3 Marks Question

Question 13 Marks

Solve the following question-Aftab tells his daughter, Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be. (Isn’t this interesting?) Represent this situation algebraically and graphically by the method of substitution.

Answer

Let the present age of Aftab and his daughter be x and y years respectively. Then, the pair of linear equations that represent the situation is
x - 7 = 7(y - 7), i.e., x - 7y + 42 = 0 ...(1)
and x + 3 = 3(y + 3), i.e., x - 3y = 6 ...(2)
from equation (2), we get x = 3y + 6
By putting this value of x in equation (1), we get
(3y + 6) -7y + 42 = 0,
i.e., -4y = -48, which gives y = 12
Again by putting this value of y in equation (2), we get
x = $3\times12$ + 6 = 42
So, the present age of Aftab and his daughter are 42 and 12 years respectively.

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Question 23 Marks

Solve the following pair of equations by substitution method:
7x – 15y = 2 ...(1)
x + 2y = 3 ...(2)

Answer

Step 1: By substitution method, we pick either of the equations and write one variable in terms of the other.
7x – 15y = 2 ...(1)
and x + 2y = 3 ...(2)
Let us consider the Equation (2):
x + 2y = 3 and write it as x = 3 – 2y ...(3)
Step 2: Now substitute the value of x in Equation (1)
We get 7(3 – 2y) – 15y = 2
i.e., 21 – 14y – 15y = 2
i.e., – 29y = –19
Therefore $\mathrm{y}=\frac{19}{29}$
Step 3: Substituting this value of y in Equation (3), we get
$\mathrm{x}=3-2\left(\frac{19}{29}\right)=\frac{49}{29}$
Therefore, the solution is x = $\frac{49}{29}, \mathrm{y}=\frac{19}{29}$

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Question 33 Marks

Champa went to a Sale to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased. Help her friends to find how many pants and skirts Champa bought.

Answer

Let us denote the number of pants by x and the number of skirts be y.
Then the equations formed are:
y = 2x - 2 ………… (i)
y = 4x - 4………..(ii)
From (i)
When x = 2, then y = 2
When x = 1, then y = 0

x	2	1
y	2	0

From (ii)
When x = 2, then y = 4
When x = 1, then y = 0

x	2	1
y	4	0

The graphs of two equations of line is shown below.

From the graph, the lines intersect at point (1,0)
Thus, the value of x = 1 and y = 0
Hence, the number of pants she purchased are 2 and the number of skirts she purchased are 0.

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Question 43 Marks

Check whether the pair of equations x + 3y = 6 and 2x – 3y = 12 is consistent. If so, solve them graphically.

Answer

The solution of pair of linear equations:
x + 3y = 6 and 2x – 3y = 12

x	0	6
$y=\frac{6-x}{3}$	2	0

and

x	0	3
$y=\frac{2 x-12}{3}$	-4	-2

Plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on graph paper, and join the points to form the lines AB and PQ

We observe that there is a point B (6, 0) common to both the lines AB and PQ. So, the solution of the pair of linear equations is x = 6 and y = 0, i.e., the given pair of equations is consistent.

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Question 53 Marks

Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received. Solve the pair of the linear equation obtained by the elimination method.

Answer

Let the number of ₹ 50 notes and ₹ 100 notes be x and y respectively.
According to given condition,
Meena got 25 notes in all.
⇒ x + y = 25 ...........(i)
and Meena withdraw ₹ 2000.
⇒ 50x + 100y = 2000 ............(ii)
Multiplying equation (i) by 50, we obtain:
50x + 50y = 1250 ............. (iii)
Subtracting equation (iii) from equation (ii), we obtain:
(50x + 100y ) - (50x + 50y) = 2000 - 1250
50x + 100y - 50x - 50y = 750
50y = 750
y = 15
Substituting the value of y in equation (i), we obtain:
x = 10
Hence, Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.

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Question 63 Marks

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the number. Find the number. Solve the pair of the linear equation obtained by the elimination method.

Answer

Let the unit's digit and the ten's digit in the two-digit number be x and y respectively.
Then the number = 10y + x
Also, the number obtained by reversing the order of the digits = 10x + y
According to the question,
x + y = 9...............(1)
9(10y + x) = 2(10x + y)
$\Rightarrow$ 90y + 9x = 20x + 2y
$\Rightarrow$ 11x - 88y = 0
$\Rightarrow$ x - 8y = 0 ..............(2)
Subtracting equation(2) from equation(1), we get
9y = 9
$\Rightarrow \quad y = \frac { 9 } { 9 } = 1$
Substituting this value of y in equation (1), we get
x + 1 = 9
$\Rightarrow x = 9-1=8$
Hence, the required number is 18.
Verification: substituting x = 8 and y = 1,
we find that both the equations (1) and (2) are satisfied as shown below:
x + y = 8 + 1 = 9
x - 8y = 8 - 8(1) = 0
Hence, the solution is correct.

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Question 73 Marks

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? Solve the pair of the linear equation obtained by the elimination method.

Answer

Let the present age of Nuri and Sonu be x years and y years respectively.
Then, according to the question,
x – 5 = 3(y – 5)
$\Rightarrow $ x – 5 = 3y – 15
$\Rightarrow $x – 3y = –10 ............. (1)
x + 10 = 2(y + 10)
x + 10 = 2y + 20
$\Rightarrow $ x – 2y = 10 ------------ (2)
Subtracting equation (2) from equation (1), we get
– y= – 20
$\Rightarrow $ y = 20
Subtracting equation (2) from equation (1), we get
x – 2(20) = 10
$\Rightarrow $ x – 40 = 10
$\Rightarrow $ x = 40 + 10
$\Rightarrow $ x = 50
Hence, Nuri and Sonu are 50 years and 20 years old respectively at present.
Verification.Subtracting the value of x = 50 and y = 20,we find that both the eqations (1) and (2) are satisfied as shown below:
x – 3y = 50 – 3(20) = 50 – 60 = – 10
x – 2y = 50 – 2(20) = 50 – 40=10
Hence,the solution is correct.

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Question 83 Marks

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $\frac 12$ if we only add 1 to the denominator. What is the fraction? Solve the pair of the linear equation obtained by the elimination method.

Answer

Let the fraction be $\frac xy$
Then, according to the question,
$\frac { x + 1 } { y - 1 } = 1$......(1)
$\frac { x } { y + 1 } = \frac { 1 } { 2 }$.........(2)
$\Rightarrow$x + 1 = y - 1 ...........(3)
2x = y + 1................(4)
$\Rightarrow$x - y = - 2................(5)
2x - y = 1......................(^)
Substituting equation (5) from equation (6), we get x =3
Substituting this value of x in equation (5), we get
3 - y = -2
$\Rightarrow$ y = 3 + 2
$\Rightarrow$ y = 5
Hence, the required fraction is $\frac35$

Verification: Substituting the value of x = 3 and y = 5,
we find that both the equations(1) and ( 2) are satisfied as shown below:
$\frac { x + 1 } { y - 1 } = \frac { 3 + 1 } { 5 - 1 } = \frac { 4 } { 4 } = 1$
$\frac { x } { y + 1 } = \frac { 3 } { 5 + 1 } = \frac { 3 } { 6 } = \frac { 1 } { 2 }$
Hence, the solution is correct.

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Question 93 Marks

The larger of two supplementary angles exceeds the smaller by $18$ degrees. Find them by substitution method.

Answer

Let the larger and smaller of two supplementary angles be $x^{\circ}$ and $y^{\circ}$ respectively. Then, according to the question.
The pair of linear equations formed is $x^{\circ}=y^{\circ}+18^{\circ} x^0+y^0=180^{\circ} \because$ The two angles and supplementary Substitute the value of $x^{\circ}$ from equation (1) in equation (2), we get
$y ^ { \circ } + 18 ^ { \circ } + y ^ { \circ } = 180 ^ { \circ }$
$\Rightarrow \quad 2 y ^ { 0 } + 18 ^ { \circ } = 180 ^ { \circ }$
$\Rightarrow \quad 2 y ^ { \circ } = 180 ^ { \circ } - 18 ^ { \circ }$
$\Rightarrow \quad 2 y = 162 ^ { \circ }$
$\Rightarrow \quad y ^ { \circ } = \frac { 162 ^ { \circ } } { 2 } = 81 ^ { \circ }$
Substituting this value of $y^{\circ}$ in equation (1), we get
$x^0=81^{\circ}+18^{\circ}=99^{\circ}$
Hence, the larger and smaller of the two supplementary angles are $99^{\circ}$ and $81^{\circ}$ respectively. Verification, Substituting $x^{\circ}=99^{\circ}$ and $y^{\circ}=81^{\circ}$, we find that both the equations (1) and (2) are satisfied as shown below:
$y ^ { \circ } + 18 ^ { \circ } = 81 ^ { \circ } + 18 = 99 ^ { \circ } = x ^ { \circ }$
$x ^ { 0 } + y ^ { \circ } = 99 ^ { \circ } + 81 ^ { \circ } = 180 ^ { \circ }$
This verifies the solution.

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Question 103 Marks

The difference between the two numbers is 26 and one number is three times the other. Find them by substitution method.

Answer

Let the two numbers be x and y (x > y) then, according to the question,
the pair of linear equations formed is:
x - y = 26.........(1)
x = 3y.............(2)
Substitute the value of x from equation (2) in equation (1), we get
3y - y = 26
$\Rightarrow \quad 2 y = 26$
$\Rightarrow \quad y = \frac { 26 } { 2 }$
$\Rightarrow \quad y = 13$
Substituting this value of y in equation (2), we get
x = 3(13) = 39
Hence, the required numbers are 39 and 13.
verification: Substituting x = 39 and y = 13, we find that both
the equation (1) and (2) are satisfied as shown below:
x - y = 39 - 13 = 26
3y = 3(13) = 39 = x.
This verifies the solution.

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Question 113 Marks

Solve the pair of linear equations by substitution method: 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3

Answer

0.2 x + 0.3 y = 1.3 ; 0.4 x + 0.5 y = 2.3
The given system of linear equations is:
0.2 x + 0.3 y = 1.3..............(1)
0.4 x + 0.5 y = 2.3...................(2)
From equation (1),
0.3 y = 1.3 - 0.2 x
$\Rightarrow \quad y = \frac { 1.3 - 0.2 x } { 0.3 }$.........................(3)
Substituting this value of y in equation(2), we get
$0.4 x + 0.5 \left( \frac { 1.3 - 0.2 x } { 0.3 } \right) = 2.3$
$\Rightarrow$0.12 x + 0.65 - 0.1 x = 0.69
$\Rightarrow$0.12 x - 0.1 x = 0.69 - 0.65
$\Rightarrow$0.02 x = 0.04
$\Rightarrow$$\mathrm { x } = \frac { 0.04 } { 0.02 } = 2$
Substituting this value of x in equation(3), we get
$y = \frac { 1.3 - 0.2 ( 2 ) } { 0.3 } = \frac { 1.3 - 0.4 } { 0.3 } = \frac { 0.9 } { 0.3 } = 3$
Therefore, the solution is x = 2, y = 3, we find that both equation (1) and (2) are satisfied as shown below:
0.2 x + 0.3 y = ( 0.2 )( 2 )+( 0.3)( 3 ) = 0.4 + 0.9 = 1.3
0.4 x + 0.5 y= ( 0.4 )( 2 ) + ( 0.5 )( 3 ) } = 0.8 + 1.5 = 2.3
This verifies the solution.

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Question 123 Marks

Solve the pair of linear equations by substitution method: x + y = 14; x – y = 4

Answer

x + y = 14; x - y = 4
the given pair of linear equations is
x + y = 14.................(1)
x - y = 4....................(2)
From equation(1),
y = 14 - x...................(3)
Substitute this value of y in equation(2), we get
x - (14 - x) = 4
$\Rightarrow$ x - 14 + x = 4
$\Rightarrow$2x - 14 = 4
$\Rightarrow$2x = 4 + 14
$\Rightarrow$2x = 18
$\Rightarrow x = \frac { 18 } { 2 } = 9$
Substituting this value of x in equation (3), we get y = 14 - 9 = 5
Therefore, the solution is x = , y = 5
verification: Substituting x = 9 and y = 5, we find that both the equations (1) and (2) are satisfied as shown below:
x + y = 9 + 5 = 14
x - y = 9 - 5=4
This verifies the solution.

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Question 133 Marks

Given the linear equation 2x + 3y - 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

intersecting lines
parallel lines
coincident lines

Answer

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:

x	10	0
y	0	10

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.

x	4	0
y	0	-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).

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Question 143 Marks

The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save ₹ 2000 per month, then find their monthly incomes.

Answer

Let us denote the incomes of the two-person by ₹ 9x and ₹ 7x and their expenditures by ₹ 4y and ₹ 3y respectively.
Then the equations formed in the situation is given by :
9x – 4y = 2000 ...(i)
and 7x – 3y = 2000 ...(2)
Step 1: Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of y equal. Then, we get the equations:
27x – 12y = 6000 ...(3)
28x – 12y = 8000 ...(4)
Step 2: Subtract Equation (3) from Equation (4) to eliminate y, because the coefficients of y are the same. So, we get
(28x – 27x) – (12y – 12y) = 8000 – 6000
i.e., x = 2000
Step 3: Substituting this value of x in (1), we get
9(2000) – 4y = 2000
i.e., y = 4000
So, the solution of the equations is x = 2000, y = 4000. Therefore, the monthly incomes of the persons are ₹18,000 and ₹14,000 respectively.

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