MAK LABS — Std 9 Maths Part 1 Chapter 4 Ratio and Proportion — Practice Sets 4.1 to 4.5 and Problem Set 4 Solutions

1From the following pairs of numbers, find the reduced form of ratio of first number to second number.

(i) 72, 60 — 7260 = 72 ÷ 1260 ÷ 12 = 65 ⇒ 6 : 5

(ii) 38, 57 — 3857 = 38 ÷ 1957 ÷ 19 = 23 ⇒ 2 : 3

(iii) 52, 78 — 5278 = 52 ÷ 2678 ÷ 26 = 23 ⇒ 2 : 3

2Find the reduced form of the ratio of the first quantity to second quantity.

Both quantities must be in the same unit before forming the ratio.

(i) 700 ₹, 308 ₹ — 700308 = 700 ÷ 28308 ÷ 28 = 2511 ⇒ 25 : 11

(ii) 14 ₹, 12 ₹ 40 paise — 14 ₹ = 1400 paise, 12 ₹ 40 paise = 1240 paise
14001240 = 3531 ⇒ 35 : 31

(iii) 5 litre, 2500 ml — 5 litre = 5000 ml. 50002500 = 21 ⇒ 2 : 1

(iv) 3 years 4 months, 5 years 8 months — 3 years 4 months = 40 months, 5 years 8 months = 68 months
4068 = 1017 ⇒ 10 : 17

(v) 3.8 kg, 1900 gm — 3.8 kg = 3800 gm. 38001900 = 21 ⇒ 2 : 1

(vi) 7 minutes 20 seconds, 5 minutes 6 seconds — 7 min 20 s = 440 s, 5 min 6 s = 306 s
440306 = 220153 ⇒ 220 : 153

3Express the following percentages as ratios in the reduced form.

(i) 75 : 100 = 3 : 4

(ii) 44 : 100 = 11 : 25

(iii) 6.25% = 6.25100 = 62510000 = 1 : 16

(iv) 52 : 100 = 13 : 25

(v) 0.64% = 0.64100 = 6410000 = 4 : 625

4Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?

Fewer days need more persons, so the number of persons and the number of days are in inverse proportion: persons × days = constant.

3 × 8 = x × 6

x = 246 = 4

4 persons are required.

5Convert the following ratios into percentage.

(i) 15 : 25 = 1525 = 15 × 425 × 4 = 60100 = 60%

(ii) 47 : 50 = 47 × 250 × 2 = 94100 = 94%

(iii) 710 = 70100 = 70%

(iv) 546600 = 91100 = 91%

(v) 716 = 716 × 100% = 43.75%

6The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha’s birth her mother’s age was 27 years. Find the present ages of Abha and her mother.

Let Abha’s age = 2x and mother’s age = 5x years.

The difference of their ages never changes, and at Abha’s birth it was 27 years. So 5x − 2x = 27 ⇒ 3x = 27 ⇒ x = 9

Abha: 18 years Mother: 45 years

7Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5 : 4?

After x years: 14 + x10 + x = 54

4(14 + x) = 5(10 + x) ⇒ 56 + 4x = 50 + 5x ⇒ x = 6

After 6 years (their ages will then be 20 and 16).

8The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana’s present age?

Let the present ages be 2x and 7x years.

2x + 27x + 2 = 13 ⇒ 6x + 6 = 7x + 2 ⇒ x = 4

Rehana’s present age = 8 years (mother: 28 years)

1Using the property ab = akbk, fill in the blanks substituting proper numbers.

(i) 57 = 2028 = 3549 = 2.53.5 (k = 4, 7 and 0.5)

(ii) 914 = 4.57 = 2742 = 2.253.5 (k = 0.5, 3 and 0.25)

2Find the following ratios.

(i) The ratio of radius to circumference of the circle.
r2πr = 12π ⇒ 1 : 2π

(ii) The ratio of circumference of circle with radius r to its area.
2πrπr² = 2r ⇒ 2 : r

(iii) The ratio of diagonal of a square to its side, if side is 7 cm.
Diagonal = 7√2 cm. 7√27 = √21 ⇒ √2 : 1

(iv) Sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of its perimeter to area.
Perimeter = 2(5 + 3.5) = 17 cm; Area = 5 × 3.5 = 17.5 cm²
1717.5 = 170175 = 3435 ⇒ 34 : 35

Perimeter and area have different units, so strictly this compares only their numerical values, as the textbook intends.

3Compare the following pairs of ratios.

For positive numbers: if ad > bc then ab > cd; if ad < bc then ab < cd.

(i) √53, 3√7
√5 × √7 = √35 and 3 × 3 = 9 = √81. √35 < √81
√53 < 3√7

(ii) 3√55√7, √63√125
3√5 × √125 = 3√625 = 75 and 5√7 × √63 = 5√441 = 105. 75 < 105
3√55√7 < √63√125

(iii) 518, 17121
5 × 121 = 605 and 18 × 17 = 306. 605 > 306
518 > 17121

(iv) √80√48, √45√27
√80 × √27 = √2160 and √48 × √45 = √2160. They are equal.
√80√48 = √45√27 (both equal √5√3)

(v) 9.25.1, 3.47.1
9.2 × 7.1 = 65.32 and 5.1 × 3.4 = 17.34. 65.32 > 17.34
9.25.1 > 3.47.1

4Solve.

(i) □ABCD is a parallelogram. The ratio of ∠A and ∠B is 5 : 4. Find the measure of ∠B.
Let ∠A = 5x and ∠B = 4x. Adjacent angles of a parallelogram are supplementary.
5x + 4x = 180° ⇒ 9x = 180° ⇒ x = 20° ∴ ∠B = 80°

(ii) The ratio of present ages of Albert and Salim is 5 : 9. Five years hence the ratio will be 3 : 5. Find their present ages.
Let the ages be 5x and 9x. 5x + 59x + 5 = 35 ⇒ 25x + 25 = 27x + 15 ⇒ x = 5
Albert: 25 years Salim: 45 years

(iii) The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth.
Let length = 3x, breadth = x. 2(3x + x) = 36 ⇒ 8x = 36 ⇒ x = 4.5
Length = 13.5 cm, Breadth = 4.5 cm

(iv) The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers.
31x + 23x = 216 ⇒ 54x = 216 ⇒ x = 4
124 and 92

(v) The product of two numbers is 360 and their ratio is 10 : 9. Find the numbers.
10x × 9x = 360 ⇒ 90x² = 360 ⇒ x² = 4 ⇒ x = 2 (taking the positive value)
20 and 18 (−20 and −18 also satisfy the conditions)

5*If a : b = 3 : 1 and b : c = 5 : 1 then find the value of (i) (a³15b²c)³ (ii) a²7bc

a = 3b and b = 5c, so a = 15c and b = 5c.

(i) a³15b²c = (15c)³15 × (5c)² × c = 3375c³375c³ = 9 ∴ (a³15b²c)³ = 9³ = 729

(ii) a²7bc = (15c)²7 × 5c × c = 225c²35c² = 457

6*If √0.04 × 0.4 × a = 0.4 × 0.04 × √b then find the ratio ab.

Squaring both sides: 0.04 × 0.4 × a = (0.4 × 0.04)² × b

0.016a = 0.000256b

ab = 0.0002560.016 = 0.016 = 161000 = 2125

a : b = 2 : 125

7(x + 3) : (x + 11) = (x − 2) : (x + 1) then find the value of x.

(x + 3)(x + 1) = (x − 2)(x + 11)

x² + 4x + 3 = x² + 9x − 22

3 + 22 = 9x − 4x ⇒ 25 = 5x

x = 5

1If ab = 73 then find the values of the following ratios.

(i) 5a + 3b5a − 3b
5a3b = 5 × 73 × 3 = 359. By componendo-dividendo, 5a + 3b5a − 3b = 35 + 935 − 9 = 4426 = 2213

(ii) 2a² + 3b²2a² − 3b²
a²b² = 499 ⇒ 2a²3b² = 9827. By componendo-dividendo, = 98 + 2798 − 27 = 12571

(iii) a³ − b³b³
a³b³ = 34327. By dividendo, a³ − b³b³ = 343 − 2727 = 31627

(iv) 7a + 9b7a − 9b
7a9b = 4927. By componendo-dividendo, = 49 + 2749 − 27 = 7622 = 3811

2If 15a² + 4b²15a² − 4b² = 477 then find the values of the following ratios.

(i) ab
By componendo-dividendo, 15a²4b² = 47 + 747 − 7 = 5440 = 2720
a²b² = 27 × 420 × 15 = 108300 = 925 ∴ ab = 35

(ii) 7a − 3b7a + 3b
7a3b = 2115 = 75. By componendo-dividendo, 7a + 3b7a − 3b = 122 = 6, so 7a − 3b7a + 3b = 16

(iii) b² − 2a²b² + 2a²
b²2a² = 2518. By componendo-dividendo, b² + 2a²b² − 2a² = 437, so b² − 2a²b² + 2a² = 743

(iv) b³ − 2a³b³ + 2a³
a³b³ = 27125 ⇒ b³2a³ = 12554. By componendo-dividendo, b³ + 2a³b³ − 2a³ = 17971, so b³ − 2a³b³ + 2a³ = 71179

3If 3a + 7b3a − 7b = 43 then find the value of the ratio 3a² − 7b²3a² + 7b².

By componendo-dividendo: 6a14b = 4 + 34 − 3 = 7 ⇒ 3a7b = 7 ⇒ ab = 493

3a²7b² = 37 × a²b² = 37 × 24019 = 3433

By componendo-dividendo: 3a² + 7b²3a² − 7b² = 343 + 3343 − 3 = 346340

3a² − 7b²3a² + 7b² = 340346 = 170173

4Solve the following equations.

(i) x² + 12x − 203x − 5 = x² + 8x + 122x + 3

Using dividendo with the multiplier 4 (subtract 4 × denominator from each numerator):
x² + 12x − 20 − 4(3x − 5)3x − 5 = x² + 8x + 12 − 4(2x + 3)2x + 3

x²3x − 5 = x²2x + 3

x = 0 satisfies this (both sides of the original equation equal 4).

If x ≠ 0, divide by x²: 3x − 5 = 2x + 3 ⇒ x = 8

x = 0 or x = 8

(ii) 10x² + 15x + 635x² − 25x + 12 = 2x + 3x − 5

x = 0 does not satisfy the equation (6312 ≠ 3−5), so x ≠ 0.

Multiply the terms of the right side by 5x: each ratio = 10x² + 15x + 635x² − 25x + 12 = 10x² + 15x5x² − 25x

By the theorem of equal ratios (subtracting): each ratio = 6312 = 214

2x + 3x − 5 = 214 ⇒ 8x + 12 = 21x − 105 ⇒ 13x = 117 ⇒ x = 9

x = 9

(iii) (2x + 1)² + (2x − 1)²(2x + 1)² − (2x − 1)² = 178

By componendo-dividendo: 2(2x + 1)²2(2x − 1)² = 17 + 817 − 8 = 259

2x + 12x − 1 = 53 or 2x + 12x − 1 = −53

6x + 3 = 10x − 5 ⇒ x = 2 or 6x + 3 = −10x + 5 ⇒ x = 18

x = 2 or x = 18 (both values satisfy the given equation)

(iv*) √4x + 1 + √x + 3√4x + 1 − √x + 3 = 41

By componendo-dividendo: 2√4x + 12√x + 3 = 4 + 14 − 1 = 53

Squaring: 4x + 1x + 3 = 259 ⇒ 36x + 9 = 25x + 75 ⇒ 11x = 66 ⇒ x = 6

x = 6 (check: 5 + 35 − 3 = 4)

(v) (4x + 1)² + (2x + 3)²4x² + 12x + 9 = 6136

4x² + 12x + 9 = (2x + 3)², so (4x + 1)² + (2x + 3)²(2x + 3)² = 6136

By dividendo: (4x + 1)²(2x + 3)² = 61 − 3636 = 2536

4x + 12x + 3 = 56 ⇒ 24x + 6 = 10x + 15 ⇒ x = 914

or 4x + 12x + 3 = −56 ⇒ 24x + 6 = −10x − 15 ⇒ x = −2134

x = 914 or x = −2134

(vi) (3x − 4)³ − (x + 1)³(3x − 4)³ + (x + 1)³ = 61189

By componendo-dividendo: 2(3x − 4)³−2(x + 1)³ = 61 + 18961 − 189 = 250−128

(3x − 4)³(x + 1)³ = 12564 ⇒ 3x − 4x + 1 = 54

12x − 16 = 5x + 5 ⇒ 7x = 21 ⇒ x = 3

x = 3

In (iii) and (v) the square gives two possible signs, so there are two solutions. Many guides list only the positive case (x = 2 in (iii), x = 914 in (v)); both values given here satisfy the equations.

1Fill in the blanks of the following.

(i) x7 = y3 = 3x + 5y36 = 7x − 9y22
(3 × 7 + 5 × 3 = 36 and 7 × 7 − 9 × 3 = 22)

(ii) a3 = b4 = c7 = a − 2b + 3c16 = 2a − 2b + 2c6 − 8 + 14
(3 − 8 + 21 = 16; and 6, 8, 14 are 2 × 3, 2 × 4, 2 × 7)

25m − n = 3m + 4n then find the values of the following expressions. (i) m² + n²m² − n² (ii) 3m + 4n3m − 4n

5m − 3m = 4n + n ⇒ 2m = 5n ⇒ mn = 52

(i) m²n² = 254. By componendo-dividendo, m² + n²m² − n² = 25 + 425 − 4 = 2921

(ii) 3m4n = 158. By componendo-dividendo, 3m + 4n3m − 4n = 15 + 815 − 8 = 237

3Prove the following.

(i) If a(y + z) = b(z + x) = c(x + y) and no two of a, b, c are equal, show that y − za(b − c) = z − xb(c − a) = x − yc(a − b).

Let a(y + z) = b(z + x) = c(x + y) = k. Then y + z = ka, z + x = kb, x + y = kc

y − z = (x + y) − (z + x) = kc − kb = k(b − c)bc ⇒ y − za(b − c) = kabc

z − x = (y + z) − (x + y) = ka − kc = k(c − a)ac ⇒ z − xb(c − a) = kabc

x − y = (z + x) − (y + z) = kb − ka = k(a − b)ab ⇒ x − yc(a − b) = kabc

All three ratios equal kabc. Hence proved.

(ii) If x3x − y − z = y3y − z − x = z3z − x − y and x + y + z ≠ 0, show that each ratio is 1.

By the theorem of equal ratios, each ratio = x + y + z(3x − y − z) + (3y − z − x) + (3z − x − y) = x + y + zx + y + z = 1. Hence proved.

(iii) If ax + byx + y = bx + azx + z = ay + bzy + z and x + y + z ≠ 0, show that each ratio = a + b2.

By the theorem of equal ratios, each ratio = ax + by + bx + az + ay + bz(x + y) + (x + z) + (y + z) = a(x + y + z) + b(x + y + z)2(x + y + z) = a + b2. Hence proved.

(iv) If y + za = z + xb = x + yc then show that xb + c − a = yc + a − b = za + b − c.

Let each ratio = k. Then y + z = ak, z + x = bk, x + y = ck

Adding: 2(x + y + z) = k(a + b + c) ⇒ x + y + z = k(a + b + c)2

x = (x + y + z) − (y + z) = k(a + b + c)2 − ak = k(b + c − a)2 ⇒ xb + c − a = k2

Similarly y = k(c + a − b)2 and z = k(a + b − c)2, so yc + a − b = za + b − c = k2

Hence proved.

(v) If 3x − 5y5z + 3y = x + 5zy − 5x = y − zx − z then show that every ratio = xy.

Multiply the terms of the third ratio by 5 and use the theorem of equal ratios on all three:

each ratio = (3x − 5y) + (x + 5z) + 5(y − z)(5z + 3y) + (y − 5x) + 5(x − z) = 4x4y = xy. Hence proved.

4Solve.

(i) 16x² − 20x + 98x² + 12x + 21 = 4x − 52x + 3

x = 0 does not satisfy the equation (921 ≠ −53), so x ≠ 0.

Multiply the terms of the right side by 4x: each ratio = 16x² − 20x8x² + 12x

By the theorem of equal ratios (subtracting): each ratio = 921 = 37

4x − 52x + 3 = 37 ⇒ 28x − 35 = 6x + 9 ⇒ 22x = 44 ⇒ x = 2

x = 2

(ii) 5y² + 40y − 125y + 10y² − 4 = y + 81 + 2y

y = 0 does not satisfy the equation (3 ≠ 8), so y ≠ 0.

Multiply the terms of the right side by 5y: each ratio = 5y² + 40y5y + 10y²

By the theorem of equal ratios (subtracting): each ratio = −12−4 = 3

y + 81 + 2y = 3 ⇒ y + 8 = 3 + 6y ⇒ 5y = 5 ⇒ y = 1

y = 1

1Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?

Let x be subtracted. Then (16 − x)² = (12 − x)(21 − x)

256 − 32x + x² = 252 − 33x + x² ⇒ x = −4

−4 should be subtracted (that is, 4 added). The numbers become 16, 20, 25, and 20² = 400 = 16 × 25.

2If (28 − x) is the mean proportional of (23 − x) and (19 − x) then find the value of x.

(28 − x)² = (23 − x)(19 − x)

784 − 56x + x² = 437 − 42x + x²

784 − 437 = 56x − 42x ⇒ 347 = 14x

x = 34714

Check: 28 − x = 4514, 23 − x = −2514, 19 − x = −8114, and (4514)² = 2025196 = (−2514)(−8114). The answer is correct but not a whole number, which suggests a possible printing change in the question.

3Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.

Let the numbers be a, 12, c. Then ac = 12² = 144 and a + c = 26

a(26 − a) = 144 ⇒ a² − 26a + 144 = 0 ⇒ (a − 8)(a − 18) = 0 ⇒ a = 8 or 18

8, 12, 18 (or 18, 12, 8)

4If (a + b + c)(a − b + c) = a² + b² + c² show that a, b, c are in continued proportion.

(a + b + c)(a − b + c) = [(a + c) + b][(a + c) − b] = (a + c)² − b² = a² + 2ac + c² − b²

a² + 2ac + c² − b² = a² + b² + c² ⇒ 2ac = 2b² ⇒ b² = ac

∴ a, b, c are in continued proportion.

5If ab = bc and a, b, c > 0 then show that (i) (a + b + c)(b − c) = ab − c² (ii) (a² + b²)(b² + c²) = (ab + bc)² (iii) a² + b²ab = a + cb

Let ab = bc = k. Then b = ck and a = bk = ck².

(i) LHS = (ck² + ck + c)(ck − c) = c²(k² + k + 1)(k − 1) = c²(k³ − 1)
RHS = ck² × ck − c² = c²k³ − c² = c²(k³ − 1). LHS = RHS.

(ii) LHS = (c²k⁴ + c²k²)(c²k² + c²) = c²k²(k² + 1) × c²(k² + 1) = c⁴k²(k² + 1)²
RHS = (ck² × ck + ck × c)² = (c²k³ + c²k)² = [c²k(k² + 1)]² = c⁴k²(k² + 1)². LHS = RHS.

(iii) LHS = c²k⁴ + c²k²ck² × ck = c²k²(k² + 1)c²k³ = k² + 1k
RHS = ck² + cck = c(k² + 1)ck = k² + 1k. LHS = RHS.

6Find mean proportional of x + yx − y, x² − y²x²y².

Let m be the mean proportional. m² = x + yx − y × (x + y)(x − y)x²y² = (x + y)²x²y²

m = x + yxy

1Select the appropriate alternative answer for the following questions.

(i) If 6 : 5 = y : 20 then what will be the value of y? — (B) 24
y = 6 × 205 = 24
(ii) What is the ratio of 1 mm to 1 cm? — (C) 1 : 10
(iii*) Ages are 16, 24 and 36 years. Ratio of Nitin’s age to Mohasin’s age? — (B) 2 : 3
24 : 36 = 2 : 3
(iv) 24 bananas were distributed between Shubham and Anil in the ratio 3 : 5. How many did Shubham get? — (D) 9
38 × 24 = 9
(v) What is the mean proportional of 4 and 25? — (C) 10
√4 × 25 = √100 = 10

2For the following numbers write the ratio of first number to second number in the reduced form.

(i) 21 : 48 = 7 : 16 (ii) 36 : 90 = 2 : 5 (iii) 65 : 117 = 5 : 9 (iv) 138 : 161 = 6 : 7 (v) 114 : 133 = 6 : 7

Common factors used: 3, 18, 13, 23 and 19 respectively.

3Write the following ratios in the reduced form.

(i) Radius to the diameter of a circle. — r : 2r = 1 : 2

(ii) Diagonal to length of a rectangle with length 4 cm and breadth 3 cm. — Diagonal = √4² + 3² = 5 cm. 5 : 4

(iii) Perimeter to area of a square with side 4 cm. — Perimeter = 16, Area = 16. 1 : 1 (numerical values)

4Check whether the following numbers are in continued proportion.

a, b, c are in continued proportion if b² = ac.

(i) 2, 4, 8: 4² = 16 = 2 × 8 — Yes
(ii) 1, 2, 3: 2² = 4, 1 × 3 = 3 — No
(iii) 9, 12, 16: 12² = 144 = 9 × 16 — Yes
(iv) 3, 5, 8: 5² = 25, 3 × 8 = 24 — No

5a, b, c are in continued proportion. If a = 3 and c = 27 then find b.

b² = ac = 3 × 27 = 81 ∴ b = 9 (b = −9 also satisfies b² = ac)

6Convert the following ratios into percentages.

(i) 37 : 500 = 7.4100 = 7.4%
(ii) 58 = 62.5100 = 62.5%
(iii) 2230 = 1115 × 100% = 7313% (≈ 73.33%)
(iv) 516 = 31.25%
(v) 1441200 = 12100 = 12%

7Write the ratio of first quantity to second quantity in the reduced form.

(i) 1024 MB, 1.2 GB — 1024 MB = 1 GB. 1 : 1.2 = 10 : 12 = 5 : 6

(ii) 17 Rupees, 25 Rupees 60 paise — 1700 paise : 2560 paise = 85 : 128

(iii) 5 dozen, 120 units — 60 : 120 = 1 : 2

(iv) 4 sq.m, 800 sq.cm — 4 sq.m = 40000 sq.cm. 40000 : 800 = 50 : 1

(v) 1.5 kg, 2500 gm — 1500 : 2500 = 3 : 5

8If ab = 23 then find the values of the following expressions.

(i) 4a + 3b3b — 4a3b = 89. By componendo, 4a + 3b3b = 8 + 99 = 179

(ii) 5a² + 2b²5a² − 2b² — 5a²2b² = 5 × 42 × 9 = 109. By componendo-dividendo, = 10 + 910 − 9 = 19

(iii) a³ + b³b³ — a³b³ = 827. By componendo, = 8 + 2727 = 3527

(iv) 7b − 4a7b + 4a — 7b4a = 7 × 34 × 2 = 218. By componendo-dividendo, 7b + 4a7b − 4a = 2913, so the value is 1329

9If a, b, c, d are in proportion, then prove that (i) 11a² + 9ac11b² + 9bd = a² + 3acb² + 3bd (ii*) √a² + 5c²b² + 5d² = ab (iii) a² + ab + b²a² − ab + b² = c² + cd + d²c² − cd + d²

a, b, c, d are in proportion, so ab = cd = k. Then a = bk and c = dk.

(i) LHS = 11b²k² + 9bk × dk11b² + 9bd = k²(11b² + 9bd)11b² + 9bd = k²
RHS = b²k² + 3bk × dkb² + 3bd = k²(b² + 3bd)b² + 3bd = k². LHS = RHS.

(ii) LHS = √b²k² + 5d²k²b² + 5d² = √k² = k = ab = RHS (the terms are taken as positive).

(iii) LHS = b²k² + b²k + b²b²k² − b²k + b² = k² + k + 1k² − k + 1
RHS = d²k² + d²k + d²d²k² − d²k + d² = k² + k + 1k² − k + 1. LHS = RHS.

10If a, b, c are in continued proportion, then prove that (i) aa + 2b = a − 2ba − 4c (ii) bb + c = a − ba − c

Let ab = bc = k. Then b = ck and a = ck².

(i) LHS = ck²ck² + 2ck = kk + 2
RHS = ck² − 2ckck² − 4c = ck(k − 2)c(k − 2)(k + 2) = kk + 2. LHS = RHS.

(ii) LHS = ckck + c = kk + 1
RHS = ck² − ckck² − c = ck(k − 1)c(k − 1)(k + 1) = kk + 1. LHS = RHS.

11Solve: 12x² + 18x + 4218x² + 12x + 58 = 2x + 33x + 2

x = 0 does not satisfy the equation (4258 ≠ 32), so x ≠ 0.

Multiply the terms of the right side by 6x: each ratio = 12x² + 18x18x² + 12x

By the theorem of equal ratios (subtracting): each ratio = 4258 = 2129

2x + 33x + 2 = 2129 ⇒ 58x + 87 = 63x + 42 ⇒ 5x = 45 ⇒ x = 9

x = 9

12If 2x − 3y3z + y = z − yz − x = x + 3z2y − 3x then prove that every ratio = xy.

Multiply the terms of the second ratio by −3 and use the theorem of equal ratios on all three:

each ratio = (2x − 3y) − 3(z − y) + (x + 3z)(3z + y) − 3(z − x) + (2y − 3x) = 3x3y = xy. Hence proved.

13*If by + czb² + c² = cz + axc² + a² = ax + bya² + b² then prove that xa = yb = zc.

Let each ratio = k. By the theorem of equal ratios (adding all three):
k = 2(ax + by + cz)2(a² + b² + c²) = ax + by + cza² + b² + c²

Subtracting the first ratio from this one (theorem of equal ratios):
k = (ax + by + cz) − (by + cz)(a² + b² + c²) − (b² + c²) = axa² = xa

In the same way, using the second and third ratios, k = byb² = yb and k = czc² = zc

∴ xa = yb = zc. Hence proved.

4. Ratio and Proportion

Practice Set 4.1

Practice Set 4.2

Practice Set 4.3

Practice Set 4.4

Practice Set 4.5

Problem Set 4