Elementary Mathematics 2020 (I)

Set A

Questions 21–40 of 100

21Algebra • Logarithm • Equations

EasyIdeal time: about 50 s

If $lo g x = 1.2500$ and $y = x^{l o g x}$ , then what is $lo g y$ equal to ?

4.2500

2.5625

1.5625

1.2500

Show answer & solution

Option C

1.5625

Approach

Take logarithms of $y = x^{l o g x}$ and use the power rule, which turns the exponent into a multiplier.

lo g a^{n} = n lo g a

Step-by-step solution

1
$y = x^{l o g x} \Rightarrow lo g y = lo g (x^{l o g x})$
Why: The unknown sits in an exponent, so a logarithm is the move that brings it down.
$lo g a^{n} = n lo g a$
2
$lo g y = (lo g x) (lo g x) = (lo g x)^{2}$
Why: The power rule $lo g a^{n} = n lo g a$ applies with $n = lo g x$ ; the index becomes a factor.
3
$lo g y = (1.2500)^{2} = 1.5625$
Why: Substituting the given value; $1.2 5^{2} = (\frac{5}{4})^{2} = \frac{25}{16} = 1.5625$ .

Shortcut

$1.25 = \frac{5}{4}$ , so the square is $\frac{25}{16}$ , which is $1.5625$ without any decimal multiplication.

Common mistakes

Option A

No route reaches $4.25$ from $lo g x = 1.25$ .

Option B

Added $1$ to the correct $1.5625$ , typically by writing $lo g y = lo g x + (lo g x)^{2}$ instead of $(lo g x)^{2}$ alone.

Option D

Treated the exponent as $1$ , so $lo g y = lo g x = 1.25$ .

Skills used

lo g a^{n} = n lo g a

22Algebra • Polynomial • Factor and Remainder Theorem

HardIdeal time: about 2 min

If $f (x)$ is divided by $(x - α) (x - β)$ where $α \neq = β$ , then what is the remainder ?

\frac{( x - α ) f ( α ) - ( x - β ) f ( β )}{α - β}

\frac{( x - α ) f ( β ) - ( x - β ) f ( α )}{α - β}

\frac{( x - β ) f ( α ) - ( x - α ) f ( β )}{α - β}

\frac{( x - β ) f ( β ) - ( x - α ) f ( α )}{α - β}

Show answer & solution

Option C

\frac{( x - β ) f ( α ) - ( x - α ) f ( β )}{α - β}

Approach

The remainder on dividing by a quadratic is linear, so write it as $A x + B$ , evaluate at $x = α$ and $x = β$ by the remainder theorem, and solve for the linear expression.

Factor and remainder theorem

Step-by-step solution

1
$f (x) = (x - α) (x - β) Q (x) + R (x)$ with $R$ of degree at most $1$ , say $R (x) = A x + B$ .
Why: A remainder must be of lower degree than the divisor, and the divisor here is a quadratic.
Factor and remainder theorem Division algorithm for polynomials
2
Putting $x = α$ : $f (α) = R (α)$ . Putting $x = β$ : $f (β) = R (β)$ .
Why: At $x = α$ and $x = β$ the product $(x - α) (x - β)$ vanishes, so $Q$ drops out whatever it is.
3
$R$ is the unique linear function through $(α, f (α))$ and $(β, f (β))$ , which since $α \neq = β$ is $R (x) = f (α) \cdot \frac{x - β}{α - β} + f (β) \cdot \frac{x - α}{β - α}$ .
Why: Each of the two fractions equals $1$ at its own point and $0$ at the other, so the sum takes the required value at both -- this is the quickest way to write the line through two points without solving for $A$ and $B$ .
Eliminate one variable from a linear pair
4
$R (x) = \frac{( x - β ) f ( α ) - ( x - α ) f ( β )}{α - β}$
Why: Writing $β - α = - (α - β)$ puts both terms over the single denominator $α - β$ .

Shortcut

Rather than deriving it, test each option at $x = α$ : the correct remainder must return $f (α)$ exactly.

Common mistakes

Option A

Paired each bracket with the wrong value: at $x = α$ this returns $- f (β)$ , not $f (α)$ .

Option B

This is the negative of the correct expression; at $x = α$ it gives $- f (α)$ .

Option D

Paired $(x - β)$ with $f (β)$ and $(x - α)$ with $f (α)$ -- each bracket with its own value, so both check-points fail.

Skills used

Factor and remainder theorem Eliminate one variable from a linear pair Division algorithm for polynomials

23Algebra • Expressions • Factorisation

EasyIdeal time: about 55 s

If the area of a square is $2401 x^{4} + 196 x^{2} + 4$ , then what is its side length ?

49 x^{2} + 3 x + 2

49 x^{2} - 3 x + 2

49 x^{2} + 2

59 x^{2} + 2

Show answer & solution

Option C

49 x^{2} + 2

Approach

The side is the square root of the area, so test whether the trinomial is a perfect square of the form $(a x^{2} + b)^{2}$ .

Square of a sum and of a difference

Step-by-step solution

1
Side $= 2401 x^{4} + 196 x^{2} + 4$
Why: For a square of area $A$ the side is $A$ , so the trinomial must be a perfect square.
Perimeter and area of a square
2
$2401 x^{4} = (49 x^{2})^{2}$ and $4 = 2^{2}$ , so try $(49 x^{2} + 2)^{2}$ .
Why: The first and last terms of a perfect-square trinomial are the squares of the two parts, which names the candidate before any expansion.
Square of a sum and of a difference
3
$(49 x^{2} + 2)^{2} = (49 x^{2})^{2} + 2 (49 x^{2}) (2) + 2^{2} = 2401 x^{4} + 196 x^{2} + 4$
Why: The middle term must come out as $2 ab$ ; here $2 (49) (2) = 196$ , which matches exactly -- so no term in $x$ is needed.
4
Side $= 49 x^{2} + 2$
Why: Lengths are positive, so the non-negative square root is taken.
denotes the non-negative root

Common mistakes

Option A

Assumed a square root must carry a term in $x$ and guessed $(49 x^{2} + 3 x + 2)$ ; squaring it produces an $x^{3}$ term and an $x^{2}$ coefficient of $205$ , neither of which the area has.

Option B

The same assumption with a negative middle term; squaring gives $- 294 x^{3}$ , which the area does not contain.

Option D

Mis-rooted $2401$ as $59$ rather than $49$ ; $5 9^{2} = 3481$ .

Skills used

Square of a sum and of a difference Perimeter and area of a square

denotes the non-negative root

24Arithmetic • Ratio and Proportion • Direct and Inverse Variation

ModerateIdeal time: about 1 min

If $x$ varies as $y z$ , then $y$ varies inversely as

x z

\frac{x}{z}

\frac{z}{x}

\frac{1}{( x z )}

Show answer & solution

Option C

\frac{z}{x}

Approach

Write the variation as an equation with a constant, solve it for $y$ , and read off which quantity $y$ is inversely proportional to.

Direct variation as a power:

y \propto x^{n}

Step-by-step solution

1
$x \propto y z \Rightarrow x = k y z$ for a constant $k \neq = 0$ .
Why: 'Varies as' always means 'equals a constant times', and writing the constant is what makes the algebra possible.
Direct variation as a power: $y \propto x^{n}$
2
$y = \frac{x}{k z} = \frac{1}{k} \cdot \frac{x}{z}$
Why: Rearranging for $y$ , the variable the question asks about.
3
$y = \frac{1}{k} \cdot \frac{1}{( \frac{z}{x} )}$ , so $y$ varies inversely as $\frac{z}{x}$ .
Why: ' $y$ varies inversely as $W$ ' means $y = \frac{c}{W}$ ; here $W = \frac{z}{x}$ , because $\frac{x}{z}$ is the reciprocal of $\frac{z}{x}$ .
Negative exponent is a reciprocal

Common mistakes

Option A

Wrote the product of the two other letters by reflex; $y \propto \frac{1}{x z}$ would mean $x y z$ is constant, but the relation makes $\frac{x}{y z}$ constant.

Option B

Gave the quantity $y$ varies DIRECTLY as, since $y = \frac{1}{k} \cdot \frac{x}{z}$ ; the question asks for the inverse.

Option D

Took the reciprocal one time too many: $y$ inversely as $\frac{1}{x z}$ would mean $y \propto x z$ .

Skills used

Direct variation as a power:

y \propto x^{n}

Negative exponent is a reciprocal

25Arithmetic • Number System • Rational Numbers

EasyIdeal time: about 1 min

If the points $P$ and $Q$ represent real numbers $0.7 3$ and $0.5 6$ on the number line, then what is the distance between $P$ and $Q$ ?

\frac{1}{6}

\frac{1}{5}

\frac{16}{45}

\frac{11}{90}

Show answer & solution

Option A

\frac{1}{6}

Approach

Convert both recurring decimals to fractions, then subtract.

Convert a recurring decimal to a fraction

Step-by-step solution

1
$0.7 3 = \frac{73 - 7}{90} = \frac{66}{90} = \frac{11}{15}$
Why: For a decimal with one non-recurring and one recurring digit after the point, the numerator is 'all the digits minus the non-recurring ones' and the denominator is one $9$ per recurring digit followed by one $0$ per non-recurring digit.
Convert a recurring decimal to a fraction
2
$0.5 6 = \frac{56 - 5}{90} = \frac{51}{90} = \frac{17}{30}$
Why: The same rule, with digits $5$ and $6$ .
3
$P Q = \frac{11}{15} - \frac{17}{30} = \frac{22}{30} - \frac{17}{30} = \frac{5}{30} = \frac{1}{6}$
Why: Distance on a number line is the difference of the two coordinates, and $0.733 \dots > 0.566 \dots$ , so this order is already positive.
Comparing two fractions

Shortcut

Subtract the decimals directly: $0.7333 \dots - 0.5666 \dots = 0.1666 \dots = \frac{1}{6}$ .

Common mistakes

Option B

No route reaches $\frac{1}{5} = 0.2$ ; even a rough decimal subtraction gives about $0.17$ .

Option C

No route reaches $\frac{16}{45} = 0.356$ , which is more than twice the gap between the two points.

Option D

Converted $0.5 6$ as $\frac{56 - 1}{90} = \frac{55}{90}$ , subtracting the wrong digit, which leaves $\frac{66 - 55}{90} = \frac{11}{90}$ .

Skills used

Convert a recurring decimal to a fraction Comparing two fractions

26Algebra • Equations • Linear Equations

ModerateIdeal time: about 1.5 min

What is the point on the $x y$ -plane satisfying $5 x + 2 y = 7 x y$ and $10 x + 3 y = 8 x y$ ?

(- 1, \frac{1}{6})

(\frac{1}{6}, - 1)

(1, \frac{1}{6})

(- \frac{1}{6}, - 1)

Show answer & solution

Option B

(\frac{1}{6}, - 1)

Approach

Divide each equation by $x y$ so that only $\frac{1}{x}$ and $\frac{1}{y}$ appear, solve the resulting linear pair, then invert.

Linearise a system by substituting reciprocals

Step-by-step solution

1
Neither $x$ nor $y$ can be $0$ (the equations would force the other to be $0$ too, and $(0, 0)$ is not offered), so divide both by $x y$ : $\frac{5}{y} + \frac{2}{x} = 7$ and $\frac{10}{y} + \frac{3}{x} = 8$ .
Why: Every term of each equation carries $x$ , $y$ or $x y$ , so dividing by $x y$ turns a pair of bilinear equations into a pair of linear ones.
Linearise a system by substituting reciprocals
2
With $u = \frac{1}{x}$ , $v = \frac{1}{y}$ : $2 u + 5 v = 7$ and $3 u + 10 v = 8$ .
Why: Naming the reciprocals makes the system an ordinary linear pair.
3
$2 \times (2 u + 5 v = 7)$ gives $4 u + 10 v = 14$ ; subtracting $3 u + 10 v = 8$ leaves $u = 6$ .
Why: Doubling the first equation matches the coefficients of $v$ , so one subtraction removes $v$ .
Eliminate one variable from a linear pair
4
$2 (6) + 5 v = 7 \Rightarrow 5 v = - 5 \Rightarrow v = - 1$
Why: Back-substitution. A negative $v$ is fine -- nothing in the problem requires positive coordinates.
5
$x = \frac{1}{u} = \frac{1}{6}$ and $y = \frac{1}{v} = - 1$ , so the point is $(\frac{1}{6}, - 1)$ .
Why: The question asks for the point, not for $u$ and $v$ -- the reciprocals must be undone.

Common mistakes

Option A

Wrote the coordinates the wrong way round: $(- 1, \frac{1}{6})$ puts $x = - 1$ , and then $5 (- 1) + 2 (\frac{1}{6}) = - \frac{14}{3}$ while $7 x y = - \frac{7}{6}$ .

Option C

No route reaches $(1, \frac{1}{6})$ ; it fails both equations.

Option D

Dropped a sign when inverting, reporting $x = - \frac{1}{6}$ in place of $\frac{1}{6}$ .

Skills used

Linearise a system by substituting reciprocals Eliminate one variable from a linear pair

27Arithmetic • Percentage • Successive Increase and Decrease

ModerateIdeal time: about 2 min

The price of an article $X$ increases by $20%$ every year and price of article $Y$ increases by $10%$ every year. In the year $2010$ , the price of article $X$ was Rs. $5000$ and price of article $Y$ was Rs. $2000$ . In which year the difference in their prices exceeded Rs. $5000$ for the first time ?

2012

2013

2014

2015

Show answer & solution

Option B

2013

Approach

Apply the growth multipliers year by year and tabulate the difference until it passes Rs. $5000$ .

Percentage change as a multiplier

Step-by-step solution

1
After $n$ years from $2010$ : $X = 5000 (1.2)^{n}$ and $Y = 2000 (1.1)^{n}$ .
Why: An increase of $p %$ every year multiplies the price by $1 + \frac{p}{100}$ each year, so $n$ years of growth is that factor to the $n$ -th power.
Percentage change as a multiplier
2
$2010$ : $5000 - 2000 = 3000$ . $2011$ : $6000 - 2200 = 3800$ .
Why: Stepping year by year is quicker here than solving an inequality, because only a handful of years need testing.
3
$2012$ : $7200 - 2420 = 4780$ , still below $5000$ .
Why: $X$ : $6000 \times 1.2 = 7200$ ; $Y$ : $2200 \times 1.1 = 2420$ .
4
$2013$ : $8640 - 2662 = 5978 > 5000$ .
Why: $X$ : $7200 \times 1.2 = 8640$ ; $Y$ : $2420 \times 1.1 = 2662$ . This is the first year the difference exceeds Rs. $5000$ .

Shortcut

Only the year of crossing matters, so track the two prices as a running pair -- $(5000, 2000) \to (6000, 2200) \to (7200, 2420) \to (8640, 2662)$ -- and stop at the first difference above $5000$ .

Common mistakes

Option A

Stopped at $2012$ , where the difference is $4780$ -- close to $5000$ but short of it.

Option C

Counted $2010$ itself as the first year of growth, so the crossing came out one year late.

Option D

No route reaches $2015$ ; by then the difference is over Rs. $9200$ .

Skills used

Percentage change as a multiplier

28Arithmetic • Time, Speed and Distance • Races

ModerateIdeal time: about 1.5 min

The ratio of speeds of $X$ and $Y$ is $5 : 6$ . If $Y$ allows $X$ a start of $70$ m in a $1.2$ km race, then who will win the race and by what distance ?

X

wins the race by

30

Y

wins the race by

90

Y

wins the race by

130

mDThe race finishes in a dead heat

Show answer & solution

Option C

Y

wins the race by

130

Approach

Read the head start as a shorter distance for $X$ , give the two runners speeds $5 k$ and $6 k$ , and see where the slower one stands when the faster one finishes.

Head starts and winning margins in a race

Step-by-step solution

1
Let the speeds be $X = 5 k$ and $Y = 6 k$ . The race is $1.2$ km $= 1200$ m.
Why: A ratio fixes the speeds only up to a common factor; $k$ will cancel, so it never has to be found.
Introduce a common factor $k$ for a ratio
2
A start of $70$ m means $X$ has only $1200 - 70 = 1130$ m to run, while $Y$ runs the full $1200$ m.
Why: A start measured in metres shortens the distance of the runner who receives it; it does not change anyone's speed.
Head starts and winning margins in a race
3
$Y$ 's time $= \frac{1200}{6 k} = \frac{200}{k}$ ; $X$ 's time $= \frac{1130}{5 k} = \frac{226}{k}$ . So $Y$ finishes first.
Why: Comparing times settles the winner; the common $k$ cancels out of the comparison.
Speed, distance and time
4
In $Y$ 's winning time, $X$ covers $5 k \times \frac{200}{k} = 1000$ m, so $X$ stands $70 + 1000 = 1070$ m along the track.
Why: The margin has to be measured from the finish line, so $X$ 's $70$ m start must be added back to the distance he has actually run.
5
Margin $= 1200 - 1070 = 130$ m, so $Y$ wins by $130$ m.
Why: 'Wins by $d$ metres' means the loser still has $d$ metres to run when the winner finishes.

Shortcut

In the time $Y$ runs $1200$ m, $X$ runs $\frac{5}{6}$ of it, namely $1000$ m. $X$ needed only $1130$ m, so he is $130$ m short -- and for a dead heat the start would have had to be $1200 - 1000 = 200$ m, not $70$ m.

Common mistakes

Option A

No route makes $X$ the winner: he would need a start of at least $200$ m, and $70$ m is less than that.

Option B

No clean route reaches $90$ m; the margin is fixed at $1200 - (1000 + 70)$ .

Option D

Equated the two times without checking, or found the dead-heat start of $200$ m and read the given $70$ m as sufficient.

Skills used

Head starts and winning margins in a race Introduce a common factor

k

for a ratio Speed, distance and time

29Arithmetic • Time, Speed and Distance

EasyIdeal time: about 1.5 min

A train takes two hours less for a journey of $300$ km if its speed is increased by $5$ km/hr from its usual speed. What is its usual speed ?

50

km/hrB

40

km/hrC

35

km/hrD

25

km/hr

Show answer & solution

Option D

25

km/hr

Approach

Write both journey times with the usual speed as the unknown, set their difference to $2$ hours, and solve the quadratic.

Speed, distance and time

Step-by-step solution

1
Let the usual speed be $v$ km/hr. Times: $\frac{300}{v}$ and $\frac{300}{v + 5}$ hours.
Why: Time $= \frac{distance}{speed}$ , and the distance is the same $300$ km in both cases.
Speed, distance and time
2
$\frac{300}{v} - \frac{300}{v + 5} = 2$
Why: The faster journey is the shorter one, so the saving is the slower time minus the faster time.
Turn a worded condition into a linear equation in one unknown
3
$300 [(v + 5) - v] = 2 v (v + 5) \Rightarrow 1500 = 2 v^{2} + 10 v \Rightarrow v^{2} + 5 v - 750 = 0$
Why: Multiplying through by $v (v + 5)$ clears the fractions; $v > 0$ , so the multiplier is not zero.
Clear the denominators of an equation
4
$v^{2} + 30 v - 25 v - 750 = 0 \Rightarrow (v + 30) (v - 25) = 0$
Why: Splitting the middle term: $30 \times (- 25) = - 750$ and $30 - 25 = 5$ .
Factorise a quadratic by splitting the middle term
5
$v = 25$ (rejecting $v = - 30$ , as a speed is positive).
Why: A negative root has no meaning for a train's speed.
A product is zero only when a factor is zero

Shortcut

The equation reduces to $v (v + 5) = 750$ , so look for two numbers five apart with product $750$ : $25 \times 30$ .

Common mistakes

Option A

At $50$ km/hr the saving would be $6 - \frac{300}{55} \approx 0.55$ hours, nowhere near $2$ .

Option B

At $40$ km/hr the saving would be $7.5 - \frac{300}{45} \approx 0.83$ hours.

Option C

At $35$ km/hr the saving would be $\frac{300}{35} - 7.5 \approx 1.07$ hours.

Skills used

Speed, distance and time Turn a worded condition into a linear equation in one unknown Clear the denominators of an equation Factorise a quadratic by splitting the middle term A product is zero only when a factor is zero

30Arithmetic • Time and Work

EasyIdeal time: about 1.5 min

If $6$ men and $8$ women can do a piece of work in $10$ days; and $13$ men and $24$ women can do the same work in $4$ days, then what is the ratio of daily work done by a man to that of a woman ?

2 : 1

1 : 2

4 : 3

3 : 4

Show answer & solution

Option A

2 : 1

Approach

Both teams finish the same job, so their total man-day counts are equal; equate them and compare the two daily rates.

The work-equivalence chain rule

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

Step-by-step solution

1
Let a man do $m$ units of work a day and a woman $w$ units.
Why: Naming the daily rates is what lets the two statements be compared; the job itself never has to be measured.
The work-equivalence chain rule $M_{1} D_{1} H_{1} = M_{2} D_{2} H_{2}$
2
$10 (6 m + 8 w) = 4 (13 m + 24 w)$
Why: Both sides are the same complete job, counted as daily output times number of days.
3
$60 m + 80 w = 52 m + 96 w \Rightarrow 8 m = 16 w \Rightarrow m = 2 w$
Why: Collecting the $m$ terms on one side and the $w$ terms on the other.
Eliminate one variable from a linear pair
4
$m : w = 2 : 1$
Why: The question asks for a man's daily work against a woman's, in that order.
Introduce a common factor $k$ for a ratio

Shortcut

Take the first team's ten days as $\frac{10}{4} = 2.5$ times the second's four, so $13 m + 24 w = 2.5 (6 m + 8 w) = 15 m + 20 w$ , giving $4 w = 2 m$ straight away.

Common mistakes

Option B

Reported woman to man instead of man to woman.

Option C

Mis-subtracted the coefficients, reaching $6 m = 8 w$ and so $m : w = 4 : 3$ .

Option D

Mis-subtracted and then also inverted, reaching $m : w = 3 : 4$ .

Skills used

The work-equivalence chain rule

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

Eliminate one variable from a linear pair Introduce a common factor

k

for a ratio

31Algebra • Equations • Linear Equations

ModerateIdeal time: about 1.5 min

Students of a class are made to sit in rows of equal number of chairs. If number of students is increased by $2$ in each row, then the number of rows decreases by $3$ . If number of students is increased by $4$ in each row, then the number of rows decreases by $5$ . What is the number of students in the class ?

100

105

110

120

Show answer & solution

Option D

120

Approach

Let the arrangement be $r$ rows of $c$ chairs; each of the two conditions keeps the total unchanged, giving two linear equations.

Eliminate one variable from a linear pair

Step-by-step solution

1
Let there be $r$ rows of $c$ students, so the class is $N = r c$ .
Why: Both re-seatings hold the same students, so the product $r c$ is the quantity that must be preserved.
Turn a worded condition into a linear equation in one unknown
2
$(c + 2) (r - 3) = r c \Rightarrow r c - 3 c + 2 r - 6 = r c \Rightarrow 2 r - 3 c = 6$
Why: The $r c$ terms cancel, which is what makes a product condition collapse into a linear one.
3
$(c + 4) (r - 5) = r c \Rightarrow r c - 5 c + 4 r - 20 = r c \Rightarrow 4 r - 5 c = 20$
Why: The second condition, treated the same way.
4
$2 \times (2 r - 3 c = 6)$ gives $4 r - 6 c = 12$ ; subtracting from $4 r - 5 c = 20$ leaves $c = 8$ .
Why: Doubling the first equation matches the $r$ coefficients, so one subtraction removes $r$ .
Eliminate one variable from a linear pair
5
$2 r - 3 (8) = 6 \Rightarrow 2 r = 30 \Rightarrow r = 15$ , so $N = 15 \times 8 = 120$ .
Why: Back-substitution, then the product the question actually asks for.

Common mistakes

Option A

$100$ has no factorisation $r \times c$ meeting either condition.

Option B

$105$ likewise fails both conditions.

Option C

$110$ likewise fails both conditions.

Skills used

Eliminate one variable from a linear pair Turn a worded condition into a linear equation in one unknown

32Arithmetic • Interest • Simple Interest

EasyIdeal time: about 45 s

A sum was put at simple interest at certain rate for $2$ years. Had it been put at $1%$ higher rate of interest, it would have fetched Rs. $24$ more. What is the sum ?

ARs.

500

BRs.

600

CRs.

800

DRs.

1200

Show answer & solution

Option D

Rs.

1200

Approach

Write the simple interest at both rates and subtract; the principal is all that survives.

Simple interest

= \frac{P R T}{100}

Step-by-step solution

1
At $R %$ for $2$ years: $SI = \frac{P \times R \times 2}{100}$ . At $(R + 1) %$ : $SI^{'} = \frac{P \times ( R + 1 ) \times 2}{100}$ .
Why: Only the rate changes, so writing both with the same $P$ and $T$ sets up a subtraction in which $R$ cancels.
Simple interest $= \frac{P R T}{100}$
2
$SI^{'} - SI = \frac{2 P [ ( R + 1 ) - R ]}{100} = \frac{2 P}{100}$
Why: The unknown rate drops out -- which is why the question is answerable without it.
3
$\frac{2 P}{100} = 24 \Rightarrow P = \frac{24 \times 100}{2} = 1200$
Why: One division gives the principal.

Shortcut

$1%$ extra for one year is $1%$ of the sum; over two years it is $2%$ . So $2%$ of the sum is Rs. $24$ , and the sum is Rs. $1200$ .

Common mistakes

Option A

No route reaches Rs. $500$ ; $2%$ of it is Rs. $10$ .

Option B

Treated the extra as $2%$ per year, so $4%$ of the sum $= 24$ and $P = 600$ .

Option C

No route reaches Rs. $800$ ; $2%$ of it is Rs. $16$ .

Skills used

Simple interest

= \frac{P R T}{100}

33Arithmetic • Ratio and Proportion

ModerateIdeal time: about 1.5 min

The population of two villages is $1525$ and $2600$ respectively. If the ratio of male to female population in the first village is $27 : 34$ and the ratio of male to female population in the second village is $6 : 7$ , then what is the ratio of male to female population of these two villages taken together ?

\frac{33}{41}

\frac{85}{82}

\frac{71}{90}

\frac{5}{6}

Show answer & solution

Option D

\frac{5}{6}

Approach

Split each village into males and females using its own ratio, add the two male totals and the two female totals, then form the combined ratio.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
Village 1: males $= 27 k$ , females $= 34 k$ , so $61 k = 1525 \Rightarrow k = 25$ .
Why: A ratio fixes the two parts only up to a common factor, and the village total determines it.
Introduce a common factor $k$ for a ratio
2
Males $= 27 (25) = 675$ , females $= 34 (25) = 850$ .
Why: Substituting $k$ back.
3
Village 2: males $= 6 j$ , females $= 7 j$ , so $13 j = 2600 \Rightarrow j = 200$ ; males $= 1200$ , females $= 1400$ .
Why: A second, independent ratio needs its own letter -- the two villages share no term.
4
Combined males $= 675 + 1200 = 1875$ ; combined females $= 850 + 1400 = 2250$ .
Why: 'Taken together' means the counts add, not the ratios.
5
$\frac{1875}{2250} = \frac{1875 \div 375}{2250 \div 375} = \frac{5}{6}$
Why: Both are divisible by $375$ ; equivalently divide by $75$ to get $\frac{25}{30}$ and then by $5$ .
Comparing two fractions

Common mistakes

Option A

Added the ratio terms instead of the populations: $(27 + 6) : (34 + 7) = 33 : 41$ . Ratios cannot be added -- the villages are of different sizes.

Option B

No route reaches $\frac{85}{82}$ ; the combined ratio must lie between $\frac{27}{34}$ and $\frac{6}{7}$ , both below $1$ .

Option C

No route reaches $\frac{71}{90} = 0.789$ , which is below both village ratios.

Skills used

Introduce a common factor

k

for a ratio Comparing two fractions

34Statistics • Mean • Weighted Mean

EasyIdeal time: about 1.5 min

In a class room the ratio of number of girls to that of boys is $3 : 4$ . The average height of students in the class is $4.6$ feet. If the average height of the boys in the class is $4.8$ feet, then what is the average height of the girls in the class ?

ALess than

4.2

feetBMore than

4.2

feet but less than

4.3

feetCMore than

4.3

feet but less than

4.4

feetDMore than

4.4

feet but less than

4.5

feet

Show answer & solution

Option C

More than

4.3

feet but less than

4.4

feet

Approach

Use the class average as a weighted mean of the two group averages, with the girl-boy ratio as the weights.

Mean of a frequency distribution

Step-by-step solution

1
Take the numbers as $3 k$ girls and $4 k$ boys, so the class has $7 k$ students.
Why: Only the ratio is given, and a weighted mean depends on the proportions alone, so $k$ will cancel.
Introduce a common factor $k$ for a ratio
2
Total height $= 7 k (4.6) = 32.2 k$ feet.
Why: The class mean times the number of students recovers the total.
Mean of a frequency distribution
3
Boys' total $= 4 k (4.8) = 19.2 k$ feet, so the girls' total $= 32.2 k - 19.2 k = 13.0 k$ feet.
Why: Heights add, so the girls' total is what is left after the boys' total is removed.
4
Girls' average $= \frac{13.0 k}{3 k} = \frac{13}{3} = 4.333 \dots$ feet.
Why: Dividing the girls' total by the number of girls; $k$ cancels as expected.
5
$4.3 < 4.333 \dots < 4.4$
Why: The options are intervals, so the value has to be placed rather than matched.
Comparing two fractions

Shortcut

The class mean is $0.2$ feet below the boys' mean, and boys are $\frac{4}{7}$ of the class, so the girls must be $\frac{4}{3} \times 0.2 = 0.267$ feet below $4.6$ , giving $4.333$ .

Common mistakes

Option A

No route puts the girls' average below $4.2$ : that would drag the class mean under $4.46$ .

Option B

No route reaches the interval $4.2$ to $4.3$ ; the arithmetic gives $4.333$ .

Option D

Swapped the ratio, taking $4$ girls and $3$ boys: $\frac{7 ( 4.6 ) - 3 ( 4.8 )}{4} = \frac{17.8}{4} = 4.45$ , which lands in this interval.

Skills used

Mean of a frequency distribution Introduce a common factor

k

for a ratio Comparing two fractions

35Statistics • Median

Very easyIdeal time: about 25 s

What is the median of the data $3, 5, 9, 4, 6, 11, 18$ ?

6

6.5

7

7.5

Show answer & solution

Option A

6

Approach

Sort the seven values and take the middle one.

Median of a data set

Step-by-step solution

1
Sorted: $3, 4, 5, 6, 9, 11, 18$
Why: The median is defined on ordered data, so sorting is not optional.
Median of a data set
2
$n = 7$ is odd, so the median is the $(\frac{7 + 1}{2})$ th $= 4$ th value, which is $6$ .
Why: With an odd count there is a single middle observation and no averaging is needed.

Common mistakes

Option B

No route reaches $6.5$ from this data; it is not the mean of any two adjacent sorted values.

Option C

Averaged the $3$ rd and $5$ th sorted values, $\frac{5 + 9}{2} = 7$ , as though the middle had to be straddled.

Option D

Treated $n$ as even and averaged the $4$ th and $5$ th values, $\frac{6 + 9}{2} = 7.5$ .

Skills used

Median of a data set

36Statistics • Data Interpretation • Pie Chart

Very easyIdeal time: about 25 s

In a pi-diagram there are three sectors. If the ratio of the angles of the sectors is $1 : 2 : 3$ , then what is the angle of the largest sector?

20 0^{\circ}

18 0^{\circ}

15 0^{\circ}

12 0^{\circ}

Show answer & solution

Option B

18 0^{\circ}

Approach

The three sector angles make up the full $36 0^{\circ}$ , so divide it in the given ratio.

A pie chart sector's angle is proportional to its share

Step-by-step solution

1
Angles $= k, 2 k, 3 k$ with $k + 2 k + 3 k = 36 0^{\circ}$ .
Why: The sectors of a pie chart together fill the circle, so their angles sum to $36 0^{\circ}$ .
A pie chart sector's angle is proportional to its share Introduce a common factor $k$ for a ratio
2
$6 k = 36 0^{\circ} \Rightarrow k = 6 0^{\circ}$ , so the largest sector is $3 k = 18 0^{\circ}$ .
Why: The largest ratio term gives the largest angle.

Shortcut

The largest sector is $\frac{3}{1 + 2 + 3} = \frac{1}{2}$ of the circle, so $18 0^{\circ}$ .

Common mistakes

Option A

No route reaches $20 0^{\circ}$ ; it is not $36 0^{\circ}$ times any of $\frac{1}{6}, \frac{2}{6}, \frac{3}{6}$ .

Option C

No route reaches $15 0^{\circ}$ from a $1 : 2 : 3$ split of $36 0^{\circ}$ .

Option D

Gave the middle sector $2 k = 12 0^{\circ}$ , or simply divided $36 0^{\circ}$ by the number of sectors.

Skills used

A pie chart sector's angle is proportional to its share Introduce a common factor

k

for a ratio

37Arithmetic • Ratio and Proportion

EasyIdeal time: about 50 s

The maximum marks in a Test are converted from $250$ to $50$ for the purpose of an Internal Assessment. The highest marks scored were $170$ and lowest marks were $70$ . What is the difference between the maximum and minimum marks scored in the Internal Assessment?

15

17

20

24

Show answer & solution

Option C

20

Approach

Find the conversion factor from the two maxima, then apply it to the gap between the highest and lowest marks.

Unitary method: find one unit, then scale

Step-by-step solution

1
$250$ marks become $50$ , so $1$ mark becomes $\frac{50}{250} = \frac{1}{5}$ of a mark.
Why: Finding what one unit becomes is what makes the rest a single multiplication.
Unitary method: find one unit, then scale
2
Highest $= 170 \times \frac{1}{5} = 34$ ; lowest $= 70 \times \frac{1}{5} = 14$ .
Why: The conversion is a fixed multiple, so it applies to each score separately.
3
Difference $= 34 - 14 = 20$
Why: The quantity asked for.

Shortcut

Scaling is linear, so the difference scales by the same factor: $(170 - 70) \times \frac{1}{5} = \frac{100}{5} = 20$ -- the two individual marks need never be converted.

Common mistakes

Option A

No route reaches $15$ ; the scaled gap is fixed at $\frac{100}{5}$ .

Option B

Converted only the highest mark and under the wrong factor, $\frac{170}{10} = 17$ , instead of scaling the gap by $\frac{1}{5}$ .

Option D

No route reaches $24$ from a conversion factor of $\frac{1}{5}$ .

Skills used

Unitary method: find one unit, then scale

38Statistics • Data Interpretation • Table

ModerateIdeal time: about 2 min

Read the following information and answer the three items that follow:

The following data presents count of released convicts who have served prison terms ( $X$ ), those who have received some educational or technical training during their term ( $Y$ ) and those who were offered Company placement ( $Z$ ) respectively, from six different jails A, B, C, D, E and F, in the year 2010.

	$X$	$Y$	$Z$
A	86	45	25
B	1305	903	461
C	2019	940	474
D	1166	869	416
E	954	544	254
F	1198	464	174

Jails with highest and smallest percentage of trained convicts are respectively

AF and DBD and FCC and ADD and A

Show answer & solution

Option B

D and F

Approach

'Percentage of trained convicts' is $\frac{Y}{X}$ for each jail, so compute all six and pick the largest and the smallest.

Comparing two fractions

Step-by-step solution

1
The share trained in a jail is $\frac{Y}{X} \times 100%$ , where $X$ is the count released and $Y$ the count trained.
Why: A percentage has to be measured against that jail's own total, so the raw counts $Y$ cannot be compared across jails of different sizes.
$x %$ of a quantity One quantity as a percentage of another
2
$A : \frac{45}{86} = 52.3%$ ; $B : \frac{903}{1305} = 69.2%$ ; $C : \frac{940}{2019} = 46.6%$
Why: Dividing each out to one decimal place puts all six jails on one scale.
Comparing two fractions
3
$D : \frac{869}{1166} = 74.5%$ ; $E : \frac{544}{954} = 57.0%$ ; $F : \frac{464}{1198} = 38.7%$
Why: The remaining three.
4
Highest is $D$ at $74.5%$ ; smallest is $F$ at $38.7%$ . In the order asked: $D$ and $F$ .
Why: The question names highest first and smallest second, so the order of the pair is part of the answer.

Shortcut

Benchmark each against $\frac{1}{2}$ and then against $\frac{3}{4}$ : only $C$ and $F$ fall below half, and $F$ is clearly the lower of the two ( $464$ against $599$ ); at the top only $D$ approaches three-quarters ( $869$ against $874.5$ ).

Common mistakes

Option A

Found the right pair of jails but gave them in the wrong order -- the question asks for highest first.

Option C

Compared the raw counts instead of the shares: $C$ has the largest $X$ and $A$ the smallest.

Option D

Got the highest right, then took $A$ as lowest because its numbers are the smallest; $A$ in fact trains $52.3%$ , above the average.

Skills used

Comparing two fractions

x %

of a quantity One quantity as a percentage of another

39Statistics • Data Interpretation • Table

ModerateIdeal time: about 2 min

Same directions as Q38

Jail with highest placement rate of trained convicts is

AFBDCBDA

Show answer & solution

Option D

Approach

'Placement rate of trained convicts' measures placements against the trained group, so compute $\frac{Z}{Y}$ for each jail and take the largest.

Comparing two fractions

Step-by-step solution

1
The rate asked for is $\frac{Z}{Y}$ : those offered placement as a share of those who received training.
Why: The phrase fixes the base as $Y$ , not $X$ -- it is a rate among the trained, not among all released convicts.
$x %$ of a quantity
2
$A : \frac{25}{45} = 55.6%$ ; $B : \frac{461}{903} = 51.0%$ ; $C : \frac{474}{940} = 50.4%$
Why: Dividing each out to one decimal place; $A$ 's counts are small but the rate is what matters.
Comparing two fractions
3
$D : \frac{416}{869} = 47.9%$ ; $E : \frac{254}{544} = 46.7%$ ; $F : \frac{174}{464} = 37.5%$
Why: The remaining three, all below half.
4
The largest is $A$ at $55.6%$ .
Why: $A$ beats $B$ , its nearest rival, by over four percentage points.

Shortcut

Only three jails clear $\frac{1}{2}$ : $A$ ( $25$ against $22.5$ ), $B$ ( $461$ against $451.5$ ) and $C$ ( $474$ against $470$ ). $A$ clears it by more than $11%$ of its base, the others by about $2%$ and $1%$ , so $A$ wins without exact division.

Common mistakes

Option A

Picked the lowest rate instead of the highest: $F$ is last at $37.5%$ .

Option B

Used $\frac{Z}{X}$ -- placements among all released convicts -- on which $D$ at $35.7%$ does come top.

Option C

Discounted jail $A$ for its small counts and took the best of the large jails, $B$ at $51.0%$ .

Skills used

Comparing two fractions

x %

of a quantity

40Statistics • Data Interpretation • Table

ModerateIdeal time: about 2 min

Same directions as Q38

Jails from which more than half of the trained convicts are offered jobs, are

AA, B and CBA, B and DCA, D and EDA, E and F

Show answer & solution

Option A

A, B and C

Approach

Test each jail's $\frac{Z}{Y}$ against the benchmark $\frac{1}{2}$ by comparing $2 Z$ with $Y$ -- no division needed.

Comparing two fractions

Step-by-step solution

1
'More than half of the trained convicts are offered jobs' means $\frac{Z}{Y} > \frac{1}{2}$ , i.e. $2 Z > Y$ .
Why: Comparing against a benchmark of $\frac{1}{2}$ turns six divisions into six doublings, which is far faster and exact.
Comparing two fractions
2
$A : 2 (25) = 50 > 45$ (yes). $B : 2 (461) = 922 > 903$ (yes). $C : 2 (474) = 948 > 940$ (yes).
Why: All three clear the benchmark, $B$ and $C$ only narrowly.
3
$D : 2 (416) = 832 < 869$ (no). $E : 2 (254) = 508 < 544$ (no). $F : 2 (174) = 348 < 464$ (no).
Why: All three fall short, $D$ the closest.
4
The jails are $A$ , $B$ and $C$ .
Why: Exactly the three that pass.
$x %$ of a quantity

Shortcut

Double $Z$ and compare with $Y$ in the same row; the six comparisons take one pass down the table.

Common mistakes

Option B

Included $D$ at $47.9%$ by rounding it up to half, and dropped $C$ at $50.4%$ by rounding it down.

Option C

Includes $D$ and $E$ , both below half on $\frac{Z}{Y}$ , and omits $B$ and $C$ , which are above it.

Option D

Includes $E$ and $F$ , the two lowest placement rates in the table.

Skills used

Comparing two fractions

x %

of a quantity