Pdf of Previous year questions and Answers _ Click Here
Year- 2014 [Total marks =10 marks]
Question 1
If 12x + 13y= 29 and 13x + 12y =, find x + y . [2 marks]
Answer :
Given equation is,
12x + 13y= 29————- A
13x + 12y=2 1 ——————B
Adding A and B
12x + 13y= 29
13x + 12y=21
25x + 25y = 50
x+ y = 2 —————–(dividing both sides by 25)
Answer: X+ Y = 2
Question 2
Solve the following simultaneous equation equations by crammers Rule: [ 3 marks]
3x – y = 7
X + 4y =11
Answer:
Given equations are,
3x – y = 7 ———–A
X+ 4y =11 ———-B
By crammer’s Rule,
\[{\lvert}D\rvert = \left| \begin{array}{cc} 3 & -1 \\ 1 & 4 \end{array} \right| = 3(4) – (-1) (1) = 13\]
\[{\lvert}Dx\rvert = \left| \begin{array}{cc} 7 & -1 \\ 11 & 4 \end{array} \right| = 7(4) – (-1) (11) = 39\]
\[{\lvert}Dy\rvert = \left| \begin{array}{cc} 3 & 7 \\ 1 & 11 \end{array} \right| = 3(11) – (7) (1) = 26\]
\[X=\frac{Dx}{D}=\frac{39}{13}= 3\]
\[Y=\frac{Dy}{D}=\frac{26}{13}= 2\]
Answer: solution of above equation is ,
X= 3
Y= 2
Question 3
Draw the graphs representing the equation 4x + 3y = 24 and 3y = 4x + 24 on the same graph paper. Write the co-ordinates of the Point of intersection of these lines and find the area of triangle formed by these lines and x-axis. (5 marks)
Answer:
Given simultaneous equation,
Equation 1,
4x + 3y = 24
3y = 24 – 4x
\[y=\frac{24 – 4x}{3}\]
Now finding values of y by changing values of x,
When x=0
\[y=\frac{24 – 4(0)}{3} = \frac{24}{3}=8\]
Here, (x, y) = (0, 8)
When x=3
\[y=\frac{24 – 4(3)}{3} = \frac{12}{3}=4\]
Here, (x, y) = (3, 4)
When x=6
\[y=\frac{24 – 4(6)}{3} = \frac{0}{3}=0\]
Here, (x, y) = (6, 0)
Table for equation 1, 4x + 3y = 24
| x | 0 | 3 | 6 |
| y | 8 | 4 | 0 |
| (x , y) | (0 , 8) | (3 , 4) | (6 ,0) |
Equation 2,
3y = 4x + 24
\[y=\frac{4x+ 24}{3}\]
Now finding values of y by changing values of x,
When x=0
\[y=\frac{4(0)+ 24}{3}=\frac{24}{3}=8\]
Here, (x, y) = (0, 8)
When x= -3
\[y=\frac{4(-3)+ 24}{3}=\frac{12}{3}=4\]
Here, (x, y) = (-3, 4)
When x= -6
\[y=\frac{4(-6)+ 24}{3}=\frac{0}{3}=0\]
Here, (x, y) = ( -6, 0)
Table of Equation 2,
Table for Equation 2, 3y = 4x + 24
| x | 0 | -3 | -6 |
| y | 8 | 4 | 0 |
| (x , y) | (0 , 8) | ( -3, 4) | ( -6, 0) |
Plotting above points on the Graphs,
image
Year- 2015 [Total marks =12 marks]
Question 1
If 12x + 13y= 29 and 13x + 12y =, find x + y . [2 marks]
Answer :
Given equation is,
12x + 13y= 29————- A
13x + 12y=21 ——————B
Adding A and B
12x + 13y= 29
13x + 12y=21
25x + 25y = 50
x+ y = 2 —————–(dividing both sides by 25)
Answer: X+ Y = 2
Question 2
If x + y = 5 and x = 3 then find the value of y. [ 1 marks]
Answer:
Given equation is,
so,
x + y = 5
3 + y = 5
y = 5 – 3
y = 2
Answer : y =2
Question 3
If the point A (3,2) lies on the graph of the equation 5x + ay = 19, then find value of a. [ 2 marks]
Answer:
Given,
Point A(3, 2) lies on 5x + ay = 19 , where x =3 and y = 2
so,
5x + ay = 19
putting values of point A(3, 2)
5 (3) + 2a = 19
15 + 2a = 19
2a = 19 – 15
2a = 4
\[ a = \frac{4}{2}= 2\]
Answer: a = 2
Question 4
Solve the following simultaneous equation by graphical method:
x + y = 6 and x – y = 4
Answer:
Given equations are
x + y = 6 and x – y = 4
Equation 1,
x + y = 6
y = 6 – x
Now finding values of y by changing values of x
When x=0
y = 6 – 0
y = 6
Here, (x, y) = (0, 6)
When x=3
y = 6 – 3
y = 3
Here, (x, y) = (3, 3)
When x=6
y = 6 -6 = 0
Here, (x, y) = (6, 0)
Table for equation 1, x + y = 6
| x | 0 | 3 | 6 |
| y | 6 | 3 | 0 |
| (x , y) | (0 , 6) | (3 , 3) | (6 ,0) |
Equation 2,
x – y = 4
y = x – 4 (exchanging places of y and 4)
Now finding values of y by changing values of x
When x=0
y = 0 – 4
y = -4
Here, (x, y) = (0, -4)
When x= 1
y = 1 – 4
y = -3
Here, (x, y) = (1, -3)
When x= 2
y = 2 – 4
y = -2
Here, (x, y) = ( 2, -2)
Table of Equation 2,
Table for Equation 2, x – y = 4
| x | 0 | 1 | 2 |
| y | -4 | -3 | -2 |
| (x , y) | (0 , -4) | ( 1, -3) | ( 2, -2) |
Question 5
on the first day of the sale of tickets of drama, in all 35 tickets were sold. If the rates tickets were Rs.20 and Rs.40 per ticket and toral collection was Rs. 900, find the number tickers sold of each rate of tickets [5 Marks]
Answer:
Let, No. of tickets sold at Rs. 20 be ‘ x’
No. of tickets sold at Rs. 40 be ‘ y’
According to the first condition,
x + y = 35 ——————————–A
According to the second condition,
20x + 40y = 900
Dividing above equation by 20 on both sides
x + 2y = 45 —————-B
Solving equation A and B by elimination method,
x + y = 35
x + 2y = 45
0x -y = – 10
hence, y = 10
Put y =10 in equation A,
x + y = 35
x + 10 =35
x = 35-10
x = 25
Answer:
No of tickets sold for Rs. 20 = 25, and
No. of tickets sold for Rs. 40 = 10
Year- 2016 [Total marks =12 marks]
Question 1
Find the value of determinant: [1 mark]
\[{\lvert}D\rvert = \left| \begin{array}{cc} 4 & -2 \\ 3 & 1 \end{array} \right|\]
Answer:
\[{\lvert}D\rvert = \left| \begin{array}{cc} 4 & -2 \\ 3 & 1 \end{array} \right|= 4(1) – (3) (-2) = 10\]
Answer : 10
Question 2
If the value of determinant \[{\lvert}D\rvert = \left| \begin{array}{cc} m & 2 \\ -5 & 7 \end{array} \right|\]
is 31 .Find the value of m.
Answer:
\[{\lvert}D\rvert = \left| \begin{array}{cc} m & 2 \\ -5 & 7 \end{array} \right|= 31\]
(m x 7) – (-5 x 2) = 31
7m – (-10) = 31
7m + 10 =31
7m = 31 -10
7m = 21
\[ m = \frac {21}{7} = 3\]
Answer: m =3
Question 3
Solve the following simultaneous equations:
\[\frac{7}{2x + 1} + \frac{13}{y +2} = 27\]
and
\[\frac{13}{2x + 1} + \frac{7}{y + 2} = 33\]
Answer:
Let ,
\[\frac {1}{2x + 1}= m\] and \[\frac{1}{y + 2}= n\]
now, equations become,
7m + 13n = 27 ——–A
13m + 7n = 33 ——– B
adding A and B,
7m + 13n = 27
13m + 7n = 33
20m + 20n = 60
dividing above equation by 20 on both sides
now, m + n = 3 —————C
subtracting equation B from A
7m + 13n = 27
13m + 7n = 33
-6m + 6n = -6
Diving above equation by (-6 ) on both sides,
m – n = 1 ————-D
Solving equation C and D,
m + n = 3
m – n = 1
2m = 4
m =2
put m =2 in equation C,
m + n = 3
2 + n = 3
n = 3 – 2
n = 1
we have
\[\frac {1}{2x + 1}= m\]
\[\frac {1}{2x + 1}= 2\]
2(2x + 1) = 1
4x + 2 = 1
4x = 1-2
4x = -1
\[x= \frac{-1}{4}\]
we have,
\[\frac{1}{y + 2}= n\]
\[\frac{1}{y + 2}= 1\]
y + 2= 1
y= 1 – 2
y = -1
Answer: \[x= \frac{-1}{4} and y= -1 \]
Question 4
A three digit number is equal to 17 times the sum of its digits. If 198 is added to the number , the digits are interchanged . The addition of first and third digit is 1 less than middle digit. Find the number. [5 marks]
Answer:
Let the ones place digit be z
tens place digit be y
hundredth place digit be x
the , the original number = 100x + 10y + z
Interchanged number = 100z+ 10y + x
According to the 1st condition,
100x +10y + z =17(x + y +z)
100x +10y + z =17x + 17y +17z
100x – 17x + 10y- 17y + z- 17z=0
83x -7y -16z =0 ——————A
According to 2nd condition,
100x +10y + z +198 = 100z+ 10y + x
100x +10y + z-100z – 10y – x = -198
99x – 99z = -198
Dividing by 99 both sides,
x – z = -2 i.e x +2 = z —————B
According to 3rd condition,
x +z = y – 1
x + x +2 = y – 1 (From— B)
2x + 2+ 1 =y
2x + 3 = y ————————-C
Substitute values of Z and Y from equation B and C in equation A,
83x – 7y -16z = 0
83x – 7(2x + 3) – 16(x + 2) = 0
83x – 14x -21 -16x -32 =0
83x -30x -53 =0
53x = 53
x =1
Put x = 1 in equation B
x + 2 = z
1 + 2 = z
z = 3
Put x= 1 in equation C
2x + 3 = y
2(1) + 3 =y
y = 5
We have x= 1; y =5 ; z=3
three digit number = 100x + 10y + z
= 100 (1) + 10 (5) + 3
= 100 + 50 + 3
= 153
Year- 2017 [Total marks =12 marks]
Question 1
Write the equation of x-axis. Hence find the point of intersection of the graph of the equation x+ y= 3 with x-axis.[2 marks]
Answer:
Equation of x-axis is y =0
to find the intersection of equation x + y = 3 with x -axis,
put y =0
x + y =3
x +0 = 3
x = 3
Point of intersection will be (x, y) = (3,0)
Question 2
Find the value of k for which the given simultaneous equation have infinitely many solutions: [2 marks]
Kx + 2y =6
9x + 6y = 18
Answer: Given equation are,
Kx + 2y = 6 ———A
9x + 6y = 18
divide above equation by 3
3x + 2y = 6 ———–B
Subtracting B from A
Kx + 2y = 6
_3x +_ 2y = _6
K – 3 = 0
k = 3
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