Proportion and Variation

Proportion: Extremes and Means

When two ratios are equal, the four numbers are said to be in proportion. If a : b = c : d, we write

$$a : b :: c : d$$

and read it as "a is to b as c is to d". The outer terms a and d are the extremes, and the inner terms b and c are the means.

The test. Four numbers are in proportion exactly when

$$\text{product of extremes} = \text{product of means}, \quad ad = bc$$

This is just cross-multiplying $\frac{a}{b} = \frac{c}{d}$.

• 3, 8, 12, 32: 3 × 32 = 96 and 8 × 12 = 96. In proportion.
• 4, 9, 6, 15: 4 × 15 = 60 but 9 × 6 = 54. Not in proportion.

Order matters. 2, 5, 4, 10 are in proportion, since 2 × 10 = 5 × 4. The same numbers in the order 2, 5, 10, 4 are not, because 2 × 4 = 8 while 5 × 10 = 50. When a question says "a, b, c, d are in proportion", it means a : b = c : d in exactly that order.

Finding a missing term. Cross-multiply and solve.

• x : 12 :: 15 : 36 gives 36x = 180, so x = 5.
• (x − 1) : (x + 5) :: 1 : 3 gives 3x − 3 = x + 5, so x = 4. Check: 3 : 9 = 1 : 3.

Units in a proportion. In a : b :: c : d, a and b must be of the same kind, and so must c and d, but the two pairs can be different things. "5 kg : 8 kg :: ₹300 : ₹480" is a perfectly good proportion. The unitary method rests on this: if 5 kg of rice costs ₹300, then 8 kg costs 300 × $\frac{8}{5}$ = ₹480.

Ratio and proportion. A ratio compares two quantities; a proportion says that two ratios are equal. So 3 : 4 is a ratio, and 3 : 4 :: 15 : 20 is a proportion.

Quick consequences. If a : b :: c : d, the means can swap places (a : c :: b : d) and both sides can be turned upside down (b : a :: d : c). These and a few more rules are in Properties of Proportion below.

Key Point: a : b :: c : d means ad = bc. The proportionals and componendo-dividendo later in this section are all worked out from it.

Fourth, Third and Mean Proportional

These three come straight from a : b :: c : d, and CGL asks them often.

Fourth proportional. To 4, 10 and 6: x = $\frac{10 \times 6}{4}$ = 15. Check: 4 : 10 and 6 : 15 are both 2 : 5.

Third proportional. To 9 and 15: x = $\frac{15^2}{9} = \frac{225}{9}$ = 25. The second number is used twice, so 9, 15, 25 grow by the same multiple ($\frac{5}{3}$) each time.

Mean proportional. Between 6 and 24: x = $\sqrt{6 \times 24} = \sqrt{144}$ = 12. Check: 6 : 12 and 12 : 24 are both 1 : 2. The mean proportional is the geometric mean of the two numbers, and it always lies between them.

Order matters for the fourth and the third. The third proportional to 9 and 15 is 25, but the third proportional to 15 and 9 is $\frac{81}{15}$ = 5.4. The fourth proportional to 4, 10, 6 is 15, but to 6, 10, 4 it is $\frac{40}{6} = \frac{20}{3}$. The mean proportional does not depend on order. Write the proportion out (a : b :: b : x) before you calculate.

Decimals and fractions. The same formulas work.

• Mean proportional between 0.32 and 0.08: $\sqrt{0.0256}$ = 0.16.
• Third proportional to 0.4 and 0.6: $\frac{0.36}{0.4}$ = 0.9.
• Mean proportional between $\frac{4}{9}$ and $\frac{1}{4}$: $\sqrt{\frac{1}{9}} = \frac{1}{3}$.

Working backwards. Often one of the given numbers is the unknown.

• 15 is the fourth proportional to x, 10 and 6: x : 10 :: 6 : 15, so 15x = 60 and x = 4.
• 25 is the third proportional to 9 and x: 9 : x :: x : 25, so $x^2 = 225$ and x = 15.
• 36 is the third proportional to x and 24: x : 24 :: 24 : 36, so 36x = 576 and x = 16.
• 10 is the mean proportional between 4 and x: 4x = 100, so x = 25.

Traps in the options. For the mean proportional between 6 and 24, expect 15, the average (6 + 24) ÷ 2, and 144, the product without the square root. For the third proportional to 9 and 15, expect 21 (adding the gap of 6 again) and 135 (9 × 15).

Learn the three together: fourth = $\frac{bc}{a}$, third = $\frac{b^2}{a}$, mean = $\sqrt{ab}$. The third proportional is just the fourth proportional with b repeated.

Mixed Proportional Questions

CGL papers from 2023 to 2025 often chained two or three proportionals in one question. The method never changes: work out the innermost phrase first, write each value down, then do the outer step.

1. A third proportional built on a mean proportional. "Find the third proportional to the mean proportional between 4 and 9, and 12." The mean proportional between 4 and 9 is 6. The third proportional to 6 and 12 is $\frac{144}{6}$ = 24.

2. A mean proportional of a third and a fourth. "Find the mean proportional between the third proportional to 2 and 6, and the fourth proportional to 3, 4 and 6." The third proportional is $\frac{36}{2}$ = 18, the fourth proportional is $\frac{4 \times 6}{3}$ = 8, and the mean proportional between 18 and 8 is $\sqrt{144}$ = 12.

3. Sums and ratios. The sum of the third proportional to 8 and 12 and the mean proportional between 4 and 25 is 18 + 10 = 28. The ratio of the third proportional to 12 and 30 to the mean proportional between 9 and 25 is 75 : 15 = 5 : 1.

4. A fourth proportional built from others. "Find the fourth proportional to the third proportional to 3 and 6, the mean proportional between 2 and 8, and 18." The three numbers are 12, 4 and 18, so the answer is $\frac{4 \times 18}{12}$ = 6.

5. An unknown inside. "If the fourth proportional to 3, 8 and 15 is the mean proportional between 16 and x, find x." The fourth proportional is 40, so 16x = 1600 and x = 100. In the same way, "the mean proportional between 4 and 16 is the third proportional to 18 and x" gives $\frac{x^2}{18}$ = 8, so $x^2 = 144$ and x = 12.

6. A shortcut worth knowing. If m is the mean proportional between a and b, then the third proportional to a and m is b itself, because $\frac{m^2}{a} = \frac{ab}{a} = b$. So if x is the mean proportional between 12.8 and 64.8, and y is the third proportional to 12.8 and x, then y = 64.8 at once, with no square roots. (x itself is 28.8.)

7. Algebraic terms. The same formulas hold.

• Third proportional to $x^2 - y^2$ and x + y: $\frac{(x + y)^2}{x^2 - y^2} = \frac{x + y}{x - y}$.
• Fourth proportional to $a^2 - ab + b^2$, $a^3 + b^3$ and a − b: $\frac{(a^3 + b^3)(a - b)}{a^2 - ab + b^2} = (a + b)(a - b) = a^2 - b^2$.

8. Surds. Simplify the product before taking the root. For the mean proportional between $3 + \sqrt{2}$ and $12 - \sqrt{32}$: since $\sqrt{32} = 4\sqrt{2}$, the second number is $4(3 - \sqrt{2})$. The product is $4(9 - 2) = 28$, so the mean proportional is $\sqrt{28} = 2\sqrt{7}$.

Continued Proportion

Three numbers a, b, c are in continued proportion when

$$a : b = b : c, \quad \text{so } b^2 = ac$$

The middle number is the mean proportional of the outer two, and c is the third proportional to a and b. For example, 4, 6, 9 are in continued proportion because $6^2 = 36 = 4 \times 9$. Such numbers form a geometric progression: each term is the one before it times the same number (here $\frac{3}{2}$).

So if x, 12 and 18 are in continued proportion, $12^2 = 18x$ and x = 8.

Four terms. a, b, c, d are in continued proportion when

$$\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k$$

Working back from d: c = dk, b = $dk^2$ and a = $dk^3$. Two results follow, and SSC uses both.

• a : c = $k^2$, the duplicate ratio of a : b.
• a : d = $k^3$, the triplicate ratio of a : b.

So if a, b, c, d are in continued proportion and a : b = 2 : 5, then a : c = 4 : 25 and a : d = 8 : 125.

With a value given. If $\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{3}{2}$ and d = 16, then c = 24, b = 36 and a = 54.

Filling in missing terms. "8, x, y, 27 are in continued proportion." Going forward, each term is the last one times some number r, so $27 = 8r^3$. That gives $r^3 = \frac{27}{8}$ and r = $\frac{3}{2}$. The terms are 8, 12, 18, 27, so x = 12 and y = 18. Check: $12^2 = 144 = 8 \times 18$ and $18^2 = 324 = 12 \times 27$.

[Tier 2] Identities. Questions sometimes ask you to simplify an expression when a, b, c are in continued proportion. Use $b^2 = ac$, or put a = $ck^2$ and b = ck. Some results that come up:

• $(a + b + c)(a - b + c) = a^2 + b^2 + c^2$
• $\frac{a^2 + b^2}{b^2 + c^2} = \frac{a}{c}$
• for four terms a, b, c, d: $\frac{a^3 + b^3 + c^3}{b^3 + c^3 + d^3} = \frac{a}{d}$

For the first one: $(a + c)^2 - b^2 = a^2 + 2ac + c^2 - b^2$, and since ac = $b^2$ this is $a^2 + b^2 + c^2$.

Check with numbers. In the hall, test such an identity with an easy set like 1, 2, 4 (or 1, 2, 4, 8 for four terms). For 1, 2, 4: (1 + 2 + 4)(1 − 2 + 4) = 7 × 3 = 21, and 1 + 4 + 16 = 21. Any option that fails the test is out.

Adding or Subtracting to Make Numbers Proportional

"What number must be added to (or subtracted from) each of a, b, c, d so that they are in proportion?" This type appears almost every year.

Adding x. The new numbers a + x, b + x, c + x, d + x must have product of extremes = product of means:

$$(a + x)(d + x) = (b + x)(c + x)$$

The $x^2$ terms cancel on both sides, which leaves

$$x = \frac{bc - ad}{(a + d) - (b + c)}$$

For 1, 3, 7 and 12: x = (3 × 7 − 1 × 12) ÷ (13 − 10) = 9 ÷ 3 = 3. The new numbers are 4, 6, 10, 15, and indeed 4 × 15 = 60 = 6 × 10.

Subtracting x. You do not need a second formula. Use the same one; if x comes out negative, that number has to be subtracted instead. For 7, 10, 9 and 14: x = (90 − 98) ÷ (21 − 19) = −4. So 4 must be subtracted, giving 3, 6, 5, 10, and 3 × 10 = 30 = 6 × 5.

The option method. With four options it is often faster just to test them. Adding 3 to 1, 3, 7, 12 gives 4, 6, 10, 15, and 4 × 15 = 6 × 10, so 3 is the answer. Testing all four options takes under a minute even when the formula slips your mind.

Three numbers in continued proportion. For "what must be added to each of a, b, c so that they are in continued proportion", the condition is $(b + x)^2 = (a + x)(c + x)$, which gives

$$x = \frac{b^2 - ac}{a + c - 2b}$$

For 1, 5 and 13: x = (25 − 13) ÷ (14 − 10) = 3, giving 4, 8, 16. Check: $8^2 = 64 = 4 \times 16$. As before, a negative answer means subtract.

Why the order matters. In the formula, a and d are the first and last numbers and b and c are the middle two, in the order the question gives them. Mixing them up gives a wrong value, and that value is often one of the options.

[SSC Tip] If the formula gives a fraction or a negative number when the question says "added", recheck the order of the four numbers before anything else. SSC chooses its numbers so that the answer is a small whole number.

Properties of Proportion

If $\frac{a}{b} = \frac{c}{d}$, each of these is also true:

Componendo-dividendo (C-D for short) is the one SSC uses most. It also works backwards: if $\frac{a + b}{a - b} = \frac{p}{q}$, then $\frac{a}{b} = \frac{p + q}{p - q}$.

Using C-D to find a ratio. If $\frac{x + y}{x - y} = \frac{7}{3}$, then $\frac{x}{y} = \frac{7 + 3}{7 - 3} = \frac{10}{4}$, so x : y = 5 : 2. Check: $\frac{5 + 2}{5 - 2} = \frac{7}{3}$.

It works with coefficients too. If $\frac{3a + 4b}{3a - 4b} = \frac{5}{2}$, then $\frac{3a}{4b} = \frac{7}{3}$, so $\frac{a}{b} = \frac{28}{9}$.

Going the other way. If a : b = 5 : 3, then by C-D $\frac{a + b}{a - b} = \frac{5 + 3}{5 - 3}$ = 4, with no need to find a and b.

Equations with square roots. C-D clears the roots in one step:

$$\frac{\sqrt{x + 6} + \sqrt{x - 6}}{\sqrt{x + 6} - \sqrt{x - 6}} = 3$$

By C-D, $\frac{\sqrt{x + 6}}{\sqrt{x - 6}} = \frac{3 + 1}{3 - 1} = 2$. Squaring, $\frac{x + 6}{x - 6} = 4$, so x + 6 = 4x − 24 and x = 10. Check: $\sqrt{16} = 4$ and $\sqrt{4} = 2$, and $\frac{4 + 2}{4 - 2} = 3$.

[Tier 2] A well-known one. If $x = \frac{4ab}{a + b}$ (with a ≠ b), find $\frac{x + 2a}{x - 2a} + \frac{x + 2b}{x - 2b}$.

Write $\frac{x}{2a} = \frac{2b}{a + b}$ and apply C-D: $\frac{x + 2a}{x - 2a} = \frac{3b + a}{b - a}$. In the same way, $\frac{x + 2b}{x - 2b} = \frac{3a + b}{a - b}$. Adding the two:

$$\frac{3b + a}{b - a} - \frac{3a + b}{b - a} = \frac{2(b - a)}{b - a} = 2$$

[SSC Tip] C-D works only when one single fraction equals another single fraction. If there is an extra term on either side, tidy it into one fraction first.

The k-Method and Values of Expressions

Putting in the ratio. If a : b = 5 : 3, the numbers are 5k and 3k. In an expression where every term has the same degree (a homogeneous expression), k cancels, so you can simply put a = 5 and b = 3:

$$\frac{3a + 2b}{4a - b} = \frac{15 + 6}{20 - 3} = \frac{21}{17}$$

$$\frac{a^2 + b^2}{a^2 - b^2} = \frac{25 + 9}{25 - 9} = \frac{34}{16} = \frac{17}{8}$$

The warning. This works only when the terms balance. $\frac{a + 2}{b + 2}$ or $\frac{a^2 + b}{ab}$ cannot be found from a : b alone, because k does not cancel; you would need another fact from the question to fix k.

Finding the ratio from an expression. If $\frac{2x + 3y}{3x - y} = \frac{7}{5}$, cross-multiply: 10x + 15y = 21x − 7y, so 22y = 11x and x = 2y. Thus x : y = 2 : 1. (You can also divide the top and bottom by y and solve for $\frac{x}{y}$.)

Quadratic forms. If $2x^2 - 5xy + 2y^2 = 0$, divide by $y^2$ and put r = $\frac{x}{y}$: $2r^2 - 5r + 2 = 0$, so (2r − 1)(r − 2) = 0 and r = 2 or $\frac{1}{2}$. If the question also says x > y, then x : y = 2 : 1.

Equal ratios: the k-method. If $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k$, put a = bk, c = dk and e = fk. Then

• $\frac{a + c + e}{b + d + f} = k$
• $\frac{2a + 3c}{2b + 3d} = k$ (any multipliers, as long as they match top and bottom)
• $\frac{a^2 + c^2}{b^2 + d^2} = k^2$
• $\frac{ace}{bdf} = k^3$

So if $\frac{a}{3} = \frac{b}{4} = \frac{c}{7}$, then $\frac{a + b + c}{c} = \frac{3 + 4 + 7}{7}$ = 2.

Two ratios at once. If a : b = 2 : 3 and b : c = 4 : 5, combine them first (a : b : c = 8 : 12 : 15, as shown in the ratio-basics section) and then substitute. For instance, $\frac{a + b + c}{a - b + c} = \frac{35}{11}$.

A favourite. If $\frac{x}{y} = \frac{4}{5}$, find $\frac{4}{7} + \frac{2y - x}{2y + x}$. Put x = 4 and y = 5: $\frac{10 - 4}{10 + 4} = \frac{6}{14} = \frac{3}{7}$, and $\frac{4}{7} + \frac{3}{7}$ = 1.

Direct and Inverse Variation

Direct variation. "x varies directly as y", written $x \propto y$, means x = ky for some constant k. The ratio $\frac{x}{y}$ stays fixed, so

$$\frac{x_1}{y_1} = \frac{x_2}{y_2}$$

If y doubles, x doubles; if y is cut to a third, so is x. The cost of identical items varies directly as the number bought: if 12 pens cost ₹180, then 20 pens cost 180 × $\frac{20}{12}$ = ₹300.

Inverse variation. "x varies inversely as y", written $x \propto \frac{1}{y}$, means xy = k. Now the product stays fixed:

$$x_1 y_1 = x_2 y_2$$

If y doubles, x halves. For a fixed distance, time varies inversely as speed: a trip that takes 6 hours at 40 km/h covers 240 km, so at 48 km/h it takes 240 ÷ 48 = 5 hours. Men and days on a fixed job behave the same way.

Finding the constant. Turn the statement into an equation with k, use the given pair to find k, then use k for the new value. "x varies directly as y, and x = 21 when y = 3. Find x when y = 8." Here k = 7, so x = 56. Write k down before you use the second pair; mixing the two pairs in one step is where most mistakes happen.

Powers and roots. The wording tells you the power.

• "x varies as the square of y": x = $ky^2$. If x = 18 when y = 3, then k = 2, and at y = 5, x = 50.
• "x varies inversely as the square of y": $x = \frac{k}{y^2}$.
• "x varies as the square root of y": x = $k\sqrt{y}$.
• "$x^2$ varies inversely as $y^3$": $x^2y^3 = k$. If x = 3 when y = 4, then k = 9 × 64 = 576. At y = 9, $x^2 = \frac{576}{729} = \frac{64}{81}$, so x = $\frac{8}{9}$.

The percentage effect. If $x \propto y^2$ and y rises by 20%, x is multiplied by $1.2^2$ = 1.44, so x rises by 44%. If $x \propto \frac{1}{y}$ and y rises by 25%, x is multiplied by $\frac{4}{5}$ and falls by 20%. These are the product-rule ideas from the Percentage chapter.

Spotting the type. Ask yourself: when one goes up, does the other go up (direct) or down (inverse)? More workers, fewer days: inverse. More distance, more fare: direct, or partly direct, as the next part shows. More speed, less time: inverse.

Joint and Partial Variation

Joint variation. "x varies jointly as y and z" means $x \propto yz$, so x = kyz. "x varies directly as y and inversely as z" means $x = \frac{ky}{z}$.

If x varies jointly as y and z, and x = 30 when y = 3 and z = 5, then k = 2. When y = 4 and z = 7, x = 2 × 4 × 7 = 56.

In ratio form, when x = kyz, $\frac{x_1}{x_2} = \frac{y_1 z_1}{y_2 z_2}$. The area of a triangle varies jointly as its base and height, and simple interest varies jointly as the principal, the rate and the time.

Varying as a sum. "x varies as (y + 4)" means x = k(y + 4). If x = 21 when y = 3, then k = 3, and at y = 9, x = 3 × 13 = 39. Treat the bracket as one quantity.

Partial variation. Many real costs have a fixed part and a part that varies:

$$x = a + by$$

Here a is the fixed part and b the rate. Two pairs of values give two equations, and subtracting them removes a.

The monthly expenses of a hostel are partly fixed and partly vary with the number of students. With 40 students the bill is ₹28,000, and with 60 students it is ₹39,000. The extra 20 students add ₹11,000, so b = ₹550 per student. The fixed part is 28,000 − 40 × 550 = ₹6,000. For 75 students: 6,000 + 75 × 550 = ₹47,250.

The trap is to treat the whole bill as varying directly: 28,000 × $\frac{75}{40}$ = ₹52,500. That is wrong because the fixed ₹6,000 does not grow with the number of students.

Value that varies as a square. The price of a diamond often varies as the square of its weight. If a diamond breaks into pieces whose weights are in the ratio 1 : 3, the original value is proportional to $4^2$ = 16 and the pieces together to $1^2 + 3^2$ = 10. The loss is $\frac{6}{16} = \frac{3}{8}$ of the original value; on a ₹1,60,000 diamond that is ₹60,000. The more evenly it breaks, the bigger the loss.

[Tier 2] Variation between expressions. If $(x + y) \propto (x - y)$, then $\frac{x + y}{x - y}$ is constant, and by C-D so is $\frac{x}{y}$. That means $x \propto y$. It follows that any two expressions of the same degree in x and y vary as each other; for instance $(x^2 + y^2) \propto xy$.

Solved Examples

Question 1

If (x + 2) : (3x − 2) :: 3 : 5, then x = ? (a) 4 (b) 2 (c) 6 (d) 8

Answer. Product of extremes = product of means: 5(x + 2) = 3(3x − 2). So 5x + 10 = 9x − 6, which gives 4x = 16 and x = 4. Check: 6 : 10 = 3 : 5. The answer is (a) 4.

Question 2

The fourth proportional to 0.12, 0.21 and 8 is: (a) 1.4 (b) $\frac{32}{7}$ (c) 14 (d) 16

Answer. The fourth proportional x satisfies 0.12 : 0.21 :: 8 : x, so x = $\frac{0.21 \times 8}{0.12} = \frac{1.68}{0.12}$ = 14. The answer is (c) 14. Putting the numbers in the wrong order (0.21 : 0.12 :: 8 : x) gives $\frac{32}{7}$.

Question 3

The third proportional to 18 and 24 is: (a) 30 (b) 32 (c) 13.5 (d) 432

Answer. 18 : 24 :: 24 : x, so x = $\frac{24^2}{18} = \frac{576}{18}$ = 32. The answer is (b) 32.

Question 4

The mean proportional between 0.18 and 0.5 is: (a) 0.34 (b) 0.03 (c) 0.09 (d) 0.3

Answer. Mean proportional = $\sqrt{0.18 \times 0.5} = \sqrt{0.09}$ = 0.3. The answer is (d) 0.3.

Question 5

If 49 is the third proportional to x and 28, then x = ? (a) 16 (b) 7 (c) 21 (d) 36

Answer. x : 28 :: 28 : 49, so 49x = $28^2$ = 784 and x = 16. Check: 16 : 28 = 4 : 7 and 28 : 49 = 4 : 7. The answer is (a) 16.

Question 6

If the mean proportional between x and 3.2 is 4.8, then x = ? (a) 6.4 (b) 7.2 (c) 1.5 (d) 15.36

Answer. $4.8^2 = 3.2x$, so 23.04 = 3.2x and x = 7.2. Check: 3.2 : 4.8 = 2 : 3 and 4.8 : 7.2 = 2 : 3. The answer is (b) 7.2.

Question 7

The third proportional to the mean proportional between 3 and 27, and 15, is: (a) 15 (b) 5.4 (c) 25 (d) 75

Answer. Inner part first: the mean proportional between 3 and 27 is $\sqrt{81}$ = 9. Now the third proportional to 9 and 15: 9 : 15 :: 15 : x, so x = $\frac{225}{9}$ = 25. The answer is (c) 25.

Question 8

The mean proportional between the third proportional to 3 and 9 and the fourth proportional to 4, 6 and 8 is: (a) 19.5 (b) 15 (c) 24 (d) 18

Answer. Third proportional to 3 and 9: $\frac{81}{3}$ = 27. Fourth proportional to 4, 6 and 8: $\frac{6 \times 8}{4}$ = 12. Mean proportional between 27 and 12: $\sqrt{324}$ = 18. The answer is (d) 18.

Question 9

The ratio of the fourth proportional to 4, 9 and 12 to the third proportional to 25 and 45 is: (a) 3 : 1 (b) 1 : 3 (c) 3 : 4 (d) 1 : 9

Answer. Fourth proportional to 4, 9 and 12: $\frac{9 \times 12}{4}$ = 27. Third proportional to 25 and 45: $\frac{45^2}{25} = \frac{2025}{25}$ = 81. The ratio is 27 : 81 = 1 : 3. The answer is (b) 1 : 3.

Question 10

The mean proportional between $4 + \sqrt{7}$ and $16 - 4\sqrt{7}$ is: (a) 6 (b) 36 (c) $2\sqrt{7}$ (d) 3

Answer. $16 - 4\sqrt{7} = 4(4 - \sqrt{7})$. So the product is $4(4 + \sqrt{7})(4 - \sqrt{7}) = 4(16 - 7) = 36$, and the mean proportional is $\sqrt{36}$ = 6. The answer is (a) 6. 36 is the product before the square root.

Question 11

[Tier 2] The third proportional to $\frac{x}{y} + \frac{y}{x}$ and $\sqrt{x^2 + y^2}$ is: (a) $\frac{x}{y}$ (b) $x^2 + y^2$ (c) $\sqrt{xy}$ (d) $xy$

Answer. The third proportional is $\frac{b^2}{a}$. Here $b^2 = x^2 + y^2$, and taking the LCM, $a = \frac{x^2 + y^2}{xy}$. Dividing: $(x^2 + y^2) \times \frac{xy}{x^2 + y^2} = xy$. The answer is (d) $xy$. A quick check with x = y = 1: a = 2 and $b^2 = 2$, so the third proportional is 1, which matches xy.

Question 12

x is the mean proportional between 7.2 and 16.2, and y is the third proportional to 7.2 and x. Then x + y = ? (a) 15.6 (b) 21.6 (c) 27 (d) 32.4

Answer. x = $\sqrt{7.2 \times 16.2} = \sqrt{116.64}$ = 10.8. For y I use the shortcut: the third proportional to a and the mean proportional of a and b is b itself, so y = 16.2. (Check: $\frac{10.8^2}{7.2} = \frac{116.64}{7.2}$ = 16.2.) So x + y = 10.8 + 16.2 = 27. The answer is (c) 27. Taking the third proportional the wrong way round gives $\frac{7.2^2}{10.8}$ = 4.8, and 10.8 + 4.8 = 15.6.

Question 13

If 27, x, y and 125 are in continued proportion, then x + y = ? (a) 152 (b) 120 (c) 100 (d) 130

Answer. In continued proportion each term is the one before it times the same number r. So $125 = 27r^3$, $r^3 = \frac{125}{27}$ and r = $\frac{5}{3}$. Then x = 45 and y = 75, so x + y = 120. The answer is (b) 120. Check: 27 : 45, 45 : 75 and 75 : 125 are all 3 : 5.

Question 14

If a, b, c, d are in continued proportion and a : b = 3 : 2, then a : d = ? (a) 9 : 4 (b) 8 : 27 (c) 3 : 2 (d) 27 : 8

Answer. Each of a : b, b : c and c : d is 3 : 2. So $\frac{a}{d} = \frac{a}{b} \times \frac{b}{c} \times \frac{c}{d} = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$. The answer is (d) 27 : 8. 9 : 4 is a : c, the duplicate ratio.

Question 15

What number must be added to each of 4, 7, 8 and 13 so that the numbers are in proportion? (a) 2 (b) 1 (c) 3 (d) 4

Answer. I need (4 + x)(13 + x) = (7 + x)(8 + x). The $x^2$ cancels on both sides: 52 + 17x = 56 + 15x, so 2x = 4 and x = 2. The new numbers are 6, 9, 10 and 15, and 6 × 15 = 90 = 9 × 10. The answer is (a) 2.

Question 16

What number must be subtracted from each of 20, 27, 46 and 67 so that the remainders are in proportion? (a) 3 (b) 5 (c) 7 (d) 9

Answer. Using the formula, x = $\frac{bc - ad}{(a + d) - (b + c)}$. Here bc = 27 × 46 = 1242, ad = 20 × 67 = 1340, and (20 + 67) − (27 + 46) = 87 − 73 = 14. So x = $\frac{1242 - 1340}{14}$ = −7, which means 7 must be subtracted. The new numbers are 13, 20, 39 and 60, and 13 × 60 = 780 = 20 × 39. The answer is (c) 7.

Question 17

If $\frac{5x + 3y}{5x - 3y} = \frac{7}{3}$, then x : y = ? (a) 2 : 3 (b) 3 : 2 (c) 5 : 2 (d) 7 : 3

Answer. By componendo-dividendo, $\frac{5x}{3y} = \frac{7 + 3}{7 - 3} = \frac{10}{4} = \frac{5}{2}$. So 10x = 15y, and x : y = 3 : 2. The answer is (b) 3 : 2. Stopping at $\frac{5x}{3y} = \frac{5}{2}$ and reading off 5 : 2 is the trap.

Question 18

If $\frac{\sqrt{x + 4} + \sqrt{x - 4}}{\sqrt{x + 4} - \sqrt{x - 4}} = 2$, then x = ? (a) 8 (b) 4 (c) 6 (d) 5

Answer. Componendo-dividendo gives $\frac{\sqrt{x + 4}}{\sqrt{x - 4}} = \frac{2 + 1}{2 - 1} = 3$. Squaring, $\frac{x + 4}{x - 4} = 9$, so x + 4 = 9x − 36 and x = 5. Check: $\sqrt{9} = 3$ and $\sqrt{1} = 1$, and $\frac{3 + 1}{3 - 1} = 2$. The answer is (d) 5.

Question 19

If $\frac{4x - 3y}{2x + 5y} = \frac{1}{3}$, then x : y = ? (a) 7 : 5 (b) 5 : 7 (c) 2 : 1 (d) 7 : 2

Answer. Cross-multiplying: 12x − 9y = 2x + 5y, so 10x = 14y and $\frac{x}{y} = \frac{14}{10}$. So x : y = 7 : 5. The answer is (a) 7 : 5. Check with x = 7 and y = 5: $\frac{28 - 15}{14 + 25} = \frac{13}{39} = \frac{1}{3}$.

Question 20

x varies inversely as the square of y. If x = 9 when y = 4, then the value of x when y = 6 is: (a) 6 (b) 13.5 (c) 4 (d) 20.25

Answer. $x = \frac{k}{y^2}$, so k = 9 × 16 = 144. When y = 6, x = $\frac{144}{36}$ = 4. The answer is (c) 4.

Question 21

The fare of a taxi is partly fixed and partly varies with the distance travelled. A 10 km ride costs ₹260 and a 16 km ride costs ₹380. What is the fare for 25 km? (a) ₹650 (b) ₹560 (c) ₹500 (d) ₹620

Answer. The extra 6 km cost 380 − 260 = ₹120, so the rate is ₹20 per km. For 10 km that is ₹200, so the fixed part is 260 − 200 = ₹60. For 25 km: 60 + 25 × 20 = ₹560. The answer is (b) ₹560. Scaling ₹260 straight up to 25 km gives ₹650, which wrongly lets the fixed part grow too.

Question 22

The value of a diamond varies as the square of its weight. A diamond worth ₹5,00,000 breaks into three pieces whose weights are in the ratio 2 : 3 : 5. The loss in value is: (a) ₹1,90,000 (b) ₹2,50,000 (c) ₹3,80,000 (d) ₹3,10,000

Answer. I take the weights as 2, 3 and 5, so the whole diamond weighed 10. Values go with squares: the whole is worth 100 units and the pieces 4 + 9 + 25 = 38 units. The loss is 62 units out of 100, which is 62% of ₹5,00,000 = ₹3,10,000. The answer is (d) ₹3,10,000.