NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations Exercise 4.1

Exercise- 4.1

1. Complete the last column of the table:

Ans:

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

a. \[n+5=19\left( n=1 \right)\]

Ans: $ n+5=19 $ 

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) $ 

Ans: $ 7n+5=19 $ 

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) $ 

Ans: $ 7n+5=19 $ 

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) $ 

Ans: $ 4p-3=13 $ 

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) $ 

Ans: $ 4p-3=13 $ 

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) $ 

Ans: $ 4p-3=13 $ 

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.

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

a. $ 5p+2=17 $ 

Ans: Putting  $ p=-3 $ in L.H.S.  $ 5(-3)+2=-15+2=-13 $     $ \because -13\ne 17 $ 

Therefore, \[p=-3\text{ }\] is not the solution. 

Putting  $ p=-2 $ in L.H.S.  $ 5(-2)+2=-10+2=-8 $    $ \because -8\ne 17\quad  $ 

Therefore,  $ p=-2 $  is not the solution. 

Putting  $ p=-1 $ in L.H.S. $ 5(-1)+2=-5+2=-3 $   $ \because -3\ne 17 $ 

Therefore,  $ p=-1 $ is not the solution.

Putting  $ p=0 $ in L.H.S.  $ 5(0)+2=0+2=2 $    $ \because 2\ne 17\quad  $ 

Therefore,  $ p=0 $ is not the solution.

Putting  $ p=1 $ in L.H.S.  $ 5(1)+2=5+2=7 $    $ \because 7\ne 17\quad  $ 

Therefore,  $ p=1 $ is not the solution. 

Putting  $ p=2 $  in L.H.S.  $ 5(2)+2=10+2=12 $   $ \because 12\ne 17\quad  $ 

Therefore,  $ p=2 $ is not the solution.

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

Therefore,  $ p=3 $ is the solution.

b. $ 3m-14=4 $ 

Ans: Putting  $ m=-2 $ in L.H.S.  $ \quad 3(-2)-14=-6-14=-20\,\,\,\because -20\ne 4\quad  $ 

Therefore,  $ m=-2 $  is not the solution. 

Putting  $ m=-1 $ in L.H.S.  $ \quad 3(-1)-14=-3-14=-17\,\,\,\,\because -17\ne 4\quad  $ 

Therefore,  $ m=-1 $  is not the solution.

Putting  $ m=0 $  in L.H.S.  $ 3(0)-14-0-14=-14 $   $ \because -14\ne 4\quad  $ 

Therefore,  $ m=0 $  is not the solution.

Putting  $ m=1 $ in L.H.S.  $ 3(1)-14=3-14=-11 $ 

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

Putting  $ m=2 $ in L.H.S.  $ \quad 3(2)-14=6-14=-8 $ 

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

Putting  $ m=3 $ in L.H.S.  $ 3(3)-14=9-14=-5 $ 

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

Putting  $ m=4 $ in L.H.S.  $ \quad 3(4)-14=12-14=-2 $ 

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

Putting  $ m=5 $ in L.H.S.  $ \quad 3(5)-14=15-14=1 $    $ \because 1\ne 4 $ 

Therefore,  $ m=5 $  is not the solution.

4. Write equations for the following statements: 

i. The sum of numbers x and 4 is 9. 

Ans: $ x+4=9 $ 

ii. 2 subtracted from y is 8. 

Ans: $ y-2=8 $ 

iii. Ten times a is 70. 

Ans: $ 10a=70 $ 

iv. The number b divided by 5 gives 6. 

Ans: $ \dfrac{b}{5}=6 $ 

v. Three-fourth of t is 15. 

Ans: $ \dfrac{3}{4}t=15 $ 

vi. Seven times m plus 7 gets you 77. 

Ans: $ 7m+7=77 $ 

vii. One-fourth of a number x minus 4 gives 4. 

Ans: $ \dfrac{x}{4}-4=4 $ 

viii. If you take away 6 from 6 times y, you get 60. 

Ans: $ 6y-6=60 $ 

ix. If you add 3 to one-third of z, you get 30.

Ans: $ \dfrac{z}{3}+3=30 $ 

5. Write the following equations in statement form:

i. $ p+4=15 $ 

Ans: The sum of numbers p and 4 is 15.

ii. $ m-7=3 $ 

Ans: 7 subtracted from m is 3

iii. $ 2m=7 $ 

Ans: Two times m is 7.

iv. $ \dfrac{m}{5}=3 $ 

Ans: The number m is divided by 5 gives 3.

v. $ \dfrac{3m}{5}=6 $ 

Ans: Three-fifth of the number m is 6.

vi. $ 3p+4=25 $ 

Ans: Three times p plus 4 gets 25.

vii. $ 4p-2=18 $ 

Ans: If you take away 2 from 4 times p, you get 18.

viii. $ \dfrac{p}{2}+2=8 $ 

Ans: If you added 2 to half is p, you get 8.

6. Set up an equation in the following cases:

i. Irfan says that he has \[\mathbf{7}\] marbles more than five times the marbles Permit has. Irfan has \[\mathbf{37}\] marbles. (Tale m to be the number of Permit’s marbles.) 

Ans: Let  $ m $ be the number of Permit’s marbles.

$ \therefore \quad 5m+7=37 $ 

ii. Laxmi’s father is \[\mathbf{49}\] years old. He is \[\mathbf{4}\] years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.) 

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

$ \therefore \quad 3y+4=49 $ 

iii. The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus\[\mathbf{7}\]. The highest score is\[\mathbf{87}\]. (Take the lowest score to be l. ) 

Ans: Let the lowest score be \[l.\]   

$ \therefore \quad 2l+7=87 $ 

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 \[\mathbf{180}{}^\circ \]

Ans: Let the base angle of the isosceles triangle be \[b,\]so vertex angle \[=2\text{ }b.\] $ \therefore 2b+b+b={{180}^{{}^\circ }}\Rightarrow 4b={{180}^{{}^\circ }}\quad  $Angle sum property of a $ \Delta$

Conclusion

Exercise 4.1 of Class 7 Maths Chapter 4 focuses on the foundational concepts of solving simple equations. This exercise is crucial as it introduces students to the process of balancing equations, an essential skill in algebra. By practising these problems, students learn how to isolate variables and perform operations on both sides of an equation to maintain equality. Class 7 Maths Ch 4 Ex 4.1 helps in building a strong understanding of equations, variables, and constants, which are fundamental to more advanced mathematical concepts. Mastery of these basics will greatly benefit students in their future studies.

Class 7 Maths Chapter 4: Exercises Breakdown

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