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Class 9 SELINA Solutions Maths Chapter 6 - Simultaneous (Linear) Equations (Including Problems)

Rational and Irrational Numbers Exercise Ex. 6(E)

Solution 5

Let the two numbers be x and y.

Then,

x + y = 8 …..(i)

x2 - y2 = 32 …..(ii)

(x - y)(x + y) = 32

x - y = 4 ….. (iii)

Adding (i) and (iii), we get

2x = 12 x = 6

From (i), y = 8 - x = 8 - 6 = 2

Therefore, the numbers are 6 and 2.

Solution 1

Let the two numbers be x and y

According to the question,

 

3x - 2y = 0 ...(1)

Also,

2x - 3y = -20 ...(2)

Multiplying equation no. (1) by 2 and (2) by 3and substracting

 

From (1), we get

3x - 2(12) = 0

 

x = 8

Thus, the numbers are 8 and 12.

Solution 2

Let the smaller number be x

and the larger number be y.

According to the question,

7x -; 4y = 0...(1)

and,3y + 2x = 59...(2)

Multiplying equation no. (1) by 3 and (2) by 4.and adding them

x =

 

From (1)

Hence, the number are

Solution 3

Let the two numbers be a and b respectively such that b > a.

According to given condition,

Solution 4

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.

Solution 6

Two numbers are x and y respectively such that x > y.

Then,

x - y = 4 ….(i)

x = 4 + y

And,

  

Solution 7

  

Solution 8

Let the numerator and denominator a fraction be x and y respectively .

According to the question,

 

3x - 2y = -8...(1)

And,

 

 

Now subtracting, 

 

 

 

From (1) ,

3x - 2 (7) = -8

3x = -8 + 14

x = 2

Required fraction =

Solution 9

Let the numerator and denominator of a fraction be x and y respectively .Then the fraction will be

According to the question,

x + y = 7...(1)

5y - 4x = 8...(2)

Multiplying equation no. (1) by 4 and add with (2),

 

 

From (1)

x + 4 = 7

x = 3

Required fraction =

Solution 10

  

Solution 11

Let space the space numerator space of space the space fraction space be space straight x space and space denominator space of space the space fraction space be space straight y.
Then comma space the space fraction equals space straight x over straight y
According space to space given space condition comma space we space have
fraction numerator straight x minus 5 over denominator straight y minus 3 end fraction equals 1 half
rightwards double arrow 2 straight x minus 10 equals straight y minus 3
rightwards double arrow 2 straight x minus straight y equals 7 space space space space space space space space space space.... left parenthesis straight i right parenthesis
And comma
straight x plus 5 equals straight y
rightwards double arrow straight x minus straight y equals negative 5 space space space space space space space space space space space space.... left parenthesis ii right parenthesis
Subtracting space left parenthesis ii right parenthesis space from space left parenthesis straight i right parenthesis comma space we space get
straight x equals 12
rightwards double arrow straight y equals straight x plus 5 equals 12 plus 5 equals 17
hence comma space the space fraction space is space 12 over 17.

Solution 12

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

 

Now adding the two equation,

 

From (1)

x + 4 = 5

x = 1

Require no is

10 (4) + 1 = 41

Solution 13

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.

 

Now adding equation(2) and (3)

 

x = 5

From (1)

5 +y = 7

y = 2

Required number is

10(2) + 5

= 25.

Solution 14

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

 

Now adding (3)and(2)

 

From (1),we get

y = 3(1) = 3

Required no is

10(3) + 1 = 31

Solution 15

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.

 

Now adding (2)and(3)

 

From (1) ,we get

-2(3) + y = 2

y = 8

Required number is

10(8) + 3 = 83.

Solution 16

 

 

 

 

Solution 17

begin mathsize 14px style Let space the space tens space digit space of space the space number space be space straight x space and space the space units space digit space be space straight y.
So comma space the space number space is space 10 straight x plus straight y.
The space number space obtained space by space interchanging space the space digits space will space be space 10 straight y plus straight x.
According space to space question comma space we space have
10 straight x plus straight y plus 10 straight y plus straight x equals 121
rightwards double arrow 11 straight x plus 11 straight y equals 121
rightwards double arrow 11 left parenthesis straight x plus straight y right parenthesis equals 121
rightwards double arrow straight x plus straight y equals 11 space space space space space space space space space space.... left parenthesis straight i right parenthesis
And comma
straight x minus straight y equals 3 space space space space space space space space space space.... left parenthesis ii right parenthesis
Adding space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get
2 straight x equals 14
rightwards double arrow straight x equals 7
rightwards double arrow straight y equals 11 minus straight x equals 11 minus 7 equals 4
Hence comma space the space number space is space 74. end style

Solution 18

  

Simultaneous Linear Equations Exercise Ex. 6(A)

Solution 1

8x + 5y = 9...(1)

3x + 2y = 4...(2)

 

(2)y =

 

Putting this value of y in (2)

3x + 2

 

15x + 18 - 16x = 20

x = -2

From (1) y = =

y = 5

 

 

 

Solution 2

2x - 3y = 7...(1)

5x + y = 9...(2)

(2)y = 9 - 5x

Putting this value of y in (1)

2x - 3 (9 - 5x) = 7

2x - 27 + 15x = 7

17x = 34

x = 2

From (2)

y = 9 - 5(2)

y = -1

 

 

Solution 3

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

2x = 2 + 3y...(2)

 

(2) 2x = 2 + 3y

Putting this value of 2x in (1)

2 + 3y + 3y = 8

6y = 6

y = 1

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

x =

x = 2.5

 

Solution 4

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 5

6x = 7y + 7...(1)

7y - x = 8...(2)

(2) x = 7y - 8

Putting this value of x in (1)

6(7y - 8) = 7y + 7

42y - 48 = 7y + 7

35y = 55

 

From (2) x =

x = 3

Solution 6

y = 4x -7...(1)

16x- 5y = 25...(2)

(1) y = 4x - 7

Putting this value of y in (2)

16x - 5 (4x - 7) = 25

16x - 20x + 35 = 25

-4x = -10

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

From (1)

Syntax error from line 1 column 49 to line 1 column 69. Unexpected '<mi '.

y = 10-7=3

Solution is Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '..

Solution 7

2x + 7y = 39...(1)

3x + 5y = 31...(2)

(1) x =

Putting this value of x in (2)

 

117 - 21y + 10y = 62

-11y = -55

y = 5

From (1) x =

 

 

Solution 8

Solution 9

  

Solution 10

  

Solution 11

Solution 12

Solution 13

Solution 14

Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(B)

Solution 1

13 + 2y = 9x...(1)

3y = 7x...(2)

Multiplying equation no. (1) by 3 and (2) by 2, we get,

 

 

From (2)

3y = 7(3)

y = 7

Solution 2

3x - y = 23...(1)

4x + 3y = 48...(2)

Multiplying equation no. (1) by 3

 

From (1)

3(9) - y = 23

y = 27 - 23

y = 4

 

Solution 3

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 4

Multiplying equation no. (1) by 3 and(2) by 5.

 

 

From (1)

Solution 5

y = 2x - 6

y = 0

 

x = 3 ; y = 0

Solution 6

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 7

3 - (x - 5) = y + 2

2(x + y) = 4 - 3y

-x - y = -6

x + y = 6...(1)

2x + 5y = 4...(2)

 

Multiplying equation no. (1) by 2.

From (1)

 

 

Solution 8

2x -3y - 3 = 0

 

 

2x - 3y = 3...(1)

4x + 24y = -3...(2)

 

 

Multiplying equation no. (1) by 8.

 

 

From (1)

 

 

 

Solution 9

13x + 11y = 70...(1)

11x+ 13y = 74...(2)

Adding(1) and (2)

24x + 24y = 144

x + y = 6...(3)

 

subtracting (2) from (1)

2x - 2y = -4

x - y = -2...(4)

x + y = 6...(3)

 

From (3)

2 + y = 6y = 4

 

Solution 10

41x + 53y = 135...(1)

53x + 41y = 147...(2)

Adding (1) and (2)

94x + 94y = 282

x + y = 3...(3)

Subtracting (2) from (1)

 

 

From (3)

x + 1 = 3x = 2

Solution 11

 

2x + y = 23...(1)

 

4x - y = 19...(2)  


Adding equation (1) and (2) we get,

2x + y = 23

4x - y = 19     

 

From (1)

2(7) + y = 23

y = 23 - 14

y = 9

x - 3y = 7 - 3(9) = -20

 

And 5y - 2x = 5(9) - 2(7) = 45- 14 = 31

Solution 12

10 y = 7x - 4

-7x + 10y = -4...(1)

12x + 18y = 1...(2)

 

 

Multiplying equation no. (1) by 12 and (2) by 7.

 

 

 

 

From (1)

-7x + 10

 

-7x = -4 +

4and 8y - x = 8

 

Solution 13

(i)


  


(ii)


 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 14

Solution 15

 

Solution 16

Solution 17

  

Solution 18

  

Solution 19

  

Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(C)

Solution 1

   

Solution 2

   

Solution 3

   

Solution 4

   

Solution 5

   

Solution 6

   

Solution 7

   

Solution 8

   

Solution 9

   

Solution 10

   

Solution 11

Solution 12

Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(D)

Solution 1

 

Multiplying equation no. (1) by 7 and (2) by 4.

 

 

From (1)

 

Solution 2

 begin mathsize 12px style 3 over straight x plus 2 over straight y equals 10 space space space space space space space space space.... left parenthesis straight i right parenthesis
9 over straight x minus 7 over straight y equals 10.5 space space space space space.... left parenthesis ii right parenthesis
Multiplying space equation space left parenthesis straight i right parenthesis space by space 3 comma space we space get
9 over straight x plus 6 over straight y equals 30 space space space space space space space space.... left parenthesis iii right parenthesis
Subtracting space left parenthesis ii right parenthesis space from space left parenthesis iii right parenthesis comma space we space get
13 over straight y equals 19.5
rightwards double arrow straight y equals fraction numerator 13 over denominator 19.5 end fraction equals 2 over 3
From space left parenthesis straight i right parenthesis comma
3 over straight x plus fraction numerator 2 cross times 3 over denominator 2 end fraction equals 10
rightwards double arrow 3 over straight x plus 3 equals 10
rightwards double arrow 3 over straight x equals 7
rightwards double arrow straight x equals 3 over 7 end style

 

Solution 3

5 straight x plus 8 over straight y equals 19 space space space space space space space space.... left parenthesis straight i right parenthesis
3 straight x minus 4 over straight y equals 7 space space space space space space space space space space space.... left parenthesis ii right parenthesis
Multiplying space equation space left parenthesis ii right parenthesis space by space 2 comma space we space get
6 straight x minus 8 over straight y equals 14 space space space space space space space space space space.... left parenthesis iii right parenthesis
Adding space left parenthesis straight i right parenthesis space and space left parenthesis iii right parenthesis comma space we space get
11 straight x equals 33
rightwards double arrow straight x equals 3
Substituting space straight x equals 3 space in space equation space 91 right parenthesis comma space we space get
5 left parenthesis 3 right parenthesis plus 8 over straight y equals 19
rightwards double arrow 8 over straight y equals 19 minus 15
rightwards double arrow straight y equals 8 over 4 equals 2

Solution 4

4 straight x plus 6 over straight y equals 15 space space space space space space space.... left parenthesis straight i right parenthesis
3 straight x minus 4 over straight y equals 7 space space space space space space space space space.... left parenthesis ii right parenthesis
Multiplying space left parenthesis straight i right parenthesis space by space 4 space and space left parenthesis ii right parenthesis space by space 6
16 straight x plus 24 over straight y equals 60 space space space space space space space space space.... left parenthesis iii right parenthesis
18 straight x minus 24 over straight y equals 42 space space space space space space space space space space.... left parenthesis iv right parenthesis
Adding space left parenthesis iii right parenthesis space and space left parenthesis iv right parenthesis comma space we space get
34 straight x equals 102
rightwards double arrow straight x equals 3
Substituting space straight x equals 3 space in space left parenthesis straight i right parenthesis comma space we space get
4 left parenthesis 3 right parenthesis plus 6 over straight y equals 15
rightwards double arrow 6 over straight y equals 15 minus 12
rightwards double arrow straight y equals 6 over 3 equals 2
Now comma space straight y equals ax minus 2
rightwards double arrow 2 equals straight a left parenthesis 3 right parenthesis minus 2
rightwards double arrow 2 equals 3 straight a minus 2
rightwards double arrow 3 straight a equals 4
rightwards double arrow straight a equals 4 over 3 equals 1 1 third

Solution 5

 

 

 

Multiplying equation no. (1) by 5 and (2) by 2.

From (1) 3

y = ax + 3

 

Solution 6


left parenthesis straight i right parenthesis space
fraction numerator 20 over denominator straight x plus straight y end fraction plus fraction numerator 3 over denominator straight x minus straight y end fraction equals 7 space space space space space... left parenthesis 1 right parenthesis
fraction numerator 8 over denominator straight x minus straight y end fraction minus fraction numerator 15 over denominator straight x plus straight y end fraction equals 5 space space space space space... left parenthesis 2 right parenthesis
Multiplying space equation space no. space left parenthesis 1 right parenthesis space by space 8 space and space left parenthesis 2 right parenthesis space by space 3.
fraction numerator 160 over denominator straight x plus straight y end fraction plus fraction numerator 24 over denominator straight x minus straight y end fraction equals 56 space space space space space... left parenthesis 3 right parenthesis
fraction numerator negative 45 over denominator straight x plus straight y end fraction plus fraction numerator 24 over denominator straight x minus straight y end fraction equals 15 space space space space space... left parenthesis 4 right parenthesis
bottom enclose negative space space space space space space space minus space space space space space space space space space space space minus end enclose
space space space space space space space space space space space fraction numerator 205 over denominator straight x plus straight y end fraction equals space 41
straight x plus space straight y space equals space 5 space space space space space space space space space space space space space... left parenthesis 5 right parenthesis
From space left parenthesis 1 right parenthesis
20 over 5 plus fraction numerator 3 over denominator straight x minus straight y end fraction equals 7
fraction numerator 3 over denominator straight x minus straight y end fraction equals 3
straight x minus straight y equals 1 space space space space space space space space space... left parenthesis 6 right parenthesis

straight x plus space straight y space equals space 5 space space space space space... left parenthesis 5 right parenthesis
bottom enclose straight x minus space straight y space equals space 1 end enclose space space space space space... left parenthesis 6 right parenthesis
space space space 2 straight x space space equals space 6
straight x space equals space 3

from space left parenthesis 5 right parenthesis
3 plus space straight y space equals space 5 space rightwards double arrow space straight y space equals space 2


(ii)

Let space straight a space equals space 3 straight x space plus 4 straight y space and space straight b space equals space 3 straight x space minus space 2 straight y
therefore fraction numerator 34 over denominator 3 straight x plus 4 straight y end fraction plus fraction numerator 15 over denominator 3 straight x minus 2 straight y end fraction equals 5
rightwards double arrow 34 over straight a plus 15 over straight b equals 5.......... left parenthesis straight i right parenthesis
fraction numerator 25 over denominator 3 straight x minus 2 straight y end fraction minus fraction numerator 8.50 over denominator 3 straight x plus 4 straight y end fraction equals 4.5
rightwards double arrow minus fraction numerator 8.50 over denominator straight a end fraction plus 25 over straight b equals 4.5.......... left parenthesis ii right parenthesis
Multiply space equation space left parenthesis ii right parenthesis by space 4 comma space we space get space :
minus 34 over straight a plus 100 over straight b equals 18
space space space 34 over straight a plus 15 over straight b equals 5 space space space space space space space space space space space space space space space open square brackets Equation space left parenthesis straight i right parenthesis close square brackets
bottom enclose plus space space space space space space space plus space space space space space space space space space space space space space space plus end enclose space space space space space space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space space open square brackets Adding close square brackets
thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space thin space 115 over straight b equals 23
rightwards double arrow straight b equals 5
rightwards double arrow 3 straight x minus 2 straight y equals 5.......... left parenthesis iii right parenthesis

Substituting space straight b space equals space 5 space in space equation space left parenthesis straight i right parenthesis comma space we space get
34 over straight a plus 15 over 5 equals 5
rightwards double arrow 2 straight a equals 34
rightwards double arrow straight a equals 17
rightwards double arrow 3 straight x plus space 4 straight y space equals space 17.......... left parenthesis iv right parenthesis
Subtracting space equation space left parenthesis iv right parenthesis from space equation space left parenthesis iii right parenthesis comma space we space get : :
3 straight x minus space 2 straight y space equals space 5
3 straight x plus space 4 straight y space equals space 17
bottom enclose minus space space minus space space space space space thin space thin space thin space thin space minus thin space thin space thin space thin space end enclose
thin space thin space thin space thin space thin space thin space minus 6 straight y space equals space minus 12
rightwards double arrow space straight y space equals space 2
SubStituting space space straight y space equals space 2 space in space equation space left parenthesis iii right parenthesis comma space we space get
3 straight x minus 2 left parenthesis 2 right parenthesis equals 5
rightwards double arrow 3 straight x equals 9
rightwards double arrow straight x equals 3

therefore Solution space is space straight x space equals space 3 space and space straight y space equals space 2.

Solution 7

(i)

From (1)

(ii)

x + y = 7xy...(1)

2x - 3 = -xy...(2)

Multiplying equation no. (1) by 3.

From (1)

 

 

 

Solution 8

  

Solution 9

  

Solution 10

  

Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(F)

Solution 1

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)

 

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.

Solution 2

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)

Subtracting (1)from(2),

 

y = 15

From (1)

x = 15 + 20

x = 35

Thus, present ages of A and B are 35 years and 15 years.

Solution 3

Solution 4

Solution 5

Let A’s annual in come = Rs.x

and B’s annual income = Rs. y

According to the question,


4x - 3y = 0...(1)

and,

 

7x - 5y = 10000...(2)

Multiplying equation no. (1) by 7 and (2) by 4.and subtracting (4) from (3)

 

From (1)

4x - 3 (40000) = 0

x = 30000

Thus, A’s income in Rs. 30,000 and B’s income is Rs. 40,000.

Solution 6

Let the no. of pass candidates be x

and the no. of fail candidates be y.

According to the question,

x - 4y = 0...(1)

and

 

x - 5y = -30...(2)

 

From (1)

- 4(30) = 0

x = 120

Total students appeared = x + y

= 120 + 30

= 150

Solution 7

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

 

From (1)

2(50) - y = 30

y = 70

Thus, A has 50 pencils and B has 70 pencils.

Solution 8

Let the number of adults = x

and the number of children = y

According to the question,

x + y = 1250...(1)

and 75x + 25y = 61250

 

x = 6000

From (1)

600 + y = 1250

y = 650

Thus, number of adults = 600

and the number of children = 650.

Solution 9

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

 

105x + 107y = 1167...(1)

and

 

107x + 105y = 116500...(2)

Adding(1) and (2)

212x + 212y = 233200

x + y = 1100...(3)

subtracting (2)from (1)

-2x + 2y = 200

 

 

y = 600

from (3)

x +600 = 1100

x = 500

Thus, cost price of article A is Rs. 500.

and that of article B is Rs. 600.

Solution 10

Let Pooja’s 1 day work =

and Ritu’s 1 day work =

According the question,

 

and,

 

 

Using the value of y from (2) in (1)

From (2)

y = 30

Pooja will complete the work in 40 days and Ritu will complete the work in 30 days.

Simultaneous (Linear) Equations (Including Problems) Exercise Ex. 6(G)

Solution 1

Let Rohit has Rs. x

and Ajay has Rs. y

When Ajay gives Rs. 100 to Rohit

x + 100 = 2(y - 100)

x - 2y = -300...(1)

When Rohit gives Rs. 10 to Ajay

6(x-10) = y + 10

6x - y = 70...(2)

Multiplying equation no. (2) By 2.

x = 40


From (1)

40 - 2y = -300

-2y = -340

y = 170

 

Thus, Rohit has Rs. 40

and Ajay has Rs. 170

 

Solution 2

Solution 3

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)

66x = 33y

2x - y = 0...(1)

Also,

 

 

From (1) 2(3) - y = 0

y = 6

Thus, Required number = 10(3) + 6 = 36

 

Solution 4

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)

and3x+5y = 124...(2)

Multiplying equation no. (1) by 3 and (2) by 2.

 

 

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.

 

Solution 5

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 6

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) by90, we get,

 

From (1)x + 15 = 21

x = 6

Hence, 90% acid solution is 6 litres and 97% acid solution is 15 litres.

Solution 7

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 8

Weight of Mr. Ahuja = x kg

and weight of Mrs. Ahuja = y kg.

After the dieting,

x - 5 = y

x - y = 5...(1)

and,

7x - 8y = -32...(2)

Multiplying equation no. (1) by 7, we get

 

Now subtracting (2) from (3)

 

 

From (1)

x - 67 = 5x = 72

Thus, weight of Mr. Ahuja = 72 kg.

and that of Mr. Anuja = 67 kg.

Solution 9

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 10

  

Solution 11

  

Solution 12

  

Solution 13

  

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