Elementary Mathematics 2017 (II)

Set A

Questions 21–40 of 100

21Arithmetic • Number System • Divisibility

ModerateIdeal time: about 1.5 min

$a, b, c, d$ are non-zero integers such that $(ab)$ divides $(c d)$ . If $a$ and $c$ are coprime, then which one of the following is correct ?

a

is a factor of

c

a

is a factor of

b

a

is a factor of

d

d

is a factor of

a

Show answer & solution

Option C

a

is a factor of

d

Approach

$a$ divides $ab$ , hence $c d$ ; coprimality with $c$ leaves only $d$ for its prime factors to sit in.

Euclid's lemma

Step-by-step solution

1
$a ∣ ab$ and $ab ∣ c d \Rightarrow a ∣ c d$
Why: Divisibility is transitive: $a$ divides $ab$ , which divides $c d$ .
2
$HCF (a, c) = 1$ , so $a$ and $c$ share no prime factor
Why: Coprime means the two prime factorisations are disjoint.
Coprime means no shared prime factor
3
every prime power in $a$ must divide $d$ , so $a ∣ d$
Why: Each prime power dividing $a$ divides $c d$ ; it cannot come from $c$ , so it must come from $d$ . Hence $a$ is a factor of $d$ .
Prime factorisation Euclid's lemma

Shortcut

$a ∣ c d$ with $HCF (a, c) = 1$ forces $a ∣ d$ - Euclid's lemma in its general form.

Common mistakes

Option A

Confused coprimality with divisibility; $HCF (a, c) = 1$ makes $a ∣ c$ impossible unless $a = \pm 1$ .

Option B

$b$ is unconstrained by the hypotheses: $a = 4$ , $b = 3$ has $a ∤ b$ while every condition holds.

Option D

Reversed the direction of divisibility between $a$ and $d$ .

Skills used

Euclid's lemma Coprime means no shared prime factor Prime factorisation

22Algebra • Quadratic Equations • Roots

HardIdeal time: about 2 min

If the roots of the equation $a (b - c) x^{2} + b (c - a) x + c (a - b) = 0$ are equal, then which one of the following is correct ?

2 b = a + c

b^{2} = a c

\frac{2}{b} = \frac{1}{a} + \frac{1}{c}

\frac{1}{b} = \frac{1}{a} + \frac{1}{c}

Show answer & solution

Option C

\frac{2}{b} = \frac{1}{a} + \frac{1}{c}

Approach

The coefficients sum to zero, so $x = 1$ is always a root; equal roots then force the other root to be $1$ too.

Sum and product of roots (Vieta)

Step-by-step solution

1
$a (b - c) + b (c - a) + c (a - b) = ab - a c + b c - ab + a c - b c = 0$
Why: The sum of the three coefficients is identically zero, whatever $a, b, c$ are.
2
so $x = 1$ is a root of the equation
Why: Substituting $x = 1$ gives exactly the sum of the coefficients, which is $0$ ; equivalently $(x - 1)$ is a factor.
Factor and remainder theorem
3
equal roots $\Rightarrow$ both roots are $1 \Rightarrow$ product of roots $= 1$
Why: If the two roots are equal and one of them is $1$ , the other is $1$ as well.
Sum and product of roots (Vieta)
4
$\frac{c ( a - b )}{a ( b - c )} = 1 \Rightarrow c a - c b = ab - a c \Rightarrow 2 a c = ab + b c$
Why: The product of the roots of $A x^{2} + B x + C = 0$ is $\frac{C}{A}$ .
5
divide by $ab c$ : $\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$
Why: $\frac{2 a c}{ab c} = \frac{2}{b}$ , $\frac{ab}{ab c} = \frac{1}{c}$ and $\frac{b c}{ab c} = \frac{1}{a}$ ; so $a, b, c$ are in harmonic progression.

Shortcut

Spot that the coefficients add to $0$ : the root $x = 1$ is free, and 'equal roots' then says the product of the roots is $1$ .

Common mistakes

Option A

Took $a, b, c$ to be in arithmetic progression instead of harmonic.

Option B

Took $a, b, c$ to be in geometric progression instead of harmonic.

Option D

Lost the factor $2$ when dividing $2 a c = ab + b c$ through by $ab c$ .

Skills used

Sum and product of roots (Vieta)Factor and remainder theorem

23Algebra • Equations

HardIdeal time: about 2.5 min

The non-zero solution of the equation $\frac{a - x ^{2}}{b x} - \frac{b - x}{c} = \frac{c - x}{b} - \frac{b - x ^{2}}{c x}$ , where $b \neq = 0, c \neq = 0$ is

\frac{b ^{2} + a c}{b ^{2} + c ^{2}}

\frac{b ^{2} - a c}{b ^{2} - c ^{2}}

\frac{b ^{2} - a c}{b ^{2} + c ^{2}}

\frac{b ^{2} + a c}{b ^{2} - c ^{2}}

Show answer & solution

Option A

\frac{b ^{2} + a c}{b ^{2} + c ^{2}}

Approach

Group the two terms carrying $x^{2}$ on one side, clear denominators, and watch the $x^{2}$ terms cancel.

Clear the denominators of an equation

Step-by-step solution

1
$\frac{a - x ^{2}}{b x} + \frac{b - x ^{2}}{c x} = \frac{c - x}{b} + \frac{b - x}{c}$
Why: Move the two fractions with $x^{2}$ in the numerator to the left and the two linear ones to the right; this is the move that makes the cancellation visible.
2
$\frac{c ( a - x ^{2} ) + b ( b - x ^{2} )}{b c x} = \frac{c ( c - x ) + b ( b - x )}{b c}$
Why: Combine each side over its common denominator, $b c x$ on the left and $b c$ on the right.
Comparing two fractions Clear the denominators of an equation
3
$\frac{a c + b ^{2} - x ^{2} ( b + c )}{b c x} = \frac{b ^{2} + c ^{2} - x ( b + c )}{b c}$
Why: Collect like terms in each numerator.
4
$a c + b^{2} - x^{2} (b + c) = x (b^{2} + c^{2}) - x^{2} (b + c)$
Why: Multiply both sides by $b c x$ , which is legitimate since $b, c \neq = 0$ and the non-zero solution has $x \neq = 0$ .
5
$b^{2} + a c = x (b^{2} + c^{2}) \Rightarrow x = \frac{b ^{2} + a c}{b ^{2} + c ^{2}}$
Why: The $x^{2} (b + c)$ terms are identical on both sides and cancel, leaving a linear equation.

Shortcut

Pair the fractions so that both $x^{2}$ terms land on the same side: they then cancel outright and only a linear equation is left.

Common mistakes

Option B

Sign slips in both the numerator and the denominator when collecting terms.

Option C

Sign slip in the numerator only, giving $b^{2} - a c$ .

Option D

Sign slip in the denominator only, giving $b^{2} - c^{2}$ .

Skills used

Clear the denominators of an equation Comparing two fractions

24Algebra • Quadratic Equations • Roots

HardIdeal time: about 2 min

If $k$ is an integer, then $x^{2} + 7 x - 14 (k^{2} - \frac{7}{8}) = 0$ has

ABoth integral rootsBAt least one integral rootCNo integral rootDBoth positive integral roots

Show answer & solution

Option C

No integral root

Approach

Compute the discriminant; it is $56 k^{2}$ , whose square root is irrational for every non-zero integer $k$ .

Quadratic formula

Step-by-step solution

1
$x^{2} + 7 x - 14 (k^{2} - \frac{7}{8}) = 0 \Rightarrow x^{2} + 7 x + (\frac{49}{4} - 14 k^{2}) = 0$
Why: Expand the bracket: $- 14 k^{2} + 14 \times \frac{7}{8} = - 14 k^{2} + \frac{49}{4}$ .
2
$D = 7^{2} - 4 (\frac{49}{4} - 14 k^{2}) = 49 - 49 + 56 k^{2} = 56 k^{2}$
Why: The discriminant of $x^{2} + 7 x + C$ is $49 - 4 C$ , and the constant $49$ cancels exactly.
Quadratic formula
3
$D = 56 k^{2} = 2∣ k ∣ 14$
Why: $56 = 4 \times 14$ , so the square root splits as $2∣ k ∣ 14$ .
$ab = a b$ and $a / b = a / b$
4
$x = \frac{- 7 \pm 2∣ k ∣ 14}{2}$
Why: Apply the quadratic formula with $a = 1$ , $b = 7$ .
5
$k = 0$ gives $x = - \frac{7}{2}$ ; $k \neq = 0$ gives an irrational $2∣ k ∣ 14$ , so $x$ is never an integer
Why: $14$ is irrational, so $- 7 \pm 2∣ k ∣ 14$ is irrational for $k \neq = 0$ ; and for $k = 0$ the repeated root $- \frac{7}{2}$ is not an integer either.

Shortcut

Multiply through by $4$ : $(2 x + 7)^{2} = 56 k^{2}$ , so $2 x + 7 = \pm 2 k 14$ - irrational unless $k = 0$ , and $k = 0$ gives $2 x + 7 = 0$ .

Common mistakes

Option A

Assumed the tidy-looking coefficients must give integer roots without testing the discriminant.

Option B

Noticed that the roots straddle an integer for small $k$ and concluded one of them lands on it.

Option D

No route reaches two positive integral roots; the sum of the roots is $- 7$ , so they cannot both be positive.

Skills used

Quadratic formula

ab = a b

and

a / b = a / b

25Arithmetic • Number System • Divisibility

EasyIdeal time: about 1 min

How many numbers between $500$ and $1000$ are divisible by $13$ ?

36

37

38

39

Show answer & solution

Option C

38

Approach

Bracket the multiples of $13$ by $500$ and $1000$ and count the admissible multipliers.

Enumerate a residue class within a bound

Step-by-step solution

1
$13 \times 38 = 494$ and $13 \times 39 = 507$ , so the first multiple above $500$ is $507$
Why: The multiples of $13$ form an arithmetic progression of common difference $13$ ; dividing $500$ by $13$ locates its start.
Enumerate a residue class within a bound
2
$13 \times 76 = 988$ and $13 \times 77 = 1001$ , so the last multiple below $1000$ is $988$
Why: Dividing $1000$ by $13$ gives $76.9$ , so $76$ is the largest admissible multiplier.
3
count $= 76 - 39 + 1 = 38$
Why: The multipliers run over the consecutive integers from $39$ to $76$ inclusive.

Shortcut

$⌊ \frac{999}{13} ⌋ - ⌊ \frac{500}{13} ⌋ = 76 - 38 = 38$ .

Common mistakes

Option A

Started the count at $13 \times 40 = 520$ , overlooking $507$ , and counted $76 - 40 = 36$ .

Option B

Counted $76 - 39 = 37$ , forgetting to add $1$ for an inclusive range.

Option D

Ran the count up to $13 \times 77 = 1001$ , which is above $1000$ : $77 - 39 + 1 = 39$ .

Skills used

Enumerate a residue class within a bound

26Arithmetic • Ratio and Proportion

EasyIdeal time: about 1 min

To maintain $8$ cows for $60$ days, a milkman has to spend Rs. $6, 400$ . To maintain $5$ cows for $n$ days, he has to spend Rs. $4, 800$ . What is the value of $n$ ?

46

daysB

50

daysC

58

daysD

72

days

Show answer & solution

Option D

72

days

Approach

Find the cost of maintaining one cow for one day, then scale it to the second situation.

Unitary method: find one unit, then scale

Step-by-step solution

1
cost per cow-day $= \frac{6400}{8 \times 60} = \frac{6400}{480} = \frac{40}{3}$ rupees
Why: Eight cows for sixty days is $480$ cow-days; dividing the total cost by that gives the rate for one unit.
Unitary method: find one unit, then scale
2
$5 \times n$ cow-days cost $5 n \times \frac{40}{3} = 4800$
Why: Scale the unit rate up to the second situation, whose cost is given.
3
$n = \frac{4800 \times 3}{5 \times 40} = \frac{14400}{200} = 72$
Why: Solve the linear equation for $n$ .

Shortcut

Cost is proportional to cow-days, so $\frac{6400}{480} = \frac{4800}{5 n}$ gives $5 n = \frac{4800 \times 480}{6400} = 360$ and $n = 72$ .

Common mistakes

Option A

No route reaches $46$ days.

Option B

Scaled only the cost, $\frac{4800}{6400} \times 60 = 45$ , then adjusted loosely to $50$ without the cow count.

Option C

No route reaches $58$ days.

Skills used

Unitary method: find one unit, then scale

27Arithmetic • Percentage

Very easyIdeal time: about 40 s

A student has to secure $40%$ of marks to pass an examination. He gets only $45$ marks and fails by $5$ marks. The maximum marks are

120

125

130

150

Show answer & solution

Option B

125

Approach

The pass mark is his score plus his shortfall; it is also $40%$ of the maximum.

x %

of a quantity

Step-by-step solution

1
pass mark $= 45 + 5 = 50$
Why: Failing by $5$ marks means he was $5$ marks short of the pass mark.
2
$\frac{40}{100} \times M = 50$
Why: The pass mark is $40%$ of the maximum marks $M$ .
$x %$ of a quantity
3
$M = \frac{50 \times 100}{40} = 125$
Why: Solve the linear equation for the maximum.
Turn a worded condition into a linear equation in one unknown

Shortcut

$5$ marks is the gap to $40%$ , so $50$ marks is $40%$ and $M = \frac{50}{0.4} = 125$ .

Common mistakes

Option A

Divided $50$ by $0.4$ as $50 \times 2.4$ instead of $50 \times 2.5$ .

Option C

No route reaches $130$ .

Option D

No route reaches $150$ .

Skills used

x %

of a quantity Turn a worded condition into a linear equation in one unknown

28Algebra • Equations • Simultaneous Equations

ModerateIdeal time: about 1.5 min

What is the value of $u$ in the system of equations $3 (2 u + v) = 7 uv, 3 (u + 3 v) = 11 uv$ ?

0

\frac{1}{4}

\frac{1}{2}

1

Show answer & solution

Option D

1

Approach

Divide each equation by $uv$ to turn the pair into linear equations in $\frac{1}{u}$ and $\frac{1}{v}$ .

Linearise a system by substituting reciprocals

Step-by-step solution

1
$\frac{3 ( 2 u + v )}{uv} = 7 \Rightarrow \frac{6}{v} + \frac{3}{u} = 7$
Why: Dividing by $uv$ splits each numerator term into a reciprocal; this is legitimate for the non-degenerate solution, where $u \neq = 0$ and $v \neq = 0$ .
Negative exponent is a reciprocal Linearise a system by substituting reciprocals
2
$\frac{3 ( u + 3 v )}{uv} = 11 \Rightarrow \frac{3}{v} + \frac{9}{u} = 11$
Why: The same division applied to the second equation.
3
put $p = \frac{1}{u}$ , $q = \frac{1}{v}$ : $3 p + 6 q = 7$ and $9 p + 3 q = 11$
Why: The substitution makes the system linear in $p$ and $q$ .
4
$3 p = 7 - 6 q \Rightarrow 9 p = 3 (7 - 6 q)$ , so $9 p + 3 q = 11$ becomes $3 (7 - 6 q) + 3 q = 11$
Why: Eliminate $p$ by substitution, using $9 p = 3 \times 3 p$ .
5
$21 - 18 q + 3 q = 11 \Rightarrow - 15 q = - 10 \Rightarrow q = \frac{2}{3}$ , so $3 p = 7 - 4 = 3$ , $p = 1$ and $u = \frac{1}{p} = 1$
Why: Solve for $q$ , back-substitute for $p$ , and invert to recover $u$ .

Shortcut

Divide both equations by $uv$ at sight: they become $3 p + 6 q = 7$ and $9 p + 3 q = 11$ in the reciprocals, and eliminating $q$ gives $p = 1$ .

Common mistakes

Option A

Took the degenerate solution $u = v = 0$ , which satisfies both equations trivially but is not the intended non-zero one.

Option B

Solved for $q = \frac{2}{3}$ and then inverted the wrong unknown, or halved $p$ twice.

Option C

Reported $\frac{1}{2}$ , halving $p = 1$ or confusing $u$ with $\frac{1}{v}$ 's partner.

Skills used

Linearise a system by substituting reciprocals Negative exponent is a reciprocal

29Arithmetic • Age Problems

EasyIdeal time: about 1 min

Five years ago, Ram was three times as old as Shyam. Four years from now, Ram will be only twice as old as Shyam. What is the present age of Ram ?

30

yearsB

32

yearsC

36

yearsD

40

years

Show answer & solution

Option B

32

years

Approach

Let Shyam's present age be the single unknown and write Ram's age twice, once from each condition.

Turn a worded condition into a linear equation in one unknown

Step-by-step solution

1
let Shyam be $S$ now; five years ago he was $S - 5$ and Ram was $3 (S - 5)$
Why: Give the unknown a letter and express every other quantity in terms of it.
Turn a worded condition into a linear equation in one unknown
2
Ram now $= 3 (S - 5) + 5 = 3 S - 10$
Why: Ram's present age is his age five years ago plus five.
3
four years from now: $3 S - 10 + 4 = 2 (S + 4)$
Why: The second condition says Ram's age then is twice Shyam's age then.
4
$3 S - 6 = 2 S + 8 \Rightarrow S = 14$
Why: Collect like terms and solve the linear equation.
5
Ram $= 3 (14) - 10 = 32$
Why: Substitute back to get the quantity actually asked for.

Common mistakes

Option A

Reported an age that satisfies neither condition; $30$ for Ram would need Shyam at $13. 3$ five years ago.

Option C

Answered with Ram's age four years from now, $36$ , rather than his present age.

Option D

No route reaches $40$ .

Skills used

Turn a worded condition into a linear equation in one unknown

30Arithmetic • Profit and Loss

EasyIdeal time: about 1 min

Ram buys $4$ chairs and $9$ stools for Rs. $1, 340$ . If he sells chairs at $10%$ profit and stools at $20%$ profit, he earns a total profit of Rs. $188$ . How much money did he have to pay for the chairs ?

ARs.

200

BRs.

400

CRs.

800

DRs.

1, 600

Show answer & solution

Option C

Rs.

800

Approach

Work with the total cost of the chairs and of the stools, not with single-item prices.

Turn a worded condition into a linear equation in one unknown

Step-by-step solution

1
let $C$ and $S$ be the total cost of the $4$ chairs and the $9$ stools: $C + S = 1340$
Why: Only the totals are constrained, and only a total is asked for, so individual prices never need to be found.
Turn a worded condition into a linear equation in one unknown
2
$0.10 C + 0.20 S = 188$
Why: Profit is a percentage OF THE COST PRICE, so the two profits are $10%$ of $C$ and $20%$ of $S$ .
Gain percent is measured on the cost price
3
$0.10 C + 0.20 (1340 - C) = 188$
Why: Substitute $S = 1340 - C$ to leave one unknown.
4
$0.10 C + 268 - 0.20 C = 188 \Rightarrow - 0.10 C = - 80 \Rightarrow C = 800$
Why: Collect like terms and solve.

Shortcut

If everything earned $10%$ the profit would be $134$ ; the extra $54$ comes from the stools at an extra $10%$ , so $S = 540$ and $C = 1340 - 540 = 800$ .

Common mistakes

Option A

Divided the chairs' total by $4$ and gave the price of one chair.

Option B

No route reaches Rs. $400$ .

Option D

Doubled the chairs' total, or took the stools' share as the complement of $800$ in a mis-set equation.

Skills used

Turn a worded condition into a linear equation in one unknown Gain percent is measured on the cost price

31Algebra • Sets • Types of Sets

EasyIdeal time: about 45 s

Which one of the following is a correct statement ?

{x : x + 5 = 5} = ϕ

{x : x + 5 = 5} = {0}

{x : x + 5 = 5} = 0

{x : x + 5 = 5} = {ϕ}

Show answer & solution

Option B

{x : x + 5 = 5} = {0}

Approach

Solve the defining equation $x + 5 = 5$ , then write the solution set in correct set notation.

A set is not its element, and

{ϕ}

is not

ϕ

Step-by-step solution

1
$x + 5 = 5 \Rightarrow x = 0$
Why: The set is given in set-builder form, so its members are exactly the solutions of the stated condition.
Turn a worded condition into a linear equation in one unknown
2
${x : x + 5 = 5} = {0}$
Why: There is exactly one solution, $x = 0$ , so the set has exactly one member and is written with braces around that member.
A set is not its element, and ${ϕ}$ is not $ϕ$

Common mistakes

Option A

Read ' $x + 5 = 5$ ' as imposing no condition that any number meets, or confused the solution $x = 0$ with 'no element', and called the set empty.

Option C

Dropped the braces and equated a set to a number: ${x : x + 5 = 5}$ is a set, and $0$ is an element, so the two can never be equal.

Option D

Identified the number $0$ with the empty set $ϕ$ , giving the set whose only member is $ϕ$ instead of the set whose only member is $0$ .

Skills used

A set is not its element, and

{ϕ}

is not

ϕ

Turn a worded condition into a linear equation in one unknown

32Algebra • Expressions • Simplification

ModerateIdeal time: about 1.5 min

If $ab + b c + c a = 0$ , then what is the value of $\frac{a ^{2}}{a ^{2} - b c} + \frac{b ^{2}}{b ^{2} - c a} + \frac{c ^{2}}{c ^{2} - ab}$ ?

3

0

1

- 1

Show answer & solution

Option C

1

Approach

Use $ab + b c + c a = 0$ to rewrite each denominator as a multiple of $a + b + c$ , then add the three fractions.

Complete each equation to the common sum

Step-by-step solution

1
$ab + b c + c a = 0 \Rightarrow b c = - ab - c a = - a (b + c)$
Why: The constraint lets one product be traded for the other two, which is what makes the denominators factorise.
2
$a^{2} - b c = a^{2} + a (b + c) = a (a + b + c)$
Why: Substituting for $b c$ makes the full symmetric sum $a + b + c$ appear as a factor of the denominator.
Complete each equation to the common sum
3
$b^{2} - c a = b (a + b + c)$ and $c^{2} - ab = c (a + b + c)$
Why: The constraint is symmetric in $a, b, c$ , so the same substitution works on the other two denominators.
4
$\frac{a ^{2}}{a ( a + b + c )} + \frac{b ^{2}}{b ( a + b + c )} + \frac{c ^{2}}{c ( a + b + c )} = \frac{a + b + c}{a + b + c} = 1$
Why: Each fraction reduces to one term over the common factor $a + b + c$ , and the three numerators then add to that same factor.

Shortcut

Each term collapses to $\frac{a}{a + b + c}$ , and the three such fractions obviously add to $1$ .

Common mistakes

Option A

Assumed each fraction equals $1$ because its numerator and denominator look alike, and added three ones.

Option B

Took the constraint $ab + b c + c a = 0$ to make the whole expression vanish, by analogy with expressions that genuinely do collapse to $0$ under it.

Option D

Sign slip in the substitution, writing $a^{2} - b c = a^{2} - a (b + c) = a (a - b - c)$ , which makes the three fractions sum to $- 1$ .

Skills used

Complete each equation to the common sum

33Algebra • Sets • Operations

EasyIdeal time: about 1 min

In an examination, $35%$ students failed in Hindi, $45%$ students failed in English and $20%$ students failed in both the subjects. What is the percentage of students passing in both the subjects ?

0

20

30

40

Show answer & solution

Option D

40

Approach

Count the percentage failing at least one subject by inclusion-exclusion, then take the complement.

Inclusion-exclusion for two sets

Step-by-step solution

1
Take $100$ students. Failed Hindi $= 35$ , failed English $= 45$ , failed both $= 20$ .
Why: Reading percentages as counts out of $100$ turns the problem into plain counting.
$x %$ of a quantity
2
Failed at least one $= 35 + 45 - 20 = 60$
Why: The $20$ who failed both are counted once in each of $35$ and $45$ , so one copy must be removed.
Inclusion-exclusion for two sets
3
Passed both $= 100 - 60 = 40$
Why: Passing both subjects is exactly failing neither, which is the complement of failing at least one.

Shortcut

Failed Hindi only $= 15$ , failed English only $= 25$ , failed both $= 20$ ; these $60$ are all the failures, leaving $40%$ .

Common mistakes

Option A

Added all three figures, $35 + 45 + 20 = 100$ , treating the $20%$ who failed both as a fourth group instead of an overlap already inside the other two.

Option B

Computed $100 - 35 - 45 = 20$ , subtracting the two failure percentages without adding back the $20%$ counted twice.

Option C

No natural route reaches $30$ .

Skills used

Inclusion-exclusion for two sets

x %

of a quantity

34Algebra • Expressions • Simplification

ModerateIdeal time: about 1.5 min

What is $\frac{( x - y ) ( y - z ) ( z - x )}{( x - y ) ^{3} + ( y - z ) ^{3} + ( z - x ) ^{3}}$ equal to ?

- \frac{1}{3}

\frac{1}{3}

3

- 3

Show answer & solution

Option B

\frac{1}{3}

Approach

Name the three brackets $a, b, c$ , note that they sum to zero, and use $a^{3} + b^{3} + c^{3} = 3 ab c$ .

a^{3} + b^{3} + c^{3} - 3 ab c

, and the case

a + b + c = 0

Step-by-step solution

1
$a = x - y, b = y - z, c = z - x$
Why: Naming the brackets exposes the structure; the expression is then $\frac{ab c}{a ^{3} + b ^{3} + c ^{3}}$ .
2
$a + b + c = (x - y) + (y - z) + (z - x) = 0$
Why: Every variable appears once positively and once negatively, so the three differences cancel identically.
3
$a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 ab c$
Why: The identity $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ has a zero first factor here.
Sum and difference of two cubes $a^{3} + b^{3} + c^{3} - 3 ab c$ , and the case $a + b + c = 0$
4
$\frac{ab c}{3 ab c} = \frac{1}{3}$
Why: The common factor $ab c$ cancels, which is why the answer is a pure number independent of $x, y, z$ .

Shortcut

Recognise at sight that the three cubed brackets sum to zero, so the cube sum is three times the product.

Common mistakes

Option A

Sign slip in the identity, using $a^{3} + b^{3} + c^{3} = - 3 ab c$ .

Option C

Inverted the fraction, reporting $\frac{a ^{3} + b ^{3} + c ^{3}}{ab c} = 3$ instead of its reciprocal.

Option D

Inverted the fraction and slipped the sign as well.

Skills used

a^{3} + b^{3} + c^{3} - 3 ab c

, and the case

a + b + c = 0

Sum and difference of two cubes

35Algebra • Surds • Nested Radicals

ModerateIdeal time: about 1 min

The value of $1 + 1 + 1 + \dots$

AEquals to

1

BLies between

0

and

1

CLies between

1

and

2

DIs greater than

2

Show answer & solution

Option C

Lies between

1

and

2

Approach

Call the value $x$ , exploit the self-similarity of the infinite radical to get $x = 1 + x$ , and solve.

Value of an infinite nested radical

Step-by-step solution

1
$x = 1 + 1 + 1 + \dots$ , so $x = 1 + x$
Why: Everything under the outermost radical after the $1$ is the same infinite expression again, which is what turns an infinite process into one equation.
Value of an infinite nested radical
2
$x^{2} = 1 + x \Rightarrow x^{2} - x - 1 = 0$
Why: Squaring is legitimate because $x \geq 0$ by the principal-root convention, so no spurious sign is introduced.
denotes the non-negative root
3
$x = \frac{1 + 5}{2}$
Why: The quadratic formula gives $\frac{1 \pm 5}{2}$ , and the negative root is rejected because $x \geq 0$ .
Quadratic formula
4
$5 \approx 2.236 \Rightarrow x \approx 1.618$ , so $1 < x < 2$
Why: The value is placed against the options only after it has been found.

Shortcut

$x = 1 + x$ gives $x^{2} = x + 1$ ; since $1^{2} < 1 + 1$ and $2^{2} > 2 + 1$ , the root lies strictly between $1$ and $2$ with no need to solve.

Common mistakes

Option A

Stopped at the innermost $1$ and assumed the tower of square roots of $1$ leaves the value at $1$ .

Option B

Reasoned that repeated square-rooting shrinks a number, forgetting that a $1$ is added at every level so the argument never drops below $1$ .

Option D

Assumed the infinitely repeated addition makes the value grow without bound, ignoring that each addition is followed by a square root.

Skills used

Value of an infinite nested radical

denotes the non-negative root Quadratic formula

36Algebra • Logarithm • Laws of Logarithms

EasyIdeal time: about 1 min

If $lo g_{10} 6 = 0 \cdot 7782$ and $lo g_{10} 8 = 0 \cdot 9031$ , then what is the value of $lo g_{10} 8000 + lo g_{10} 600$ ?

4 \cdot 6813

5 \cdot 5813

1 \cdot 5813

6 \cdot 6813

Show answer & solution

Option D

6 \cdot 6813

Approach

Split each logarithm into a given value plus a power of ten, then add.

lo g (ab) = lo g a + lo g b

and

lo g \frac{a}{b} = lo g a - lo g b

Step-by-step solution

1
$lo g_{10} 8000 = lo g_{10} (8 \times 1 0^{3}) = lo g_{10} 8 + 3$
Why: The product rule separates the given value from the power of ten, whose logarithm is its exponent.
$lo g (ab) = lo g a + lo g b$ and $lo g \frac{a}{b} = lo g a - lo g b$ Logarithm to exponential form
2
$lo g_{10} 600 = lo g_{10} (6 \times 1 0^{2}) = lo g_{10} 6 + 2$
Why: The same split applies to the second term.
3
$(0 \cdot 9031 + 3) + (0 \cdot 7782 + 2) = 3 \cdot 9031 + 2 \cdot 7782$
Why: Substitute the two given mantissas.
4
$3 \cdot 9031 + 2 \cdot 7782 = 6 \cdot 6813$
Why: Add.

Shortcut

Add the mantissas first: $0 \cdot 9031 + 0 \cdot 7782 = 1 \cdot 6813$ , then add the characteristics $3 + 2 = 5$ , giving $6 \cdot 6813$ .

Common mistakes

Option A

Split $8000$ correctly but treated $600$ as $6$ , adding $3 \cdot 9031 + 0 \cdot 7782$ .

Option B

Mis-added the characteristics or the mantissas at the last step.

Option C

Dropped both powers of ten and added only the two given mantissas.

Skills used

lo g (ab) = lo g a + lo g b

and

lo g \frac{a}{b} = lo g a - lo g b

Logarithm to exponential form

37Arithmetic • Time and Work

EasyIdeal time: about 1.5 min

$30$ men can complete a job in $40$ days. However, after $24$ days some men out of the assigned $30$ left the job. The remaining people took another $40$ days to complete the job. The number of men who left the job is

24

18

12

6

Show answer & solution

Option B

18

Approach

Measure the job in man-days, subtract what the full team did in $24$ days, and see how many men the remaining man-days imply over $40$ days.

The work-equivalence chain rule

M_{1} D_{1} H_{1} = M_{2} D_{2} H_{2}

Step-by-step solution

1
Total work $= 30 \times 40 = 1200$ man-days
Why: With men of equal efficiency the product (men) $\times$ (days) is an invariant measure of the job.
The work-equivalence chain rule $M_{1} D_{1} H_{1} = M_{2} D_{2} H_{2}$ Take total work as the LCM of the times
2
Done in the first $24$ days $= 30 \times 24 = 720$ man-days
Why: All $30$ men were still on the job for that stretch.
3
Remaining $= 1200 - 720 = 480$ man-days
Why: What is left of the job, in the same units.
4
$m \times 40 = 480 \Rightarrow m = 12$
Why: The men who stayed did the remaining $480$ man-days over $40$ days.
5
Men who left $= 30 - 12 = 18$
Why: The question asks how many left, not how many stayed.

Shortcut

The last $\frac{480}{1200} = \frac{2}{5}$ of the job took $40$ days instead of the $16$ it should have, so the workforce fell to $\frac{16}{40} = \frac{2}{5}$ of $30$ , i.e. $12$ men, and $18$ left.

Common mistakes

Option A

Lifted $24$ straight from the question, treating the number of days already worked as the number of men who left.

Option C

Found the $12$ men who remained and stopped there, without subtracting from $30$ .

Option D

Computed $30 - 24$ , subtracting a number of days from a number of men.

Skills used

The work-equivalence chain rule

M_{1} D_{1} H_{1} = M_{2} D_{2} H_{2}

Take total work as the LCM of the times

38Arithmetic • Time and Work

EasyIdeal time: about 1 min

$4$ goats or $6$ sheep can graze a field in $50$ days. $2$ goats and $3$ sheep will graze it in

200

daysB

150

daysC

100

daysD

50

days

Show answer & solution

Option D

50

days

Approach

Convert each herd into a daily grazing rate, halve both, and add.

Take total work as the LCM of the times

Step-by-step solution

1
$4$ goats: rate $= \frac{1}{50}$ field/day; $6$ sheep: rate $= \frac{1}{50}$ field/day
Why: Each group finishes the field in $50$ days, so each group's rate is one field per $50$ days.
Take total work as the LCM of the times
2
$1$ goat $= \frac{1}{200}$ , $1$ sheep $= \frac{1}{300}$ field/day
Why: Animals within a kind are identical, so a group's rate divides evenly among its members.
Unitary method: find one unit, then scale
3
$2$ goats $+ 3$ sheep $= \frac{2}{200} + \frac{3}{300} = \frac{1}{100} + \frac{1}{100} = \frac{1}{50}$
Why: Rates of animals grazing together add.
4
Time $= \frac{1}{1/50} = 50$ days
Why: Time is the reciprocal of the rate when the job is one whole field.

Shortcut

$2$ goats are half of the $4$ goats and $3$ sheep are half of the $6$ sheep, so together they are one whole equivalent group: $50$ days.

Common mistakes

Option A

Reported one goat's $200$ days instead of the combined herd's time.

Option B

No natural route reaches $150$ days.

Option C

Halved only one group: $2$ goats alone would take $100$ days, and the $3$ sheep were forgotten.

Skills used

Take total work as the LCM of the times Unitary method: find one unit, then scale

39Arithmetic • Time and Work • Pipes and Cisterns

ModerateIdeal time: about 1.5 min

A tap can fill a tub in $10$ hours. After opening the tap for $5$ hours it was found that a small outlet at the bottom of the tub was open and water was leaking through it. It was then immediately closed. It took $7$ hours to fill the tub after closing the outlet. What time will be taken by the outlet to empty the full tub of water ?

35

hoursB

25

hoursC

20

hoursD

17

hours

Show answer & solution

Option B

25

hours

Approach

Work out how much the tap alone filled in the final $7$ hours, deduce how little got in during the leaking $5$ hours, and turn the shortfall into the leak's rate.

Take total work as the LCM of the times

Step-by-step solution

1
Tap rate $= \frac{1}{10}$ tub/hour
Why: It fills the whole tub in $10$ hours.
Take total work as the LCM of the times
2
Last $7$ hours (outlet closed): $7 \times \frac{1}{10} = \frac{7}{10}$ filled
Why: After the outlet was closed only the tap was acting.
3
First $5$ hours therefore held $1 - \frac{7}{10} = \frac{3}{10}$
Why: The tub ended up full, so the two stretches must account for the whole tub between them.
4
Tap alone would have put in $\frac{5}{10}$ , so the leak removed $\frac{5}{10} - \frac{3}{10} = \frac{1}{5}$ in $5$ hours
Why: The shortfall over those $5$ hours is exactly what the outlet drained.
5
Leak rate $= \frac{1/5}{5} = \frac{1}{25}$ tub/hour $\Rightarrow 25$ hours to empty a full tub
Why: Time to empty is the reciprocal of the emptying rate.

Shortcut

The job took $12$ hours instead of $10$ , so the leak wasted $\frac{2}{10}$ of a tub while it was open for $5$ hours: $\frac{1}{25}$ per hour, i.e. $25$ hours.

Common mistakes

Option A

No natural route reaches $35$ hours.

Option C

Took the shortfall over the leaking period as $\frac{1}{4}$ of a tub rather than $\frac{1}{5}$ , giving a leak rate of $\frac{1}{20}$ .

Option D

Added the two stretches to the fill time, $10 + 7$ , instead of working with rates.

Skills used

Take total work as the LCM of the times

40Arithmetic • Time, Speed and Distance • Average Speed

EasyIdeal time: about 1 min

A boy went to his school at a speed of $12$ km/hr and returned to his house at a speed of $8$ km/hr. If he has taken $50$ minutes for the whole journey, what was the total distance walked ?

4

kmB

8

kmC

16

kmD

20

Show answer & solution

Option B

8

Approach

Let the one-way distance be $d$ , add the two leg times, set the total to $\frac{5}{6}$ hour, and then double $d$ .

Speed, distance and time

Step-by-step solution

1
$50$ minutes $= \frac{50}{60} = \frac{5}{6}$ hour
Why: The speeds are in km/hr, so the time must be in hours before it is used.
Speed, distance and time
2
$\frac{d}{12} + \frac{d}{8} = \frac{5}{6}$
Why: Time is distance over speed on each leg, and the two legs cover the same distance $d$ .
3
$d (\frac{2 + 3}{24}) = \frac{5}{6} \Rightarrow \frac{5 d}{24} = \frac{5}{6} \Rightarrow d = 4$
Why: Combine over the common denominator $24$ and solve.
4
Total distance walked $= 2 d = 8$ km
Why: The question asks for the whole journey, which is there and back.

Shortcut

The average speed over equal distances is $\frac{2 \times 12 \times 8}{12 + 8} = 9 \cdot 6$ km/hr, and $9 \cdot 6 \times \frac{5}{6} = 8$ km.

Common mistakes

Option A

Solved for the one-way distance $d = 4$ km and stopped, without doubling for the return leg.

Option C

No natural route reaches $16$ km.

Option D

Added the two speeds, $12 + 8 = 20$ , and reported that as a distance.

Skills used

Speed, distance and time