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CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 3 - MCQExams.com
CBSE
Class 12 Commerce Applied Mathematics
Linear Equations
Quiz 3
Solve the following simultaneous equations:
37
x
+
29
y
=
13
;
29
x
+
37
y
=
53
Report Question
0%
x
=
1
,
y
=
4
0%
x
=
3
,
y
=
−
1
0%
x
=
−
2
,
y
=
3
0%
x
=
1
,
y
=
−
2
Explanation
Consider the equations:
37
x
+
29
y
=
13........
(
1
)
and
29
x
+
37
y
=
53........
(
2
)
Multiplying equation 1 by
29
and
equation 2 by
37
, we get:
1073
x
+
841
y
=
377........
(
3
)
and
1073
x
+
1369
y
=
1961........
(
4
)
Now subtracting equations 1 from equation 2 we get:
1073
x
+
1369
y
−
1073
x
−
841
y
=
1961
−
377
or
528
y
=
1584
or
y
=
3
Substitute the value of
y
in equation 1:
37
x
+
87
=
13
37
x
=
−
74
x
=
−
2
Hence,
x
=
−
2
,
y
=
3
.
Find the values of the
x
+
y
and
x
−
y
from the examples given below without solving for x and y
4
x
+
3
y
=
24
;
3
x
+
4
y
=
25
Report Question
0%
7
,
−
1
0%
1
,
−
7
0%
1
,
7
0%
−
7
,
−
1
Explanation
The given equations are
4
x
+
3
y
=
24
...(i)
3
x
+
4
y
=
25
...(ii)
On adding (i) and (ii), we get
7
x
+
7
y
=
49
⇒
x
+
y
=
7
On subtracting (ii) from (i), we get
x
−
y
=
−
1
∴
x
+
y
=
7
a
n
d
x
−
y
=
−
1
Find the values of the
x
+
y
and
x
−
y
from the examples given below without solving for x and y
7
x
−
5
y
=
−
1
;
5
x
−
7
y
=
−
11
Report Question
0%
−
1
,
−
5
0%
1
,
−
5
0%
5
,
1
0%
5
,
−
1
Explanation
The given equations are
7
x
−
5
y
=
−
1
...(i)
5
x
−
7
y
=
−
11
...(ii)
On adding (i) and (ii), we get
12
x
−
12
y
=
−
12
⇒
x
−
y
=
−
1
On subtracting (ii) from (i), we get
2
x
+
2
y
=
10
⇒
x
+
y
=
5
∴
x
+
y
=
5
a
n
d
x
−
y
=
−
1
Solve the following simultaneous equations.
3
x
+
4
y
=
18
;
4
x
+
3
y
=
17
Report Question
0%
x
=
2
,
y
=
3
0%
x
=
−
2
,
y
=
5
0%
x
=
4
,
y
=
3
0%
x
=
−
2
,
y
=
−
3
Explanation
The given equations are
3
x
+
4
y
=
18
...(i)
4
x
+
3
y
=
17
...(ii)
On multiplying (i) by 4 and (ii) by 3 and subtracting, we get
12
x
+
16
y
=
72
1
−
2
x
+
−
9
y
=
5
−
1
_
7
y
=
21
⇒
y
=
3
On putting
y
=
3
in (i), we get
3
x
+
12
=
18
⇒
3
x
=
6
⇒
x
=
2
∴
x
=
2
,
y
=
3
Solve the following simultaneous equations:
15
x
−
17
y
=
28
;
15
y
−
17
x
+
36
=
0
Report Question
0%
x
=
3
,
y
=
1
0%
x
=
−
3
,
y
=
1
0%
x
=
6
,
y
=
2
0%
x
=
2
,
y
=
3
Explanation
The given equations are
15
x
−
17
y
=
28
...(i)
17
x
−
15
y
=
36
...(ii)
On multiplying (i) by 17 and (ii) by 15 and subtracting, we get
255
x
−
289
y
=
476
2
−
55
x
−
+
225
y
=
5
−
40
_
−
64
y
=
−
64
⇒
y
=
1
On putting
y
=
1
in (i), we get
15
x
−
17
=
28
⇒
15
x
=
45
⇒
x
=
3
∴
x
=
3
,
y
=
1
Solve following pair of equations by equating the coefficient method:
2
x
−
y
=
9
3
x
−
7
y
=
19
Report Question
0%
x
=
2
,
y
=
1
0%
x
=
4
,
y
=
−
1
0%
x
=
3
,
y
=
2
0%
x
=
7
,
y
=
−
8
Explanation
Multiply the equation
2
x
−
y
=
9
by
7
to make the coefficients of
y
equal. Then we get the equations:
14
x
−
7
y
=
63.........
(
1
)
3
x
−
7
y
=
19.........
(
2
)
Subtract Equation (2) from Equation (1) to eliminate
y
, because the coefficients of
y
are the same. So, we get
(
14
x
−
3
x
)
+
(
−
7
y
+
7
y
)
=
63
−
19
i.e.
11
x
=
44
i.e.
x
=
4
Substituting this value of
x
in (2), we get
12
−
7
y
=
19
i.e.
−
7
y
=
19
−
12
i.e.
−
7
y
=
7
i.e.
y
=
−
1
Hence, the solution of the equations is
x
=
4
,
y
=
−
1
.
Solve the following simultaneous equation:
4
x
+
5
y
=
10
;
5
x
+
4
y
=
17
Report Question
0%
x
=
3
y
=
2
0%
x
=
1
y
=
4
0%
x
=
0
y
=
3
0%
x
=
5
y
=
−
2
Explanation
4
x
+
5
y
=
10
.....(1)
5
x
+
4
y
=
17
.......(2)
Adding (1) and (2). we get
9
x
+
9
y
=
27
⇒
x
+
y
=
3
......(3)
Now, subtracting (1) from (2). we get,
x
−
y
=
7
.......(4)
Adding (3) and (4). we get
2
x
=
10
⇒
x
=
5
Now, subtracting (3) from (4). we get,
−
2
y
=
4
⇒
y
=
−
2
∴
x
=
5
,
y
=
−
2
Option D is correct.
Solve the given pair of equations by substitution method:
x
+
5
y
=
18
3
x
+
2
y
=
41
Report Question
0%
(
13
,
1
)
0%
(
1
,
3
)
0%
(
12
,
17
)
0%
(
9
,
16
)
Explanation
We have,
x
+
5
y
=
18
⇒
x
=
18
−
5
y
..........(1)
and
3
x
+
2
y
=
41
.......(2)
Now substitute value x in terms of y from (1) to (2)
⇒
3
(
18
−
5
y
)
+
2
y
=
41
⇒
54
−
15
y
+
2
y
=
41
⇒
15
y
−
2
y
=
54
−
41
⇒
13
y
=
13
⇒
y
=
13
13
=
1
So
x
=
18
−
5
y
=
18
−
5
(
1
)
=
13
Hence solution set is
(
x
,
y
)
=
(
13
,
1
)
Solve the given pair of equations by substitution method:
x
+
y
=
11
x
−
y
=
−
3
Report Question
0%
(
4
,
6
)
0%
(
3
,
11
)
0%
(
4
,
7
)
0%
(
6
,
2
)
Explanation
We pick either of the equations and write one variable in terms of the other.
Let us consider the equation
x
−
y
=
−
3
and write it as
x
=
−
3
+
y
.
.
.
.
.
.
.
.
.
.
.
(
1
)
Substitute the value of
x
in Equation
x
+
y
=
11
. We get
−
3
+
y
+
y
=
11
i.e.
−
3
+
y
+
y
=
11
i.e.
−
3
+
2
y
=
11
i.e.
2
y
=
14
Therefore,
y
=
7
Substituting this value of
y
in Equation (1), we get
x
=
−
3
+
7
=
4
Hence, the solution is
x
=
4
,
y
=
7
.
Solve the given pair of equations by substitution method:
4
a
−
b
=
10
2
a
+
3
b
=
12
Report Question
0%
a
=
0
,
b
=
5
0%
a
=
7
,
b
=
6
0%
a
=
1
,
b
=
7
0%
a
=
3
,
b
=
2
Explanation
We have,
4
a
−
b
=
10
⇒
b
=
4
a
−
10
.........(1)
and
2
a
+
3
b
=
12
........(2)
Now substitute value of b in terms of a in (2)
⇒
2
a
+
3
(
4
a
−
10
)
=
12
⇒
2
a
+
12
a
−
30
=
12
⇒
14
a
=
12
+
30
=
42
⇒
a
=
42
14
=
3
So
b
=
4
a
−
10
=
4
(
3
)
−
10
=
12
−
10
=
2
Hence option 'D' is correct choice
Solve following pair of equations by equating the coefficient method:
8
x
=
5
y
13
x
=
8
y
+
1
Report Question
0%
x
=
3
and
y
=
1
0%
x
=
5
and
y
=
8
0%
x
=
6
and
y
=
2
0%
x
=
9
and
y
=
−
5
Explanation
Rewriting the given equations, we get
8
x
−
5
y
=
0
...(i)
13
x
−
8
y
=
1
...(ii)
On multiplying (i) by 13 and (ii) by 8, we get
104
x
−
65
y
=
0
1
−
04
x
−
+
64
y
=
8
−
_
−
y
=
−
8
∴
y
=
8
On putting
y
=
8
in (i), we get
x
=
5
y
8
⇒
x
=
5
∴
x
=
5
,
y
=
8
Solve the given pair of equations by substitution method:
x
−
4
y
=
−
8
x
−
2
y
=
0
Report Question
0%
(
2
,
−
1
)
0%
(
7
,
−
6
)
0%
(
8
,
4
)
0%
(
−
3
,
6
)
Explanation
We have,
x
−
4
y
=
−
8
⇒
x
=
4
y
−
8
..........(1)
and
x
−
2
y
=
0
.......(2)
Now substitute value of x in terms of y from (1) to (2)
⇒
4
y
−
8
−
2
y
=
0
⇒
2
y
=
8
⇒
y
=
8
2
=
4
So
x
=
4
y
−
8
=
4
(
4
)
−
8
=
8
Hence solution set is
(
x
,
y
)
=
(
8
,
4
)
Solve the given pair of equations by substitution method:
2
a
+
3
b
=
6
3
a
+
5
b
=
15
Report Question
0%
a
=
3
,
b
=
2
0%
a
=
−
15
,
b
=
12
0%
a
=
8
,
b
=
15
0%
a
=
−
7
,
b
=
3
Explanation
We pick either of the equations and write one variable in terms of the other.
Let us consider the equation
2
a
+
3
b
=
6
and write it as
b
=
6
−
2
a
3
.
.
.
.
.
.
.
.
.
(
1
)
Substitute the value of
b
in Equation
3
a
+
5
b
=
15
. We get
3
a
+
5
(
6
−
2
a
3
)
=
15
⇒
9
a
+
5
(
6
−
2
a
)
=
45
⇒
9
a
+
30
−
10
a
=
45
⇒
−
a
=
45
−
30
⇒
a
=
−
15
Therefore,
a
=
−
15
Substituting this value of
a
in Equation (1), we get
b
=
6
−
(
−
30
)
3
b
=
36
3
=
12
Hence, the solution is
a
=
−
15
,
b
=
12
.
Solve the following pairs of equations:
x
+
1
4
=
2
3
(
1
−
2
y
)
2
+
5
y
3
=
x
7
−
2
Report Question
0%
x
=
8
,
y
=
−
1
0%
x
=
3
,
y
=
−
1
0%
x
=
7
,
y
=
−
1
0%
x
=
9
,
y
=
−
1
Solve the given pair of equations by substitution method:
x
+
y
=
0
y
−
x
=
6
Report Question
0%
(
2
,
7
)
0%
(
−
3
,
3
)
0%
(
4
,
9
)
0%
(
−
6
,
2
)
Explanation
We have,
x
+
y
=
0
⇒
x
=
−
y
..........(1)
and
y
−
x
=
6
.......(2)
Now substitute value of x in terms of y from (1) to (2)
⇒
y
−
(
−
y
)
=
6
⇒
y
+
y
=
6
⇒
2
y
=
6
⇒
y
=
6
2
=
3
x
=
−
y
=
−
(
3
)
=
−
3
Hence solution set is
(
x
,
y
)
=
(
−
3
,
3
)
Solve the following pair of simultaneous equations:
x
3
+
x
+
y
6
=
3
;
y
3
−
x
−
y
2
=
6
Report Question
0%
x
=
2
,
y
=
−
5
0%
x
=
7
,
y
=
4
0%
x
=
6
,
y
=
3
0%
x
=
3
,
y
=
9
Explanation
x
3
+
x
+
y
6
=
3
⇒
2
x
+
x
+
y
6
=
3
⇒
3
x
+
y
=
18
...(i)
y
3
−
x
−
y
2
=
6
⇒
2
y
−
3
x
+
3
y
6
=
6
⇒
−
3
x
+
5
y
=
36
...(ii)
On adding (i) and (ii), we get
3
x
+
y
=
18
−
3
x
+
5
y
=
36
_
6
y
=
54
∴
y
=
9
On putting
y
=
9
in (i), we get
3
x
+
9
=
18
⇒
3
x
=
9
⇒
x
=
3
∴
x
=
3
,
y
=
9
Solve the following pair of simultaneous equations:
4
x
−
3
y
=
8
3
x
−
4
y
=
−
1
Report Question
0%
x
=
3
,
y
=
−
1
0%
x
=
5
,
y
=
4
0%
x
=
−
4
,
y
=
2
0%
x
=
7
,
y
=
−
6
Explanation
Multiply the equation
4
x
−
3
y
=
8
by
3
and
equation
3
x
−
4
y
=
−
1
by
4
to make the coefficients of
x
equal. Then we get the equations:
12
x
−
9
y
=
24.........
(
1
)
12
x
−
16
y
=
−
4.........
(
2
)
Subtract Equation (2) from Equation (1) to eliminate
x
, because the coefficients of
x
are the same. So, we get
(
12
x
−
12
x
)
+
(
−
9
y
+
16
y
)
=
24
+
4
i.e.
7
y
=
28
i.e.
y
=
4
Substituting this value of
y
in
the equation
4
x
−
3
y
=
8
, we get
4
x
−
12
=
8
i.e.
4
x
=
20
i.e.
x
=
5
Hence, the solution of the equations is
x
=
5
,
y
=
4
.
Solve the following pair of simultaneous equations:
8
a
−
7
b
=
1
4
a
=
3
b
+
5
Report Question
0%
a
=
8
,
b
=
9
0%
a
=
3
,
b
=
−
8
0%
a
=
7
,
b
=
−
6
0%
a
=
1
,
b
=
−
1
Explanation
Multiply the equation
4
a
=
3
b
+
5
or
4
a
−
3
b
=
5
by
2
to make the coefficients of
a
equal. Then we get the equations:
8
a
−
7
b
=
1.........
(
1
)
8
a
−
6
b
=
10.........
(
2
)
Subtract Equation (2) from Equation (1) to eliminate
a
, because the coefficients of
a
are the same. So, we get
(
8
a
−
8
a
)
+
(
−
7
b
+
6
b
)
=
1
−
10
i.e.
−
b
=
−
9
i.e.
b
=
9
Substituting this value of
b
in
the equation
8
a
−
7
b
=
1
, we get
8
a
−
63
=
1
i.e.
8
a
=
64
i.e.
a
=
8
Hence, the solution of the equations is
a
=
8
,
b
=
9
.
Solve the following pair of linear (simultaneous) equations by the method of elimination:
y
=
4
x
−
7
16
x
−
5
y
=
25
Report Question
0%
x
=
3
2
,
y
=
6
0%
x
=
5
2
,
y
=
3
0%
x
=
7
4
,
y
=
2
0%
x
=
9
2
,
y
=
0
Explanation
4
x
−
y
=
7
...
(
1
)
16
x
−
5
y
=
25
...
(
2
)
Multiplying
(
1
)
by
4
, we get
16
x
−
4
y
=
28
...
(
3
)
16
x
−
5
y
=
25
...
(
2
)
Subtracting
(
3
)
and
(
2
)
,
y
=
3
Substituting
y
=
3
in
(
1
)
, we get
x
as
4
x
−
3
=
7
4
x
=
10
x
=
10
4
or
x
=
5
2
x
=
5
2
,
y
=
3
Option
B
Solve the following pair of simultaneous equations:
5
x
−
6
y
=
8
7
y
−
15
x
=
9
Report Question
0%
(1 , 3)
0%
(6 , -1)
0%
(-2 , -3)
0%
(3 , -5)
Explanation
Multiply the equation
5
x
−
6
y
=
8
by
3
to make the coefficients of
x
equal.
Another
equation is
7
y
−
15
x
=
9
or
−
15
x
+
7
y
=
9
.
Then we get the equations:
15
x
−
18
y
=
24.........
(
1
)
−
15
x
+
7
y
=
9.........
(
2
)
Add Equations 1 and 2 to eliminate
x
, because the coefficients of
x
are the same. So, we get
(
15
x
−
15
x
)
+
(
−
18
y
+
7
y
)
=
24
+
9
i.e.
−
11
y
=
33
i.e.
y
=
−
3
Substituting this value of
y
in
the equation
5
x
−
6
y
=
8
, we get
5
x
+
18
=
8
i.e.
5
x
=
−
10
i.e.
x
=
−
2
Hence, the solution of the equations is
x
=
−
2
,
y
=
−
3
.
Solve following pair of equations by equating the coefficient method:
3
x
−
7
y
=
35
2
x
+
5
y
=
4
Report Question
0%
x
=
5
,
y
=
−
1
0%
x
=
4
,
y
=
−
1
0%
x
=
5
,
y
=
−
7
0%
x
=
7
,
y
=
−
2
Explanation
Multiply the equation
3
x
−
7
y
=
35
by
2
and
equation
2
x
+
5
y
=
4
by
3
to make the coefficients of
x
equal. Then we get the equations:
6
x
−
14
y
=
70.........
(
1
)
6
x
+
15
y
=
12.........
(
2
)
Subtract Equation (2) from Equation (1) to eliminate
x
, because the coefficients of
x
are the same. So, we get
(
6
x
−
6
x
)
+
(
−
14
y
−
15
y
)
=
70
−
12
i.e.
−
29
y
=
58
i.e.
y
=
−
2
Substituting this value of
y
in (2), we get
6
x
−
30
=
12
i.e.
6
x
=
42
i.e.
x
=
7
Hence, the solution of the equations is
x
=
7
,
y
=
−
2
.
Solve following pair of equations by equating the coefficient method:
x
+
2
y
=
11
2
x
−
y
=
2
Report Question
0%
x
=
3
,
y
=
4
0%
x
=
5
,
y
=
4
0%
x
=
−
1
,
y
=
−
1
0%
x
=
6
,
y
=
−
4
Explanation
Multiply the equation
x
+
2
y
=
11
by
2
to make the coefficients of
x
equal. Then we get the equations:
2
x
+
4
y
=
22.........
(
1
)
2
x
−
y
=
2.........
(
2
)
Subtract Equation (2) from Equation (1) to eliminate
x
, because the coefficients of
x
are the same. So, we get
(
2
x
−
2
x
)
+
(
4
y
+
y
)
=
22
−
2
i.e.
5
y
=
20
i.e.
y
=
4
Substituting this value of
y
in (2), we get
2
x
−
4
=
2
i.e.
2
x
=
6
i.e.
x
=
3
Hence, the solution of the equations is
x
=
3
,
y
=
4
.
Solve the following pair of simultaneous equations:
x
3
=
y
2
;
2
x
3
−
y
2
=
2
Report Question
0%
(
−
3
,
4
)
0%
(
1
,
9
)
0%
(
6
,
4
)
0%
(
3
,
8
)
Explanation
The equation
x
3
=
y
2
can be rewritten as:
x
3
=
y
2
⇒
2
x
=
3
y
⇒
2
x
−
3
y
=
0
.
.
.
.
.
.
.
.
.
(
1
)
T
he equation
2
x
3
−
y
2
=
2
can be solved as:
2
x
3
−
y
2
=
2
⇒
4
x
−
3
y
6
=
2
⇒
4
x
−
3
y
=
12
.
.
.
.
.
.
.
.
.
(
2
)
Subtract Equation 2 from equation 1 to eliminate
y
, because the coefficients of
y
are the same. So, we get
(
4
x
−
2
x
)
+
(
−
3
y
+
3
y
)
=
12
−
0
i.e.
2
x
=
12
i.e.
x
=
6
Substituting this value of
x
in
the equation
1
, we get
12
−
3
y
=
0
i.e.
3
y
=
12
i.e.
y
=
4
Hence, the solution of the equations is
x
=
6
,
y
=
4
.
Solve the following pair of simultaneous equations:
x
2
−
y
3
=
2
;
x
5
+
y
3
=
15
Report Question
0%
x
=
24
2
7
,
y
=
30
3
7
0%
x
=
12
1
3
,
y
=
10
2
3
0%
x
=
16
7
6
,
y
=
13
4
3
0%
x
=
2
22
7
,
y
=
12
13
17
Explanation
The equation
x
2
−
y
3
=
2
can be solved as:
x
2
−
y
3
=
2
⇒
3
x
−
2
y
6
=
2
⇒
3
x
−
2
y
=
12
.
.
.
.
.
.
.
.
.
(
1
)
T
he equation
x
5
+
y
3
=
15
can be solved as:
x
5
+
y
3
=
15
⇒
3
x
+
5
y
15
=
15
⇒
3
x
+
5
y
=
225
.
.
.
.
.
.
.
.
.
(
2
)
Subtract Equation 2 from equation 1 to eliminate
x
, because the coefficients of
x
are the same. So, we get
(
3
x
−
3
x
)
+
(
−
2
y
−
5
y
)
=
12
−
225
i.e.
−
7
y
=
−
213
i.e.
y
=
213
7
=
30
3
7
Substituting this value of
y
in
the equation
1
, we get
3
x
−
2
(
213
7
)
=
12
⇒
3
x
−
426
7
=
12
⇒
21
x
−
426
=
84
⇒
21
x
=
510
⇒
x
=
510
21
=
170
7
=
24
2
7
Hence, the solution of the equations is
x
=
24
2
7
,
y
=
30
3
7
.
Solve for
x
and
y
:
4
x
=
17
−
x
−
y
8
2
y
+
x
=
2
+
5
y
+
2
3
Report Question
0%
x
=
4
;
y
=
−
4
0%
x
=
1
;
y
=
−
6
0%
x
=
−
6
;
y
=
−
3
0%
x
=
2
;
y
=
−
7
Explanation
Given equations are,
4
x
=
17
−
x
−
y
8
⇒
32
x
=
136
−
x
+
y
⇒
33
x
−
y
=
136
....(1)
and
2
y
+
x
=
2
+
5
y
+
2
3
⇒
3
(
2
y
+
x
)
=
6
+
5
y
+
2
⇒
6
y
+
3
x
=
6
+
5
y
+
2
⇒
3
x
+
y
=
8
....(2)
Add equations (1) and (2), we get
36
x
=
144
⇒
x
=
4
Put this value in equation (1), we get
33
(
4
)
−
y
=
136
⇒
132
−
y
=
136
⇒
y
=
−
4
Therefore, the solution is
x
=
4
,
y
=
−
4
.
Solve the following pair of simultaneous equations:
3
x
+
5
(
y
+
2
)
=
1
3
x
+
8
y
=
0
Report Question
0%
(
−
8
,
3
)
0%
(
−
4
,
9
)
0%
(
1
,
−
3
)
0%
(
0
,
−
9
)
Explanation
The equation
3
x
+
5
(
y
+
2
)
=
1
can be written as
3
x
+
5
y
+
10
=
1
or
3
x
+
5
y
=
−
9
.
Another
equation is
3
x
+
8
y
=
0
.
Then we get the equations:
3
x
+
5
y
=
−
9.........
(
1
)
3
x
+
8
y
=
0.........
(
2
)
Subtract Equation 2 from equation 1 to eliminate
x
, because the coefficients of
x
are the same. So, we get
(
3
x
−
3
x
)
+
(
5
y
−
8
y
)
=
−
9
−
0
i.e.
−
3
y
=
−
9
i.e.
y
=
3
Substituting this value of
y
in
the equation
3
x
+
8
y
=
0
, we get
3
x
+
24
=
0
i.e.
3
x
=
−
24
i.e.
x
=
−
8
Hence, the solution of the equations is
x
=
−
8
,
y
=
3
.
Solve the following pair of simultaneous equations
3
x
+
2
y
=
−
1
6
y
=
5
(
1
−
x
)
Report Question
0%
x
=
−
2
and
y
=
2.5
0%
x
=
−
1
and
y
=
3
0%
x
=
6
and
y
=
1.5
0%
x
=
−
7
and
y
=
−
4
Explanation
3
x
+
2
y
=
−
1
...(i)
6
y
=
5
(
1
−
x
)
⇒
5
x
+
6
y
=
5
...(ii)
On multiplying (i) by 5 and (ii) by 3, we get
15
x
+
10
y
=
−
5
1
−
5
x
+
−
18
y
=
1
−
5
_
−
8
y
=
−
20
∴
y
=
20
8
=
2.5
On putting
y
=
2.5
in (i), we get
3
x
+
2
(
2.5
)
=
−
1
⇒
3
x
=
−
6
⇒
x
=
−
2
∴
x
=
−
2
,
y
=
2.5
Solve the following pair of simultaneous equations:
x
−
1
2
+
y
+
1
5
=
4
1
5
;
x
+
y
3
=
y
−
1
Report Question
0%
x
=
7
,
y
=
5
0%
x
=
2
,
y
=
4
0%
x
=
0
,
y
=
6
0%
x
=
5
,
y
=
3
Explanation
x
−
1
2
+
y
+
1
5
=
4
1
5
⇒
5
x
−
5
+
2
y
+
2
10
=
21
5
⇒
5
x
+
2
y
=
45
...(i)
x
+
y
3
=
y
−
1
⇒
x
+
y
=
3
y
−
3
⇒
x
−
2
y
=
−
3
...(ii)
On multiplying (i) by 1 and (ii) by 5, we get
5
x
+
2
y
=
45
5
−
x
−
+
10
y
=
−
+
15
_
12
y
=
60
∴
y
=
5
On putting
y
=
5
in (ii), we get
x
−
2
(
5
)
=
−
3
⇒
x
=
7
∴
x
=
7
,
y
=
5
Find two numbers such that twice of the first added to the second gives
21
, and twice the second added to the first gives
27
.
Report Question
0%
5
and
11
0%
9
and
16
0%
3
and
18
0%
3
and
17
Explanation
Let the first number is
x
and second is
y
Given twice of first added with second gives
21
∴
2
x
+
y
=
21
...
(
1
)
And given twice of second added with first gives
27
∴
x
+
2
y
=
27
...
(
2
)
Multiply
(
1
)
by
2
Then
4
x
+
2
y
=
42
...
(
3
)
Subtract
(
3
)
with
(
2
)
Then
3
x
=
15
Or
x
=
5
Put the value of
x
=
5
in
(
1
)
Then
10
+
y
=
21
Or
y
=
21
−
10
Or
y
=
11
Then first number is
5
and second is
11
.
A man's age is three times that of his son and in twelve years he will be twice as old as his son would be. What are their present ages?
Report Question
0%
Present age of man is
60
years and
Present age of son is
20
years
0%
Present age of man is
45
years and
Present age of son is
15
years
0%
Present age of man is
54
years and
Present age of son is
18
years
0%
Present age of man is
36
years and
Present age of son is
12
years
Explanation
Let the present age of son be
x
years
As per question the
present
age of father is three times
then
present age of father is
3
x
As per question after
12
years the age of father will be twice the age of son
3
x
+
12
=
2
(
x
+
12
)
⇒
3
x
+
12
=
2
x
+
24
⇒
3
x
−
2
x
=
24
−
12
⇒
x
=
12
Then present age of son is
12
years
So
present age of father
=
3
×
12
=
36
years
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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