ICSE Selina Solutions for Class 10

Mathematics
Chapter 7 – Ratio and Proportion (Including Properties and
Uses)

Exercise – 7(A)
5a  b
−3
1. If a :b=5 :3 , find
5a  b
+3
Ans: Here, we have a : b = 5 : 3
a 5
Therefore, =
b 3
 5a   − 3b 
5a   − 3b  b 
Now,   
=
5a   + 3b  5a   + 3b 
 
 b 
 5a 
 − 3
b
= 
 5a 
   3+ 
 b 
 5 
 5 ×   −3 
3 
=
 5 
 5 ×    + 3 
 3 

 25 
 −3 
3
= 
 25 
   3+ 
 3 
 25 − 9 
 
 3 
=
 25 + 9 
 
 3 

Class X Mathematics www.vedantu.com 1

  16 
 
= 3 
34
 
 3 
16
=
34
8
=
17
5a   − 3b 8
Hence, Value of is .
5a   + 3b 17

2. If x :y=4 :7 , Find the value of ( 3x+2y ) :( 5x+y ) .

Ans: Here, we have x : y = 4 : 7

x 4
Therefore, =
y 7

 3 x +2 y 
 
3 x   + 2 y  y 
Now,   
=
5 x   + y  5 x    +y
 y 
 
 3x 
 y   2
+ 
= 
 5x 
 + 1
 y 
 4 
 3 ×   + 2 
7 
=
 4 
 5 × + 1
 7 
 12 
   2+ 
 7 
=
 20 
 + 1
 7 

Class X Mathematics www.vedantu.com 2

  12 +14 
 
 7 
=
 20  7
+ 
 
 7 

 26 
 
= 7 
27
 
 7 
26
=
27
3 x   + 2 y 26
Hence, Value of is .
5 x   +y 27

4a + 3b
3. If a :b=3 :8, find the value of .
6a- b
Ans: Here, we have a : b = 3 :8
a 3
Therefore, =
b 8
 4a   + 3b 
4a   + 3b  b 
Now,   
=
6a − b  6a    −b 
 
 b 
 4a 
   3+ 
b
= 
 6a 
 −1
 b 
 3 
 4 ×   +3 
8 
=
 3 
 6 × −1 
 8 

Class X Mathematics www.vedantu.com 3

  12 
   3+ 
 8 
=
 18 
 −1 
8 
 12  24
+ 
 
 8 
=
 18  8
− 
 
 8 

 36 
 
= 8 
10
 
 8 
36
=
10
18
=
5
4a   + 3b 18
Hence, Value of is .
6a − b 5

4. If ( a-b ) :( a+b ) =1 :11, find the ratio:

( 5a+4b+15 ) :( 5a-4b+3 ) .
Ans: Here, we have ( a − b ) : ( a + b ) =
1 :11

⇒
(a − b) = 1
( a + b ) 11
⇒ 11a − 11b =a+b
⇒ 11a − a = b + 11b
⇒ 10a = 12b
a 12
⇒ =
b 10

Class X Mathematics www.vedantu.com 4

 a 6
⇒ =
b 5
Let a = 6 x and b = 5 x

Now,
( 5a + 4b + 15) = ( 5  6 + × 5 x +15 )
× x   4
( 5a − 4b + 3) ( 5  6 − × 5 x +3)
× x   4

  
=
( 30 x   20
+ x +15 )
( 30 x   20
− x +3)

=
( 50 x +15)
(10 x +3)
5 (10 x +3)
=
(10 x +3)
=5

Hence, Value of
( 5a + 4b + 15) is 5.
( 5a − 4b + 3)

7 8 4
5. Find the number which bears the same ratio to that does to .
33 21 9
Ans: Let the required number be x .
Therefore,
7 8 4
⇒ x: = :
33 21 9
8
x
⇒ = 21
7 4
33 9
x 8  9
×
⇒ =
7 21  4×
33

Class X Mathematics www.vedantu.com 5

 7 8  9
×
x
⇒= ×
33 21  4
×
7 8  9
×
x
⇒= ×
33 21  4
×
2
⇒ x=
11
2
Hence, the required number is .
11

m+n 2 2n 2
6. If = , find .
m + 3n 3 3m 2 + mn
m   
+n 2
Ans: Here, we have =
  
m + 3n 3
⇒ 3 ( m + n )= 2 ( m + 3n )

⇒ 3m + 3n = 2m + 6n
⇒ 3m − 2m =6n − 3n
⇒ m = 3n
2n 2 2n 2
Now, =
3m 2 + mn 3 × ( 3n ) + 3n.n
2

2n 2
=
3 × 9n 2 + 3n 2
2n 2
=
27 n 2 + 3n 2
2n 2
=
30n 2
1
=
15
2n 2 1
Hence, Value of is .
3m + mn
2
15

Class X Mathematics www.vedantu.com 6

 x
7. Find ; when x 2 +6y 2 =5xy .
y

Ans: Here, we have x 2 + 6 y 2 =

5 xy

Divide by y 2 in above equation

x 2 6 y 2 5 xy
⇒ 2+ 2 =
y y y2
2
x x
⇒   + 6 =5×
 y y

x
Let a =
y

⇒ a2 + 6 =5a
⇒ a 2 − 5a + 6 =0
⇒ a 2 − 2a − 3a + 6 =0
⇒ a ( a − 2 ) − 3( a − 2 ) =
0

⇒ ( a − 2 )( a − 3) =
0

or ( a − =
2 ) 0  
⇒ (a −= 3) 0

⇒ a 2  
= or a 3
=
x x
Thus, = 2 or = 3.
y y
x
Hence, Value of is 2 or 3.
y

8. If the ratio between 8 and 11 is the same as the ratio of 2 x − y to x + 2 y,

7x
find the value of   .
9y
Ans: here, the ratio between 8 and 11 is the same as the ratio of 2x − y to x + 2 y
.

Class X Mathematics www.vedantu.com 7

 Therefore,
8 2x − y
⇒ =
11 x   2
+ y
⇒ 22 x − 11 y =8 x + 16 y

⇒ 22 x − 8 x = 16 y + 11 y

⇒ 14 x = 27 y

x 27
⇒ =
y 14
7
Multiply by   in the above equation, we get
9
7 x 7 27
⇒ = ×
9 y 9 14
7x 3
⇒ =
9y 2
7x 3
Hence, value of is .
9y 2

2
9. Divide Rs. 1290 into A, B and C such that A is of B and B:C = 4 :3 .
5
2 B 4
Ans: here, we have A = B and =
5 C 3
2
⇒ A= B and 4C = 3B
5
2 3
⇒ A= B and C = B
5 4
1290
Now, A + B + C =
2 3
⇒ 1290
B+B+ B=
5 4

Class X Mathematics www.vedantu.com 8

 8 B + 20 B + 15 B
⇒ = 1290
20
43B
⇒ = 1290
20
⇒ 43
= B 1290 × 20
1290 × 20
⇒ B=
43
= 30 × 20
⇒ B
⇒ B = 600
2
Now, Amount of A = B
5
2
= × 600
5
= 2 × 120
= 240
3
And amount of C = B
4
3
= × 600
4
= 3 × 150
= 450
Hence, amount of A = Rs. 240
Amount of B = Rs. 600
Amount of C = Rs. 450

10. A school has 630 students. The ratio of the number of boys to the number
of girls is 3:2. This ratio changes to 7:5 after the admission of 90 new
students. Find the number of newly admitted boys.
Ans: Let the number of boys be 3x .

Class X Mathematics www.vedantu.com 9

And the number of girls be 2x .
Now, Total student = 630
⇒ 3x + 2 x =
630
⇒ 5 x = 630
⇒ x = 126
Thus, number of boys before new admission = 3x
= 3 × 126
= 378
And number of girls before new admission = 2x
= 2 × 126
= 252.
Now, Let the number of newly admitted boys be y.

And the number of newly girls admitted be ( 90 − y ) .

According to the question,

378  
+y 7
⇒ =
252  90
+ −y 5
378  
+y 7
⇒ =
342 − y 5
⇒ 1890 + 5 y= 2394 − 7 y .

⇒ 5 y + 7 y= 2394 − 1890

⇒ 12 y = 504

504
⇒ y=
12
⇒ y = 42

Hence, the number of newly admitted boys is 42.

Class X Mathematics www.vedantu.com 10

11. What quantity must be subtracted from each term of the ratio 9:17 to
make it equal to 1:3?
Ans: Let the required quantity be x .
w, according to question
9-x 1
⇒ =
17-x 3
⇒ 17-x=27-3x
⇒ -x+3x=27-17
⇒ 2 x = 10
10
⇒ x=
2
⇒ x=5
Hence, the required quantity is 5.

12. The monthly pocket money of Ravi and Sanjeev are in the ratio 5:7. Their
expenditures are in the ratio 3:5. If each saves Rs. 80 every month, find their
monthly pocket money.
Ans: Let the pocket money of Ravi be Rs. 5x .
And the pocket money of Sanjeev is Rs. 7x .
Expenditure of Ravi is 3 y.

Expenditure of Sanjeev is 5y .

Now, monthly saving of Ravi = Rs. 80

80 ……….. eq (i)
⇒ 5x − 3 y =

And monthly saving of Sanjeev = Rs. 80

80 ……….. eq (ii)
⇒ 7x − 5y =

Now, multiplying by 5 in eq(i) and 3 in eq (ii), we get

400 ……….. eq (iii)
⇒ 25 x − 15 y =

240 ……….. eq (iv)

⇒ 21x − 15 y =

Class X Mathematics www.vedantu.com 11

Now, subtracting eq (iv) from eq (iii)
⇒ 4 x = 160
160
⇒ x=
4
⇒ x = 40
Thus, pocket money of Ravi = Rs. 5x
= Rs. 5 × 40
= Rs. 200
And , pocket money of Sanjeev = Rs. 7x
= Rs. 7 × 40
= Rs. 280

13. The work done by ( x-2 ) men in ( 4x+1) days and the work done by
( 4x+1) men in ( 2x-3 ) days are in the ratio 3:8. Find the value of x.

Ans: here, we have, work done by ( x − 2 ) men in ( 4 x + 1) days and the work
done by ( 4 x + 1) men in ( 2 x − 3) days are in the ratio 3 :8

Thus, the amount of work done by ( x − 2 ) men in ( 4 x + 1) days is

(x-2)(4x+1).
And the amount of work done by ( 4 x + 1) men in ( 2 x − 3) days is

( 4 x + 1)( 2 x − 3) .
Now, according to question

⇒
( x − 2 )( 4 x + 1) = 3
( 4 x + 1)( 2 x − 3) 8

⇒
( x − 2) = 3
( 2 x − 3) 8
⇒ 8 x − 16 = 6 x − 9

Class X Mathematics www.vedantu.com 12

 ⇒ 8 x − 6 x =−9 + 16
⇒ 2x = 7
7
⇒ x=
2
7
Hence, Value of x is   .
2

14. The bus fare between two cities is increased in the ratio 7:9. Find the
increase in the fare, if:
The original fare is Rs. 245.
Ans: here, bus fare between two cities is increased in the ratio 7 : 9
And original fare is Rs. 245
Now, according to question
9
⇒ increased bus fare = × original fare
7
9
                                       
= ×245
7
                                        = 9 × 35
= Rs. 315
Thus, increase in fare = increased fare – original fare
= Rs. 315 – Rs. 245
= Rs. 70.
The increased fare is Rs. 207.
Ans: here, bus fare between two cities is increased in the ratio 7 : 9
And increased fare is Rs. 207
Now, according to question
9
⇒ increased bus fare of= × original fare
7

Class X Mathematics www.vedantu.com 13

 9
⇒ 207     
=× original fare
7
7
⇒    207
original fare =×
9
⇒    23 ×7
original fare =
⇒     Rs.161
original fare =

Thus, increase in fare = increased fare – original fare

= Rs. 207 – Rs. 161
= Rs. 46.

15. By increasing the cost of entry ticket to a fair in the ratio 10:13, the
number of visitors to the fair has decreased in the ratio 6:5. In what ratio has
the total collection increased or decreased?
Ans: Let the initial entry ticket cost be Rs.10 x .
And initial number of visitors be 6 y .

Thus, initial collection = Rs.10 x ×6 y

= Rs. 60 xy

Now, let the increased entry ticket cost be Rs. 13x

And decreased visitors are 5y .

Thus, final collection = Rs.13 x ×5 y

= Rs. 65 xy

Here, we have a final collection that is greater than initial collection

60 xy
Thus, required increased ratio =
65 xy
60
= .
65

Class X Mathematics www.vedantu.com 14

16. In a basket, the ratio between the number of oranges and the number of
apples is 7:13. If 8 oranges and 11 apples are eaten, the ratio between the
number of oranges and the number of apples becomes 1: 2. Find the original
number of oranges and the original number of apples in the basket.
Ans: Let the original number of oranges is 7x .
And the original number of apples is 13 x.
Now, according to question
7x − 8 1
⇒ =
13 x − 11 2
⇒ 14 x − 16 = 13 x − 11
⇒ 14 x − 13 x =−11 + 16
⇒ x=5
Thus, the original number of oranges = 7x
= 7×5
  = 35
And the original number of apples = 13x
= 13 × 5
= 65 .

17. In a mixture of 126 kg of milk and water, milk and water are in the ratio
5:2. How much water must be added to the mixture to make this ratio 3:2?
Ans: Let the initial quantity of milk be 5x kg
And the initial quantity of water be 2x kg.
Now, total quantity of mixture = 126 kg
⇒ 5x + 2 x =
126
⇒ 7 x = 126
126
⇒ x=
7
⇒ x = 18

Class X Mathematics www.vedantu.com 15

Thus, the initial quantity of milk = 5x
= 5 × 18
= 90 kg
And the initial quantity of water = 2x
= 2 × 18
= 36 kg
Now, let y kg water be added to the mixture to get ratio 3 : 2.

90 3
⇒ =
36 + y    2
⇒ 108 + 3 y =
180

⇒ 3 y = 72

72
⇒y=
3
⇒ y = 24

Hence, 24 kg water must be added to the mixture to make ratio 3 : 2.

17. (a) If A:B=3:4 and B:C=6:7, find A : B : C

Ans: here, we have A : B = 3 : 4 and B : C = 6 : 7
Therefore,
A 3 B 6
⇒ = and =
B 4 C 7
A 3× 3 B 6× 2
⇒ = and =
B 4×3 C 7×2
A 9 B 12
⇒ = and =
B 12 C 14
On comparing above both equations, we get
⇒ A : B : C = 9 :12 :14
Hence, value of A : B : C   
is 9 :12 :14 .

Class X Mathematics www.vedantu.com 16

Here, we have A : B = 3 : 4 and B : C = 6 : 7 .
A A B
Now, we can write as × .
C C B
A A B
⇒ = ×
C C B
A A B
⇒ = ×
C B C
A 3 6
⇒ = ×
C 4 7
A 18
⇒ =
C 28
A 9
⇒ =
C 14
Thus, value of A : C is 9 :14 .
(b) If A: B = 2: 5 and A: C = 3: 4, find A  B  C
: : .
Ans: here, we have A : B = 2 : 5 and A : C = 3 : 4
Therefore,
A 2 A 3
⇒ = and =
B 5 C 4
A 2×3 A 3× 2
⇒ = and =
B 5×3 C 4× 2
A 6 A 6
⇒ = and =
B 15 C 8
On comparing above both equations, we get
⇒ A : B : C = 6 :15 :8
Hence, value of A : B : C   6
is :15 :8 .
(i) If 3=
A 4=
B 6C ; find A: B: C
Ans: here, we have 3= B 6C .
A 4=
Therefore,

Class X Mathematics www.vedantu.com 17

⇒ 3 A = 4 B and 4 B = 6C
A 4 B 6
⇒ = and =
B 3 C 4
A 4 B 3
⇒ = and =
B 3 C 2
On comparing above both equations, we get
⇒ A : B :C = 4 :3 :2
Hence, value of A : B : C   4
is : 3 : 2 .
(ii) If 2a = 3b and 4b = 5c, Find a: c.
Ans: here, we have 2a = 3b and 4b = 5c .
Therefore,
a 3 b 5
⇒ = and =
b 2 c 4
a a b
Now, we can write as × .
c c b
a a b
⇒ = ×
c c b
a a b
⇒ = ×
c b c
a 3 5
⇒ = ×
c 2 4
a 15
⇒ =
c 8
Thus, value of a : c is 15 :8 .

18. Find the compound ratio of:

(a) 2:   
3,9:14 and 14: 27.
ace
Ans: We know that the compound ratio of a : b,  :
c d and e : f is .
bdf

Class X Mathematics www.vedantu.com 18

 2 × 9 × 14
Therefore, the compound ratio of 2 : 3, 9 :14 and 14 : 27 =
3 × 14 × 27
2
=
3× 3
2
=
9
Thus, the required compound ratio is 2 : 9 .
: 2 and x: n .
(b) 2a: 3b,   mn  x
ace
Ans: We know that the compound ratio of a : b,  :
c d and e : f is .
bdf
2a × mn × x
mn x 2 and x : n =
Therefore, the compound ratio of 2a : 3b,   :  
3b     
× x2 × n
2am
=
3bx
Thus, the required compound ratio is 2am : 3bx .

(c) :1, 3: 5 and

2       20 : 9.
ace
Ans: We know that the compound ratio of a : b,  :
c d and e : f is .
bdf

2 × 3 × 20
Therefore, the compound ratio of 2 :1,  3 : 5 and 20 : 9 =
1× 5 × 9

2×2
=
3

2 2
=
3

Thus, the required compound ratio is 2 2 : 3 .

19. Find duplicate ratio of:

(a) 3:4

Class X Mathematics www.vedantu.com 19

Ans: We know that duplicate ratio of a : b is a 2 : b 2
Therefore, duplicate ratio of 3 : 4 is 32 : 42
= 32 : 42
32
= 2
4
9
=
16
= 9 :16
Hence, duplicate ratio of 3 : 4 is 9 :16.

(b) 3 3 :2 5
Ans: We know that duplicate ratio of a : b is a 2 : b 2

( ) ( )
2 2
Therefore, duplicate ratio of 3 3 : 2 5 is 3 3 : 2 5

( ) ( )
2 2
= 3 3 : 2 5

( )
2
3 3
=
(2 5 )
2

9  3
×
=
4  5
×
27
=
20
= 27 : 20

Hence, duplicate ratio of 3 3 : 2 5 is 27 : 20.

20. Find triplicate ratio of:

(a) 1: 3
Ans: We know that triplicate ratio of a : b is a 3 : b3

Class X Mathematics www.vedantu.com 20

Therefore, triplicate ratio of 1 : 3 is 13 : ( 3b )
3

= 13 : 33
13
= 3
3
1
=
27
= 1 : 27
Hence, triplicate ratio of 1 : 3 is 1 : 27.
m n
(b) :
2 3
Ans: We know that triplicate ratio of a : b is a 3 : b3
3 3
m n m n
Therefore, triplicate ratio of : is   :  
2 3  2  3
3 3
m n
=   : 
 2  3
m3 n3
= :
8 27
m3
= 83
n
27
= 27 m3 :8n3
m n
Hence, triplicate ratio of : is 27 m3 :8n3 .
2 3

21. Find sub – duplicate ratio of:

(a) 9:16

Ans: We know that sub - duplicate ratio of a : b is a: b

Class X Mathematics www.vedantu.com 21

Therefore, sub - duplicate ratio of 9 :16 is 9 : 16

= 9 : 16

9
=
16
3
=
4
=3 :4
Hence, sub- duplicate ratio of 9 :16 is 3 : 4.

(b) ( x − y )   :( x + y )
4 6

Ans: We know that sub - duplicate ratio of a : b is a: b

( x − y ) :  
( x + y)
4 6
Therefore, sub - duplicate ratio of is

( x − y) ( x + y)
4 6
:

( x y) : ( x + y)
4 6
=−

( x − y)
4

=
( x + y)
6

( x − y)
2

=
( x + y)
3

( x − y ) :( x + y )
2 3
=

( x − y ) :  
( x + y ) is
4 6
Hence, sub- duplicate ratio of

( x − y ) :( x + y ) .
2 3

22. Find the sub – triplicate ratio of:

(a) 64:27

Ans: We know that sub - triplicate ratio of a : b is 3

a :3b

Class X Mathematics www.vedantu.com 22

Therefore, sub - triplicate ratio of 64 : 27 is 3
64 : 3 27

= 3 64 : 3 27
3
64
= 3
27
3
4  4  4
× ×
= 3
3  3  3
× ×
3
43
= 3
33
4
=
3
= 4 :3
Hence, sub - triplicate ratio of 64 : 27 is 4 : 3.
(b) x 3 :125 y 3

Ans: We know that sub - triplicate ratio of a : b is 3

a :3b

Therefore, sub - triplicate ratio of x 3 :125 y 3 is 3

x 3 : 3 125 y 3

= 3 x 3 : 3 125 y 3
3
x3
=
3
125 y 3
3
x3
=
(5 y )
3
3

x
=
5y
= x :5y

Hence, sub - triplicate ratio of x 3 :125 y 3 is x : 5 y.

23: Find the reciprocal ratio of:

Class X Mathematics www.vedantu.com 23

(a) 5 : 8
1 1
Ans: We know that reciprocal ratio of a : b is : .
a b
1 1
Therefore, reciprocal ratio of 5 : 8 is : .
5 8
1 1
= :
5 8
1
=5
1
8
8
=
5
= 8 :5
Hence, reciprocal ratio of 5 :8 is 8 : 5 .
x y
(b) :
  
3 7
1 1
Ans: We know that reciprocal ratio of a : b is : .
a b
x y 1 1
Therefore, reciprocal ratio of : is : .
3 7 x y
3 7
1 1
= :
x y
3 7
3 7
= :
x y
3
= x
7
y

Class X Mathematics www.vedantu.com 24

 3y
=
7x
= 3y : 7x

x y
Hence, reciprocal ratio of : is 3 y : 7 x
3 7

23. If ( x+3 ) :( 4x+1) is the duplicate ratio of 3 :5. Find the value of x.

Ans: here, we have ( x + 3) : ( 4 x + 1) is the duplicate ratio of 3 : 5 .

Therefore,

⇒
( x + 3) = 3
2

( 4 x + 1) 52

⇒
( x + 3) = 9
( 4 x + 1) 25
⇒ 36 x + 9= 25 x + 75
⇒ 36 x − 25 x =75 − 9
⇒ 11x = 66
66
⇒ x=
11
⇒ x=6
Hence, value of x is 6.

24. If m :n is the duplicate ratio of ( m+x ) :( n+x ) , prove that x 2 =mn.

Ans: here, we have m : n is the duplicate ratio of ( m + x ) : ( n + x ) .

Therefore,

m ( m   
+ x)
2

⇒ =
n ( n   
+ x)
2

Class X Mathematics www.vedantu.com 25

 m m 2 + 2mx + x 2
⇒ =
n n 2 + 2nx + x 2
⇒ m ( n 2 + 2nx + x 2 )  
= n ( m 2 + 2mx + x 2 )

⇒ mn 2 + 2mnx +   
mx 2 =nm 2 + 2mnx + nx 2
⇒ mn 2 + mx 2 = nm 2 + nx 2
⇒ mx 2 − nx 2 = nm 2 − mn 2
) mn ( m − n )
⇒ x 2 ( m − n=

mn ( m − n )
⇒ x2 =
(m − n)
⇒ x 2 = mn
Hence proved.

25. If ( 3x-9 ) :( 5x+4 ) is the triplicate ratio of 3 :4, Find the value of x.

Ans: Here, we have ( 3 x − 9 ) : ( 5 x + 4 ) is the triplicate ratio of 3 : 4 .

Therefore,

⇒
( 3x − 9 ) = 33

(5x + 4) 4 3

⇒
( 3x − 9 ) = 27
( 5 x + 4 ) 64

⇒
( 3x − 9 ) = 27
( 5 x + 4 ) 64
⇒ 192 x − 576 = 135 x + 108
⇒ 192 x − 135 x =108 + 576
⇒ 192 x − 135 x =108 + 576
⇒ 57 x = 684

Class X Mathematics www.vedantu.com 26

 684
⇒x=
57
⇒ x = 12
Hence, Value of x is 12.

26. Find the ratio compounded of the reciprocal ratio of 15:28 the sub –
duplicate ratio of 36:49 and the triplicate ratio of 5:4.
15 28
Ans: here, the reciprocal ratio of    is .
28 15

36 36 6
And the sub-duplicate ratio of is =
49 49 7
5 53 125
The triplicate ratio of is 3 =
4 4 64
28  6
× ×125
Now, the required compound ratio =
15  7  64
× ×
4 × 6 × 25
=
3 × 64
25
=
8
25
Hence, required compound ratio is .
8

27. (a) If r 2 = pq, show that p: q is the duplicate ratio of ( p + r ) : ( q + r ) .

Ans: here, we have r 2 = pq

Therefore, duplicate ratio of ( p + r ) : ( q + r ) is ( p + r ) : ( q + r )

2 2

( p + r)
2

=
(q + r )
2

Class X Mathematics www.vedantu.com 27

 p 2 + 2 pr + r 2
=
q 2 + 2qr + r 2

p 2 + 2 pr + pq
=
q 2 + 2qr + pq

p ( p + 2r + q )
=
q ( q + 2r + p )

p
=
q
Thus, p : q is the duplicate ratio of ( p + r ) : ( q + r ) .

Hence proved.
(b) If ( p − x ): ( q − x ) be the duplicate ratio of p: q then, show that
1 1 1
+ =.
p q x
Ans: here, we have ( p − x ) : ( q − x ) be the duplicate ratio of p : q .

Therefore,

⇒
( p − x) = p 2

(q − x) q 2

) p2 ( q − x )
⇒ q 2 ( p − x=

⇒ q 2 p − q 2 x = p 2q − p 2 x

⇒ p 2 x − q 2 x = p 2q − q 2 p

⇒ x ( p 2 − q 2 )= pq ( p − q )

⇒ x ( p − q )( p + q=
) pq ( p − q )
⇒ x( p + q) =
pq

⇒
( p + q) = 1
pq x

Class X Mathematics www.vedantu.com 28

 1 1 1
⇒ + =
q p x
1 1 1
⇒ + =
p q x
Hence proved.

Exercise- 7(B)
1. Find the fourth proportional to:
1.5, 4.5 and 3.5
Ans: Let the fourth proportional be x.
Therefore, 1.5 : 4.5 = 3.5 : x
1.5 3.5
⇒ =
4.5 x
1 3.5
⇒ =
3 x
⇒ x = 10.5
Thus, fourth proportional of 1.5, 4.5 and 3.5 is 10.5 .

3a, 6a 2 and 2ab 2

Ans: Let the fourth proportional be x .
Therefore, 3a : 6a 2 = 2ab 2 : x
3a 2ab 2
⇒ =
6a 2 x
1 2ab 2
⇒ =
2a x
⇒ x = 4 a 2b 2
Thus, fourth proportional of 3a, 6a 2 and 2ab 2 is 4a 2b 2 .

Class X Mathematics www.vedantu.com 29

2. Find the third proportional to:
2
2 and 4
3
Ans: Let the required third proportional be x .
2
Thus, 2 , 4 and x are in continued proportion.
3
Therefore,
2
⇒2 :4 = 4 : x
3
8
⇒ :4 = 4 : x
3
8
4
⇒ 3=
4 x
8 4
⇒ =
12 x
⇒ 8 x = 48
48
⇒ x=
8
⇒ x=6
2
Hence, required third proportional to 2 and 4 is 6.
3

3. ( a − b ) and ( a 2 − b 2 )

Ans: Let the required third proportional be x .

Thus, ( a − b ) , ( a 2 − b 2 ) and x are in continued proportion.

Therefore,
⇒ ( a − b ) : ( a 2 − b2 ) = ( a 2 − b2 ) : x

Class X Mathematics www.vedantu.com 30

⇒
(a − b) =
(a 2
− b2 )
(a 2
− b2 ) x

⇒ x (a − b) = (a 2
− b 2 )( a 2 − b 2 )

⇒ x ( a − b ) = ( a − b )( a + b ) ( a 2 − b 2 )

( a b ) ( a 2 − b2 )
⇒ x =+

Hence, required third proportional is ( a + b ) ( a 2 − b 2 ) .

4. Find the mean proportional between:

(a) 6 + 3 3 and 8 − 4 3
Ans: Let the required mean proportional be x .

Thus, 6 + 3 3,  x and 8 − 4 3 are in continued proportion.

Therefore,

(
⇒ 6+3 3 :x = )
x : 8−4 3 ( )
⇒
( 6 + 3 3)
=
x
x (8 − 4 3 )
(
6+3 3 8−4 3
⇒ x2 = )( )
48 − 24 3 + 24 3 − 36
⇒ x2 =
⇒ x 2 = 12

⇒ x = 12

⇒ x = ±2 3
Since, the mean proportional to positive number is positive.

Thus, x = 2 3

Hence, required mean proportional is 2 3.

Class X Mathematics www.vedantu.com 31

(b) ( a − b ) and ( a 3 − a 2b ) .

Ans: Let the required mean proportional be x .

Thus, ( a − b ) ,  x and ( a 3 − a 2b ) are in continued proportion.

Therefore,
⇒ ( a − b ) : x = x : ( a 3 − a 2b )

⇒
(a − b) = x
x ( a − a 2b )
3

( a b ) ( a 3 − a 2b )
⇒ x 2 =−

( a − b ). a 2 ( a − b )
⇒ x2 =

x2 a2 ( a − b )
2
⇒=

a2 ( a − b)
2
⇒ x
=

⇒=x a (a − b)

Hence, required mean proportional is a ( a − b ) .

5. If x + 5 is the mean proportion between x + 2 and x + 9 ; find the value of

x.
Ans: Here, x + 5 is the mean proportion between x + 2 and x + 9
Thus, ( x + 2 ) ,  ( x + 5 ) and ( x + 9 ) in continued proportion.

Therefore,
⇒ ( x + 2 ) : ( x + 5) = ( x + 5) : ( x + 9 )

⇒
( x + 2 ) = ( x + 5)
( x + 5) ( x + 9 )
⇒ ( x + 5 ) =( x + 2 )( x + 9 )
2

⇒ x 2 + 10 x + 25 = x 2 + 9 x + 2 x + 18

Class X Mathematics www.vedantu.com 32

⇒ 10 x + 25 = 11x + 18
⇒ 10 x − 11x =18 − 25
⇒ − x =−7
⇒ x=7
Thus, Value of x is 7 .

6. If x 2 , 4 and 9 are in continued proportion, find x .

Ans: Here, x 2 ,  4  9   

and in continued proportion
Therefore,
⇒ x 2   : 4 = 4 : 9
x2 4
⇒ =
4 9
16
⇒ x2 =
9

16
⇒ x=
9
4
⇒ x= ±
3
4
Hence, Value of x is ± .
3

7. What least number must be added to each of the numbers 6, 15, 20 and 43
to make them proportional?
Ans: Let the least number is x to be added to the numbers 6, 15, 20 and 43 to
make them proportional
Therefore,
6 + x 20 + x
⇒ =
15 + x 43 + x

Class X Mathematics www.vedantu.com 33

⇒ ( 6 + x )( 43 + x ) = ( 20 + x )(15 + x )
⇒ 258 + 6 x + 43 x + x 2 = 300 + 20 x + 15 x + x 2
⇒ 258 + 49 x = 300 + 35 x
⇒ 49 x − 35 x = 300 − 258
⇒ 14 x = 42
42
⇒ x=
14
⇒ x=3
Thus, the required least number is 3 .

(i) If a , b, c are in continued proportion,

a 2 + b2 b(a + c )
Show that: = 2
b(a + c ) b + c2

Ans: Here, a, b, c are in continued proportion

Therefore,
a b
⇒ =
b c
⇒ b 2 = ac ……… (i)
Now,
⇒ ( a 2 + b 2 )( b 2 + c 2 ) = ( a 2 + ac )( ac + c 2 ) [from eq (i)]

⇒ ( a 2 + b 2 )( b 2 + c 2 ) = a ( a + c ) .c ( a + c )

⇒ ( a 2 + b 2 )( b 2 + c 2 ) = ac ( a + c )
2

⇒ ( a 2 + b 2 )( b 2 + c 2 ) = b 2 ( a + c )
2
[ b 2 = ac]

⇒ ( a 2 + b 2 )( b 2 + c 2 ) = b ( a + c ) . b ( a + c )

Class X Mathematics www.vedantu.com 34

 a 2 + b2 b(a + c)
⇒ = 2
b(a + c) b + c2

Hence proved.
(ii) If a , b, c are in continued proportion and a ( b − c ) =
2b,

2(a + b)
Prove that: a − c = .
a
Ans: Here, a, b, c are in continued proportion and a ( b − c ) =
2b

a b
Therefore, =
b c
⇒ b 2 = ac and a ( b − c ) =
2b

⇒ b 2 = ac and ab − ac =
2b
2b
⇒ b 2 = ac and ab − b 2 =
⇒ b 2 = ac and b ( a − b ) =
2b

⇒ b 2 = ac and ( a − b ) =
2

Now, L.H.S = a − c
a(a − c)
  =
a
a 2 − ac
=
a
a 2 − b2
= [ b 2 = ac ]
a

=
( a + b )( a − b ) [ a 2 − b 2 = ( a + b )( a − b )]
a
2(a + b)
= [a −b =2]
a
= R.H .S
Therefore, L.H.S = R.H.S

Class X Mathematics www.vedantu.com 35

 Hence proved.

a 3c + ac 3 ( a + c )
4
a c
(iii). If = , show that: 3 =
b d + bd 3 ( b + d )
4
b d

a c
Ans: Here, we have =
b d
a c
Let = = k , then we get
b d
⇒ a kb
= =   
and c kd
a 3c + ac 3
Now, L.H.S =
b3d + bd 3

( kb ) .kd + kb.( kd )
3 3

=
b3d + bd 3
k 3b3 .kd + kb.k 3d 3
=
b3d + bd 3
k 4b3d + k 4bd 3
=
b3d + bd 3
k 4 ( b3d + bd 3 )
=
( b d + bd )
3 3

= k4

(a + c)
4

⇒ R.H.S =
(b + d )
4

( kb + kd )
4

=
(b + d )
4

 k ( b + d ) 
4

=
(b + d )
4

k 4 (b + d )
4

=
(b + d )
4

Class X Mathematics www.vedantu.com 36

 = k4
Therefore, L.H.S = R.H.S
Hence proved.

8. What least number must be subtracted from each of the numbers 7, 17

and 47 so that the remainders are in continued proportion?
Ans: Let the least number is x to be subtracted from each of the numbers 7, 17
and 47 to make them in continued proportional
Therefore,
7 − x 17 − x
⇒ =
17 − x 47 − x
⇒ ( 7 − x )( 47 − x ) = (17 − x )(17 − x )

⇒ 329 − 7 x − 47 x + x 2 = 289 − 17 x − 17 x + x 2
⇒ 329 − 54 x = 289 − 34 x
⇒ −54 x + 34 x = 289 − 329
⇒ −20 x =
−40
−40
⇒ x=
−20
⇒ x=2
Thus, the required least number is 2 .

9. If y is the mean proportional between x and z; show that xy + yz is the

mean proportional between x 2 + y 2 and y 2 + z 2 .

Ans: Here, y is the mean proportional between x and z

Therefore,
⇒ y 2 = xz ……… (i)

Now, the mean proportional between x 2 + y 2 and y 2 + z 2 will be,

Class X Mathematics www.vedantu.com 37

 =( x 2 + y 2 )( y 2 + z 2 )

=( x 2 + xz ) ( xz + z 2 ) [ y 2 = xz ]

= x ( x + z ) .z ( x + z )

xz ( x + z )
2
=    

y2 ( x + z )
2
= [    y 2 = xz ]

= y( x + z)
= xy + yz

Hence, xy + yz is the mean proportional between x 2 + y 2 and y 2 + z 2 .

10. If q is the mean proportional between p and r , show that:

pqr ( p + q + r ) = ( pq + qr + pr ) .
3 3

Ans: Here, q is the mean proportional between x and z

Therefore,
⇒ q 2 = pr ……… (i)

Now, show that pqr ( p + q + r ) = ( p + q + r)

3 3

= pqr ( p + q + r )
3
L.H.S

= q. pr ( p + q + r )
3

= q.q 2 ( p + q + r )
3

= q3 ( p + q + r )
3

=  q ( p + q + r ) 
3

3
=  pq + q 2 + qr 

[ pq + pr + qr ]
3
= [   q 2 = pr ]

Class X Mathematics www.vedantu.com 38

 = R.H .S

Therefore, pqr ( p + q + r ) = ( pq + qr + pr ) .
3 3

Hence proved.

11. If three quantities are in continued proportion; show that the ratio of the
first to the third is the duplicate ratio of the first to the second.
Ans: Let x, y and z are in continued proportion.

Therefore,
⇒ y 2 = xz

Now, multiply by x in the above equation, we get

⇒ xy 2 = x 2 z

x x2
⇒ =
z y2
Hence, the ratio of the first to the third is the duplicate ratio of the first to the
second.

12. If y is the mean proportional between x and z , Prove that:

x2 − y2 + z2
= y4.
x − y +z
−2 −2 −2

Ans: Here, y is the mean proportional between x and z

Therefore,
⇒ y 2 = xz ……… (i)

x2 − y 2 + z 2
Now, L.H.S = −2
x − y −2 + z −2
2 2 2
 y2   y2   y2 
 z  − y  + x 
=   −2  −2 −2 
x −y +z

Class X Mathematics www.vedantu.com 39

 4
y4 y4 y
− +
z 2 y 2 x2
= −2
x − y −2 + z −2

y 4 z −2 − y 4 y −2 + y 4 x −2
=
x −2 − y −2 + z −2

y 4 ( z −2 − y −2 + x −2 )
=
x −2 − y −2 + z −2

y 4 ( x −2 − y −2 + z −2 )
=
(x −2
− y −2 + z −2 )

= y4
L.H.S = R.H .S
x2 − y 2 + z 2
Therefore, = y4
x −y +z
−2 −2 −2

Hence proved.

13. Given four quantities a , b, c and d are in the proportion. Show that:

( a − c ) b : ( b − d ) cd = ( a
2 2
− b 2 − ab ) : ( c 2 − d 2 − cd )

Ans: Here a, b, c and d are in the proportion

a c
Therefore, =
b d
a c
Let = = k , then we get
b d
⇒ a kb
= =   
and c kd

Now, L.H.S =
(a − c)b 2

( b − d ) cd

=
( kb − kd ) b 2

( b − d ) kd .d

Class X Mathematics www.vedantu.com 40

 k ( b − d ) b2
=
k (b − d ) d 2

b2
= 2
d

⇒ R.H.S =
(a 2
− b 2 − ab )
  
(c 2
− d 2 − cd )

=
(k b 2 2
− b 2 − kb.b )
(k d 2 2
− d 2 − kd .d )

=
(k b 2 2
− b 2 − kb 2 )
(k d 2 2
− d 2 − kd 2 )

b2 ( k 2 − 1 − k )
=
d 2 (k 2 −1− k )

b2
= 2
d
Therefore, L.H.S = R.H.S
Hence proved.

14. Find two numbers such that the mean proportionality between them is
12 and the third proportional to them is 96.
Ans: Let the two numbers are x and y.

According to 1st condition,

The mean proportionality between x and y is 12.

⇒ xy = 12

On squaring both sides, we get

⇒ xy = 144

144
⇒ x= ……………. Eq(i)
y

Class X Mathematics www.vedantu.com 41

According to the 2nd condition,
x y
⇒ =
y 96

⇒ y 2 = 96 x

144
⇒ y=
2
96 ×
y

⇒ y=
3
96 × 144

⇒ y 3 = 12 × 8 × 12 × 12

⇒ y=
3
123 × 23

⇒ y 3 = ( 24 )
3

⇒ y = 24

Now, put the value of y in eq(i), we get

144
⇒ x=
24
⇒ x=6
Hence, two required numbers are 6 and 24.

x y
15. Find the third proportional to + and x2 + y2 .
y x
Ans: Let the required third proportional be z.
x y
Therefore, + , x 2 + y 2 and z are in continued proportion.
y x
x y
+
y x x2 + y 2
⇒ =
x +y
2 2
z

x y
( )
2
⇒ z + = x2 + y 2
y x

Class X Mathematics www.vedantu.com 42

  x2 + y 2 
⇒ z =  ( x2 + y 2 )
 xy 
xy ( x 2 + y 2 )
⇒ z=
( x2 + y 2 )
⇒ z = xy

Hence, the required third proportional is xy .

16. If p: q = r: s; then show that: ( mp + nq ) : q =( mr + ns ): s .

p r
Ans: Here, we have =
q s
p r
⇒ =
q s
Multiply by m in the above equation, we get
mp mr
⇒ =
q s
Now, adding n to both sides
mp mr
⇒ + n= +n
q s
mp + nq mr + ns
⇒ =
q s

⇒ ( mp + nq ) : q =
( mr + ns ) : s
Hence proved.

1 1 m
17. If p + r =mq and + = ; then prove that p: q = r: s .
q s r
1 1 m
Ans: Here, we have p + r =mq and + =
q s r

Class X Mathematics www.vedantu.com 43

⇒ mq= p + r

p+r
⇒ m=
q
1 1 m
Now, put the value of m in + = , we get
q s r
p+r
1 1 q
⇒ + =
q s r
s+q p+r
⇒ =
qs qr
s+q p+r
⇒ =
s r
⇒ r (s + q) = s( p + r )

⇒ rs + qr = sp + rs

⇒ qr = sp

r p
⇒ =
s q
p r
⇒ =
q s
Hence proved.

Exercise- 7(C)

1. If a: b = c: d , prove that:
( 5a + 7b ): ( 5a − 7b ) = ( 5c + 7d ): ( 5c − 7d )
Ans: Here, we have
   a : b = c : d
a c
Therefore, =
b d

Class X Mathematics www.vedantu.com 44

 5
Now, multiplying by in the above equation, we get
7
5 a  5 c 
⇒  =  
7b  7 d 
5a 5c
⇒ =
7b 7 d
Applying componendo and dividendo rule
5a + 7b 5c + 7 d
⇒ =
5a − 7b 5c − 7 d
⇒ ( 5a + 7b ) : ( 5a − 7b ) = ( 5c + 7 d ) : ( 5c − 7 d )
     

Hence proved.

2. ( 9a + 13b )( 9c − 13d ) =( 9c + 13d )( 9a − 13b ) .

Ans: Here, we have

   a : b = c : d
a c
Therefore, =
b d
9
Now, multiplying by in the above equation, we get
13
9 a 9  c 
⇒  =  
13  b  13  d 
9a 9c
⇒ =
13b 13d
Applying componendo and dividendo rule
9a + 13b 9c + 13d
⇒ =
9a − 13b 9c − 13d
⇒ ( 9a + 13b )( 9c − 13d ) = ( 9c + 13d )( 9a − 13b )
     

Hence proved.

Class X Mathematics www.vedantu.com 45

3. ( xa + yb ) : ( xc + yd ) =
b: d

Ans: Here, we have

   a : b = c : d
a c
Therefore, =
b d
x
Now, multiplying by in the above equation, we get
y

xa x c 
⇒  =  
yb yd 
xa xc
⇒ =
yb yd
Adding 1 to both sides, we get
xa xc
⇒ + 1= +1
yb yd
xa + yb xc + yd
⇒ =
yb yd
xa + yb yb
⇒ =
xc + yd yd
xa + yb b
⇒ =
xc + yd d

⇒ ( xa + yb ) : ( xc + yd ) =
b :d

Hence proved.

4. If a: b = c: d , prove that:
( 6a + 7b )( 3c − 4d ) = ( 6c + 7d )( 3a − 4b ) .
Ans: Here, we have
   a : b = c : d
a c
Therefore, = ……… eq (i)
b d

Class X Mathematics www.vedantu.com 46

 6
Now, multiplying by in the above equation, we get
7
6 a  6 c 
⇒  =  
7b  7 d 
6a 6c
⇒ =
7b 7 d
Adding 1 to both sides, we get
6a 6c
⇒ + 1= +1
7b 7d
6 a + 7 b 6c + 7 d
⇒ =
7b 7d
6 a + 7b 7b
⇒ =
6c + 7 d 7 d
6 a + 7b b
⇒ = ………… eq (ii)
6c + 7 d d
3
Now, multiplying by in eq (i), we get
4
3 a  3 c 
⇒  =  
4 b  4 d 
3a 3c
⇒ =
4b 4d
Subtracting 1 from both sides, we get
3a 3c
⇒ − 1= −1
4b 4d
3a − 4b 3c − 4d
⇒ =
4b 4d
3a − 4b 4b
⇒ =
3c − 4d 4d
3a − 4b b
⇒ = ………….. eq (iii)
3c − 4d d

Class X Mathematics www.vedantu.com 47

 Now, from eq (ii) and (iii), we get
6a + 7b 3a − 4b
⇒ =
6c + 7 d 3c − 4d
⇒ ( 6a + 7b )( 3c − 4d ) =( 6c + 7 d )( 3a − 4b )

Hence proved.

a c
5. Given , prove that:
=   
b d
3a − 5b 3c − 5d
= .
3a + 5b 3c + 5d
a c
Ans: Here, we have = .
b d
3
Now, multiplying by in the above equation, we get
5
3 a  3 c 
⇒  =  
5 b  5 d 
3a 3c
⇒ =
5b 5d
Applying componendo and dividendo rule
3a + 5b 3c + 5d
⇒ =
3a − 5b 3c − 5d
3a − 5b 3c − 5d
⇒ =
3a + 5b 3c + 5d
Hence proved.

5x + 6 y 5x − 6 y
6. If = ; then prove that x: y = u: v.
5u + 6v 5u − 6v
5x + 6 y 5x − 6 y
Ans: Here, we have = .
5u + 6v 5u − 6v

Class X Mathematics www.vedantu.com 48

 5 x + 6 y 5u + 6v
⇒ =
5 x − 6 y 5u − 6v
Applying componendo and dividendo rule
5x + 6 y + 5x − 6 y 5u + 6v + 5u − 6v
⇒ =
5 x + 6 y − ( 5 x − 6 y ) 5u + 6v − ( 5u − 6v )

10 x 10u
⇒ =
5 x + 6 y − 5 x + 6 y 5u + 6v − 5u + 6v
10 x 10u
⇒ =
12 y 12v
x u
⇒ =
y v
⇒ x : y = u :v

Hence proved.

7. If ( 7a + 8b )( 7c − 8d ) = ( 7a − 8b )( 7c + 8d ) ; prove that a: b = c: d .

Ans: Here, we have ( 7 a + 8b )( 7c − 8d ) = ( 7 a − 8b )( 7c + 8d ) .

⇒
( 7a + 8b ) = ( 7c + 8d )
( 7a − 8b ) ( 7c − 8d )
Applying componendo and dividendo rule

⇒
( 7a + 8b ) + ( 7a − 8b ) = ( 7c + 8d ) + ( 7c − 8d )
( 7a + 8b ) − ( 7a − 8b ) ( 7c + 8d ) − ( 7c − 8d )
7 a + 8b + 7 a − 8b 7c + 8d + 7c − 8d
⇒ =
7 a + 8b − 7 a + 8b 7c + 8d − 7c + 8d
14a 14c
⇒ =
16b 16d
a c
⇒ =
b d
⇒ a :b = c :d

Class X Mathematics www.vedantu.com 49

Hence proved.

6ab x + 3a x + 3b
8. (i) If x = , find the value of + .
a+b x − 3a x − 3b
6ab
Ans: Here, we have x = ……… eq (i)
a+b
Now, dividing by 3a in the above equation, we get
x 6ab
⇒ =
3a 3a ( a + b )

x 2b
⇒ =
3a ( a + b )

Applying componendo and dividendo rule

x + 3a 2b + a + b
⇒ =
x − 3a 2b − ( a + b )

x + 3a 3b + a
⇒ =
x − 3a 2b − a − b
x + 3a 3b + a
⇒ = ……….. eq (ii)
x − 3a b − a
Now, dividing by 3b in eq (i), we get
x 6ab
⇒ =
3b 3b ( a + b )

x 2a
⇒ =
3b ( a + b )

Applying componendo and dividendo rule

x + 3b 2a + a + b
⇒ =
x − 3b 2a − ( a + b )

x + 3b 3a + b
⇒ =
x − 3b 2a − a − b

Class X Mathematics www.vedantu.com 50

 x + 3b 3a + b
⇒ = ……… eq (iii)
x − 3b a − b
Adding eq (ii) and (iii)
x + 3a x + 3b 3b + a 3a + b
⇒ + = +
x − 3a x − 3b b − a a −b
3b + a 3a + b
= −
b−a b−a
3b + a − 3a − b
=
b−a
2b − 2a
=
b−a
2(b − a )
=
(b − a )
=2
x + 3a x + 3b
Hence value of + is 2.
x − 3a x − 3b

4 6 a+2 2 a+2 3
(ii) If = , find the value of + .
2+ 3 a−2 2 a−2 3

4 6
Ans: Here, we have a = ……… eq (i)
2+ 3

Now, dividing by 2 2 in the above equation, we get

a 4 6
⇒ =
2 2 2 2 ( 2+ 3 )
a 2 3
⇒ =
2 2 ( 2+ 3 )
Applying componendo and dividendo rule

Class X Mathematics www.vedantu.com 51

⇒
a+2 2 2 3+ 2+ 3
=
( )
a−2 2 2 3−( 2 + 3 )

a+2 2 2 3+ 2+ 3
⇒ =
a−2 2 2 3− 2 − 3

a+2 2 3 3+ 2
⇒ = ……….. eq (ii)
a−2 2 3− 2

Now, dividing by 2 3 in eq (i), we get

a 4 6
⇒ =
2 3 2 3 ( 2+ 3 )
a 2 2
⇒ =
2 3 ( 2+ 3 )
Applying componendo and dividendo rule

⇒
a+2 3 2 2+ 2+ 3
=
( )
a−2 3 2 2 −( 2 + 3 )

a+2 3 2 2+ 2+ 3
⇒ =
a−2 3 2 2 − 2 − 3

a+2 3 3 2+ 3
⇒ = ……….. eq (iii)
a−2 3 2− 3
Adding eq (ii) and (iii)

a+2 2 a+2 3 3 3+ 2 3 2+ 3
⇒ + = +
a−2 2 a−2 3 3− 2 2− 3

3 3+ 2 3 2+ 3
= −
3− 2 3− 2

3 3 + 2 −3 2 − 3
=
3− 2

Class X Mathematics www.vedantu.com 52

 2 3−2 2
=
3− 2

=
2 ( 3 − 2)
( 3 − 2)
=2
a+2 2 a+2 3
Hence value of + is 2.
a−2 2 a−2 3

9. If ( a + b + c + d )( a − b − c + d ) = ( a + b − c − d )( a − b + c − d ) ; Prove that
a: b = c: d .
Ans: Here, we have
( a + b + c + d )( a − b − c + d ) = ( a + b − c − d )( a − b + c − d )
Therefore,

⇒
(a + b + c + d ) = (a − b + c − d )
(a + b − c − d ) (a − b − c + d )
Applying componendo and dividendo rule

⇒
(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 )
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
2a + 2b 2a − 2b
⇒ =
2c + 2d 2c − 2d
2(a + b) 2(a − b)
⇒ =
2(c + d ) 2(c − d )

⇒
(a + b) = (a − b)
(c + d ) (c − d )

Class X Mathematics www.vedantu.com 53

⇒
(a + b) = (c + d )
(a − b) (c − d )
Applying componendo and dividendo rule

⇒
(a + b) + (a − b) = (c + d ) + (c − d )
(a + b) − (a − b) (c + d ) − (c − d )
a +b+a −b c+d +c−d
⇒ =
a+b−a+b c+d −c+d
2a 2c
⇒ =
2b 2d
a c
⇒ =
b d
⇒ a :b = c :d
Hence proved.

a − 2b − 3c + 4d a − 2b + 3c − 4d
10. If = , show that: 2ad = 3bc
a + 2b − 3c − 4d a + 2b + 3c + 4d
a − 2b − 3c + 4d a − 2b + 3c − 4d
Ans: Here, we have =
a + 2b − 3c − 4d a + 2b + 3c + 4d
a − 2b − 3c + 4d a − 2b + 3c − 4d
⇒ =
a + 2b − 3c − 4d a + 2b + 3c + 4d
Applying componendo and dividendo rule
⇒
( a − 2b − 3c + 4d ) + ( a + 2b − 3c − 4d ) = ( a − 2b + 3c − 4d ) + ( a + 2b + 3c + 4d )
( a − 2b − 3c + 4d ) − ( a + 2b − 3c − 4d ) ( a − 2b + 3c − 4d ) − ( a + 2b + 3c + 4d )
a − 2b − 3c + 4d + a + 2b − 3c − 4d a − 2b + 3c − 4d + a + 2b + 3c + 4d
⇒ =
a − 2b − 3c + 4d − a − 2b + 3c + 4d a − 2b + 3c − 4d − a − 2b − 3c − 4d
2 a − 6c 2a   + 6c
⇒ =
−4b + 8d −4b − 8d

Class X Mathematics www.vedantu.com 54

 2 ( a − 3c ) 2 ( a   + 3c )
⇒ =
−4 ( b − 2d ) −4 ( b + 2d )

⇒
( a − 3c )  
=
( a + 3c )
( b − 2d ) ( b + 2d )

⇒
( a − 3c ) = ( b − 2d )
( a + 3c ) ( b + 2d )
Applying componendo and dividendo rule

⇒
( a − 3c ) + ( a + 3c ) = ( b − 2d ) + ( b + 2d )
( a − 3c ) − ( a + 3c ) ( b − 2d ) − ( b + 2d )
a − 3c   
+ a + 3c b − 2d + b + 2d
⇒ =
a − 3c −   a − 3c b − 2d − b − 2d
2a 2b
⇒ =
−6c −4d
a b
⇒ =
3c 2d
⇒ 2ad = 3bc
Hence proved.

a b
11. If ( a 2 + b 2 )( x 2 + y 2 ) = ( ax + by ) ; prove that
2
= .
x y

Ans: Here, we have ( a 2 + b 2 )( x 2 + y 2 ) = ( ax + by )

2

On simplifying, we get
⇒ a 2 x 2 + a 2 y 2 + b 2 x 2 + b 2 y 2 = a 2 x 2 + b 2 y 2 + 2abxy

2abxy
⇒ a 2 y 2 + b2 x2 =

⇒ a 2 y 2 + b 2 x 2 − 2abxy =
0

⇒ ( ay − bx ) =
2
0

Class X Mathematics www.vedantu.com 55

 0
⇒ ay − bx =

⇒ ay = bx

a b
⇒ =
x y
Hence proved.

12. If a , b and c are in continued proportions, prove that:

a 2 + ab  b
+ 2 a
(a) 2 =
b   bc  c
+ + 2 c
Ans: Here, a, b and c are in continued proportions
a b
Therefore, =
b c
⇒ b 2 = ac ……… eq (i)
a 2 +    
ab + b 2
Now, L.H.S =
b 2     
+ bc + c 2
a 2 +    
ab + ac
= [From eq (i)]
ac     
+ bc + c 2
a ( a     
+ b + c)
=
c ( a     
+ b + c)

a
=
c
= R.H .S
Thus, L.H.S = R.H.S
Hence proved.

a 2   b
+ 2   c
+ 2 a −b  c+
(b) 2 =
( a +b  c
+ ) a  b  c
+ +

Ans: Here, a, b and c are in continued proportions

Class X Mathematics www.vedantu.com 56

 a b
Therefore, =
b c
⇒ b 2 = ac ……… eq (i)
a 2     
+ b2 + c2
Now, L.H.S =
( a     
+b + c )
2

+ b 2 + c 2 + 2ab    + 2bc + 2ac − 2 ( ab + bc + ac )

a 2      
=
( a     
+b + c )
2

a + b + c ) − 2 ( ab + bc + b )
(       
2 2

=
( a     
+b + c )
2

(       
a + b + c ) − 2b ( a + c + b )
2

=
( a     
+b + c )
2

a + b + c ) [ a + b + c − b]
(             2
=
( a     
+b + c )
2

                       =
(       
a −b + c )
( a      
+b+c )

= R.H .S
Thus, L.H.S = R.H.S
Hence proved.

13. Using properties of proportion, solve for x:

+ 5 + x − 16 7
x    
(a) =
+ 5 − x − 16 3
x   

x   
+ 5   
+ x − 16 7
Ans: Here, we have =
+ − x − 16 3
x   5
Applying componendo and dividendo rule in above equation

Class X Mathematics www.vedantu.com 57

⇒
x   
+ 5    (
+ x − 16   
+ x + 5 − x − 16 ) = 7  3
+
x   
+ 5    − ( x+5 −
+ x − 16    x − 16 ) 7 − 3

x   
+ 5   
+ x − 16    
+ x + 5 − x − 16 10
⇒ =
x   
+ 5   
+ x − 16    + x − 16 4
− x + 5  

2   5
x+ 5
⇒ =
2 x − 16 2

x   5
+ 5
⇒ =
x − 16 2
Now, squaring both sides, we get
2
 x   5
+  5
2

⇒  = 
 x − 16  2
x   5
+ 25
⇒ =
x − 16 4
⇒ 25 ( x − 16 ) = 4 ( x + 5 )

⇒ 25 x − 400 =4 x + 20
⇒ 25 x − 4 x = 20 + 400
⇒ 21x = 420
420
⇒ x=
21
⇒ x = 20
Hence value of x is 20.

+1 + x − 1 4 x − 1
x    
(b) =
+1 − x − 1
x    2

x +1   
+ x − 1 4x − 1
Ans: Here, we have =
x +1  
− x −1 2
Applying componendo and dividendo rule in above equation

Class X Mathematics www.vedantu.com 58

⇒
x +1   
+ x − 1  (
+ x +1  
− x −1 ) = 4 x − 1  2
+
x +1   
+ x − 1   − x − 1 ) 4 x −1 − 2
− ( x +1  

x +1   
+ x − 1   
+ x +1  
− x − 1 4 x +1
⇒ =
x +1   
+ x − 1   + x − 1 4 x   − 3
− x +1   

2  
x +1   4 x +1
⇒ =
2 x − 1 4 x   − 3

x +1   4 x +1
⇒ =
x −1 4 x   − 3
Now, squaring both sides, we get
2
 x +1    4 x +1 2
⇒  = 
 x − 1   4 x   − 3 
x +1 16 x 2 +1   
+ 8x
⇒ =
x −1 16 x   9
2
+ − 24 x
Applying componendo and dividendo rule in above equation
x +1   
+ x −1 16 x 2 +1   
+ 8 x +16 x 2   
+ 9 − 24 x
⇒ =
+ 8 x − (16 x 2   
x + 1 − ( x −1) 16 x 2 +1    + 9 − 24 x )

2x 32 x 2 +10 − 16 x
⇒ =
x + 1 − x +1 16 x 2 +1   
+ 8 x − 16 x 2 − 9 + 24 x
2 x 32 x 2 +10 − 16 x
⇒ =
2 −8  32
+ x
32 x 2 +10 − 16 x
⇒ x=
−8  32
+ x
⇒ −8 x + 32 x 2= 32 x 2 +10 − 16 x
⇒ −8 x =10 − 16 x
⇒ −8 x + 16 x =
10
⇒ 8 x =10

Class X Mathematics www.vedantu.com 59

 10
⇒x=
8
5
⇒x=
4
5
Hence value of x is .
4

3 x  
+ 9x2 − 5
(c) = 5.
3x − 9 x2 − 5

3 x   
+ 9x2 − 5 5
Ans: Here, we have =
3x − 9 x 2 − 5 1
Applying componendo and dividendo rule in above equation

3 x   + 9 x 2 − 5   + 3 x − 9 x 2 − 5 5 +1
⇒ =
(
3 x   + 9 x 2 − 5 − 3 x − 9 x 2 − 5 ) 5 −1

6x 6
⇒ =
3x   + 9 x 2 − 5 − 3x + 9 x 2 − 5 4
6x 6
⇒ =
2 9x2 − 5 4
x 1
⇒ =
9x2 − 5 2
Now, squaring both sides, we get
2 2
 x  1
⇒  = 
 9x − 5   2 
2

x2 1
⇒ 2 =
9x − 5 4
⇒ 9x2 − 5 =4x2
⇒ 9x2 − 4x2 =
5
⇒ 5x2 = 5

Class X Mathematics www.vedantu.com 60

⇒ x2 = 1

⇒x= 1
⇒ x = ±1
Hence value of x is ±1 .

+ 3b + a − 3b
a  
14. If x = , prove that 3bx 2 − 2ax + 3b =
0.
+ 3 − a − 3b
a  b

a   + 3b   
+ a − 3b
Ans: Here, we have x =
a   + 3b   
− a − 3b
Applying componendo and dividendo rule in above equation

x +1    a + 3b   + a − 3b     

+ a + 3b   
− a − 3b
⇒ =
x −1 a   + 3b    (
+ a − 3b − a   + 3b   
− a − 3b )
x +1 2 a   
+ 3b   
⇒ =
x −1 a   + 3b   
+ a − 3b − a   + 3b + a − 3b

x +1 2 a   
+ 3b   
⇒ =
x −1 2 a − 3b

x +1   a + 3b   
⇒ =
x −1 a − 3b
Now, squaring both sides, we get
2
  a + 3b   
2
 x +1   
⇒  = 
 x −1   a − 3b 

x 2 + 2 x +1   a + 3b
⇒ 2 =
x − 2 x +1    a − 3b

( x 2 − 2 x +1  )
⇒ ( a − 3b ) ( x 2 + 2 x +1) = ( a + 3b )   

⇒ ax 2 + 2ax + a − 3bx 2 − 6bx − 3b = ax 2 − 2ax + a + 3bx 2 − 6bx + 3b

⇒ 2ax − 3bx 2 − 3b =
−2ax + 3bx 2 + 3b

Class X Mathematics www.vedantu.com 61

⇒ 6bx 2 − 4ax + 6b   =
0
⇒ 2 ( 3bx 2 − 2ax + 3b )   =
0

⇒ 3bx 2 − 2ax + 3b   =

0
Hence proved.

+1 17
x 4   
15. Using the properties of proportion, solve for x, given = .
2 x2 8
x 4 +1 17
Ans: Here, we have =
2x2 8
Applying componendo and dividendo rule in above equation
x 4 +1   
+ 2 x 2 17  
+8
⇒ 4 =
x +1   − 2 x 17 − 8
2

(x +1)
2 2
25
⇒ =
(x −1)
2
2
9
2 2
 x 2 +1   5 
⇒ 2 = 
− 1   3 
 x   
x 2 +1 5
⇒ 2 =
x   
−1 3
⇒ 5 x 2 − 5= 3 x 2 + 3
⇒ 5 x 2 − 3 x 2 =+
3 5
⇒ 2x2 = 8
8
⇒ x2 =
2
⇒ x2 = 4

⇒ x= 4
⇒ x = ±2
Hence value of x is ± 2 .

Class X Mathematics www.vedantu.com 62

 m  n
+   + m  n
−
16. If = , express n in terms of x and m .
m  n
+   − m  n
−

m        
+n + m−n
Ans: Here, we have x =
m        
+n − m−n
Applying componendo and dividendo rule in above equation

x +1               
m+n + m−n + m+n − m−n
⇒ =
x −1 m        
+ n + m − n − ( m         
+n − m−n )

x +1 2 m    
+n
⇒ =
x −1 m                
+n + m−n − m+n + m−n

x +1 2 m    
+n
⇒ =
x −1 2     
m−n

x +1    
m+n
⇒ =
x −1      
m−n
Now, squaring both sides, we get
2
  m+n 
2
 x +1    
⇒  = 
 x −1        
m−n 

x 2 + 2 x +1   m+n
⇒ 2 =
x − 2 x +1    m−n
Applying componendo and dividendo rule in above equation
x 2 + 2 x +1   
+ x 2 − 2 x +1      
m+n +m−n
⇒ 2 =
− ( x 2 − 2 x +1     
x + 2 x +1     ) m + n − (m − n)
2x2 + 2 2m
⇒ =
x 2 + 2 x +1    
− x 2 + 2 x −1     
m+n − m+n
2 x 2 + 2 x 2m
⇒ =
4x 2n
2( x 2 + 1) m
⇒ =
4x n

Class X Mathematics www.vedantu.com 63

 ( x 2 + 1) m
⇒ =
2x n
2mx
⇒n=
( x 2 + 1)
2mx
Hence value of n is .
( x 2 + 1)

+ 3 2 m 3   mn
x 3   xy +3 2
17. If = , show that nx = my .
3 x 2 y  y
+ 3 3m 2 n  n
+ 3

   x 3 + 3 xy   2
m3 + 3mn 2
Ans: Here, we have =
3 x 2 y   + y 3 3m 2 n   + n3
Applying componendo and dividendo rule in above equation
x  3 + 3 xy 2 + 3 x 2 y     
+ y3 m3 + 3mn 2 + 3m 2  
n + n3
⇒ 3 =
x    + 3 xy 2 − ( 3 x 2 y     
+ y 3 ) m3 + 3mn 2 − ( 3m 2 n   + n3 )

x 3    
+ y 3 + 3 x 2 y    + 3 xy 2       
m3 + n3 + 3m 2 n    + 3mn 2
⇒ 3 =
  
x + 3 xy 2 − 3 x 2 y −     y 3 m3 + 3mn 2 − 3m 2 n − n3

x 3    
+ y 3 + 3 x 2 y    + 3 xy 2       
m3 + n3 + 3m 2 n    + 3mn 2
⇒ =
x 3    − y 3 −3 x 2 y + 3 xy 2 m3 − n3 − 3m   2
n + 3mn 2

+ y 3 + 3 xy ( x         
x 3     ( m + n)
+ y ) m3 + n3 + 3mn     
⇒ 3 3 = 3 3
x    
− y −3 xy      ( m − n)
( x − y ) m − n − 3mn     
( x     
+ y ) (m  n)
3 3

⇒ =
( x     
− y ) (m − n)
3 3

3 3
 x     
+ y   m+n 
⇒ = 
− y   m − n 
 x     
x     
+ y m+n
⇒ =
x     
− y m−n
Applying componendo and dividendo rule in above equation

Class X Mathematics www.vedantu.com 64

 x          
+ y+ x−y m+n+m − n
⇒ =
+ y − ( x − y ) m + n − (m − n)
x         

2x 2m
⇒ =
x           
+ y − x+ y m+n − m+n
2 x 2m
⇒ =
2 y 2n
x m
⇒ =
y n
⇒ nx = my

Hence proved.

Exercise- 7(D)

1. If : b = 3: 5 , find (10a + 3b ) : ( 5a + 2b ) .

Ans: Here, we have   :

a b = 3 :5
a 3
Therefore, = ……… (i)
b 5
10
Now, multiplying by in the above equation, we get
3
10  a  10  3 
⇒  =  
3 b  3 5
10a
⇒ =2
3b
Adding 1 to both sides, we get
10a
⇒ +1= 2 +1
3b
10a + 3b
⇒ =3
3b

Class X Mathematics www.vedantu.com 65

 9b ……………… eq (ii)
⇒ 10a + 3b =
5
Now, multiplying by in the above equation, we get
2
5 a  53
⇒  =  
2 b  25
5a 3
⇒ =
2b 2
Adding 1 to both sides, we get
5a 3
⇒ +1 = +1
2b 2
5a + 2b 3  2
+
⇒ =
2b 2
5a + 2b
⇒ =5
b
5b ……………. eq (iii)
⇒ 5a + 2b =
Now, divide eq (i) by (ii), we get
10a   3
+ b 9b
⇒ =
5a   + 2b 5b
10a   3
+ b 9
⇒ =
5a   + 2b 5
Hence value of (10a + 3b ) : ( 5a + 2b ) is 9 : 5 .

2. If ( 5 x + 6 y ) : ( 8 x + 5 y ) =
8: 9 , find x: y .

Ans: Here, we have ( 5 x + 6 y ) : ( 8 x + 5 y ) =

8 :9

Therefore,
(5x + 6 y ) = 8
(8x + 5 y ) 9
⇒ 8 (8x + 5 y ) = 9 ( 5x + 6 y )

⇒ 64 x + 40 y =45 x + 54 y

Class X Mathematics www.vedantu.com 66

⇒ 64 x − 45 x = 54 y − 40 y

⇒ 19 x = 14 y

x 14
⇒ =
y 19
⇒ x : y = 14 :19

Hence, value of x : y is 14 :19 .

3. If ( 3 x − 4 y ) : ( 2 x − 3 y ) = ( 5 x − 6 y ) : ( 4 x − 5 y ) . find x: y

Ans: Here, we have ( 3 x − 4 y ) : ( 2 x − 3 y ) = ( 5 x − 6 y ) : ( 4 x − 5 y )

Therefore,
( 3x − 4 y ) = ( 5 x − 6 y )
( 2x − 3y ) ( 4x − 5 y )
⇒ ( 3 x − 4 y )( 4 x − 5 y ) = ( 5 x − 6 y )( 2 x − 3 y )

⇒ 12 x 2 − 15 xy − 16 xy + 20 y 2 = 10 x 2 − 15 xy − 12 xy + 18 y 2

⇒ 12 x 2 − 31xy + 20 y 2 = 10 x 2 − 27 xy + 18 y 2

⇒ 2 x 2 − 4 xy + 2 y 2 =
0

⇒ 2 ( x 2 − 2 xy + y 2 ) =
0

⇒ ( x 2 − 2 xy + y 2 ) =
0

⇒ ( x − y) =
2
0

0
⇒ x− y=

⇒ x= y

x
⇒ =1
y
⇒ x : y = 1 :1

Hence, value of x : y is 1  :1 .

Class X Mathematics www.vedantu.com 67

4. Find the:

(a) Duplicate ratio of 2 2 : 3 5

Ans: We know that duplicate ratio of a : b is a 2 : b 2

( ) ( )
2 2
Therefore, duplicate ratio of 2 2 : 3 5 is 2 2 : 3 5

( ) ( )
2 2
= 2 2 : 3 5

( )
2
2 2
=
(3 5 )
2

4  2
×
=
9  5
×
8
=
45
= 8 : 45

Hence, duplicate ratio of 2 2 : 3 5 is 8 : 45.

(b) Triplicate ratio of 2a: 3b.

Ans: We know that triplicate ratio of a : b is a 3 : b3

Therefore, triplicate ratio of 2a : 3b is ( 2a ) : ( 3b )

3 3

= ( 2a ) : ( 3b )
3 3

( 2a )
3

=
( 3b )
3

8 a3
=
27b3
= 8a 3 : 27b3
Hence, triplicate ratio of 2a : 3b is 8a 3 : 27b3 .

Class X Mathematics www.vedantu.com 68

(c) Sub – duplicate ratio of 9 x 2a 4 : 25 y 6b 2 .

Ans: We know that sub - duplicate ratio of a : b is a: b

Therefore, sub - duplicate ratio of 9 x 2 a 4 : 25 y 6b 2 is 9 x 2 a 4 : 25 y 6b 2

= 9 x 2 a 4 : 25 y 6b 2

9 x2a 4
=
25 y 6b 2

3 xa 2
= 3
5y b

= 3 xa 2 : 5 y 3b

Hence, sub- duplicate ratio of 9 x 2 a 4 : 25 y 6b 2 is 3 xa 2 : 5 y 3b.

(d) Sub – triplicate ratio of 216: 343.

Ans: We know that sub - triplicate ratio of a : b is 3

a :3b

Therefore, sub - triplicate ratio of 216 : 343 is 3

216 : 3 343

= 3 216 : 3 343
3
216
= 3
343
3
6  6  6
× ×
= 3
7  7  7
× ×
3
63
= 3
73
6
=
7
= 6 :7
Hence, sub - triplicate ratio of 216 : 343 is 6 : 7.

Class X Mathematics www.vedantu.com 69

(e) Reciprocal ratio of 3: 5.
1 1
Ans: We know that reciprocal ratio of a : b is : .
a b
1 1
Therefore, reciprocal ratio of 3 : 5 is : .
3 5
1 1
= :
3 5
1
=3
1
5
5
=
3
= 5 :3
Hence, reciprocal ratio of 3 : 5 is 5 : 3.

5. Ratio compounded of the duplicate ratio of 5:6, the reciprocal ratio of

25: 42 and the sub – duplicate ratio of 36: 49.
Ans: The duplicate ratio of 5 : 6 is 52 : 62
52
= 2
6
25
=
36
And the reciprocal ratio of 25 : 42 is 42 : 25
42
=
25

Now, sub – duplicate ratio of 36 : 49 is 36 : 49

36
=
49

Class X Mathematics www.vedantu.com 70

 62
=
72
6
=
7
25 42 6
Thus, the compound ratio of    ,    and will be,
36 25 7
25 × 42 × 6
=
36 × 25 × 7
6×6
=
36
36
=
36
=1
Hence, the required compound ratio is 1 :1 .

6. Find the value of x , if:

(a) ( 2 x + 3) : ( 5 x − 38 ) is the duplicate ratio of 5  : 6 .

Ans: We have ( 2 x + 3) : ( 5 x − 38 ) is the duplicate ratio of 5  : 6

Therefore,

( 2 x + 3) = ( )
2
5
⇒
( 5 x − 38) ( 6)
2

⇒
( 2 x + 3) = 5
( 5 x − 38) 6
⇒ 5 ( 5 x − 38 ) = 6 ( 2 x + 3)

⇒ 25 x − 190 = 12 x + 18
⇒ 25 x − 12 x =18 + 190
⇒ 13 x = 208

Class X Mathematics www.vedantu.com 71

 208
⇒ x=
13
⇒ x = 16
Hence, value of x is 16.

(b) ( 2 x + 1) : ( 3 x + 13) is the sub - duplicate ratio of 9: 25.

Ans: We have ( 2 x + 1) : ( 3 x + 13) is the sub - duplicate ratio of 9 : 25

Therefore,

⇒
( 2 x + 1) = 9
( 3x + 13) 25

⇒
( 2 x + 1) = 32
( 3x + 13) 52

⇒
( 2 x + 1) = 3
( 3x + 13) 5
) 3( 3x + 13)
⇒ 5 ( 2 x + 1=

⇒ 10 x + 5 = 9 x + 39
⇒ x = 34
Hence, value of x is 34.

(c) ( 3 x − 7 ) : ( 4 x + 3) is the sub - triplicate ratio of 8: 27.

Ans: We have ( 3 x − 7 ) : ( 4 x + 3) is the sub - triplicate ratio of 8 : 27

Therefore,

⇒
( 3x − 7 ) = 3
8
( 4 x + 3) 3
27

⇒
( 3x − 7 ) = 3
23
( 4 x + 3) 3
33

Class X Mathematics www.vedantu.com 72

⇒
( 3x − 7 ) = 2
( 4 x + 3) 3
⇒ 3 ( 3 x − 7=
) 2 ( 4 x + 3)
⇒ 9 x − 21 = 8 x + 6
⇒ 9 x − 8 x =6 + 21
⇒ x = 27
Hence, value of x is 27.

7. What quantity must be added to each term of the ratio x: y so that it may
become equal to c: d .
Ans: Let the quantity must be added is p .

x   
+p c
Therefore, =
y   
+p d
⇒ dx + dp = cy + cp

⇒ dp − cp = cy − dx

⇒ p ( d − c ) = cy − dx

cy − dx
⇒ p=
d −c
cy − dx
Hence, required quantity is .
d −c

8. A woman reduces her weight in the ratio 7:5. What does her weight
become if originally it was 84 kg?
Ans: Let the reduced weight of the woman be x kg
84 7
Therefore, =
x 5
⇒ 7=
x 84 × 5

Class X Mathematics www.vedantu.com 73

 84  5
×
⇒ x=
7
x 12 × 5
⇒ =
⇒ x = 60 kg
Hence, the reduced weight of the woman is 60 kg.

9. If 15 ( 2 x 2 − y 2 ) =
7 xy, find x: y; if x and y both are positive.

Ans: Here, we have 15 ( 2 x 2 − y 2 ) =

7 xy

⇒ 30 x 2 − 15 y 2 =
7 xy

Divide by y 2 in the above equation,

30 x 2 15 y 2 7 xy
⇒ − 2 =
   
y2 y y2
2
x x
⇒ 30   − 15 =
7 
 y  y
x
Now, let a =
y

⇒ 30 a 2 − 15 =
7a
⇒ 30 a 2 − 7 a − 15 =
0
⇒ 30 a 2 − 25a + 18a − 15 =
0
⇒ 5a ( 6 a − 5 ) + 3 ( 6 a − 5 ) =
0

⇒ ( 6a − 5 )( 5a + 3) =
0

⇒ ( 6a −=
5 ) 0  or ( 5a +=
3) 0

5 3
⇒ a=    
or a = −
6 5
x
Now, put a = , we get
y

Class X Mathematics www.vedantu.com 74

 x 5 x 3
⇒ =   
or = −
y 6 y 5
⇒ x : y = 5 : 6   :
or x y = −3 : 5

Since, x and y Both are positive. Thus, value of x : y will be positive.

Hence, Value of x : y is 5 : 6.

10. Find the:

(a) Fourth proportional to 2 xy,   x 2 and y 2 .

Ans: Let the fourth proportional be z.

Therefore, 2 xy, x 2 , y 2 and z are in continued proportion.

⇒ 2 xy :    :
x2 = y 2 z

2xy y 2
⇒ 2 =
x z
2 y
⇒ =
x z
xy
⇒ z=
2
xy
Hence, the required fourth proportion is .
2

(b) Third proportional to a 2 − b 2 and a + b .

Ans: Let the third proportional be x.
Therefore, a 2 − b 2 , a + b and x are in continued proportion.

⇒ a 2 − b 2 :    :
a + b =a + b x
a 2 − b 2 a   
+b
⇒ =
a   
+b x

Class X Mathematics www.vedantu.com 75

 ( a   
+ b)
2

⇒ x=
a 2 − b2

⇒ x=
( a + b )( a + b ) [ a 2 − b 2 = ( a + b )( a − b )]
( a + b )( a − b )

⇒ x=
(a + b)
(a − b)
a+b 
Hence, the required third proportion is  .
 a −b 

(c) Mean proportional to ( x − y ) and ( x 3 − x 2 y ).

Ans: Let the third proportional be x.

Therefore, a 2 − b 2 , a + b and x are in continued proportion.

⇒ a 2 − b 2 :    :
a + b =a + b x
a 2 − b 2 a   
+b
⇒ =
a   
+b x

( a   
+ b)
2

⇒ x=
a 2 − b2

⇒ x=
( a + b )( a + b ) [ a 2 − b 2 = ( a + b )( a − b )]
( a + b )( a − b )

⇒ x=
(a + b)
(a − b)
a+b 
Hence, the required third proportion is  .
 a −b 

11. Find two numbers such that the mean proportional between them is 14
and third proportional to them 112.
Ans: Let two required numbers be x and y .

Class X Mathematics www.vedantu.com 76

Now, according to the 1st condition,
The mean proportional between x and y is 14

⇒ xy = 14

Squaring both sides,

⇒ xy = 196

196
⇒ x= ………….. eq (i)
y
Now, according to 2nd condition,
Third proportional of x and y is 112.

Therefore,
x y
⇒ =
y 112

⇒ y 2 = 112 x

196
y 2 112 ×
⇒=
y

y 3 112 × 196
⇒=

⇒ y3 = 7 × 4 × 4 × 7 × 7 × 4

⇒ y 3 = 73.43

⇒ y 3 = ( 28 )
3

⇒ y = 28

Now, put the value of y in equation (i),

196
⇒ x=
28
⇒ x=7
Hence, two required numbers are 7 and 28.

Class X Mathematics www.vedantu.com 77

12. If x and y . be unequal and x: y is the duplicate ratio of x + z and y+z,
prove that z is mean proportional between x and y.
Ans: We have x : y is the duplicate ratio of ( x + z ) and ( y + z ) .

Therefore,

x ( x + z)
2

⇒ =
y ( y + z)
2

⇒ x( y + z) = y( x + z)
2 2

⇒ x ( y 2 + 2 yz + z 2 ) = y ( x 2 + 2 xz + z 2 )

⇒ xy 2 + 2 xyz + xz 2 =yx 2 + 2 xyz + yz 2

⇒ xy 2 + xz 2 = yx 2 + yz 2

⇒ xz 2 − yz 2 = yx 2 − xy 2

⇒ z 2 ( x − y )= xy ( x − y )

⇒ z 2 = xy

⇒ z = xy

Thus, z is mean proportional between x and y.

Hence proved.

2ab x  a
+ x  b
+
13. If x = , find the value of + .
a  b
+ x  a
− x  b
−
2ab
Ans: Here, we have x = ……… eq (i)
a+b
Now, dividing by a in the above equation, we get
x 2ab
⇒ =
a a (a + b)

x 2b
⇒ =
a (a + b)

Class X Mathematics www.vedantu.com 78

Applying componendo and dividendo rule
x   
+a 2b + a + b
⇒ =
− a 2b − ( a + b )
x   

x   
+a 3b + a
⇒ =
x   
− a 2b − a − b
x   
+ a 3b + a
⇒ = ……….. eq (ii)
x   
−a b − a
Now, dividing by b in eq (i), we get
x 2ab
⇒ =
b b(a + b)

x 2a
⇒ =
b (a + b)

Applying componendo and dividendo rule

x +b 2a + a + b
⇒ =
x − b 2a − ( a + b )

x +b 3a + b
⇒ =
x − b 2a − a − b
x + b 3a + b
⇒ = ……… eq (iii)
x −b a − b
Adding eq (ii) and (iii)
x +   
a x + b 3b + a 3a + b
⇒ + = +
x −   
a x −b b − a a −b
3b + a 3a + b
= −
b−a b−a
3b + a − 3a − b
=
b−a
2b − 2a
=
b−a

Class X Mathematics www.vedantu.com 79

 2(b − a )
=
(b − a )
=2
x +   
a x +b
Hence value of + is 2.
x −   
a x −b

14. If ( 4a + 9b )( 4c − 9d ) = ( 4a − 9b )( 4c + 9d ) , prove that a: b = c: d .

Ans: Here, we have ( 4a + 9b )( 4c − 9d ) = ( 4a − 9b )( 4c + 9d )

Therefore,

⇒
( 4a + 9b ) = ( 4c + 9d )
( 4a − 9b ) ( 4c − 9d )
Applying componendo and dividendo rule

⇒
( 4a + 9b ) + ( 4a − 9b ) = ( 4c + 9d ) + ( 4c − 9d )
( 4a + 9b ) − ( 4a − 9b ) ( 4c   + 9d ) − ( 4c − 9d )
4a + 9b    4
+ a − 9b 4c   9+ d   4+ c − 9d
⇒ =
4a + 9b   − 4a   + 9b 4c   + 9d − 4c   + 9d
8a 8c
⇒ =
18b 18d
a c
⇒ =
b d
⇒ a :b = c :d
Hence proved.

a c
15. If = , show that ( a + b ) : ( c + d ) = : c2 + d 2 .
a 2 + b 2   
b d
a c
Ans: Here, we have = ……… eq (i)
b d
Adding 1 to both sides in the above equation,

Class X Mathematics www.vedantu.com 80

 a c
⇒ +1= +1
b d
a     
+b c+d
⇒ =
b d
a   
+b b
⇒ = ……… eq (ii)
c   
+d d
Now, squaring both sides in eq (i), we get
2 2
a  c 
⇒   = 
b d 
a2 c2
⇒ =
b2 d 2
Adding 1 to both sides in the above equation,
a2 c2
⇒ 2 + 1= 2 + 1
b d
a 2 +   
b2 c2 + d 2
⇒ =
b2 d2
a 2 + b2 b2
⇒ 2 =
c +d2 d2
Taking square root to both sides, we get,

a 2 + b2 b
⇒ =
c2 + d 2 d

a 2 + b2 b
⇒ = ……….. eq (iii)
c 2 +   d 2 d
Now, from eq (ii) and (iii),

a    
+b a 2 + b2
⇒ = 2
c   
+d c +   d 2

( a + b ) ( c + d ) =   :
⇒   : a 2 + b2 c2 + d 2

Hence proved.

Class X Mathematics www.vedantu.com 81

16. There are 36 members in a student council in a school and the ratio of
the number of boys to the number of girls is 3:1. How many more girls should
be added to the council so that the ratio of number of boys to the number of
girls may be 9:5?
Ans: here, we have the ratio of the number of boys to the number of girls is 3 :1
.
Let the number of boys be 3 x.
And number of girls be x.
Now, according to question
⇒ Total no. of members in council = 36
⇒ 3x + x =
36
⇒ 4 x = 36
36
⇒ x=
4
⇒ x=9
Thus, the number of boys = 3 × 9 = 27
And the number of girls = 9
Let y more girls should be added to get required ratio 9 : 5. .

27 9
⇒ =
9  
+y 5
⇒ 81 + 9 y =
135

⇒ 9=
y 135 − 81

⇒ 9 y = 54

54
⇒ y=
9
⇒ y=6

Thus, 6 girls should be added.

Class X Mathematics www.vedantu.com 82

7. If 7 x − 15 y =4 x + y , find the value of x: y . Hence, use componendo and
dividendo to find the values of:
9 x  y
+5
(a)
9 x −5 y
Ans: here, we have 7 x − 15 y =4 x + y .

⇒ 7 x − 4 x =y + 15 y

⇒ 3 x = 16 y

x 16
⇒ = …………. Eq (i)
y 3
Thus, value of x : y is 16 : 3 .

9
Now, multiplying by   in eq (i), we get
5

9  x  9  16 
⇒ =  
5  y  5  3 

9 x 48
⇒ =
5y 5
Applying componendo and dividendo rule
9 x   5
+ y 48  5
+
⇒ =
9 x   5
− y 48 −5
9 x   5
+ y 53
⇒ =
9 x   5
− y 43
9 x   5
+ y 53
Hence, value of      is .
9 x   5
− y 43

3x2 + 2 y2
(b)
3x2 − 2 y2
x 16
Ans: here, we have = …………. Eq (i)
y 3

Class X Mathematics www.vedantu.com 83

Now, squaring both sides in eq (i),
2 2
 x   16 
⇒   = 
 y  3 
x 2 256
⇒ =
y2 9
3
Multiplying by in the above equation, we get
2
3  x 2  3  256 
⇒  2 =  
2 y  2 9 

3 x 2 128
⇒ =
2 y2 3
Applying componendo and dividendo rule
3 x 2   
+ 2 y 2 128  +3
⇒ 2 =
3 x   −2y 2
128  
−3

3 x 2   
+ 2 y 2 131
⇒ 2 =
3 x   2
− y 2 125

3 x 2   
+ 2 y 2 131
Hence, Value of   
is .
3 x 2   2
− y 2 125

4m  n
+3 7
18. If = , use properties of proportion to find:
4m  n
−3 4
(a) m : n
4m   + 3n 7
Ans: Here, we have =
4m   − 3n 4
⇒ 7 ( 4m − 3n )= 4 ( 4m + 3n )

⇒ 28m − 21n = 16m + 12n

⇒ 28m − 16m =12n + 21n
⇒ 12m = 33n

Class X Mathematics www.vedantu.com 84

 m 33
⇒ =
n 12
m 11
⇒ =
n 4
Hence, value of m : n is 11 : 4 .

2m 2 − 11n 2
(b)
2m 2 + 11n 2
m 11
Ans: Here, we have =
n 4
Squaring both sides, we get
2 2
 m   11 
⇒   = 
n 4
m 2 121
⇒ 2 =
n 16
2
Multiplying by    in the above equation, we get
11
2  m 2  2  121 
⇒  2 =  
11  n  11  16 

2m 2 11
⇒ =
11n 2 8
Applying componendo and dividendo rule
2m 2 +11n 2 11  
+8
⇒ =
2m −11n 11  
2 2
−8
2m 2 +11n 2 19
⇒ =
2m 2 −11n 2 3
2m 2 −11n 2 3
⇒ =
2m 2 +11n 2 19

Class X Mathematics www.vedantu.com 85

 2m 2 −11n 2 3
Hence, Value of is .
  
2m 2 +11n 2 19

( x + y)
2
x
19. If x , y, z are in continued proportion, prove that = .
( y+z )
2
z

Ans: Here, x, y, z are in continued proportion

Therefore,
x y
⇒ =
y z

⇒ y 2 = xz ………….. eq (i)

( x + y)
2

Now, L.H.S =
( y +z)
2

x 2   + 2 xy   + y 2
= 2
y   + 2 yz   + z 2

x 2   + 2 xy   + xz
=    [from eq (i)]
xz   + 2 yz   + z 2

x ( x   2 + z)
+ y   
=
( x + y   
z    2 + z)

x
=
z
= R.H.S
Thus, L.H.S = R.H.S
Hence proved.

a 2 + b2 + a 2 − b2
20. Given, x = . Use componendo and dividendo to prove
a 2 + b2 − a 2 − b2
2a 2 x
that b 2 = .
+1
x 2   

Class X Mathematics www.vedantu.com 86

 a 2 + b2 + a 2 − b2
Ans: Here, we have x =
a 2 + b2 − a 2 − b2
Applying componendo and dividendo rule

x +1    a 2 + b2 + a 2 − b2 + a 2 + b2 − a 2 − b2
⇒ =
x −1 a 2 + b 2 +      (
a 2 − b2 − a 2 + b2 − a 2 − b2 )
x +1 2 a 2 + b2
⇒ =
x −1 a 2 + b 2 +      
a 2 − b2 − a 2 + b2 + a 2 − b2

x +1 2 a 2 + b 2
⇒ =
x −1 2  
a 2 − b2

x +1 a 2 + b2
⇒ =
x −1 a 2 − b2
Squaring both sides, we get
2
 x +1   a + b 
2 2 2

⇒  = 
 x −1   a 2 − b 2 
x 2   
+ 2 x +1 a 2 + b 2
⇒ 2 =
x − 2 x +1 a 2 − b 2
Applying componendo and dividendo rule
x 2   
+ 2 x +1   
+ x 2 − 2 x +1    a 2 + b2 + a 2 − b2
⇒ 2 =
x  2 x +1    ) a 2 + b2 − ( a 2 − b2 )
− ( x 2 − 2 x +1  

2x2 + 2 2a 2
⇒ 2 = 2 2
x +   2 x +1   
− x 2 + 2 x −1    
a + b − a 2 + b2
2 x 2 + 2 2a 2
⇒ = 2
4x 2b
2 ( x 2 +1) a2
⇒ =
4x b2

⇒
(x 2
+1) a2
= 2
2x b

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 2a 2 x
⇒ b2 =
( x 2 +1)
Hence proved.

x2 + y2 1
21. If 2 = 2 , find:
x −y 2
8
x
(a)
y

x2 + y 2 1
Ans: Here, we have 2 =2
x −y 2
8

x 2 + y 2 17
⇒ 2 =
x − y2 8

( x 2 − y 2 ) = 8 ( x 2 +   
⇒ 17   y2 )

⇒ 17 x 2 − 17 y 2 =8 x 2 + 8 y 2

⇒ 17 x 2 − 8 x 2 = 8 y 2 + 17 y 2

⇒ 9 x 2 = 25 y 2

x 2 25
⇒ =
y2 9
2 2
 x  5
⇒   = 
 y  3
x 5
⇒ = ±
y 3
x 5
Hence, Value of   
is ± .
y 3

x 3   + y 3
(b) 3
x   − y 3

Class X Mathematics www.vedantu.com 88

 x 5
Ans: Here, we have = from part (i)
y 3
Now, cubing both sides in above equation
3 3
 x  5
⇒   = 
 y  3
x 3 125
⇒ 3=
y 27
Applying componendo and dividendo rule
x 3    
+ y 3 125  
+ 27
⇒ 3 3 =
x   − y 125 − 27

x 3    
+ y 3 152
⇒ 3 3 =
x   −y 98

x 3    
+ y 3 76
⇒ 3 3 =
x   −y 49

x 3    
+ y 3 76
Hence, Value of 3 3    is .
x   
− y 49

22. Using componendo and dividendo, find the value of x :

3x + 4 + 3x − 5
=9 .
3x + 4 − 3x − 5

3x + 4 + 3x − 5 9
Ans: Here, we have =
3x + 4 − 3x − 5 1
Applying componendo and dividendo rule in above equation

3 x + 4 + 3 x − 5    + 3 x + 4 − 3 x − 5 9 +1
⇒ =
3x + 4 + 3x − 5 − ( 3x + 4 − 3x − 5 ) 9 −1

2 3x + 4 10
⇒ =
3 x + 4 + 3 x − 5 − 3 x + 4    + 3 x − 5 8

Class X Mathematics www.vedantu.com 89

 2 3x + 4 5
⇒ =
2 3x − 5 4

3x + 4 5
⇒ =
3x − 5 4
Squaring both sides, we get
2
 3x + 4   5 2
⇒  = 
 3 x − 5  4
3 x + 4 25
⇒ =
3 x − 5 16
⇒ 25 ( 3 x − 5=
) 16 ( 3x + 4  
)
⇒ 75 x − 125 = 48 x + 64
⇒ 75 x − 48 x =64 + 125
⇒ 27 x = 189
189
⇒ x=
27
⇒ x=7
Hence, Value of x is 7.

+1 + a   
a     −1
23. If x = , using properties of proportion show that:
+ 1 − −1
a    a  
x 2 − 2ax + 1 =0.

a +1    
+ a −1  
Ans: Here, we have x =
a +1    
− a −1
Applying componendo and dividendo rule in above equation

x +1   a +1    
+ a −1    
+ a +1    
− a −1  
⇒ =
x −1 a +1    
+ a −1   (
− a +1    
− a −1 )

Class X Mathematics www.vedantu.com 90

 x +1 2 a +1   
⇒ =
x −1 a +1    
+ a −1  
− a +1   
+ a −1

x +1 2 a +1   
⇒ =
x −1 2  
a –1

x +1  a +1   
⇒ =
x −1 a −1
Squaring both sides, we get
2
  a +1   

2
 x +1  
⇒  =  
 x −1   a −1 

x 2   
+ 2 x +1   a +1
⇒ =
x 2 − 2 x +1   a −1
Applying componendo and dividendo rule
x 2   
+ 2 x +1   
+ x 2 − 2 x +1   a +1   
+ a −1
⇒ =
− ( x 2 − 2 x +1  
x 2  2 x +1    ) a +1 − ( a −1)
2x2 + 2 2a
⇒ 2 =
x +   2 x +1   
− x + 2 x −1  
2
a +1 − a +1
2 x 2 + 2 2a
⇒ =
4x 2
2 ( x 2 +1)
⇒ =a
4x

⇒
(x 2
+1)
=a
2x
⇒ x2 + 1 =2ax
⇒ x 2 − 2ax + 1 =0
Hence proved.

x 3 + 12 x y 3   
+ 27 y
24. Given = 2 . Using componendo and dividendo, find x: y.
6 x   
2
+ 8 9 y   
+ 27

Class X Mathematics www.vedantu.com 91

 x 3 + 12 x y 3   27
+ y
Ans: Here, we have =
6 x 2   
+ 8 9 y 2   
+ 27
Applying componendo and dividendo rule in above equation
x 3 + 12 x   
+ 6 x 2   
+8 y 3   
+ 27 y   
+ 9 y 2   
+ 27
⇒ 3 = 3
x + 12 x − ( 6 x   
2
+ 27 y − ( 9 y   
+ 8 ) y    2
+ 27 )

x 3 + 12 x   
+ 6 x 2   
+ 8 y 3   
+ 27 y   
+ 9 y 2   
+ 27
⇒ 3 =
x + 12 x − 6 x 2   8
− y 3   27
+ y −9 y 2 − 27

x 3 + 8 + 6 x 2 +12 x    27  9
y 3 + + y 2 + 27 y
⇒ 3 =
x − 8 − 6 x 2 +12 x y 3 − 27  9
− y 2 + 27 y

⇒ 3
( x + 2 )    
x 3 +   23 + 6 x     y 3 + 33    
= 3 3
( y + 3)
+ 9 y   
x −   2 − 6 x   
3
( x − 2 ) y − 3 −   9 y    ( y −3)
( y + 3)
( x + 2 ) =  
3 3
  
⇒
( x − 2 ) ( y − 3)
3 3

3 3
  
 x + 2    
 y +3 
⇒  = 
 x − 2   y −3 

x   + 2 y   + 3
⇒ =
x − 2 y −3
Applying componendo and dividendo rule in above equation
x   
+ 2    
+ x −2 y   
+ 3   
+ y −3
⇒ =
  x + 2 − ( x − 2 ) y − 3 − ( y − 3)

2x 2y
⇒ =
  x + 2 −  x + 2 y − 3 − y  3
2x 2 y
⇒ =
4 6
x y
⇒ =
2 3
x 2
⇒ =
y 3

Class X Mathematics www.vedantu.com 92

Hence, Value of x : y is 2 : 3 .

x y z x 3 y 3 z 3 3 xyz
25. If = = , show that 3 + 3 + 3 = .
a b c a b c abc
x y z
Ans: Here, we have = = ………. Eq(i)
a b c
x3 y 3 z 3
Now, L.H.S = 3 + 3 + 3
a b c
3 3 3
x  y z
=  +  + 
a b c
3 3 3
x x  z
=  +  +  [from eq (i) ]
a a a
x3 x3 x3
= + +
a3 a3 a3
x3
= 3× 3
a
x x x
=3 × × ×
a a a
x y z
=3 × × × [from eq (i)]
a b c
3xyz
=
abc
= R.H.S
Thus, L.H.S = R.H.S
Hence proved.

26. If b is the mean proportion between a and c , show that:

a 4 + a 2b 2 + b 4 a 2
= .
b 4 + b 2c 2 + c 4 c 2

Class X Mathematics www.vedantu.com 93

Ans: Here, we have b is the mean proportion between a and c
Therefore,
⇒ b 2 = ac
Now, squaring both sides
⇒ b 4 = a 2c 2
a 4 + a 2b 2 + b 4
Thus, L.H.S   
= 4
b + b 2c 2 + c 4
a 4 + a 2b 2 + a 2 c 2
= 2 2 [ b 4 = a 2c 2 ]
a c + b 2c 2 + c 4
a 2 ( a 2 +b 2 + c 2 )
=
c2 ( a 2 + b2 + c2 )

a2
= 2
b
= R.H.S
Thus, L.H.S = R.H.S
Hence proved.

7 m  n
+2 5
27. If    = , use properties of proportion to find:
7 m  n
−2 3
(a) m : n
7 m   + 2n 5
Ans: Here, we have =
7 m   − 2n 3
Applying componendo and dividendo rule in above equation
7 m   
+ 2n   
+ 7 m   − 2n 5  
+3
⇒ =
7 m + 2n − ( 7 m   2
− n ) 5 −3

14m 8
⇒ =
7 m + 2n − 7 m   + 2n 2

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 14m
⇒ =4
4n
7m
⇒ =4
2n
m 8
⇒ =
n 7
Hence, Value of m : n is 8 : 7 .

m 2   + n 2
(b) 2
m − n2
m 8
Ans: Here, we have = from the part (i)
n 7
Now, squaring both sides in above equation,
2 2
m 8
⇒   = 
 n  7
m 2 64
⇒ 2 =
n 49
Applying componendo and dividendo rule in above equation
m 2   
+ n 2 64  
+ 49
⇒ 2 2=
m   −n 64  
− 49
m 2   
+ n 2 113
⇒ 2 2=
m   −n 15
m 2   
+ n 2 113
Hence, Value of     is .
m 2   
− n 2 15

28. (i) If x and y both are positive and ( 2 x 2 − 5 y 2 ) : xy =

1: 3, find x: y .

Ans: Here, we have ( 2 x 2 − 5 y 2 ) : xy =

1 :3

Therefore,

Class X Mathematics www.vedantu.com 95

⇒
( 2x 2
− 5 y2 )
=
1
xy 3

⇒ 6 x 2 − 15 y 2 =
xy

Divide by y 2 in above equation,

6 x 2 15 y 2 xy
⇒ 2 − 2 =
   
y y y2
2
x x
⇒ 6   − 15 =
 y
 y  
x
Now, let a =
y

⇒ 6a 2 − 15 =
a
⇒ 6a 2 − a − 15 =0
⇒ 6a 2 − 10a + 9a − 15 =
0
⇒ 2a ( 3a − 5 ) + 3 ( 3a − 5 ) =
0

⇒ ( 3a − 5 )( 2a + 3) =
0

⇒ ( 3a −=
5 ) 0  2
or ( a +=
3) 0

5 3
⇒ a=    
or a = −
3 2
x
Now, put a = , we get
y
x 5 x 3
⇒ =   
or = −
y 3 y 2
⇒ x : y = 5 : 3   :
or x y = −3 : 2

Since, x and y both are positive. Thus, value of x : y will be positive.

Hence, Value of x : y is 5 : 3.

Class X Mathematics www.vedantu.com 96

 3
 a  x
−  a   x+
(ii). Find x , if 16   = .
 a  x
+  a  x
−
3
 a     
−x  a+ x
Ans: Here, we have 16   =
 a     
+ x  a−x
4
a+x
⇒  = 16
a−x
4
a+x
 =2
4
⇒
a−x
a+x
⇒ = ±2
a−x
a+x
Case 1: When =2
a−x
⇒ a + x = 2a − 2 x
⇒ x + 2 x = 2a − a
⇒ 3x = a
a
⇒ x =   
3
a+x
Now, Case 2: = −2
a−x
⇒ −2a + 2 x =a + x
⇒ 2 x − x = a + 2a
⇒ x = 3a
a
Hence, Value of x is or 3a.
3

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