Elementary Mathematics 2020 (II)

Set A

Questions 21–40 of 100

21Arithmetic • Profit and Loss • False Weights

ModerateIdeal time: about 1.5 min

A shopkeeper sells his articles at their cost price but uses a faulty balance which reads $1000$ gm for $800$ gm. What is the actual profit percentage ?

20%

25%

30%

40%

Show answer & solution

Option B

25%

Approach

Price a single transaction: the shopkeeper charges for $1000$ gm but parts with only $800$ gm, so compare what he spends with what he takes, with the cost of the goods he actually gives away as the base.

Gain percent is measured on the cost price

Step-by-step solution

1
The balance reads $1000$ gm when the goods on it weigh $800$ gm
Why: This is what a faulty balance 'reading $1000$ gm for $800$ gm' does: the customer is billed for a kilogram and receives $800$ gm.
2
Take the cost price as Rs. $1$ per gm. Cost to the shopkeeper $=$ Rs. $800$ ; money received $=$ Rs. $1000$
Why: He sells 'at cost price', so the rate he charges is the cost rate - but he charges it on the false weight. Choosing Rs. $1$ per gm makes the two figures the weights themselves.
Gain percent is measured on the cost price
3
Profit $= 1000 - 800 =$ Rs. $200$
Why: The gain is the money received less what the goods actually cost him.
4
Profit $% = \frac{200}{800} \times 100 = 25%$
Why: Gain percent is always measured on the cost price - here Rs. $800$ , the cost of the goods he handed over - and never on the Rs. $1000$ he collected.

Shortcut

He gives $800$ gm and is paid for $1000$ : gain $= \frac{1000 - 800}{800} = \frac{1}{4} = 25%$ .

Common mistakes

Option A

Divided the Rs. $200$ gain by the Rs. $1000$ collected instead of the Rs. $800$ spent - a percentage on the selling price, not on the cost.

Option C

No route reaches $30%$ .

Option D

Compared the $200$ gm shortfall with the $500$ gm half-kilogram, or doubled the $20%$ figure.

Skills used

Gain percent is measured on the cost price

22Mensuration • Solids • Cuboid

ModerateIdeal time: about 1.5 min

A river $3$ m deep and $40$ m wide is flowing at the rate of $2$ km/hr and falls into the sea. What is the amount of water in litres that will fall into the sea from this river in a minute ?

40, 00, 000

litresB

4, 00, 000

litresC

40, 000

litresD

4, 000

litres

Show answer & solution

Option A

40, 00, 000

litres

Approach

Water in a minute forms a cuboid whose cross-section is the river's mouth and whose length is the distance the water travels in that minute; find its volume and convert to litres.

Volume and wall area of a cuboid

Step-by-step solution

1
In one minute the water advances $\frac{2000}{60} = \frac{100}{3}$ m
Why: $2$ km/hr is $2000$ m in $60$ minutes, so $\frac{2000}{60}$ m in one minute. Keeping it as the fraction $\frac{100}{3}$ avoids a recurring decimal.
Speed, distance and time
2
Volume $= depth \times width \times length = 3 \times 40 \times \frac{100}{3}$ m $^{3}$
Why: The slab of water that passes the mouth in a minute is a cuboid: the $3$ m $\times$ $40$ m cross-section swept along the distance travelled. The $3$ s cancel.
Volume and wall area of a cuboid
3
Volume $= 4000$ m $^{3}$
Why: $40 \times 100 = 4000$ after the depth cancels against the denominator.
4
$1$ m $^{3} = 1000$ litres, so $4000$ m $^{3} = 40, 00, 000$ litres
Why: A cubic metre is $100 \times 100 \times 100$ cubic centimetres, i.e. $1 0^{6}$ ml, which is $1000$ litres. $4000 \times 1000 = 4 \times 1 0^{6}$ .
Unitary method: find one unit, then scale

Common mistakes

Option B

Took $1$ m $^{3}$ as $100$ litres instead of $1000$ .

Option C

Stopped at the volume $4000$ m $^{3}$ and appended a factor of $10$ , or converted with $1$ m $^{3} = 10$ litres.

Option D

Answered with the volume in cubic metres, $4000$ , and called it litres.

Skills used

Volume and wall area of a cuboid Speed, distance and time Unitary method: find one unit, then scale

23Arithmetic • Profit and Loss

EasyIdeal time: about 55 s

If a television set is sold at Rs. x, a loss of $28%$ would be incurred. If it is sold at Rs. y, a profit of $12%$ would be incurred. What is the ratio of y to x ?

41 : 9

31 : 9

23 : 9

14 : 9

Show answer & solution

Option D

14 : 9

Approach

Express both selling prices as multiples of the same cost price; the cost price then cancels out of the ratio.

Percentage change as a multiplier

Step-by-step solution

1
Let the cost price be Rs. $C$
Why: Both $x$ and $y$ are percentages of the one unknown cost price, so naming it lets them be compared.
Gain percent is measured on the cost price
2
A $28%$ loss: $x = C (1 - \frac{28}{100}) = 0.72 C$
Why: A loss of $28%$ multiplies the cost price by $1 - 0.28$ .
Percentage change as a multiplier
3
A $12%$ profit: $y = C (1 + \frac{12}{100}) = 1.12 C$
Why: A profit of $12%$ multiplies the cost price by $1 + 0.12$ . Both percentages are taken on the same base, which is why the ratio is clean.
4
$\frac{y}{x} = \frac{1.12 C}{0.72 C} = \frac{112}{72} = \frac{14}{9}$
Why: $C$ cancels; divide numerator and denominator by $8$ .

Shortcut

Take the cost price as Rs. $100$ : $y : x = 112 : 72 = 14 : 9$ .

Common mistakes

Option A

No route reaches $41 : 9$ .

Option B

No route reaches $31 : 9$ .

Option C

No route reaches $23 : 9$ .

Skills used

Percentage change as a multiplier Gain percent is measured on the cost price

24Arithmetic • Time, Speed and Distance

ModerateIdeal time: about 1.5 min

By increasing the speed of his car by $15$ km/hr, a person covers a distance of $300$ km by taking an hour less than before. What was the original speed of the car ?

45

km/hrB

50

km/hrC

60

km/hrD

75

km/hr

Show answer & solution

Option C

60

km/hr

Approach

Write both journey times for the fixed $300$ km, set their difference to one hour, and clear the denominators into a quadratic.

Speed, distance and time

Step-by-step solution

1
Let the original speed be $v$ km/hr; the two times are $\frac{300}{v}$ and $\frac{300}{v + 15}$ hours
Why: Time is distance over speed, and the distance is the same on both runs.
Speed, distance and time
2
$\frac{300}{v} - \frac{300}{v + 15} = 1$
Why: The slower run takes the longer time, so the original time minus the new time is the hour saved. Getting this subtraction the right way round is what keeps $v$ positive.
3
$300 (v + 15) - 300 v = v (v + 15) \Rightarrow 4500 = v^{2} + 15 v$
Why: Multiply through by $v (v + 15)$ . The $300 v$ terms cancel on the left, which is why the equation stays manageable.
Clear the denominators of an equation
4
$v^{2} + 15 v - 4500 = 0 \Rightarrow v^{2} + 75 v - 60 v - 4500 = 0 \Rightarrow (v + 75) (v - 60) = 0$
Why: Split the middle term: $75 \times (- 60) = - 4500$ and $75 - 60 = 15$ .
Factorise a quadratic by splitting the middle term
5
$v = 60$ (rejecting $v = - 75$ )
Why: A product is zero only when a factor is zero, and a speed cannot be negative.
A product is zero only when a factor is zero

Shortcut

The two times differ by an hour and multiply the same $300$ km, so look for a factor pair of $300$ differing by $1$ : $5$ and $4$ hours, giving $60$ and $75$ km/hr. The original speed is $60$ .

Common mistakes

Option A

No route reaches $45$ km/hr; it would save $\frac{300}{45} - \frac{300}{60} = 1 \frac{2}{3}$ hours.

Option B

No route reaches $50$ km/hr; it would save $\frac{300}{50} - \frac{300}{65} \approx 1.38$ hours.

Option D

Answered with the increased speed $v + 15 = 75$ instead of the original speed.

Skills used

Speed, distance and time Clear the denominators of an equation Factorise a quadratic by splitting the middle term A product is zero only when a factor is zero

25Arithmetic • Profit and Loss • Partnership

HardIdeal time: about 2.5 min

Three persons start a business with capitals in the ratio $\frac{1}{3} : \frac{1}{4} : \frac{1}{5}$ . The first person withdraws half his capital after $4$ months. What is his share of profit if the business fetches an annual profit of Rs. $96, 800$ ?

ARs.

32, 000

BRs.

34, 500

CRs.

36, 000

DRs.

36, 800

Show answer & solution

Option A

Rs.

32, 000

Approach

Clear the fractional capital ratio to whole numbers, convert each partner's stake into capital-months (allowing for the mid-year withdrawal), and share the profit in that ratio.

Profit is shared in the ratio of capital

\times

time

Step-by-step solution

1
$\frac{1}{3} : \frac{1}{4} : \frac{1}{5} = 20 : 15 : 12$
Why: Multiply every term by $60$ , the LCM of the denominators; a ratio is unchanged by a common multiplier, and whole numbers are far easier to carry through the next step.
Least multiplier that makes a fraction a whole number
2
Take the capitals as $20 k$ , $15 k$ and $12 k$ over a $12$ -month year
Why: A ratio fixes the capitals only up to a common factor $k$ , and $k$ will cancel when the shares are formed.
Introduce a common factor $k$ for a ratio
3
First partner: $20 \times 4 + 10 \times 8 = 80 + 80 = 160$ capital-months
Why: He holds $20$ for the first $4$ months, then half of it, $10$ , for the remaining $8$ . Profit follows capital $\times$ time, so a withdrawal has to be split into two spells rather than averaged.
Profit is shared in the ratio of capital $\times$ time
4
Second: $15 \times 12 = 180$ ; third: $12 \times 12 = 144$ ; total $= 160 + 180 + 144 = 484$
Why: Neither of the others withdraws, so each runs for the full twelve months.
5
First partner's share $= \frac{160}{484} \times 96800$
Why: Profit is divided in the ratio of the capital-months.
6
$\frac{96800}{484} = 200$ , so the share is $160 \times 200 =$ Rs. $32, 000$
Why: Divide before multiplying: $484 \times 200 = 96800$ exactly, which keeps the arithmetic in two-digit steps.

Common mistakes

Option B

No clean route reaches Rs. $34, 500$ .

Option C

No clean route reaches Rs. $36, 000$ ; that is the second partner's share, reached only by also mis-timing the withdrawal.

Option D

No clean route reaches Rs. $36, 800$ .

Skills used

Profit is shared in the ratio of capital

\times

time Least multiplier that makes a fraction a whole number Introduce a common factor

k

for a ratio

26Arithmetic • Ratio and Proportion • Direct and Inverse Variation

HardIdeal time: about 2.5 min

If x varies as y, then which of the following is/are correct ?

1: $x^{2} + y^{2}$ varies as $x^{2} - y^{2}$

2: $\frac{x}{y ^{2}}$ varies inversely as y

3: $n x^{2} y$ varies as $2 n x^{4} y^{2}$

Select the correct answer using the code given below :

A1 and 2 onlyB2 and 3 onlyC3 onlyD1, 2 and 3

Show answer & solution

Option D

1, 2 and 3

Approach

' $x$ varies as $y$ ' means $x = k y$ for a fixed $k$ ; substitute that into each statement and check whether the ratio of the two sides is constant.

Direct variation as a power:

y \propto x^{n}

Step-by-step solution

1
$x$ varies as $y$ means $x = k y$ for a constant $k$ ; 'P varies as Q' means $\frac{P}{Q}$ is constant
Why: Everything here is a test of one thing: does the quotient come out free of $x$ and $y$ ?
Direct variation as a power: $y \propto x^{n}$
2
Statement $1$ : $\frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}} = \frac{k ^{2} y ^{2} + y ^{2}}{k ^{2} y ^{2} - y ^{2}} = \frac{k ^{2} + 1}{k ^{2} - 1}$ , a constant - correct
Why: $y^{2}$ cancels top and bottom, leaving only $k$ . So $x^{2} + y^{2}$ does vary as $x^{2} - y^{2}$ .
3
Statement $2$ : $\frac{x}{y ^{2}} = \frac{k y}{y ^{2}} = \frac{k}{y}$ , so $\frac{x}{y ^{2}}$ varies inversely as $y$ - correct
Why: 'Varies inversely as $y$ ' means the quantity is a constant over $y$ , which is exactly what $\frac{k}{y}$ is.
4
Statement $3$ : $2 n x^{4} y^{2} = (x^{4} y^{2})^{\frac{1}{2 n}} = ((x^{2} y)^{2})^{\frac{1}{2 n}} = (x^{2} y)^{\frac{1}{n}} = n x^{2} y$
Why: $x^{4} y^{2}$ is the square of $x^{2} y$ , so the $2 n$ -th root of one is the $n$ -th root of the other. The two expressions are identical.
Power of a power, and the move to a common base
5
Their ratio is $1$ , a constant - statement $3$ is correct. All three hold.
Why: A quantity trivially varies as itself; no appeal to $x = k y$ is even needed here.

Common mistakes

Option A

Rejected statement $3$ without simplifying, taking an $n$ -th root and a $2 n$ -th root to be different things.

Option B

Rejected statement $1$ , reading 'varies as' as requiring the two expressions to be equal or to be proportional to $y$ itself rather than to each other.

Option C

Rejected both statements $1$ and $2$ while accepting only the identity in statement $3$ .

Skills used

Direct variation as a power:

y \propto x^{n}

Power of a power, and the move to a common base

27Arithmetic • Age Problems

EasyIdeal time: about 1 min

Ena was born $4$ years after her parents marriage. Her mother is $3$ years younger than her father and $24$ years older than Ena, who is $13$ years old. At what age did Ena's father get married ?

25

yearsB

24

yearsC

23

yearsD

22

years

Show answer & solution

Option C

23

years

Approach

Work up the chain of present ages from Ena to her father, then step back by the number of years since the marriage.

The age difference between two people never changes

Step-by-step solution

1
Ena is $13$ ; her mother is $24$ years older, so the mother is $13 + 24 = 37$
Why: ' $24$ years older than Ena' is given directly.
The age difference between two people never changes
2
The mother is $3$ years younger than the father, so the father is $37 + 3 = 40$
Why: 'Younger than' points the other way: the father is the older of the two.
3
Ena was born $4$ years after the marriage, so the marriage was $13 + 4 = 17$ years ago
Why: Her age counts the years since her birth; the extra $4$ reaches back to the wedding.
4
Father's age at marriage $= 40 - 17 = 23$ years
Why: An age gap never changes, so subtracting the elapsed $17$ years from his present age gives his age then.

Common mistakes

Option A

No route reaches $25$ ; it needs the marriage to be $15$ years ago.

Option B

Used the $3$ -year age gap in place of the $4$ years between the marriage and Ena's birth, making the marriage $16$ years ago.

Option D

Pushed the marriage back $18$ years instead of $17$ , counting Ena's birth year twice.

Skills used

The age difference between two people never changes

28Arithmetic • Age Problems

EasyIdeal time: about 50 s

Mahesh is $60$ years old. Ram is $5$ years younger to Mahesh and $4$ years elder to Raju. Babu is a younger brother of Raju and he is $6$ years younger. What is the age difference between Mahesh and Babu?

18

yearsB

15

yearsC

13

yearsD

11

years

Show answer & solution

Option B

15

years

Approach

Walk down the chain of ages from Mahesh to Babu, then take the difference - or simply add the three gaps.

The age difference between two people never changes

Step-by-step solution

1
Ram $= 60 - 5 = 55$
Why: Ram is $5$ years younger than Mahesh.
The age difference between two people never changes
2
Raju $= 55 - 4 = 51$
Why: Ram is $4$ years elder to Raju, so Raju is the younger by $4$ .
3
Babu $= 51 - 6 = 45$
Why: Babu is Raju's younger brother by $6$ years. Note the $6$ is measured from Raju, not from Ram.
4
Mahesh $-$ Babu $= 60 - 45 = 15$ years
Why: Equivalently, add the gaps along the chain: $5 + 4 + 6 = 15$ . The individual ages were never needed.

Shortcut

The gaps simply add: $5 + 4 + 6 = 15$ years. Mahesh's age of $60$ is not needed at all.

Common mistakes

Option A

No route reaches $18$ years.

Option C

Dropped two years somewhere in the chain, for instance taking Babu as $6$ years younger than Ram's junior by $2$ .

Option D

Measured Babu's $6$ years from Ram rather than from Raju, giving $55 - 6 = 49$ and a gap of $11$ .

Skills used

The age difference between two people never changes

29Arithmetic • Percentage • Successive Increase and Decrease

EasyIdeal time: about 55 s

The number of items in a booklet is N. In the first year there is an increase of $x %$ in this number and in the subsequent year there is a decrease of $x %$ . At the end of the two years, what will be the number of items in the booklet?

ALess than NBEqual to NCMore than NDIt depends on the value of N

Show answer & solution

Option A

Less than N

Approach

Turn each percentage change into a multiplier and multiply them; the product is a difference of squares and so is always below $1$ .

Percentage change as a multiplier

Step-by-step solution

1
After the rise: $N (1 + \frac{x}{100})$
Why: An increase of $x %$ multiplies by $1 + \frac{x}{100}$ .
Percentage change as a multiplier
2
After the fall: $N (1 + \frac{x}{100}) (1 - \frac{x}{100})$
Why: The second year's $x %$ decrease is taken on the new, larger number, not on $N$ - which is exactly why the two changes do not cancel.
3
$(1 + \frac{x}{100}) (1 - \frac{x}{100}) = 1 - \frac{x ^{2}}{1 0 ^{4}}$
Why: Difference of two squares. The result is a single multiplier, and the $x$ terms have cancelled while the $x^{2}$ term has not.
Difference of two squares
4
$\frac{x ^{2}}{1 0 ^{4}} > 0$ for any $x > 0$ , so the final number is $N (1 - \frac{x ^{2}}{1 0 ^{4}}) < N$
Why: The multiplier is strictly less than $1$ whatever positive $x$ is, and whatever $N$ is - so the answer never depends on $N$ .

Shortcut

Try $N = 100$ , $x = 10$ : $100 \to 110 \to 99$ . Less than $N$ .

Common mistakes

Option B

Added the percentages instead of multiplying the factors, so $+ x %$ and $- x %$ appeared to cancel.

Option C

Applied the decrease to $N$ rather than to the increased figure, and then took the increase as the larger of the two.

Option D

Read the dependence as being on $N$ ; the multiplier $1 - \frac{x ^{2}}{1 0 ^{4}}$ depends on $x$ but is below $1$ for every $N$ .

Skills used

Percentage change as a multiplier Difference of two squares

30Algebra • Expressions • Algebraic Identities

ModerateIdeal time: about 1.5 min

If $ab + x y - x b = 0$ and $b c + y z - cy = 0$ , then what is $\frac{x}{a} + \frac{c}{z}$ equal to?

\frac{y}{b}

\frac{b}{y}

1

0

Show answer & solution

Option C

1

Approach

Rearrange each equation so that the quantity asked for becomes the subject, taking out the common bracket $(b - y)$ in both - it is the shared denominator that makes the two fractions add to something simple.

Clear the denominators of an equation

Step-by-step solution

1
$ab + x y - x b = 0 \Rightarrow ab = x b - x y = x (b - y)$
Why: Move the two terms carrying $x$ to the other side and factor $x$ out; $b - y$ is what is left.
Clear the denominators of an equation
2
$\frac{x}{a} = \frac{b}{b - y}$
Why: Divide both sides by $a (b - y)$ . This is the first of the two quantities wanted.
3
$b c + y z - cy = 0 \Rightarrow c (b - y) = - y z$
Why: Group the two terms carrying $c$ : $b c - cy = c (b - y)$ , and the remaining $y z$ moves across with a minus sign.
4
$\frac{c}{z} = \frac{- y}{b - y}$
Why: Divide both sides by $z (b - y)$ . The same bracket $b - y$ appears, which is the point of the question.
5
$\frac{x}{a} + \frac{c}{z} = \frac{b}{b - y} + \frac{- y}{b - y} = \frac{b - y}{b - y} = 1$
Why: The common denominator is already in place, and the numerators add to exactly the denominator.

Shortcut

Substitute convenient numbers for $a, b, y, z$ that satisfy both equations and evaluate; the sum comes out $1$ whatever they are.

Common mistakes

Option A

No route reaches $\frac{y}{b}$ ; the $b - y$ brackets cancel completely.

Option B

No route reaches $\frac{b}{y}$ .

Option D

Took $\frac{c}{z} = \frac{y}{y - b}$ with the bracket the other way round, or dropped the minus sign in step $3$ , so the numerators appeared to cancel to $0$ .

Skills used

Clear the denominators of an equation

31Algebra • Polynomial • HCF and LCM

ModerateIdeal time: about 1.5 min

What is the HCF of the polynomials $x^{6} - 3 x^{4} + 3 x^{2} - 1$ and $x^{3} + 3 x^{2} + 3 x + 1$ ?

(x + 1)

(x + 1)^{2}

x^{2} + 1

(x + 1)^{3}

Show answer & solution

Option D

(x + 1)^{3}

Approach

Recognise the first polynomial as a cube in $x^{2}$ and the second as a cube in $x$ , factor both completely, then take each common factor to its lowest power.

HCF and LCM of polynomials from their complete factorisations

Step-by-step solution

1
$x^{6} - 3 x^{4} + 3 x^{2} - 1 = (x^{2})^{3} - 3 (x^{2})^{2} (1) + 3 (x^{2}) (1)^{2} - 1^{3} = (x^{2} - 1)^{3}$
Why: The coefficients $1, - 3, 3, - 1$ are the expansion of $(A - B)^{3}$ with $A = x^{2}$ and $B = 1$ . Spotting that avoids any division.
Cube of a sum and of a difference
2
$(x^{2} - 1)^{3} = [(x - 1) (x + 1)]^{3} = (x - 1)^{3} (x + 1)^{3}$
Why: Difference of squares inside, then the cube distributes over the product. This is the complete factorisation.
Difference of two squares
3
$x^{3} + 3 x^{2} + 3 x + 1 = (x + 1)^{3}$
Why: Coefficients $1, 3, 3, 1$ are the expansion of $(x + 1)^{3}$ .
4
HCF $= (x + 1)^{3}$
Why: The only common irreducible factor is $x + 1$ , and it appears to the power $3$ in both, so the lowest power is $3$ . $(x - 1)$ is absent from the second polynomial, so it cannot appear in the HCF.
HCF and LCM of polynomials from their complete factorisations

Common mistakes

Option A

Took the HCF as the common factor to the first power, stopping before matching the exponents.

Option B

Matched the exponents as $2$ rather than $3$ , for instance by reading the first polynomial as $(x^{2} - 1)^{2}$ .

Option C

Read the signs as all positive, making the first polynomial $(x^{2} + 1)^{3}$ , which puts $x^{2} + 1$ in the HCF.

Skills used

HCF and LCM of polynomials from their complete factorisations Cube of a sum and of a difference Difference of two squares

32Algebra • Polynomial • HCF and LCM

ModerateIdeal time: about 1.5 min

The HCF and the LCM of two polynomials are $3 x + 1$ and $30 x^{3} + 7 x^{2} - 10 x - 3$ respectively. If one polynomial is $6 x^{2} + 5 x + 1$ , then what is the other polynomial?

15 x^{2} + 4 x + 3

15 x^{2} + 4 x - 3

15 x^{2} - 4 x + 3

15 x^{2} - 4 x - 3

Show answer & solution

Option D

15 x^{2} - 4 x - 3

Approach

Use HCF $\times$ LCM $=$ product of the two polynomials, then divide out the known polynomial - factorising it first so the division is by a linear factor.

HCF

\times

LCM = product of the two numbers

Step-by-step solution

1
$6 x^{2} + 5 x + 1 = (3 x + 1) (2 x + 1)$
Why: Split the middle term: $6 \times 1 = 6$ and $3 + 2 = 5$ . The HCF $3 x + 1$ duly appears as a factor, which is a first consistency check on the data.
Factorise a quadratic by splitting the middle term
2
$HCF \times LCM = (first) \times (second)$ , so the other polynomial is $\frac{( 3 x + 1 ) ( 30 x ^{3} + 7 x ^{2} - 10 x - 3 )}{( 3 x + 1 ) ( 2 x + 1 )}$
Why: For two polynomials, as for two numbers, the product of the HCF and the LCM equals the product of the polynomials. The factor $3 x + 1$ cancels at once.
HCF $\times$ LCM = product of the two numbers
3
Other polynomial $= \frac{30 x ^{3} + 7 x ^{2} - 10 x - 3}{2 x + 1}$
Why: What is left is a single division by a linear factor rather than by a quadratic.
4
Dividing: $30 x^{3} \div 2 x = 15 x^{2}$ , remainder $- 8 x^{2} - 10 x - 3$ ; $- 8 x^{2} \div 2 x = - 4 x$ , remainder $- 6 x - 3$ ; $- 6 x \div 2 x = - 3$ , remainder $0$
Why: Long division, one term at a time. The zero remainder confirms $2 x + 1$ really is a factor of the LCM.
Quotient on dividing by a linear factor
5
Other polynomial $= 15 x^{2} - 4 x - 3$
Why: The quotient is the answer.

Shortcut

The other polynomial must be $(3 x + 1) \times (linear)$ , and its product with $6 x^{2} + 5 x + 1$ has leading term $30 x^{3}$ , so the linear factor is $5 x - 3$ : $(3 x + 1) (5 x - 3) = 15 x^{2} - 4 x - 3$ .

Common mistakes

Option A

Both signs wrong in the quotient - a sign slip carried through the long division.

Option B

Sign slip on the middle term, giving $+ 4 x$ instead of $- 4 x$ .

Option C

Sign slip on the constant, giving $+ 3$ instead of $- 3$ ; then $(3 x + 1)$ is not a factor at all.

Skills used

HCF

\times

LCM = product of the two numbers Factorise a quadratic by splitting the middle term Quotient on dividing by a linear factor

33Algebra • Equations • Linear Equations

ModerateIdeal time: about 1.5 min

If $(p + 2) (2 q - 1) = 2 pq - 10$ and $(p - 2) (2 q - 1) = 2 pq - 10$ , then what is $pq$ equal to?

- 10

- 5

5

10

Show answer & solution

Option C

5

Approach

The two equations have the same right-hand side, so subtracting them kills $pq$ and leaves a product that must be zero.

Eliminate one variable from a linear pair

Step-by-step solution

1
$(p + 2) (2 q - 1) - (p - 2) (2 q - 1) = 0$
Why: Both left-hand sides equal $2 pq - 10$ , so their difference is zero. Subtracting is cheaper than expanding either equation.
Eliminate one variable from a linear pair
2
$[(p + 2) - (p - 2)] (2 q - 1) = 0 \Rightarrow 4 (2 q - 1) = 0$
Why: Take out the common factor $(2 q - 1)$ ; the $p$ terms cancel inside the bracket.
3
$2 q - 1 = 0 \Rightarrow q = \frac{1}{2}$
Why: A product is zero only when a factor is zero, and $4$ is not zero.
A product is zero only when a factor is zero
4
Substituting in the first equation: $(p + 2) (0) = 2 p (\frac{1}{2}) - 10$ , i.e. $0 = p - 10$
Why: With $2 q - 1 = 0$ the left side vanishes, which turns the first equation into a linear equation in $p$ alone.
5
$p = 10$ , so $pq = 10 \times \frac{1}{2} = 5$
Why: Both unknowns are now fixed, and their product is what was asked for.

Shortcut

The two left-hand sides are equal, so $4 (2 q - 1) = 0$ and $q = \frac{1}{2}$ ; the first equation then collapses to $p = 10$ , giving $pq = 5$ .

Common mistakes

Option A

Read the constant $- 10$ on the right-hand side as the value of $pq$ .

Option B

Sign slip at the last step, reporting $- 5$ .

Option D

Answered with $p = 10$ instead of the product $pq$ .

Skills used

Eliminate one variable from a linear pair A product is zero only when a factor is zero

34Algebra • Expressions • Simplification

HardIdeal time: about 2.5 min

What is the value of $\frac{a ^{2} + a c}{a ^{2} c - c ^{3}} - \frac{a ^{2} - c ^{2}}{a ^{2} c + 2 a c ^{2} + c ^{3}} - \frac{2 c}{a ^{2} - c ^{2}} + \frac{3}{a + c}$ ?

0

1

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

\frac{6}{a + c}

Show answer & solution

Option D

\frac{6}{a + c}

Approach

Factorise every denominator, simplify each of the four fractions on its own, then put the lot over the single common denominator $c (a - c) (a + c)$ .

Combine algebraic fractions over one denominator

Step-by-step solution

1
$\frac{a ^{2} + a c}{a ^{2} c - c ^{3}} = \frac{a ( a + c )}{c ( a ^{2} - c ^{2} )} = \frac{a ( a + c )}{c ( a - c ) ( a + c )} = \frac{a}{c ( a - c )}$
Why: Take out $a$ above and $c$ below, then split $a^{2} - c^{2}$ as a difference of squares; the factor $a + c$ cancels.
Difference of two squares
2
$\frac{a ^{2} - c ^{2}}{a ^{2} c + 2 a c ^{2} + c ^{3}} = \frac{( a + c ) ( a - c )}{c ( a ^{2} + 2 a c + c ^{2} )} = \frac{( a + c ) ( a - c )}{c ( a + c ) ^{2}} = \frac{a - c}{c ( a + c )}$
Why: The denominator is $c$ times a perfect square; cancel one $a + c$ .
Square of a sum and of a difference
3
The four terms are $\frac{a}{c ( a - c )} - \frac{a - c}{c ( a + c )} - \frac{2 c}{( a - c ) ( a + c )} + \frac{3}{a + c}$
Why: All four denominators divide $c (a - c) (a + c)$ , so that is the common denominator.
Combine algebraic fractions over one denominator
4
Numerator over $c (a - c) (a + c)$ : $a (a + c) - (a - c)^{2} - 2 c^{2} + 3 c (a - c)$
Why: Each fraction is multiplied above and below by whatever its denominator is missing: $(a + c)$ , $(a - c)$ , $c$ and $c (a - c)$ respectively.
5
$= a^{2} + a c - (a^{2} - 2 a c + c^{2}) - 2 c^{2} + 3 a c - 3 c^{2} = 6 a c - 6 c^{2}$
Why: Expand and collect. The $a^{2}$ terms cancel; the $a c$ terms give $1 + 2 + 3 = 6$ and the $c^{2}$ terms $- 1 - 2 - 3 = - 6$ .
6
$\frac{6 c ( a - c )}{c ( a - c ) ( a + c )} = \frac{6}{a + c}$
Why: $6 a c - 6 c^{2} = 6 c (a - c)$ , and that bracket cancels against the denominator.

Shortcut

Substitute $a = 2, c = 1$ into the expression; it evaluates to $2$ , and $\frac{6}{a + c}$ is the only option that gives $2$ . (A second pair, $a = 3, c = 1$ , confirms it.)

Common mistakes

Option A

A sign slip in collecting the numerator wipes out the $a c$ and $c^{2}$ terms together, leaving $0$ .

Option B

Collected the numerator as $c (a - c)$ rather than $6 c (a - c)$ and then mis-cancelled the $a + c$ as well.

Option C

No route reaches $\frac{a c}{a ^{2} + c ^{2}}$ ; $a^{2} + c^{2}$ never appears as a denominator.

Skills used

Combine algebraic fractions over one denominator Difference of two squares Square of a sum and of a difference

35Algebra • Expressions • Factorisation

ModerateIdeal time: about 1.5 min

What is the square root of $4 x^{4} + 8 x^{3} - 4 x + 1$ ?

2 x^{2} - 2 x - 1

2 x^{2} - x - 1

2 x^{2} + 2 x + 1

2 x^{2} + 2 x - 1

Show answer & solution

Option D

2 x^{2} + 2 x - 1

Approach

Assume the square root is a quadratic $a x^{2} + b x + c$ , square it, and match coefficients with the given quartic.

Equate coefficients of like powers

Step-by-step solution

1
Let $4 x^{4} + 8 x^{3} - 4 x + 1 = a x^{2} + b x + c$
Why: A quartic whose square root is a polynomial must have a quadratic square root, since degrees halve.
Equate coefficients of like powers
2
$(a x^{2} + b x + c)^{2} = a^{2} x^{4} + 2 ab x^{3} + (b^{2} + 2 a c) x^{2} + 2 b c x + c^{2}$
Why: Square of a sum of three terms: the squares, plus twice each product two at a time, sorted by power of $x$ .
Square of a sum of several terms
3
Matching $x^{4}$ and $x^{3}$ : $a^{2} = 4 \Rightarrow a = 2$ , then $2 ab = 8 \Rightarrow b = 2$
Why: Take the positive root for $a$ ; the negative choice just flips every sign of the answer. Note the given quartic has no $x^{2}$ term, so its coefficient is $0$ .
4
Matching the constant and the $x$ term: $c^{2} = 1$ and $2 b c = - 4 \Rightarrow 4 c = - 4 \Rightarrow c = - 1$
Why: $c^{2} = 1$ allows $c = \pm 1$ ; the linear coefficient $- 4$ picks out $c = - 1$ .
5
Consistency on $x^{2}$ : $b^{2} + 2 a c = 4 + 2 (2) (- 1) = 0$ , matching the missing $x^{2}$ term
Why: The fifth coefficient was not used to find $a, b, c$ , so it is a genuine check: it had to come out $0$ , and it does.
6
$4 x^{4} + 8 x^{3} - 4 x + 1 = 2 x^{2} + 2 x - 1$
Why: Taking the non-negative root for the leading coefficient is the usual convention.
denotes the non-negative root

Common mistakes

Option A

Sign slip on $b$ : $2 x^{2} - 2 x - 1$ squares to $4 x^{4} - 8 x^{3} - 4 x + 1$ , with the cubic term negative.

Option B

Took $2 ab = 8$ as $ab = 8$ or similar, landing on $b = 1$ instead of $b = 2$ .

Option C

Took $c = + 1$ from $c^{2} = 1$ without checking the $x$ term; that squares to $4 x^{4} + 8 x^{3} + 8 x^{2} + 4 x + 1$ .

Skills used

Equate coefficients of like powers Square of a sum of several terms

denotes the non-negative root

36Algebra • Equations • Linear Equations

EasyIdeal time: about 1 min

The sum of the digits of a two digit number is $13$ and the difference between the number and that formed by reversing the digits is $27$ . What is the product of the digits of the number?

35

40

45

54

Show answer & solution

Option B

40

Approach

Write the number in place-value form; the sum condition and the reversal condition become two linear equations in the digits.

Place-value expansion of a numeral

Step-by-step solution

1
Let the number be $10 a + b$ , so the reversed number is $10 b + a$
Why: Place-value expansion is what makes 'reversing the digits' an algebraic operation.
Place-value expansion of a numeral
2
$a + b = 13$
Why: The sum of the digits, given directly.
3
$(10 a + b) - (10 b + a) = 9 (a - b) = 27 \Rightarrow a - b = 3$
Why: The difference between a two-digit number and its reversal is always nine times the difference of the digits - which is why $27$ gives such a clean condition.
4
Adding: $2 a = 16 \Rightarrow a = 8$ , so $b = 5$
Why: Add the two equations to eliminate $b$ , then back-substitute.
Eliminate one variable from a linear pair
5
Product of the digits $= 8 \times 5 = 40$
Why: The question asks for the product, not for the number.

Common mistakes

Option A

Used a digit sum of $12$ , giving the digits $7$ and $5$ and a product of $35$ .

Option C

Used a digit sum of $14$ , giving the digits $9$ and $5$ and a product of $45$ .

Option D

No route reaches $54$ ; the digits $9$ and $6$ sum to $15$ , not $13$ .

Skills used

Place-value expansion of a numeral Eliminate one variable from a linear pair

37Arithmetic • Ratio and Proportion • Proportion

EasyIdeal time: about 1 min

If $\frac{x}{b + c} = \frac{y}{c + a} = \frac{z}{b - a}$ , then which one of the following is correct?

x + y + z = 0

x - y - z = 0

x + y - z = 0

x + 2 y + 3 z = 0

Show answer & solution

Option B

x - y - z = 0

Approach

Set the three equal ratios to a common value $k$ , write $x$ , $y$ and $z$ in terms of $k$ , and see which combination of them cancels identically.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
Let $\frac{x}{b + c} = \frac{y}{c + a} = \frac{z}{b - a} = k$
Why: Equal ratios are handled by naming the common value; every numerator then becomes $k$ times its denominator.
Introduce a common factor $k$ for a ratio
2
$x = k (b + c), y = k (c + a), z = k (b - a)$
Why: Now any combination of $x$ , $y$ and $z$ is $k$ times the same combination of the brackets, so $k$ never needs a value.
3
$x - y - z = k [(b + c) - (c + a) - (b - a)]$
Why: The brackets to subtract are chosen so that each of $a$ , $b$ and $c$ appears with coefficients that can cancel: $b$ once and once, $c$ once and once, $a$ once and once.
4
$(b + c) - (c + a) - (b - a) = b + c - c - a - b + a = 0$ , so $x - y - z = 0$
Why: Every letter cancels, so the identity holds for all $a, b, c$ - which is what 'correct' means here.

Shortcut

Put $a = 1, b = 2, c = 3$ and $k = 1$ : $x = 5, y = 4, z = 1$ . Only $x - y - z$ comes out zero.

Common mistakes

Option A

Expected the symmetric pattern that would hold if the third denominator were $a + b$ or $- (a + b)$ ; with the printed $b - a$ it is $x - y - z$ that vanishes, not $x + y + z$ .

Option C

Sign slip on one bracket, giving $x + y - z = k (2 c + 2 a)$ mistaken for zero.

Option D

No route reaches $x + 2 y + 3 z = 0$ ; the substitution leaves $k (4 b + 3 c - a)$ .

Skills used

Introduce a common factor

k

for a ratio

38Arithmetic • Time, Speed and Distance • Relative Speed

HardIdeal time: about 2 min

X, Y and Z travel from the same place with uniform speeds $4$ km/hr, $5$ km/hr and $6$ km/hr respectively. Y starts $2$ hours after X. How long after Y must Z start in order that they overtake X at the same instant?

\frac{3}{2}

hoursB

\frac{4}{3}

hoursC

\frac{9}{8}

hoursD

\frac{11}{8}

hours

Show answer & solution

Option B

\frac{4}{3}

hours

Approach

Find when and where Y overtakes X; that fixes the meeting point, and Z's own travel time to that point then fixes when Z must leave.

Speed, distance and time

Step-by-step solution

1
Taking X's start as time $0$ : X is at $4 t$ km and Y at $5 (t - 2)$ km
Why: Y starts two hours later, so Y has been travelling for $t - 2$ hours when X has been travelling for $t$ .
Speed, distance and time
2
$4 t = 5 (t - 2) \Rightarrow t = 10$ hours, at $4 \times 10 = 40$ km
Why: Equivalently by relative speed: Y gains $5 - 4 = 1$ km/hr on X and starts $2 \times 4 = 8$ km behind, so Y needs $8$ hours to close the gap, i.e. $t = 10$ .
Relative speed: add when opposed, subtract when alike
3
Z must also be at $40$ km when $t = 10$ ; at $6$ km/hr that takes $\frac{40}{6} = \frac{20}{3}$ hours
Why: 'Overtake X at the same instant' means all three are at the same place at $t = 10$ , so Z's journey is a fixed $40$ km.
4
Z therefore starts at $t = 10 - \frac{20}{3} = \frac{10}{3}$ hours
Why: Work backwards from the meeting time by Z's travel time.
5
After Y: $\frac{10}{3} - 2 = \frac{4}{3}$ hours
Why: The question asks how long after Y, not after X. Y started at $t = 2$ , so subtract that.

Shortcut

Y catches X after $\frac{2 \times 4}{5 - 4} = 8$ hours of Y's travel, at $40$ km. Z covers $40$ km in $\frac{20}{3}$ hours, so Z leaves $\frac{20}{3}$ hours before the meeting - that is $8 - \frac{20}{3} = \frac{4}{3}$ hours after Y set off.

Common mistakes

Option A

Rounded $\frac{40}{6}$ to $6.5$ hours, making Z start at $3.5$ and giving $1.5$ hours after Y.

Option C

No route reaches $\frac{9}{8}$ hours.

Option D

No route reaches $\frac{11}{8}$ hours.

Skills used

Speed, distance and time Relative speed: add when opposed, subtract when alike

39Algebra • Expressions • Factorisation

EasyIdeal time: about 1 min

$1 - x - x^{n} + x^{n + 1}$ , where n is a natural number, is divisible by

(1 + x)^{2}

(1 - x)^{2}

1 - 2 x - x^{2}

1 + 2 x - x^{2}

Show answer & solution

Option B

(1 - x)^{2}

Approach

Group the four terms in pairs; both pairs share the factor $1 - x$ , and the second bracket then supplies another copy of it.

a - b

divides

a^{n} - b^{n}

for every

n

Step-by-step solution

1
$1 - x - x^{n} + x^{n + 1} = (1 - x) - x^{n} (1 - x)$
Why: Group the first two terms and the last two. Taking $x^{n}$ out of $- x^{n} + x^{n + 1}$ leaves $- x^{n} (1 - x)$ , matching the first bracket.
Factorise a quadratic by splitting the middle term
2
$= (1 - x) (1 - x^{n})$
Why: Both groups carry the factor $1 - x$ , so it comes out.
3
$1 - x^{n} = (1 - x) (1 + x + x^{2} + \dots + x^{n - 1})$
Why: $a - b$ divides $a^{n} - b^{n}$ for every $n$ ; here $a = 1$ and $b = x$ , so a second factor of $1 - x$ appears.
$a - b$ divides $a^{n} - b^{n}$ for every $n$
4
$1 - x - x^{n} + x^{n + 1} = (1 - x)^{2} (1 + x + \dots + x^{n - 1})$
Why: So the expression is divisible by $(1 - x)^{2}$ for every natural $n$ .

Shortcut

Put $n = 1$ : the expression is $1 - 2 x + x^{2} = (1 - x)^{2}$ , so only $(1 - x)^{2}$ can be the general answer.

Common mistakes

Option A

Took the repeated factor as $1 + x$ , grouping the terms as $(1 + x)$ -type pairs.

Option C

Expanded $(1 - x)^{2}$ as $1 - 2 x - x^{2}$ , with the sign of the $x^{2}$ term wrong.

Option D

No route reaches $1 + 2 x - x^{2}$ .

Skills used

a - b

divides

a^{n} - b^{n}

for every

n

Factorise a quadratic by splitting the middle term

40Algebra • Quadratic Equations • Word Problems

ModerateIdeal time: about 1.5 min

A person sold an article for Rs. $75$ which cost him Rs. x. He finds that he realised $x %$ profit on his outlay. What is x equal to?

20%

25%

50%

100%

Show answer & solution

Option C

50%

Approach

Write the profit two ways - as selling price minus cost, and as $x %$ of the outlay $x$ - and equate them into a quadratic.

Gain percent is measured on the cost price

Step-by-step solution

1
Cost $=$ Rs. $x$ , selling price $=$ Rs. $75$ , so the profit is Rs. $(75 - x)$
Why: Profit is simply selling price less cost price.
Gain percent is measured on the cost price
2
A profit of $x %$ on the outlay $x$ is $\frac{x}{100} \times x = \frac{x ^{2}}{100}$
Why: Gain percent is measured on the cost price, and here the cost price and the percentage happen to be the same number $x$ - which is what makes the equation quadratic.
3
$\frac{x ^{2}}{100} = 75 - x \Rightarrow x^{2} + 100 x - 7500 = 0$
Why: Multiply by $100$ and bring everything to one side.
4
$x^{2} + 150 x - 50 x - 7500 = 0 \Rightarrow (x + 150) (x - 50) = 0$
Why: Split the middle term: $150 \times (- 50) = - 7500$ and $150 - 50 = 100$ .
Factorise a quadratic by splitting the middle term
5
$x = 50$ (rejecting $x = - 150$ )
Why: A cost price cannot be negative.
A product is zero only when a factor is zero

Common mistakes

Option A

No route reaches $20$ ; an outlay of $20$ at $20%$ sells for Rs. $24$ .

Option B

Mis-split the middle term, pairing $- 7500$ as $300 \times (- 25)$ and reading off $25$ without checking that $300 - 25 \neq = 100$ .

Option D

No route reaches $100$ ; an outlay of $100$ at $100%$ sells for Rs. $200$ .

Skills used

Gain percent is measured on the cost price Factorise a quadratic by splitting the middle term A product is zero only when a factor is zero