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2a – 3 = 5 

2a = 5 + 3 (transposing – 3)

2a = 8 (transposing x 2) 

a = 4 

Check 

LHS = 2a – 3 = 2 x 4 – 3 = 8 – 3 = 5 

RHS =5 

∴ L.H.S = R.H.S

3(x – 3)=5(2x+ 1) 

3(x – 3)= 5(2x ÷ 1) 

3x – 9= 10x+5 

3x = 10x + 5 ÷ 9 (transposing – 9) 

3x = 10x + 14

3x – 10x = 14(transposing+ 10 x) – 7x =14 

x = 14/-7 (transposing x ( – 7))

Check 

LHS = 3(x – 3) = 3[( – 2) – 3] = 3x( – 5) = – 15 

RHS = 5(2x + 1) = 5 x [2( – 2) + 1] = 5 x [ – 4 + 1] 

= 5 x ( – 3)= – 15 

∴ LH.S = R.H.S

(i) 2 + y = 7 

if y = 1; LHS = 2 + y = 2 + 1 = 3 ≠ 7 

If y = 2; LHS = 2 + 2 = 4 ≠ 7 

If y = 3; LHS = 2 + 3 = 5 ≠ 7 

If y = 4; LHS = 2 + 4= 6 ≠ 7 

If y = 5; LHS = 2 + 5 = 7 = 7 = RHS 

∴ y = 5 is the solution of 2 + y = 7

(ii) a – 2 = 6 

As a – 2 = 6 ; the value of ‘a’ must be greater than 6. 

If a = 7; LHS = 7 – 2 = 5 ≠ 6 

If a = 8; LHS = 8 – 2 = 6 = RHS 

∴ a = 8 is the solution of a – 2 = 6 

(iii) 5m = 15 

If m = 1 then LHS = 5 × 1 = 5 ≠ b15 

m = 2 then LHS = 5 × 2 = 10 ≠ 15 

m = 3 then LHS = 5 × 3 = 15 = 15 

∴ m = 3 is the solution of 5m = 15

(iv) 2n = 14 

If n = 1 then LHS = 2 × 1 = 2 ≠ 14 

n = 2 then LHS = 2 × 2 = 4 ≠ 14 

n = 3 then LHS = 2 × 3 = 6 ≠ 14 

n = 4 then LHS = 2 × 4 = 8 ≠ 14

n = 5 then LHS = 2 × 5 = 10 ≠ 14 

n = 6 then LHS = 2 × 6 = 12 ≠ 14 

n = 7 then LHS = 2 × 7 = 14 ≠ 14

Let the Lowest mark of the class = x 

Twice the least mark = 2x 

On adding 7 = 2x + 7 

By problem; 2x + 7 = 87 

2x = 87 – 7 

2x = 80 

x = 80/2 = 40 

x = 40 

∴ The lowest mark = 40

Let one number be = x. 

Then the other number x – 3 

Sum of the numbers x + x – 3 = 2x – 3 

By problem, 2x – 3 = 95 

2x = 95 + 3 (transposing – 3) 

2x = 98 

x = 98/2 (transposing x 2) 

x = 49 

∴ if one number x 49 then the other number x – 3 = 49 – 3 = 46

Let the number be x 

Multiplied by 7 

⇒ 7x Then reduced by 3 

⇒ 7x – 3 

By problem, 7x – 3 = 53 

7x = 53 + 3 (transposing – 3) 

7x = 56 

x = 56/7 (transposing x 7) 

x = 8 . 

∴ The required number = 8

i) x + 5 = 9 

x + 5 – 5 9 – 5 (subtract 5 from both sides) 

x = 4 

Check LHS = x + 5 (substituting x = 4) 

= 4 + 5 = 9 

RHS = 9 

∴ L.H.S = R.H.S

ii) y – 12 = – 5 

y – 12 = – 5 

y – 12 + 12= – 5 + 12 (add l2on both sides) 

y = 7 

Check 

LHS = y – 12

= 7 – 12= – 5 

RHS = -5 

∴ L.H.S = R.H.S

iii) 3x+4= 19 

3x + 4 = 19 

3x + 4 – 4 = 19 – 4 (subtract 4 from both sides) 

3x = 15

\(\frac {3x}{3} = \frac {15}{3}\) (Divide both sides by 3)

x = 5 

Check 

LHS

 = 3x + 4 

= 3 x 5 + 4 

= 15 + 4 = 19 

RHS = 19 

∴ L.,H.S = R.H.S

iv) 9z = 81

\(\frac {9z}{9} = \frac {81}{9}\) (Divide both sides by 9)

z = 9 

Check 

LHS = 9z = 9 x 9 = 81 

RHS = 81 

∴ LHS = RHS

v) 3x + 8 = 5x + 2 

3x + 8 = 5x + 2 

3x + 8 – 8 = 5x + 2 – 8 

(Adding -8 on both sides) 

3x = 5x – 6 

3x – 5x = 5x – 6 – 5x 

(Subtract 5x from both sides)

-2x = -6

\(\frac {-2x}{-2} = \frac {-6}{-2}\) (Divide both sides by -2)

Check 

LHS = 3x + 8 = 3(3) + 8 = 9 + 8 = 17 

RHS = 5x + 2 = 5(3) + 2 = 15 + 2 = 17 

∴ LHS = RHS

(vi) 5y + 10 = 4y – 10 

5y + 10 = 4y – 10 

5y + 10 – 1o = 4y – 10 – 10 (Subtract 10 from both sides) 

5y =4y – 20

5y – 4y = 4y – 20 – 4y (Substract ty from both sides) 

y = – 20 

Check 

LHS = 5y + 10 = 5 x( – 20) + 10 = – 100+ 10= – 90

RHS = 4y – 10 = 4 x ( – 20) – 10 = – 80 – 10 = -90

∴ LHS = RHS

Understudies can get to the Class 7 Maths MCQ Questions for Simple Equations with Answers supports your test preparation and you can get a decent hold of the part. Use MCQ Questions for Class 7 Maths with Answers during preparation and score more marks in the test. Practicing the MCQ Questions on Simple Equations Class 7 with answers will support your certainty subsequently assisting you with scoring more marks in the test. 

Understudies can check their exam preparation level by solving MCQ Questions for Class 7 Maths Simple Equations. Examine various MCQ Questions of Simple Equations Class 7 with answers furnished with definite arrangements by looking underneath.

Practice MCQ Questions for Class 7 Maths

1. The solution of the equation x + 3 = 0 is

(a) 3
(b) -3
(c) 0
(d) 1

2. The solution of the equation x – 6 = 1 is

(a) 1
(b) 6
(c) -7
(d) 7

3. The solution of the equation 5x = 10 is

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

4. The solution of the equation m/2 = 3 is

(a) 2
(b) 3
(c) 12
(d) 6

5. The solution of the equation 7n + 5 = 12 is

(a) 0
(b) – 1
(c) 1
(d) 5

6. The solution of the equation 3p – 2 = 4 is

(a) 0
(b) 1
(c) 2
(d) 3

7. The solution of the equation m/3 = 3 is

(a) 3
(b) 6
(c) 9
(d) 12

8. The solution of the equation = 6 is

(a) 6
(b) 7
(c) 14
(d) 3

9. The solution of the equation 4p – 2 = 10 is

(a) 1
(b) 2
(c) 3
(d) 4

10. The solution of the equation p/2 + 1 = 3 is

(a) 1
(b) 2
(c) 3
(d) 4

11. The solution of the equation 10 y – 20 = 30 is

(a) 1
(b) 2
(c) 3
(d) 5

12. The solution of the equation 2s = 0 is

(a) 2
(b) -2
(c) 0
(d) 1/2

13. The solution of the equation 12p – 11 = 13 is

(a) 1
(b) 2
(c) 3
(d) 4

14. The solution of the equation – 2(x + 3) = 4 is

(a) -2
(b) -3
(c) -4
(d) -5

15. Write the statements “Seven times a number plus 7 gets you 77”in the form of equations:

(a) 7x + 7 = 77
(b) 7x – 7 = 77
(c) 7x + 6 = 66
(d) None of these

16. 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

17. In an isosceles triangle, the base angles are equal to 50º. The vertex angle is

(a) 45º
(b) 80º
(c) 75º
(d) 85º

18. The letters which are used to represent numbers are called

(a)  variables
(b) constant
(c) binomial
(d) monomial

19. The value of the variable which satisfies the equation is called the _____

(a) coefficient
(b) degree
(c) binomial
(d) solution of the equation

20. A number when added to its half gives 36. Find the number.

(a) 24
(b) 28
(c) 20
(d) 18

21. Ila read 25 pages of a book containing 100 pages. Lalita read 2/5 of the same book. Who reads less?

(a) Ila
(b) Lalita
(c) Both read same pages
(d) None of these

22. If four-fifths of a number is greater than three-fourths of the number by 4. Find the number

(a) 70
(b) 80
(c) 90
(d) None of these

23. After 12 years I shall be 3 times as old as I was 4 years ago. Find my present age.

(a) 12
(b) 13
(c) 14
(d) 15

24. The sum of two consecutive multiples of 6 is 66. Find these multiples. 

(a) 30,36
(b) 40,50
(c) 50,60
(d) 40,70

25. 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

Answer:

1. Answer: (b) -3

Explanation: x + 3 = 0 

⇒ x = -3.

2. Answer: (d) 7

Explanation: x – 6 = 1 

⇒ x = 1 + 6 

= 7.

3. Answer: (b) 2

Explanation: 5x = 10 

⇒ x = 10/5 

= 2

4. Answer: (d) 6

Explanation: m/2 = 3 

⇒ m = 3 × 2 

= 6.

5. Answer: (c) 1

Explanation: 7n + 5 = 12 

⇒ 7n = 12 – 5

⇒ 7n = 7 

⇒ n = 7/7 

= 1

6. Answer: (c) 2

Explanation: 3p – 2 = 4 

⇒ 3p = 4 + 2 

= 6

⇒ P = 6/3

= 2

7. Answer: (c) 9

Explanation: m/3 = 3 

⇒ m = 3 × 3 

= 9.

8. Answer: (c) 14

Explanation: 3m/7  = 6 

⇒ 3m = 7 × 6

3m = 42

⇒ m = 42/3 

= 14

9. Answer: (c) 3

Explanation: 4p – 2 = 10

⇒ 4p = 10 + 2 

= 12

⇒ p = 12/4 

= 3

10. Answer: (b) 2

Explanation: p/2  + 1 = 3 

⇒ p/2 = 3 – 1 

= 2

p = 2 × 2 

= 4

11. Answer: (d) 5

Explanation: 10y – 20 = 30 

⇒ 10y = 30 + 20 

= 50

⇒ y = 50/10 

= 5

12. Answer: (c) 0

Explanation: 2s = 0 

⇒ s = 0/2 

= 0.

13. Answer: (b) 2

Explanation: 12p – 11 = 13 

⇒ 12p = 13 + 11 

= 24

⇒ p = 24/12 

= 2

14. Answer: (d) -5

Explanation: 2 (x + 3) = 4 

⇒ x + 3 = 4−2 

x + 3 = -2

⇒ x = -2 – 3

= -5.

15. Answer: (a) 7x + 7 = 77

Explanation: Seven Times a Number Plus Seven is 77

=> 7x+7 

= 77

16. Answer: (b) 80º

Explanation: \(\frac{(x+2)\times60+x\times120+(x-2)\times 180}{x+2+x+x-2}=100\)

\(\frac{60x+120+120x+180x-360}{3x}=100\)

360x−240 = 300x

60x = 240

x = 4

17. Answer: (b) 80º

Explanation: The sum of all the angles of a triangle is always equal to 180°. Moreover, the isosceles triangle always has two equal sides, hence, the two base angles will also be equal.

Let us assume the vertex angle to be x.

Now, representing the sum of triangles in equation form -

50 + 50 + x = 180

100 + x = 180

x = 180 - 100

x = 80º

18. Answer: (a)  variables

Explanation: A variable is a letter or symbol used as a placeholder for an unknown value.

19. Answer: (d) solution of the equation

Explanation: The value of the variable which satisfies the equation is called the solution of the equation. For example the value of variable x=−5 satisfies the equation x+5=0, therefore, x=−5 is the solution of the equation x+5=0.

20. Answer: (a)  24

Explanation: Let the number be x

⇒ x+ x/2 =36

= 3x/2 = 36

⇒ x = 24

21. Answer: (a) Ila

Explanation: Numbers of pages read by Lalita  \(=\frac{2}{5}\times100=40\)

Number of pages read by Ila =25

Hence, Ila has read less number of pages.

22. Answer: (b) 80

Explanation: Let the required number be 'x', then four fifths of the number = 4/5x

And three fourths of the number = 3/4x

It is given that 4/5x is greater than 3/4x by 4

\(\frac{4}{5}x-\frac{3}{4}x=4\)

\(\frac{16x-15x}{20}=4\)

\(\frac{x}{20}=4\)

x = 80

hence the required number is 80.

23. Answer: (a) 12

Explanation: Let my present age=x years

After 12 years my age=x+12 years

4 years ago my age =x−4 years

According to the question,

x+12=3(x−4)

⇒ x +12 = 3x−12

3x−x=12+12

⇒ 2x =24 or x = 24/2 = 12 years

∴ my present age is 12 years.

24. Answer: (a) 30,36

Explanation: Let the two consecutive multiples of 6 be 6x and 6x+6

According to the given condition,

6x+6x+6=66

12x+6=66

12x=60

x=5

Thus required multiples are 30 and 36.

25. Answer: (a) -1/11

Explanation: 0.2(2x−1)−0.5(3x−1)=0.4

0.4x−0.2−1.5x+0.5=0.4

−10.1x=0.1

x = -0.1/10.1

= -1/11

Click here for Practice MCQ Questions for Simple Equations Class 7

5(x + 4) = 35 

x + 4 = 35/5 (transposing x 5)

x + 4 = 7 

x = 7 – 4 (transposing + 4) 

x = 3 

Check 

LHS = 5(3 + 4) = 5 x 7 = 35 

RHS = 35 

∴ L.H.S = R.H.S

14 = 27 – x

0 = 27 – x – 14 (transposing + 14) 

0 = 13 – x (transposing – x) 

x = 13 

Check

 LHS = 14 

RHS = 27 – x = 27 – 13 = 14 

∴ L.H.S = R.FIS

Let the number of ₹ 100 prizes be x 

Then the number of ₹ 25 prizes be = 63 – x 

Value of the prizes = 100x + (63 – x) x 25 

= 100x+ 1575 – 25x 

= 75x + 1575 

By problem, 75x + 1575 = 3000 

75x = 3000 – 1575 

75x = 1425

x = 1425/75

x = 19 

∴ ₹ 100 prizes = 19 

₹25 prizes= 63 – x =63 – 19 = 44

Let the number be x 

Then three times the number = 3x 

On subtracting 22 

⇒ 3x – 22 

By problem, 3x – 22 = 68 

3x = 68 + 22 (transposing -22) 

3x = 90 

x = 90/3 (transposing x 3) 

x = 30 

∴ The required number 30

Let the present age of Hema be x years 

After 15 years Hema Age = 4x 

By problem, x + 15 = 4x 

x + 15 – 4x = 4x – 4x (Subtracting 4x from both sides) 

– 3x + 15 = 0 

– 3x = – 15 

x = -15/-3 transposing x ( – 3)] 

x = 5 

∴ Her present age is 5 years. 

∴ Her present age is 5 years.

Let the number be x 

Then twice the number = 2x 

On adding = 2x + 7

By problem, 2x + 7 = 49 

2x = 49 – 7 (transposing + 7) 

2x = 42

x =42/2 (transposing x 2) 

x = 21

Let the three integers be = x, x + 1, x 2

Sum of the integers = x + x + 1 + x + 2 = 3x + 3 

By problem, 3x + 3 = 24’ 

3x = 24 – 3 (Transposing + 3) 

3x = 21 

x = 21/3 (transposing x 3) 

x = 7 

∴ The integers x = 7 

x + 1 = 7 + 1 = 8 

x + 2 = 7 + 2 = 9

Length of the rectangle = 5x + 4

Breadth of the rectangle = x – 4 

Perimeter of the rectangle = 2 x (length + breadth) 

= 2x[(5x + 4)+(x – 4)] 

= 2[5x + 4 + x – 4] 

= 2(6x) 

= 12x 

By problem, 12x = 72

x = 72/12 (transposing x 12) 

x = 6 

∴ Length of the rectangle = 5x + 4 = 5 x 6.4 = 34cm 

Breadth of the rectangle = x – 4 = 6 – 4 = 2cm