Exercise 4.3 (Revised) - Chapter 4 Simple Equations class 7 ncert solutions Maths - SaraNextGen [2024-2025]

Updated On August 15, 2024

By SaraNextGen

Chapter 4 - Simple Equations - NCERT Solutions for Class 7 Maths | Comprehensive Guide

Ex 4.3 Question 1.

Set up equations and solve them to find the unknown numbers in the following cases:
1. Add 4 to eight times a number; you get 60 .
2. One-fifth of a number minus 4 gives 3 .
3. If I take three-fourth of a number and add 3 to it, I get 21 .
4. When I subtracted 11 from twice a number, the result was 15 .
5. Munna subtracts thrice the number of notebooks he has from 50 , he finds the result to be 8.
6. Ibenhal thinks of a number. If she adds 19 to it divides the sum by 5 , she will get 8 .
7. Answer thinks of a number. If he takes away 7 from $\frac{5}{2}$ of the number, the result is $\frac{11}{2}$.

Answer:

(a) Let the number be $x$.
According to the question, $8 x+4=60$
$
\begin{aligned}
& \Rightarrow 8 x=60-4 \Rightarrow 8 x=56 \\
& \Rightarrow x=\frac{56}{8} \Rightarrow x=7
\end{aligned}
$
(b) Let the number be $y$.

According to the question, $\frac{y}{5}-4=3$

$
\begin{aligned}
& \Rightarrow \frac{y}{5}=3+4 \Rightarrow \frac{y}{5}=7 \\
& \Rightarrow y=7 \times 5 \Rightarrow y=35
\end{aligned}
$
(c) Let the number be $z$.

According to the question, $\frac{3}{4} z+3=21$
$
\Rightarrow \frac{3}{4} z=21-3 \Rightarrow \frac{3}{4} z=18 \Rightarrow 3 z=18 \times 4
$

$
\Rightarrow 3 z=72 \Rightarrow z=\frac{72}{3} \Rightarrow z=24
$
(d) Let the number be $x$.

According to the question, $2 x-11=15$
$
\begin{aligned}
& \Rightarrow 2 x=15+11 \Rightarrow 2 x=26 \\
& \Rightarrow x=\frac{26}{2} \Rightarrow x=13
\end{aligned}
$
(e) Let the number be $m$.

According to the question, $50-3 m=8$
$
\begin{aligned}
& \Rightarrow-3 m=8-50 \Rightarrow-3 m=-42 \\
& \Rightarrow m=\frac{-42}{-3} \Rightarrow m=14
\end{aligned}
$
(f) Let the number be $n$.

According to the question, $\frac{n+19}{5}=8$
$
\begin{aligned}
& \Rightarrow n+19=8 \times 5 \Rightarrow n+19=40 \\
& \Rightarrow n=40-19 \Rightarrow n=21
\end{aligned}
$
(g) Let the number be $x$.

According to the question, $\frac{5}{2} x-7=\frac{11}{2}$
$
\begin{aligned}
& \Rightarrow \frac{5}{2} x=\frac{11}{2}+7 \Rightarrow \frac{5}{2} x=\frac{11+14}{2} \\
& \Rightarrow \frac{5}{2} x=\frac{25}{2} \Rightarrow 5 x=\frac{25 \times 2}{2} \Rightarrow 5 x=25 \\
& \Rightarrow x=\frac{25}{5} \Rightarrow x=5
\end{aligned}
$

Ex 4.3 Question 2.

Solve the following:
1. 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?
2. 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 $180^{\circ}$.)
3. 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?

Answer:

(a) Let the lowest marks be $y$.
According to the question, $2 y+7=87$
$
\begin{aligned}
& \Rightarrow 2 y=87-7 \Rightarrow 2 y=80 \Rightarrow y=\frac{80}{2} \\
& \Rightarrow y=40
\end{aligned}
$

Thus, the lowest score is 40 .
(b) Let the base angle of the triangle be $b$.

Given, $a=40^{\circ}, b=c$
Since, $a+b+c=180^{\circ}$ [Angle sum property of a triangle]
$
\Rightarrow 40^{\circ}+b+b=180^{\circ}
$

$
\begin{aligned}
& \Rightarrow 40^{\circ}+2 b=180^{\circ} \\
& \Rightarrow 2 b=180^{\circ}-40^{\circ} \Rightarrow 2 b=140^{\circ} \\
& \Rightarrow b=\frac{140^{\circ}}{2} \Rightarrow b=70^{\circ}
\end{aligned}
$

Thus, the base angles of the isosceles triangle are $70^{\circ}$ each.
(c) Let the score of Rahul be $x$ runs and Sachin's score is $2 x$.

According to the question, $x+2 x=198$
$
\Rightarrow 3 x=198 \Rightarrow x=\frac{198}{3}
$

$
\Rightarrow x=66
$

Thus, Rahul's score $=66$ runs
And Sachin's score $=2 \times 66=132$ runs.

Ex 4.3 Question 3.

Solve the following:
1. Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
2. Laxmi's father is 49 years old. He is 4 years older than three times Laxmi's age. What is Laxmi's age?
3. 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?

Answer:

(i) Let the number of marbles Parmit has be $m$.
According to the question, $5 m+7=37$
$
\begin{aligned}
& \Rightarrow 5 m=37-7 \Rightarrow 5 m=30 \\
& \Rightarrow m=\frac{30}{5} \Rightarrow m=6
\end{aligned}
$

Thus, Parmit has 6 marbles.

(ii) Let the age of Laxmi be $y$ years.

Then her father's age $=(3 y+4)$ years
According to question, $3 y+4=49$
$
\begin{aligned}
& \Rightarrow 3 y=49-4 \Rightarrow 3 y=45 \\
& \Rightarrow y=\frac{45}{3} \Rightarrow y=15
\end{aligned}
$

Thus, the age of Laxmi is 15 years.
(iii) Let the number of fruit trees be $t$.

Then the number of non-fruits tree $=3 t+2$
According to the question, $t+3 t+2=102$
$
\begin{aligned}
& \Rightarrow 4 t+2=102 \Rightarrow 4 t=102-2 \\
& \Rightarrow 4 t=100 \Rightarrow t=\frac{100}{4} \\
& \Rightarrow t=25
\end{aligned}
$

Thus, the number of fruit trees are 25.

Ex 4.3 Question 4.

Solve the following riddle:
I am a number, Tell my identity!
Take me seven times over, And add a fifty!
To reach a triple century, You still need forty!
Answer:

Let the number be $n$.
According to the question, $7 n+50+40=300$
$
\begin{aligned}
& \Rightarrow 7 n+90=300 \Rightarrow 7 n=300-90 \\
& \Rightarrow 7 n=210 \Rightarrow n=\frac{210}{7} \\
& \Rightarrow n=30
\end{aligned}
$

$\text { Thus, the required number is } 30 \text {. }$