Exercise 4.1 (Revised) - Chapter 4 - Simple Equations - Ncert Solutions class 7 - Maths

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

Ex 4.1 Question 1.

Complete the last column of the table:

Answer:

Ex 4.1 Question 2

.Check whether the value given in the brackets is a solution to the given equation or not:
(a) $n+5=19(n=1)$
(b) $7 n+5=19(n=-2)$
(c) $7 n+5=19(n=2)$
(d) $4 p-3=13(p=1)$
(e) $4 p-3=13(p=-4)$
(f) $4 p-3=13(p=0)$

Answer:

(a) $n+5=19(n=1)$
Putting $n=1$ in L.H.S.,
$
1+5=6
$
$\because$ L.H.S. $\neq$ R.H.S.,
$\therefore n=1$ is not the solution of given equation.
(b) $7 n+5=19(n=-2)$

Putting $n=-2$ in L.H.S.,
$
7(-2)+5=-14+5=-9
$

$\because$ L.H.S. $\neq$ R.H.S.,
$\therefore n=-2$ is not the solution of given equation.
(c) $7 n+5=19(n=2)$

Putting $n=2$ in L.H.S.,
$7(2)+5=14+5=19$
$\because$ L.H.S. $=$ R.H.S.,
$\therefore n=2$ is the solution of given equation.
(d) $4 p-3=13(p=1)$

Putting $p=1$ in L.H.S.,
$4(1)-3=4-3=1$
$\because$ L.H.S. $\neq$ R.H.S.,
$\therefore p=1$ is not the solution of given equation.
(e) $4 p-3=13(p=-4)$

Putting $p=-4$ in L.H.S.,
$4(-4)-3=-16-3=-19$
$\because$ L.H.S. $\neq$ R.H.S.,
$\therefore p=-4$ is not the solution of given equation.
(f) $4 p-3=13(p=0)$

Putting $p=0$ in L.H.S.,

$
\begin{aligned}
& 4(0)-3=0-3=-3 \\
& \because \text { L.H.S. } \neq \text { R.H.S., }
\end{aligned}
$
$\therefore p=0$ is not the solution of given equation.

Ex 4.1 Question3.

Solve the following equations by trial and error method:
(i) $5 p+2=17$
(ii) $3 m-14=4$

Answer:

(i) $5 p+2=17$
Putting $p=-3$ in L.H.S. $5(-3)+2=-15+2=-13$
$\because-13 \neq 17$ Therefore, $p=-3$ is not the solution.
Putting $p=-2$ in L.H.S. $5(-2)+2=-10+2=-8$
$\because-8 \neq 17$ Therefore, $p=-2$ is not the solution.
Putting $p=-1$ in L.H.S. $5(-1)+2=-5+2=-3$
$\because-3 \neq 17$ Therefore, $p=-1$ is not the solution.
Putting $p=0$ in L.H.S. $5(0)+2=0+2=2$
$\because 2 \neq 17$ Therefore, $p=0$ is not the solution.
Putting $p=1$ in L.H.S. $5(1)+2=5+2=7$
$\because 7 \neq 17$ Therefore, $p=1$ is not the solution.
Putting $p=2$ in L.H.S. $5(2)+2=10+2=12$
$\because 12 \neq 17$ Therefore, $p=2$ is not the solution.
Putting $p=3$ in L.H.S. $5(3)+2=15+2=17$
$\because 17=17$ Therefore, $p=3$ is the solution.

(ii) $3 m-14=4$

Putting $m=-2$ in L.H.S. $3(-2)-14=-6-14=-20$
$\because-20 \neq 4$ Therefore, $m=-2$ is not the solution.
Putting $m=-1$ in L.H.S. $3(-1)-14=-3-14=-17$
$\because-17 \neq 4$ Therefore, $m=-1$ is not the solution.
Putting $m=0$ in L.H.S. $3(0)-14=0-14=-14$

$\because-14 \neq 4$ Therefore, $m=0$ is not the solution. Putting $m=1$ in L.H.S. $3(1)-14=3-14=-11$ $\because-11 \neq 4$ Therefore, $m=1$ is not the solution. Putting $m=2$ in L.H.S. $3(2)-14=6-14=-8$ $\because-8 \neq 4$ Therefore, $m=2$ is not the solution. Putting $m=3$ in L.H.S. $3(3)-14=9-14=-5$ $\because-5 \neq 4$ Therefore, $m=3$ is not the solution. Putting $m=4$ in L.H.S. $3(4)-14=12-14=-2$ $\because-2 \neq 4$ Therefore, $m=4$ is not the solution. Putting $m=5$ in L.H.S. $3(5)-14=15-14=1$ $\because 1 \neq 4$ Therefore, $m=5$ is not the solution. Putting $m=6$ in L.H.S. $3(6)-14=18-14=4$ $\because 4=4$ Therefore, $m=6$ is the solution.

Putting $m=6$ in L.H.S. $3(6)-14=18-14=4$
$\because 4=4$ Therefore, $m=6$ is the solution.

Ex 4.1 Question 4.

Write equations for the following statements:
(i) The sum of numbers $x$ and 4 is 9 .
(ii) 2 subtracted from $y$ is 8 .
(iii) Ten times $a$ is 70 .
(iv) The number $b$ divided by 5 gives 6 .
(v) Three-fourth of $t$ is 15 .
(vi) Seven times $m$ plus 7 gets you 77 .
(vii) One-fourth of a number $x$ minus 4 gives 4 .
(viii) If you take away 6 from 6 times $y$, you get 60 .
(ix) If you add 3 to one-third of $z$, you get 30 .

Answer:

(i) $x+4=9$

(ii) $y-2=8$
(iii) $10 a=70$
(iv) $\frac{b}{5}=6$
(v) $\frac{3}{4} t=15$
(vi) $7 m+7=77$
(vii) $\frac{x}{4}-4=4$
(viii) $6 y-6=60$
(ix) $\frac{z}{3}+3=30$

Ex 4.1 Question 5.

Write the following equations in statement form:
(i) $p+4=15$
(ii) $m-7=3$
(iii) $2 m=7$
(iv) $\frac{m}{5}=3$
(v) $\frac{3 m}{5}=6$
(vi) $3 p+4=25$
(vii) $4 p-2=18$
(viii) $\frac{p}{2}+2=8$

Answer:

(i) The sum of numbers $p$ and 4 is 15 .
(ii) 7 subtracted from $m$ is 3 .
(iii) Two times $m$ is 7 .
(iv) The number $m$ is divided by 5 gives 3 .
(v) Three-fifth of the number $m$ is 6 .
(vi) Three times $p$ plus 4 gets 25 .

(vii) If you take away 2 from 4 times $p$, you get 18 .
(viii) If you added 2 to half is $p$, you get 8 .

Ex 4.1 Question 6.

Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Tale $m$ to be the number of Parmit's marbles.)
(ii) Laxmi's father is 49 years old. He is 4 years older than three times Laxmi's age. (Take Laxmi's age to be $y$ years.)
(iii) 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 . (Take the lowest score to be $l$.)
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be $b$ in degrees. Remember that the sum of angles of a triangle is $180^{\circ}$.)

Answer:

(i) Let $m$ be the number of Parmit's marbles.
$
\therefore 5 m+7=37
$
(ii) Let the age of Laxmi be $y$ years.
$
\therefore 3 y+4=49
$

$
\therefore 3 y+4=49
$
(iii) Let the lowest score be $l$.
$
\therefore 2 l+7=87
$
(iv) Let the base angle of the isosceles triangle be $b$, so vertex angle $=2 b$.
$
\therefore 2 b+b+b=180^{\circ} \Rightarrow 4 b=180^{\circ} \text { [Angle sum property of a } \Delta \text { ] }
$