Chapter 3 Pair of Linear Equations in Two Variables R.D. Sharma Solutions for Class 10th Math MCQ's
Multiple Choice Questions
1. The value of k for which the system of equations has a unique solution, is
Solution
Solution
The given system of equations are
3. The value of k for which the system of equations x + 2y - 3 = 0 and 5x + ky + 7 = 0 has no solution, is
Solution
4. The value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0 has non-zero solution, is
The given system of equations are,
5. If the system of equations has infinitely many solutions, then
Solution
For the equations to have infinite number of solutions , a1/a2 = b1/b
Let us take a1/a2 = b1/b2
By cross multilication we get ,
Now take b1/b2 = c1/c2
Solution
8. If the system of equations
Solution
The given system of equation is
10. If 2x – 3y = 7 and (a+b) x – (a+b-3)y = 4a + b represent coincident lines, then a and b satisfy the equation
12. The area of the triangle formed by the line x/a + y/b = 1 with the coordinate axes is
Solution
14. If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions, then k =
The given system of equations
The given systems of equations are
Solution
19. If x = a, y = b is the solution of the systems of equations x – y = 2 and x + y = 4 , then the values of a and b are respectively
21. Arun has only 1 and 2 coins with her . If the toatl number of coins that she has is 50 and the amonut of money with her is 75 , then the number of 1 and 2 coins are , respectively
Solution
Let the number of ₹1 coins be x and that of ₹2 coins be y.
kx − y = 2
6x − 2y = 3
(a) = 3
(b) ≠ 3
(c) ≠ 0
(d) = 0
Solution
The given system of equation are
kx – y = 2
6x – 2y = 3
a1/a2 ≠ b1/b2 for unique solution
Here a1 = k, a2 = 6, b1 = -1, b2 = -2
k/6 ≠ -1/-2
By cross multiply we get
2k ≠ 6
k ≠ 6/2
k ≠ 3
Hence, the correct choice is b .
2. The value of k for which the system, of equations has infinite number of solutions, is
2x + 3y = 5
4x + ky = 10
(a) 1
(b) 3
(c) 6
(d) 0
Solution
The given system of equations are
2x + 3y = 5
4x + ky = 10
For the equation to have infinite number of solutions a1/a2 = b1/b2 = c1/c2
Here, a1 = 2, a2 = 4, b1 = 3, b2 = k
Therefore 2/4 = 3/k = 5/10
By cross multiplication of a1,a2 = b1/b2 we get ,
2/4 = 3/k
2k = 12
k = 12/2
k = 6
And
b1/b2 = c1/c2
3/k = 5/10
30 = 5k
30/5 = k
6 = k
Therefore the value of k is 6
Hence the correct choice is c .
3. The value of k for which the system of equations x + 2y - 3 = 0 and 5x + ky + 7 = 0 has no solution, is
(a) 10
(b) 6
(c) 3
(d) 1
Solution
The given system of equations are
x + 2y – 3 = 0
5x + ky + 7 = 0
For the equations to have no solutions,
a1/a2 = b1/b2 ≠ c1/c2
1/5 = 2/k = -3/7
If we take
a1a2 = b1/b2
1/5 = 2/k
Therefore te value of k is 10
Hence , correct choice is a .
4. The value of k for which the system of equations 3x + 5y = 0 and kx + 10y = 0 has non-zero solution, is
(a) 0
(b) 2
(c) 6
(d) 8
Solution
The given system of equations are,
3x + 5y = 0
kx + 10y = 0
Here a1 = 3, a2 = k, b1 = 5, b2 = 10
a1/a2 = b1/b ≠ 0
3/k = 5/10 ≠ 0
By cross multiply we get
30 = 5k
30/5 = k
6 = k
Therefore the value of k is 6,
Hence , the correct choice is c .
5. If the system of equations has infinitely many solutions, then
2x + 3y = 7
(a + b)x + (2a − b)y = 21
(a) a = 1, b = 5
(b) a = 5, b = 1
(c) a = −1, b = 5
(d) a = 5, b = −1
Solution
The given systems of equations are
2x + 3y = 7
(a+b) x + (2a-b) y = 21
For the equations to have infinite number of solutions , a1/a2 = b1/b
2 = c1/c2
Here a1 = 2, a2 = (a+b), b1 = 3, b2 = 2a – b, c1 = 7, c2 = 21,
2/a+b = 3/2a-b = 7/21
Let us take a1/a2 = b1/b2
2/a+b = 3/2a-b
By cross multilication we get ,
2(2a – b) = 3(a+b)
4a – 2b = 3a + 3b
4a – 3a = 3b + 2b
a = 5b …(i)
Now take b1/b2 = c1/c2
3/2a – b = 7/21
3/2a – b = 1/3
By cross multiplication we get,
3 × 3 = 1 × 2a – b
9 = 2a – b …(ii)
Substitute a = 5b in the above equation
9 = 2 × 5b – b
9 = 10b – b
9 = 9b
9/9 = b
1 = b
Substitute b = 1 in equation (i) we get , a = 5b
a = 5 × 1
a = 5
Therefore a = 5 and b = 1
Hence, the correct choice is b .
6. If the system of equations 3x + y = 1 and, (2k – 1) x + (k-1) y = 2k + 1 is inconsistent, then k = 1
(a) 1
(b) 0
(c) -1
(d) 2
Solution
The given system of equations is inconsistent,
3x + y = 1
(2k – 1) x + (k-1) y = 2k + 1
If the system of equations is in consistent, we have
a1b2 – a2b1 = 0
3×(k-1)-1(2k-1) = 0
3k – 3 – 2k + 1 = 0
1k – 2 = 0
1k = 2
Therefore, the value of k is 2
Hence, the correct choice is d
7. If am ≠ bl, then the system of equations ax + by = c and, lx + my = n
(a) has a unique solution
(b) has no solution
(c) has infinitely many solutions
(d) may or may not have a solution
Solution
Given am ≠ bl , the system of equation has
ax + by = c
lx + my = n
We know that intersecting lines have unique solution a1/a2 ≠ b1/b2
a1 × b2 = a2 × b1
Here a1 = a, a2 = 1, b1 = b, b2 = m
a/l ≠ b/m
a × m ≠ l×b
Therefore intersecting lines, have unique solution
Hence, the correct choice is a .
8. If the system of equations
2x+ 3y = 7
2ax + (a+b)y = 28
has infinitely many solutions , then
(a) a = 2b
(b) b = 2a
(c) a + 2b = 0
(d) 2a + b = 0
Solution
Given the systems of equations are
2x + 3y = 7
2ax + (a+b)y = 28
For the equations to have infinite number of solutions ,
a1/a2 = b1/b2 = c1/c2
a1 = 2, a2 = 2a , b1 = 3, b2 = a + b
2/2a = 3/a+b = 7/28
By cross multiplication we have
2/2a = 3/a+b
2(a+b) = 2a(3)
2a + 2b = 6a
2b = 6a – 2a
2b = 4a
Divide both sides by 2. We get b = 21
Hence, the correct choice is b .
9. The value of k for which the system of equations
x + 2y = 5
3x + ky + 15 = 0
has no solution is
(a) 6
(b) -6
(c) 3/2
(d) None of these
Solution
The given system of equation is
x + 2y = 5
3x + ky + 15 = 0
If a1a2 = b1/b2 ≠ c1/c2 then the equation have no solution .
1/3 = 2/k = -5/15
By cross multiply we get
k × 1 = 3 × 2
k = 6
Hence, the correct choice is a .
10. If 2x – 3y = 7 and (a+b) x – (a+b-3)y = 4a + b represent coincident lines, then a and b satisfy the equation
(a) a+5b = 0
(b) 5a + b = 0
(c) a – 5b = 0
(d) 5a – b = 0
Solution
The given system of equations are
2x – 3y = 7
(a+b) x – (a+b-3)y = 4a + b
For coincident lines, infinite number of solution
a1a2 = b1/b2 = c1c2
⇒ 2/(a+b) = -3/-(a+b-3) = 7/4a + b
⇒ 2/(a+b) = 3(a+b-3) = 7/(4a+b)
⇒ 2(a+b-3) = 3(a+b)
⇒ 2a + 2b – 6 = 3a + 3b
⇒ 2a + 2b – 3a – 3b = 6
⇒ - a – b = 6
⇒ a + b = -6 …(i)
3(4a + b) = 7 (a+b-3)
⇒ 12a + 3b = 7a + 7b – 21
⇒ 5a – 4b = -21 …(ii)
Multiply equation (i) by 5, we get 5a + 5b = -30 …(iii)
Subtract (ii) from (iii),
(5a + 5b) – (5a – 4b) = -30 + 21
⇒ 5a + 5b – 5a + 4b = -9
⇒ 9b = -9
⇒ b = -1
Substitute b = -1 in equation (1)
a + (-1) = -6
⇒ a = -6 + 1 = -5
Option A :
a + 5b = 0
-5 + 5(-1) = -5-5 = -10 ≠ 0
Option B :
5a + b = 0
5(-5) + (-1) = - 25 – 1 = -26 ≠ 0
Option C :
a –b = 0
-5-(-1) = -4 ≠ 0
None of the option satisfies the values .
11. If a pair of linear equations in two variables is consistent, then the lines represented by two equations are
(a) intersecting
(b) parallel
(c) always coincident
(d) intersecting or coincident
Solution
If a pair of linear equations in two variables is consistent, then its solution exists.
∴The lines represented by the equations are either intersecting or coincident.
Hence, the correct choice is d .
12. The area of the triangle formed by the line x/a + y/b = 1 with the coordinate axes is
(a) ab
(b) 2ab
(c) 1/2 ab
(d) 1/4 ab
Solution
Given the area of the triangle formed by the line x/a + y/b = 1
If in the equation x/a + y/b = 1 either A and B approaches infinity, the line become parallel to either x axis or y axis respectively ,
Therefore
x = a;
y = b
Area of triangle = 1/2 × x × y = 1/2 × a × b
Hence , the correct choice is c .
13. The area of the triangle formed by the lines y = x, x = 6 and y = 0 is
(a) 36 sq. units
(b) 18 sq. units
(c) 9 sq. units
(d) 72 sq. units
Solution
Given x = 6 , y = 0 and x = y
We have plotting points as (6,0)(0,0)(6,6) when x = y
Therefore , area of ΔABC = 1/2 (Base × Height) = 1/2 (CA × AB) = 1/2 (6×6) = 1/2 × 36 = 18
Area of triangle ABC is 18 square units
Hence, the correct choice is b .
14. If the system of equations 2x + 3y = 5, 4x + ky = 10 has infinitely many solutions, then k =
(a) 1
(b) 1/2
(c) 3
(d) 6
Solution
The given system of equations
2x + 3y = 5
4x + ky = 10
a1a2 = 2/4, b1/b2 = 3/k , c1c2 = 5/10
For the equations to have infinite number of solutions
a1a2 = b1b2 = c1c2
2/4 = 3/k = 5/10
If we take
2/4 = 3/4
2k = 12
k = 12/2
k = 6
And
3/k = 5/10
30 = 5k
30/5 = k
6 = k
Therefore, the value of k is 6 .
Hence, the correct choice is d .
15. If the system of equations kx − 5y = 2, 6x + 2y = 7 has no solution, then k =
(a) −10
(b) −5
(c) −6
(d) −15
Solution
The given systems of equations are
kx – 5y = 2
6x + 2y = 7
a1a2 = b1b2 ≠ c1/c2
Here a1 = k , a2 = 6, b1 = -5 , b2 = 2
k/6 = -5/2
2k = -30
k = -30/2
k = -15
Hence , the correct choice is d .
16. The area of the triangle formed by the lines x = 3, y = 4 and x = y is
(a) 1/2 sq. unit
(b) 1 sq. unit
(c) 2 sq . unit
(d) None of these
Solution
Given x = 3, y = 4 and x = y
We have plotting points as (3,4)(3,3)(4,4) when x = y
Therefore, area of ΔABC = 1/2 (Base × Height) = 1/2 (AB × AC) = 1/2 (1×1) = 1/2
Area of triangle ABC is 1/2 square units
Hence, the correct choice is a .
17. The area of the triangle formed by the lines 2x + 3y = 12, x − y − 1 = 0 and x = 0 (as shown in Fig. 3.23), is
(a) 7 sq. units
(b) 7.5 sq. units
(c) 6.5 sq. units
(d) 6 sq. units
Solution
Given, 2x + 3y = 12, x – y – 1 = 0 and x = 0
Given, 2x + 3y = 12, x – y – 1 = 0 and x = 0
If x = 0 we have plotting points as D(0,-1) B(0,4) P(3,2)
Therefore, area of CBPD = 1/2 (Base × Height) = 1/2 (BP×PM) = 1/2 (5×3) = 1/2(15) = 7.5
Area of triangle ABC is 7.5 square units
Hence, the correct choice is b.
18. The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the number get reversed . The number is
(a) 25
(b) 72
(c) 63
(d) 36
Solution
Let the digits at the tens and the ones place be x and y, respectively. So, the two digit number is 10x + y.
Let the digits at the tens and the ones place be x and y, respectively. So, the two digit number is 10x + y.
Now,
x + y = 9 .....(i)
Also,
10x + y + 27 = 10y + x
⇒ 9x − 9y = −27
⇒ x − y = −3 .....(ii)
Adding (i) and (ii), we get
2x = 6
⇒ x = 3
Putting x = 3 in (i), we get
3 + y = 9
⇒ y = 6
Thus, the required number is 10 × 3 + 6 = 36.
Hence, the correct answer is option (d).
19. If x = a, y = b is the solution of the systems of equations x – y = 2 and x + y = 4 , then the values of a and b are respectively
(a) 3 and 1
(b) 3 and 5
( c) 5 and 3
(d) -1 and -3
Solution
The given equation are
The given equation are
x – y = 2 ….(1)
x + y = 4 ….(2)
Adding (1) and (2) , we get
2x = 6
⇒ x = 3
Putting x = 3 in (1) , we get
3 + y = 4
⇒ y = 1
So, x = a = 3 and y = 1.
20. For what the value k, do the equation 3x – y + 8 = 0 and 6x – ky + 16 = represent coincident lines ?
Solution
The given system of equations is
The given system of equations is
3x – y + 8 = 0
6x – ky + 16 = 0
We know that the lines
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
are coinciden if
a1/a2 = b1/b2 = c1/c2
∴ 3/6 = -1/-k = 8/16
⇒ 1/2 = 1/k = 1/2
⇒ k = 2
Thus, the value of k = 2.
Hence the orrect answer is option c.
21. Arun has only 1 and 2 coins with her . If the toatl number of coins that she has is 50 and the amonut of money with her is 75 , then the number of 1 and 2 coins are , respectively
(a) 35 and 15
(b) 35 and 20
(c) 15 and 35
(d) 25 and 25
Solution
Let the number of ₹1 coins be x and that of ₹2 coins be y.
Now,
Total number of coins = 50
So, x + y = 50 .....(i)
Also,
₹1 × x + ₹2 × y = ₹75
∴ x + 2y = 75 .....(ii)
Subtracting (i) from (ii), we get
y = 25
Putting y = 25 in (i), we get
x + 25 = 50
⇒ x = 25
So, the number of ₹1 coins and ₹2 coins are 25 and 25, respectively.
Hence, the correct answer is option (d).
Note: The answer given in the book does not match with the one obtained.