Simple Equations - Solutions

 CSBE Class –VII Mathematics

NCERT Solutions
Simple Equations (Ex. 4.1)


Question 1. Complete the last column of the table:

S. No.

Equation

Value

Say, whether the Equation is satisfied. (Yes / No)

1

x+3=0

x=3

 

2

x+3=0

x=0

 

3

x+3=0

x=3

 

4

x7=1

x=7

 

5

x7=1

x=8

 

6

5x=25

x=0

 

7

5x=25

x=5

 

8

5x=25

x=5

 

9

m3=2

m=6

 

10

m3=2

m=0

 

11

m3=2

m=6

 

Answer:

S. No.

Equation

Value

Say, whether the Equation is satisfied. (Yes / No)

1

x+3=0

x=3

No

2

x+3=0

x=0

No

3

x+3=0

x=3

Yes

4

x7=1

x=7

No

5

x7=1

x=8

Yes

6

5x=25

x=0

No

7

5x=25

x=5

Yes

8

5x=25

x=5

No

9

m3=2

m=6

No

10

m3=2

m=0

No

11

m3=2

m=6

Yes


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) 7n+5=19(n=2)
(c) 7n+5=19(n=2)
(d) 4p3=13(p=1)
(e) 4p3=13(p=4)
(f) 4p3=13(p=0)

Answer: (a) n+5=19(n=1)

Putting n=1 in L.H.S.,

1 + 5 = 6

 L.H.S.  R.H.S.,

 n=1 is not the solution of given equation.

(b) 7n+5=19(n=2)

Putting n=2 in L.H.S.,

7(2)+5=14+5=9

 L.H.S.  R.H.S.,

 n=2 is not the solution of given equation.

(c) 7n+5=19(n=2)

Putting n=2 in L.H.S.,

7(2)+5=14+5=19

 L.H.S. = R.H.S.,

 n=2 is the solution of given equation.

(d) 4p3=13(p=1)

Putting p=1 in L.H.S.,

4(1)3=43=1

 L.H.S.  R.H.S.,

 p=1 is not the solution of given equation.

(e) 4p3=13(p=4)

Putting p=4 in L.H.S.,

4(4)3=163=19

 L.H.S.  R.H.S.,

 p=4 is not the solution of given equation.

(f) 4p3=13(p=0)

Putting p=0 in L.H.S.,

4(0)3=03=3

 L.H.S.  R.H.S.,

 p=0 is not the solution of given equation.


Question3.Solve the following equations by trial and error method:
(i) 5p+2=17
(ii) 3m14=4

Answer: (i) 5p+2=17

Putting p=3 in L.H.S. 5(3)+2 = 15+2=13

1317 Therefore, p=3 is not the solution.

Putting p=2 in L.H.S. 5(2)+2=10+2=8

817 Therefore, p=2 is not the solution.

Putting p=1 in L.H.S. 5(1)+2=5+2=3

317 Therefore, p=1 is not the solution.

Putting p=0 in L.H.S. 5(0)+2=0+2=2

217 Therefore, p=0 is not the solution.

Putting p=1 in L.H.S. 5(1)+2=5+2=7

717 Therefore, p=1 is not the solution.

Putting p=2 in L.H.S. 5(2)+2=10+2=12

1217 Therefore, p=2 is not the solution.

Putting p=3 in L.H.S. 5(3)+2=15+2=17

17=17 Therefore, p=3 is the solution.

(ii) 3m14=4

Putting m=2 in L.H.S. 3(2)14=614=20

204 Therefore, m=2 is not the solution.

Putting m=1 in L.H.S. 3(1)14=314=17

174 Therefore, m=1 is not the solution.

Putting m=0 in L.H.S. 3(0)14=014=14

144 Therefore, m=0 is not the solution.

Putting m=1 in L.H.S. 3(1)14=314=11

114 Therefore, m=1 is not the solution.

Putting m=2 in L.H.S. 3(2)14=614=8

84 Therefore, m=2 is not the solution.

Putting m=3 in L.H.S. 3(3)14=914=5

54 Therefore, m=3 is not the solution.

Putting m=4 in L.H.S. 3(4)14=1214=2

24 Therefore, m=4 is not the solution.

Putting m=5 in L.H.S. 3(5)14=1514=1

14 Therefore, m=5 is not the solution.

Putting m=6 in L.H.S. 3(6)14=1814=4

4=4 Therefore, m=6 is the solution.


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) y2=8

(iii) 10a=70

(iv) b5=6

(v) 34t=15

(vi) 7m+7=77

(vii) x44=4

(viii) 6y6=60

(ix) z3+3=30


Question 5. Write the following equations in statement form:

(i) p+4=15
(ii) m7=3
(iii) 2m=7
(iv) m5=3
(v) 3m5=6
(vi) 3p+4=25
(vii) 4p2=18
(viii) p2+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.


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.)

Answer: (i) Let m be the number of Parmit’s marbles.

 5m+7=37

(ii) Let the age of Laxmi be y years.

 3y+4=49

(iii) Let the lowest score be l.

 2l+7=87

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

 2b+b+b=180 4b=180 [Angle sum property of a Δ]