NCERT Solution class 7 Mathematics EXERCISE- 4.1
CHAPTER – 4
EXERCISE – 4.1
Q1. Complete the last column of the table.
| S. No | Equation | Value | Say, whether the equation is satisfied. ( Yes/No) |
| (i) | X + 3 = 0 | X = 3 | No |
| (ii) | X +3 = 0 | X = 0 | No |
| (iii) | X + 3 = 0 | X = -3 | Yes |
| (iv) | X – 7 = 1 | X = 7 | No |
| (v) | X – 7 = 1 | X = 8 | Yes |
| (vi) | 5x = 25 | X = 0 | No |
| (vii) | 5x = 25 | X = 5 | Yes |
| (viii) | 5x = 25 | X = -5 | No |
| (ix) | m/3 = 2 | m = -6 | No |
| (x) | m/3 = 2 | m = 0 | No |
| (xi) | m/3 = 2 | m = 6 | Yes |
Q2. Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)
SOLUTION:-
= 1 + 5 = 19
= 6 = 19
RHS ≠LHS
(b) 7n + 5 = 19 ( n = -2)
SOLUTION:-
= 7(-2) + 5 = 19
= (-14) + 5 = 19
= (-9) = 19
= LHS ≠ RHS
(C) 7n + 5 = 19 ( n = 2)
SOLUTION:-
= 7(2) + 5 = 19
= 14 + 5 = 19
= 19=19
= LHS=RHS
(D) 4p – 3 = 13 ( p = 1)
SOLUTION:-
= 4(1) – 3 = 13
= 4 – 3 = 13
= 1 = 13
RHS ≠ LHS
(E) 4p – 3 = 13 ( p = -4)
SOLUTION:-
= 4(-4) – 3 = 13
= (-16) – 3 = 13
= (-19) = 13
RHS ≠LHS
(F) 4p -3 = 13 (p=0)
SOLUTION:-
= 4(0) – 3 = 13
= (-3) = 13
Q3. Solve the following equations by trial and error method:
(a) 5p + 2 = 17
SOLUTION:-
p = 0
5(0) + 2 = 17
0 + 2 = 17
2 ≠ 17 LHS ≠ RHS
p = 1
5(1) + 2 = 17
5 + 2 = 17
7 ≠ 17 LHS ≠ RHS
p = 2
5(2) + 2 = 17
10 + 2 = 17
12 ≠ 17 LHS ≠ RHS
p = 3
5(3) + 2 = 17
15 + 2 = 17
17 = 17 LHS = RHS
(B) 3m – 14 = 4
SOLUTION:-
m = 1
3(1) – 14 = 4
3 -14 = 4
-11 ≠ 4 LHS ≠ RHS
m = 6
3(6) – 14 = 4
18 – 14 = 4
4 = 4 LHS = RHS
Q4. Write equations for the following statements:
(i) The sum of numbers x and 4 is 9
SOLUTION:-
X + 4 = 9
(ii) 2 subtracted from y is 8.
SOLUTION:-
Y – 2 = 8
(iii) Ten times a is 70.
SOLUTION:-
10A = 70
(iv) The numbers b divided by 5 gives 6 .
SOLUTION:-
B/5 = 6
(v) Three fourth of t is 15.
SOLUTION:-
3T /4 = 15
(vi) Seven times m plus 7 gets you 77.
SOLUTION:-
7m + 7 = 77
(vii) One – fourth of a numbers x minus 4 gives 4.
SOLUTION:-
X/4 – 4 = 4
(viii) If you take away 6 from 6 times y , you get 60.
SOLUTION:-
6Y – 6 = 60
(ix) If you add 3 to one – third of z, you get 30.
SOLUTION:-
Z/3 + 3 = 30
Q5. Write the following equations in statements forms:
(i) p + 4 = 15
SOLUTION:-
The sum of numbers p and 4 is 15.
(ii) m – 7 = 3
SOLUTION:-
7 SUBTRACTED FROM M IS 3.
(iii) 2m = 7
SOLUTION:-
Twice of number m is 7.
(iv) m/5 = 3
SOLUTION:-
The numbers m divided by 5 gives 3.
(v) (3m)/5 = 6
Three – fifth of m is 6.
(vi) 3p + 4 = 25
SOLUTION:-
Three times p plus 4 gives you 25.
(vii) 4p – 2 = 18
SOLUTION:-
Four times p minus 2 gives you 18.
(viii) p/2 + 2 = 8
SOLUTION:-
If you add half of a number p to 2, you get 8.
Q6. Set up an equation in the following cases:
(i) Ifran says that he has 7 marbles more than five times the marbels. Parmit has. Ifran has 37 marbles. (Take m to be the numbers of Parmit marbles.)
SOLUTION:-
From the question it is given that,
Number of Parmit marbles = m
Then,
Ifran has 7 marbles more than five times the marbles Parmit has
= ( 5 x m ) + 7 = 37
= 5m + 7 = 37
(ii) Laxmi father is 49 years old. He is 4 years older than three times Laxmi age. (Take Laxmi age to be y years.)
SOLUTION:-
From the question it is given that,
Let Laxmi age to be = y years old
Then,
Laxmi father is 4 years older than the three times of her age
= 3 x Laxmi father age + 4 = Age of Laxmi father
= (3 x y )+4 = 49
= 3y + 4 = 49
(iii) The teacher tells the class that the highest marks obtained by a students in her class is twice the lowest marks plus 7. The highest score is 87.(Take the lowest score to be l.)
SOLUTION:-
From the question it is given that,
Highest score in the class = 87
Let lowest score be l
= 2 x Lowest score + 7 = Highest score in the class
= (2 x 1 ) + 7 = 87
= 2 + 7 = 87
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let us the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).
SOLUTION:-
From the question it is given that,
We know that, the sum of angles of a triangle is 1800
Let base angle be b
Then,
Vertex angle = 2 x base angle = 2b
= b + b + 2b = 1800
= 4b = 1800