Elementary Mathematics 2017 (I)

Set A

Questions 21–40 of 100

21Arithmetic • Number System • Prime Numbers

EasyIdeal time: about 45 s

The number of prime numbers which are less than $100$ is

24

25

26

27

Show answer & solution

Option B

25

Approach

Count the primes decade by decade, sieving out multiples of $2, 3, 5, 7$ .

Prime factorisation

Step-by-step solution

1
$2, 3, 5, 7$ ; $11, 13, 17, 19$ ; $23, 29$ ; $31, 37$ ; $41, 43, 47$
Why: Below $100$ a number is prime unless it has a factor among $2, 3, 5, 7$ , since $100 = 10$ .
Prime factorisation
2
$53, 59$ ; $61, 67$ ; $71, 73, 79$ ; $83, 89$ ; $97$
Why: The second half of the range, taken the same way.
3
$4 + 4 + 2 + 2 + 3 + 2 + 2 + 3 + 2 + 1 = 25$
Why: Adding the per-decade tallies gives the count.

Common mistakes

Option A

Dropped one prime, most often $2$ (as the only even prime) or $97$ at the end of the range.

Option C

Counted $1$ as a prime.

Option D

Counted $1$ and also kept a composite such as $91$ or $51$ .

Skills used

Prime factorisation

22Arithmetic • Ratio and Proportion • Direct and Inverse Variation

ModerateIdeal time: about 1.5 min

The cost of a diamond varies directly as the square of its weight. A diamond broke into four pieces with their weights in the ratio of $1 : 2 : 3 : 4$ . If the loss in total value of the diamond was Rs. $70, 000$ , what was the price of the original diamond?

ARs.

1, 00, 000

BRs.

1, 40, 000

CRs.

1, 50, 000

DRs.

1, 75, 000

Show answer & solution

Option A

Rs.

1, 00, 000

Approach

With value proportional to the square of the weight, compare the whole diamond's value with the total value of the four pieces.

Direct variation as a power:

y \propto x^{n}

Step-by-step solution

1
Weights $k, 2 k, 3 k, 4 k$ , total $10 k$ ; value $= c w^{2}$ for a constant $c$
Why: A ratio fixes the four weights only up to a common factor, and the stated law fixes the value of any piece from its weight.
Introduce a common factor $k$ for a ratio Direct variation as a power: $y \propto x^{n}$
2
original value $= c (10 k)^{2} = 100 c k^{2}$
Why: Before breaking, the whole weight is a single stone.
3
value after breaking $= c (k^{2} + 4 k^{2} + 9 k^{2} + 16 k^{2}) = 30 c k^{2}$
Why: Each piece is now valued on its own weight, and squares do not add the way weights do.
4
loss $= 100 c k^{2} - 30 c k^{2} = 70 c k^{2} = 70000 \Rightarrow c k^{2} = 1000$ , so the original value $= 100 \times 1000 =$ Rs. $1, 00, 000$
Why: The unknowns $c$ and $k$ only ever appear as the product $c k^{2}$ , so the ratio alone is enough.

Common mistakes

Option B

Doubled the loss, or added the loss to a value of Rs. $70, 000$ .

Option C

No route reaches Rs. $1, 50, 000$ .

Option D

Treated the loss as $40%$ of the original — for instance by comparing $\frac{16}{40}$ of the weight rather than the squares.

Skills used

Introduce a common factor

k

for a ratio Direct variation as a power:

y \propto x^{n}

23Arithmetic • Time, Speed and Distance • Races

ModerateIdeal time: about 1.5 min

In a $100$ m race, A runs at a speed of $\frac{5}{3}$ m/s. If A gives a start of $4$ m to B and still beats him by $12$ seconds, what is the speed of B?

\frac{5}{4}

m/sB

\frac{7}{5}

m/sC

\frac{4}{3}

m/sD

\frac{6}{5}

m/s

Show answer & solution

Option C

\frac{4}{3}

m/s

Approach

A's time over the full $100$ m fixes B's time; B has only $96$ m to cover in it.

Speed, distance and time

Step-by-step solution

1
A's time $= \frac{100}{5/3} = 60$ s
Why: Time is distance over speed, and A runs the whole course.
Speed, distance and time
2
A gives B a start of $4$ m, so B runs $100 - 4 = 96$ m
Why: A start of $4$ m means B begins $4$ m up the track and has that much less to run.
3
A beats B by $12$ s, so B's time $= 60 + 12 = 72$ s
Why: Beating by a time means B finishes that many seconds after A.
4
B's speed $= \frac{96}{72} = \frac{4}{3}$ m/s
Why: Distance over time again, now for B.

Common mistakes

Option A

Applied the start to the wrong runner or read it as $10$ m, giving $\frac{90}{72} = \frac{5}{4}$ .

Option B

Ignored the $4$ m start and rounded $\frac{100}{72} = \frac{25}{18} \approx 1.39$ to $\frac{7}{5}$ .

Option D

No exact route reaches $\frac{6}{5}$ ; it needs $96$ m over $80$ s, and B's time is $72$ s.

Skills used

Speed, distance and time

24Arithmetic • Time and Work

ModerateIdeal time: about 1.5 min

If $15$ men take $21$ days of $8$ hours each to do a piece of work, then what is the number of days of $6$ hours each that $21$ women would take, if $3$ women would do as much work as $2$ men?

18

20

25

30

Show answer & solution

Option D

30

Approach

Measure the job in man-hours, convert the women to their equivalent in men, and divide.

The work-equivalence chain rule

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

Step-by-step solution

1
total work $= 15 \times 21 \times 8 = 2520$ man-hours
Why: Men, days and hours per day multiply to the total labour the job takes.
Unitary method: find one unit, then scale The work-equivalence chain rule $M_{1} D_{1} H_{1} = M_{2} D_{2} H_{2}$
2
$3$ women do the work of $2$ men, so $21$ women do the work of $21 \times \frac{2}{3} = 14$ men
Why: The exchange rate is one woman to $\frac{2}{3}$ of a man, applied to all $21$ .
3
$14$ men working $6$ hours a day supply $14 \times 6 = 84$ man-hours a day
Why: The working day is shorter now, so the daily output must be recomputed.
4
days $= \frac{2520}{84} = 30$
Why: Total work divided by daily output gives the number of days.

Common mistakes

Option A

No route reaches $18$ .

Option B

No route reaches $20$ .

Option C

No route reaches $25$ .

Skills used

Unitary method: find one unit, then scale The work-equivalence chain rule

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

25Arithmetic • Ratio and Proportion

Very easyIdeal time: about 40 s

What number must be subtracted from both the numerator and the denominator of the fraction $\frac{27}{35}$ so that it becomes $\frac{2}{3}$ ?

6

8

9

11

Show answer & solution

Option D

11

Approach

Call the number $n$ , subtract it from both parts of the fraction, and solve the resulting linear equation.

Turn a worded condition into a linear equation in one unknown

Step-by-step solution

1
$\frac{27 - n}{35 - n} = \frac{2}{3}$
Why: The same $n$ comes off the numerator and the denominator, which is what makes the equation linear.
Turn a worded condition into a linear equation in one unknown
2
$3 (27 - n) = 2 (35 - n) \Rightarrow 81 - 3 n = 70 - 2 n$
Why: Cross-multiplying clears both denominators at once.
3
$81 - 70 = 3 n - 2 n \Rightarrow n = 11$
Why: Collecting the $n$ terms on one side leaves the value directly.

Common mistakes

Option A

No route reaches $6$ : $\frac{21}{29}$ is not $\frac{2}{3}$ .

Option B

No route reaches $8$ : $\frac{19}{27}$ is not $\frac{2}{3}$ .

Option C

No route reaches $9$ : $\frac{18}{26}$ is not $\frac{2}{3}$ .

Skills used

Turn a worded condition into a linear equation in one unknown

26Arithmetic • Interest • Compound Interest

HardIdeal time: about 2 min

A sum of Rs. $8, 400$ was taken as a loan. This is to be paid in two equal instalments. If the rate of interest is $10%$ per annum, compounded annually, then the value of each instalment is

ARs.

4, 200

BRs.

4, 480

CRs.

4, 840

DNone of the above

Show answer & solution

Option C

Rs.

4, 840

Approach

Discount each instalment back to today at $10%$ and set the two present values against the loan.

A loan repaid in equal instalments under compound interest

Step-by-step solution

1
Let each instalment be $P$ , paid at the end of year $1$ and of year $2$
Why: 'Two equal instalments' under annual compounding means one payment at each year-end.
Percentage change as a multiplier A loan repaid in equal instalments under compound interest
2
$\frac{P}{1.1} + \frac{P}{( 1.1 ) ^{2}} = 8400$
Why: A sum one year away is worth its value divided by the growth factor $1.1$ today; two years away, divided twice.
3
$P (\frac{10}{11} + \frac{100}{121}) = P \times \frac{110 + 100}{121} = P \times \frac{210}{121} = 8400$
Why: Keeping the factors as fractions avoids decimals entirely.
Negative exponent is a reciprocal
4
$P = 8400 \times \frac{121}{210} = 40 \times 121 = 4840$
Why: $8400 \div 210 = 40$ , so the instalment is Rs. $4, 840$ .

Common mistakes

Option A

Halved the principal and ignored the interest altogether.

Option B

No route reaches Rs. $4, 480$ — it is Rs. $4, 840$ with two digits transposed.

Option D

Chosen by anyone whose arithmetic missed Rs. $4, 840$ , which is printed.

Skills used

Percentage change as a multiplier Negative exponent is a reciprocal A loan repaid in equal instalments under compound interest

27Arithmetic • Age Problems

EasyIdeal time: about 55 s

Leela got married $6$ years ago. Today her age is $1 \frac{1}{4}$ times her age at the time of her marriage. Her son's age is $\frac{1}{10}$ times her age. What is the present age of her son?

1

yearB

2

yearsC

3

yearsD

4

years

Show answer & solution

Option C

3

years

Approach

Let her age at marriage be $a$ ; today she is $a + 6$ , and the stated multiple fixes $a$ .

Turn a worded condition into a linear equation in one unknown

Step-by-step solution

1
$a + 6 = \frac{5}{4} a$
Why: $1 \frac{1}{4} = \frac{5}{4}$ , and her present age is her age at marriage plus the six intervening years.
Turn a worded condition into a linear equation in one unknown
2
$6 = \frac{1}{4} a \Rightarrow a = 24$ , so her present age is $24 + 6 = 30$
Why: The six years are exactly the quarter by which her age has grown.
3
son's age $= \frac{1}{10} \times 30 = 3$ years
Why: The fraction is of her present age, not of her age at marriage.

Common mistakes

Option A

No route reaches $1$ year.

Option B

Took a tenth of her age at marriage, $\frac{24}{10} = 2.4$ , and rounded to $2$ .

Option D

No route reaches $4$ years.

Skills used

Turn a worded condition into a linear equation in one unknown

28Arithmetic • Time and Work

Very easyIdeal time: about 40 s

A and B working together can finish a piece of work in $12$ days while B alone can finish it in $30$ days. In how many days can A alone finish the work?

18

daysB

20

daysC

24

daysD

25

days

Show answer & solution

Option B

20

days

Approach

Rates add: A's daily rate is the joint rate minus B's.

Take total work as the LCM of the times

Step-by-step solution

1
Take the total work as $LCM (12, 30) = 60$ units
Why: A total divisible by both times keeps every rate a whole number.
Take total work as the LCM of the times
2
A and B together: $\frac{60}{12} = 5$ units/day; B alone: $\frac{60}{30} = 2$ units/day
Why: Each rate is the total divided by that party's time.
3
A alone: $5 - 2 = 3$ units/day, so A needs $\frac{60}{3} = 20$ days
Why: Rates subtract when one worker is removed from the pair; times never do.

Common mistakes

Option A

Subtracted the times, $30 - 12 = 18$ , instead of the rates.

Option C

No route reaches $24$ days.

Option D

No route reaches $25$ days.

Skills used

Take total work as the LCM of the times

29Algebra • Equations • Exponential Equations

EasyIdeal time: about 1 min

The values of $x$ which satisfy the equation $5^{1 + x} + 5^{1 - x} = 26$ are

- 1, 1

0, 1

1, 2

- 1, 0

Show answer & solution

Option A

- 1, 1

Approach

Put $u = 5^{x}$ ; the equation becomes a quadratic whose two roots are reciprocals.

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

Step-by-step solution

1
$5^{1 + x} = 5 \cdot 5^{x} = 5 u$ and $5^{1 - x} = \frac{5}{5 ^{x}} = \frac{5}{u}$ , where $u = 5^{x} > 0$
Why: Splitting the exponents into a constant part and the $x$ part puts the whole equation in terms of one quantity.
Power of a power, and the move to a common base Negative exponent is a reciprocal
2
$5 u + \frac{5}{u} = 26 \Rightarrow 5 u^{2} - 26 u + 5 = 0 \Rightarrow (5 u - 1) (u - 5) = 0$
Why: Multiplying through by $u$ (non-zero) clears the fraction; the middle term splits as $- 25 u - u$ .
Factorise a quadratic by splitting the middle term
3
$u = 5$ or $u = \frac{1}{5}$ , i.e. $5^{x} = 5^{1}$ or $5^{x} = 5^{- 1}$
Why: Both roots are positive, so both are admissible values of $5^{x}$ .
4
$x = 1$ or $x = - 1$
Why: Equal powers of the same base force equal exponents.

Shortcut

The equation is symmetric in $x \to - x$ , so the roots come in a $\pm$ pair; $x = 1$ works, so the pair is $- 1, 1$ .

Common mistakes

Option B

Found $x = 1$ and added $x = 0$ without testing it; $x = 0$ gives $5 + 5 = 10$ , not $26$ .

Option C

Found $x = 1$ and read the second root of $5 u^{2} - 26 u + 5 = 0$ as $u = 25$ , giving $x = 2$ .

Option D

Found $x = - 1$ and added $x = 0$ , again untested.

Skills used

Power of a power, and the move to a common base Negative exponent is a reciprocal Factorise a quadratic by splitting the middle term

30Arithmetic • Time and Work

EasyIdeal time: about 1 min

If $5$ men can do a piece of work in $10$ days and $12$ women can do the same work in $15$ days, the number of days required to complete the work by $5$ men and $6$ women is

7 \frac{1}{2}

daysB

8

daysC

9 \frac{1}{2}

daysD

12

days

Show answer & solution

Option A

7 \frac{1}{2}

days

Approach

Turn each group's stated time into a daily rate, scale the women to the number actually working, and add.

Take total work as the LCM of the times

Step-by-step solution

1
$5$ men finish in $10$ days, so those five men together do $\frac{1}{10}$ of the work per day
Why: The rate belongs to the group as stated; there is no need to find one man's rate.
Take total work as the LCM of the times
2
$12$ women finish in $15$ days, so $6$ women — half as many — do $\frac{1}{2} \times \frac{1}{15} = \frac{1}{30}$ per day
Why: Halving the workforce halves the daily output.
3
together: $\frac{1}{10} + \frac{1}{30} = \frac{3 + 1}{30} = \frac{2}{15}$ per day
Why: Rates add when the two groups work at the same time.
4
time $= \frac{1}{2/15} = \frac{15}{2} = 7 \frac{1}{2}$ days
Why: Time is the whole job divided by the combined daily rate.

Common mistakes

Option B

No route reaches $8$ days.

Option C

No route reaches $9 \frac{1}{2}$ days.

Option D

No route reaches $12$ days.

Skills used

Take total work as the LCM of the times

31Arithmetic • Time, Speed and Distance • Relative Speed

ModerateIdeal time: about 2 min

A passenger train departs from Delhi at 6 p.m. for Mumbai. At 9 p.m., an express train, whose average speed exceeds that of the passenger train by $15$ km/hour leaves Mumbai for Delhi. Two trains meet each other mid-route. At what time do they meet, given that the distance between the cities is $1080$ km?

A4 p.m.B2 a.m.C12 midnightD6 a.m.

Show answer & solution

Option D

6 a.m.

Approach

Each train covers half of $1080$ km; turn the three-hour head start into an equation in the passenger train's speed, solve it, then add the journey time to $6$ p.m.

Speed, distance and time

Step-by-step solution

1
$each train covers \frac{1080}{2} = 540 km$
Why: "Meet each other mid-route" fixes the meeting point at the midpoint, so both trains have run $540$ km when they meet.
Speed, distance and time
2
$v = passenger speed, v + 15 = express speed$
Why: The express's average speed is given as $15$ km/hour more than the passenger train's, so one letter carries both.
Turn a worded condition into a linear equation in one unknown
3
$\frac{540}{v} = \frac{540}{v + 15} + 3$
Why: They are at the midpoint at the same instant. The passenger train left at $6$ p.m. and the express at $9$ p.m., so the passenger train has been running exactly $3$ hours longer.
4
$540 (v + 15) - 540 v = 3 v (v + 15) \Rightarrow 8100 = 3 v^{2} + 45 v \Rightarrow v^{2} + 15 v - 2700 = 0$
Why: Clearing the fractions across the common denominator $v (v + 15)$ leaves a quadratic in $v$ .
5
$(v + 60) (v - 45) = 0 \Rightarrow v = 45 km/hour$
Why: $- 2700 = 60 \times (- 45)$ and $60 - 45 = 15$ , so the middle term splits cleanly; a speed cannot be negative, so $v = 45$ .
Factorise a quadratic by splitting the middle term
6
$t = \frac{540}{45} = 12 hours, 6 p.m. + 12 h = 6 a.m.$
Why: The passenger train needs $12$ hours to reach the midpoint, and it started at $6$ p.m.

Shortcut

Hunt for two speeds differing by $15$ whose $540$ -km times differ by $3$ hours: $45$ and $60$ give $12$ h and $9$ h at once.

Common mistakes

Option A

$4$ p.m. is earlier than the passenger train's own $6$ p.m. departure, so no route reaches it.

Option B

No consistent pair of speeds differing by $15$ km/hour puts the midpoint meeting at $2$ a.m.

Option C

Arithmetic slip while clearing the fractions (for instance $\frac{540 \times 3}{15} = 18$ instead of $108$ in the time form $t^{2} - 3 t = 108$ ), giving a $6$ -hour run and a midnight meeting.

Skills used

Speed, distance and time Turn a worded condition into a linear equation in one unknown Factorise a quadratic by splitting the middle term

32Arithmetic • Ratio and Proportion

EasyIdeal time: about 45 s

In a class of $49$ students, the ratio of girls to boys is $4 : 3$ . If $4$ girls leave the class, the ratio of girls to boys would be

11 : 7

8 : 7

6 : 5

9 : 8

Show answer & solution

Option B

8 : 7

Approach

Turn the ratio into actual head counts using the class total, remove the four girls, then re-form the ratio.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$girls = 4 k, boys = 3 k, 4 k + 3 k = 49 \Rightarrow k = 7$
Why: A ratio fixes the two counts only up to a common factor; the total of $49$ pins $k$ down.
Introduce a common factor $k$ for a ratio
2
$girls = 28, boys = 21$
Why: Substituting $k = 7$ gives the actual numbers, and $28 + 21 = 49$ confirms them.
3
$\frac{28 - 4}{21} = \frac{24}{21} = \frac{8}{7}$
Why: Only the girls' count changes; dividing both parts by their HCF $3$ puts the new ratio in lowest terms.

Common mistakes

Option A

$11 : 7$ does not divide the surviving $45$ students into whole numbers; no route reaches it.

Option C

$6 : 5$ does not divide the surviving $45$ students into whole numbers; no route reaches it.

Option D

$9 : 8$ does not divide the surviving $45$ students into whole numbers; no route reaches it.

Skills used

Introduce a common factor

k

for a ratio

33Algebra • Expressions • Simplification

EasyIdeal time: about 50 s

If $a + b = 5$ and $ab = 6$ , then what is the value of $a^{3} + b^{3}$ ?

35

40

90

125

Show answer & solution

Option A

35

Approach

Get $a^{2} + b^{2}$ from the square of the sum, then feed it into the sum-of-cubes identity.

Sum and difference of two cubes

Step-by-step solution

1
$a^{2} + b^{2} = (a + b)^{2} - 2 ab = 5^{2} - 2 \times 6 = 13$
Why: Expanding $(a + b)^{2}$ shows $a^{2} + b^{2}$ is the square of the sum less twice the product, both of which are given.
Square of a sum and of a difference
2
$a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$
Why: The sum-of-cubes factorisation expresses the target in exactly the two symmetric quantities that are known.
Sum and difference of two cubes
3
$a^{3} + b^{3} = 5 \times (13 - 6) = 5 \times 7 = 35$
Why: Substituting $a + b = 5$ , $a^{2} + b^{2} = 13$ and $ab = 6$ finishes it.

Shortcut

$a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b) = 125 - 90 = 35$ in one line.

Common mistakes

Option B

No natural mis-step produces $40$ from $a + b = 5$ and $ab = 6$ .

Option C

Computed the correction term $3 ab (a + b) = 3 \times 6 \times 5 = 90$ and reported it instead of $(a + b)^{3} - 3 ab (a + b)$ .

Option D

Took $a^{3} + b^{3} = (a + b)^{3} = 125$ , forgetting the $- 3 ab (a + b)$ correction.

Skills used

Sum and difference of two cubes Square of a sum and of a difference

34Arithmetic • Profit and Loss • Discount

EasyIdeal time: about 40 s

Rajendra bought a mobile with $25%$ discount on the selling price. If the mobile cost him Rs. $4, 875$ , what is the original selling price of the mobile?

ARs.

6, 300

BRs.

6, 400

CRs.

6, 500

DRs.

6, 600

Show answer & solution

Option C

Rs.

6, 500

Approach

A $25%$ discount multiplies the selling price by $0.75$ ; divide the amount paid by that multiplier to reverse it.

Percentage change as a multiplier

Step-by-step solution

1
$price paid = (1 - \frac{25}{100}) \times SP = \frac{3}{4} SP$
Why: The discount is stated on the selling price, so the selling price is the base and the multiplier is $1 - \frac{25}{100}$ .
Percentage change as a multiplier
2
$\frac{3}{4} SP = 4875 \Rightarrow SP = 4875 \times \frac{4}{3} = 6500$
Why: A percentage change is undone by dividing by its multiplier, never by adding the same percentage back.

Common mistakes

Option A

Rs. $6, 300$ is not $\frac{4}{3}$ of Rs. $4, 875$ and no discount mis-step lands on it.

Option B

Rs. $6, 400$ is not $\frac{4}{3}$ of Rs. $4, 875$ and no discount mis-step lands on it.

Option D

Rs. $6, 600$ is not $\frac{4}{3}$ of Rs. $4, 875$ and no discount mis-step lands on it.

Skills used

Percentage change as a multiplier

35Arithmetic • Time, Speed and Distance • Trains

EasyIdeal time: about 50 s

A $225$ m long train is running at a speed of $30$ km/hour. How much time does it take to cross a man running at $3$ km/hour in the same direction?

40

secondsB

30

secondsC

25

secondsD

15

seconds

Show answer & solution

Option B

30

seconds

Approach

Same direction means the speeds subtract; crossing a man means covering just the train's own length at that relative speed.

Relative speed: add when opposed, subtract when alike

Step-by-step solution

1
$v_{rel} = 30 - 3 = 27 km/hour$
Why: Train and man move the same way, so the man's speed is taken off the train's to get the closing speed.
Relative speed: add when opposed, subtract when alike
2
$27 \times \frac{5}{18} = 7.5 m/second$
Why: The length is in metres and the answer is wanted in seconds, so the speed must be converted with the factor $\frac{5}{18}$ .
km/hour to m/second
3
$t = \frac{225}{7.5} = 30 seconds$
Why: A man has no length, so crossing him means the train covers exactly its own $225$ m relative to him.
What a train covers when it crosses something Speed, distance and time

Common mistakes

Option A

$40$ seconds needs a relative speed of $20.25$ km/hour, which no combination of $30$ and $3$ gives.

Option C

Added the speeds as though the man ran towards the train: $\frac{225}{33 \times \frac{5}{18}} \approx 24.5$ seconds, rounded to $25$ .

Option D

$15$ seconds needs a relative speed of $54$ km/hour, which no combination of $30$ and $3$ gives.

Skills used

Relative speed: add when opposed, subtract when alike km/hour to m/second What a train covers when it crosses something Speed, distance and time

36Arithmetic • Number System • Rational Numbers

EasyIdeal time: about 1 min

Which one among the following is the largest?

\frac{7}{9}

\frac{11}{14}

\frac{3}{4}

\frac{10}{13}

Show answer & solution

Option B

\frac{11}{14}

Approach

Reduce each fraction to its decimal value per unit and take the biggest.

Comparing two fractions

Step-by-step solution

1
$\frac{7}{9} = 0.7 7, \frac{11}{14} = 0.7857 \dots, \frac{3}{4} = 0.75, \frac{10}{13} = 0.7692 \dots$
Why: Dividing numerator by denominator puts all four on the same scale, which is what comparison needs; the denominators $9, 14, 4, 13$ share no convenient common multiple.
Comparing two fractions
2
$0.7857 \dots > 0.7 7 > 0.7692 \dots > 0.75$
Why: Ordering the four decimals puts $\frac{11}{14}$ on top.

Shortcut

Compare each with $\frac{3}{4}$ : $\frac{7}{9}, \frac{11}{14}, \frac{10}{13}$ all beat it, then cross-multiply the survivors pairwise.

Common mistakes

Option A

Used the rule of thumb "smaller gap between numerator and denominator wins" ( $9 - 7 = 2$ against $14 - 11 = 3$ ); it is only valid when the denominators agree.

Option C

$\frac{3}{4} = 0.75$ is the smallest of the four; picked as the familiar-looking fraction.

Option D

Cross-multiplied $\frac{10}{13}$ against $\frac{11}{14}$ the wrong way round: $10 \times 14 = 140$ is less than $11 \times 13 = 143$ , so $\frac{10}{13}$ is in fact smaller.

Skills used

Comparing two fractions

37Arithmetic • Interest • Difference of SI and CI

ModerateIdeal time: about 1.5 min

The difference between the simple and the compound interest on a certain sum of money at $4%$ per annum in $2$ years is Rs. $10$ . What is the sum?

ARs.

5, 000

BRs.

6, 000

CRs.

6, 250

DRs.

7, 500

Show answer & solution

Option C

Rs.

6, 250

Approach

Write both two-year amounts as multipliers of the principal and set their difference to Rs. $10$ .

Two-year difference between compound and simple interest

Step-by-step solution

1
$SI for 2 years = P \times \frac{4}{100} \times 2 = 0.08 P$
Why: Simple interest is charged on the original principal each year, so two years give twice $4%$ of $P$ .
$x %$ of a quantity
2
$CI for 2 years = P (1.04)^{2} - P = P (1.0816 - 1) = 0.0816 P$
Why: Compound interest applies the multiplier $1 + \frac{4}{100}$ once per year, so the second year's interest is charged on the grown amount.
Percentage change as a multiplier
3
$0.0816 P - 0.08 P = 0.0016 P = 10 \Rightarrow P = \frac{10}{0.0016} = 6250$
Why: The difference is the interest earned in year two on year one's interest, and it is given as Rs. $10$ .
Two-year difference between compound and simple interest

Shortcut

For $2$ years the difference is always $P (\frac{r}{100})^{2}$ , so $P = 10 \div (\frac{4}{100})^{2} = 6250$ .

Common mistakes

Option A

Rs. $5, 000$ gives a two-year difference of Rs. $8$ , not Rs. $10$ .

Option B

Rs. $6, 000$ gives a difference of Rs. $9.60$ , rounded up to Rs. $10$ .

Option D

Rs. $7, 500$ gives a two-year difference of Rs. $12$ , not Rs. $10$ .

Skills used

Two-year difference between compound and simple interest

x %

of a quantity Percentage change as a multiplier

38Arithmetic • Percentage

ModerateIdeal time: about 1 min

If $a %$ of $a$ $+$ $b %$ of $b$ $=$ $2%$ of $ab$ , then what percent of $a$ is $b$ ?

50%

75%

100%

DCannot be determined

Show answer & solution

Option C

100%

Approach

Write each percentage as a hundredth, clear the $100$ s, and recognise the perfect square that results.

x %

of a quantity

Step-by-step solution

1
$\frac{a}{100} \times a + \frac{b}{100} \times b = \frac{2}{100} \times ab$
Why: " $a %$ of $a$ " is $\frac{a}{100}$ of $a$ ; every term carries the same $\frac{1}{100}$ .
$x %$ of a quantity
2
$a^{2} + b^{2} = 2 ab$
Why: Multiplying through by $100$ removes the percentages entirely.
3
$a^{2} - 2 ab + b^{2} = 0 \Rightarrow (a - b)^{2} = 0$
Why: Everything to one side is the square-of-a-difference identity.
Square of a sum and of a difference
4
$a = b \Rightarrow \frac{b}{a} \times 100 = 100%$
Why: A square is zero only when its base is zero, so $a$ and $b$ are equal and $b$ is the whole of $a$ .
A product is zero only when a factor is zero

Common mistakes

Option A

$a = b$ read as " $b$ is half of the pair", reporting $50%$ .

Option B

$75%$ satisfies no version of $a^{2} + b^{2} = 2 ab$ .

Option D

Counted one equation in two unknowns and concluded the ratio was free, missing that $a^{2} + b^{2} = 2 ab$ forces $a = b$ .

Skills used

x %

of a quantity Square of a sum and of a difference A product is zero only when a factor is zero

39Arithmetic • Percentage

ModerateIdeal time: about 1.5 min

$\frac{5}{9}$ th part of the population in a village are males. If $30%$ of the males are married, the percentage of unmarried females in the total population is

20 \frac{2}{9}

27 \frac{2}{9}

27 \frac{7}{9}

29 \frac{2}{9}

Show answer & solution

Option C

27 \frac{7}{9}

Approach

Take the population as a whole, count married males, pair them with married females, and subtract from the female share.

x %

of a quantity

Step-by-step solution

1
$males = \frac{5}{9} P, females = P - \frac{5}{9} P = \frac{4}{9} P$
Why: Everyone is male or female, so the female share is the complement of the male share.
Introduce a common factor $k$ for a ratio
2
$married males = \frac{30}{100} \times \frac{5}{9} P = \frac{1}{6} P$
Why: $30%$ is taken of the males, not of the population.
$x %$ of a quantity
3
$married females = married males = \frac{1}{6} P$
Why: Each married man in the village is married to one woman of the village, so the two counts match.
4
$\frac{4}{9} P - \frac{1}{6} P = \frac{8 P - 3 P}{18} = \frac{5}{18} P$
Why: Unmarried females are the females who are not married.
5
$\frac{5}{18} \times 100 = \frac{250}{9} = 27 \frac{7}{9} %$
Why: The question asks for a percentage of the TOTAL population, so the share is multiplied by $100$ .

Common mistakes

Option A

$20 \frac{2}{9} %$ corresponds to no share of the $\frac{4}{9}$ female population.

Option B

Slipped in the fractional part when converting $\frac{5}{18}$ to a mixed percentage, writing $\frac{2}{9}$ for $\frac{7}{9}$ .

Option D

$29 \frac{2}{9} %$ corresponds to no share of the $\frac{4}{9}$ female population.

Skills used

x %

of a quantity Introduce a common factor

k

for a ratio

40Arithmetic • Number System

ModerateIdeal time: about 1.5 min

Sunil wants to spend Rs. $200$ on two types of sweets, costing Rs. $7$ and Rs. $10$ respectively. What is the maximum number of sweets he can get so that no money is left over?

25

26

27

28

Show answer & solution

Option B

26

Approach

Solve $7 x + 10 y = 200$ in non-negative integers, then note that the count is largest when as much as possible goes on the cheaper sweet.

Enumerate a residue class within a bound

Step-by-step solution

1
$7 x + 10 y = 200, x, y non-negative integers$
Why: $x$ sweets at Rs. $7$ and $y$ at Rs. $10$ must use up exactly Rs. $200$ — "no money left over".
Turn a worded condition into a linear equation in one unknown
2
$7 x \equiv 200 \equiv 0 (mod 10) \Rightarrow x \equiv 0 (mod 10)$
Why: $10 y$ is a multiple of $10$ , so $7 x$ must end in $0$ ; since $7$ is coprime to $10$ , $x$ itself is a multiple of $10$ .
Enumerate a residue class within a bound
3
$x \in {0, 10, 20} \Rightarrow (x, y) = (0, 20), (10, 13), (20, 6)$
Why: $7 x \leq 200$ caps $x$ at $28$ , so the residue class $10 k$ offers only these three, each giving a whole $y$ .
4
$x + y = 20, 23, 26 \Rightarrow max = 26$
Why: Every Rs. $70$ moved from the dearer sweet to the cheaper buys $10$ instead of $7$ , so the total count rises with $x$ ; $x = 20$ is the largest admissible value.

Shortcut

Only $x \equiv 0 (mod 10)$ works, so take the largest such $x$ with $7 x \leq 200$ : $x = 20$ , $y = 6$ , total $26$ .

Common mistakes

Option A

$x + y = 25$ forces $3 x = 50$ , so no whole-number purchase totals $25$ sweets; stopped at a non-maximal guess.

Option C

$x + y = 27$ forces $3 x = 70$ , so no whole-number purchase totals $27$ sweets; overshot the true maximum by one.

Option D

$x + y = 28$ would need $3 x = 80$ ; reached by dividing $200$ by $7$ and rounding.

Skills used

Enumerate a residue class within a bound Turn a worded condition into a linear equation in one unknown