Class 7 Maths Chapter 4 Practice Question Answers - Simple Equations

Q1. Solve for x:  3x+9=33
(a) 8
(b) 3
(c) 2
(d) 5
Ans: (a)
Sol: Given, 3x+9=33
Subract 9 on both side
⇒3x+9−9=33−9
⇒3x=24
On dividing throughout by 3, we get
⇒ x= 24/3
=8
Hence, required solution is x=8.

Q2. Solve the following equation: y+3=10
(a) 3
(b) 7
(c) 10
(d) 13
Ans: (b)
Sol: Given, y + 3 = 10
Subtracting 3 on both the sides, we get:
y + 3 − 3 = 10 − 3
⇒ y = 7
Hence, option B is correct.

Q3. If  t/5 =10, then t is equal to
(a) 25
(b) 10
(c) 50
(d) 100
Ans: (c)
Sol: Given, t/5 =10
To eliminate 5 from the denominator, we multiply both sides by 5
So, 5 x t/5 = 10 x 5
⇒ t = 50

Q4. If 2x/3 = 18, then x is equal to
(a) 36
(b) 54
(c) 32
(d) 27
Ans: (d)
Sol: Given, 2x/3 = 18
Multiply 3 on both the sides, we get
2x = 18 × 3
⇒ 2x = 54
Divide both side by 2
⇒ x = 27

Q5. Solve the following equation: x−2=7
(a) 7
(b) 2
(c) 9
(d) 11
Ans: (c)
Sol: Given, x − 2 = 7
Add 2 on both the sides, we get
x − 2 + 2 = 7 + 2
⇒ x = 9
Hence, option C is correct.

Q6. If 6x=12, then x is
(a) 12
(b) 2
(c) 72
(d) 6
Ans: (b)
Sol: Given, 6x=12
Dividing both sides by 6, we get:
6x/6 = 12/6
∴ x = 2
Hence, option B is correct.

Q7. The solution of the equation 7+3(x+5)=31 is
(a) 4
(b) 3
(c) 0
(d) 2
Ans: (b)
Sol: Given, 7 + 3(x + 5) = 31
⇒ 7 + 3x + 15 = 31
⇒ 22 + 3x = 31
⇒ 3x = 31 − 22 = 9
⇒  x =  9/3
= 3.
Hence, required solution is x=3.

Q8. When 6 is subtracted from four times a number, the result is 54. What is the number?
(a) 60
(b) 30
(c) 15
(d) 10
Ans: (c)
Sol: Let the number be x.
Then, 4x − 6 = 54
Add 6 on both side
⇒ 4x − 6 + 6 = 54 + 6
⇒ 4x = 60
Divide throughout by 4
⇒ x = 60/4
= 15
Hence, the required number is 15.

Q9. Solve the following equation: 6=z+2
(a) 4
(b) 2
(c) 6
(d) 8
Ans: (a)
Sol: Given, 6 = z + 2
Subtract 2 on both the sides,
6 − 2 = z + 2 − 2
⇒ z = 4
Hence, option A is correct.

Q10. If 7x−9=16, then x is equal to:
(a) x = 16/2
(b) x = 7/7
(c) x = 25/7
(d) x = 2/16
Ans: (c)
Sol: Given, 7x − 9 = 16
Add 9 on both the sides, we get
7x − 9 + 9 = 16 + 9
∴ 7x = 25
Divide 7 on both the sides,
∴  x = 25/7
Therefore, C is the correct answer.

Q11. Find the value of x:  x−7=−8
(a) 1
(b) −1
(c) 0
(d) None of the above
Ans: (b)
Sol: x − 7 = −8
Add 7 to both sides,
⇒ x − 7 + 7 = −8 + 7
⇒ x= −8 + 7
⇒ x = −1
Hence, answer is option B. 

Q12. Solve the following equations:
3(y – 2) = 2(y – 1) – 3
Sol:
3(y – 2) = 2(y – 1) – 3
⇒ 3y – 6 = 2y – 2 – 3 (Removing the brackets)
⇒ 3y – 6 = 2y – 5
⇒ 3y – 2y = 6 – 5 (Transposing 6 to RHS and 2y to LHS)
⇒ y = 1
Thus y = 1

Q13. The length of a rectangle is twice its breadth. If its perimeter is 60 cm, find the length and the breadth of the rectangle.
Sol:
Let the breadth of the rectangle be x cm.
its length = 2x
Perimeter = 2 (length + breadth) = 2(2x + x) = 2 × 3x = 6x
As per the condition of the question, we have
6x = 60 ⇒ x = 10
Thus the required breadth = 10 cm
and the length = 10 × 2 = 20 cm.

Q14. Frame the statement into an equation: Adding 16 to 3 times x is 39.
(a) 3x − 16 = 39
(b) 3x + 16 = −39
(c) −3x + 16 = 39
(d) 3x + 16 = 39
Ans: (d)
Sol: 3 × x = 3x
As per the given condition,
Adding 16 to 3x results in 16 + 3x = 39
So, option D is correct.

Q15. The method of finding solution by trying out various values for the variable is called
(a) Error method
(b) Trial and error method
(c) Testing method
(d) Checking method
Ans: (b)
Sol: The required method is called "Trial and error method"

Q16. Sunita's mother is 36 years old. She is 3 years older than 3  times sunita's age. What is sunita's age?
(a) 6
(b) 7
(c) 8
(d) 11
Ans: (d)
Sol: We do not know sunita’s age.
Let us take it to be y years. .
Sunita’s mother’s age is 3years older than 3y;
It is also given that Sunita’s mother is 36 years old.
Therefore, 3y + 3 = 36
Subract 3 from both side
3y + 3 − 3 = 36 − 3
3y = 33
Divide both side by 3
y = 11

Q17. Convert the statement into an equation : Adding 14 to 9 times y is 89.
(a) 14y + 9 = 89
(b) 14 + 9y = 89
(c) 14 − 9y = −89
(d) 14 + y = 89
Ans: (b)
Sol: According to the question,
9 times y means 9 × y = 9y.
Adding 14 to it means  9y + 14.
∴ the equation is 9y + 14=89.

Q18. Algebraic expression for the statement: 6 times a taken away from 40
(a) 6a − 40
(b) 40a − 6
(c) 40 − 6a
(d) 0
Ans: (c)
Sol: Given, 6 times a takes away from 40.
Required algebraic equation is given by,
40 − 6 × a = 40 − 6a

Q19. Frame the statement "Twice a number decreased by 119 equals 373" into an equation.
(a) 2y + 119 = 373
(b) 3y − 119 = 373
(c) 2y − 119 = 373
(d) None of the above.
Ans: (c)
Sol: Let y be the number.
According to the question,
Twice a number y, decreased by 119 means
2y − 119 = 373.
So, option C is correct.

Q20. Frame the given statement into a mathematical expression: "Thrice of a number increased by 150."
(a) 150 − x
(b) 3x − 150
(c) 3x + 150
(d) None of these
Ans: (c)
Sol: Let x be the required number.
∴ thrice of a number shall be 3 × x = 3x.
According to the question, if it is increased by 150,
The number would be (3x + 150).

Q21. What is an equation?
(a) A statement in which the values of two mathematical expressions are equal.
(b) An expression that computes the values of variables.
(c) A mathematical expression.
(d) None of these
Ans: (a)
Sol: An equation is a mathematical statement which represents two things are equal. It consists of two expressions, one on either side of equal to symbol. In other words, an equation is a statement in which the values of two mathematical expressions are equal.

Q22. If the average of 3,4, and x is 2, then find x.
(a) −15
(b) −1
(c) −8
(d) −6
Ans: (b)
Sol: Given: The average of 3,4 and x is 2
we have to find x
⇒ 3 + 4 + x /3 = 2
⇒ 7 + x = 6
⇒ x = −1
x is −1
Hence, the answer is option B.

Q23. The value of x in the equation 5x−35=0 is:
(a) 2
(b) 7
(c) 8
(d) 11
Ans: (b)
Sol: Given, 5x − 35 = 0
Transpose 35 to R.H.S.,
5x = 35
Divide throughout by 5
5x/5 = 35/5
⇒ x = 7

Q24. 12 times a number is 4 less than the 5 times the number. Write an equation representing the statement given.
(a) 12x − 4 = 5x
(b) 12x + 4 = 5x
(c) 12 + 4x = 5
(d) 12 − 4x = 5
Ans: (B)
Sol: Let the number be x
⇒ 12 × x = 5 × x − 4
⇒ 12x + 4 = 5x

Q25. Three times the number is 300. Convert this statement in mathematical form.
(a) 3x = 100
(b) 3+x = 300
(c) 3 = 300 + x
(d) 3x = 300
Ans: (d)
Sol: 3 times the number is 300.
⇒ Let the number = x
⇒ 3 times the number = 3x
⇒ This is equal to 3x = 300.
Hence, the answer is 3x = 300.

Q26. If the mean of 6, 8, 5, x and 4 is 7, then the value of x is ______.
(a) 11
(b) 12
(c) 13
(d) 14
Ans: (b)
Sol: Given, mean =7 and data is 6,8,5,x,4
Therefore, 6 + 8 + 5 + x + 4 / 5 = 7
⇒  x + 23 = 35
⇒  x = 12

Q27. Which of the sign should always be there in an equation?
(a) =
(b) ≠
(c) >
(d) <
Ans: (a)
Sol: An equation is a mathematical statement that two things are equal. It consists of two expressions, one on each side of an 'equals' sign. For example x=y is an equation where two expressions x and y are equal.
Hence, = sign should always be there in an equation.

Q28. 12, a ,6, 8, 2, 14 The average (arithmetic mean) of the numbers listed above is 6. Calculate the value of a.
(a) −36
(b) −6
(c) 0
(d) 6
Ans: (b)
Sol: We know that  average of a bunch of numbers is the sum of the numbers divided by how many numbers in the bunch. We can set up an equation for the average from the given question,
⇒ 12 + a + 6 + 8 + 2 + 14 / 6  = 6
⇒ 42 + a / 6 = 6
⇒ 42 + a = 6
⇒ a = -6