NCERT Solutions for Class 7 Maths Chapter 4 - Exercise

Exercise 4.4

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 $\dfrac{{11}}{2}$.

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

According to the question,

$\dfrac{5}{2}x - 7 = \dfrac{{11}}{2}$

Add 7 to both sides.

$\dfrac{5}{2}x = \dfrac{{11}}{2} + 7$

$\dfrac{5}{2}x = \dfrac{{11 + 14}}{2}$

Multiply both sides by $\dfrac{2}{5}$.

$x = \dfrac{{25}}{2} \times \dfrac{2}{5}$

$x = 5$

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 is 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 Sundergram planted a total of 102 trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted?

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

Then, according to the question, $t + 3t + 2 = 102$.

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

$4t + 2 = 102$

$4t = 102 - 2$

$4t = 100$

Divide both sides by 4.

$t = \dfrac{{100}}{4}$

$t = 25$

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.

What do you learn from Class 7 Maths exercise 4.4?

Algebraic simple equations give us a lesson about using a set of variables and constants and the relation between them to deduce the value of the variable. Algebra makes use of alphabets to represent variables. You will find the active usage of letters such as x, y, z or a, b, and is used to represent variables in the lesson. Linear equations in themselves may not seem so daunting a lesson at first, but the exercise introduces students to a plethora of different problems. There are various patterns to the problems and you will have to commandeer knowledge of the most effective techniques to deduce the value of ‘x.’

The NCERT Maths book Class 7 Chapter 4 exercise 4.4 provides students with four questions of different patterns. All four questions have multiple Each of the patterns come with their methods for the deduction. You will need to practice them several times to answer the questions in the exam within time. For instance, you may be asked to write up an equation based on an unknown variable and a given constant value. The paper can also ask you anecdotal questions, where you will have to both set-up the equation and then derive the value of the variable.

The best way to proceed is to try and finish the exercise by yourself at first. Should your initial attempt to produce unfavorable results, you can always resort to consulting a solution paper, and then try solving the exercise without help.

Class 7 exercise 4.4 solution PDF

The NCERT Maths for Class 7, exercise 4.4 Solutions PDF from Vedantu contains all the answers to the lesson on problem-based Simple Equations. The solution guide consists of all the answers to the most recent version of the NCERT exercise book. With all of the corresponding answers provided next to the questions, it may be easier for students to compare patterns, and choose the one more convenient for them. You have the with the assistance of the Vedantu guide, then you need to try the same after studying the solutions.

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Benefits of Vedantu’s NCERT Solutions for Class 7 Maths Chapter 4.4 PDF

Given the vast nature of the 4th chapter, let alone the whole Maths syllabus, students have to retain a lot of formulae and patterns on their own. Unfortunately, Maths is not one of those subjects where you can simply memorize and put it in your own words. You need to solve the problems in the exam with clearly defined steps. Algebra goes on to play a big role in advanced, college-level Mathematics, Engineering, Architecture, Physics, Chemistry, and so many other subjects.

Thus, if you want to guarantee confidence in one of the most important lessons in your Class 7 syllabus, you must download the NCERT Solutions for Class 7 Maths Chapter 4 exercise 4.4 from Vedantu. There are many benefits to seeking assistance from Vedantu NCERT Solutions:

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Advantages of Vedantu’s Maths Class 7 Chapter 4 exercise 4.4 Solutions PDF

Students who wish to outperform themselves in their Class 7 Maths paper, must go the extra mile to practice the NCERT exercise multiple times. They must stress on exercise 4.4 because it serves as the building blocks of a student’s understanding of Algebra. Armed with the NCERT Solutions of Class 7 Maths Chapter 4 exercise 4.4 from Vedantu, students can attempt any question they want from this section. For those who are still wondering if the solution guide from Vedantu may make any sort of difference at all, here’s how you can make your Maths preparations easier:

• Head over to Vedantu’s website and download the NCERT Solutions for exercise 4.4. The file is free to download just like all of the other NCERT study materials for all classes, on the site. You will find the PDF categorized under the subjects for Class 7.

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Vedantu’s solutions not only provide you with easy steps to solving an equation that felt so difficult on your own; it will also ensure you have a proper understanding of the formulae and the pattern so you can solve the problem on your own no matter how much they tweak the question.

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