Elementary Mathematics 2018 (II)

Set A

Questions 21–40 of 100

21Algebra • Quadratic Equations • Maximum and Minimum Values

EasyIdeal time: about 50 s

The minimum value of the expression $2 x^{2} + 5 x + 5$ is

5

15/8

- 15/8

0

Show answer & solution

Option B

15/8

Approach

Complete the square: the squared bracket cannot go below zero, so what is left beside it is the minimum.

Range of a quadratic by completing the square

Step-by-step solution

1
$2 x^{2} + 5 x + 5 = 2 (x^{2} + \frac{5}{2} x) + 5$
Why: Take the leading coefficient out first, so the bracket is a plain $x^{2} + \dots$ and the square can be completed in the usual way.
Range of a quadratic by completing the square
2
$= 2 [(x + \frac{5}{4})^{2} - \frac{25}{16}] + 5 = 2 (x + \frac{5}{4})^{2} - \frac{25}{8} + 5$
Why: Half of $\frac{5}{2}$ is $\frac{5}{4}$ , and $(\frac{5}{4})^{2} = \frac{25}{16}$ has to be subtracted back out.
3
$= 2 (x + \frac{5}{4})^{2} + \frac{15}{8}$ , least value $\frac{15}{8}$ at $x = - \frac{5}{4}$
Why: $5 - \frac{25}{8} = \frac{40 - 25}{8} = \frac{15}{8}$ , and the square term is never negative, so the smallest the whole expression can be is $\frac{15}{8}$ - reached when the bracket is zero.

Shortcut

For $a x^{2} + b x + c$ with $a > 0$ the least value is $c - \frac{b ^{2}}{4 a} = 5 - \frac{25}{8} = \frac{15}{8}$ .

Common mistakes

Option A

Evaluated at $x = 0$ and took the constant term $5$ as the minimum, without checking that negative $x$ makes the expression smaller.

Option C

Subtracted where the completed square adds: wrote $- \frac{25}{8} - 5$ or dropped the sign on $\frac{15}{8}$ , giving $- \frac{15}{8}$ - impossible, since the expression has no real root and so never reaches zero.

Option D

Assumed a quadratic's least value is $0$ , which holds only when the discriminant is zero; here $25 - 40 < 0$ , so the expression is always positive.

Skills used

Range of a quadratic by completing the square

22Statistics • Mean

EasyIdeal time: about 1 min

If H is the harmonic mean of P and Q, then the value of $\frac{H}{P} + \frac{H}{Q}$ is

1

2

\frac{P + Q}{P Q}

\frac{P Q}{P + Q}

Show answer & solution

Option B

2

Approach

Write $H$ in terms of $P$ and $Q$ , then take $H$ out as a common factor - the rest is the reciprocal sum that $H$ was built from.

Average speed over two equal distances

Step-by-step solution

1
$H = \frac{2 P Q}{P + Q}$
Why: The harmonic mean of two numbers is the reciprocal of the average of their reciprocals, which rearranges to $\frac{2 P Q}{P + Q}$ - the same expression as the average speed over two equal distances.
Average speed over two equal distances
2
$\frac{H}{P} + \frac{H}{Q} = H (\frac{1}{P} + \frac{1}{Q}) = H \cdot \frac{P + Q}{P Q}$
Why: Taking $H$ outside turns the question into the reciprocal sum, and the reciprocal sum is exactly what the harmonic mean is defined against - so the two will cancel.
Evaluate a nested fraction from the inside out
3
$= \frac{2 P Q}{P + Q} \times \frac{P + Q}{P Q} = 2$
Why: $(P + Q)$ and $P Q$ both cancel, leaving the $2$ from the numerator of $H$ - and nothing depending on $P$ or $Q$ survives.

Shortcut

The definition itself says $\frac{1}{H} = \frac{1}{2} (\frac{1}{P} + \frac{1}{Q})$ ; multiply both sides by $2 H$ and the left becomes $2$ while the right becomes $\frac{H}{P} + \frac{H}{Q}$ .

Common mistakes

Option A

Used $H = \frac{P Q}{P + Q}$ , dropping the factor $2$ from the harmonic mean, which halves the answer to $1$ .

Option C

Stopped at $\frac{1}{P} + \frac{1}{Q} = \frac{P + Q}{P Q}$ and never multiplied by $H$ .

Option D

Reported $H$ itself, up to the factor $2$ , instead of the expression asked for.

Skills used

Average speed over two equal distances Evaluate a nested fraction from the inside out

23Algebra • Expressions • Algebraic Identities

ModerateIdeal time: about 1.5 min

The sum of all possible products taken two at a time out of the numbers $\pm 1$ , $\pm 2$ , $\pm 3$ , $\pm 4$ is

0

- 30

30

55

Show answer & solution

Option B

- 30

Approach

Group the $28$ pairs: a number with its own negative, and everything else in fours that cancel.

Square of a sum and of a difference

Step-by-step solution

1
The eight numbers are $1, - 1, 2, - 2, 3, - 3, 4, - 4$ , giving $\frac{8 \times 7}{2} = 28$ pairs
Why: ' $\pm 1$ ' stands for two numbers, not one, so there are eight numbers in play - getting this wrong halves the problem.
Square of a sum and of a difference
2
Pairs of the form ${k, - k}$ : products $- 1, - 4, - 9, - 16$ , summing to $- 30$
Why: There are four such pairs, one for each size, and each contributes $- k^{2}$ .
Sums of the first $n$ natural numbers, their squares and their cubes
3
For two different sizes $k \neq = m$ the four pairs give $k m$ , $- k m$ , $- k m$ , $k m$ , summing to $0$
Why: Each of $\pm k$ meets each of $\pm m$ , and the two positive products cancel the two negative ones - so every one of the $(2 4) = 6$ size-pairs contributes nothing.
4
Total $= - 30 + 6 \times 0 = - 30$
Why: $4 + 6 \times 4 = 28$ pairs are accounted for, so nothing has been missed.

Shortcut

Use $(\sum a_{i})^{2} = \sum a_{i}^{2} + 2 \sum_{i < j} a_{i} a_{j}$ . Here $\sum a_{i} = 0$ and $\sum a_{i}^{2} = 2 (1 + 4 + 9 + 16) = 60$ , so the required sum is $\frac{0 - 60}{2} = - 30$ .

Common mistakes

Option A

Argued that because the eight numbers add to $0$ their pairwise products must too - but $(\sum a_{i})^{2} = 0$ gives $2 \sum_{i < j} a_{i} a_{j} = - \sum a_{i}^{2}$ , not $0$ .

Option C

Dropped the minus sign from $\frac{0 - 60}{2}$ , or summed the four products $k \times (- k)$ as $+ k^{2}$ .

Option D

No route reaches $55$ ; it is not produced by any grouping of these products.

Skills used

Square of a sum and of a difference Sums of the first

n

natural numbers, their squares and their cubes

24Algebra • Polynomial • Factor and Remainder Theorem

EasyIdeal time: about 50 s

The remainder when $3 x^{3} - 2 x^{2} y - 13 x y^{2} + 10 y^{3}$ is divided by $(x - 2 y)$ is equal to

AZeroB

y

y - 5

y + 3

Show answer & solution

Option A

Zero

Approach

The remainder on dividing by $(x - 2 y)$ is the value of the polynomial at $x = 2 y$ , so substitute.

Factor and remainder theorem

Step-by-step solution

1
Set $x - 2 y = 0$ , i.e. $x = 2 y$
Why: The remainder theorem works with two letters exactly as with one: the divisor is linear in $x$ , so substitute the value of $x$ that makes it vanish and treat $y$ as a constant.
Factor and remainder theorem
2
$3 (2 y)^{3} - 2 (2 y)^{2} y - 13 (2 y) y^{2} + 10 y^{3} = 24 y^{3} - 8 y^{3} - 26 y^{3} + 10 y^{3} = 0$
Why: $24 - 8 - 26 + 10 = 0$ , so the remainder is zero and $(x - 2 y)$ is in fact a factor.

Common mistakes

Option B

No route reaches $y$ : the polynomial is homogeneous of degree $3$ in $x$ and $y$ , so its remainder on division by a homogeneous linear divisor is a multiple of $y^{3}$ or nothing at all.

Option C

No route reaches $y - 5$ ; it is not homogeneous and cannot be a remainder here.

Option D

No route reaches $y + 3$ ; it is not homogeneous and cannot be a remainder here.

Skills used

Factor and remainder theorem

25Algebra • Expressions • Simplification

HardIdeal time: about 2.5 min

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

- 1

0

1

2

Show answer & solution

Option B

0

Approach

Recognise the fraction as a sum of three reciprocals, then use $ab + b c + c a = 0$ to factorise each denominator as one letter times $(a + b + c)$ .

Complete each equation to the common sum

Step-by-step solution

1
With $A = a^{2} - b c$ , $B = b^{2} - c a$ , $C = c^{2} - ab$ , the expression is $\frac{B C + A C + A B}{A B C} = \frac{1}{A} + \frac{1}{B} + \frac{1}{C}$
Why: The numerator is the three denominators taken two at a time, which is exactly what appears when $\frac{1}{A} + \frac{1}{B} + \frac{1}{C}$ is put over a common denominator - so reading it backwards removes the whole fraction.
Clear the denominators of an equation
2
$ab + b c + c a = 0 \Rightarrow b c = - a (b + c)$ , so $a^{2} - b c = a^{2} + ab + a c = a (a + b + c)$
Why: The condition lets $b c$ be traded for terms that each carry an $a$ , and the whole symmetric sum $a + b + c$ then falls out as a factor.
Complete each equation to the common sum
3
Likewise $b^{2} - c a = b (a + b + c)$ and $c^{2} - ab = c (a + b + c)$
Why: The same trade works cyclically: $c a = - b (c + a)$ and $ab = - c (a + b)$ .
4
Sum $= \frac{1}{a + b + c} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = \frac{1}{a + b + c} \cdot \frac{b c + c a + ab}{ab c} = \frac{0}{ab c ( a + b + c )} = 0$
Why: The common factor $(a + b + c)$ comes out, and the reciprocal sum brings back $ab + b c + c a$ - which the given condition sets to zero.

Common mistakes

Option A

Got the reciprocal sum but mis-signed the trade, writing $a^{2} - b c = a (a - b - c)$ , which leaves a non-zero numerator and a sign that points to $- 1$ .

Option C

Assumed the numerator and denominator were arranged to cancel to $1$ - a reasonable guess at the shape of the fraction, but the reciprocal sum is what it actually equals.

Option D

No route reaches $2$ from this condition.

Skills used

Clear the denominators of an equation Complete each equation to the common sum

26Arithmetic • Interest • Compound Interest

EasyIdeal time: about 1 min

What is the principal amount which earns Rs. $210$ as compound interest for the second year at $5%$ per annum ?

ARs.

2000

BRs.

3200

CRs.

4000

DRs.

4800

Show answer & solution

Option C

Rs.

4000

Approach

The second year's interest is charged on the first year's closing amount, so write it as $5%$ of $P$ plus $5%$ of that $5%$ .

Compound interest amount

A = P (1 + \frac{r}{100})^{n}

Step-by-step solution

1
$first year’s interest = P \times \frac{5}{100} = \frac{P}{20}$
Why: Under compound interest the first year is no different from simple interest: the base is still $P$ .
Simple interest $= \frac{P R T}{100}$
2
$amount after one year = P + \frac{P}{20} = \frac{21 P}{20}$
Why: This closing amount, not $P$ , is what the second year charges on. That is the whole difference between simple and compound interest here.
Compound interest amount $A = P (1 + \frac{r}{100})^{n}$
3
$second year’s interest = \frac{21 P}{20} \times \frac{5}{100} = \frac{21 P}{400}$
Why: Apply $5%$ to the new base.
$x %$ of a quantity
4
$\frac{21 P}{400} = 210 \Rightarrow P = \frac{210 \times 400}{21} = 4000$
Why: Equate to the given Rs $210$ and solve; the $21$ cancels, so no long division is needed.
Turn a worded condition into a linear equation in one unknown

Shortcut

The second year's interest is the first year's interest grown by one more year at the same rate: $210 = P \times \frac{5}{100} \times \frac{105}{100}$ , so $P = 210 \times \frac{20 \times 100}{105} = 4000$ .

Common mistakes

Option A

Treated Rs $210$ as the total interest of two years and used a remembered two-year rate of $10.5%$ instead of the correct $10.25%$ , giving $P = \frac{210}{0.105} = 2000$ .

Option B

No route reaches Rs $3200$ : $5.25%$ of it is Rs $168$ .

Option D

No route reaches Rs $4800$ : $5.25%$ of it is Rs $252$ .

Skills used

Compound interest amount

A = P (1 + \frac{r}{100})^{n}

Simple interest

= \frac{P R T}{100}

x %

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

27Algebra • Sets • Operations

EasyIdeal time: about 50 s

In an examination, $50%$ of the candidates failed in English, $40%$ failed in Hindi and $15%$ failed in both the subjects. The percentage of candidates who passed in both English and Hindi is

20%

25%

60%

75%

Show answer & solution

Option B

25%

Approach

Count the candidates who failed at least one subject by inclusion-exclusion, then take the complement.

Inclusion-exclusion for two sets

Step-by-step solution

1
$n (E \cup H) = n (E) + n (H) - n (E \cap H) = 50 + 40 - 15 = 75$
Why: Percentages behave like counts out of $100$ , so work with $100$ candidates. The $15%$ who failed both have been counted in both the $50$ and the $40$ , so subtract them once.
Inclusion-exclusion for two sets
2
$failed at least one = 75%$
Why: A candidate who fails English, or Hindi, or both, is exactly a member of $E \cup H$ .
3
$passed both = 100% - 75% = 25%$
Why: Passing both subjects means failing neither, which is the complement of the union.
Set difference $A - B$

Common mistakes

Option A

Added the two failure rates to $90%$ and took $100 - 90 = 10$ , then adjusted; any route that forgets to add the overlap back lands below $25%$ .

Option C

Answered $100 - 40 = 60$ , the percentage who passed Hindi, instead of the percentage who passed both.

Option D

Stopped at $n (E \cup H) = 75%$ , which is the percentage who failed at least one subject, not the percentage who passed both.

Skills used

Inclusion-exclusion for two sets Set difference

A - B

28Arithmetic • Time, Speed and Distance • Trains

EasyIdeal time: about 45 s

A train $100$ m long passes a platform $100$ m long in $10$ seconds. The speed of the train is

36

kmphB

45

kmphC

54

kmphD

72

kmph

Show answer & solution

Option D

72

kmph

Approach

A train crossing a platform covers its own length plus the platform's length; divide that by the time, then convert to km/hour.

What a train covers when it crosses something

Step-by-step solution

1
$distance = 100 + 100 = 200 m$
Why: The train has cleared the platform only when its rear end leaves it, so the front travels the two lengths added.
What a train covers when it crosses something
2
$speed = \frac{200}{10} = 20 m/s$
Why: Metres with seconds give metres per second.
Speed, distance and time
3
$20 \times \frac{18}{5} = 72 kmph$
Why: The options are in km/hour, so convert with the factor $\frac{18}{5}$ .
km/hour to m/second

Common mistakes

Option A

Used only the train's own $100$ m, getting $10$ m/s $= 36$ kmph — the answer to "passes a pole", not "passes a platform".

Option B

$45$ kmph is $12.5$ m/s, which no combination of $100$ m, $100$ m and $10$ s produces.

Option C

$54$ kmph is $15$ m/s, which would need a distance of $150$ m.

Skills used

What a train covers when it crosses something Speed, distance and time km/hour to m/second

29Arithmetic • Time, Speed and Distance • Average Speed

ModerateIdeal time: about 1.5 min

A cyclist covers his first $20$ km at an average speed of $40$ kmph, another $10$ km at an average speed of $10$ kmph and the last $30$ km at an average speed of $40$ kmph. Then the average speed of the entire journey is

20

kmphB

26.67

kmphC

28.24

kmphD

30

kmph

Show answer & solution

Option B

26.67

kmph

Approach

Average speed is total distance over total time, never the average of the speeds, so find the three times and add them.

Speed, distance and time

Step-by-step solution

1
$total distance = 20 + 10 + 30 = 60 km$
Why: The three legs are travelled one after another.
Speed, distance and time
2
$t_{1} = \frac{20}{40} = \frac{1}{2}, t_{2} = \frac{10}{10} = 1, t_{3} = \frac{30}{40} = \frac{3}{4} hours$
Why: Each leg has its own speed, so each needs its own time; keep them as fractions and the sum stays exact.
3
$t = \frac{1}{2} + 1 + \frac{3}{4} = \frac{9}{4} hours$
Why: Over quarters: $\frac{2}{4} + \frac{4}{4} + \frac{3}{4}$ .
Unitary method: find one unit, then scale
4
$average speed = \frac{60}{9/4} = \frac{240}{9} = \frac{80}{3} = 26.67 kmph$
Why: Divide total distance by total time. The $10$ kmph leg is short but eats $1$ of the $2 \frac{1}{4}$ hours, which is why the average sits well below $40$ .

Common mistakes

Option A

Took the harmonic mean of $40$ , $10$ and $40$ as if the three legs were of equal length: $\frac{3}{\frac{1}{40} + \frac{1}{10} + \frac{1}{40}} = 20$ .

Option C

$28.24$ kmph corresponds to a total time of $2.125$ hours, which none of the three legs produces.

Option D

Averaged the three speeds: $\frac{40 + 10 + 40}{3} = 30$ , which would be correct only if the three speeds were held for equal times.

Skills used

Speed, distance and time Unitary method: find one unit, then scale

30Arithmetic • Time, Speed and Distance • Races

HardIdeal time: about 2.5 min

In a race of $1000$ m, A beats B by $150$ m, while in another race of $3000$ m, C beats D by $400$ m. Speed of B is equal to that of D. (Assume that A, B, C and D run with uniform speed in all the events). If A and C participate in a race of $6000$ m, then which one of the following is correct ?

AA beats C by

250

mBC beats A by

250

mCA beats C by

115.38

mDC beats A by

115.38

Show answer & solution

Option C

A beats C by

115.38

Approach

Turn each "beats by" into a speed ratio, bridge the two ratios on the shared runner (B and D have equal speeds), then see how far C has gone when A finishes $6000$ m.

Head starts and winning margins in a race

Step-by-step solution

1
$A : B = 1000 : 850 = 20 : 17$
Why: "A beats B by $150$ m" means B has covered $1000 - 150 = 850$ m when A finishes $1000$ m; same time, so distances are in the ratio of speeds.
Head starts and winning margins in a race
2
$C : D = 3000 : 2600 = 15 : 13$
Why: The same reading applied to the second race.
3
$take B = D = 1 : A = \frac{20}{17}, C = \frac{15}{13}$
Why: B and D run at the same speed, so that common speed is the bridge between the two races. Setting it to $1$ puts A and C on one scale.
Join two ratios on their shared term
4
$A : C = \frac{20}{17} : \frac{15}{13} = 260 : 255 = 52 : 51$
Why: Multiply both sides by $17 \times 13 = 221$ . A is the faster of the two, so A wins.
Introduce a common factor $k$ for a ratio
5
$when A runs 6000, C runs 6000 \times \frac{51}{52} = \frac{76500}{13} = 5884.62 m$
Why: Equal times again, so C's distance is A's scaled by the speed ratio.
Over a fixed distance, time varies inversely with speed
6
$6000 - 5884.62 = 115.38 m$
Why: The margin is what C still has to run when A crosses the line.

Shortcut

$A : C = 52 : 51$ , so over any race A finishes $\frac{1}{52}$ of the distance ahead: $\frac{6000}{52} = 115.38$ m.

Common mistakes

Option A

Right winner, but $250$ m is not the margin; scaling the two given margins to $6000$ m gives $900$ and $800$ , whose difference of $100$ m is itself a wrong answer and still not $250$ .

Option B

No route reaches a $250$ m margin, and the direction is wrong as well.

Option D

Correct margin, wrong winner: this is what comes of comparing $\frac{17}{20}$ with $\frac{13}{15}$ , or of reading the ratios as times instead of distances.

Skills used

Head starts and winning margins in a race Join two ratios on their shared term Introduce a common factor

k

for a ratio Over a fixed distance, time varies inversely with speed

31Arithmetic • Age Problems

ModerateIdeal time: about 1.5 min

The sum of ages of a father, a mother, a son Sonu and daughters Savita and Sonia is $96$ years. Sonu is the youngest member of the family. The year Sonu was born, the sum of the ages of all the members of the family was $66$ years. If the father's age now is $6$ times that of Sonu's present age, then $12$ years hence, the father's age will be

44

yearsB

45

yearsC

46

yearsD

48

years

Show answer & solution

Option D

48

years

Approach

The family's age total falls by $5$ years for every year you go back, because Sonu's own age is counted in the total and he was $0$ at the start; use that to find Sonu's age, then the father's.

Turn a worded condition into a linear equation in one unknown

Step-by-step solution

1
$let Sonu’s present age = s years$
Why: The father's age is tied to Sonu's, so Sonu is the unknown worth naming.
Turn a worded condition into a linear equation in one unknown
2
$sum s years ago = 96 - 5 s$
Why: Each of the five members was $s$ years younger then, so the total was $5 s$ less. The year Sonu was born he contributed $0$ , which is consistent: his $s - s = 0$ is already inside the $5 s$ .
3
$96 - 5 s = 66 \Rightarrow 5 s = 30 \Rightarrow s = 6$
Why: The year Sonu was born is exactly $s$ years ago, so that is the year whose total is given as $66$ .
4
$father now = 6 s = 36 years$
Why: Given: the father is $6$ times Sonu's present age.
Introduce a common factor $k$ for a ratio
5
$36 + 12 = 48 years$
Why: The question asks for $12$ years hence, not now.

Shortcut

The drop of $30$ years over the five members gives Sonu's age at once: $s = \frac{96 - 66}{5} = 6$ . Father $= 36$ , so $48$ twelve years hence.

Common mistakes

Option A

No route gives the father $32$ years now; the names Savita and Sonia carry no information and inventing a relation for them leads nowhere.

Option B

Divided the $30$ -year drop by $4$ (the members who "really" aged) to get $s = 7.5$ , then quoted the father's present age $6 s = 45$ instead of his age $12$ years hence.

Option C

No route gives the father $34$ years now.

Skills used

Turn a worded condition into a linear equation in one unknown Introduce a common factor

k

for a ratio

32Arithmetic • Time and Work

EasyIdeal time: about 1 min

'A' is thrice as good a workman as 'B' and takes $10$ days less to do a piece of work than 'B' takes. The number of days taken by 'B' alone to finish the work is

12

15

20

30

Show answer & solution

Option B

15

Approach

"Thrice as good a workman" fixes the ratio of times at $1 : 3$ , so write the two times through one multiplier and use the $10$ -day gap.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$rate_{A} = 3 \times rate_{B} \Rightarrow t_{A} : t_{B} = 1 : 3$
Why: Three times the output per day means one third of the time for the same job: rates and times are reciprocal.
Over a fixed distance, time varies inversely with speed
2
$t_{A} = k, t_{B} = 3 k$
Why: A ratio fixes the two times only up to a common factor.
Introduce a common factor $k$ for a ratio
3
$3 k - k = 10 \Rightarrow 2 k = 10 \Rightarrow k = 5$
Why: A takes $10$ days less than B, so the gap is $t_{B} - t_{A}$ .
Turn a worded condition into a linear equation in one unknown
4
$t_{B} = 3 k = 15 days$
Why: The question asks for B alone, not A.

Common mistakes

Option A

$12$ days for B would put A at $4$ days, a gap of $8$ , not $10$ ; no route reaches it.

Option C

Read "thrice as good" as twice as good: $t_{B} - \frac{t _{B}}{2} = 10$ gives $t_{B} = 20$ .

Option D

$30$ days for B would put A at $10$ days, a gap of $20$ — the $10$ -day gap was applied to A's time instead of to the difference.

Skills used

Over a fixed distance, time varies inversely with speed Introduce a common factor

k

for a ratio Turn a worded condition into a linear equation in one unknown

33Algebra • Sets • Operations

HardIdeal time: about 3 min

Out of $85$ children playing badminton or table tennis or both, the total number of girls in the group is $70%$ of the total number of boys in the group. The number of boys playing only badminton is $50%$ of the number of boys and the total number of boys playing badminton is $60%$ of the total number of boys. The number of children playing only table tennis is $40%$ of the total number of children and a total of $12$ children play badminton and table tennis both. The number of girls playing only badminton is

14

16

17

35

Show answer & solution

Option A

14

Approach

Split the $85$ children into boys and girls from the $70%$ relation, build the boys' badminton figures from the percentages of boys, then use the children-level totals to get the girls' only-badminton count.

x %

of a quantity

Step-by-step solution

1
$b + 0.7 b = 85 \Rightarrow 1.7 b = 85 \Rightarrow b = 50, girls = 35$
Why: The $70%$ is measured on the boys, so let the boys be $b$ ; every child plays at least one game, so the two groups make up the whole $85$ .
$x %$ of a quantity
2
$boys, badminton only = 50% of 50 = 25; boys, badminton total = 60% of 50 = 30$
Why: Both of these percentages are of the number of boys, not of the $85$ children — that is the shift the question is built on.
3
$boys, both games = 30 - 25 = 5$
Why: A boy's badminton total splits into badminton only and both games.
Set difference $A - B$
4
$children, table tennis only = 40% of 85 = 34$
Why: This percentage is of the total children, as stated.
5
$children, badminton only = 85 - 34 - 12 = 39$
Why: Only badminton, only table tennis and both are three disjoint groups covering all $85$ children.
Inclusion-exclusion for two sets
6
$girls, badminton only = 39 - 25 = 14$
Why: Take the boys' only-badminton count out of the children's.

Common mistakes

Option B

$16$ is $51 - 35$ , the badminton total less the number of girls — a subtraction that mixes a games count with a gender count.

Option C

$17$ is $20%$ of $85$ and corresponds to no group in the table.

Option D

$35$ is the total number of girls, not the girls who play badminton only.

Skills used

x %

of a quantity Set difference

A - B

Inclusion-exclusion for two sets

34Arithmetic • Percentage

EasyIdeal time: about 50 s

A person bought two articles X and Y from a departmental store. The sum of prices before sales tax was Rs. $130$ . There was no sales tax on the article X and $9%$ sales tax on the article Y. The total amount the person paid, including the sales tax was Rs. $136.75$ . What was the price of the article Y before sales tax ?

ARs.

75

BRs.

85

CRs.

122

DRs.

125

Show answer & solution

Option A

Rs.

75

Approach

The whole excess over Rs $130$ is the tax on Y alone, so divide the excess by $9%$ .

x %

of a quantity

Step-by-step solution

1
$tax paid = 136.75 - 130 = 6.75$
Why: X carries no tax, so every rupee above the pre-tax total of Rs $130$ is tax on Y.
Turn a worded condition into a linear equation in one unknown
2
$\frac{9}{100} \times Y = 6.75$
Why: The tax is $9%$ of Y's pre-tax price, so Y's price is the base.
$x %$ of a quantity
3
$Y = \frac{6.75 \times 100}{9} = 75$
Why: $675 \div 9 = 75$ , so the division stays in small numbers.

Common mistakes

Option B

Transposed the tax as Rs $7.65$ instead of Rs $6.75$ , giving $Y = 85$ .

Option C

No route reaches Rs $122$ .

Option D

Divided the whole bill by $1.09$ — $\frac{136.75}{1.09} = 125.46$ — as if the $9%$ tax applied to both articles.

Skills used

Turn a worded condition into a linear equation in one unknown

x %

of a quantity

35Arithmetic • Ratio and Proportion • Division into Parts

ModerateIdeal time: about 2 min

According to Mr. Sharma's will, half of his property goes to his wife and the rest is equally divided between his two sons, Ravi and Raj. Some years later, Ravi dies and leaves half of his property to his widow and rest to his brother Raj. When Raj dies he leaves half of his property to his widow and remaining to his mother, who is still alive. The mother now owns Rs. $88, 000$ worth of the property. The total worth of the property of Mr. Sharma was

ARs.

1, 00, 000

BRs.

1, 24, 000

CRs.

1, 28, 000

DRs.

1, 32, 000

Show answer & solution

Option C

Rs.

1, 28, 000

Approach

Track the mother's holding as a fraction of the whole estate through the three deaths, then set that fraction equal to Rs $88, 000$ .

Turn a worded condition into a linear equation in one unknown

Step-by-step solution

1
$wife \frac{P}{2}, Ravi \frac{P}{4}, Raj \frac{P}{4}$
Why: Half to the wife, the other half split equally between the two sons.
Unitary method: find one unit, then scale
2
$Ravi dies: widow \frac{P}{8}, Raj gains \frac{P}{8}$
Why: Half of Ravi's $\frac{P}{4}$ to his widow, the rest to his brother.
3
$Raj now holds \frac{P}{4} + \frac{P}{8} = \frac{3 P}{8}$
Why: Raj's own share plus what he inherited from Ravi. Missing this inheritance is the commonest slip in the chain.
Turn a worded condition into a linear equation in one unknown
4
$Raj dies: widow \frac{3 P}{16}, mother \frac{3 P}{16}$
Why: "Half to his widow and remaining to his mother" splits $\frac{3 P}{8}$ into two equal parts.
5
$mother holds \frac{P}{2} + \frac{3 P}{16} = \frac{8 P + 3 P}{16} = \frac{11 P}{16}$
Why: The mother is Mr Sharma's wife, so she still has her original half as well as what Raj left her.
6
$\frac{11 P}{16} = 88000 \Rightarrow P = 8000 \times 16 = 128000$
Why: $88000 \div 11 = 8000$ , so the arithmetic stays whole.
Clear the denominators of an equation

Common mistakes

Option A

Rs $1, 00, 000$ would leave the mother $\frac{11}{16} \times 100000 =$ Rs $68, 750$ , not Rs $88, 000$ .

Option B

Rs $1, 24, 000$ leaves the mother Rs $85, 250$ .

Option D

Rs $1, 32, 000$ corresponds to the mother holding $\frac{2}{3}$ of the estate rather than $\frac{11}{16}$ .

Skills used

Unitary method: find one unit, then scale Turn a worded condition into a linear equation in one unknown Clear the denominators of an equation

36Arithmetic • Ratio and Proportion

HardIdeal time: about 2 min

X bought $4$ bottles of lemon juice and Y bought one bottle of orange juice. Orange juice per bottle costs twice the cost of lemon juice per bottle. Z bought nothing but contributed Rs. $50$ for his share of the drink which they mixed together and shared the cost equally. If Z's Rs. $50$ is covered from his share, then what is the cost of one bottle of orange juice ?

ARs.

75

BRs.

50

CRs.

46

DRs.

30

Show answer & solution

Option B

Rs.

50

Approach

Price everything in terms of one lemon bottle, add up the cost of the whole drink, and use the fact that each of the three paid a third of it.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$lemon = L per bottle, orange = 2 L per bottle$
Why: One unknown is enough: the orange price is tied to the lemon price.
Introduce a common factor $k$ for a ratio
2
$total cost = 4 L + 2 L = 6 L$
Why: Four lemon bottles from X and one orange bottle from Y went into the mixture.
Turn a worded condition into a linear equation in one unknown
3
$each share = \frac{6 L}{3} = 2 L$
Why: Three people shared the drink and its cost equally, so a share is a third of the whole cost.
Unitary method: find one unit, then scale
4
$2 L = 50 \Rightarrow L = 25$
Why: Z paid Rs $50$ and that exactly covered his share, so his share is Rs $50$ .
5
$orange = 2 L = 50$
Why: The question asks for the orange bottle, which costs two lemon bottles.

Shortcut

Z's Rs $50$ is one third of the cost, so the whole drink costs Rs $150$ ; that is six lemon-bottles' worth, so a lemon bottle is Rs $25$ and the orange bottle Rs $50$ .

Common mistakes

Option A

Took the Rs $150$ total as the cost of the four lemon bottles alone, giving $L = 37.5$ and orange $= 75$ ; the orange bottle's own cost was left out of the total.

Option C

No route reaches Rs $46$ .

Option D

Divided the Rs $150$ total equally over the five bottles to get Rs $30$ each, ignoring that the orange bottle costs twice as much as a lemon one.

Skills used

Introduce a common factor

k

for a ratio Turn a worded condition into a linear equation in one unknown Unitary method: find one unit, then scale

37Arithmetic • Age Problems

ModerateIdeal time: about 1.5 min

Ten ( $10$ ) years before, the ages of a mother and her daughter were in the ratio $3 : 1$ . In another $10$ years from now, the ratio of their ages will be $13 : 7$ . What are their present ages ?

39

years,

21

yearsB

55

years,

25

yearsC

75

years,

25

yearsD

49

years,

31

years

Show answer & solution

Option B

55

years,

25

years

Approach

Write the ages of ten years ago through the $3 : 1$ ratio, push both forward twenty years, and solve the second ratio.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$ten years ago: mother = 3 k, daughter = k$
Why: A ratio fixes the two ages only up to a common factor $k$ .
Introduce a common factor $k$ for a ratio
2
$ten years from now: mother = 3 k + 20, daughter = k + 20$
Why: From ten years ago to ten years hence is twenty years for both of them.
Turn a worded condition into a linear equation in one unknown
3
$\frac{3 k + 20}{k + 20} = \frac{13}{7}$
Why: The second condition, written on the same $k$ .
Clear the denominators of an equation
4
$21 k + 140 = 13 k + 260 \Rightarrow 8 k = 120 \Rightarrow k = 15$
Why: Cross-multiply and collect.
5
$present: mother = 3 (15) + 10 = 55, daughter = 15 + 10 = 25$
Why: The question asks for the present ages, so add the ten years back on.

Common mistakes

Option A

$39 : 21$ is $13 : 7$ — the second ratio was applied to the present ages instead of to the ages ten years hence.

Option C

$75 : 25$ is $3 : 1$ — the first ratio was applied to the present ages instead of to the ages ten years ago.

Option D

$49$ and $31$ were $39$ and $21$ ten years ago, i.e. $13 : 7$ : the two ratios were attached to the wrong time points.

Skills used

Introduce a common factor

k

for a ratio Turn a worded condition into a linear equation in one unknown Clear the denominators of an equation

38Algebra • Sets • Operations

EasyIdeal time: about 45 s

In a class of $60$ boys, there are $45$ boys who play chess and $30$ boys who play carrom. If every boy of the class plays at least one of the two games, then how many boys play carrom only ?

30

20

15

10

Show answer & solution

Option C

15

Approach

Every boy plays at least one game, so the union is the whole class; inclusion-exclusion gives the overlap, and carrom only is carrom minus the overlap.

Inclusion-exclusion for two sets

Step-by-step solution

1
$n (C \cup K) = 60$
Why: "Every boy plays at least one of the two games" says nobody is outside the union.
Inclusion-exclusion for two sets
2
$60 = 45 + 30 - n (C \cap K) \Rightarrow n (C \cap K) = 15$
Why: The boys who play both have been counted twice in $45 + 30 = 75$ , and the excess over $60$ is exactly that double count.
3
$carrom only = 30 - 15 = 15$
Why: Take the both-games boys out of the carrom total.
Set difference $A - B$

Common mistakes

Option A

$30$ is the number who play chess only (and also the carrom total before the overlap is removed), not the number who play carrom only.

Option B

No route gives $20$ ; the overlap is forced to be $15$ by the fact that the union is the full $60$ .

Option D

No route gives $10$ .

Skills used

Inclusion-exclusion for two sets Set difference

A - B

39Arithmetic • Interest • Simple Interest

EasyIdeal time: about 1 min

Two equal amounts were borrowed at $5%$ and $4%$ simple interest. The total interest after $4$ years amounted to Rs. $405$ . What was the total amount borrowed ?

ARs.

1075

BRs.

1100

CRs.

1125

DRs.

1150

Show answer & solution

No option is correct

Approach

Each equal part earns $20%$ over four years at $5%$ and $16%$ at $4%$ , so the two interests together are $36%$ of one part.

Simple interest

= \frac{P R T}{100}

Step-by-step solution

1
$let each amount be P, so the total borrowed is 2 P$
Why: "Two equal amounts" means two equal principals, one at each rate; the total borrowed is their sum.
Turn a worded condition into a linear equation in one unknown
2
$SI_{1} = \frac{P \times 5 \times 4}{100} = \frac{20 P}{100}, SI_{2} = \frac{P \times 4 \times 4}{100} = \frac{16 P}{100}$
Why: Simple interest is charged on the principal every year, so four years at $5%$ is $20%$ of $P$ .
Simple interest $= \frac{P R T}{100}$
3
$\frac{20 P}{100} + \frac{16 P}{100} = \frac{36 P}{100} = 405$
Why: The two interests add to the stated total.
4
$P = \frac{405 \times 100}{36} = 1125$
Why: $405 \div 36 = 11.25$ , so each amount is Rs $1125$ .
5
$total borrowed = 2 P = 2250$
Why: The question asks for the total, which is both parts together.

Skills used

Turn a worded condition into a linear equation in one unknown Simple interest

= \frac{P R T}{100}

40Arithmetic • Time and Work

EasyIdeal time: about 45 s

Twelve ( $12$ ) men work $8$ hours per day and require $10$ days to build a wall. If $8$ men are available, how many hours per day must they work to finish the work in $8$ days ?

10

hoursB

12

hoursC

15

hoursD

18

hours

Show answer & solution

Option C

15

hours

Approach

Measure the wall in man-hours, then divide by the man-hours now available per day.

The work-equivalence chain rule

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

Step-by-step solution

1
$job = 12 \times 8 \times 10 = 960 man-hours$
Why: With equally efficient workers the product $M \times D \times H$ measures the job and does not change when the workforce does.
The work-equivalence chain rule $M_{1} D_{1} H_{1} = M_{2} D_{2} H_{2}$
2
$8 \times 8 \times H = 960$
Why: The same job, now by $8$ men over $8$ days, working $H$ hours a day.
3
$64 H = 960 \Rightarrow H = 15 hours$
Why: $960 \div 64 = 15$ .

Common mistakes

Option A

Adjusted only for the shorter deadline: $8 \times \frac{10}{8} = 10$ hours, forgetting that there are fewer men as well.

Option B

Adjusted only for the smaller workforce: $8 \times \frac{12}{8} = 12$ hours, forgetting that the deadline also shortened from $10$ days to $8$ .

Option D

No route reaches $18$ hours; it would correspond to a job of $1152$ man-hours.

Skills used

The work-equivalence chain rule

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