Simple Equations Ex-4.1

NCERT solutions for Maths Simple Equations

NCERT Solutions for Class 7 Maths Simple Equations

Class –VII Mathematics (Ex. 4.1)

Question 1.Complete the last column of the table:

Answer:

NCERT Solutions for Class 7 Maths Exercise 4.1

Question 2.Check whether the value given in the brackets is a solution to the given equation or not:

(a) $n + 5 = 19\left( {n = 1} \right)$

(b) $7n + 5 = 19\left( {n = – 2} \right)$

(c) $7n + 5 = 19\left( {n = 2} \right)$

(d) $4p – 3 = 13\left( {p = 1} \right)$

(e) $4p – 3 = 13\left( {p = – 4} \right)$

(f) $4p – 3 = 13\left( {p = 0} \right)$

Answer:

(a) $n + 5 = 19\left( {n = 1} \right)$

Putting $n = 1$ in L.H.S.,

1 + 5 = 6

$\because$ L.H.S. $\ne$ R.H.S.,

$\therefore$ $n = 1$ is not the solution of given equation.

(b) $7n + 5 = 19\left( {n = – 2} \right)$

Putting $n = – 2$ in L.H.S.,

$7\left( { – 2} \right) + 5 = – 14 + 5 = – 9$

$\because$ L.H.S. $\ne$ R.H.S.,

$\therefore$ $n = – 2$ is not the solution of given equation.

(c) $7n + 5 = 19\left( {n = 2} \right)$

Putting $n = 2$ in L.H.S.,

$7\left( 2 \right) + 5 = 14 + 5 = 19$

$\because$ L.H.S. $=$ R.H.S.,

$\therefore$ $n = 2$ is the solution of given equation.

(d) $4p – 3 = 13\left( {p = 1} \right)$

Putting $p = 1$ in L.H.S.,

$4\left( 1 \right) – 3 = 4 – 3 = 1$

$\because$ L.H.S. $\ne$ R.H.S.,

$\therefore$ $p = 1$ is not the solution of given equation.

(e) $4p – 3 = 13\left( {p = – 4} \right)$

Putting $p = – 4$ in L.H.S.,

$4\left( { – 4} \right) – 3 = – 16 – 3 = – 19$

$\because$ L.H.S. $\ne$ R.H.S.,

$\therefore$ $p = – 4$ is not the solution of given equation.

(f) $4p – 3 = 13\left( {p = 0} \right)$

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

$4\left( 0 \right) – 3 = 0 – 3 = – 3$

$\because$ L.H.S. $\ne$ R.H.S.,

$\therefore$ $p = 0$ is not the solution of given equation.

NCERT Solutions for Class 7 Maths Exercise 4.1

Question 3.Solve the following equations by trial and error method:

(i) $5p + 2 = 17$

(ii) $3m – 14 = 4$

Answer:

(i) $5p + 2 = 17$

Putting $p = – 3$ in L.H.S. $5\left( { – 3} \right) + 2$ = $– 15 + 2 = – 13$

$\because$$– 13 \ne 17$ Therefore, $p = – 3$ is not the solution.

Putting $p = – 2$ in L.H.S. $5\left( { – 2} \right) + 2 =$$– 10 + 2 = – 8$

$\because$$– 8 \ne 17$ Therefore, $p = – 2$ is not the solution.

Putting $p = – 1$ in L.H.S. $5\left( { – 1} \right) + 2 =$$– 5 + 2 = – 3$

$\because$$– 3 \ne 17$ Therefore, $p = – 1$ is not the solution.

Putting $p = 0$ in L.H.S. $5\left( 0 \right) + 2 =$$0 + 2 = 2$

$\because$$2 \ne 17$ Therefore, $p = 0$ is not the solution.

Putting $p = 1$ in L.H.S. $5\left( 1 \right) + 2 =$$5 + 2 = 7$

$\because$$7 \ne 17$ Therefore, $p = 1$ is not the solution.

Putting $p = 2$ in L.H.S. $5\left( 2 \right) + 2 =$$10 + 2 = 12$

$\because$$12 \ne 17$ Therefore, $p = 2$ is not the solution.

Putting $p = 3$ in L.H.S. $5\left( 3 \right) + 2 =$$15 + 2 = 17$

$\because$$17 = 17$ Therefore, $p = 3$ is the solution.

(ii) $3m – 14 = 4$

Putting $m = – 2$ in L.H.S. $3\left( { – 2} \right) – 14 = – 6 – 14 = – 20$

$\because$$– 20 \ne 4$ Therefore, $m = – 2$ is not the solution.

Putting $m = – 1$ in L.H.S. $3\left( { – 1} \right) – 14 = – 3 – 14 = – 17$

$\because$$– 17 \ne 4$ Therefore, $m = – 1$ is not the solution.

Putting $m = 0$ in L.H.S. $3\left( 0 \right) – 14 = 0 – 14 = – 14$

$\because$$– 14 \ne 4$ Therefore, $m = 0$ is not the solution.

Putting $m = 1$ in L.H.S. $3\left( 1 \right) – 14 = 3 – 14 = – 11$

$\because$$– 11 \ne 4$ Therefore, $m = 1$ is not the solution.

Putting $m = 2$ in L.H.S. $3\left( 2 \right) – 14 = 6 – 14 = – 8$

$\because$$– 8 \ne 4$ Therefore, $m = 2$ is not the solution.

Putting $m = 3$ in L.H.S. $3\left( 3 \right) – 14 = 9 – 14 = – 5$

$\because$$– 5 \ne 4$ Therefore, $m = 3$ is not the solution.

Putting $m = 4$ in L.H.S. $3\left( 4 \right) – 14 = 12 – 14 = – 2$

$\because$$– 2 \ne 4$ Therefore, $m = 4$ is not the solution.

Putting $m = 5$ in L.H.S. $3\left( 5 \right) – 14 = 15 – 14 = 1$

$\because$$1 \ne 4$ Therefore, $m = 5$ is not the solution.

Putting $m = 6$ in L.H.S. $3\left( 6 \right) – 14 = 18 – 14 = 4$

$\because$$4 = 4$ Therefore, $m = 6$ is the solution.

NCERT Solutions for Class 7 Maths Exercise 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) $10a = 70$

(iv) $\frac{b}{5} = 6$

(v) $\frac{3}{4}t = 15$

(vi) $7m + 7 = 77$

(vii) $\frac{x}{4} – 4 = 4$

(viii) $6y – 6 = 60$

(ix) $\frac{z}{3} + 3 = 30$

NCERT Solutions for Class 7 Maths Exercise 4.1

Question 5. Write the following equations in statement form:

(i) $p + 4 = 15$

(ii) $m – 7 = 3$

(iii) $2m = 7$

(iv) $\frac{m}{5} = 3$

(v) $\frac{{3m}}{5} = 6$

(vi) $3p + 4 = 25$

(vii) $4p – 2 = 18$

(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.

NCERT Solutions for Class 7 Maths Exercise 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.

$\therefore$ $5m + 7 = 37$

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

$\therefore$ $3y + 4 = 49$

(iii) Let the lowest score be $l.$

$\therefore$ $2l + 7 = 87$

(iv) Let the base angle of the isosceles triangle be $b,$ so vertex angle = $2b.$

$\therefore$ $2b + b + b = 180^\circ$$\Rightarrow$ $4b = 180^\circ$ [Angle sum property of a $\Delta$]