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 = 2π √(L/G)

Answer

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

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


17. If A = pr2 and C = 2pr, then express r in terms of A and C.

Answer


18. If V = pr2h and S = 2pr2 + 2prh, then express V in terms of S, p and r.

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


8. Make y the subject of the formula x/a + y/b = 1. Find y, when a = 2, b = 8 and x = 5.

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 π 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, π = 22/7, h = 14 cm.

Answer


16. “The volume of a cone V is equal to the product of one third of π 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)

(b)

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

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