Linear Equations and Maximum or Minimum Values

Linear Equations in One Variable

A linear equation in one variable has the unknown only to the first power, like $3x + 7 = 22$. Whatever you do to one side, do to the other. In practice that means moving a term across the $=$ sign and changing its sign.

• $5x - 8 = 2x + 13$: move $2x$ to the left and $-8$ to the right to get $3x = 21$, so $x = 7$.
• Brackets: $4(x - 3) - 2(x + 1) = 6$ opens to $4x - 12 - 2x - 2 = 6$, so $2x = 20$ and $x = 10$.

Fractions: multiply by the LCM. For $\frac{x}{3} - \frac{x - 4}{5} = 2$, multiply every term by 15:

$$5x - 3(x - 4) = 30 \quad\Rightarrow\quad 5x - 3x + 12 = 30 \quad\Rightarrow\quad x = 9$$

Keep the bracket round $x - 4$. Writing $5x - 3x - 4$ is the classic slip; the minus sign in front has to reach the $-4$ as well, which turns it into $+12$.

Cross-multiplying. When each side is a single fraction, multiply across. For $\frac{4x - 3}{2x + 1} = \frac{9}{7}$:

$$7(4x - 3) = 9(2x + 1) \quad\Rightarrow\quad 28x - 21 = 18x + 9 \quad\Rightarrow\quad 10x = 30 \quad\Rightarrow\quad x = 3$$

Decimals. Multiply through by 10 or 100 first. $0.3x + 1.2 = 0.5x - 0.4$ becomes $3x + 12 = 5x - 4$, so $2x = 16$ and $x = 8$.

The unknown in a denominator. $\frac{5}{x - 2} = \frac{3}{x + 2}$ gives $5x + 10 = 3x - 6$, so $x = -8$. Check that no denominator becomes zero: here they are $-10$ and $-6$, so $x = -8$ stands.

No solution, or every number works. Sometimes $x$ cancels out completely. $3(x + 2) = 3x + 6$ is true for every $x$ (it is an identity), while $3(x + 2) = 3x + 5$ is never true, so it has no solution.

Simple inequalities. They are solved the same way, with one extra rule: multiplying or dividing both sides by a negative number turns the sign round. So $-2x > 6$ gives $x < -3$, and $5 - 3x \le 2x - 10$ gives $15 \le 5x$, that is $x \ge 3$.

Two Variables: Substitution and Elimination

Two unknowns need two equations. There are two standard ways to solve them.

Substitution. Make one letter the subject of one equation and put it into the other. From $x + 2y = 11$ and $3x - y = 5$: the second gives $y = 3x - 5$. Then $x + 2(3x - 5) = 11$, so $7x = 21$, $x = 3$ and $y = 4$.

Elimination. Multiply so that one letter has the same coefficient in both equations, then add or subtract. From $3x + 4y = 18$ and $5x - 2y = 4$: double the second to get $10x - 4y = 8$. Adding it to the first, $13x = 26$, so $x = 2$, and then $6 + 4y = 18$ gives $y = 3$. Add when the matched terms have opposite signs, subtract when they have the same sign.

The cross-multiplication formula. For $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$:

$$x = \frac{c_1b_2 - c_2b_1}{a_1b_2 - a_2b_1} \qquad y = \frac{a_1c_2 - a_2c_1}{a_1b_2 - a_2b_1}$$

For the pair above, the bottom is $3(-2) - 5(4) = -26$, so $x = \frac{18(-2) - 4(4)}{-26} = \frac{-52}{-26} = 2$ and $y = \frac{3(4) - 5(18)}{-26} = \frac{-78}{-26} = 3$.

Equations in $\frac{1}{x}$ and $\frac{1}{y}$. Put $u = \frac{1}{x}$ and $v = \frac{1}{y}$, and they become ordinary linear equations. From $\frac{2}{x} + \frac{3}{y} = 2$ and $\frac{4}{x} - \frac{9}{y} = -1$ we get $2u + 3v = 2$ and $4u - 9v = -1$. Tripling the first and adding, $10u = 5$, so $u = \frac{1}{2}$ and then $v = \frac{1}{3}$. Flip back: $x = 2$ and $y = 3$.

[Tier 2] The $\frac{xy}{x + y}$ form. Turn each equation upside down, because $\frac{x + y}{xy} = \frac{1}{x} + \frac{1}{y}$. If $\frac{xy}{x + y} = \frac{6}{5}$ and $\frac{xy}{x - y} = 6$, then $\frac{1}{y} + \frac{1}{x} = \frac{5}{6}$ and $\frac{1}{y} - \frac{1}{x} = \frac{1}{6}$. Adding, $\frac{2}{y} = 1$, so $y = 2$; subtracting, $\frac{2}{x} = \frac{2}{3}$, so $x = 3$.

[SSC Tip] Check any answer pair in both equations. SSC often builds a wrong option that satisfies only one of them. And with $\frac{1}{x}$ equations, do not stop at $u$ and $v$: the question wants $x$ and $y$.

Swapped Coefficients, Ratios and Letter Coefficients

Swapped coefficients: add, then subtract. When the numbers in front of $x$ and $y$ swap places between the two equations,

$$ax + by = p, \quad bx + ay = q \quad\Rightarrow\quad x + y = \frac{p + q}{a + b}, \quad x - y = \frac{p - q}{a - b}$$

Take $47x + 31y = 63$ and $31x + 47y = 15$. Adding: $78x + 78y = 78$, so $x + y = 1$. Subtracting: $16x - 16y = 48$, so $x - y = 3$. Then $x = 2$ and $y = -1$, with no big multiplications at all. Check: $94 - 31 = 63$ and $62 - 47 = 15$.

Often the question wants only $x + y$, $x - y$ or $x^2 - y^2$, and then you can stop even earlier: here $x^2 - y^2 = (x + y)(x - y) = 1 \times 3 = 3$.

Given a ratio. "If $x : y = 3 : 4$ and $2x + 3y = 36$, find $x$ and $y$." Write $x = 3m$ and $y = 4m$. Then $6m + 12m = 36$, so $m = 2$, $x = 6$ and $y = 8$. That is one equation in one unknown instead of two in two.

Asked for a combination. If the question wants something like $2x + y$, see whether it is the sum or difference of the given equations before solving. From $5x + 3y = 29$ and $3x + 2y = 18$, subtracting gives $2x + y = 11$ at once. (Solving fully gives $x = 4$, $y = 3$, and $8 + 3 = 11$ agrees.)

[Tier 2] Letters as coefficients. A pair such as $\frac{x}{a} + \frac{y}{b} = 2$ and $ax - by = a^2 - b^2$ looks hard but usually has a neat answer like $x = a$, $y = b$. Try it: $\frac{a}{a} + \frac{b}{b} = 2$, and $a \cdot a - b \cdot b = a^2 - b^2$. Both hold, and a pair like this has only one solution, so that is it. Testing the options is the fastest method here.

Three Variables: Add All Three, and Cost Questions

Pair sums. "$x + y = 11$, $y + z = 15$ and $z + x = 14$." Add all three: every letter appears twice, so $2(x + y + z) = 40$ and $x + y + z = 20$. Now take away each pair: $z = 20 - 11 = 9$, $x = 20 - 15 = 5$ and $y = 20 - 14 = 6$. The usual trap is forgetting to halve, which gives 40 for $x + y + z$.

A full system. For $x + y + z = 6$, $2x - y + z = 3$ and $x + 2y - z = 2$, remove the same letter twice. Adding the first and third removes $z$: $2x + 3y = 8$. Adding the second and third also removes $z$: $3x + y = 5$. From these two, $y = 5 - 3x$ and $2x + 15 - 9x = 8$, so $x = 1$, $y = 2$ and $z = 3$. In the exam it is often faster to test the options in all three equations.

Cyclic bills: add all three. "2 biscuits, 3 chocolates and 1 coffee cost ₹68; 3 biscuits, 1 chocolate and 2 coffees cost ₹70; 1 biscuit, 2 chocolates and 3 coffees cost ₹90." Call the prices $b$, $c$ and $f$. Each item appears $2 + 3 + 1 = 6$ times across the three bills, so adding them gives $6(b + c + f) = 228$, and one of each costs ₹38. To get the separate prices, take $2(b + c + f) = 76$ away from each bill: $c - f = -8$, $b - c = -6$ and $f - b = 14$. So $c = b + 6$ and $f = b + 14$, and $3b + 20 = 38$ gives $b = 6$, $c = 12$, $f = 20$.

Two bills, three items: the combination trick. "3 pens, 5 notebooks and 2 erasers cost ₹166; 4 pens, 7 notebooks and 3 erasers cost ₹233. Find the cost of 2 pens, 3 notebooks and 1 eraser." With three prices and only two bills, the separate prices cannot be found, but the asked bill can. Look for multipliers $m$ and $n$ with $m(3, 5, 2) + n(4, 7, 3) = (2, 3, 1)$. Pens: $3m + 4n = 2$; erasers: $2m + 3n = 1$. These give $m = 2$, $n = -1$, and the notebooks agree: $10 - 7 = 3$. So the asked bill is $2 \times 166 - 233 = 99$ rupees.

The common special case: if the second bill has exactly 2 more of each item than the first (like 3, 5, 4 and 5, 7, 6), the difference of the bills pays for 2 of each item, so one of each costs half the difference.

Consistency: One, None or Infinitely Many

Write the pair as $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, and compare the ratios of matching coefficients.

Condition	Number of solutions	The two lines
$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$	exactly one (consistent)	cross at one point
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$	none (inconsistent)	parallel
$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$	infinitely many	the same line

• $x + 2y = 4$ and $3x - y = 5$: $\frac{1}{3} \neq \frac{2}{-1}$, so exactly one solution.
• $2x + 3y = 7$ and $4x + 6y = 9$: $\frac{2}{4} = \frac{3}{6} = \frac{1}{2}$ but $\frac{7}{9} \neq \frac{1}{2}$, so no solution.
• $3x - y = 5$ and $6x - 2y = 10$: every ratio is $\frac{1}{2}$, so infinitely many solutions. The second equation is just the first one doubled.

It does not matter whether you keep the constants on the right ($= c$) or move them to the left ($+ c = 0$), as long as both equations are written the same way.

Find $k$ for no solution. "$kx + 3y = 5$ and $8x + 6y = 7$ have no solution." We need $\frac{k}{8} = \frac{3}{6}$, so $k = 4$. Then check the constants: $\frac{5}{7} \neq \frac{1}{2}$, so $k = 4$ really gives no solution.

Find $k$ for infinitely many. "$3x + 2y = 8$ and $(k + 1)x + (k - 1)y = 4k - 4$ have infinitely many solutions." From $\frac{3}{k + 1} = \frac{2}{k - 1}$: $3k - 3 = 2k + 2$, so $k = 5$. The ratios are then $\frac{3}{6}$, $\frac{2}{4}$ and $\frac{8}{16}$, all equal to $\frac{1}{2}$, so $k = 5$ works.

Unique solution. "$kx + 2y = 5$ and $3x + y = 1$ have a unique solution." That needs $\frac{k}{3} \neq \frac{2}{1}$, so it holds for every $k$ except 6.

[SSC Tip] After finding $k$ from the $x$ and $y$ ratios, always look at the constants. If they give the same ratio too, the answer is "infinitely many", not "no solution", and SSC keeps both as options.

Word Problems: Numbers, Digits and Fractions

Every word problem goes the same way: name the unknowns, turn each sentence into an equation, solve, and check the answer against the words.

Words	Algebra
a number exceeds another by 7	$x = y + 7$
twice a number, decreased by 5	$2x - 5$
three consecutive numbers	$x$, $x + 1$, $x + 2$
three consecutive odd (or even) numbers	$x$, $x + 2$, $x + 4$
two-digit number with tens digit $a$ and units digit $b$	$10a + b$
the same number with its digits reversed	$10b + a$

Numbers. "Two numbers add up to 74 and differ by 18." The larger is $\frac{74 + 18}{2} = 46$ and the smaller is $\frac{74 - 18}{2} = 28$. "One number is 5 less than three times the other, and they add up to 43." Then $x + (3x - 5) = 43$, so $x = 12$ and the other number is 31.

Consecutive numbers. "Three consecutive even numbers add up to 102." So $x + (x + 2) + (x + 4) = 102$, giving $3x + 6 = 102$ and $x = 32$: the numbers are 32, 34 and 36. Quicker: the middle number is the average, $102 \div 3 = 34$.

Two-digit numbers. "The digits of a two-digit number add up to 11. Reversing the digits gives a number 45 more than the original." With tens digit $a$ and units digit $b$: $a + b = 11$, and $(10b + a) - (10a + b) = 9(b - a) = 45$, so $b - a = 5$. Then $a = 3$, $b = 8$ and the number is 38. Check: $83 - 38 = 45$. In general, a number minus its reverse is $9(a - b)$ and a number plus its reverse is $11(a + b)$.

Fractions. "If 2 is added to both the numerator and the denominator, a fraction becomes $\frac{2}{3}$. If 1 is taken from both, it becomes $\frac{1}{2}$." Call it $\frac{x}{y}$. Then $3(x + 2) = 2(y + 2)$, so $3x - 2y = -2$; and $2(x - 1) = y - 1$, so $y = 2x - 1$. Putting this in: $3x - 4x + 2 = -2$, so $x = 4$, $y = 7$ and the fraction is $\frac{4}{7}$. Check: $\frac{6}{9} = \frac{2}{3}$ and $\frac{3}{6} = \frac{1}{2}$.

In digit questions, check that what you get are real digits: whole numbers from 0 to 9, with the tens digit not 0. And when testing options, an option has to pass every condition in the question, not just the first one.

Word Problems: Costs, Tickets, Coins and Marks

Costs of two items. "4 chairs and 3 tables cost ₹6,000; 5 chairs and 2 tables cost ₹5,400. Find the price of a table." With chair price $c$ and table price $t$: $4c + 3t = 6000$ and $5c + 2t = 5400$. To remove $c$, multiply by 5 and by 4: $20c + 15t = 30000$ and $20c + 8t = 21600$. Subtracting, $7t = 8400$, so a table costs ₹1,200 and a chair ₹600.

Tickets: the "all of one kind" shortcut. "500 tickets were sold, adults at ₹120 and children at ₹70, for ₹48,500 in all. How many children's tickets were sold?" If all 500 had been adult tickets, they would have brought ₹60,000, which is ₹11,500 too much. Each child's ticket in place of an adult's brings ₹50 less, so there were $11500 \div 50 = 230$ children's tickets and 270 adult ones. The equations $p + q = 500$ and $120p + 70q = 48500$ (adults $p$, children $q$) give the same answer.

Coins, in brief. "40 coins of ₹5 and ₹2 are worth ₹134." All ₹2 coins would make ₹80; the extra ₹54 comes ₹3 at a time, so there are 18 coins of ₹5 and 22 of ₹2.

Marks with negative marking. "A test has 50 questions: 3 marks for a right answer and 1 mark off for a wrong one. A student answers all of them and scores 102." With $r$ right and $w$ wrong: $r + w = 50$ and $3r - w = 102$. Adding, $4r = 152$, so $r = 38$ and $w = 12$.

A fixed charge plus a rate. "A taxi charges a fixed amount plus a rate per km. A 12 km ride costs ₹220 and a 20 km ride costs ₹340." The extra 8 km cost ₹120, so the rate is ₹15 per km, and the fixed charge is $220 - 12 \times 15 = 40$ rupees. A 30 km ride would cost $40 + 30 \times 15 = 490$ rupees.

Money passing between two people. "If Asha gives Bina ₹50, they have equal amounts. If Bina gives Asha ₹50, Asha has twice as much as Bina." With amounts $p$ (Asha) and $q$ (Bina): $p - 50 = q + 50$, so $p - q = 100$; and $p + 50 = 2(q - 50)$, so $p - 2q = -150$. Subtracting, $q = 250$ and $p = 350$. Notice that giving ₹50 makes them equal only because the gap was twice ₹50.

The "all of one kind" idea settles any two-type count (tickets, coins, animals with 2 and 4 legs) in one line: work out the total as if everything were the cheaper kind, take it from the real total, and divide the gap by the difference in price.

Least and Greatest Values by Completing the Square

A square is never negative: $(x - h)^2 \ge 0$, and it is 0 only when $x = h$. Writing an expression as "a square plus a number" shows its least value straight away.

Completing the square. Take half the coefficient of $x$, square it, and add and subtract it:

$$x^2 - 8x + 21 = (x^2 - 8x + 16) + 5 = (x - 4)^2 + 5$$

The square is at least 0, so the expression is at least 5. Its least value is 5, reached at $x = 4$. It has no greatest value, since it grows without limit.

A negative $x^2$ term gives a greatest value. $-x^2 + 6x + 2 = 11 - (x - 3)^2$, so the greatest value is 11, at $x = 3$, and there is no least value.

With a number in front of $x^2$. Take it out of the $x$ terms first: $2x^2 - 12x + 23 = 2(x^2 - 6x) + 23 = 2(x - 3)^2 - 18 + 23 = 2(x - 3)^2 + 5$, so the least value is 5, at $x = 3$.

[Tier 2] The formula. For $ax^2 + bx + c$, the turning point is at $x = -\frac{b}{2a}$, and the value there is

$$c - \frac{b^2}{4a} = \frac{4ac - b^2}{4a}$$

It is the least value when $a > 0$ and the greatest value when $a < 0$. For $2x^2 - 12x + 23$: $23 - \frac{144}{8} = 23 - 18 = 5$, at $x = \frac{12}{4} = 3$.

Fractions. The greatest value of $\frac{1}{x^2 - 4x + 7}$ comes when the bottom is smallest. Since $x^2 - 4x + 7 = (x - 2)^2 + 3$, the bottom is at least 3, and the greatest value is $\frac{1}{3}$, at $x = 2$.

Two variables. Complete the square in each letter: $x^2 + y^2 - 6x + 4y + 20 = (x - 3)^2 + (y + 2)^2 + 7$, so the least value is 7, at $x = 3$, $y = -2$.

Sums of squares. $(x - 2)^2 + (x - 8)^2$ is least when $x$ is halfway between 2 and 8, at $x = 5$, where it is $9 + 9 = 18$. With three such squares, the best $x$ is the average of the three numbers.

When $x$ is restricted. The least value of $x^2 - 8x + 21$ is 5 at $x = 4$, but if the question says $x \ge 6$, then $x = 4$ is not allowed. The expression keeps rising after $x = 4$, so for $x \ge 6$ the least value is at $x = 6$: $36 - 48 + 21 = 9$.

[SSC Tip] The least value is the number left over, not the $x$ where it happens. For $x^2 - 8x + 21$ the options will carry both 5 and 4, and only 5 is the least value.

$x + \frac{1}{x} \ge 2$, Fixed Sums and Fixed Products

The reciprocal rule. For every positive $x$,

$$x + \frac{1}{x} \ge 2, \quad \text{with equality only at } x = 1$$

The reason takes one line: $x + \frac{1}{x} - 2 = \frac{x^2 - 2x + 1}{x} = \frac{(x - 1)^2}{x}$, which cannot be negative when $x > 0$.

• In the same way $x^2 + \frac{1}{x^2} \ge 2$ for every $x \neq 0$, with equality at $x = \pm 1$.
• For negative $x$ the signs flip: $x + \frac{1}{x} \le -2$.
• So for real $x$, the value of $x + \frac{1}{x}$ never lies strictly between $-2$ and $2$. A question that says $x + \frac{1}{x} = 1$ is not about a real $x$; the Reciprocal Forms (x + 1/x) and Surd Values section shows what to do with it.

The scaled form. For positive $x$, $ax + \frac{b}{x} \ge 2\sqrt{ab}$, with the least value where the two parts are equal, at $x = \sqrt{\frac{b}{a}}$. The least value of $4x + \frac{9}{x}$ is $2\sqrt{36} = 12$, at $x = \frac{3}{2}$, where both parts are 6. The least value of $x^2 + \frac{16}{x^2}$ is $2\sqrt{16} = 8$, at $x^2 = 4$.

AM-GM. For positive numbers, the average is at least the geometric mean: $\frac{a + b}{2} \ge \sqrt{ab}$, with equality only when $a = b$. For three numbers, $\frac{a + b + c}{3} \ge \sqrt[3]{abc}$. Two uses:

• A fixed sum gives the greatest product when the numbers are equal. If $x + y = 18$, then $xy \le 81$, at $x = y = 9$. If $x + y + z = 12$, then $xyz \le 64$, at 4, 4 and 4.
• A fixed product gives the least sum when the numbers are equal. If $xy = 49$ with both positive, then $x + y \ge 14$.

Reciprocals with a fixed sum. For positive $a$, $b$, $c$, $(a + b + c)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \ge 9$. So if $a + b + c = 1$, the least value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is 9, at $a = b = c = \frac{1}{3}$.

Squares with a fixed sum. A sum of squares is least when the numbers are equal. If $x + y = 10$, then $x^2 + y^2 \ge 50$, because $x^2 + y^2 = \frac{(x + y)^2 + (x - y)^2}{2}$. If $x + y + z = 9$, then $x^2 + y^2 + z^2 \ge 27$. These two hold for negative numbers too.

[Tier 2] Unequal weights. "If $2x + 3y = 24$ with $x$ and $y$ positive, find the greatest value of $xy$." Make the two parts equal: $2x = 3y = 12$, so $x = 6$, $y = 4$ and $xy = 24$. Completing the square agrees: $xy = \frac{x(24 - 2x)}{3} = 24 - \frac{2}{3}(x - 6)^2$.

These rules need positive numbers, and the equal case must be allowed. If $x$ and $y$ must be whole numbers with $x + y = 15$, the greatest product is $7 \times 8 = 56$, not $7.5 \times 7.5$.

Solved Examples

Question 1

Solve: $\frac{x + 2}{3} - \frac{x - 1}{4} = 2$ (a) $x = 19$ (b) $x = 5$ (c) $x = 13$ (d) $x = 11$

Answer. Clear the fractions by multiplying every term by 12: $4(x + 2) - 3(x - 1) = 24$. Open the brackets carefully: $4x + 8 - 3x + 3 = 24$, because minus times minus is plus. So $x + 11 = 24$ and (c) $x = 13$. Check: $\frac{15}{3} - \frac{12}{4} = 5 - 3 = 2$. If you write $-3x - 3$ instead of $-3x + 3$, you get $x = 19$, which is the trap.

Question 2

Solve: $\frac{3x - 4}{2x + 1} = \frac{5}{4}$ (a) $x = \frac{21}{2}$ (b) $x = \frac{11}{2}$ (c) $x = -\frac{21}{2}$ (d) $x = \frac{9}{2}$

Answer. Cross-multiply: $4(3x - 4) = 5(2x + 1)$, so $12x - 16 = 10x + 5$. Then $2x = 21$ and (a) $x = \frac{21}{2}$. Check: $3 \times 10.5 - 4 = 27.5$ and $2 \times 10.5 + 1 = 22$, and $\frac{27.5}{22} = \frac{5}{4}$.

Question 3

If $\frac{3}{x} + \frac{2}{y} = 2$ and $\frac{9}{x} - \frac{4}{y} = 1$, what is $x + y$? (a) $\frac{5}{6}$ (b) 5 (c) 6 (d) $\frac{1}{5}$

Answer. Put $u = \frac{1}{x}$ and $v = \frac{1}{y}$: $3u + 2v = 2$ and $9u - 4v = 1$. Double the first: $6u + 4v = 4$. Add it to the second: $15u = 5$, so $u = \frac{1}{3}$. Then $1 + 2v = 2$, so $v = \frac{1}{2}$. Now flip back: $x = 3$ and $y = 2$, so $x + y$ = (b) 5. $\frac{5}{6}$ is $u + v$, which is what you get if you don't flip back.

Question 4

If $29x + 23y = 110$ and $23x + 29y = 98$, what is the value of $x^2 - y^2$? (a) 4 (b) 2 (c) 10 (d) 8

Answer. The coefficients have swapped places, so add and subtract. Adding: $52x + 52y = 208$, so $x + y = 4$. Subtracting: $6x - 6y = 12$, so $x - y = 2$. Then $x^2 - y^2 = (x + y)(x - y) = 4 \times 2$ = (d) 8. (Here $x = 3$ and $y = 1$.)

Question 5

If $x : y = 5 : 3$ and $4x - 3y = 33$, what is $x + y$? (a) 24 (b) 8 (c) 15 (d) 33

Answer. Write $x = 5m$ and $y = 3m$. Then $4(5m) - 3(3m) = 20m - 9m = 11m = 33$, so $m = 3$. That gives $x = 15$ and $y = 9$, and $x + y$ = (a) 24.

Question 6

[Tier 2] If $\frac{xy}{x + y} = \frac{12}{7}$ and $\frac{xy}{x - y} = 12$, what is the value of $x$? (a) 3 (b) 4 (c) $\frac{1}{4}$ (d) 12

Answer. Turn both equations upside down. $\frac{x + y}{xy} = \frac{7}{12}$ means $\frac{1}{y} + \frac{1}{x} = \frac{7}{12}$, and $\frac{x - y}{xy} = \frac{1}{12}$ means $\frac{1}{y} - \frac{1}{x} = \frac{1}{12}$. Subtract the second from the first: $\frac{2}{x} = \frac{6}{12} = \frac{1}{2}$, so $x$ = (b) 4. Adding instead gives $\frac{2}{y} = \frac{8}{12}$, so $y = 3$. Check: $\frac{12}{7}$ and $\frac{12}{1}$, both right. Don't mark $\frac{1}{4}$: that is $\frac{1}{x}$, not $x$.

Question 7

If $a + b = 23$, $b + c = 30$ and $c + a = 27$, what is the value of $b$? (a) 10 (b) 17 (c) 13 (d) 20

Answer. Add all three: each letter appears twice, so $2(a + b + c) = 80$ and $a + b + c = 40$. The pair without $b$ is $c + a = 27$, so $b = 40 - 27$ = (c) 13. In the same way $a = 10$ and $c = 17$. If you don't halve and take $a + b + c = 80$, you get $b = 53$, which is not even an option, a hint that something slipped.

Question 8

Solve: $x + y + z = 9$, $2x + y - z = 3$ and $x - y + 2z = 7$. (a) $x = 3$, $y = 2$, $z = 4$ (b) $x = 1$, $y = 4$, $z = 4$ (c) $x = 2$, $y = 4$, $z = 3$ (d) $x = 2$, $y = 3$, $z = 4$

Answer. Every option adds up to 9, so the first equation cannot tell them apart; use the others. Add the first two equations to remove $z$: $3x + 2y = 12$. Double the second and add the third, which also removes $z$: $5x + y = 13$. From these, $y = 13 - 5x$ and $3x + 26 - 10x = 12$, so $x = 2$, $y = 3$ and $z = 9 - 5 = 4$. The answer is (d) $x = 2$, $y = 3$, $z = 4$. Check in the third: $2 - 3 + 8 = 7$.

Question 9

2 cakes, 3 pastries and 1 juice cost ₹145; 3 cakes, 1 pastry and 2 juices cost ₹160; 1 cake, 2 pastries and 3 juices cost ₹145. What is the price of one cake? (a) ₹30 (b) ₹25 (c) ₹20 (d) ₹75

Answer. Each item appears $2 + 3 + 1 = 6$ times across the three bills, so adding them gives $6 \times (\text{cake} + \text{pastry} + \text{juice}) = 450$, and one of each costs ₹75. Now take twice that, ₹150, from each bill. Second bill: $160 - 150 = 10$, which leaves one cake minus one pastry, so a cake costs ₹10 more than a pastry. Third bill: $145 - 150 = -5$, so a juice costs ₹5 less than a cake. If a pastry costs $p$, then $p + (p + 10) + (p + 5) = 75$, so $p = 20$ and a cake costs (a) ₹30.

Question 10

3 pens, 5 pencils and 4 erasers cost ₹97, and 5 pens, 7 pencils and 6 erasers cost ₹149. What is the cost of 1 pen, 1 pencil and 1 eraser together? (a) ₹52 (b) ₹26 (c) ₹13 (d) ₹24

Answer. There are three prices and only two bills, so do not try to find each price. Compare the bills: the second has exactly 2 more pens, 2 more pencils and 2 more erasers. So the extra $149 - 97 = 52$ rupees pays for 2 of each item, and one of each costs $52 \div 2$ = (b) ₹26. ₹52 is the cost of two of each.

Question 11

For what value of $k$ do the equations $2x + ky = 5$ and $6x + 15y = 11$ have no solution? (a) 45 (b) 3 (c) 5 (d) $\frac{5}{2}$

Answer. No solution needs $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. So $\frac{2}{6} = \frac{k}{15}$, which gives $k$ = (c) 5. Then check the constants: $\frac{5}{11}$ is not $\frac{1}{3}$, so the lines really are parallel and there is no solution.

Question 12

The equations $(a - 1)x + 3y = 2$ and $6x + (1 - 2b)y = 6$ have infinitely many solutions. Find $a$ and $b$. (a) $a = 3$, $b = -4$ (b) $a = 3$, $b = 4$ (c) $a = 19$, $b = -4$ (d) $a = -1$, $b = 4$

Answer. Infinitely many solutions means all three ratios are equal. The constants give $\frac{2}{6} = \frac{1}{3}$, so every ratio must be $\frac{1}{3}$. From $\frac{a - 1}{6} = \frac{1}{3}$: $a - 1 = 2$, so $a = 3$. From $\frac{3}{1 - 2b} = \frac{1}{3}$: $1 - 2b = 9$, so $b = -4$. The answer is (a) $a = 3$, $b = -4$.

Question 13

For the pair $kx - 4y = 3$ and $6x - 12y = 9$, which statement is true? (a) No solution when $k = 2$ (b) Infinitely many solutions when $k = 2$, exactly one otherwise (c) Exactly one solution for every value of $k$ (d) No solution when $k = 6$

Answer. The $y$ ratio is $\frac{-4}{-12} = \frac{1}{3}$ and the constant ratio is $\frac{3}{9} = \frac{1}{3}$, the same. If $\frac{k}{6} = \frac{1}{3}$, that is $k = 2$, all three ratios match, so there are infinitely many solutions (the second equation is the first one times 3). For any other $k$ the $x$ ratio differs, so there is exactly one solution. That is (b) Infinitely many solutions when $k = 2$, exactly one otherwise. "No solution when $k = 2$" is wrong, because at $k = 2$ the constants have the same ratio too.

Question 14

Two numbers add up to 85, and the larger one is 13 more than twice the smaller. What is the larger number? (a) 24 (b) 49 (c) 57 (d) 61

Answer. Let the smaller be $y$; then the larger is $2y + 13$. Their sum: $y + 2y + 13 = 85$, so $3y = 72$ and $y = 24$. The larger is $2 \times 24 + 13$ = (d) 61. Check: $61 + 24 = 85$.

Question 15

A two-digit number is 7 times the sum of its digits. The number formed by reversing its digits is 27 less than the original number. What is the number? (a) 36 (b) 84 (c) 63 (d) 42

Answer. Let the tens digit be $a$ and the units digit $b$. First condition: $10a + b = 7(a + b)$, so $3a = 6b$ and $a = 2b$. Second condition: the original minus the reverse is $9(a - b) = 27$, so $a - b = 3$. With $a = 2b$, this gives $b = 3$ and $a = 6$, so the number is (c) 63. Check: $6 + 3 = 9$ and $7 \times 9 = 63$; reversed it is 36, which is $63 - 27$.

Question 16

The denominator of a fraction is 3 more than its numerator. If 4 is added to both, the fraction becomes $\frac{3}{4}$. What is the fraction? (a) $\frac{5}{8}$ (b) $\frac{9}{12}$ (c) $\frac{8}{11}$ (d) $\frac{2}{5}$

Answer. Let the fraction be $\frac{x}{x + 3}$. Adding 4 to both: $\frac{x + 4}{x + 7} = \frac{3}{4}$. Cross-multiply: $4x + 16 = 3x + 21$, so $x = 5$, and the fraction is (a) $\frac{5}{8}$. Check: $\frac{9}{12} = \frac{3}{4}$.

Question 17

A test has 80 questions. Each right answer gets 2 marks and each wrong answer loses 0.5 mark. A student attempts every question and scores 115. How many answers were wrong? (a) 62 (b) 20 (c) 15 (d) 18

Answer. If all 80 were right, the score would be 160. Each wrong answer costs $2 + 0.5 = 2.5$ marks compared with a right one (he loses the 2 he could have had and another 0.5). The score is $160 - 115 = 45$ short, so there were $45 \div 2.5$ = (d) 18 wrong answers. By equations: $r + w = 80$ and $2r - 0.5w = 115$ give $r = 62$, $w = 18$.

Question 18

A plumber charges a fixed visit fee plus an hourly rate. A 3-hour job costs ₹950 and a 5-hour job costs ₹1,450. How much does an 8-hour job cost? (a) ₹2,000 (b) ₹2,200 (c) ₹2,400 (d) ₹2,450

Answer. The 2 extra hours cost $1450 - 950 = 500$ rupees, so the rate is ₹250 an hour. The fixed fee is $950 - 3 \times 250 = 200$ rupees. An 8-hour job costs $200 + 8 \times 250$ = (b) ₹2,200. Simply adding the two given bills gives ₹2,400, but that counts the fixed fee twice.

Question 19

What is the least value of $x^2 + 10x + 32$? (a) $-5$ (b) 32 (c) 7 (d) 57

Answer. Complete the square: half of 10 is 5, and $5^2 = 25$. So $x^2 + 10x + 32 = (x + 5)^2 + 7$. The square is never negative, and it is 0 at $x = -5$, so the least value is (c) 7.

Question 20

What is the greatest value of $5 + 8x - 2x^2$? (a) 13 (b) 2 (c) 5 (d) 21

Answer. Take $-2$ out of the $x$ terms: $5 - 2(x^2 - 4x) = 5 - 2[(x - 2)^2 - 4] = 13 - 2(x - 2)^2$. The part being taken away is never negative, so the expression is at most 13, reached at $x = 2$. The greatest value is (a) 13. The formula agrees: $c - \frac{b^2}{4a} = 5 - \frac{64}{-8} = 5 + 8 = 13$.

Question 21

What is the least value of $x^2 + y^2 + 4x - 10y + 35$? (a) 35 (b) 6 (c) 64 (d) $-6$

Answer. Complete the square in each letter. $x^2 + 4x = (x + 2)^2 - 4$ and $y^2 - 10y = (y - 5)^2 - 25$. So the expression is $(x + 2)^2 + (y - 5)^2 + 35 - 4 - 25 = (x + 2)^2 + (y - 5)^2 + 6$. Both squares can be 0 together, at $x = -2$ and $y = 5$, so the least value is (b) 6.