Elementary Mathematics 2018 (II)

Set A

Questions 41–60 of 100

41Arithmetic • Mixtures and Alligation

ModerateIdeal time: about 1.5 min

A milk vendor bought $28$ litres of milk at the rate of Rs. $8.50$ per litre. After adding some water he sold the mixture at the same price. If his gain is $12.5%$ , how much water did he add ?

4.5

litresB

4

litresC

3.5

litresD

3

litres

Show answer & solution

Option C

3.5

litres

Approach

The water costs nothing, so the whole gain comes from selling water at milk price; compare cost with receipts.

Gain percent is measured on the cost price

Step-by-step solution

1
$cost = 28 \times 8.50 = 238$
Why: Only the milk was paid for.
Unitary method: find one unit, then scale
2
$receipts = 238 \times \frac{112.5}{100} = 267.75$
Why: A gain of $12.5%$ is measured on the cost price, so the selling total is $\frac{9}{8}$ of Rs $238$ .
Gain percent is measured on the cost price
3
$volume sold = \frac{267.75}{8.50} = 31.5 litres$
Why: The mixture went out at the same Rs $8.50$ a litre, so the receipts divided by that price give the quantity sold.
4
$water = 31.5 - 28 = 3.5 litres$
Why: The extra litres over the $28$ bought are the water added.
Turn a worded condition into a linear equation in one unknown

Shortcut

When the mixture sells at the milk price, the gain percent is the water-to-milk ratio: water $= 12.5%$ of $28 = 3.5$ litres. The price Rs $8.50$ never has to be used.

Common mistakes

Option A

No route reaches $4.5$ litres; it would give a gain of $\frac{4.5}{28} = 16.07%$ .

Option B

Measured the gain on the selling total instead of the cost: $\frac{w}{28 + w} = 12.5%$ gives $w = 4$ litres.

Option D

$3$ litres gives a gain of $\frac{3}{28} = 10.7%$ , not $12.5%$ .

Skills used

Unitary method: find one unit, then scale Gain percent is measured on the cost price Turn a worded condition into a linear equation in one unknown

42Arithmetic • Clocks • Gaining and Losing Clocks

HardIdeal time: about 2.5 min

The minute hand of a clock overtakes the hour hand after every $72$ minutes of correct time. How much time does the clock lose or gain in a day of normal time ?

ALose

121 \frac{9}{11}

minutesBLose

157 \frac{1}{11}

minutesCGain

121 \frac{9}{11}

minutesDGain

157 \frac{1}{11}

minutes

Show answer & solution

No option is correct

Approach

A correct clock's hands meet every $\frac{720}{11}$ minutes of its own time; compare that with the $72$ minutes of true time the faulty clock takes, and scale the shortfall to a day.

Angle between the hands of a clock

Step-by-step solution

1
$minute hand gains \frac{11}{2} degrees per minute on the hour hand$
Why: The minute hand moves at $6^{\circ}$ a minute and the hour hand at $\frac{1}{2}^{\circ}$ a minute, which is what the hands-separation formula $30 H - \frac{11 M}{2}$ encodes.
Angle between the hands of a clock
2
$coincidence interval = \frac{360}{11/2} = \frac{720}{11} = 65 \frac{5}{11} minutes$
Why: From one overtaking to the next the minute hand must gain a full $36 0^{\circ}$ . This interval is in the clock's OWN minutes, whatever rate the clock runs at, because coincidence is a condition on the dial.
3
$\frac{720}{11} clock-minutes = 72 true minutes$
Why: The overtaking that costs the clock $\frac{720}{11}$ of its own minutes took $72$ minutes of correct time, so the clock runs slow.
Unitary method: find one unit, then scale
4
$clock shows 1440 \times \frac{720/11}{72} = 1440 \times \frac{10}{11} = 1309 \frac{1}{11} minutes in a day$
Why: Scale the rate up to the $1440$ true minutes of a day.
5
$loss = 1440 - 1309 \frac{1}{11} = \frac{1440}{11} = 130 \frac{10}{11} minutes$
Why: The shortfall between true time and shown time is the loss.

Skills used

Angle between the hands of a clock Unitary method: find one unit, then scale

43Arithmetic • Time, Speed and Distance • Relative Speed

ModerateIdeal time: about 1.5 min

A thief steals a car parked in a house and goes away with a speed of $40$ kmph. The theft was discovered after half an hour and immediately the owner sets off in another car with a speed of $60$ kmph. When will the owner meet the thief ?

55

km from the owner's house and one hour after the theftB

60

km from the owner's house and

1.5

hours after the theftC

60

km from the owner's house and

1.5

hours after the discovery of the theftD

55

km from the owner's house and

1.5

hours after the theft

Show answer & solution

Option B

60

km from the owner's house and

1.5

hours after the theft

Approach

The half-hour head start is a $20$ km lead; the owner closes it at the difference of the speeds.

Relative speed: add when opposed, subtract when alike

Step-by-step solution

1
$thief’s lead = 40 \times \frac{1}{2} = 20 km$
Why: The theft was discovered half an hour after it happened, and the owner left immediately.
Speed, distance and time
2
$closing speed = 60 - 40 = 20 kmph$
Why: Both cars go the same way, so the gap shrinks at the difference of the speeds.
Relative speed: add when opposed, subtract when alike
3
$time to close = \frac{20}{20} = 1 hour after the owner starts$
Why: Gap divided by closing speed.
4
$distance = 60 \times 1 = 60 km; time from the theft = \frac{1}{2} + 1 = 1.5 hours$
Why: The distance from the house is what the owner covered. The clock in the options runs from the theft, not from the owner's start, so add the half hour back.

Common mistakes

Option A

Both halves wrong: $55$ km matches no position, and "one hour after the theft" measures the catching-up hour from the theft instead of from the owner's start.

Option C

Right distance, clock started in the wrong place: $1.5$ hours after the DISCOVERY is $2$ hours after the theft, by which time both cars are well past $60$ km.

Option D

Right time, but $55$ km is not where they meet — at $1.5$ hours after the theft the thief has done $40 \times 1.5 = 60$ km.

Skills used

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

44Arithmetic • Time and Work

Very easyIdeal time: about 30 s

X and Y together can finish a job in $6$ days. X can alone do the same job in $12$ days. How long will Y alone take to do the same job ?

16

daysB

12

daysC

10

daysD

8

days

Show answer & solution

Option B

12

days

Approach

Rates add: subtract X's daily rate from the pair's rate to get Y's.

Take total work as the LCM of the times

Step-by-step solution

1
$take the job as 12 units; pair = \frac{12}{6} = 2, X = \frac{12}{12} = 1 unit per day$
Why: Taking the total work as the LCM of $6$ and $12$ keeps every rate a whole number.
Take total work as the LCM of the times
2
$Y = 2 - 1 = 1 unit per day$
Why: Working together, the two rates add, so Y's rate is the pair's less X's.
3
$Y alone = \frac{12}{1} = 12 days$
Why: Total work divided by Y's rate.

Common mistakes

Option A

No route gives $16$ days; Y would then contribute only $\frac{1}{16}$ a day and the pair would need $6 \frac{6}{7}$ days.

Option C

$10$ days for Y would give the pair $\frac{1}{12} + \frac{1}{10} = \frac{11}{60}$ , i.e. $5 \frac{5}{11}$ days, not $6$ .

Option D

$8$ days for Y would give the pair $\frac{1}{12} + \frac{1}{8} = \frac{5}{24}$ , i.e. $4.8$ days.

Skills used

Take total work as the LCM of the times

45Arithmetic • Time and Work

EasyIdeal time: about 1 min

Twelve ( $12$ ) persons can paint $10$ identical rooms in $16$ days. In how many days can $8$ persons paint $20$ such rooms ?

12

24

36

48

Show answer & solution

Option D

48

Approach

Measure one room in person-days, scale to twenty rooms, then divide by the eight people.

The work-equivalence chain rule

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

Step-by-step solution

1
$10 rooms = 12 \times 16 = 192 person-days$
Why: With equally efficient painters the product (people $\times$ days) measures the work done.
The work-equivalence chain rule $M_{1} D_{1} H_{1} = M_{2} D_{2} H_{2}$
2
$1 room = \frac{192}{10} = 19.2 person-days$
Why: The rooms are identical, so the work splits evenly between them.
Unitary method: find one unit, then scale
3
$20 rooms = 20 \times 19.2 = 384 person-days$
Why: Twice as many rooms is twice the work.
4
$days = \frac{384}{8} = 48$
Why: Divide the work by the daily output of eight people.

Shortcut

$16 \times \frac{20}{10} \times \frac{12}{8} = 48$ — more rooms multiplies the days, fewer painters multiplies them again.

Common mistakes

Option A

Inverted the room ratio, using $\frac{10}{20}$ instead of $\frac{20}{10}$ : $16 \times \frac{10}{20} \times \frac{12}{8} = 12$ .

Option B

Adjusted only for the smaller crew: $16 \times \frac{12}{8} = 24$ , leaving the room count at $10$ .

Option C

No route reaches $36$ days.

Skills used

The work-equivalence chain rule

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

Unitary method: find one unit, then scale

46Algebra • Logarithm • Characteristic and Mantissa

HardIdeal time: about 2 min

There are $n$ zeros appearing immediately after the decimal point in the value of $(0.2)^{25}$ . It is given that the value of $lo g_{10} 2 = 0.30103$ . The value of $n$ is

25

19

18

17

Show answer & solution

Option D

17

Approach

Count the digits of $2^{25}$ from its logarithm, then place $2^{25}$ over $1 0^{25}$ to see where the first significant figure lands.

Number of digits from a common logarithm

Step-by-step solution

1
$(0.2)^{25} = (\frac{2}{10})^{25} = \frac{2 ^{25}}{1 0 ^{25}}$
Why: Splitting the fraction turns the question into one about the size of $2^{25}$ , which is what the given logarithm is for.
Negative exponent is a reciprocal
2
$lo g_{10} 2^{25} = 25 lo g_{10} 2 = 25 \times 0.30103 = 7.52575$
Why: The power rule converts a $25$ th power into a multiplication.
$lo g a^{n} = n lo g a$
3
$digits of 2^{25} = ⌊ 7.52575 ⌋ + 1 = 8$
Why: Take the floor and add one; so $1 0^{7} < 2^{25} < 1 0^{8}$ , i.e. $2^{25}$ is an $8$ -digit number.
Number of digits from a common logarithm
4
$1 0^{- 18} < \frac{2 ^{25}}{1 0 ^{25}} < 1 0^{- 17}$
Why: Dividing an $8$ -digit number by $1 0^{25}$ shifts it $25$ places right, leaving a value between $1 0^{- 18}$ and $1 0^{- 17}$ .
5
$first significant digit sits in the 18 th decimal place \Rightarrow n = 17$
Why: A value just below $1 0^{- 17}$ has its leading digit in the $18$ th place, so the first $17$ decimal places are zeros.

Shortcut

The number of zeros is one less than the number of places the value is shifted: $25$ places right, $8$ digits of $2^{25}$ , so $25 - 8 = 17$ zeros.

Common mistakes

Option A

Answered the exponent $25$ itself, which is the number of decimal places the division shifts, not the number of leading zeros.

Option B

$19$ is two too many; it comes of shifting by $25$ and then counting the zeros as $25 - ⌊ 7.52575 ⌋ + 1$ .

Option C

Read the characteristic of $lo g (0.2)^{25} = - 18 + 0.52575$ as the zero count: $18$ is the position of the first significant digit, so the zeros number one fewer.

Skills used

Negative exponent is a reciprocal

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

Number of digits from a common logarithm

47Arithmetic • Age Problems

ModerateIdeal time: about 2 min

The ratio of the sum and difference of the ages of the father and the son is $11 : 3$ . Consider the following statements :

1: The ratio of their ages is $8 : 5$ .

2: The ratio of their ages after the son attains twice the present age will be $11 : 8$ .

Which of the statements given above is/are correct ?

1

onlyB

2

onlyCBoth

1

and

2

DNeither

1

nor

2

Show answer & solution

Option B

2

only

Approach

Turn the sum-to-difference ratio into the age ratio by taking half the sum and half the difference, then test each statement against it.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$\frac{F + S}{F - S} = \frac{11}{3} \Rightarrow 3 F + 3 S = 11 F - 11 S$
Why: Cross-multiply the given ratio.
Clear the denominators of an equation
2
$14 S = 8 F \Rightarrow \frac{F}{S} = \frac{14}{8} = \frac{7}{4}$
Why: Collect the $F$ and $S$ terms. Equivalently: if $F + S = 11 k$ and $F - S = 3 k$ then $F = 7 k$ and $S = 4 k$ , the half-sum and half-difference.
Introduce a common factor $k$ for a ratio
3
$Statement 1: 8 : 5 would give \frac{8 + 5}{8 - 5} = \frac{13}{3} \neq = \frac{11}{3}$
Why: Test the claimed ratio against the one condition the question gives. It fails, so Statement 1 is wrong.
4
$son doubles: 4 k \to 8 k, so 4 k years pass; F = 7 k + 4 k = 11 k, S = 8 k$
Why: "After the son attains twice the present age" fixes how many years pass: as many as his present age. Both ages advance by that same amount.
Turn a worded condition into a linear equation in one unknown
5
$11 k : 8 k = 11 : 8 \Rightarrow Statement 2 is correct$
Why: The common factor cancels, so the ratio holds whatever $k$ is.

Common mistakes

Option A

Accepted Statement 1 and rejected Statement 2; $8 : 5$ is the ratio that would follow from a sum-to-difference ratio of $13 : 3$ , not $11 : 3$ .

Option C

Verified Statement 2 and took Statement 1 on trust without testing $8 : 5$ against the given $11 : 3$ .

Option D

Read "after the son attains twice the present age" as "after twice his present age has passed", i.e. $8 k$ years, giving $15 k : 12 k = 5 : 4$ and so rejecting Statement 2 as well.

Skills used

Clear the denominators of an equation Introduce a common factor

k

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

48Algebra • Equations • Linear Equations

ModerateIdeal time: about 1.5 min

The solution of linear inequalities $x + y \geq 5$ and $x - y \leq 3$ lies

AOnly in the first quadrantBIn the first and second quadrantsCIn the second and third quadrantsDIn the third and fourth quadrants

Show answer & solution

Option B

In the first and second quadrants

Approach

Each inequality is a half-plane; find the region they share and check which quadrants it reaches.

A linear inequality in two variables is a half-plane

Step-by-step solution

1
$x + y \geq 5 is the side of the line x + y = 5 away from the origin$
Why: Test the origin: $0 + 0 = 0$ , which is not $\geq 5$ , so the region is the far side of the line through $(5, 0)$ and $(0, 5)$ .
A linear inequality in two variables is a half-plane
2
$x - y \leq 3 is the side of the line y = x - 3 containing the origin$
Why: Test the origin again: $0 - 0 = 0 \leq 3$ , so the origin's side is kept. The two lines cross at $(4, 1)$ .
3
$third quadrant: x < 0, y < 0 \Rightarrow x + y < 0 < 5$
Why: No point with both coordinates negative can have $x + y \geq 5$ , so the third quadrant is ruled out entirely.
4
$fourth quadrant: y < 0 and x \leq y + 3 < 3 \Rightarrow x + y < 3$
Why: In the fourth quadrant the second inequality caps $x$ below $3$ while $y$ is negative, so $x + y$ can never reach $5$ . The fourth quadrant is out too.
5
$(3, 3) and (- 1, 7) both satisfy the pair$
Why: One point in the first quadrant and one in the second settle that both of those quadrants are reached, so the region lies in the first and second quadrants.

Common mistakes

Option A

Stopped at the first quadrant, missing that large positive $y$ allows $x$ to be negative — $(- 1, 7)$ satisfies both inequalities.

Option C

The third quadrant is impossible: $x + y$ is negative there and cannot reach $5$ .

Option D

Both the third and fourth quadrants are impossible, so this pair cannot be right.

Skills used

A linear inequality in two variables is a half-plane

49Algebra • Equations

HardIdeal time: about 2 min

It is given that the equations $x^{2} - y^{2} = 0$ and $(x - a)^{2} + y^{2} = 1$ have single positive solution. For this, the value of 'a' is

2

2

- 2

1

Show answer & solution

Option A

2

Approach

Substitute $y^{2} = x^{2}$ into the circle to get a quadratic in $x$ , then force it to have equal roots.

Quadratic formula

Step-by-step solution

1
$x^{2} - y^{2} = 0 \Rightarrow y^{2} = x^{2}$
Why: The first equation is the pair of lines $y = x$ and $y = - x$ ; either way $y^{2}$ equals $x^{2}$ , which is all the circle needs.
Difference of two squares
2
$(x - a)^{2} + x^{2} = 1 \Rightarrow 2 x^{2} - 2 a x + a^{2} - 1 = 0$
Why: Substituting leaves one quadratic in $x$ ; every solution of the system comes from a root of it.
A product is zero only when a factor is zero
3
$single solution \Rightarrow discriminant = 0 : 4 a^{2} - 8 (a^{2} - 1) = 0$
Why: Two distinct roots would give two distinct values of $x$ and so more than one solution; a single solution forces equal roots.
Quadratic formula
4
$- 4 a^{2} + 8 = 0 \Rightarrow a^{2} = 2 \Rightarrow a = \pm 2$
Why: Solve for $a$ .
5
$x = \frac{a}{2} > 0 \Rightarrow a = 2$
Why: The repeated root is $x = \frac{2 a}{4} = \frac{a}{2}$ , so the solution is positive only for the positive value of $a$ .

Common mistakes

Option B

Stopped at $a^{2} = 2$ and read off $a = 2$ without taking the square root.

Option C

Took the other root of $a^{2} = 2$ ; $a = - 2$ does give a single solution, but it is $x = - \frac{1}{2}$ , which is not positive.

Option D

Set $a^{2} - 1 = 0$ instead of the discriminant, i.e. made the circle pass through the origin; $a = 1$ gives the three solutions $(0, 0)$ , $(1, 1)$ and $(1, - 1)$ , not one.

Skills used

Difference of two squares A product is zero only when a factor is zero Quadratic formula

50Algebra • Polynomial • Zeros and Coefficients

ModerateIdeal time: about 1 min

If $α$ , $β$ and $γ$ are the zeros of the polynomial $f (x) = a x^{3} + b x^{2} + c x + d$ , then $α^{2} + β^{2} + γ^{2}$ is equal to

\frac{b ^{2} - a c}{a ^{2}}

\frac{b ^{2} - 2 a c}{a}

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

\frac{b ^{2} - 2 a c}{a ^{2}}

Show answer & solution

Option D

\frac{b ^{2} - 2 a c}{a ^{2}}

Approach

Read the sum and the pairwise-product sum off the coefficients, then use $(α + β + γ)^{2}$ to reach the sum of squares.

Sum and product of the zeros of a cubic

Step-by-step solution

1
$α + β + γ = - \frac{b}{a}, α β + β γ + γ α = \frac{c}{a}$
Why: For a cubic the coefficients give the symmetric sums directly, with alternating signs: $- \frac{b}{a}$ , $\frac{c}{a}$ , $- \frac{d}{a}$ . The constant $d$ is not needed here.
Sum and product of the zeros of a cubic
2
$α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + γ α)$
Why: Squaring a sum of three terms produces each square once and each cross product twice, so subtracting twice the pairwise sum leaves the squares.
Square of a sum of several terms
3
$= (- \frac{b}{a})^{2} - 2 \cdot \frac{c}{a} = \frac{b ^{2}}{a ^{2}} - \frac{2 c}{a} = \frac{b ^{2} - 2 a c}{a ^{2}}$
Why: Put both terms over $a^{2}$ : the second needs multiplying by $\frac{a}{a}$ , which is where the $a c$ comes from.

Common mistakes

Option A

Dropped the factor $2$ on the pairwise-product sum, giving $\frac{b ^{2} - a c}{a ^{2}}$ .

Option B

Left the denominator as $a$ instead of $a^{2}$ — the $\frac{b ^{2}}{a ^{2}}$ term was not brought to a common denominator properly.

Option C

Took $+ 2 a c$ instead of $- 2 a c$ and squared the wrong quantity into the denominator.

Skills used

Sum and product of the zeros of a cubic Square of a sum of several terms

51Algebra • Sets • Operations

ModerateIdeal time: about 1.5 min

Consider the following for the next 04 (four) items that follow :

In an examination of Class XII, $55%$ students passed in Biology, $62%$ passed in Physics, $60%$ passed in Chemistry, $25%$ passed in Physics and Biology, $30%$ passed in Physics and Chemistry, $28%$ passed in Biology and Chemistry. Only $2%$ failed in all the subjects.

What percentage of students passed in all the three subjects ?

6

5

4

3

Show answer & solution

Option C

4

Approach

Only $2%$ failed everything, so $98%$ passed at least one subject; put that into inclusion-exclusion for three sets and solve for the triple overlap.

Inclusion-exclusion for three sets

Step-by-step solution

1
$n (B \cup P \cup C) = 100 - 2 = 98$
Why: "Only $2%$ failed in all the subjects" means $2%$ passed none, so the rest passed at least one. Work with $100$ students.
Inclusion-exclusion for two sets
2
$98 = (55 + 62 + 60) - (25 + 30 + 28) + x$
Why: For three sets, add the singles, subtract every pair, add the triple back: each student in all three was counted three times and removed three times, so one addition restores them.
Inclusion-exclusion for three sets
3
$98 = 177 - 83 + x = 94 + x \Rightarrow x = 4$
Why: Solve for the triple overlap.
Turn a worded condition into a linear equation in one unknown

Common mistakes

Option A

Took the union as the full $100%$ , forgetting the $2%$ who passed nothing: $100 = 94 + x$ gives $x = 6$ .

Option B

Arithmetic slip in $177 - 83$ , taking it as $93$ and so getting $x = 5$ .

Option D

No consistent route gives $3%$ ; it would need the union to be $97%$ .

Skills used

Inclusion-exclusion for three sets Inclusion-exclusion for two sets Turn a worded condition into a linear equation in one unknown

52Algebra • Sets • Operations

HardIdeal time: about 2 min

Same directions as Q51

What percentage of students passed in exactly one subject ?

21

23

25

27

Show answer & solution

Option B

23

Approach

With the triple overlap known, the exactly-one count is the singles less twice the pairs plus three times the triple.

Exactly one, exactly two and at least two of three sets

Step-by-step solution

1
$n (B \cup P \cup C) = 98 \Rightarrow all three = 98 - 177 + 83 = 4$
Why: First settle the triple overlap, as in the previous item: $2%$ passed nothing, so the union is $98%$ .
Inclusion-exclusion for three sets
2
$exactly one = \sum n (A) - 2 \sum n (A \cap B) + 3 n (A \cap B \cap C)$
Why: A student in exactly two subjects is counted twice in the singles and once in the pairs, so subtracting twice the pairs removes them; a student in all three is counted $3 - 6 = - 3$ times and must be added back three times.
Exactly one, exactly two and at least two of three sets
3
$= 177 - 2 (83) + 3 (4) = 177 - 166 + 12 = 23$
Why: Substitute the three totals.
Turn a worded condition into a linear equation in one unknown

Shortcut

Build the three regions in order: all three $= 4$ , exactly two $= 83 - 12 = 71$ , so exactly one $= 98 - 71 - 4 = 23$ . No formula to remember.

Common mistakes

Option A

$21$ is the percentage passing Biology and Physics but not Chemistry, a two-subject region, not the exactly-one total.

Option C

No consistent route gives $25%$ .

Option D

No consistent route gives $27%$ .

Skills used

Exactly one, exactly two and at least two of three sets Inclusion-exclusion for three sets Turn a worded condition into a linear equation in one unknown

53Algebra • Sets • Operations

ModerateIdeal time: about 1.5 min

Same directions as Q51

If the number of students is $360$ , then how many passed in at least two subjects ?

270

263

265

260

Show answer & solution

Option A

270

Approach

At least two subjects is the pair total less twice the triple overlap; convert that percentage to a count out of $360$ .

Exactly one, exactly two and at least two of three sets

Step-by-step solution

1
$all three = 98 - 177 + 83 = 4%$
Why: As before: $2%$ passed nothing, so the union is $98%$ .
Inclusion-exclusion for three sets
2
$at least two = \sum n (A \cap B) - 2 n (A \cap B \cap C) = 83 - 8 = 75%$
Why: The pair total counts a two-subject student once and a three-subject student three times; removing twice the triple leaves each student once.
Exactly one, exactly two and at least two of three sets
3
$75% of 360 = \frac{75 \times 360}{100} = 270$
Why: Only now bring in the roll of $360$ students.
$x %$ of a quantity

Common mistakes

Option B

$263$ is $73.1%$ of $360$ and matches no region of the diagram.

Option C

$265$ is $73.6%$ of $360$ and matches no region.

Option D

$260$ is $72.2%$ of $360$ and matches no region.

Skills used

Exactly one, exactly two and at least two of three sets Inclusion-exclusion for three sets

x %

of a quantity

54Algebra • Sets • Operations

ModerateIdeal time: about 1.5 min

Same directions as Q51

What is the ratio of number of students who passed in both Physics and Chemistry to number of students who passed in both Biology and Physics but not Chemistry ?

7 : 10

10 : 7

9 : 7

7 : 9

Show answer & solution

Option B

10 : 7

Approach

The first quantity is given outright; the second needs the triple overlap removed from the Biology-Physics figure.

Set difference

A - B

Step-by-step solution

1
$n (P \cap C) = 30%$
Why: Read straight from the stem: $30%$ passed in Physics and Chemistry. Nothing excludes those who also passed Biology, and the question does not ask for it to be excluded.
Inclusion-exclusion for two sets
2
$all three = 98 - 177 + 83 = 4%$
Why: Needed for the second quantity; as in item 51, the $2%$ who passed nothing make the union $98%$ .
Inclusion-exclusion for three sets
3
$n (B \cap P) - n (B \cap P \cap C) = 25 - 4 = 21%$
Why: "Both Biology and Physics but not Chemistry" is the Biology-Physics overlap with the all-three region taken out.
Set difference $A - B$
4
$30 : 21 = 10 : 7$
Why: Divide both parts by $3$ .
Introduce a common factor $k$ for a ratio

Common mistakes

Option A

Wrote the ratio the wrong way round: $7 : 10$ is $21 : 30$ .

Option C

No pair of regions gives $9 : 7$ .

Option D

No pair of regions gives $7 : 9$ .

Skills used

Set difference

A - B

Inclusion-exclusion for two sets Inclusion-exclusion for three sets Introduce a common factor

k

for a ratio

55Statistics • Data Classification

ModerateIdeal time: about 35 s

Data on ratings of hotels in a city is measured on

ANominal scaleBOrdinal scaleCInterval scaleDRatio scale

Show answer & solution

Option B

Ordinal scale

Approach

Ask what can be done with a hotel rating: the categories have a clear order, but the gaps between them carry no fixed size.

Nominal, ordinal, interval and ratio scales

Step-by-step solution

1
$hotel ratings are labels: one star, two star, \dots, five star$
Why: They are categories, not measurements, so the data are qualitative rather than quantitative.
Qualitative, discrete and continuous data Nominal, ordinal, interval and ratio scales
2
$the labels are ordered: five star ranks above four star$
Why: An ordering is present, which is more than a bare name such as a brand or a blood group carries. That rules out the nominal scale.
3
$differences are not fixed: four star minus three star \neq = two star minus one star in any measurable sense$
Why: Interval and ratio scales both need differences (and the ratio scale a true zero as well) to mean something numerically. Ratings give rank only, so the scale is ordinal.

Common mistakes

Option A

Nominal treats the ratings as unordered names, which throws away the fact that five star outranks three star.

Option C

Interval would require equal gaps between consecutive ratings, which a star rating does not guarantee.

Option D

Ratio would require both equal gaps and a meaningful zero, so that one rating could be twice another.

Skills used

Nominal, ordinal, interval and ratio scales Qualitative, discrete and continuous data

56Statistics • Mean • Weighted Mean

EasyIdeal time: about 45 s

The average marks of section A are $65$ and that of section B are $70$ . If the average marks of both the sections combined are $67$ , then the ratio of number of students of section A to that of section B is

3 : 2

1 : 3

3 : 1

2 : 3

Show answer & solution

Option A

3 : 2

Approach

The combined average is the weighted mean of the two section averages; write that equation and the ratio drops out.

Mean of a frequency distribution

Step-by-step solution

1
$\frac{65 a + 70 b}{a + b} = 67$
Why: With $a$ students in A and $b$ in B, the combined average is the total marks over the total students.
Mean of a frequency distribution
2
$65 a + 70 b = 67 a + 67 b$
Why: Clear the denominator.
Clear the denominators of an equation
3
$3 b = 2 a \Rightarrow \frac{a}{b} = \frac{3}{2}$
Why: Collect: $70 b - 67 b = 67 a - 65 a$ . The combined mean sits $2$ above A and $3$ below B, so A must carry the larger weight.
Introduce a common factor $k$ for a ratio

Shortcut

Alligation: the distances from the combined mean are $67 - 65 = 2$ and $70 - 67 = 3$ , and the numbers are in the inverse ratio of those distances, so $a : b = 3 : 2$ .

Common mistakes

Option B

$1 : 3$ would pull the combined average to $68.75$ , above $67$ .

Option C

$3 : 1$ gives $66.25$ , below $67$ .

Option D

Took the distances in the order they appear rather than inversely, giving $2 : 3$ ; that ratio produces a combined average of $68$ .

Skills used

Mean of a frequency distribution Clear the denominators of an equation Introduce a common factor

k

for a ratio

57Statistics • Median

HardIdeal time: about 1.5 min

The median of $19$ observations is $30$ . Two more observations are made and the values of these are $8$ and $32$ . What is the median of the $21$ observations ?

32

30

20

DCannot be determined due to insufficient data

Show answer & solution

Option B

30

Approach

The median is a position, so track where the middle value sits: one new observation lands below $30$ and one above, which leaves $30$ in the middle.

Median of a data set

Step-by-step solution

1
$n = 19 \Rightarrow median is the 10 th value = 30$
Why: For an odd count the median is the $\frac{n + 1}{2}$ th value in order, so nine values sit at or below $30$ before it and nine at or above it after it.
Median of a data set
2
$8 < 30 \Rightarrow it joins the lower block, which becomes 10 values$
Why: Wherever exactly $8$ falls among the nine smaller values does not matter; what matters is that it is below $30$ .
3
$32 > 30 \Rightarrow it joins the upper block, which becomes 10 values$
Why: Likewise $32$ only has to be above $30$ .
4
$n = 21 \Rightarrow median is the 11 th value, and 30 now stands 11 th$
Why: Ten values below, $30$ , ten values above: the old median has been pushed one place up the list and is still the middle one.

Shortcut

One new value below the median and one above cancel out, so the median does not move.

Common mistakes

Option A

Took the larger new observation as the new median.

Option C

Averaged the two new observations, $\frac{8 + 32}{2} = 20$ , as though they replaced the median rather than flanking it.

Option D

Concluded that the individual observations are needed; they are not, because the median depends only on position and the two new values fall on opposite sides of it.

Skills used

Median of a data set

58Statistics • Frequency Distribution

EasyIdeal time: about 30 s

As the number of observations and classes increases, the shape of a frequency polygon

ATends to become jaggedBTends to become increasingly smoothCStays the sameDVaries only if data become more reliable

Show answer & solution

Option B

Tends to become increasingly smooth

Approach

A frequency polygon joins the class midpoints by straight segments, so think about what happens to those segments as the classes get narrower and more numerous.

Frequency polygon and the frequency curve it tends to

Step-by-step solution

1
$more classes \Rightarrow narrower classes \Rightarrow midpoints closer together$
Why: Splitting the same range into more classes puts the plotted points closer along the horizontal axis.
Inclusive and exclusive class intervals Frequency polygon and the frequency curve it tends to
2
$more observations \Rightarrow steadier class frequencies$
Why: With few observations a single class can be unusually full or empty, which is what makes the outline jagged; more observations smooth those fluctuations out.
Cumulative frequency
3
$short segments between steady neighbouring heights \Rightarrow the outline approaches a smooth curve$
Why: The polygon tends to the frequency curve, which is exactly how the frequency curve is defined as a limit of polygons.

Common mistakes

Option A

Jagged is what a polygon looks like with FEW observations spread over few classes; adding data has the opposite effect.

Option C

The shape cannot stay the same, since the number of plotted points is itself changing.

Option D

Reliability of the data is not what governs the outline; the number of classes and observations is.

Skills used

Frequency polygon and the frequency curve it tends to Inclusive and exclusive class intervals Cumulative frequency

59Statistics • Mean • Weighted Mean

ModerateIdeal time: about 1 min

Let $x ˉ_{1}$ and $x ˉ_{2}$ (where $x ˉ_{2} > x ˉ_{1}$ ) be the means of two sets comprising $n_{1}$ and $n_{2}$ (where $n_{2} < n_{1}$ ) observations respectively. If $x ˉ$ is the mean when they are pooled, then which one of the following is correct ?

x ˉ_{1} < x ˉ < x ˉ_{2}

x ˉ > x ˉ_{2}

x ˉ < x ˉ_{1}

(x ˉ_{1} - x ˉ) + (x ˉ_{2} - x ˉ) = 0

Show answer & solution

Option A

x ˉ_{1} < x ˉ < x ˉ_{2}

Approach

The pooled mean is a weighted average of the two means, so it must lie between them; then check whether the extra conditions force anything stronger.

Mean of a frequency distribution

Step-by-step solution

1
$x ˉ = \frac{n _{1} x ˉ _{1} + n _{2} x ˉ _{2}}{n _{1} + n _{2}}$
Why: Pooling two sets adds their totals and adds their counts.
Mean of a frequency distribution
2
$x ˉ - x ˉ_{1} = \frac{n _{2} ( x ˉ _{2} - x ˉ _{1} )}{n _{1} + n _{2}} > 0 and x ˉ_{2} - x ˉ = \frac{n _{1} ( x ˉ _{2} - x ˉ _{1} )}{n _{1} + n _{2}} > 0$
Why: Both differences are positive because $x ˉ_{2} > x ˉ_{1}$ and both counts are positive. So $x ˉ_{1} < x ˉ < x ˉ_{2}$ always.
3
$n_{2} < n_{1} \Rightarrow x ˉ is nearer x ˉ_{1}, but still strictly between$
Why: The smaller group pulls less hard, which moves the pooled mean towards $x ˉ_{1}$ without ever taking it past either end.

Common mistakes

Option B

A pooled mean cannot exceed the larger of the two means; every observation lies in sets whose averages are $x ˉ_{1}$ and $x ˉ_{2}$ .

Option C

Likewise it cannot fall below the smaller mean.

Option D

$(x ˉ_{1} - x ˉ) + (x ˉ_{2} - x ˉ) = 0$ says $x ˉ$ is the plain average of the two means, which holds only when $n_{1} = n_{2}$ — and the question states $n_{2} < n_{1}$ .

Skills used

Mean of a frequency distribution

60Statistics • Median

ModerateIdeal time: about 1 min

Consider the following statements :

Statement I: Median can be computed even when the end intervals of a frequency distribution are open.

Statement II: Median is a positional average.

Which one of the following is correct in respect of the above statements ?

ABoth Statement I and Statement II are true and Statement II is the correct explanation of Statement IBBoth Statement I and Statement II are true but Statement II is not the correct explanation of Statement ICStatement I is true but Statement II is falseDStatement I is false but Statement II is true

Show answer & solution

Option A

Both Statement I and Statement II are true and Statement II is the correct explanation of Statement I

Approach

Decide each statement on its own, then ask whether the second is the reason the first is true.

Median of a data set

Step-by-step solution

1
$median is located at the \frac{n}{2} th observation, found from cumulative frequency$
Why: The median is fixed by position in the ordered data, not by the values of the observations. That is what "positional average" means, so Statement II is true.
Median of a data set
2
$an open end class contributes its frequency to the cumulative total without needing its limits$
Why: Only the class containing the middle position needs boundaries for interpolation, and the middle position does not fall in an open end class in practice. So Statement I is true.
Cumulative frequency
3
$positional \Rightarrow independent of the extreme values \Rightarrow open ends do not matter$
Why: The mean, by contrast, needs every class midpoint and so cannot be computed with open ends. The median escapes precisely because it is positional, so Statement II explains Statement I.

Common mistakes

Option B

Both statements are true, but the link is not accidental: being positional is exactly why open ends do not obstruct the median.

Option C

Statement II is true — positional average is the standard description of the median.

Option D

Statement I is true: the open end classes only need to contribute their frequencies.

Skills used

Median of a data set Cumulative frequency