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Published on: 05/09/2019
Linear Equations in One Variable
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Questions + Answers key
Take MCQ Mathematics Test
1.
Find the root the following equations \(3x-\frac { 1 }{ 3 } =\frac { 2 }{ 3 } \)
2.
Solve \(\frac { t }{ 2 } =10\)
3.
Solve the following equations \(\frac { x }{ 3 } +1=\frac { 7 }{ 15 } \)
4.
Solve the following equations 17 + 6p =9
5.
Solve the following equations 1.6 = \(\frac { y }{ 1.5 } \)
6.
Solve the following equations and check your result \(\frac { x }{ 5 } +11=\frac { 1 }{ 15 } \)
7.
Divide Rs. 1380 among Atul, Ravi and Kishan, so that the amount Atul receives is 5 times as much as Kishan's share and is 3 times as much as Ravi's share.
8.
Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
9.
Baichung's father is 26 yr younger than Baichung's grandfather and 29 yr older than Baichung. The sum of the ages of all the three is 135yr.What is the age of each one of them?
10.
Three consecutive integers add upto 51. What are these integers?
11.
Arpita's present age is thrice of Shilpa. If Shilpa's age 3 yr ago was x. Then, Arpita's present age is
3(x - 3)
3x + 3
3x - 9
3(x + 3)
12.
The digit in the ten's place of a two-digit number is 3 more than the digit in the unit's place. If the digit at unit's place be b. Then, the number is
11b + 30
10b+30
11b+3
10b+3
13.
A linear equation in one variable has
only one solution
no solution
two solutions
more than two solutions
14.
Which of the following is a linear expression?
x2 +2 + y
y + y2 + 3
4
1 + z
15.
The Solution of the equation ax+b = 0 is
\(x=\frac { a }{ b } \)
x = -b
\(x=\frac { -b }{ a } \)
\(x=\frac { b }{ a } \)
16.
If 4t - 3 - (3t + 1)= 5t - 4, then the root of t is.....
17.
The sum of two consecutive multiples of 10 is 210. The smaller multiple is........
18.
After 18 yr, Saurabh will be 4 times as old as he is now. His present age is...........
19.
19 is subtracted from the product of P and 14. The result is 21. The value of P is
20.
If on dividing a number by 18, the result is -144, then the number is.................
21.
Two different equations can never have the same answer.
22.
The number of boys and girls in a class in the ratio 5 : 4. If the number of boys is 9 more than the number of girls, then the number of boys is 9.
23.
Two numbers differ by 40, when each number is increased by 8, the bigger becomes thrice the lesser number. If one number is x, then the other number is (40 - x).
24.
If x is an even number, then the next even number is 2(x + 1).
25.
If 16x = 80, then 18x = 90
26.
The length of a rectangle exceeds its breadth by 4 cm. If length and breadth are each increased by 3 cm then area of new rectangle will be 81 cm2 more than that of the given rectangle. Find the length and breadth of the given rectangle.
27.
Divide 34 into two parts in such a way that \(\left( \frac { 4 }{ 7 } \right) \) th of one part is equal to \(\left( \frac { 2 }{ 5 } \right) \) th of the other.
1.
\(x=\frac { 1 }{ 3 } \)
2.
t = 20
3.
\(x=\frac { -8 }{ 5 } \)
4.
p = \(\frac { -4 }{ 3 } \)
5.
We have 1.6 =\(\frac { y }{ 1.5 } \) \(\Rightarrow\) 1.6 x 1. 5 = \(\frac { y }{ 1.5 } \) x 1.5 [multiplying both sides by 1.5]
1.6 x 1.5 = y \(\Rightarrow\) 2.40 = y
y = 2.4, Which is the required solution
6.
\(x=\frac { -164 }{ 3 } \)
7.
Rs.900, Rs 300, Rs.180
8.
Let the present age of son be x yr.
Then, the present age of Arnan =3x yr
10 yr ago, age of Arnan = (3x - 10) yr and age of son
= (x -10)yr
According to the question, (3x -10) = 5 (x -10)
\(\Rightarrow\) 3x - 10= 5x - 50
\(\Rightarrow\) 3x - 5x = - 50 + 10 [transposing -10 to RHS and 5x to LHS]
\(\Rightarrow\) - 2x = - 40 \(\Rightarrow\) 2x = 40 \(\Rightarrow\) x = \(\frac { 40 }{ 2 } \)=20 [dividing both sides by 2]
\(\therefore\) Age of son = x = 20 yr
and age of Arnan = 3x = 3 x 20 = 60 yr
Hence, the present ages of Arnan and his son are 60 yr and 20 yr, respectively
9.
Let the age of Baichung be x yr.
Then, age of Baichung's father = (x + 29) yr
and age of Baichung's grandfather = (x + 29) + 26 = (x + 55)yr
According to the question,
x + (x + 29) + (x + 55) = 135
\(\Rightarrow\) x + x + 29 + x + 55 = 135
\(\Rightarrow\) 3x +84 =135
\(\Rightarrow\) 3x = 135 - 84 [transposing 84 to RHS]
\(\Rightarrow\) 3x = 51 \(\Rightarrow\) x = \(\frac { 51 }{ 3 } \) = 17
Age of Baichung = x = 17 yr
Age of Baichung's father = (x + 29) = (17 + 29) = 46 yr
and age of Baichung's grandfather = (x + 55)
= (17 + 55) = 72 yr
10.
Let the three consecutive integers be x, (x + 1)and (x + 2).
According to the question,
Sum of consecutive integers = 51
\(\Rightarrow\) x + (x + 1)+ (x + 2) = 51
\(\Rightarrow\) x + x + l + x + 2 = 51
\(\Rightarrow\) 3x + 3 = 51 \(\Rightarrow\) 3x = 51 - 3 [transposing 3 to RHS]
\(\Rightarrow\) 3x = 48 \(\Rightarrow\) x = \(\frac { 48 }{ 3 } =16\) [dividing both sides by 3]
\(\therefore\) First integer = x = 16
second integer = x + 1= 16 + 1= 17
and third integer = x + 2 = 16 + 2 = 18
Hence, the required integers are 16, 17 and 18.
11.
(d)
3(x + 3)
12.
(a)
11b + 30
13.
(a)
only one solution
14.
(d)
1 + z
15.
(c)
\(x=\frac { -b }{ a } \)
16.
( )
4t - 3 - (3t + 1)= 5t - 4
\(\Rightarrow\) 4t - 3 - 3t - 1= 5t - 4 \(\Rightarrow\) t - 4 = 5t - 4
\(\Rightarrow\) t - 5t =-4 +4 -4t = 0 \(\Rightarrow\) t = 0
17.
( )
Let two consective multiples of 10 be x and (x + 1)
10 xx + 10 x (x + 1) = 210 \(\Rightarrow\) 10x + 10x +10 = 210
\(\Rightarrow\) 20x = 210 - 10 = 200 \(\Rightarrow\)x = 200 \(\div\) 20 = 10
So the smaller multiple is 10
18.
( )
\(\because\) x + 18 = 4x \(\Rightarrow\) 4x - x = 18
3x = 18 \(\Rightarrow\) x = 6yr
19.
( )
\(\because\) (p x 14) - 19 =21 \(\Rightarrow\)149 -19 = 21
14p = 40 \(\Rightarrow\) p=\(\frac { 40 }{ 14 } =\frac { 20 }{ 7 } =2\frac { 6 }{ 7 } \)
20.
( )
\(\because \frac { -2592 }{ 18 } =-144\)
21.
(b)
22.
(b)
23.
(b)
24.
(b)
25.
(a)
26.
Length of rectangle = 14 cm and breadth of rectangle = 10 cm
27.
First Part = 14, Second Part = 20
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