Frank Solutions for Class 9 Maths Chapter 6 Changing the Subject of a Formula explain important formulas mentioned in this chapter, according to the current syllabus of the ICSE Board. These solutions are prepared by subject-matter experts in simple language to assist students in exam preparation. Students who practise these solutions on a regular basis will undoubtedly, achieve good marks in the final examination. Besides, it will enhance their problem-solving skills, which is vital from an exam point of view.
Chapter 6, Changing the Subject of a Formula, contains problems based on formulas with clear explanations. Students can cross-check their answers while revising textbook questions using these solutions. Moreover, it will help them to obtain an idea of important questions that would appear in the annual examination. For more conceptual knowledge, students can download Frank Solutions for Class 9 Maths Chapter 6 Changing the Subject of a Formula in PDF from the link provided below and practise regularly.
Frank Solutions for Class 9 Maths Chapter 6 Changing the Subject of a Formula Download PDF
Access Frank Solutions for Class 9 Maths Chapter 6 Changing the Subject of a Formula
1. The simple interest on a sum of money is the product of the sum of money, the number of years and the rate percentage. Write the formula to find the simple interest on Rs A for T years at R% per annum.
Solution:
Let the simple interest = I
Now,
Simple interest on sum of money = product of sum of money, number of years and rate percentage = (A × I × R) / 100
As per the data: I = (A × I × R) / 100
Therefore,
The required formula is,
I = (A × I × R) / 100
2. The volume, V, of a cone is equal to one-third of π times the cube of the radius. Find a formula for it.
Solution:
Let radius = r
Hence,
Cube of radius = r3
One-third of π times the cube of the radius = (1 / 3) πr3
As per the data: V = (1 / 3) πr3
Therefore,
The required formula is,
V = (1 / 3) πr3
3. The Fahrenheit temperature, F, is 32, more than nine-fifths of the centigrade temperature C. Express this relation by a formula.
Solution:
Centigrade temperature = C
Nine–fifths of the centigrade temperature = (9 / 5) C
32 more than nine–fifths of the centigrade temperature C = (9 / 5) C + 32
As per the data: F = (9 / 5) C + 32
Therefore,
The required formula is,
F = (9 / 5) C + 32
4. The arithmetic mean M of the five numbers a, b, c, d, e is equal to their sum divided by the number of quantities. Express it as a formula.
Solution:
Sum of a, b, c, d, e = (a + b + c + d + e)
Number of quantities = 5
Sum divided by the number of quantities = (a + b + c + d + e) / 5
As per the data: M = (a + b + c + d + e) / 5
Therefore,
The required formula is,
M = (a + b + c + d + e) / 5
5. Make a formula for the statement: “The reciprocal of focal length f is equal to the sum of reciprocals of the object distance u and the image distance v”.
Solution:
Object distance = u
Image distance = v
Reciprocal of Object distance = (1 / u)
Reciprocal of Image distance = (1 / v)
Sum of reciprocals = (1 / u) + (1 / v)
Reciprocal of focal length = (1 / f)
As per the data: (1 / f) = (1 / u) + (1 / v)
Therefore,
The required formula for the given statement is,
(1 / f) = (1 / u) + (1 / v)
6. Make R the subject of formula A = P {1 + (R / 100)}N
Solution:
A = P {1 + (R / 100)}N
(A / P) = {1 + (R / 100)}N
Taking Nth root both sides,
We get,
(A / P)1 / N = {1 + (R / 100)}
(A / P)1 / N – 1 = (R / 100)
On calculating further, we get,
100 {(A / P)1 / N – 1} = R
Hence,
7. Make L the subject of formula T = 2π √ (L / G)
Solution:
Given
T = 2π √ (L / G)
(T / 2π) = √ (L / G)
Squaring on both sides,
We get,
(T / 2π)2 = (L / G)
G (T / 2π)2 = L
We get,
L = (GT2 / 4π2)
8. Make a the subject of formula S = ut + (1 / 2) at2
Solution:
Given
S = ut + (1 / 2) at2
On further calculation, we get,
S – ut = (1 / 2) at2
2 (S – ut) = at2
{2 (S – ut)} / t2 = a
Therefore,
a = {2 (S – ut)} / t2
9. Make x the subject of formula (x2 / a2) + (y2 / b2) = 1
Solution:
Given
(x2 / a2) + (y2 / b2) = 1
On calculating further, we get,
(x2 / a2) = 1 – (y2 / b2)
x2 = a2 {1 – (y2 / b2)}
On taking L.C.M., we get,
x2 = a2 {(b2 – y2) / b2}
Now,
Taking square root on both sides, we get,
x = {√a2 (b2 – y2) / b2}
Hence,
x = (a / b) {√ (b2 – y2)}
10. Make a the subject of formula S = {a (rn – 1)} / (r – 1)
Solution:
Given
S = {a (rn – 1)} / (r – 1)
On further calculation, we get,
S (r – 1) = a (rn – 1)
{S (r – 1)} / (rn – 1) = a
Therefore,
a = {S (r – 1)} / (rn – 1)
11. Make h the subject of the formula R = (h / 2) (a – b). Find h when R = 108, a = 16 and b = 12.
Solution:
Given
R = (h / 2) (a – b)
On calculating further, we get,
2R = h (a – b)
h = 2R / (a – b)
Now,
Substituting R = 108, a = 16 and b = 12,
We get,
h = (2 × 108) / (16 – 12)
h = (2 × 108) / 4
We get,
h = 54
12. Make s the subject of the formula v2 = u2 + 2as. Find s when u = 3, a = 2 and v = 5.
Solution:
Given
v2 = u2 + 2as
v2 – u2 = 2as
s = (v2 – u2) / 2a
Now,
Substituting u = 3, a = 2 and v = 5,
We get,
s = (52 – 32) / (2 × 2)
s = (25 – 9) / 4
s = 16 / 4
We get,
s = 4
13. Make y the subject of the formula x = (1 – y2) / (1 + y2). Find y if x = (3 / 5)
Solution:
Given
x = (1 – y2) / (1 + y2)
On further calculation, we get,
x (1 + y2) = 1 – y2
x + xy2 = 1 – y2
xy2 + y2 = 1 – x
Taking y2 as common, we get,
y2 (x + 1) = 1 – x
y2 = (1 – x) / (1 + x)
y = √ (1 – x) / (1 + x)
Now,
Substituting x = (3 / 5), we get,
y = [√ {1 – (3 / 5)} / {1 + (3 / 5)}]
y = √ (2 / 8)
y = √ (1 / 4)
We get,
y = (1 / 2)
14. Make a the subject of the formula S = (n / 2) {2a + (n – 1) d}. Find a when S = 50, n = 10 and d = 2.
Solution:
Given
S = (n / 2) {2a + (n – 1) d}
On further calculation, we get,
2S = n {2a + (n – 1) d}
(2S / n) = 2a + (n – 1) d
(2S / n) – (n – 1) d = 2a
We get,
a = (S / n) – {(n – 1) d / 2}
Now,
Substituting S = 50, n = 10 and d = 2,
We get,
a = (S / n) – {(n – 1) d / 2}
a = (50 / 10) – {(9 × 2) / 2}
a = 5 – 9
We get,
a = – 4
15. Make x the subject of the formula a = 1 – {(2b) / (cx – b)}. Find x, when a = 5, b =12 and c = 2
Solution:
Given
a = 1 – {(2b) / (cx – b)}
On further calculation, we get,
a – 1 = – {(2b) / (cx – b)}
(a – 1) (cx – b) + 2b = 0
acx – ab – cx + b + 2b = 0
acx – ab – cx + 3b = 0
acx – cx + 3b – ab = 0
x (ac – c) + b (3 – a) = 0
xc (a – 1) = – b (3 – a)
x = {b (a – 3)} / {c (a – 1)}
Now,
Substituting a = 5, b = 12 and c = 2,
We get,
x = {12 (5 – 3)} / {2 (5 – 1)}
x = (12 × 2) / (2 × 4)
We get,
x = 24 / 8
x = 3
Comments