10th Grade > Mathematics
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES MCQs
:
A
Let's assume that the speed of the boat in still water is x km/hr and speed of the stream is y km/hr.
So, the speed of the boat in upstream will be (x-y) km/hr.
Similarly, the speed of the boat downstream will be (x+y) km/hr.
We know time=(distancespeed).
Using the above formula we can form the equations in two variables.
Taking the first case,
30x - y+44x + y=14.
Taking the second case,
60x - y+55x + y=25.
Now, we have the equations in two variables but the equations are not linear.
So, we will assume 1x - y=u and 1x + y=v.
So on substituting u and v in the above two equations, we get
30u+44v=14 ...(1)
60u+55v=25 ...(2)
We can solve the above two equations using the elimination method.
60u+88v=28 ...(3)
(by multiplying equation (1) by 2)
On subtracting equation (2) from (3), we get v=111
On substituting v in equation (2) we get u=13
Now as we have assumed
1x - y=u and 1x + y= v
On substituting the values of u and v,
we get a pair of linear equations in x and y
x - y=3...(4)
x + y=11...(5)
On adding (5) from (4), we have
2x=14
x=7
On subsituting the value of x in x−y=3, we get y = 4.
So, the speed of the boat in still water is 7 km/hr and the speed of the stream is 4 km/hr.
:
A
We have,
2x+y=7 ...(1)
3x+2y=12...(2)
Multiply equation (1) by 2, we get:
2(2x+y)=2(7)
⇒4x+2y=14...(3)
Subtracting (2) from (3) we get ,
x=2
Substituting the value of x in (1) we get,
2(2)+y=7⟹y=3
Thus, the solution for the given pair of linear equations is (2,3).
:
D
Given equations gives infinitely many solutions if,
a1a2=b1b2=c1c2
The given linear equations are:
3x+ky=9;6x+4y=18.
⇒a1=3,b1=k,c1=−9 and a2=6,b2=4,c2=−18
⇒36=k4=−9−18
⇒12=k4
⇒k=2
:
B
Take first two components,
x+y−82=x+2y−148
⇒8(x+y−8)=2(x+2y−14)
⇒8x+8y−64=2x+4y−28
⇒6x+4y−36=0
⇒3x+2y−18=0.....(i)
Take last two components,
x+2y−148=3x+y−1211
⇒11(x+2y−14)=8(3x+y−12)
⇒11x+22y−154=24x+8y−96
⇒−13x+14y−58=0.....(ii)
On multiplying equation (i) by 7, we get
⇒21x+14y−126=0...(iii),
On subtracting equation (ii) from (iii), we get
⇒34x=68
⇒x=2
On substituting value of x=2 in equation (i), we get
3×2+2y−18=0
⇒y=6
∴ The solution is (2, 6).
:
A
Let, the cost of a bike be x.
The cost of scooter be y.
According to question x=2y
Therefore linear equation will be:
⇒x−2y=0
:
B
In the given graph, lines of the equations 2x + 4y - 12 = 0 and x + 2y - 4 = 0 are parallel to each other which gives no solution; hence it is an inconsistent pair of linear equation.
:
If the point (1,3k) lies on the given equation, then it should satisfy the equation.
Substituting the values of x and y in the given equation:
⇒k(1)+4(3k)=26
⇒13k=26
⇒k=2
Thus, the value of k is 2.
:
A and D
Option A:
x+y= –1−−−−−(i)
2x+3y= –5−−−−(ii)
(i)×2→ 2x+2y=−2−−−−(iii)
Solving by elimination((iii)−(ii)) we get,
x=2 and y=−3.
Option D:
x−4y=14−−−−(i)
5x−y=13−−−−−(ii)
(ii)×4→20x−4y=52−−−−(iii)
Solving by elimination method ((iii)−(i)) we get,
x=2 and y=−3
:
B
Let starting salary be ₹ x
and annual increment be ₹ y
According to the first condition:
x + 4y = 15000 ......(i)
According to the second condition:
x + 10y = 18000 ......(ii)
Subtracting (i) from (ii), we get
6y = 3000
y = 500
Substituiting y = 500 in equation (i), we get
x + (4×500) = 15000
x = 15000 - 2000
x = 13000
Hence starting salary is ₹ 13000 and annual increment is ₹ 500.
:
A
Let the fraction be xy
∴ x+2y−1=23 and x+1y+2=13
⇒ 3x−2y=−8....(i)
and 3x−y=−1......(ii)
⇒3x−2y=−8
−(3x−y=−1)
⇒3x−2y=−8
−3x+y=1
______________________
−y=−7
______________________
⇒y=7
Substituting y = 7 in equation (ii), we get
3x−(14)=−8
3x=−8+14=6
∴x=2
⇒x=2 and y=7