3ab2b3+3a2b=6362. Using componendo and dividendo find a: b. (2009)

Sol :
Given that a3+3ab2b3+3a2b=6362
By componendo and dividendo 
a3+3ab2+b3+3a2ba3+3ab2b33a2b=63+626362=1251

(a+b)3(ab)3=(51)3

a+bab=5a+b=5a5b

⇒5a-a-5b-b=0

⇒4a-6b=0

⇒4a=60

ab=64

ab=32

a : b = 3 : 2


Question 18

Give x3+12x6x2+8=y3+27y9y2+27 Using componendo and dividendo find x : y.

Sol :
Given : 
x3+12x6x2+8=y3+27y9y2+27

Using componendo-dividendo, we have

x3+12x+6x2+8x3+12x6x28=y3+27y+9y2+27y3+27y9y227

(x+2)3(x2)3=(y+3)3(9y3)3

(x+2x2)3=(y+3y3)3

x+2x2=y+3y3

Again using componendo-dividendo, we get

x+2+x2x+2x+2=y+3+y3y+3y+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

x3+3x3x2+1=34191
Sol :
x3+3x3x2+1=34191

Applying componendo and dividendo

x3+3x+3x2+1x3+3x3x21=341+9134191

x3+3x2+3x+1x33x2+3x1=432250=216125

(x+1)3(x1)3=216125=(65)3

x+1x1=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

Sol :

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+z0

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