Comprehensive Notes of Exercise 3.2
These notes of exercise 3.2 of class 10 math are about solving problems related to variation. Variation explains how two or more things are connected, like when one increases, the other may also increase (direct variation) or decrease (inverse variation). In this exercise, you will learn how to form equations using these relationships and solve them step by step. The questions cover different types of variation, such as direct, inverse, and joint variation. This post provides complete notes of Exercise 3.2 with easy-to-understand solutions to help you solve these problems without confusion.
Question 1: If y varies directly as x, and y=8 when x=2, find:
(i) y in terms of x
(ii) y when x = 5
(iii) x when y = 28
Solution:
Since y ∝ x, we can write y = kx …(1)
(i) To find y in terms of x:
Using y = 8 and x = 2, substitute into equation (1):
8 = k(2)
k = 8 ÷ 2 = 4
So, the equation becomes:
y = 4x …(2)
(ii) To find y when x = 5:
Substitute x = 5 into equation (2):
y = 4x
y = 4(5) = 20
(iii) To find x when y = 28:
Substitute y = 28 into equation (2):
28 = 4x
x = 28 ÷ 4 = 7
Final Results:
(i) y = 4x
(ii) y = 20
(iii) x = 7
Question 2: If y ∝ x, and y = 7 when x = 3, find:
(i) y in terms of x
(ii) x when y = 35 and y when x = 18
Solution:
Since y ∝ x, we can write y = kx …(1)
(i) To find y in terms of x:
Using y = 7 and x = 3, substitute into equation (1):
7 = k(3)
k = 7 ÷ 3 = 7/3
So, the equation becomes:
y = (7/3)x …(2)
(ii) To find x when y = 35:
Substitute y = 35 into equation (2):
35 = (7/3)x
x = 35 × (3/7)
x = 15
To find y when x = 18:
Substitute x = 18 into equation (2):
y = (7/3)x
y = (7/3)(18)
y = 42
Final Results:
(i) y = (7/3)x
(ii) x = 15, y = 42
Question 3: If R ∝ T, and R = 5 when T = 8, find:
(i) The equation connecting R and T
(ii) R when T = 64
(iii) T when R = 20
Solution:
Since R ∝ T, we can write R = kT …(1)
(i) To find the equation connecting R and T:
Using R = 5 and T = 8, substitute into equation (1):
5 = k(8)
k = 5 ÷ 8 = 5/8
So, the equation becomes:
R = (5/8)T …(2)
(ii) To find R when T = 64:
Substitute T = 64 into equation (2):
R = (5/8)T
R = (5/8)(64)
R = 40
(iii) To find T when R = 20:
Substitute R = 20 into equation (2):
20 = (5/8)T
T = 20 × (8/5)
T = 32
Final Results:
(i) R = (5/8)T
(ii) R = 40
(iii) T = 32
Question 4: If R ∝ T², and R = 8 when T = 3, find R when T = 6.
Solution:
Since R ∝ T², we can write:
R = kT² …(1)
To find k:
Using R = 8 and T = 3, substitute into equation (1):
8 = k(3)²
8 = 9k
k = 8 ÷ 9 = 8/9
The equation becomes:
R = (8/9)T² …(2)
To find R when T = 6:
Substitute T = 6 into equation (2):
R = (8/9)T²
R = (8/9)(6)²
R = (8/9)(36)
R = 32
Final Result:
R = 32
Question 5: V ∝ R³ and V = 5 when R = 3,
find R, when V = 625.
Solution:
V ∝ R³
V = kR³ ………………(i)
To find k,
Put V = 5 and R = 3 in equation (i)
5 = k(3³)
5 = k(27)
5/27 = k ⇒ k = 5/27
Put k = 5/27 in equation (i)
V = kR³
V = 5/27 R³ ………………………(ii)
To find R,
Now put V = 625 in equation (ii)
(625) = 5/27 R³
125 = 5/27 × 625
27 × 125 = R³
R³ = 3³ × 5³
R³ = (3 × 5)³
R³ = (15)³
Taking cube root of both sides
∛R³ = ∛15³
R = 15
Question 6: if w varies directly as u³ and w=81 when u=3 find w when u=5.
If w varies directly as u³, we can express the relationship as:
w = k × u³
where k is the constant of proportionality.
Step 1: Find k
We are given w = 81 when u = 3. Substitute these values into the equation:
81 = k × 3³
81 = k × 27
k = 81 ÷ 27 = 3
Step 2: Find w when u = 5
Now substitute k = 3 and u = 5 into the equation:
w = 3 × 5³
w = 3 × 125
w = 375
Final Answer:
When u = 5, w = 375.
Question 7: if y varies inversely as x and y=7 when x=2 find y when x=126.
If y varies inversely as x, the relationship is expressed as:
y × x = k
where k is the constant of proportionality.
Step 1: Find k
We are given y = 7 when x = 2. Substitute these values into the equation:
7 × 2 = k
k = 14
Step 2: Find y when x = 126
Using the formula y × x = k, substitute k = 14 and x = 126:
y × 126 = 14
y = 14 ÷ 126
y = 1 ÷ 9
Final Answer:
When x = 126, y = 1/9.
Question 8: If y ∝ 1/x and y = 4 when x = 3, find x when y = 24.
If y ∝ 1/x, the relationship is expressed as:
y = k / x
where k is the constant of proportionality.
Step 1: Find k
We are given y = 4 when x = 3. Substitute these values into the equation:
4 = k / 3
k = 4 × 3 = 12
Step 2: Find x when y = 24
Using the formula y = k / x, substitute k = 12 and y = 24:
24 = 12 / x
Multiply both sides by x:
24x = 12
Divide by 24:
x = 12 ÷ 24 = 1/2
Final Answer:
When y = 24, x = 1/2.
Question 9: If w ∝ 1/z and w = 5 when z = 7, find w when z = 175/4.
Step 1: Express the proportionality as w = k / z, where k is the constant of proportionality.
Given w = 5 when z = 7:
5 = k / 7
k = 5 × 7 = 35
Step 2: Find w when z = 175/4.
Substitute k = 35 and z = 175/4 into w = k / z:
w = 35 / (175/4)
w = 35 × (4/175)
w = (35 × 4) / 175
w = 140 / 175
w = 4/5
Final Answer: w = 4/5.
Question 10: A ∝ 1/r² and A = 2 when r = 3, find r when A = 72.
Step 1: Express the proportionality as A = k / r².
Given A = 2 when r = 3:
2 = k / 3²
2 = k / 9
k = 2 × 9 = 18
Step 2: Find r when A = 72.
Substitute k = 18 and A = 72 into A = k / r²:
72 = 18 / r²
r² = 18 / 72
r² = 1 / 4
r = ±√(1 / 4)
r = ±1 / 2
Final Answer: r = ±1/2.
Question 11: a ∝ 1/b² and a = 3 when b = 4, find a when b = 8.
Step 1: Express the proportionality as a = k / b².
Given a = 3 when b = 4:
3 = k / 4²
3 = k / 16
k = 3 × 16 = 48
Step 2: Find a when b = 8.
Substitute k = 48 and b = 8 into a = k / b²:
a = 48 / 8²
a = 48 / 64
a = 3 / 4
Final Answer: a = 3/4.
Question 12: V ∝ 1/r³ and V = 5 when r = 3, find V when r = 6 and r when V = 320.
Step 1: Express the proportionality as V = k / r³.
Given V = 5 when r = 3:
5 = k / 3³
5 = k / 27
k = 5 × 27 = 135
Step 2: Find V when r = 6.
Substitute k = 135 and r = 6 into V = k / r³:
V = 135 / 6³
V = 135 / 216
V = 5 / 8
Step 3: Find r when V = 320.
Substitute k = 135 and V = 320 into V = k / r³:
320 = 135 / r³
r³ = 135 / 320
r³ = 27/64
r = ∛(27 / 64)
r= 3/4
Final Answer: V = 5/8 when r = 6, and r = 3/4 when V = 320.
Question 13:
Step 1: Express the proportionality as m = k / n³.
Given m = 2 when n = 4:
2 = k / 4³
2 = k / 64
k = 2 × 64 = 128
Step 2: Find m when n = 6.
Substitute k = 128 and n = 6 into m = k / n³:
m = 128 / 6³
m = 128 / 216
m = 16 / 27
Step 3: Find n when m = 432.
Substitute k = 128 and m = 432 into m = k / n³:
432 = 128 / n³
n³ = 128 / 432
n³ = 8 / 27
n = ∛(8 / 27)
n = 2 / 3
Final Answer:
- m = 16 / 27 when n = 6
- n = 2 / 3 when m = 432
These notes of Exercise 3.2 explain each question with simple steps and clear solutions. By practicing these problems, you will better understand the concepts of variation and improve your problem-solving skills. If you want to review the previous exercise, check out the notes of Exercise 3.1. To continue learning, move on to the notes of Exercise 3.3. For more helpful resources, visit our homepage. Keep practicing and stay confident in your math journey! Explore interactive lessons on direct and inverse variation at Brilliant.org to strengthen your math fundamentals.