Class 9th MCQ Test for Chapter 1
Class 9th Maths Chapter 1st MCQ
Questions Answer of class 9th Maths Chapter 1st CBSE
Q:1 The pairs of equations x+2y-5 = 0 and -2x-4y+10=0 have:
(a) Unique solution
(b) Exactly two solutions
(c) Infinitely many solutions
(d) No solution
Answer: (c) Infinitely many solutions
Explanation:
a1/a2 = 1/-2
b1/b2 = 2/-4 = 1/-2
c1/c2 = -5/10 = -1/2
This shows:
a1/a2 = b1/b2 = c1/c2
Therefore, the pair of equations has infinitely many solutions
Q:2 If the lines 4x+8ky –7 = 0 and 2x+5y+1 = 0 are parallel, then what is the value of k?
(a) 4/5
(b) 5/4
(c) ⅘
(d) 25/4
Answer: (b) 5/4
Explanation: The condition for parallel lines is:
a1/a2 = b1/b2 ≠ c1/c2
Hence, 4/2 = 8k/5
k=5/4
Q:3 The solution of the equations 4x-y=2 and x+y=3 is:
(a) 1 and 2
(b) 3 and 3
(c) 6 and 1
(d) -3 and -8
Answer: (a) 1 and 2
Explanation: 4x-y =2
4x=2+y
X = (2+ y)/4
Substituting the value of x in the second equation we get;
(2+y)/4+y=3
Multiplying by 4 on both side
2+y+4y=12
2+5y = 12
5Y = 12 – 2
5Y = 10
Y = 10/5
y=2
Now putting the value of y, we get;
4x - y = 2
4x - 2 = 2
4x = 2+ 2
4x = 4
X = 4/4
X=1
Hence, the solutions are x=1 and y=2
Q:4 A fraction becomes 6/7 when 1 is subtracted from the numerator and it becomes 1/2 when 7 is added to its denominator. The fraction obtained is:
a) 5/12
(b) 7/12
(c) 5/7
(d) 7/7
Answer: (d) 7/7
Explanation: Let the fraction be x/y
So, as per the question given,
(x -1)/y = 6/7 => 7x – 6y = 7…………………(1)
x/(y + 7) = 1/2 => 2x –y =7 ………………..(2)
by applying either substitution or elimination method on and 2 we get
x= 7, y= 7
Therefore, the fraction is 7/7
Q:5 A pair of linear equations which has a unique solution x = 4, y = -5 is
(a) x + y = -1; 2x – 3y = 23
(b) 9x + 5y = -19; 4x + 17y = -52
(c) x –3 y = 13; 31x + 12y = 20
(d) 2x – 4y =14; 2x – 9y =13
Answer: (a) x + y = -1; 2x – 3y = 23
Explanation:
If x = 4, y = -5 is a unique solution of any pair of equations, then these values must satisfy that pair of equations.
By verifying the options, option (b) satisfies the given values.
LHS = x + y = 4 +(- 5) = 4 – 5 = -1 = RHS
LHS = 2x - 3y = 2(4) - 3(- 5)= 8 + 15 = 23 = RHS
Best questions for class 9th Maths
Q:6 Graphically, the pair of equations 7x – y = 5; 14x – 2y = 9 represents two lines which are
(a) intersecting at one point
(b) parallel
(c) intersecting at two points
(d) coincident
Answer (b) Parallel
Explanation of the parallel equation
7/14 = -1/2 ≠ 5/9
a1/a2 = b1/b2 ≠ c1/c2
Q:7 The pair of equations 2x + 4y + 5 = 0 and –3x – 6y + 1 = 0 have
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Answer: (d) no solution
Explanation:
Given pair of equations are 2x + 4y + 7 = 0 and –3x – 6y + 8 = 0.
Comparing with the standard form,
a1 = 2, b1 = 4, c1 = 7
a2 = -3, b2 = -6, c2 = 8
a1/a2 = 2/-3
b1/b2 = 4/-6 = 2/-3
c1/c2 = 7/8
Thus, a1/a2 = b1/b2 ≠ c1/c2
Q:8 The father’s age is three times his son’s age. Five years back, the age of the father was four times his son’s age. The present ages, in years, of the son and the father are, respectively
(a) 4 and 24
(b) 20 and 60
(c) 15 and 45
(d) 8 and 24
Answer: (c) 15 and 45
Explanation:
Let x years be the present age of father and y years be the present age of son.
So x = 3y….1
According to the given,
(x - 5) = 4(y - 5)
x - 5 = 4y - 20
x- 4y = -20 + 5
y = 15
Substituting y = 15 in (i),
x = 3(15) = 45
Son’s age is 15yr and father’s age is 45yr
x - 4y = -15….(ii)
Also, x = 3y….(i)
From (i) and (ii),
3y – 4y = -15
-y = -15
Substituting y = 15 in (i),
x = 3(15) = 45
Son’s age is 15yr and father’s age is 45yr
Q:9 The solution of the equations x+y=2 and x+3y=4 is:
(a) 3 and 1
(b) 1 and 1
(c) 5 and 1
(d) -1 and -3
Answer ( b) 1 and 1
Explanation: x+y =2
x=2-y
Substituting the value of x in the second equation we get;
2 – y + 3y =4
2+ 2y=4
2y = 2
y=1
Now putting the value of y, we get;
x= 2-1 = 1
Hence, the solutions are x=1 and y=1.
Q:10 The pair of equations x + 2y = 5 and 3x + 6y = 15 have
(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Answer: (c) infinitely many solutions
Explanation:
Given pair of equations are x + 2y + 5 = 0 and –3x – 6y + 1 = 0.
Comparing with the standard form,
a1 = 1, b1 = 2, c1 = 5
a2 = 3, b2 = 6, c2 = 15
a1/a2 = 1/3
b1/b2 = 2/6 = 1/3
c1/c2 = 5/15 = 1/3
Thus, a1/a2 = b1/b2 = c1/c2
Hence, the given pair of equations has infinitely many solutions