This set of Class 10 Maths Chapter 3 Multiple Choice Questions & Answers (MCQs) focuses on “Pair of Linear Equations in Two Variables”. These MCQs are created based on the latest CBSE syllabus and the NCERT curriculum, offering valuable assistance for exam preparation.
1. What will be the nature of the graph lines of the equations 5x-2y+9 and 15x-6y+1?
a) Parallel
b) Coincident
c) Intersecting
d) Perpendicular to each other
View Answer
Explanation: The given equations are 5x-2y+9 and 15x-6y+1.
Here, a1=5, b1=-2, c1=9 and a2=15, b2=-6, c2=1
Now, \(\frac {a_1}{a_2} = \frac {5}{15} = \frac {1}{3}, \frac {b_1}{b_2} = \frac {-2
}{-6}=\frac {1}{3}, \frac {c_1}{c_2} = \frac {9}{1} \)
Clearly, \(\frac {a_1}{a_2} =\frac {b_1}{b_2} \ne \frac {c_1}{c_2} \)
Therefore, the graph lines of the equations will be parallel.
2. What will be the nature of the graph lines of the equations x+3y-2 and 2x-y+5?
a) Parallel
b) Coincident
c) Intersecting
d) Perpendicular to each other
View Answer
Explanation: The given equations are x+3y-2 and 2x-y+5.
Here, a1=1, b1=3, c1=-2 and a2=2, b2=-1, c2=5
Now, \(\frac {a_1}{a_2} = \frac {1}{2}, \frac {b_1}{b_2} = \frac {3}{1}\) = 3, \(\frac {c_1}{c_2} = \frac {-2}{5} \)
Clearly, \(\frac {a_1}{a_2} \ne \frac {b_1}{b_2} \)
Therefore, the graph lines of the equations will intersect at a point.
3. What will be the nature of the graph lines of the equations 2x+5y+15 and 6x+15y+45?
a) Parallel
b) Coincident
c) Intersecting
d) Perpendicular to each other
View Answer
Explanation: The given equations are 2x+5y+15 and 6x+15y+45.
Here, a1=2, b1=5, c1=15 and a2=6, b2=15, c2=45
Now, \(\frac {a_1}{a_2} = \frac {2}{6} = \frac {1}{3}, \frac {b_1}{b_2} = \frac {5}{15} =\frac {1}{3}, \frac {c_1}{c_2} = \frac {15}{45} = \frac {1}{3} \)
Clearly, \(\frac {a_1}{a_2} =\frac {b_1}{b_2} = \frac {c_1}{c_2} \)
Therefore, the graph lines of the equations will be coincident.
4. What will be the value of k, if the lines given by (5+k)x-3y+15 and (k-1)x-y+19 are parallel?
a) 5
b) 4
c) 6
d) 7
View Answer
Explanation: The given equations are (5+k)x-3y+15 and (k-1)x-y+19.
Here, a1=5+k, b1=-3, c1=15 and a2=k-1, b2=-1, c2=19
Lines are parallel, so \(\frac {a_1}{a_2} = \frac {b_1}{b_2} \ne \frac {c_1}{c_2} \)
Now, \(\frac {a_1}{a_2} = \frac {5+k}{k-1}, \frac {b_1}{b_2} =\frac {-3}{-1}\) = 3, \(\frac {c_1}{c_2} = \frac {15}{19} \)
\(\frac {5+k}{k-1}\) = 3
5+k=3(k-1)
5+k=3k-3
5+3=3k-k
2k=8
k=4
5. What will be the value of k, if the lines given by 3x+ky-4 and 5x+(9+k)y+41 represent two lines intersecting at a point?
a) k≠\(\frac {7}{2}\)
b) k≠\(\frac {27}{8}\)
c) k=\(\frac {27}{2}\)
d) k≠\(\frac {27}{2}\)
View Answer
Explanation: The given equations are 3x+ky-4 and 5x+(9+k)y+41 .
Here, a1=3, b1=k, c1=-4 and a2=5, b2=9+k, c2=41
Lines are intersecting at a point, so \(\frac {a_1}{a_2} \ne \frac {b_1}{b_2} \)
Now, \(\frac {a_1}{a_2} =\frac {3}{5}, \frac {b_1}{b_2} = \frac {k}{9+k}, \frac {c_1}{c_2} =\frac {-4}{41}\)
\(\frac {3}{5} \ne \frac {k}{9+k}\)
3(9+k)≠5k
27+3k≠5k
27≠5k-3k
2k≠27
k≠\(\frac {27}{2}\)
6. What will be the value of k, if the lines given by x+ky+3 and 2x+(k+2)y+6 are coincident?
a) 4
b) 2
c) 6
d) 8
View Answer
Explanation: The given equations are x+ky+3 and (k-1)x+4y+6.
Here, a1=1, b1=k, c1=3 and a2=k-1, b2=4, c2=6
Lines are coincident, so \(\frac {a_1}{a_2} =\frac {b_1}{b_2} =\frac {c_1}{c_2}\)
Now, \(\frac {a_1}{a_2} = \frac {1}{k-1}, \frac {b_1}{b_2} =\frac {k}{4}, \frac {c_1}{c_2} =\frac {3}{6}\)
\(\frac {1}{k-1}=\frac {k}{4}=\frac {1}{2}\)
2k=4
k=2
7. The lines 5x-7y=13 and 10x-14y=15 are inconsistent.
a) True
b) False
View Answer
Explanation: A system of linear equations is said to be inconsistent if it has no solution at all.
The given equations are 5x-7y=13 and 10x-14y=15
Here, a1=5, b1=-7, c1=-13 and a2=10, b2=-14, c2=-15
Now, \(\frac {a_1}{a_2} = \frac {5}{10}=\frac {1}{2}, \frac {b_1}{b_2} =\frac {-7}{-14}=\frac {1}{2}, \frac {c_1}{c_2} =\frac {-13}{-15}\)
Clearly, \(\frac {a_1}{a_2} =\frac {b_1}{b_2} \ne \frac {c_1}{c_2}\)
Hence, it will have no solution.
The given equations are inconsistent.
8. The lines 2x+5y=17 and 5x+3y=14 are consistent.
a) False
b) True
View Answer
Explanation: A system of linear equations is said to be consistent if it has at least one solution.
The given equations are 2x+5y=17 and 5x+3y=14
Here, a1=2, b1=5, c1=-17 and a2=5, b2=3, c2=-14
Now, \(\frac {a_1}{a_2} =\frac {2}{5}, \frac {b_1}{b_2} =\frac {5}{3}, \frac {c_1}{c_2} =\frac {-17}{-14}\)
Clearly, \(\frac {a_1}{a_2} \ne \frac {b_1}{b_2} \)
Hence, it will have unique solution.
The given equations are consistent.
9. The sum of a two digit number and the number obtained by reversing the order of the digits is 187. If the digits differ by 1, then what will be the number?
a) 67
b) 54
c) 89
d) 67
View Answer
Explanation: Let the two digit number be 10x+y
The number obtained after reversing the digits will be 10y+x
10x+y+10y+x=187
11x+11y=187
x+y=17 (1)
Also, x-y=1 (2)
Adding (1) and (2)
2x=18
x=9
Substituting in equation (1), 9+y=17
y=8
The number is 89 or 98.
10. It takes 10 men and 6 women to finish a piece of work in 4 days, while it takes 5 men and 7 women to finish the same job in 6 days. What will be the time taken by 1 man and 1 woman to finish the job?
a) Man = 34 days, Woman = 45 days
b) Man = 45 days, Woman = 34 days
c) Man = 53 days, Woman = 96 days
d) Man = 54 days, Woman = 96 days
View Answer
Explanation: Let 1 man take x days to finish the job and 1 woman take y days to finish the same job.
Then, 1 man’s 1 day work will be \(\frac {1}{x}\) days
1 woman’s 1 day work will be \(\frac {1}{y}\) days
10 men and 6 women can finish the job in 6 days
(10 men’s 1 day work + 6 women’s 1 day work = \(\frac {1}{4}\))
\(\frac {10}{x}+\frac {6}{y}=\frac {1}{4}\)
Let, \(\frac {1}{x}\) = u, \(\frac {1}{y}\) = v
10u+6v=\(\frac {1}{4}\) (1)
5 men and 7 women can finish the job in 6 days.
(5 men’s 1 day work + 7 women’s 1 day work = \(\frac {1}{6}\))
\(\frac {5}{x}+\frac {7}{y}=\frac {1}{6}\)
Let, \(\frac {1}{x}\) = u, \(\frac {1}{y}\) = v
5u + 7v = \(\frac {1}{6}\) (2)
Multiplying equation by 2 and then subtracting both the equations we get,
10u + 14v = \(\frac {1}{6}\)
-10u + 6v = \(\frac {1}{4}\)
8v = \(\frac {1}{3}-\frac {1}{4}\)
8v = \(\frac {1}{12}\)
v = \(\frac {1}{96}\)
v = \(\frac {1}{y}=\frac {1}{96}\)
y = 96
Substituting the value of v in equation (1) we get,
10u + 6\((\frac {1}{96})=\frac {1}{4}\)
10u + \(\frac {1}{16}=\frac {1}{4}\)
10u = \(\frac {3}{16}\)
u = \(\frac {3}{160}\)
u = \(\frac {1}{x}=\frac {3}{160}\)
x = \(\frac {160}{3}\) ≈ 54
Hence, a man alone can finish the job in 54 days and a woman can finish the job in 96 days.
11. 10 years ago, a woman was thrice the age of her daughter. Two years later her daughter’s age will be 30 more than the age of the mother. What are the present ages of the woman and the daughter?
a) 70 years, 40 years
b) 60 years, 40 years
c) 55 years, 25 years
d) 45 years, 20 years
View Answer
Explanation: Let the present ages of mother be x years and daughter be y years.
10 years ago,
Age of mother = x-10 years
Age of daughter = y-10 years
Age of mother = 3(age of daughter)
x-10=3(y-10)
x-10=3y-30
x-3y+20=0
x=3y-20 (1)
Two years later,
Age of mother = x+2 years
Age of daughter = y+2 years
Age of mother will be 30 more than the age of daughter
x+2=y+2+30
x=y+30 (2)
From (1) and (2), we get,
y+30=3y-20
30+20=3y-y
50=2y
y=25
Substituting y=25 in equation (1) we get,
x=3(25)-20
x=55
The present age of mother is 55 years and that of daughter is 25 years.
12. The sum of two numbers is 13 and the sum of their reciprocals is \(\frac {13}{40}\). What are the two numbers?
a) 5, 8
b) 10, 3
c) 12, 1
d) 9, 4
View Answer
Explanation: Let the two numbers be x and y.
x+y=13 (1)
Also, \(\frac {1}{x}+\frac {1}{y}=\frac {13}{40}\)
\(\frac {y+x}{xy}=\frac {13}{40}\)
40(y+x)=13xy
xy=40
Now, x-y=\(\sqrt {(x+y)^2-4xy}\)
=\(\sqrt {13^2-4(40)}\)
=\(\sqrt {169-160}\)
=√9
=±3
x-y=3 or x-y=-3 (2)
Adding (1) and (2), we get,
x+y=13
-x-y=3
2y=16
y=8
x=5
Or
y=5
x=8
13. In a piggy bank the total number of coins of Rs. 5 and Rs. 1 is 100. If the total coins amount is 300, then what is the number of coins of each denomination?
a) 30, 70
b) 50, 50
c) 45, 55
d) 60, 40
View Answer
Explanation: Let the coins of Rs. 5 be x and that of Rs. 1 be y
Total number of coins is 100
x+y=100
y=100-x
Total amount of coins is 300
Also, 5x+y=300 (1)
Substituting y=100-x in equation (1) we get,
5x+100-x=300
4x=200
x=50
y=100-x=100-50=50
The coins of each denomination are 50.
14. A father gives Rs. 500 to his children every month. If the boy gets Rs. 100 then, the girl gets Rs. 200 and if the boy gets Rs. 100 the girl gets Rs. 150. How many children does he have?
a) 0
b) 3
c) 2
d) 1
View Answer
Explanation: Let the number of boys be x and number of girls be y
Now, if the boys get 100, the girls get 200
100x+200y=500
x+2y=5
x=5-2y
If the boys get 200, the girls get 150
200x+150y=500
4x+3y=10 (1)
Substituting x=5-2y in equation 1 we get,
4(5-2y)+3y=10
20-8y+3y=10
-5y=-10
y=2
x=5-2y=5-2(2)=1
The father has three children.
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