Maths 5 Marks Questions

1. By Elimination method
   The given system of equation is
   $\frac { x } { 2 } + \frac { 2 y } { 3 } = - 1$...............(1)
   $x - \frac { y } { 3 } = 3$....................(2)
   Multiplying equation (2) by 2, we get
   $2 x - \frac { 2 y } { 3 } = 6$......................(3)
   Adding equation(1) and equation (2), we get
   $\frac { 5 } { 2 } x = 5 \Rightarrow x = \frac { 5 \times 2 } { 5 } \Rightarrow \quad x = 2$
   Substituting this value of x in equation(2), we get
   $2 - \frac { y } { 3 } = 3 \Rightarrow \frac { y } { 3 } = 2 - 3 = - 1 \Rightarrow \quad y = - 3$
   So, the solution of the given system of equation is
   x = 2, y = -3
2. By substitution method
   The given system of equation is
   $\frac { x } { 2 } + \frac { 2 y } { 3 } = - 1$...............(1)
   $x - \frac { y } { 3 } = 3$....................(2)
   From equation (2),
   $x = \frac { y } { 3 } + 3$....................(3)
   Substituting this value of x in (1),
   $\frac { 1 } { 2 } \left( \frac { y } { 3 } + 3 \right) + \frac { 2 y } { 3 } = - 1$
   $\Rightarrow \quad \frac { y } { 6 } + \frac { 3 } { 2 } + \frac { 2 y } { 3 } = - 1 \Rightarrow \quad \frac { 5 y } { 6 } = - 1 - \frac { 3 } { 2 }$
   $\Rightarrow \quad \frac { 5 y } { 6 } - - \frac { 5 } { 2 } \Rightarrow y = - 3$
   Substituting this value of y in equation (3), we get
   $x = - \frac { 3 } { 3 } + 3 = - 1 + 3 - 2$
   So, the solution of the given system of equations is x = 2, y = -3
   Verification: Substituting x = 2, y = -3, we find that both the equation (1) and (2) are satisfied as shown below:
   $\frac { x } { 2 } + \frac { 2 y } { 3 } = \frac { 2 } { 2 } + \frac { 2 ( - 3 ) } { 3 } = 1 - 2 = - 1$

1. By Elimination method,
   The given system of equations is :
   3 x - 5 y - 4 = 0............(1)
   9 x = 2 y + 7
   9 x - 2 y - 7 = 0.............(2)
   Multiplying equation (1) by 3, we get
   9 x - 15 y - 12 = 0.............(3)
   Subtracting equation (3) from equation (2) , we get
   13 y + 5 = 0
   $\Rightarrow \quad 13 y = - 5 \Rightarrow y = \frac { - 5 } { 13 }$
   Substituting this value of y in equation (1), we get
   $3 x - 5 \left( \frac { - 5 } { 13 } \right) - 4 = 0$
   $\Rightarrow \quad 3 x + \frac { 25 } { 13 } - 4 = 0 \Rightarrow 3 x - \frac { 27 } { 13 } = 0$
   $\Rightarrow \quad 3 x = \frac { 27 } { 13 } \Rightarrow x = \frac { 9 } { 13 }$
   So, the solution of the given system of equation is
   $x = \frac { 9 } { 13 } , y = \frac { - 5 } { 13 }$
2. By Substitution method:
   The given system of equation is:
   3 x - 5 y - 4 = 0.............(1)
   9 x = 2 y + 7...................(2)
   From equation (2),
   $x = \frac { 2 y + 7 } { 9 }$..................(3)
   Substituting this value of x in equation(1), we get
   $3 \left( \frac { 2 y + 7 } { 9 } \right) - 5 y - 4 = 0$
   $\Rightarrow \quad \frac { 2 y + 7 } { 3 } - 5 y - 4 = 0$
   $\Rightarrow \quad 2 y + 7 - 15 y - 12 = 0$
   $\Rightarrow$ -13y - 5 = 0
   $\Rightarrow$ 13y = -5
   $\Rightarrow \quad y = \frac { - 5 } { 13 }$
   Substituting this value of y in equation(3), we get
   $x = \frac { 2 \left( \frac { - 5 } { 13 } \right) + 7 } { 9 } = \frac { - \frac { 10 } { 13 } + 7 } { 9 } = \frac { - 10 + 91 } { 117 } = \frac { 81 } { 117 } = \frac { 9 } { 13 }$​​​​

1. By Elimination method,
   The given system of equation is :
   3 x + 4 y = 10 ...................(1)
   2 x - 2 y = 2 ...................(2)
   Multiplying equation(2) by 2, we get
   4 x - 4 y = 4 ...................(2)
   Adding equation (1) and equation (3), we get
   7 x = 14
   $\therefore \quad x = \frac { 14 } { 7 } = 2$
   Substituting this value of x in equation (2), we get
   2(2) - 2y = 2
   $\Rightarrow \quad 4 - 2 y = 2$
   $\Rightarrow \quad 2 y = 4 - 2$
   $\Rightarrow \quad 2 y = 2$
   $\Rightarrow \quad y =\frac22=1$
   So, the solution of the given system of equation is
   x = 2, y = 1
2. By Substitution method,
   The given system of equation is:
   3 x + 4 y = 10.................(1)
   2 x - 2 y = 2....................(2)
   From equation(1)
   $3 x=10-4 y$
   $x=\left(\frac{10-4 y}{3}\right)$
   Put value of x in equation (2),
   $2 x-2 y=2$
   $2\left(\frac{10-4 y}{3}\right)-2 y=2$
   $\frac{2(10-4 y)-2 y(3)}{3}=2$
   $20-8 y-6 y=6$
   $-14 y=-14$
   y = 1
   Putting value of y = 1 in equation (2)
   2x - 2 = 2
   x = 2
   Therefore, x = 2, y = 1 is the solution.
   Verification: Substituting x = 2, y = 1, we find that both the
   equation(1) and (2) are satisfied shown below:
   $3 x + 4 y = 3 ( 2 ) + 4 ( 1 ) = 6 + 4 = 10$
   $2 x - 2 y = 2 ( 2 ) - 2 ( 1 ) = 4 - 2 = 2$
   Hence, the solution is correct.

y = 5 .......... (1)
2x - 3y=4 ............. (2)

1. Elimination method:
   Multiplying equation (1) by 2, we get equation (3)
   2x +2y =10 ............. (3)
   2x−3y =4 ........... (2)
   Subtracting equation (2) from (3), we get
   5y =6 ⇒ y = $\frac{6}{5}$
   Putting value of y in (1), we get
   x + $\frac{6}{5}$=5
   ⇒ x =5− $\frac{6}{5}$= $\frac{{19}}{5}$
   Therefore, x = $\frac{{19}}{5}$and y = $\frac{6}{5}$
2. Substitution method:
   x +y =5 .......................... (1)
   2x−3y =4 ......................... (2)
   From equation (1), we get,
   x =5−y
   Putting this in equation (2), we get
   2(5−y )−3y =4
   ⇒ 10−2y−3y =4
   ⇒ 5y =6
   ⇒ y = $\frac{6}{5}$
   Putting value of y in (1), we get
   x =5−$\frac{6}{5}$=$\frac{{19}}{5}$
   Therefore, x = $\frac{{19}}{5}$and y = $\frac{6}{5}$

Let the cost of each bat and each ball be Rs.x and Rs. y respectively. Then, according to the equation, The pair of linear equations formed is
7x + 6y = 3800 ....... (1)
3x + 5y = 1750 ...... (2)
From equation (2), 5y = 1750 - 3x
$y = \frac { 1750 - 3 x } { 5 }$ ......... (3)
Substitute this value of y in equation (1), we get
$7 x + 6 \left( \frac { 1750 - 3 x } { 5 } \right) = 3800$

$\Rightarrow \quad 35 x + 10500 - 18 x = 19000$
$\Rightarrow \quad 17 x + 10500 = 19000$
$\Rightarrow \quad 17 x = 19000 - 10500$
$\Rightarrow \quad 17 x = 8500$
$\Rightarrow \quad x = \frac { 8500 } { 17 } = 500$

Substituting this value of x in equation (3), we get
$y = \frac { 1750 - 3 ( 500 ) } { 5 } = \frac { 1750 - 1500 } { 5 } = \frac { 250 } { 5 } = 50$
Hence, the cost of each bat and each ball is Rs.500 and Rs.50 respectively.
Verification,
Substituting x = 500 and y = 50, we find that both the equations (1) and (2) are satisfied as shown below:
$7 x + 6 y = 7 ( 500 ) + 6 ( 50 )$
$= 3500 + 300 = 3800$
$3 x + 5 y = 3 ( 500 ) + 5 ( 50 )$
$= 1500 + 250 = 1750$. This verifies the solution.

Let x (in years) be the present age of Jacob's son and y (in years) be the present age of Jacob. 5 years hence, it has relation:

(y + 5) = 3(x + 5)
or, y + 5 = 3x + 15

3x + 15 - y - 5 = 0

or, 3 x - y + 10 = 0 .......(i)

5 years ago, it has relation

(y - 5) = 7(x - 5)

y - 5 = 7x - 35

or, 7x - 35 - y + 5 = 0

or, 7 x - y - 3 0 = 0 ....(ii)

From eqn. (i), y = 3x + 10 ....(iii)

On substituting the value of y in eqn. (ii), we get

7x-(3x + 10) - 30 = 0

7x - 3x - 10 - 30 = 0

or, 4x - 40 = 0

or, 4x = 40

x=10

On substituting x = 10 in eqn. (iii),

$y = 3 \times 10 + 10$

y = 30 + 10

$\therefore $ y = 40

Hence, the present age of Jacob = 40 years and son's age = 10 years

Let the numerator be x and denominator be y

$$if 2 is added to both numerator and denominator, the fraction becomes $\frac { 9 } { 11 }$

$\frac { x + 2 } { y + 2 } = \frac { 9 } { 11 }$

11(x + 2) = 9( y + 2)

= 11x + 22 = 9y + 18

or, 11x + 22 - 9y - 18 = 0

or, 11 x - 9 y + 4 = 0 ........(i)

If 3 is added to both numerator and denominator the fraction becomes $\frac { 5 } { 6 }$

and $\frac { x + 3 } { y + 3 } = \frac { 5 } { 6 }$

6(x+3) = 5(y+3)

6x +18 = 5y + 15

or, 6x + 18 - 5y - 15 = 0

or, 6x -5y + 3 = 0 ...(ii)

On comparing with ax + by + c = 0

we get $a _ { 1 } = 11 , b _ { 1 } = -9 , c _ { 1 } = 4$

$a _ { 2 } = 6 , b _ { 2 } = - 5 \text { and } c _ { 2 } = 3$

Now, $\frac { x } { b _ { 1 } c _ { 2 } - b _ { 2 } c _ { 1 } } = \frac { y } { c _ { 1 } a _ { 2 } - c _ { 2 } a _ { 1 } } = \frac { 1 } { a _ { 1 } b _ { 2 } - a _ { 2 } b _ { 1 } }$

$\frac { x } { (-9 ) ( 3 ) - ( - 5 ) ( 4 ) } = \frac { y } { ( 4 ) ( 6 ) - ( 3 ) ( 11 ) } =\frac{1}{(11)(-5)-(6)-(9)}$
$\frac { x } { - 27 + 20 } = \frac { y } { 24 - 33 } = \frac { 1 } { - 55 + 54 }$

$\Rightarrow \quad \frac { x } { - 7 } = \frac { y } { - 9 } = \frac { 1 } { - 1 }$

$\Rightarrow \quad \frac { x } { - 7 } = -1$

or, x = 7

Hence, x=7, y=9

$\therefore$ Fraction =$\frac { 7 } { 9 }$

Let fixed charge be ₹x and the charge per km be ₹y.
For a distance of 10 km, the charge paid is ₹105 .
x + 10y = 105 ....(i)
For a journey of 15 km the charge paid is ₹155
x + 15y = 155 .... (ii)
From eqn. (i), x = 105 -10y ...(iii)
On substituting x from eqn. (iii) in eqn. (ii),
105 - 10y + 15y = 155
$\Rightarrow$ 5y = 155 - 105
$\Rightarrow$ 5y = 50
$\Rightarrow$ y = 10
Put y = 10 in (iii)
x = 105 - 10(10)
$\Rightarrow$ x = 105 - 100
$\therefore $ x = 5
Hence, fixed charges = ₹5
Rate per km = ₹10
Amount to be paid for travelling 25 km
= ₹5 + ₹10 $\times$ 25
= ₹5 + ₹250
= ₹255

The given pair of linear equations
2x + 3y = 11 ...... (1)
2x - 4y = -24 ....... (2)
From equation (1), 3y = 11 - 2x
$\Rightarrow \quad y = \frac { 11 - 2 x } { 3 }$
Substituting this value of y in equation (2), we get
$2 x - 4 \left( \frac { 11 - 2 x } { 3 } \right) = - 24$
$\Rightarrow 6 x - 44 + 8 x = - 72$
$\Rightarrow 14 x - 44 = - 72$
$\Rightarrow 14 x = 44 - 72$
$\Rightarrow 14 x = - 28$
$\Rightarrow x = - \frac { 28 } { 14 } = - 2$
Substituting this value of x in equation (3), we get
$y = \frac { 11 - 2 ( - 2 ) } { 3 } = \frac { 11 + 4 } { 3 } = \frac { 15 } { 3 } = 5$

Verification, Substituting x = -2 and y = 5, we find that both the equations (1) and (2) are satisfied as shown below:
$2 x + 3 y = 2 ( - 2 ) + 3 ( 5 ) = - 4 + 15 = 11$
$2 x - 4 y = 2 ( - 2 ) - 4 ( 5 ) = - 4 - 20 = - 24$
This verifies the solution,
Now, y = axe + 3
$\Rightarrow 5 = m ( - 2 ) + 3$
$\Rightarrow - 2 m = 5 - 3$
$\Rightarrow - 2 m = 2$
$\Rightarrow \mathrm { m } = \frac { 2 } { - 2 } = - 1$

$\frac { 3 x } { 2 } - \frac { 5 y } { 3 } = - 2 ; \frac { x } { 3 } + \frac { y } { 2 } = \frac { 13 } { 6 }$
The given system of linear equation is
$\frac { 3 x } { 2 } - \frac { 5 y } { 3 } = - 2 \quad \dots \ldots \ldots \ldots ( 1 )$
$\frac { x } { 3 } + \frac { y } { 2 } = \frac { 13 } { 6 }$....... (2)
$\Rightarrow \quad 9 x - 10 y = - 12$........ (3)
2x + 3y = 13 ...... (4)
From equation (3)
9x - 10y = -12
9x = 10y - 12
$x = \frac { 10 y - 12 } { 9 }$
Substituting the value of y in equation (4), we get
$2 \left( \frac { 10 y - 12 } { 9 } \right) + 3 y = 13$
$20 y - 24 + 27 y = 117$
$47 y = 117 + 24$
$y = \frac { 141 } { 47 }$
y = 3
Substituting the value of y in equation (4), we get
$2 x + 3 \times 3 = 13$
$2 x + 9 = 13$
$2 x = 13 - 9$
$x = \frac { 4 } { 2 } = 2$
Therefore, the solution is
x = 2, y = 3
Verification, Substituting x = 2 and y = 3, we find that both the equations (1) and (2) are satisfied as shown below:
$\frac { 3 } { 2 } x - \frac { 5 y } { 3 } = \frac { 3 } { 2 } ( 2 ) - \frac { 5 } { 3 } ( 3 ) = 3 - 5 = - 2$
$\frac { x } { 3 } + \frac { y } { 2 } = \frac { 2 } { 3 } + \frac { 3 } { 2 } = \frac { 13 } { 6 }$
This verifies the solution.

s - t = 3; $\frac { s } { 3 } + \frac { t } { 2 } = 6$
The given pair of linear equations is :
s - t = 3...............(1)
$\frac { s } { 3 } + \frac { t } { 2 } = 6$..........(2)
From equation(1),
s = t + 3..............(3)
Substitute this value of s in equation(2), we get
$\frac { t + 3 } { 3 } + \frac { t } { 2 } = 6$
$\Rightarrow \quad \frac { 2 ( t + 3 ) + 3 t } { 6 } = 6$
$\Rightarrow $2(t + 3) + 3t = 36
$\Rightarrow $ 2t + 6 + 3t = 36
$\Rightarrow $ 5t + 6 = 36
$\Rightarrow $ 5t = 30
$\Rightarrow \quad t = \frac { 30 } { 5 } = 6$
Substituting this value of t in equation (3), we get
s = 6 + 3= 9
therefore the solution is
s = 9, t = 6
Verification : Substituting s = 9 and t = 6, we find that both equation (1) and (2) are satisfied as shown below:
s - t = 9 - 6 = 3
$\frac { s } { 3 } + \frac { t } { 2 } = \frac { 9 } { 3 } + \frac { 6 } { 2 } = 3 + 3 = 6$
This verifies the solution.

The given equations are
x - y + 1 = 0 ...(1)
3x + 2y - 12 = 0 ...(2)
Let us draw the graphs of equations (1) and (2) by finding two solutions for each of these equations. These two solutions of these equations (1) and (2) are given below in table 1 and table 2 respectively.
For equation (1) x - y + 1 = 0
$\Rightarrow$ y = x + 1
Table 1 of solutions

For equation (2) 3x + 2y - 12 = 0 $\Rightarrow y = \frac{{12 - 3x}}{2}$
Table 2 of solutions

We plot the points A(0, 1) and B(-1, 0) on a graph paper and join these points to form the line AB representing the equation (1) as shown in the figure. Also, we plot the points C(4, 0) and D(0, 6) on the same graph paper and join these points to form the line CD representing the equation (2) and shown in the same figure.
In the figure, we observe that the coordinates of the vertices of the triangle formed by these given lines and the x-axis are E(2, 3 ), B(-1, 0) and C(4, 0)

The triangular region EBC has been shaded and the area of triangular region EBC = $\frac12 (5)(3)=\frac{15}{2}$

Let, cost(in RS) of one pencil = x
and cost (in RS) of one pen = y
Therefore , according to question
5x+7y = 50 ........ (1)
7x + 5y = 46 .........(2)
Multiply equation (1) by 7 and equation (2) by 5 we get
7(5x+7y)= 7 $\times$ 50
35x +49y = 350 .......(3)
and
5(7x +5y) = 5 $\times$ 46
35x +25y = 230 ....... (4)
Subtract equation (4) from equation 3 , we get
35x + 49y - 35x - 25y = 350 -230
49y -25y = 120
24y = 120
y = $\frac{120}{24}$
y= 5
Substitute y = 5 in equation 1 , we get
5x + 7 $\times$ 5 =50
5x + 35 = 50
5x = 50 - 35
5x = 15
x= $\frac{15}{5}$
x=3
Hence, Cost of One Pen = y =5

Formulation: Let the number of girls be x and the number of boys be y.
It is given that total ten students took part in the quiz.
$\therefore$ Number of girls+ Number of boys = 10
i.e. x + y =10
It is also given that the number of girls is 4 more than the number of boys.
$\therefore$ Number of girls= Number of boys + 4
i.e. x = y+4
or, x-y = 4
Thus, the algebraic representation of the given situation is
x + y=10 ........(i)
x - y =4 ..........(ii)
Add (i) and (ii) we get
x + y + x - y = 10 + 4
2x = 14
x = 7
Put x = 7 in (i)
x + y = 10
7 + y = 10
y = 10 -7
y = 3
So, value of x = 7 and y = 3
Graphical Representation: Now putting y = 0 in x + y = 10, we get
x = 10. Similarly, by putting x = 0 in x + y = 10, we get y = 10.
Thus, two solution of equation (i) are:

Similarly, two solutions of equation (ii) are:
putting y = 0 in x - y = 4, we get
x = 4. Similarly, by putting x = 0 in x + y = 10, we get y = -4.

Now, we plot the points A (10, 0), B (0, 10), P (4, 0) and Q (0, -4) corresponding to these solutions on the graph paper and draw the lines AB and PQ representing the equations x + y = 10 and x - y - 4 as shown in Fig.

We observe that the two lines representing the two equations are intersecting at the point (7, 3).

Suppose, the digit at units and tens place of the given number be x and y respectively.
$\therefore$ the number is $10y + x$
After interchanging the digits, the number becomes $10x + y$
Given: The sum of the numbers obtained by interchanging the digits and the original number is 66.
Thus, $(10x + y) + (10y + x) =66$
$\Rightarrow$ $10x + y + 10y + x = 66$
$\Rightarrow$ $11x +11y =66$
$\Rightarrow$ $11(x + y) = 66$
$\Rightarrow x + y = \frac{{66}}{{11}}$
$\Rightarrow$ $x + y = 6$ .....(i)
Also given, the two digits of the number are differing by 2.
$\therefore$ we have $x - y = ±2$....(ii)
So, we have two systems of simultaneous equations,
$x - y = 2, \;x + y = 6$
$x - y = -2, \;x + y = 6$
Here x and y are unknowns. We have to solve the above systems of equations for x and y.

1. First, we solve the system
   $x - y = 2$
   x + y = 6
   Adding the two equations,
   $\Rightarrow(x - y) + (x + y) = 2 + 6$
   $\Rightarrow$ $x - y + x + y = 8$
   $\Rightarrow$ $2x = 8$
   $\Rightarrow x = \frac{8}{2} $
   $\Rightarrow$ $x = 4$
   Substituting the value of x in the first equation, we have
   $4 - y = 2$
   $\Rightarrow$ $y = 4 - 2$
   $\Rightarrow$ $y = 2$
   Hence, the number is 10 $\times$ 2 + 4 = 24
2. Now, we solve the system
   $x - y= -2$
   $x + y = 6$
   Adding the two equations, we have
   $(x - y) + (x + y) = -2 + 6$
   $\Rightarrow$ $x - y + x + y = 4$
   $\Rightarrow$ $2x = 4$
   $\Rightarrow x = \frac{4}{2} $
   $\Rightarrow$ x = 2
   Substituting the value of x in the first equation,
   $\Rightarrow2 - y = -2$
   $\Rightarrow$ $y = 2 + 2$
   $\Rightarrow$ y = 4
   Hence, the number is 10 $\times$ 4 + 2 = 42
   
   Thus, the two numbers are 24 and 42.