Elementary Mathematics 2016 (II)

Set A

Questions 1–20 of 100

1Algebra • Logarithm • Equations

ModerateIdeal time: about 2 min

What is the solution of $lo g_{10} [1 - {1 - (1 - x^{2})^{- 1}}^{- 1}]^{- \frac{1}{2}} = 1$ ?

x = 100

x = 10

x = 1

x = 0

Show answer & solution

Option B

x = 10

Approach

Unwind the nested reciprocals from the inside out until the bracket collapses to $x^{- 2}$ , then read the logarithm as an exponent.

Negative exponent is a reciprocal

Step-by-step solution

1
$(1 - x^{2})^{- 1} = \frac{1}{1 - x ^{2}}$
Why: A negative index is a reciprocal, so the innermost bracket is one fraction.
Negative exponent is a reciprocal
2
$1 - \frac{1}{1 - x ^{2}} = \frac{( 1 - x ^{2} ) - 1}{1 - x ^{2}} = \frac{- x ^{2}}{1 - x ^{2}}$
Why: Combine over the common denominator $1 - x^{2}$ before taking the next reciprocal.
3
$(\frac{- x ^{2}}{1 - x ^{2}})^{- 1} = \frac{1 - x ^{2}}{- x ^{2}} = \frac{x ^{2} - 1}{x ^{2}}$
Why: Invert the fraction; the two minus signs cancel.
4
$1 - \frac{x ^{2} - 1}{x ^{2}} = \frac{x ^{2} - ( x ^{2} - 1 )}{x ^{2}} = \frac{1}{x ^{2}}$
Why: The $x^{2}$ terms cancel, so the whole bracket is just $x^{- 2}$ .
5
$(\frac{1}{x ^{2}})^{- \frac{1}{2}} = (x^{2})^{\frac{1}{2}} = ∣ x ∣$
Why: The outer index inverts and halves; the square root of $x^{2}$ is $∣ x ∣$ , not $x$ .
denotes the non-negative root
6
$lo g_{10} ∣ x ∣ = 1 \Rightarrow ∣ x ∣ = 1 0^{1} \Rightarrow x = 10$
Why: Rewrite the logarithm in exponential form; only $x = 10$ is offered, and $x = - 10$ is not.
Logarithm to exponential form

Shortcut

The whole bracket is $x^{- 2}$ , so the expression is $∣ x ∣$ and the equation says $lo g_{10} ∣ x ∣ = 1$ .

Common mistakes

Option A

Lost a square while combining, reaching $\frac{1}{x}$ instead of $\frac{1}{x ^{2}}$ , so the outer index gave $x$ and $lo g_{10} x = 1$ returned $x = 100$ .

Option C

Forced the innermost bracket to vanish by solving $1 - x^{2} = 0$ , ignoring that $x = 1$ makes $(1 - x^{2})^{- 1}$ undefined.

Option D

No route reaches $x = 0$ : it makes the second reciprocal $0^{- 1}$ , which is undefined.

Skills used

Negative exponent is a reciprocal

denotes the non-negative root Logarithm to exponential form

2Algebra • Quadratic Equations • Roots

HardIdeal time: about 2 min

If $λ$ is an integer and $α, β$ are the roots of $4 x^{2} - 16 x + \frac{λ}{4} = 0$ such that $1 < α < 2$ and $2 < β < 3$ , then how many values can $λ$ take?

3

9

14

15

Show answer & solution

Option D

15

Approach

Place the roots by testing the sign of the quadratic at $x = 1, 2, 3$ , which bounds $λ$ , then count the integers inside.

Locate roots by the sign of

f

at test points

Step-by-step solution

1
$f (x) = 4 x^{2} - 16 x + \frac{λ}{4}$
Why: Name the quadratic; its leading coefficient $4$ is positive, so $f$ is negative strictly between its roots and positive outside them.
2
$f (1) > 0, f (2) < 0, f (3) > 0$
Why: One root lies in $(1, 2)$ and the other in $(2, 3)$ exactly when $x = 2$ sits between the roots while $x = 1$ and $x = 3$ sit outside.
Locate roots by the sign of $f$ at test points
3
$f (1) = 4 - 16 + \frac{λ}{4} = \frac{λ}{4} - 12 > 0 \Rightarrow λ > 48$
Why: Evaluate at $x = 1$ ; this forces $λ$ above $48$ .
4
$f (2) = 16 - 32 + \frac{λ}{4} = \frac{λ}{4} - 16 < 0 \Rightarrow λ < 64$
Why: Evaluate at $x = 2$ ; this caps $λ$ below $64$ .
5
$f (3) = 36 - 48 + \frac{λ}{4} = \frac{λ}{4} - 12 > 0 \Rightarrow λ > 48$
Why: The third test repeats the first condition, so nothing new is cut off.
6
$48 < λ < 64 \Rightarrow λ \in {49, 50, \dots, 63} \Rightarrow 63 - 49 + 1 = 15$
Why: $λ$ is an integer and both endpoints are strict, so count the integers strictly inside.

Shortcut

For an upward parabola, $f (2) < 0$ with $f (1) > 0$ and $f (3) > 0$ is the whole of the root-location condition; the arithmetic is three substitutions.

Common mistakes

Option A

Counted the integer values of $\frac{λ}{4}$ rather than of $λ$ , getting $13, 14, 15$ .

Option B

No bounding of $λ$ produces nine admissible integers.

Option C

Counted $63 - 49 = 14$ , forgetting that both endpoints of the list are themselves admissible.

Skills used

Locate roots by the sign of

f

at test points

3Algebra • Surds • Simplification

ModerateIdeal time: about 1.5 min

What is $\frac{6 ^{2} + 7 ^{2} + 8 ^{2} + 9 ^{2} + 1 0 ^{2}}{7 + 4 3 - 4 + 2 3}$ equal to?

330

340

355

366

Show answer & solution

Option A

330

Approach

Denest both surds by writing each radicand as a perfect square; the denominator collapses to $1$ .

Denest

a \pm 2 b

Step-by-step solution

1
$6^{2} + 7^{2} + 8^{2} + 9^{2} + 1 0^{2} = 36 + 49 + 64 + 81 + 100 = 330$
Why: The numerator is a plain sum of squares and needs no manipulation.
2
$7 + 43 = 4 + 3 + 2 \cdot 2 \cdot 3 = (2 + 3)^{2}$
Why: Match $a^{2} + b^{2} + 2 ab$ with $a = 2$ , $b = 3$ , so the radicand is a perfect square.
Denest $a \pm 2 b$ Square of a sum and of a difference
3
$4 + 23 = 1 + 3 + 2 \cdot 1 \cdot 3 = (1 + 3)^{2}$
Why: The same match with $a = 1$ , $b = 3$ .
4
$7 + 43 - 4 + 23 = (2 + 3) - (1 + 3) = 1$
Why: Both square roots are positive as written, so each denests to the positive sum, and the $3$ terms cancel.
denotes the non-negative root
5
$\frac{330}{1} = 330$
Why: Divide by the denominator, which is $1$ .

Shortcut

Both radicands are perfect squares of $a + 3$ forms differing by $1$ , so the denominator is $1$ and the answer is just the sum of the squares.

Common mistakes

Option B

Slipped while adding the five squares, for instance carrying $49 + 64$ as $123$ .

Option C

No denesting or summation error naturally produces $355$ .

Option D

No denesting or summation error naturally produces $366$ .

Skills used

Denest

a \pm 2 b

Square of a sum and of a difference

denotes the non-negative root

4Algebra • Equations • Symmetric Equations

ModerateIdeal time: about 1.5 min

If $x^{2} = y + z$ , $y^{2} = z + x$ and $z^{2} = x + y$ , then what is the value of $\frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1}$ ?

- 1

1

2

4

Show answer & solution

Option B

1

Approach

Add the missing variable to each equation so all three share the total $s = x + y + z$ , which turns every denominator $x + 1$ into $s / x$ .

Complete each equation to the common sum

Step-by-step solution

1
$x^{2} + x = x + y + z, y^{2} + y = x + y + z, z^{2} + z = x + y + z$
Why: Adding $x$ to the first equation, $y$ to the second and $z$ to the third completes each right-hand side to the same total.
Complete each equation to the common sum
2
$x (x + 1) = s, y (y + 1) = s, z (z + 1) = s where s = x + y + z$
Why: Factorise each left-hand side; the three products are all equal to the one quantity $s$ .
3
$\frac{1}{x + 1} = \frac{x}{x ( x + 1 )} = \frac{x}{s}$
Why: Multiply numerator and denominator by $x$ and replace $x (x + 1)$ by $s$ ; the same holds for $y$ and $z$ .
4
$\frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} = \frac{x + y + z}{s} = \frac{s}{s} = 1$
Why: The three rewritten fractions share the denominator $s$ , and their numerators sum to $s$ itself.

Shortcut

Each equation says $x (x + 1) = x + y + z$ , so every term of the required sum is $\frac{x}{s}$ and the sum telescopes to $1$ .

Common mistakes

Option A

Added the missing variable with the wrong sign, obtaining $x (x - 1) = - s$ and carrying the minus into the final sum.

Option C

Found $x = y = z = 2$ and reported that common value instead of evaluating the required sum of reciprocals.

Option D

No manipulation of the symmetric system produces $4$ .

Skills used

Complete each equation to the common sum

5Arithmetic • Number System • Digit Problems

EasyIdeal time: about 1 min

What would be the maximum value of $Q$ in the equation $5 P 9 + 3 R 7 + 2 Q 8 = 1114$ ?

9

8

5

4

Show answer & solution

Option A

9

Approach

Expand the three numerals by place value to collapse the puzzle to one equation in the digit sum, then push $P$ and $R$ to their minimum.

Maximise one digit by minimising the others

Step-by-step solution

1
$5 P 9 + 3 R 7 + 2 Q 8 = (509 + 10 P) + (307 + 10 R) + (208 + 10 Q) = 1024 + 10 (P + R + Q)$ .
Why: Only the tens place carries the unknowns, so place-value expansion turns three unknown digits into a single unknown sum.
Place-value expansion of a numeral
2
$1024 + 10 (P + R + Q) = 1114 \Rightarrow 10 (P + R + Q) = 90 \Rightarrow P + R + Q = 9$ .
Why: The digit sum is pinned exactly; no carry analysis is needed once the numerals are expanded.
3
$Q$ is largest when $P$ and $R$ are smallest, and both may be $0$ since neither is a leading digit, so $Q = 9$ , which is a legal digit.
Why: With a fixed digit sum, maximising one digit means minimising the others — and the result must still be at most $9$ , which it is.
Maximise one digit by minimising the others
4
$509 + 307 + 298 = 1114$ .
Why: Exhibiting an actual sum proves $Q = 9$ is attained, not merely permitted.

Shortcut

$509 + 307 + 208 = 1024$ , which is $90$ short of $1114$ ; $90$ spread over three tens digits means $P + R + Q = 9$ , so $Q \leq 9$ .

Common mistakes

Option B

Assumed the other two digits cannot both be $0$ , forcing $P + R \geq 1$ and hence $Q \leq 8$ .

Option C

No plausible route reaches $5$ .

Option D

Read off the units digit $4$ of $1114$ instead of the maximum value of $Q$ .

Skills used

Maximise one digit by minimising the others Place-value expansion of a numeral

6Arithmetic • Number System • Divisibility

ModerateIdeal time: about 1.5 min

$A$ is a set of positive integers such that when divided by $2, 3, 4, 5$ and $6$ leaves the remainder $1, 2, 3, 4$ and $5$ respectively. How many integers between $0$ and $100$ belong to the set $A$ ?

ANo integerBOneCTwoDThree

Show answer & solution

Option B

One

Approach

Every remainder is one short of its divisor, so $N + 1$ is a common multiple of $2, 3, 4, 5, 6$ ; list the resulting $N$ below $100$ .

Prime factorisation

Step-by-step solution

1
$N = 2 a + 1 = 3 b + 2 = 4 c + 3 = 5 d + 4 = 6 e + 5$
Why: Write out what the five remainder conditions say.
2
$N + 1 = 2 (a + 1) = 3 (b + 1) = 4 (c + 1) = 5 (d + 1) = 6 (e + 1)$
Why: Each remainder is exactly one less than its divisor, so adding $1$ makes $N + 1$ divisible by all five divisors at once.
Remainders all short of their divisors by the same amount
3
$LCM (2, 3, 4, 5, 6) = 2^{2} \cdot 3 \cdot 5 = 60$
Why: Take the highest power of each prime appearing in $2 = 2$ , $3 = 3$ , $4 = 2^{2}$ , $5 = 5$ , $6 = 2 \cdot 3$ .
Prime factorisation
4
$N + 1 = 60 k \Rightarrow N = 60 k - 1 = 59, 119, 179, \dots$
Why: The common multiples of $60$ generate every member of $A$ .
5
$0 < N < 100 \Rightarrow N = 59 only$
Why: $119$ already exceeds $100$ , so exactly one member of $A$ lies in the range.

Shortcut

Remainders one less than the divisors means $N = LCM - 1 = 59$ , and the next such number is $119$ .

Common mistakes

Option A

Treated the five congruences as inconsistent without noticing that the remainders are uniformly one short of the divisors.

Option C

Kept $119$ as well, reading 'between $0$ and $100$ ' loosely.

Option D

Computed the LCM as $30$ by ignoring the $4$ , giving $29$ , $59$ and $89$ .

Skills used

Prime factorisation Remainders all short of their divisors by the same amount

7Arithmetic • Number System

EasyIdeal time: about 1 min

In an examination, a student was asked to divide a certain number by $8$ . By mistake he multiplied it by $8$ and got the answer $2016$ more than the correct answer. What was the number?

252

256

258

260

Show answer & solution

Option B

256

Approach

Set the wrong result minus the right result equal to $2016$ and solve the resulting linear equation.

Turn a worded condition into a linear equation in one unknown

Step-by-step solution

1
$8 n - \frac{n}{8} = 2016$
Why: Multiplying gave $8 n$ , the correct operation gives $\frac{n}{8}$ , and the mistake was $2016$ too large.
Turn a worded condition into a linear equation in one unknown
2
$\frac{64 n - n}{8} = 2016 \Rightarrow \frac{63 n}{8} = 2016$
Why: Put the left-hand side over the common denominator $8$ .
3
$63 n = 16128 \Rightarrow n = \frac{16128}{63} = 256$
Why: Clear the denominator and divide; $63 \cdot 256 = 16128$ .

Shortcut

The difference is $\frac{63}{8}$ of the number, and $2016 = \frac{63}{8} \cdot 256$ , so $n = \frac{2016 \cdot 8}{63} = 32 \cdot 8$ .

Common mistakes

Option A

Dropped the $\frac{n}{8}$ term and solved $8 n = 2016$ .

Option C

No reading of the difference condition gives $258$ .

Option D

No reading of the difference condition gives $260$ .

Skills used

Turn a worded condition into a linear equation in one unknown

8Algebra • Quadratic Equations • Roots

ModerateIdeal time: about 1.5 min

In the quadratic equation $x^{2} + a x + b = 0$ , $a$ and $b$ can take any value from the set ${1, 2, 3, 4}$ . How many pairs of values of $a$ and $b$ are possible in order that the quadratic equation has real roots?

6

7

8

16

Show answer & solution

Option B

7

Approach

Real roots require the discriminant $a^{2} - 4 b \geq 0$ ; for each $a \in {1, 2, 3, 4}$ count how many $b \in {1, 2, 3, 4}$ satisfy this.

Quadratic formula

Step-by-step solution

1
$x^{2} + a x + b = 0$ has real roots iff $a^{2} - 4 b \geq 0 \Leftrightarrow b \leq \frac{a ^{2}}{4}$
Why: The discriminant of the quadratic must be non-negative.
Quadratic formula
2
$a = 1 : b \leq 0.25 \Rightarrow 0$ values;\quad $a = 2 : b \leq 1 \Rightarrow b = 1$ ( $1$ value)
Why: Check each $b \in {1, 2, 3, 4}$ against the bound for $a = 1, 2$ .
3
$a = 3 : b \leq 2.25 \Rightarrow b \in {1, 2}$ ( $2$ values);\quad $a = 4 : b \leq 4 \Rightarrow b \in {1, 2, 3, 4}$ ( $4$ values)
Why: Continue the same check for $a = 3, 4$ .
4
$0 + 1 + 2 + 4 = 7$
Why: Sum the counts over all four values of $a$ .

Common mistakes

Option A

Overlooked the boundary case $a = 4, b = 4$ (discriminant exactly $0$ ) while listing valid pairs, missing one and totalling $6$ .

Option C

Rounded $b \leq 0.25$ up to allow $b = 1$ for $a = 1$ , adding one spurious pair and totalling $8$ .

Option D

Ignored the real-roots condition entirely and counted every pair, $4 \times 4 = 16$ .

Skills used

Quadratic formula

9Algebra • Quadratic Equations • Word Problems

EasyIdeal time: about 1 min

The sum of the squares of two positive integers is $208$ . If the square of the larger number is $18$ times the smaller number, then what is the difference of the larger and smaller numbers?

2

3

4

6

Show answer & solution

Option C

4

Approach

Substitute the second condition into the first to get a quadratic in the smaller number alone.

Quadratic formula

Step-by-step solution

1
$s^{2} + L^{2} = 208, L^{2} = 18 s (s < L, s, L positive integers)$
Why: Name the smaller number $s$ and the larger $L$ and write both conditions.
2
$s^{2} + 18 s = 208 \Rightarrow s^{2} + 18 s - 208 = 0$
Why: Replacing $L^{2}$ by $18 s$ removes $L$ entirely, leaving one quadratic.
3
$(s + 26) (s - 8) = 0 \Rightarrow s = 8$
Why: Factorise using $26 \cdot 8 = 208$ and $26 - 8 = 18$ ; the root $s = - 26$ is rejected because the integers are positive.
Quadratic formula
4
$L^{2} = 18 \cdot 8 = 144 \Rightarrow L = 12$
Why: Back-substitute into $L^{2} = 18 s$ and take the positive root.
denotes the non-negative root
5
$L - s = 12 - 8 = 4$
Why: The question asks for the difference, not for the numbers themselves.

Shortcut

$s^{2} + 18 s = 208$ factorises as $(s + 26) (s - 8)$ , so the pair is $8$ and $12$ .

Common mistakes

Option A

No admissible pair of positive integers differs by $2$ under both conditions.

Option B

No admissible pair of positive integers differs by $3$ under both conditions.

Option D

Found $L = 12$ but subtracted it from the $18$ in the condition instead of from the smaller number $8$ .

Skills used

Quadratic formula

denotes the non-negative root

10Algebra • Sets • Operations

EasyIdeal time: about 40 s

Let $A = {7, 8, 9, 10, 11, 12}$ and $B = {7, 10, 14, 15}$ . What is the number of elements in $(A - B)$ and $(B - A)$ respectively?

2

and

4

4

and

2

2

and

2

4

and

4

Show answer & solution

Option B

4

and

2

Approach

List the two differences element by element and count, then confirm the counts with the intersection.

Set difference

A - B

Step-by-step solution

1
$A - B$ keeps the elements of $A$ that are not in $B$ : $7$ and $10$ go, leaving ${8, 9, 11, 12}$ , so $∣ A - B ∣ = 4$ .
Why: A set difference removes only the shared elements; nothing from $B$ is ever added.
Set difference $A - B$
2
$B - A$ keeps the elements of $B$ that are not in $A$ : $7$ and $10$ go, leaving ${14, 15}$ , so $∣ B - A ∣ = 2$ .
Why: The same two shared elements are removed from the other side, and $B$ has only four members to start with.
3
'Respectively' fixes the order: $∣ A - B ∣ = 4$ then $∣ B - A ∣ = 2$ .
Why: Both $4$ and $2$ appear in two different orders among the options, so the order is the whole question.

Shortcut

$A \cap B = {7, 10}$ , so the answers are $6 - 2 = 4$ and $4 - 2 = 2$ .

Common mistakes

Option A

Gave the two counts in the reverse order, reporting $∣ B - A ∣$ first.

Option C

Read both differences as the intersection ${7, 10}$ , which has two elements each way.

Option D

Took $∣ B - A ∣ = ∣ B ∣ = 4$ , forgetting to remove the elements $B$ shares with $A$ .

Skills used

Set difference

A - B

11Arithmetic • Number System • LCM and HCF

EasyIdeal time: about 1 min

A boy saves Rs. $4.65$ daily. What is the least number of days in which he will be able to save an exact number of rupees?

10

20

21

25

Show answer & solution

Option B

20

Approach

Work in paise: the total after $n$ days is $465 n$ paise, and it is a whole number of rupees exactly when $100$ divides $465 n$ .

Least multiplier that makes a fraction a whole number

Step-by-step solution

1
$Rs. 4.65 = 465 paise, total after n days = 465 n paise$
Why: Clearing the decimal turns the condition into a divisibility statement about integers.
2
$100 ∣ 465 n$
Why: An exact number of rupees means the paise total is a multiple of $100$ .
Least multiplier that makes a fraction a whole number
3
$465 = 3 \cdot 5 \cdot 31, 100 = 2^{2} \cdot 5^{2}$
Why: Factorise both to see what $465$ already supplies and what $n$ must supply.
Prime factorisation
4
$\frac{465 n}{100} = \frac{93 n}{20} \Rightarrow 20 ∣ 93 n$
Why: The single factor $5$ shared by $465$ and $100$ cancels, leaving $20 = 2^{2} \cdot 5$ to be supplied by $n$ .
5
$g cd (93, 20) = 1 \Rightarrow 20 ∣ n \Rightarrow n_{m i n} = 20$
Why: $93 = 3 \cdot 31$ shares no factor with $20$ , so $n$ itself must carry all of $20$ .

Shortcut

$4.65 = \frac{93}{20}$ in lowest terms, so the least $n$ making $\frac{93 n}{20}$ an integer is the denominator $20$ .

Common mistakes

Option A

Assumed two decimal places always clear in $10$ days, without checking that $10 \times 4.65 = 46.50$ still carries $50$ paise.

Option C

No divisibility argument singles out $21$ .

Option D

Used the $25$ from $100 = 4 \times 25$ , as if $n$ had to supply the whole of $100$ divided by the $4$ in $4.65$ .

Skills used

Least multiplier that makes a fraction a whole number Prime factorisation

12Arithmetic • Time, Speed and Distance • Races

ModerateIdeal time: about 1.5 min

Two men, $A$ and $B$ run a $4$ km race on a course $0.25$ km round. If their speeds are in the ratio $5 : 4$ , how often does the winner pass the other?

AOnceBTwiceCThriceDFour times

Show answer & solution

Option C

Thrice

Approach

Introduce speeds $5 v$ and $4 v$ from the given ratio; find how far behind $B$ falls by the time $A$ finishes, then convert that gap into whole laps of the $0.25 km$ track.

Counting overtakes on a circular track

Step-by-step solution

1
Speeds $= 5 v, 4 v$ (ratio $5 : 4$ ); race distance $= 4 km$ , track length $= 0.25 km$
Why: Introduce a common factor $v$ for the given speed ratio.
Introduce a common factor $k$ for a ratio
2
$A$ covers $4 km$ in time $t = \frac{4}{5 v}$ ; in that time $B$ covers $4 v \times t = \frac{16}{5} = 3.2 km$
Why: Use the same elapsed time $t$ for both runners.
3
Gap when $A$ finishes $= 4 - 3.2 = 0.8 km = \frac{0.8}{0.25} = 3.2$ laps
Why: Convert the distance gap into a number of laps of the $0.25 km$ track.
4
$A$ completes $3$ full extra laps (and is $0.2$ of a lap into a $4$ th) before the race ends, so $A$ passes $B$ $3$ times
Why: A 'pass' occurs only when a full extra lap is completed; the partial $4$ th lap does not count.
Counting overtakes on a circular track

Shortcut

Equivalently, $A$ gains $1$ unit of speed for every $5$ units travelled, so a pass occurs every $0.25 \times 5 = 1.25 km$ that $A$ runs; $4/1.25 = 3.2 \Rightarrow 3$ complete passes.

Common mistakes

Option A

Interpreted 'how often does the winner pass the other' as asking only about the finish, counting a single pass.

Option B

Mistook the $0.25 km$ round for a $0.4 km$ round, halving the gap-in-laps from $3.2$ to $2$ .

Option D

Rounded the $3.2$ -lap gain up to $4$ instead of down, forgetting a partial lap does not count as a pass.

Skills used

Counting overtakes on a circular track Introduce a common factor

k

for a ratio

13Arithmetic • Number System • Rational Numbers

EasyIdeal time: about 1 min

Which one of the following rational numbers has non-terminating and repeating decimal expansion?

\frac{15}{1600}

\frac{23}{8}

\frac{35}{50}

\frac{17}{6}

Show answer & solution

Option D

\frac{17}{6}

Approach

A fraction in lowest terms terminates exactly when its denominator is of the form $2^{m} 5^{n}$ ; factorise each denominator and find the one that is not.

When a rational number's decimal expansion terminates

Step-by-step solution

1
$\frac{15}{1600} = \frac{3}{320}, 320 = 2^{6} \cdot 5$
Why: Reduce first, then factorise: only $2$ s and a $5$ remain, so this terminates.
Prime factorisation
2
$\frac{23}{8}, 8 = 2^{3}$
Why: A pure power of $2$ terminates; indeed $\frac{23}{8} = 2.875$ .
3
$\frac{35}{50} = \frac{7}{10}, 10 = 2 \cdot 5$
Why: After reduction the denominator is $10$ , so this terminates at $0.7$ .
4
$\frac{17}{6}, 6 = 2 \cdot 3$
Why: The factor $3$ is neither $2$ nor $5$ and does not cancel with $17$ , so this expansion cannot terminate.
When a rational number's decimal expansion terminates
5
$\frac{17}{6} = 2.8 3$
Why: A non-terminating expansion of a rational number is necessarily repeating, and here the repeating block is the single digit $3$ .

Shortcut

Scan the denominators for a prime other than $2$ or $5$ after cancelling: only $6 = 2 \cdot 3$ has one.

Common mistakes

Option A

Judged by the size of the denominator, assuming $1600$ must give an endless expansion.

Option B

Took any denominator that is not a power of $10$ to be non-terminating.

Option C

Tested $50$ without first reducing $\frac{35}{50}$ to $\frac{7}{10}$ , then mis-ruled it non-terminating.

Skills used

When a rational number's decimal expansion terminates Prime factorisation

14Algebra • Quadratic Equations • Roots

ModerateIdeal time: about 1.5 min

If $α$ and $β$ are the two zeros of the polynomial $25 x^{2} - 15 x + 2$ , then what is a quadratic polynomial whose zeros are $(2 α)^{- 1}$ and $(2 β)^{- 1}$ ?

x^{2} + 30 x + 2

8 x^{2} - 30 x + 25

8 x^{2} - 30 x

x^{2} + 30 x

Show answer & solution

Option B

8 x^{2} - 30 x + 25

Approach

Get the sum and product of $α, β$ from the coefficients, convert them into the sum and product of $\frac{1}{2 α}, \frac{1}{2 β}$ , and rebuild the polynomial.

Sum and product of roots (Vieta)

Step-by-step solution

1
$α + β = \frac{15}{25} = \frac{3}{5}, α β = \frac{2}{25}$
Why: For $a x^{2} + b x + c$ the zeros satisfy $α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$ .
Sum and product of roots (Vieta)
2
$\frac{1}{2 α} + \frac{1}{2 β} = \frac{α + β}{2 α β} = \frac{3/5}{2 \cdot 2/25} = \frac{3}{5} \cdot \frac{25}{4} = \frac{15}{4}$
Why: Adding the reciprocals needs only the sum and the product, so the zeros never have to be found.
Negative exponent is a reciprocal
3
$\frac{1}{2 α} \cdot \frac{1}{2 β} = \frac{1}{4 α β} = \frac{1}{4 \cdot 2/25} = \frac{25}{8}$
Why: The product of the new zeros is the reciprocal of $4 α β$ .
4
$x^{2} - \frac{15}{4} x + \frac{25}{8}$
Why: A monic quadratic with given zeros is $x^{2} - (sum) x + (product)$ .
5
$8 (x^{2} - \frac{15}{4} x + \frac{25}{8}) = 8 x^{2} - 30 x + 25$
Why: Multiplying by $8$ clears the fractions without changing the zeros, matching the form of the options.

Shortcut

Substitute $x \to \frac{1}{2 x}$ in the original polynomial and clear denominators: $\frac{25}{4 x ^{2}} - \frac{15}{2 x} + 2 = 0$ becomes $25 - 30 x + 8 x^{2} = 0$ , i.e. $8 x^{2} - 30 x + 25$ .

Common mistakes

Option A

Kept the sign of the sum as printed rather than negating it, and carried the original constant $2$ instead of the new product.

Option C

Computed the new sum but not the new product, dropping the constant term.

Option D

Made both errors at once: sign of the sum kept and constant term dropped.

Skills used

Sum and product of roots (Vieta)Negative exponent is a reciprocal

15Arithmetic • Number System • Divisibility

HardIdeal time: about 1 min

What is the remainder when $2^{100}$ is divided by $101$ ?

1

11

99

100

Show answer & solution

Option A

1

Approach

Recognise $101$ as prime and apply Fermat's little theorem directly to $2^{100} = 2^{101 - 1}$ .

Fermat's little theorem

Step-by-step solution

1
$101$ is prime and $g cd (2, 101) = 1$
Why: Fermat's little theorem requires a prime modulus and a base coprime to it.
2
$2^{100} = 2^{101 - 1} \equiv 1 (mod 101)$
Why: Fermat's little theorem: $a^{p - 1} \equiv 1 (mod p)$ for prime $p$ and $g cd (a, p) = 1$ .
Fermat's little theorem

Common mistakes

Option B

No clean route reaches this value.

Option C

No clean route reaches this value.

Option D

Confused Fermat's little theorem with a sign-flipped version, concluding $2^{100} \equiv - 1 \equiv 100 (mod 101)$ .

Skills used

Fermat's little theorem

16Arithmetic • Ratio and Proportion

EasyIdeal time: about 1 min

In an office, one-third of the workers are women, half of the women are married and one-third of the married women have children. If three-fourth of the men are married and one-third of the married men have children, then what is the ratio of married women to married men?

1 : 2

2 : 1

3 : 1

1 : 3

Show answer & solution

Option D

1 : 3

Approach

Replace the fractions by a concrete workforce that all of them divide, then count heads and form the ratio.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
The women-to-men split is $1 : 2$ , so write the counts as $k$ women and $2 k$ men; take $k = 12$ , i.e. $36$ workers in all.
Why: A ratio fixes the numbers only up to a common factor, and $36$ is chosen so that the halves, thirds and quarters below are all whole.
Introduce a common factor $k$ for a ratio
2
Women $= 12$ , and half of them are married: $6$ married women.
Why: The first fraction is applied to the women only, not to the whole office.
3
Men $= 24$ , and three-fourths are married: $18$ married men.
Why: The second fraction is applied to the men only; the two marriage fractions differ, which is the whole content of the question.
4
$6 : 18 = 1 : 3$ .
Why: Dividing both counts by $6$ puts the ratio in lowest terms, and $k$ has cancelled as expected.

Shortcut

Married women are $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$ of the office and married men $\frac{3}{4} \times \frac{2}{3} = \frac{1}{2}$ , so the ratio is $\frac{1}{6} : \frac{1}{2} = 1 : 3$ .

Common mistakes

Option A

Gave the women-to-men ratio $1 : 2$ without applying the two marriage fractions.

Option B

No plausible route reaches $2 : 1$ .

Option C

Wrote the ratio the other way round, married men to married women.

Skills used

Introduce a common factor

k

for a ratio

17Arithmetic • Time and Work

ModerateIdeal time: about 1.5 min

$A$ can do $50%$ more work than $B$ in the same time. $B$ alone can do a piece of work in $30$ hours. $B$ starts working and had already worked for $12$ hours when $A$ joins him. How many hours should $B$ and $A$ work together to complete the remaining work?

6

hoursB

12

hoursC

4.8

hoursD

7.2

hours

Show answer & solution

Option D

7.2

hours

Approach

Take the whole job as $60$ units so both rates are whole numbers, subtract what $B$ has already done, and divide the remainder by the combined rate.

Take total work as the LCM of the times

Step-by-step solution

1
$total work = 60 units \Rightarrow rate of B = \frac{60}{30} = 2 units/hour$
Why: Choosing the total as a multiple of $30$ clears the fractions from every rate.
Take total work as the LCM of the times
2
$rate of A = 1.5 \times 2 = 3 units/hour$
Why: Doing $50%$ more work in the same time means the rate is multiplied by $1.5$ .
Percentage change as a multiplier
3
$done by B alone = 12 \times 2 = 24 units \Rightarrow remaining = 60 - 24 = 36 units$
Why: Only the work left after $B$ 's first $12$ hours is shared.
4
$combined rate = 2 + 3 = 5 units/hour$
Why: Rates add when two workers work together.
5
$t = \frac{36}{5} = 7.2 hours$
Why: Time is the remaining work divided by the combined rate.

Shortcut

$B$ does $\frac{12}{30} = \frac{2}{5}$ of the job first; the pair finish a whole job in $12$ hours, so the remaining $\frac{3}{5}$ takes $\frac{3}{5} \times 12 = 7.2$ hours.

Common mistakes

Option A

Took the remaining work as $\frac{1}{2}$ of the job instead of $\frac{3}{5}$ , giving $\frac{30}{5} = 6$ hours.

Option B

Gave the time the pair would need for the whole job, ignoring the $24$ units $B$ had already finished.

Option C

Divided the completed $24$ units by the combined rate instead of the remaining $36$ .

Skills used

Take total work as the LCM of the times Percentage change as a multiplier

18Arithmetic • Time, Speed and Distance

EasyIdeal time: about 1 min

When the speed of a train is increased by $20%$ , it takes $20$ minutes less to cover the same distance. What is the time taken to cover the same distance with the original speed?

140

minutesB

120

minutesC

100

minutesD

80

minutes

Show answer & solution

Option B

120

minutes

Approach

Over a fixed distance, time is inversely proportional to speed, so a speed factor of $\frac{6}{5}$ is a time factor of $\frac{5}{6}$ ; the missing sixth of the time is the $20$ minutes saved.

Over a fixed distance, time varies inversely with speed

Step-by-step solution

1
$v^{'} = 1.2 v = \frac{6}{5} v$
Why: A $20%$ increase is multiplication by $\frac{6}{5}$ .
Percentage change as a multiplier
2
$t = \frac{d}{v}, t^{'} = \frac{d}{v ^{'}} \Rightarrow t^{'} = \frac{5}{6} t$
Why: The distance is unchanged, so the time is divided by the same factor the speed was multiplied by.
Over a fixed distance, time varies inversely with speed
3
$t - t^{'} = t - \frac{5}{6} t = \frac{t}{6} = 20$
Why: The saving of $20$ minutes is exactly the one-sixth of the original time that the faster speed removes.
4
$t = 6 \times 20 = 120 minutes$
Why: The question asks for the original time, not the reduced one.

Shortcut

A $20%$ speed rise cuts the time by $\frac{1}{6}$ ; if $\frac{1}{6}$ of the time is $20$ minutes, the time is $120$ minutes.

Common mistakes

Option A

Added the $20$ minutes to a mis-derived $120$ , or took the saving as $\frac{1}{7}$ of the original time.

Option C

Reported the reduced time $\frac{5}{6} t = 100$ minutes instead of the original.

Option D

Applied the $20%$ directly to the time, treating the saving as $20%$ of $t$ and getting $t = 100$ , then compounding it to $80$ .

Skills used

Over a fixed distance, time varies inversely with speed Percentage change as a multiplier

19Arithmetic • Percentage

ModerateIdeal time: about 1.5 min

A candidate scoring $x %$ marks in an examination fails by $a$ marks, while another candidate who scores $y %$ marks gets $b$ marks more than the minimum required pass marks. What is the maximum marks for the examination?

\frac{100 ( a + b )}{x - y}

\frac{100 ( a - b )}{x + y}

\frac{100 ( a + b )}{y - x}

\frac{100 ( a - b )}{x - y}

Show answer & solution

Option C

\frac{100 ( a + b )}{y - x}

Approach

Write the pass mark twice — once from each candidate — and equate the two expressions.

x %

of a quantity

Step-by-step solution

1
$let M = maximum marks, p = pass mark$
Why: Both candidates' scores are percentages of the same $M$ , and both are described relative to the same $p$ .
2
$\frac{x M}{100} + a = p$
Why: The first candidate falls $a$ marks short, so adding $a$ to the score reaches the pass mark.
$x %$ of a quantity
3
$\frac{y M}{100} - b = p$
Why: The second candidate is $b$ marks above the pass mark, so subtracting $b$ reaches it.
4
$\frac{y M}{100} - b = \frac{x M}{100} + a \Rightarrow \frac{( y - x ) M}{100} = a + b$
Why: Eliminating $p$ leaves one equation in $M$ ; the gap between the two percentage scores is worth $a + b$ marks.
5
$M = \frac{100 ( a + b )}{y - x}$
Why: $y > x$ because the second candidate passed and the first did not, so the denominator is positive as required.

Shortcut

The two candidates differ by $(y - x) %$ of the maximum and by $a + b$ marks, so $M = \frac{100 ( a + b )}{y - x}$ .

Common mistakes

Option A

Subtracted the equations the other way round, leaving $x - y$ in the denominator and a negative maximum.

Option B

No consistent reading of the two candidates puts a sum in the denominator: the percentages must be differenced to eliminate the pass mark.

Option D

Read $b$ as a further shortfall rather than a surplus, giving $a - b$ in the numerator, and also flipped the denominator.

Skills used

x %

of a quantity

20Algebra • Quadratic Equations • Roots

ModerateIdeal time: about 1.5 min

If $p$ and $q$ are the roots of $x^{2} + p x + q = 0$ , then which one of the following is correct?

p = 0

1

p = 1

onlyC

p = - 2

0

p = - 2

only

Show answer & solution

Option A

p = 0

1

Approach

Apply Vieta to the equation whose roots are its own coefficients, which gives two equations in $p$ and $q$ , and solve the resulting case split.

Sum and product of roots (Vieta)

Step-by-step solution

1
$p + q = - p and pq = q$
Why: For $x^{2} + p x + q = 0$ the sum of the roots is $- p$ and their product is $q$ ; here the roots are $p$ and $q$ themselves.
Sum and product of roots (Vieta)
2
$q = - 2 p$
Why: The sum condition pins $q$ to $p$ .
3
$pq - q = 0 \Rightarrow q (p - 1) = 0 \Rightarrow q = 0 or p = 1$
Why: Factorise rather than divide by $q$ , which would silently discard the case $q = 0$ .
4
$q = 0 \Rightarrow - 2 p = 0 \Rightarrow p = 0; p = 1 \Rightarrow q = - 2$
Why: Feed each case back into $q = - 2 p$ to get the complete pair.
5
$p = 0 or p = 1$
Why: Both cases survive, so the answer must name both.

Shortcut

The product condition $pq = q$ factorises immediately to $q (p - 1) = 0$ , and the sum condition then forces $(p, q) = (0, 0)$ or $(1, - 2)$ .

Common mistakes

Option B

Divided $pq = q$ through by $q$ , losing the case $q = 0$ and with it the solution $p = 0$ .

Option C

Reported the values of $q$ rather than of $p$ , since $q = - 2$ and $q = 0$ are the matching roots.

Option D

Both errors together: divided out $q$ and then quoted $q = - 2$ in place of $p = 1$ .

Skills used

Sum and product of roots (Vieta)