Frank Solutions for Chapter 6 Changing the Subject of a Formula Class 9 Mathematics ICSE
Exercise 6.1
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.
Answer
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.
Answer
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.
Answer
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.
Answer
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”.
Answer
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
Answer
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 = 2Ï€ √(L/G)
Answer
Given
T = 2Ï€ √ (L/G)
⇒ (T/2Ï€) = √(L/G)
Squaring on both sides,
We get,
(T/2Ï€)2 = (L/G)
⇒ G (T/2Ï€)2 = L
We get,
L = (GT2/4Ï€2)
8. Make a the subject of formula S = ut + (1/2) at2
Answer
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
Answer
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 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)
Answer
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.
Answer
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.
Answer
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)
Answer
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.
Answer
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
Answer
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
Exercise 6.2
1. Make R the subject of formula A = P(1 + R/100)N
Answer
2. Make L the subject of formula
Answer
3. Make a the subject of formula S = ut + 1/2at2
Answer
4. Make x the subject of formula x2/a2 + y2/b2 = 1
Answer
5. Make a the subject of formulas S = a(rn – 1)/(r – 1)
Answer
6. Make r2 the subject of formula 1/R = 1/r1 + 1/r2
Answer
7. Make a the subject of formula
Answer
8. Make y the subject of formula W = pq + ½ Wy2
Answer
9. Make N the subject of formula I = NG/(R + Ny)
Answer
10. Make V the subject of formula K = 1/2MV2
Answer
11. Make d the subject of formula S = n/2{2a + (n – 1)d}
Answer
12. Make R2 the subject of formula
Answer
13. Make A the subject of formula R = (m1B + m2A)/(m1 + m2)
Answer
14. Make c the subject of formula
Answer
15. Make k the subject of formula
Answer
16. Given: mx + ny = p and y = ax + b. Find x in terms of m, n, p, a and b.
Answer
Answer
Answer
19. If 3ax + 2b2 = 3bx + 2a2, then express x in terms of a and b. Also, express the result in the simplest form.
Answer
20. If b = 2a/(a – 2), and c = (4b – 3)/(3b + 4), then express c in terms of a.
Answer
Exercise 6.3
1. Make h the subject of the formula R = h/2(a – b). Find h when R = 108, a = 16 and b = 12.
Answer
2. Make s the subject of the formula v2 = u2 + 2as. Find s when u = 3, a = 2 and v = 5.
Answer
3. Make y the subject of the formula x = (1 – y2)/(1 + y2). Find y if x = 3/5
Answer
4. Make a the subject of the formula = S = n/2{2a + (n – 1)d}. Find a when S = 50, n = 10 and d = 2.
Answer
5. Make x the subject of the formula a = 1 – 2b/(cx – b). Find x, when a = 5, b = 12
Answer
6. Make h the subject of the formula
. Find h, when k = -2, a = -3, d = 8 and g = 32.
Answer
7. Make x the subject of the formula y = (1 – x2)/(1 + x2). Find x, when y = 1/2
Answer
Answer
9. Make m the subject of the formula x = my/(14 – mt). Find m, when x = 6, y = 10 and t = 3.
Answer
10. Make I the subject of the formula M = L + 1/F(1/2.N – C) × I. Find I, if M = 44, L = 20.
Answer
11. Make g the subject of the formula v2 = u2 – 2gh. Find g, when v = 9.8, u = 41.5 and h = 25.4
Answer
12. Make f the subject of the formula
Find f, when D = 13 and p = 21.
Answer
13. Make z the subject of the formula y = (2z + 1)/(2z – 1). If x = (y + 1)/(y – 1), express z in terms of x, and find its value when x = 34.
Answer
14. Make c the subject of the formula a = b(1 + ct). Find c, when a = 1100, b = 100 and t = 4.
Answer
15. “The volume of a cylinder V is equal to the product of Ï€ and square of radius r and the height h”. Express this statement as a formula. Make r the subject formula. Find r, when V = 44 cm3, Ï€ = 22/7, h = 14 cm.
Answer
16. “The volume of a cone V is equal to the product of one third of Ï€ and square of radius r of the base and the length h”. Express this statement as a formula. Make r the subject formula. Find r, when V = 1232 cm3.
Answer
17. The pressure P and volume V of a gas are connected by the formula PV = C, where C is a constant. If P = 4 when V = 2.1/2; find the value of P when V = 4?
Answer
18. (a) The total energy E possess by a body of Mass ‘m’, moving with a velocity ‘v’ at a height ‘h’ is given by: E = 1/2 mu2 + mgh.
Making ’m’ the subject of formula.
(b) The total energy E possess by a body of Mass ‘m’, moving with a velocity ‘v’ at a height ‘h’ is given by: E = 1/2 mu2 + mgh.
Find m, if v = 2, g = 10, h = 5 and E = 104.
Answer
(a)
19. If s = n/2[2a + (n – 1)d], then express d in terms of s, a and n. find d if n = 3, a = n + 1 and s = 18.
Answer
20. “Area A of a circular ring formed by 2 concentric circles is equal to the product of pie and the difference of the square of the bigger radius R and the square of the bigger radius R and the square of the smaller radius r. Express the above statement as a formula. Make r the subject of the formula and find r, when A = 88 sq cm and R = 8 cm.
Answer
