EXERCISE 4.2
Q 1, Ex 4.2 - Simple Equations - Chapter 4 - Maths Class 7th - NCERT
Give first the step you will use to separate the variable and then solve the equation:
(a) x+ 1 = 0
Sol :
Adding 1 to both sides of the given equation, we obtain
x - 1 + 1 = 0 + 1
x = 1
(b) x + 1 = 0
Sol :
Subtracting 1 from both sides of the given equation , we obtain
x + 1 - 1 = 0 - 1
x = - 1
(c) x - 1 = 5
Sol :
Adding 1 to both sides of the given equation, we obtain
x - 1 + 1 = 5 + 1
x = 6
(d) x + 6 = 2
Sol :
Subtracting 6 from both sides of the given equation, we obtain
x + 6 - 6 = 2 - 6
x = - 4
(e) y - 4 = -7
Sol :
adding 4 to both sides of the given equation , we obtain
y - 4 + 4= -7 + 4
y = -3
(f) y - 4 = 4
Sol :
Adding 4 to both sides of the given equation, we obtain
y - 4 + 4 = 4 + 4
y = 8
(g) y + 4 = 4
Sol :
Subtracting 4 from both sides of the given equation , we obtain
y + 4 - 4 = 4 - 4
y = 0
(h) y + 4 = -4
Sol :
Subtracting 4 from both sides of the given equation , we obtain
y + 4 - 4 = 4 - 4
y = -8
Q 2, Ex 4.2 - Simple Equations - Chapter 4 - Maths Class 7th - NCERT
Give the first step you will use to separate the variable and then solve the equation:
(a) 3l = 42
Sol :
Dividing both sides of the given equation by 3, we obtain
$\dfrac{3l}{3}=\dfrac{42}{3}$
l = 14
(b) $\dfrac{b}{2}=6$
Sol :
Multiplying both sides of the given equation by 2, we obtain
$\dfrac{b\times 2}{2}=6\times 2$
b = 12
(c) $\dfrac{p}{7}=4$
Sol :
Multiplying both sides of the given equation by 7, we obtain
$\dfrac{p\times 7}{7}=4\times 7$
p = 28
(d) 4x = 25
Sol :
Dividing both sides of the given equation by 4, we obtain
$\dfrac{4x}{4}=\dfrac{25}{4}$
$x=\dfrac{25}{4}$
(e) 8y = 36
Sol :
Dividing both sides of the given equation by 8, we obtain
$\dfrac{8y}{8}=\dfrac{36}{8}$
$y=\dfrac{9}{2}$
(f) $\dfrac{z}{3}=\dfrac{5}{4}$
Sol :
Multiplying both sides of the given equation by 3, we obtain
$\dfrac{z\times 3}{3}=\dfrac{5\times 3}{4}$
$z=\dfrac{15}{4}$
(g) $\dfrac{a}{5}=\dfrac{7}{15}$
Sol :
Multiplying both sides of the given equation by 5, we obtain
$\dfrac{a\times 5}{5}=\dfrac{7\times 5}{15}$
$a=\dfrac{7}{3}$
(h) 20t = -10
Sol :
Dividing both sides of the given equation by 20, we obtain
$\dfrac{20t}{20}=\dfrac{-10}{\phantom{-}20}$
$t=\dfrac{-1}{\phantom{-}2}$
Q 3, Ex 4.2 - Simple Equations - Chapter 4 - Maths Class 7th - NCERT
QUESTION 3
Give the steps you will use to separate the variable and then solve the equation:
(a) 3n - 2 = 26
Sol :
Adding 2 to both sides of the given equation, we obtain
3n - 2 + 2 = 46 + 2
3n = 48
Dividing both sides of the given equation by 3, we obtain
$\dfrac{3n}{3}=\dfrac{48}{3}$
n = 16
(b) 5m + 7 = 7
Sol :
Subtracting 7 from both sides of the given equation, we obtain
5m + 7 - 7 = 17 - 7
5m = 10
Dividing both sides of the given equation by 5, we obtain
$\dfrac{5}{m}=\dfrac{10}{5}$
m = 2
(c) $\dfrac{20p}{3}=40$
Sol :
Multiplying both sides of the given equation by 3, we obtain
$\dfrac{20p\times 3}{3}=40\times 3$
20p = 120
Dividing both sides of the given equation by 20, we obtain
$\dfrac{20p}{20}=\dfrac{120}{20}$
p = 6
(d) $\dfrac{3p}{10}=6$
Sol :
Multiplying both sides of the given equation by 10, we obtain
$\dfrac{3p\times 10}{10}=6\times 10$
3p = 60
Dividing both sides of the given equation by 3, we obtain
$\dfrac{3p}{3}=\dfrac{60}{3}$
p = 20
Q 4, Ex 4.2 - Simple Equations - Chapter 4 - Maths Class 7th - NCERT
QUESTION 4
Solve the following equations:
(a) 10p = 100
Sol :
$\dfrac{10p}{10}=\dfrac{100}{10}$
p = 10
(b) 10p + 10
Sol :
10p + 10 - 10 = 100 - 10
10 p = 90
$\dfrac{10p}{10}=\dfrac{90}{10}$
p = 9
(c) $\dfrac{p}{4}=5$
Sol :
$\dfrac{p\times 4}{4}=5\times 4$
p = 20
(d) $\dfrac{-p}{\phantom{-}3}=5$
Sol :
$\dfrac{-p\times (-3)}{\phantom{-}3}=5\times (-3)$
p = -15
(e) $\dfrac{3p}{4}=6$
Sol :
$\dfrac{3p\times 4}{4}=6\times 4$
3p = 24
p = 8
(f) 3s = -9
Sol :
$\dfrac{3s}{3}=\dfrac{-9}{\phantom{-}3}$
x = -3
(g) 3s + 12 = 0
Sol :
3s + 12 -12 = 0 -12
3s = -12
$\dfrac{3s}{3}=\dfrac{-12}{\phantom{-}3}$
s = -4
(h) 3s = 0
Sol :
$\dfrac{3s}{3}=\dfrac{0}{3}$
s = 0
(i) 2q = 6
Sol :
$\dfrac{2q}{2}=\dfrac{6}{2}$
q = 3
(j) 2q - 6 = 0
Sol :
2q - 6 + 6 = 0 + 6
2q = 6
$\dfrac{2q}{2}=\dfrac{6}{2}$
q = 3
(k) 2q + 6 = 0
Sol :
2q + 6 - 6 = 0
2q = -6
$\dfrac{2q}{2}=\dfrac{-6}{\phantom{-}2}$
q = -3
(l) 2q + 6 = 12
Sol :
2q + 6 - 6 = 12 - 6
2q = 6
$\dfrac{2q}{2}=\dfrac{6}{2}$
q = 3
0 comments:
Post a Comment