Simple Equations For Class 7 Maths Important Questions

Q 1 – Write the following statement in the form of an equation:
The sum of three times x and 10 is 13.
(a) 3x + 10 = 13
(b) 3x – 10 = 13
(c) 3x + 13 = 10
(d) none of these

(a) 3x + 10 = 13

Q 2 – Write the following statement in the form of an equation: Four times a number p is 8.
(a) 4P = 8
(b) P + 4 = 8
(c) p – 4 = 8
(d) p ÷ 4 = 8

(a) 4P = 8

Q 3 – Write an equation for If you take away 6 from 6 times y you get 60.
(a) 6y – 6 = 60
(b) 6y + 6 = 60
(c) 6y ÷ 6 = 60
(d) None of these

(a) 6y – 6 = 60

Q 4 – One – Fifth of a number minus 4 gives 3. Find which of the following is the number?
(a) 21
(b) 4
(c) 15
(d) 35

(d) 35

 Q 5 – Write an equation in statement form : 2m = 7.
(a) Two times of the number m is 7.
(b) Two added to m becomes 7.
(c) Two subtracted from m becomes 7.
(d) None of these. 
    

(a) Two times of the number m is 7

Q 6 – Write the following statement in the form of an equation: Add 1 to three times n to get 7
(a) 3n + 1 = 7
(b) 3n – 1 = 7
(c) 3n + 7 = 1
(d) none of these

(a) 3n + 1 = 7

Q 7 – Write the statements “The sum of three times x and 11 is 32 ” in the form of equations:

(a) 3x + 11 = 32
(b) 3x – 11 = 34
(c) 5x – 7 = 2
(d) None of these

(a) 3x + 11 = 32

Q 8 – Write an equation for 2 subtracted from y is 8.
(a) y – 2 = 8
(b) 2y = 8
(c) y + 2 = 8
(d) None of these

(a) y – 2 = 8

Q 9 – The solution of the equation 5x = 10 is

(a) 1
(b) 2
(c) 5
(d) 10

(b) 2

Q 10 – Write an equation in statement form: m minus seven gives 3
(a) m + 7 = 3
(b) 7m = 3
(c) m – 7 = 3
(d) None of these

(c) m – 7 = 3

Q 11 – The solution of the equation 7 n + 5 = 12 is
(a) 0
(b) – 1
(c) 1
(d) 5

(c) 1

Q 12 – The solution of the equation 3p + 5 = 8 is

(a) – 1

(b) 1

(c) 3

(d) 5

(b) 1

Q 13 – In a coconut grove, (x + 2) trees yield 60 coconuts per year, x trees yield 120 coconuts per year and (x 2) trees yield 180 coconuts per year. If the average yield per year per tree is 100, Find x.
(a) 4
(b) 3
(c) 2
(d) 1

(a) 4

Q 14 – One-fifth of a number minus 4 gives 3. Find which of the following is the number?
(a) 21
(b) 4
(c) 15
(d) 35

(d) 35

Q 15 – The solution of the equation P/2 + 1 = 3 is
(a) 1
(b) 2
(c) 3
(d) 4

(d) 4

Q 16 – Write an equation for Ten times a is 70.
(a) a +10 = 70
(b) a – 10 = 70
(c) 10a = 70
(d) None of these

(c) 10a = 70

Q 17 – The solution of the equation y – 4 = – 1 is
(a) 1
(b) 2
(c) 3
(d) 4

(c) 3

Q 18 – The solution of the equation 10 t = – 20 is
(a) 1
(b) – 1
(c) 2
(d) – 2

(d) – 2

Q 19 – If 0.2 (2x–1) – 0.5 (3x–1) = 0.4, what is the value of x ?
(a) 1/11
(b) – 1/11
(c) 3/11
(d) – 3/11

(a) 1/11

Q 20 – The solution of the equation 2 (m + 3) = 8 is
(a) 1
(b) 2
(c) 3
(d) 4

(a) 1

Q 21 – By solving the equation 2a  2 = 20, the value of ‘a’ will be
(a) 12
(b) 14
(c) 11
(d) 13

(c) 11

Q 22 – The solution of the equation – 4 (2 + x) = 4 is
(a)  1
(b) –
 2
(c) –
 3
(d) –
 4

(a)  1

Q 23 – The solution of the equation 0 = 4 + 4(m + 1) is
(a) 1
(b) –
 1
(c) 2
(d) –
 2

(d) – 2

Q 24 – Which is a solution of the equation 4x – 3 = 13?
(a) x = 5
(b) x = 3
(c) x = 4
(d) None of these

(c) x = 4

Q 25 – Write the following statements in the form of equations.
(a) The sum of four times a number and 5 gives a number five times of it.
(b) One-fourth of a number is 2 more than 5.

(a) Let the number be x.
Sum of 4x and 5 = 4x + 5
The sum is 5x.
The equation is 4x + 5 = 5x as required.

(b) Let the number be x.
1 / 4 x = 5 + 2
 1 / 4 x = 7 as required.

Q 26 – If k + 7 = 10, Find the value of 9k – 50.

k + 7 = 10

⇒ k = 10 – 7 = 3

Put k = 3 in 9k – 50, and we get

9 × 3 – 50

= 27 – 50

= – 23

Thus the value of k = – 23

Q 27 – Solve the following equations:
3(y – 2) = 2(y – 1) – 3

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

Q 28 – Find the value x

Q 29 – The length of a rectangle is twice its breadth. If its perimeter is 60 cm, Find the length and the breadth of the rectangle.

Let the breadth of the field = x cm

therefore, Its length = 2x

and its perimeter = 2 x (length + breadth)

 = 2 x (2x + x)

= 2 (3x)

= 6x cm

 Perimeter = 60 cm

⇒ 60 cm = 6x cm

⇒ x = 60 / 6 = 10 cm

therefore, Breadth = x = 10 cm

Length = 2x = 2 x 10 = 20 cm

Q 30 – The present age of a son is half the present age of his father. Ten years ago, the father was thrice as old as his son. What are their present age?

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x+ 10) + 10

= 2[(x+ 10) + 10]

3x+ 20 = 2x+ 40

x = 20

https://www.indiabix.com/_files/images/aptitude/1-sym-tfr.gifRequired ratio = (3x+ 10) : (x+ 10)

= 70 : 30

= 7 : 3

Q 31 – Each of the 2 equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is 30 cm, find the length of each side of the triangle.

Let the smaller side be = x cm

Then, larger side = (x + 4) cm

And, third side = (2x – 6) cm

Given that,

Perimeter = 50 cm

⇒ x + (x + 4) + (2x – 6) = 50

⇒ 4x – 2 = 50

⇒ 4x = 52

⇒ x = 13

Therefore, smaller side = 13cm

Larger side = x + 4 = 13 + 4 = 17cm

Third side = 2x – 6 = 2 × 13 – 6 = 26 – 6 = 20 cm

To find area of triangle,

Let a = 13, b = 17, c = 20

s = (a + b + c)/2

⇒ s = (13 + 17 + 20)/2 = 50/2 = 25

Area of triangle = √s (s – a)(s – b)(s – c)

= √25 (25 – 13)(25 – 17)(25 – 20)

= √25 × 12 × 8 × 5

= 20√30 cm²

Q 32 – Each of the 2 equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is 30 cm, find the length of each side of the triangle.

Let the length of the third side be x cm.
Each equal side = 2 x cm.
As per the condition of the question, we have
Perimeter = x + 2 x + 2 x = 30
5 x = 30
x = 6
Thus, the third side of the triangle = 6 cm
and other two equal sides are 2 × 6 = 12 cm each

Q 33 – Each of the 2 equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is 30 cm, find the length of each side of the triangle.

Let the length of the third side be x cm.
Each equal side = 2x cm.
As per the condition of the question, we have
Perimeter = x + 2x + 2x = 30
 5x = 30
 x = 6

Thus, the third side of the triangle = 6 cm
and other two equal sides are 2 × 6 = 12 cm each

Q 34 – Convert the following equations in statement form:
(a) 5x = 20
(b) 3y + 7 = 1

(a) Five times a number x gives 20.
(b) Add 7 to three times a number y gives 1

Q 35 – Solve the following equations:
3(y – 2) = 2(y – 1) – 3

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

Q 36 – If 5 is added to twice a number, the result is 29. Find the number.

Let the required number be x.

Step I: 2x + 5
Step II: 2x + 5 = 29
Solving the equation, we get
2x + 5 = 29
 2x = 29 – 5 (Transposing 5 to RHS)
 2x = 24
 x = 12 (Dividing both sides by 2)
 x = 12
Thus the required number is 12.

Q 37 – The length of a rectangle is twice its breadth. If its perimeter is 60 cm, find the length and the breadth of the rectangle.

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.

 

Q 38 – A man travelled two-fifth of his journey by train, one-third by bus, one-fourth by car and the remaining 3 km on foot. What is the length of his total journey?

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Thus, the required journey = 180 km.

Q 39 – If one-third of a number exceeds its one-fourth by 1, find the number.

Let the required number be x.

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Q 40 – Solve the following equations and check the answers.

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