NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations Exercise 4.3

Exercise 4.3

1. Set up equations and solve them to find the unknown numbers in the following cases.

a. Add 4 to eight times a number’ you get 60.

Ans: Let us assume that the number is $x$.

According to the question, 

$8x + 4 = 60$

To solve this equation, subtract 4 from both sides and divide by 8.

$8x + 4 - 4 = 60 - 4$

$8x = 56$

$x = \dfrac{{56}}{8}$

$x = 7$

Hence, the number is 7.

b. One-fifth of a number minus 4 gives 3.

Ans: Let the number be $y$.

According to the question, 

$\dfrac{y}{5} - 4 = 3$

To solve this equation, add 4 to both sides and multiply both sides by 5.

$\dfrac{y}{5} = 3 + 4$

$\dfrac{y}{5} = 7$

$y = 35$

Hence, the number is 35.

c. If I take three fourth of a number and add 3 to it, I get 21.

Ans: Let the number be $x$.

According to the question, 

$\dfrac{3}{4}x + 3 = 21$

To solve this equation, subtract 3 from both sides, then multiply both sides by 4.

$\dfrac{3}{4}x = 21 - 3$

$\dfrac{3}{4}x = 18$

$3x = 72$

Divide both sides by 3.

$x = \dfrac{{72}}{3}$

$x = 24$

Hence, the number is 24.

d. When I subtracted 11 from twice a number, the result was 15.

Ans: Let the number be $x$.

According to the question, 

$2x - 11 = 15$

Add 11 to both sides and divide both sides by 2.

$2x = 11 + 15$

$2x = 26$

$x = \dfrac{{26}}{2}$

$x = 13$

Hence, the number is 13.

e. Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.

Ans: Let the number of notebooks with Munna be $x$.

According to the question, 

$50 - 3m = 8$

Subtract 50 from both sides.

$ - 3m = 8 - 50$

$ - 3m =  - 42$

$m = \dfrac{{42}}{3}$

$m = 14$

Hence, the number of notebooks with Munna are 14.

f. Ibenhal thinks of a number. If she adds 19 to it divides the sum by 5, she will get 8.

Ans:  Let the number he thinks of be $x$.

$\dfrac{{x + 19}}{5} = 8$

Multiply both sides by 5.

$x + 19 = 8 \times 5$

$x + 19 = 40$

Subtract 19 from both sides.

$x = 40 - 19$

$x = 21$

Hence, the number is 21.

g. Answer thinks of a number. If he takes 7 away from $\dfrac{5}{2}$ of the number, the result is $23$.

Ans: Let the number he thinks of be $x$.

According to the question,

$\dfrac{5}{2}x - 7 = 23$

Add 7 to both sides.

$\dfrac{5}{2}x = 23 + 7$

$\dfrac{5}{2}x = 30$

Multiply both sides by $2$.

$\dfrac{5}{2}x\times 2 = 30\times 2$

$5x=60$

Hence, the number is 5.

2. Solve the Following.

a. The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. What is the lowest score?

Ans: Let the lowest score be $y$.

According to the question,

$2y + 7 = 87$

Subtract 7 from both sides.

$2y = 87 - 7$

$2y = 80$

Divide both sides by 2.

$y = \dfrac{{80}}{2}$

$y = 40$

b. In an isosceles triangle, the base angles are equal. The vertex angle is $40^\circ $. What are the base angles of the triangle? (Remember the sum of three angles of a triangle is $180^\circ $)

Ans: Let the base angle of the triangle be $b$, and $c$. Since the triangle is isosceles, then, $b = c$.

The sum of interior angles is $180^\circ $.

$40^\circ  + b + b = 180^\circ $

$40^\circ  + 2b = 180^\circ $

Subtract $40^\circ $ from both sides.

$2b = 180^\circ  - 40^\circ $

$2b = 140^\circ $

Divide both sides by 2.

$b = \dfrac{{140^\circ }}{2}$

$b = 70^\circ $

Hence, the base angle measure is $70^\circ $.

c. Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?

Ans: Let the score of Rahul be $x$ runs and Sachin’s score be $2x$.

According to the question,

$x + 2x = 198$

$3x = 198$

Divide both sides by 3.

$x = \dfrac{{198}}{3}$

$x = 66$

Thus, the runs scored by Rahul is 66.

Since, Sachin’s score is twice of Rahul then, Sachin’s score will be as follows.

$2 \times 66 = 132\,\,\,\,{\text{runs}}$.

Thus, the runs scored by Sachin are 132 runs. 

3. Solve the Following:

a. Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37marbles. How many marbles does Parmit have?

Ans: Let the number of marbles Parmit has be $m$.

According to the question,

$5m + 7 = 37$

Subtract 7 from both sides.

$5m = 37 - 7$

$5m = 30$

Divide both sides by 5.

$m = \dfrac{{30}}{5}$

$m = 6$

Hence, Parmit has 6 marbles.

b. Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

Ans: Let the age of Laxmi be $x$ years.

According to the question,

$3y + 4 = 49$

Now, to solve this, subtract 4 from both sides and divide the expression obtained by 3.

$3y = 49 - 4$

$3y = 45$

$y = \dfrac{{45}}{3}$

$y = 15$

Hence, the age of Laxmi is 15 years.

c. People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees was two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?

Ans: Let the number of fruit trees be $t$.

Then, according to the question, $3t + 2 = 77$.

To solve the expression, add the like terms first. Then, subtract 2 from both sides. 

$3t+2-2=77-2$

$3t=75$

$t=\dfrac{75}{3}$

$t=25$

Thus, the value of $t$ is 25.

4. Solve the Following Riddle:

I am a number, tell my identity!

Take me seven times over, and add fifty!

To reach a triple century, You still need forty.

Ans: Let the number that is being talked about in the riddle be $n$.

The line, take me seven times over suggests that the number is multiplied by 7. So, the number will become $7n$. Now, fifty is added. So, the number will become, $7n + 50$.

Now, according to the question,

$7n + 50 + 40 = 300$

$7n + 90 = 300$

We will solve this equation by subtracting 90 from both sides and dividing by 7.

$7n = 300 - 90$

$7n = 210$

$n = \dfrac{{210}}{7}$

$n = 30$

Therefore, the required number is 30.

Conclusion

In conclusion, Exercise 4.3 in Chapter 4 of Class 7 Maths provides a solid foundation for solving simple equations. By completing tNCERT Class 7 Maths Exercise 4.3 Solution, you have learned how to perform various operations to isolate and solve for the unknown variable. This practice is essential for building your problem-solving skills and understanding the balance required in equations. Keep practicing these methods to strengthen your mathematical abilities. The concepts you have learned here are fundamental and will be useful in more advanced topics.

Class 7 Maths Chapter 4: Exercises Breakdown

CBSE Class 7 Maths Chapter 4 Other Study Materials

Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.