Simultaneous Linear Equations Class-9th Concise Selina ICSE Maths

Simultaneous Linear Equations Class-9th Concise Selina ICSE Maths Solutions Chapter-6 (Including Problems). We provide step by step Solutions of Exercise / lesson-6 Simultaneous Equations (Including Problems)  for ICSE Class-9 Concise Selina Mathematics by RK Bansal.

Our Solutions contain all type Questions with Exe-6 A, Exe-6 B, Exe-6 C, Exe-6 D,  Exe-6 E,  Exe-6 F,  and Exe-6 G to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics .

Simultaneous Linear Equations Class-9th Concise Selina ICSE Maths Solutions Chapter-6 (Including Problems)

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–: Select Topics :–

 

Exe-6 A,

Exe-6 B,

Exe-6 C,

Exe-6 D,

Exe-6 E,

Exe-6 F,

Exe-6 G,

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Simultaneous Linear Equations (Including Problems) Class-9th Concise Selina ICSE Maths Solutions  Exercise – 6(A)

Question 1

Solve the following pair of linear (simultaneous) equation by the method of elimination by substitution :
8x + 5y = 9
3x + 2y = 4

Answer

8x + 5y = 9                                         …(1)
3x + 2y = 4                                         …(2)

8x + 5y = 9
∴ 5y = 9 – 8x

Question 2

Solve the pair of linear (simultaneous) equation by the method of elimination by substitution :
2x – 3y = 7
5x + y= 9

Answer

2x – 3y = 7                                       …(1)
5x + y = 9                                        …(2)

5x + y = 9
∴ y = 9 – 5x                                     …(3)
Putting this value of y in (1)
2x – 3 (9 – 5x) = 7
∴ 2x – 27 + 15x = 7
∴ 2x + 15x = 7 + 27
∴ 17x = 34
∴ x = 2
From (2)
y = 9 – 5(2)
y = -1

Question 3

Solve the pair of linear (simultaneous) equation by the method of elimination by substitution:
2x + 3y = 8
2x = 2 + 3y

Answer

2x + 3y = 8                                       …(1)
2x = 2 + 3y                                       …(2)

2x = 2 + 3y
Putting this value of 2x in (1)
2 + 3y + 3y = 8
∴ 6y = 8 – 2
∴ 6y = 6
∴ y = 1

From (2) 2x = 2 + 3(1)
x = $\frac{5}{2}$
x = 2.5

Question 4

Solve the following pair of linear (simultaneous) equation by the method of elimination by substitution:
0.2x + 0.1y = 25
2(x – 2) – 1.6y = 116

Answer

The given pair of linear equations are
0.2x + 0.1y = 25                        ….(1)
2( x – 2 ) – 1.6y = 116                …..(2)

Consider equation (1)
0.2x + 0.1y = 25
⇒ 0.2x = 25 – 0.1y
⇒ x = $\frac{25-0.1y}{0.2}$              ….(3)

Substitute the value of x from equation (3) in equation (2).
2( x – 2 ) – 1.6y = 116

Question 5

Solve the pair of linear (simultaneous) equation by the method of elimination by substitution:
6x = 7y + 7
7y – x = 8

Answer

6x = 7y + 7                                  …(1)
7y – x = 8                                     …(2)

7y – x = 8
∴ x = 7y – 8

Putting this value of x in (1)
6( 7y – 8 ) = 7y + 7
∴ 42y – 48 = 7y + 7
∴ 35y = 55
∴ y = $\frac{11}{7}$

From (2)

Question 6

Solve the pair of linear (simultaneous) equation by the method of elimination by substitution :
y = 4x – 7
16x – 5y = 25

Answer

y = 4x – 7                                      ….(1)
16x – 5y = 25                                ….(2)

y = 4x – 7
Putting this value of y in (2)
∴ 16x – 5 (4x – 7) = 25
∴ 16x – 20x + 35 = 25
∴ – 4x = – 10

Question 7

Solve the pair of linear (simultaneous) equation by the method of elimination by substitution:
2x + 7y = 39
3x + 5y = 31

Answer

2x + 7y = 39                           …(1)
3x + 5y = 31                           …(2)

2x + 7y = 39

Question 8

Solve the following pair of linear (simultaneous) equation by the method of elimination by substitution:
1.5x + 0.1y = 6.2
3x – 0.4y = 11.2

Answer

The given pair of linear equations are
1.5x + 0.1y = 6.2                  …(1)
3x – 0.4y = 11.2                    …(2)

Consider equation (1)
1.5x + 0.1y = 6.2
⇒ 1.5x = 6.2 – 0.1y

Question 9

Solve the following pair of linear (Simultaneous ) equation using method of elimination by substitution :
2( x – 3 ) + 3( y – 5 ) = 0
5( x – 1 ) + 4( y – 4 ) = 0

Answer

Given equations are
2( x – 3 ) + 3( y – 5 ) = 0                      …(1)
5( x – 1 ) + 4( y – 4 ) = 0                      …(2)

From (1), we get
2x – 6 + 3y – 15 = 0
⇒ 2x – 21 + 3y = 0
⇒ 2x = 21 – 3y

 

Question 10

Solve the following pair of linear (Simultaneous ) equation using method of elimination by substitution :

Answer

 

 

Question 11

Solve the following pair of linear (simultaneous) equation using method of elimination by substitution:
3x + 2y =11
2x – 3y + 10 = 0

Answer

3x + 2y = 11
⇒ 3x = 11 – 2y

Question 12

Solve the following pair of linear (simultaneous) equation using method of elimination by substitution :
2x – 3y + 6 = 0
2x + 3y – 18 = 0

Answer

2x – 3y + 6 = 0
⇒ 2x = 3y – 6

Question 13

Solve the following pair of linear (simultaneous) equation using method of elimination by substitution:

……….

Answer

Question 14

Solve the following pairs of linear (simultaneous) equation using method of elimination by substitution:

…………

Answer

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Exe-6 B Simultaneous Linear Equations (Including Problems) Class-9th Concise Selina ICSE Maths Solutions 

 Question 1

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
13 + 2y = 9x
3y = 7x

Answer

Question 2

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
3x – y = 23

……………

Answer

Question 3

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :

Answer

The given pair of linear equations are

Question 4

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :

………….

Answer

 

Question 5

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
y = 2x – 6
y = 0

Answer

Question 6

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :……..

Answer

13x – ( – 9) = 48
⇒ 13x = 39
⇒ x = 3
∴ Solution is x = 3 and y = – 9

Question 7

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
3 – (x – 5) = y + 2
2 (x + y) = 4 – 3y

Answer

3 – (x – 5) = y + 2
∴ 3 – x + 5 = y + 2
∴ – x + 8 = y + 2
∴ x + y = 6                             ….(1)

2( x + y ) = 4 – 3y
∴ 2x + 2y = 4 – 3y
∴ 2x + 5y = 4                        …..(2)

Multiplying equation no (1) by 2.
2x + 2y = 12                         …..(3)

Subtracting equation (2) from (3)

Question 8

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
2x – 3y – 3 = 0

………..

Answer

2x – 3y – 3 = 0
⇒ 2x – 3y = 3                    …..(1)

Question 9

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
13x+ 11y = 70
11x + 13y = 74

Answer

13x + 11y = 70                        …(1)
11x + 13y = 74                        …(2)

Adding (1) and (2)
13x + 11y = 70
+     11x + 13y = 74
24x + 24y = 144
Dividing by 24,
x + y = 6                             ….(3)

Subtracting (2) from (1)
13x + 11y = 70
–     11x + 13y = 74
  –     –          –       
2x – 2y = – 4

Dividing by 2
x – y = – 2                    ….(4)

Adding equation (3) and (4)
x – y = – 2
+  x + y = 6
2x = 4
x = 2

From (3)
2 + y = 6
y = 4

Question 10

For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
41x + 53y = 135
53x + 41y = 147

Answer

41x + 53y = 135                             …(1)
53x + 41y = 147                             …(2)

Adding equation (1) and (2)
41x + 53y = 135
+   53x + 41y = 147
94x + 94y = 282
Dividing by 94,
x + y = 3                             ….(3)
Subtracting equation (2) from (1)
41x + 53y = 135
–      53x + 41y = 147
–      –          –
– 12x + 12y = – 12
Dividing by 12,

– x + y = -1                          ….(4)
Adding (3) and (4)
x + y = 3
+   – x + y = -1
2y = 2
y = 1

From (3)
x + y = 3
x + 1 = 3
x = 2

 Question 11

Answer

2x + y = 23                        …(1)
4x – y = 19                         …(2)

Adding equation (1) and (2) we get,
2x + y = 23
+   4x – y = 19
6x = 42
x = 7

From (1)
2x + y = 23
⇒ 2(7) + y = 23
⇒ 14 + y = 23
⇒ y = 23 – 14

Question 12

Answer

10 y = 7x – 4
– 7x + 10y = – 4                     …(1)
12x + 18y = 1                        …(2)

Multiplying equation no. (1) by 12 and (2) by 7.
– 84x + 120y = – 48               ….(3)
84x + 126y = 7                      ….(4)

Adding equation (3) and (4)

 Question 13.1

Solve for x and y : 

…………..

Answer

Question 13.2

Solve for x and y :
4x =………

Answer

Question 14

Answer

Let x = 2 and y = 1 be a solution of the equation.
2x + 3y = m
⇒ 2(2) + 3(1) = m
⇒ 4 + 3 = m
⇒ m = 7
∴ If x = 2 and y = 1 is the solution of the equation
2x + 3y = m then the value of m is 7.

 Question 15

10% of x + 20% of y = 24
3x – y = 20

Answer

0% of x + 20% of y = 24
⇒ 0.1x + 0.2y = 24                  …..(1) [ On Simplyfying ]
3x – y = 20                               …..(2)

Multiply equation (2) by 0.2, We get :
0.6x – 0.2y = 4                        ……(3)

Adding equation (3) and (1)
0.6x – 0.2y = 4
+ 01x + 0.2y = 24
0.7x = 28
x = 40
Substituting x = 40 in equation (1), We get
0.1(40) + 0.2y = 24

⇒ 0.2y = 20
⇒ y = 100
∴ Solution is x = 40 and y = 100.

Question 16

Answer

The value of expression mx – ny is 3 when x = 5 and y = 6.
⇒ 5m – 6n = 3                              …..(1)

The value of expression mx – ny is 8 when x = 6 and y = 5.
⇒ 6m – 5n = 8                              ….(2)

Multiply equation (1) by 6 and equation (2) by 5, We get:
30m – 36n = 18                            ….(3)
30m – 25n = 40                            …..(4)

Subtracting equation (4) from (3)
30m – 36n = 18
–  30m – 25n = 40
–       +        –
– 11n = – 22

n = 2
Substituting n = 2 in equation (1), we get
5m – 6(2) = 3
⇒ 5m = 15
⇒ m = 3
∴ Solution is m = 3 and n = 2.

Question 17

Solve :
11(x – 5) + 10(y – 2) + 54 = 0
7(2x – 1) + 9(3y – 1) = 25

Answer

11( x – 5 ) + 10( y – 2 ) + 54 = 0             (given)
⇒ 11x – 55 + 10y – 20 + 54 = 0
⇒ 11x + 10y – 21 = 0
⇒ 11x + 10y = 21                                 ….(1)

7( 2x – 1 ) + 9(3y – 1) = 25                     (given)
⇒ 14x – 7 + 27y – 9 = 25
⇒ 14x + 27y – 16 = 25
⇒ 14x + 27y = 41                               …..(2)

Multiplying equation (1) by 27 and equation (2) by 10, we get,
297x + 270y = 567                            ….(3)
140x + 270y = 410                            …..(4)

Subtracting equation (4) from equation (3), we get
157x = 157
⇒ x = 1

Substituting x = 1 in equation (1), we get,
11 x 1 + 10y = 21
⇒  10y = 10

⇒  y = 1
∴ Solution set is x = 1 and y = 1.

Question 18

Solve :

………..

Answer

Substituting y = 2 in equation (1), We get
– 6x – 55 x 2 = – 128
⇒ – 6x – 110 = – 128
⇒ – 6x = – 18
⇒ x = 3
∴ Solution is x = 3 and y = 2.

Question 19

Solve :
$4x+\dots \dots \dots \dots$

Answer

4x + 3y = 17
3x – 4y + 6 = 0

Answer

Given equation are 4x + 3y = 17 and 3x – 4y + 6 = 0
Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We have

a1 = 4, b1 = 3, c1 = -17 and a2 = 3, b2 = – 4, c2 = 6

Question 2

Solve, using cross-multiplication :
3x + 4y = 11
2x + 3y = 8

Answer

Given equations are 3x + 4y = 11 and 2x + 3y = 8
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we have
a₁ = 3, b₁ = 4, c₁ = -11 and a₂ = 2, b₂ = 3, c₂ = – 8

Question 3

Solve, using cross-multiplication :
6x + 7y – 11 = 0
5x + 2y = 13

Answer

Given equation are 6x + 7y – 11 = 0 and 5x + 2y = 13
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 6, b₁ = 7, c₁ = -11 and a₂ = 5, b₂ = 2, c₂ = -13

Question 4

Solve, using cross-multiplication :
5x + 4y + 14 = 0
3x = -10 – 4y

Answer

Given equation are 5x + 4y + 14 = 0 and 3x = -10 – 4y
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have
a₁ = 5, b₁ = 4, c₁ = 14 and a₂ = 3, b₂ =  4, c₂ = 10

Question 5

Solve, using cross-multiplication :
x – y + 2 = 0
7x + 9y = 130

Answer

Given equation are x – y + 2 = 0 and 7x + 9y = 130
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 1, b₁ = -1, c₁ = 2 and a₂ = 7, b₂ = 9, c₂ = -130

Question 6

Solve, using cross-multiplication :
4x – y = 5
5y – 4x = 7

Answer

Given equation are 4x – y = 5 and 5y – 4x = 7
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 41, b₁ = – 1, c₁ = – 5 and a₂ = – 4, b₂ = 5, c₂ = – 7

Question 7

Solve, using cross-multiplication :
4x – 3y = 0
2x + 3y = 18

Answer

Given equation are 4x – 3y = 0 and 2x + 3y = 18
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 4, b₁ = – 3, c₁ = 0 and a₂ = 2, b₂ = 3, c₂ = -18

Question 8

Solve, using cross-multiplication :
8x + 5y = 9
3x + 2y = 4

Answer

Given equation are 8x + 5y = 9 and 3x + 2y = 4
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 8, b₁ = 5, c₁ = -9 and a₂ = 3, b₂ = 2, c₂ = – 4

Question 9

Solve, using cross-multiplication :
4x – 3y – 11 = 0
6x + 7y – 5 = 0

Answer

Given equation are 4x – 3y – 11 = 0 and 6x + 7y – 5 = 0
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 4, b₁ = – 3, c₁ = -11 and a₂ = 6, b₂ = 7, c₂ = – 5

Question 10

Solve, using cross-multiplication :
4x + 6y = 15
3x – 4y = 7

Answer

Given equation are 4x + 6y = 15 and 3x – 4y = 7
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 4, b₁ = 6, c₁ = -15 and a₂ = 3, b₂ = – 4, c₂ = -7

Question 11

Solve, using cross-multiplication :
0.4x – 1.5y = 6.5
0.3x + 0.2y = 0.9

Answer

Given equation are 0.4x – 1.5y = 6.5 and 0.3x + 0.2y = 0.9
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = 0.4, b₁ = -1.5, c₁ = -6.5 and a₂ = 0.3, b₂ = 0.2, c₂ = 0.9

Question 12

Solve, using cross-multiplication :
√2x – √3y = 0
√5x + √2y = 0

Answer

Given equation are √2x – √3y = 0 and √5x + √2y = 0
Comparing with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, We have

a₁ = √2, b₁ = √3, c₁ = 0 and a₂ = √5, b₂ = √2, c₂ = 0

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Exe-6 D Simultaneous Linear Equations (Including Problems) Class-9th Concise Selina ICSE Maths Solutions

Question 1

…………………….

Answer

Question 2

Solve the pairs of equations :

Answer

 

Question 3

Solve :
5x +……..

Answer

Question 4

4x + $\frac{\mathrm{6/}}{y}$ = 15 and 3x – $\frac{\mathrm{4/}}{y}$ = 7. Hence, find a if y = ax – 2.

Answer

Question 5

……………………$=\mathrm{19\; Hence,\; find\; ‘a’\; if\; y\; =\; ax\; +\; 3.}$

Answer

Question 6.1

Solve :

…….

Answer

 

Question 6.2

Solve : ………

Answer

Let a = 3x + 4y and b = 3x – 2y

Subtracting equation (5) from (4), We get :
3x – 2y = 5
–      3x + 4y = 17
–     –         –
– 6y = – 12
y = 2
Substituting y = 2 in equation (4), We get
3x – 2(2) = 5
3x = 9
x = 3

x = 3
∴ Solution is x = 3 and y = 2.

Question 7.1

Solve :
x + y = 2xy
x – y = 6xy

Answer

x + y = 2xy                       …..(1)
x – y = 6xy                        ……(2)
Adding equation (1) and (2)
x + y = 2xy
+   x – y = 6xy
2x = 8xy
2 = 8y

Question 7.2

Solve :
x+ y = 7xy
2x – 3y = – xy

Answer

x + y = 7xy                           …(1)
2x – 3 = – xy                          …(2)
Multiplying equation no. (1) by 3.
3x + 3y = 21xy                     ….(3)
Adding equation (3) and (2)
3x + 3y = 21xy
+   2x – 3y  = – xy
5x = 20xy

Question 8

Solve :
$\frac{\mathrm{a/}}{\mathrm{x\dots \dots \dots \dots }}$

Answer

Solve : 

…………..

Answer

Question 10

Solve :

……..

Answer

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Concise Selina ICSE Class-9 Maths Solutions Simultaneous Linear Equations (Including Problems)  Exercise – 6(E)

Question 1

The ratio of two numbers is $\frac{\mathrm{2/}}{3}$. If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers

Answer

Let the two numbers be x and y.
According to the question,

Question 2

Two numbers are in the ratio 4 : 7. If thrice the larger be added to twice the smaller, the sum is 59. Find the numbers.

Answer

Let the smaller number be x and the larger number be y.
According to the question,

Question 3

When the greater of the two numbers increased by 1divides the sum of the numbers, the result is $\frac{\mathrm{3/}}{2}$. When the difference of these numbers is divided by the smaller, the result $\frac{\mathrm{1/}}{2}$. Find the numbers.

Answer

Let the two numbers be a and b respectively such that b > a.
According to given condition,

Substituting a = 6 in equation (1), We get
2(6) – b = 3
⇒ 12 – b = 3
⇒ b = 9
Thus, the two numbers are 6 and 9 respectively.

Question 4

Answer

Two numbers are x and y such that x > y.
Now,
x + y = 50                          ….(i)
And,
y² – x² = 720
⇒ (y – x)(y + x) = 720
⇒ (y – x)(50) = 720
⇒ y – x = 14.4                     ….(ii)

Adding (i) and (ii), we get
2y = 64.4
⇒ y = 32.2

Substituting the value of y in (i), we have
x + 32.2 = 50
⇒ x = 17.8
Thus, the two numbers are 17.8 and 32.2 respectively.

Question 5

The sum of two numbers is 8 and the sum of their reciprocals is $\frac{\mathrm{8/}}{15}$. Find the numbers.

Answer

Let the two numbers be x and y respectively.
Then,
x + y = 8                           ….(i)
⇒ x = 8 – y
And,

⇒ xy = 15
⇒ ( 8 – y )y = 15
⇒ 8y – y² = 15
⇒ y² – 8y + 15 = 0
⇒ y² – 3y – 5y + 15 = 0
⇒ y( y – 3 ) – 5( y – 3 ) = 0
⇒ ( y – 3 )( y – 5 ) = 0
⇒ y = 3 or y = 5
⇒ x = 5 or x = 3
Thus, the two numbers are 3 and 5 respectively.

Question 6

The difference between two positive numbers x and y ( x > y ) is 4 and the difference between their reciprocals is $\frac{\mathrm{4/}}{21}$. Find the numbers.

Answer

Two numbers are x and y respectively such that x > y.
Then,
x – y = 4                                          ….(i)
⇒ x = 4 + y
And,

⇒ xy = 21
⇒ ( 4 + y )y = 21
⇒ 4y + y2 = 21
⇒ y2 + 4y – 21 = 0
⇒ y2 + 7y – 3y – 21 = 0
⇒ y( y + 7 ) – 3( y + 7 ) = 0
⇒ ( y – 3 )( y + 7 ) = 0
⇒ y = 3 and y = -7
We reject y = -7 since y is positive.
⇒ y = 3
⇒ x = 4 + y = 4 + 3 = 7
Thus, the two numbers are 7 and 3 respectively.

Question 7

Answer

Let the common multiple between the numbers be x.
So, the numbers are 4x and 5x.
According to the question,

⇒ 8x – 60 = 5x – 30
⇒ 3x = 30
⇒ x = 10
So, 4x = 4(10) = 40 and 5x = 5(10) = 50
Thus, the numbers are 40 and 50.

Question 8

If the numerator of a fraction is increased by 2 and denominator is decreased by 1, it becomes $\frac{\mathrm{2/}}{3}$. If the numerator is increased by 1 and denominator is increased by 2, it becomes $\frac{\mathrm{1/}}{3}$. Find the fraction.

Answer

Let the numerator and denominator a fraction be x and y respectively .
According to the question,

From (1) ,
3x – 2 (7) = – 8
3x = – 8 + 14
x = 2
Required fraction =  2/7

Question 9

Answer

Let the numerator and denominator of a fraction be x and y respectively .Then the fraction will be $\frac{\mathrm{x/}}{y}$.
According to the question,
x + y = 7                                    …(1)
5y – 4x = 8                                 …(2)

Multiplying equation no. (1) by 4
4x + 4y = 28                              …(3)
Adding equation (3) and (2)
4x + 4y = 28
+ – 4x + 5y = 8
9y = 36
y = 4

From (1),
x + 4 = 7
x = 3
Required fraction = $\frac{\mathrm{3/}}{4}$

Question 10

lf the numerator of a fraction is multiplied by 2 and its denominator is increased by 1, it becomes 1. However,if  the numerator is increased bu 4 and denominator is multiplied by 2, the fraction becomes $\frac{\mathrm{1/}}{2}$. Find the fraction.

Answer

Let the numerator of the fraction be x and the denominator be y.
So, the Fraction is $\frac{\mathrm{x/}}{y}$
According to the question,

Solving equations (1) and (2), We get
y = 9
Putting the value of y in (1), we get
2x – 9 = 1
⇒ 2x = 1 + 9
⇒ x = 5
So, the fraction is 5/9

Question 11

A Fraction becomes $\frac{\mathrm{1/}}{2}$ if 5 subtracted from its numerator and 3 is subtracted from its denominator. If the denominator of this fraction is 5 more than its numerator. Find the fraction.

Answer

Let the numerator of the fraction be x and denominator of the fraction be y.
Then, the fraction =$\frac{\mathrm{x/}}{y}$
According to given condition, we have

⇒ 2x – 10 = y – 3
⇒ 2x – y = 7                               …(1)
And,
x + 5 = y
⇒ x – y = – 5                             …(2)
Subtracting (2) from (1), We get
x = 12
⇒ y = x + 5
⇒ y = 12 + 5 = 17
Hence, the fraction is $\frac{\mathrm{12/}}{17}$.

Question 12

Answer

Let the digit at unit’s place be x and the digit at ten’s place y.
Required no. = 10y + x
If the digit’s are reversed,
Reversed no. = 10y + x

According to the question,
x + y = 5                             …(1)
and,
(10y + x) – (10x + y) = 27
9y – 9x = 27
y – x = 3                             …(2)
Now,
Adding equation (1) and (2),
y – x = 3                  ….(2)
+   y + x = 5                 ….(1)
2y = 8
y = 4

From (1)
x + 4 = 5
x = 1
Require no is
10 (4) + 1 = 41

Question 13

Answer

Let the digit at unit’s place be x and the digit at ten’s place be y.
Required no. = 10y + x
If the digit’s are reversed
Reversed no. = 10x + y

According to the question,
x + y = 7                        …(1)
and,
10x + y – 2 = 2(10y + x).
8x – 19y = 2                  …(2)

Multiplying equation no. (1) by 19.
19x + 19y = 133           …(3)

Now adding equation(2) and (3)

Question 14

Answer

Let the digit at unit’s place be x and the digit at ten’s place be y.
Required no. = 10y + x

According to the question,
y = 3x
⇒ 3x – y = 0                       …(1)
and,
10y + x + x = 32
10y + 2x = 32                    …(2)

Multiplying equation no. (1) by 10.
30x – 10y = 0                    ….(3)
Now,
Adding equation (3) and (2)

Question 15

Answer

Let the digit a unit’s place be x and the digit at ten’s place be y.
Required no. = 10y + x.

According to the question,
y – 2x = 2
-2x + y = 2                   …(1)
and,
(10x + y) -3 (y + x) = 5
7x – 2y = 5                   …(2)

Multiplying equation no. (1) by 2.
– 4x + 2y = 4              …(3)

Now adding (2) and (3),

From (1) ,we get
-2(3) + y = 2
y = 8

Required number is 10(8) + 3 = 83

Question 16

Answer

Let x be the number at the ten’s place.
and y be the number at the unit’s place.
So, the number is 10x + y.

Four times a certain two-digit number is seven times
the number obtained on interchanging its digits.
⇒ 4( 10x + y ) = 7( 10y + x )
⇒ 40x + 4y = 70y + 7x
⇒ 33x – 66y = 0
⇒ x – 2y = 0                  ….(1)

If the difference between the digits is 4, then
⇒ x – y = 4                     …(2)
Subtracting equation (1) from equation (2), we get :

Subtracting y = 4 in equation (1), We get
x – 2(4) = 0
⇒ x = 8
∴ The number is 10x + y = 10(8) + 4 = 84.

Question 17

Answer

Answer

Question 1

Answer

Let present age of A = x years
And present age of B = y years

According to the question,
Five years ago,
x – 5 = 4(y – 5)
x – 4y = -15                        …(1)

Five years later,
x + 5 = 2(y + 5)
x + 5 = 2y + 10
x – 2y = 5                          ….(2)

Now subtracting (1) from (2)

From (1)
x – 4 (10) = -15
x = 25
Present ages of A and B are 25 years and 10 years respectively.

Question 2

Answer

Let A’s presentage be x years
and B’s present age be y years

According to the question
x = y + 20
x – y = 20                            …(1)

Five years ago,
x – 5 = 3(y – 5)
x – 5 = 3y – 15
x  – 3y = – 10                      …(2)

Subtracting (1) and (2),

y = 15

From (1)
x = 15 + 20
x = 35
Thus, present ages of A and B are 35 years and 15 years.

Question 3

Answer

Let the present age of the mother be x years.
and the present age of the daughter be y year.
According to the queation,
x – 4 = 4( y – 4 )
⇒ x – 4 = 4y – 16
⇒  x – 4y = – 12                …(1)
And,

solving (1) and (2), We get
y = 14 and x = 44
Hence, the present age of the mother is 44 years
and the present age of the daughter is 14 years.

Question 4

Answer

Let the present age of the man be x years.
and let the sum of the ages of his two children be y years.
According to the question,
x = 2y                            …(1)
and x + 20 = y + 40      ….(2) [ Since he has two children ]

Solving (1) and (2), We get,
2y + 20 = y + 40
⇒ y = 20
So, x = 2y ⇒ x = 40.
Hence, the present age of the man is 40 years.

Question 5

Answer

Let A’s annual in come = Rs.x
and B’s annual income = Rs. y

According to the question,

From (1)
4x – 3(40,000) = 0
x = 30,000
Thus, A’s income in Rs. 30,000 and B’s income is Rs. 40,000.

Question 6

Answer

Let the no. of pass candidates be x
and the no. of fail candidates be y.

According to the question,

From (1)
– 4(30) = 0
x = 120

Total students appeared = x +  y = 120 + 30 = 150.

Question 7

Answer

Let the numberof pencils with A = x
and the number of pencils with B = y.

If A gives 10 pencils to B,
y + 10 = 2( x – 10 )
2x – y = 30                                    …(1)

If B gives to pencils to A
y – 10 = x + 10
x – y = – 20                                   …(2)

Subtracting equation (1) and (2)

From (1)
2(50) – y = 30
y = 70
Thus, A has 50 pencils and B has 70 pencils.

Question 8

Answer

Let the number of adults = x
and the number of children = y

According to the question,
x + y = 1250                                …(1)
75x + 25y = 61250                      …(2)

Subtracting equation (2) from (1)

From (1)
600 + y = 1250
y = 650
Thus, number of adults = 600 and the number of children = 650.

Question 9

Answer

Let the cost price of article A = Rs. x
and the cost price of articles B = Rs. y

According to the question,
( x + 5% of x ) +( y + 7% of y ) = 1167

from (3)
x +600 = 1100
x = 500
Thus, cost price of article A is Rs. 500 and that of article B is Rs. 600.

Question 10

Pooja and Ritu can do a piece of work in 120/7 days. If one day work of Pooja be three fourth of one day work of Ritu’ find in how many days each will do the work alone.

Answer

Exercise – 6(G) Simultaneous Linear Equations (Including Problems) Class-9th Concise Selina ICSE Maths Solutions

Question 1

Answer

Question 2

Answer

Question 3

Answer

Let the digit at ten’s place be x And the digit at unit’s place be y.
Required number = 10x + y
When the digits are interchanged,
Reversed number = 10y + x

According to the question,
7(10x + y) = 4(10y + x)
∴ 70x + 7y = 40y + 4x
∴ 66x = 33y
∴ 2x – y = 0                           …(1)

The difference between the digits is 3.
y – x = 3                                 …(2)

Adding equation (1) and (2)
y – x = 3
+ –  y + 2x = 0
x = 3

From (1)
2(3) – y = 0
y = 6
Thus, Required number = 10(3) + 6 = 36

Question 4

Answer

Let, the fare of ticket for station A be Rs. x and the fare of ticket for station B be Rs. y
According, to the question
2x + 3y = 77                           …(1) and
3x + 5y = 124                         …(2)

Multiplying equation no. (1) by 3 and (2) by 2.
6x + 9y = 231                         …(3)
6x + 10y = 248                       ….(4)

Subtracting equation (4) from (3)
6x + 9y = 231
–  6x + 10y = 248
–     –         –
– y = – 17
y = 17
From (1)
2x + 3 (17) = 77
2x = 77 – 51
2x = 26
x = 13
Thus, fare for station A = Rs. 13 and, fare for station B = Rs. 17.

Question 5

Answer

Let x be the number at the ten’s place.
and y be the number at the unit’s place.
So, the number is 10x + y.

The sum of digit of a two digit number is 11.
⇒ x + y = 11                             …(1)

lf the digit at ten’s place is ineased by 5 and the digit at unit place is decreased by 5,
the digits of the number are found to be reversed.
⇒  10( x + 5 ) + ( y – 5 ) = 10y + x
⇒  9x – 9y = -45
⇒  x – y = -5                              …(2)

Subtracting equation (1) from equation (2), we get :
x – y = – 5
–    x + y = 11
–   –      –
– 2y = – 16
⇒ y = 8

Substituting y = 8 in equation (1), we get
x + 8 = 11
⇒ x = 3
∴ The number is 10x + y = 10(3) + 8 = 38.

Question 6

Answer

Let the quantity of 90% acid solution be x litres and The quantity of 97% acid solution be y litres

According to the question,
x + y = 21                        …(1)
and  90% of x + 97% of y = 95% of 21
90x + 97y = 1995            …(2)

Multiplying equation no. (1) by 90, we get,
90x + 90y = 1890            ….(3)

Subtracting equation (2) from (3)
90x + 90y = 1890
–      90x + 97y = 1995
–       –            –
– 7y = – 105
y = 15

From (1)
x + 15 = 21
x = 6
Hence, 90% acid solution is 6 litres and 97% acid solution is 15 litres.

Question 7

Answer

Assume x kg of the first kind costing Rs. 250 per kg and y kg of the second kind costing Rs. 350 per kg sweets were bought.

It is estimated that 40 kg of sweets were needed.
⇒ x + y = 40                        ….(1)

The total budget for the sweets was Rs. 11,800.
⇒ 250x + 350y = 11,800     ….(2)

Multiply equation (1) by 250, We get :
250x + 250y = 10,000        …..(3)

Subtracting equation (2) from (3),
250x + 250y = 10,000
–    250x + 350y = 11,800
–       –             –
– 100y = – 1800
y = 18
Substituting y = 18 in equation (1), We get
x + 18 = 40
⇒ x = 22
∴ 22 kgs of the first kind costing Rs. 250 per kg and 18 kgs of the second kind costing Rs. 350 per kg sweets were bought.

Question 8

Answer

Weight of Mr. Ahuja = x kg and weight of Mrs. Ahuja = y kg.
After the dieting,
x – 5 = y
x – y = 5                      …(1)
and,

Question 9

Answer

Answer

Let the fixed charge be Rs. x and the charge per kilometer be Rs. y.
The charges for 10 km = Rs. 10y
The charges for 15 km = Rs. 15y
According to the question,
x+ 10y = 315                    …(1)
x+ 15y = 465                    ….(2)
Solving the equations, we get
– 5y = – 150
⇒ y = 30
and x = 315 – 10y = 315 – 10(30) = 15
So, the fixed charges is Rs. 15 and the charges per kilometer is Rs. 30.
To travel 32 km, a personal has to pay
= Rs. 15 + Rs. 30( 32 ) = Rs. 15 + Rs. 960 = Rs. 975.

Question 11

Answer

Let the fixed charges be Rs. x and the charge for each extra day be Rs. y.
According to the question,
x + 4y = 27                    …(1)
and x + 2y = 21            ….(2)

Solving the equations, we get
2y = 6
∴ y = 3
and x = 21 – 2y = 21 – 2(3) = 15
Hence, the fixed charges is Rs. 15 and the charge for each extra day is Rs. 3.

Question 12

Answer 

Let the length of the rectangle be x units and the breadth of the rectangle be y units.
We know that, area of a rectangle = length x breadth = xy
According to the question,
xy – 9 = ( x – 5 )( y + 3 )
⇒ xy – 9 = xy + 3x – 5y – 15
⇒  3x – 5y = 6                         ….(1)

xy + 67 = ( x + 3 )( y + 2 )
⇒ xy + 67 = xy + 2x + 3y + 6
⇒  2x + 3y = 61                      …(2)

Multiply (1) by 2 and (2) by 3, we get
6x – 10y = 12                           …(3)
6x + 9y = 183                          …(4)

Subtracting equation (4) from (3),
6x – 10y = 12
–   6x + 9y = 183
–    –         –
– 19y = – 171
y = 9
Putting y = 9 in equation (1)
3x – 5y = 6
3x – 5(9) = 6
3x = 6 + 45
x = $\frac{\mathrm{5/1}}{3}$
x = 17
Hence, the length of the rectangle is 17 units and the breadth of the rectangle is 9 units.

Question 13

Answer

Let the pipe with larger diameter and smaller diameter be pipes A and B respectively.
Also, let pipe A work at a rate of x hours/ unit and pipe B work at a rate of Y hours / unit.
According to the question,