Elementary Mathematics 2017 (I)

Set A

Questions 41–60 of 100

41Algebra • Polynomial • HCF and LCM

ModerateIdeal time: about 1.5 min

What is the LCM of $x^{3} + 8$ , $x^{2} + 5 x + 6$ and $x^{3} + 4 x^{2} + 4 x$ ?

x (x + 2)^{2} (x + 3) (x^{2} - 2 x + 4)

x (x - 2)^{2} (x - 3) (x^{2} + 2 x + 4)

(x + 2)^{2} (x + 3) (x^{2} - 2 x + 4)

(x - 2)^{2} (x - 3) (x^{2} - 2 x + 4)

Show answer & solution

Option A

x (x + 2)^{2} (x + 3) (x^{2} - 2 x + 4)

Approach

Factorise all three expressions completely, then multiply the highest power of each distinct factor that appears.

HCF and LCM of polynomials from their complete factorisations

Step-by-step solution

1
$x^{3} + 8 = x^{3} + 2^{3} = (x + 2) (x^{2} - 2 x + 4)$
Why: The sum-of-cubes identity $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ with $a = x$ , $b = 2$ ; the quadratic factor has discriminant $4 - 16 < 0$ and does not factor further.
Sum and difference of two cubes
2
$x^{2} + 5 x + 6 = (x + 2) (x + 3)$
Why: Split $5 x$ as $2 x + 3 x$ since $2 \times 3 = 6$ and $2 + 3 = 5$ .
Factorise a quadratic by splitting the middle term
3
$x^{3} + 4 x^{2} + 4 x = x (x^{2} + 4 x + 4) = x (x + 2)^{2}$
Why: Take out the common $x$ , and what remains is the perfect square $(x + 2)^{2}$ .
Square of a sum and of a difference
4
$LCM = x (x + 2)^{2} (x + 3) (x^{2} - 2 x + 4)$
Why: The LCM takes every distinct irreducible factor to the highest power in which it occurs: $x$ once, $(x + 2)$ squared, $(x + 3)$ once, $(x^{2} - 2 x + 4)$ once.
HCF and LCM of polynomials from their complete factorisations

Common mistakes

Option B

Flipped every sign, factorising $x^{3} + 8$ as $(x - 2) (x^{2} + 2 x + 4)$ and $x^{2} + 5 x + 6$ as $(x - 2) (x - 3)$ .

Option C

Correct factors but dropped the common factor $x$ pulled out of $x^{3} + 4 x^{2} + 4 x$ , so the third expression no longer divides the LCM.

Option D

Both errors at once: the signs in the linear factors flipped and the factor $x$ dropped.

Skills used

HCF and LCM of polynomials from their complete factorisations Sum and difference of two cubes Factorise a quadratic by splitting the middle term Square of a sum and of a difference

42Arithmetic • Number System • LCM and HCF

EasyIdeal time: about 40 s

The HCF of two expressions $p$ and $q$ is $1$ . What is the reciprocal of their LCM?

p + q

p - q

pq

(pq)^{- 1}

Show answer & solution

Option D

(pq)^{- 1}

Approach

Use HCF $\times$ LCM $=$ product to get the LCM, then invert it.

HCF

\times

LCM = product of the two numbers

Step-by-step solution

1
$HCF (p, q) \times LCM (p, q) = pq$
Why: The product of two expressions equals the product of their HCF and LCM.
HCF $\times$ LCM = product of the two numbers
2
$1 \times LCM (p, q) = pq \Rightarrow LCM (p, q) = pq$
Why: The HCF is given as $1$ , so $p$ and $q$ share no factor and their LCM is just their product.
Coprime means no shared prime factor
3
$\frac{1}{LCM} = \frac{1}{pq} = (pq)^{- 1}$
Why: The question asks for the reciprocal, and a reciprocal is the $- 1$ power.
Negative exponent is a reciprocal

Common mistakes

Option A

Confused the LCM of coprime expressions with their sum.

Option B

Confused the LCM of coprime expressions with their difference.

Option C

Found the LCM correctly as $pq$ but reported it instead of its reciprocal.

Skills used

HCF

\times

LCM = product of the two numbers Coprime means no shared prime factor Negative exponent is a reciprocal

43Algebra • Surds • Simplification

EasyIdeal time: about 50 s

What is the value of $3 4 \frac{12}{125}$ ?

1 \frac{3}{5}

1 \frac{2}{5}

1 \frac{4}{5}

2 \frac{2}{5}

Show answer & solution

Option A

1 \frac{3}{5}

Approach

Turn the mixed number into an improper fraction and take the cube root of numerator and denominator separately.

ab = a b

and

a / b = a / b

Step-by-step solution

1
$4 \frac{12}{125} = \frac{4 \times 125 + 12}{125} = \frac{512}{125}$
Why: A mixed number is the whole part times the denominator plus the numerator, all over the denominator.
$ab = a b$ and $a / b = a / b$
2
$512 = 8^{3}, 125 = 5^{3}$
Why: Both parts are perfect cubes, which is what makes the root exact.
Prime factorisation
3
$3 \frac{512}{125} = \frac{3 512}{3 125} = \frac{8}{5} = 1 \frac{3}{5}$
Why: The root of a quotient is the quotient of the roots.

Common mistakes

Option B

$(\frac{7}{5})^{3} = \frac{343}{125} = 2 \frac{93}{125}$ , not the given number.

Option C

$(\frac{9}{5})^{3} = \frac{729}{125} = 5 \frac{104}{125}$ , not the given number.

Option D

$(\frac{12}{5})^{3} = \frac{1728}{125} = 13 \frac{103}{125}$ , not the given number.

Skills used

ab = a b

and

a / b = a / b

Prime factorisation

44Arithmetic • Time, Speed and Distance • Relative Speed

EasyIdeal time: about 1 min

A thief is spotted by a policeman from a distance of $100$ m. When the policeman starts the chase, the thief also starts running. If the speed of the thief is $8$ km/hour and that of the policeman is $10$ km/hour, then how far will the thief have to run before he is overtaken?

200

300

400

500

Show answer & solution

Option C

400

Approach

The gap closes at the difference of the speeds; find how long that takes, then find how far the thief runs in that time.

Relative speed: add when opposed, subtract when alike

Step-by-step solution

1
$v_{rel} = 10 - 8 = 2 km/hour$
Why: Both run the same way, so the policeman gains on the thief at the difference of their speeds.
Relative speed: add when opposed, subtract when alike
2
$t = \frac{0.1}{2} = 0.05 hour = 3 minutes$
Why: The $100$ m $= 0.1$ km head start has to be eaten up at $2$ km/hour.
Speed, distance and time
3
$d_{thief} = 8 \times 0.05 = 0.4 km = 400 m$
Why: The question asks how far the THIEF runs, so the thief's own speed is used over that time.

Common mistakes

Option A

$200$ m matches neither runner's distance in the $0.05$ hour the chase lasts.

Option B

Subtracted the $100$ m head start from the policeman's $500$ m rather than taking the thief's own distance.

Option D

Answered for the policeman: he runs $10 \times 0.05 = 500$ m before the catch.

Skills used

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

45Algebra • Quadratic Equations • Roots

ModerateIdeal time: about 1.5 min

Aman and Alok attempted to solve a quadratic equation. Aman made a mistake in writing down the constant term and ended up in roots $(4, 3)$ . Alok made a mistake in writing down the coefficient of $x$ to get roots $(3, 2)$ . The correct roots of the equation are

- 4, - 3

6, 1

4, 3

- 6, - 1

Show answer & solution

Option B

6, 1

Approach

Each student got one coefficient right: take the sum of roots from the one who kept the $x$ coefficient and the product from the one who kept the constant.

Sum and product of roots (Vieta)

Step-by-step solution

1
$x^{2} - (α + β) x + α β = 0$
Why: In a monic quadratic the $x$ coefficient carries the sum of the roots and the constant term carries their product.
Sum and product of roots (Vieta)
2
$Aman erred in the constant \Rightarrow his x coefficient is correct \Rightarrow α + β = 4 + 3 = 7$
Why: A wrong constant term leaves the coefficient of $x$ untouched, so the sum of roots Aman obtained is the true sum.
3
$Alok erred in the x coefficient \Rightarrow his constant is correct \Rightarrow α β = 3 \times 2 = 6$
Why: Symmetrically, a wrong coefficient of $x$ leaves the constant term untouched, so Alok's product of roots is the true product.
4
$x^{2} - 7 x + 6 = 0 \Rightarrow (x - 6) (x - 1) = 0 \Rightarrow x = 6, 1$
Why: Rebuilding the equation from the true sum and true product and splitting the middle term gives the correct roots.
Factorise a quadratic by splitting the middle term

Common mistakes

Option A

Flipped the signs of Aman's roots.

Option C

Reported Aman's roots, treating the student who erred in the constant as having been right throughout.

Option D

Rebuilt $x^{2} + 7 x + 6 = 0$ by putting the sum of roots into the equation with the wrong sign.

Skills used

Sum and product of roots (Vieta)Factorise a quadratic by splitting the middle term

46Arithmetic • Number System • Divisibility

ModerateIdeal time: about 1.5 min

Consider the following statements:

1: Of two consecutive integers, one is even.

2: Square of an odd integer is of the form $8 n + 1$ .

Which of the above statements is/are correct?

A1 onlyB2 onlyCBoth 1 and 2DNeither 1 nor 2

Show answer & solution

Option C

Both 1 and 2

Approach

Test each statement by writing the integers in general form rather than by sampling.

Write an odd number as

2 m + 1

Step-by-step solution

1
$n, n + 1 : if n = 2 m then n is even; if n = 2 m + 1 then n + 1 = 2 (m + 1) is even$
Why: Every integer is even or odd, and in each case one of the two consecutive integers comes out even. Statement 1 holds.
Write an odd number as $2 m + 1$
2
$(2 m + 1)^{2} = 4 m^{2} + 4 m + 1 = 4 m (m + 1) + 1$
Why: Squaring the general odd integer isolates the term $4 m (m + 1)$ .
3
$m (m + 1) = 2 k \Rightarrow (2 m + 1)^{2} = 8 k + 1$
Why: By statement 1, $m$ and $m + 1$ are consecutive so their product is even; $4 \times (even)$ is a multiple of $8$ . Statement 2 holds.
4
$both statements correct$
Why: Both have been proved for every integer, not merely checked on examples.

Common mistakes

Option A

Accepted statement 1 but could not see that $4 m (m + 1)$ is divisible by $8$ , so rejected statement 2 as unproved.

Option B

Rejected statement 1, perhaps by reading "consecutive integers" as "consecutive odd integers".

Option D

Rejected both, which requires a counterexample to a statement that is true for every integer.

Skills used

Write an odd number as

2 m + 1

47Algebra • Equations • Linear Equations

EasyIdeal time: about 45 s

The system of equations $2 x + 4 y = 6$ and $4 x + 8 y = 8$ is

AConsistent with a unique solutionBConsistent with infinitely many solutionsCInconsistentDNone of the above

Show answer & solution

Option C

Inconsistent

Approach

Compare the three ratios of corresponding coefficients.

When a pair of linear equations meets, is parallel, or coincides

Step-by-step solution

1
$\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b _{1}}{b _{2}} = \frac{4}{8} = \frac{1}{2}, \frac{c _{1}}{c _{2}} = \frac{6}{8} = \frac{3}{4}$
Why: For $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ these three ratios decide between one solution, none and infinitely many.
When a pair of linear equations meets, is parallel, or coincides
2
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}} \Rightarrow parallel, no solution$
Why: Equal coefficient ratios but a different constant ratio means the lines have the same gradient and different intercepts.

Common mistakes

Option A

A unique solution needs $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , which fails here.

Option B

Saw that the coefficients are proportional and stopped, without checking that $\frac{6}{8}$ breaks the chain.

Option D

"None of the above" is only right if the system is neither consistent nor inconsistent, which cannot happen.

Skills used

When a pair of linear equations meets, is parallel, or coincides

48Arithmetic • Number System • Divisibility

EasyIdeal time: about 35 s

$(N^{p - 1} - 1)$ is a multiple of $p$ , if $N$ is prime to $p$ and $p$ is a

APrime numberBRational numberCReal numberDComposite number

Show answer & solution

Option A

Prime number

Approach

Recognise the statement as Fermat's little theorem and name the condition it places on $p$ .

Fermat's little theorem

Step-by-step solution

1
$N^{p - 1} \equiv 1 (mod p) whenever p is prime and g cd (N, p) = 1$
Why: This is exactly Fermat's little theorem; " $N$ is prime to $p$ " is the coprimality hypothesis it requires.
Fermat's little theorem
2
$N^{p - 1} - 1 \equiv 0 (mod p) \Rightarrow p ∣ (N^{p - 1} - 1), p prime$
Why: "Is a multiple of $p$ " is the same as leaving remainder $0$ , so the theorem's conclusion is the statement in the question and its hypothesis is that $p$ is prime.

Common mistakes

Option B

"Rational number" is not even a divisibility hypothesis; $p$ must be an integer for "multiple of $p$ " to mean anything.

Option C

"Real number" is not a divisibility hypothesis for the same reason.

Option D

Confused Fermat's theorem with Euler's, which handles composite $p$ but uses the exponent $φ (p)$ , not $p - 1$ .

Skills used

Fermat's little theorem

49Arithmetic • Ratio and Proportion

EasyIdeal time: about 1 min

The ratio of two numbers is $1 : 5$ and their product is $320$ . What is the difference between the squares of these two numbers?

1024

1256

1536

1640

Show answer & solution

Option C

1536

Approach

Write the two numbers with a common factor $k$ , use the product to find $k$ , then apply the difference-of-squares identity.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$a = k, b = 5 k$
Why: A ratio of $1 : 5$ fixes the numbers only up to a common factor.
Introduce a common factor $k$ for a ratio
2
$k \times 5 k = 320 \Rightarrow 5 k^{2} = 320 \Rightarrow k^{2} = 64 \Rightarrow k = 8$
Why: The given product pins $k$ down; taking the positive root gives the numbers $8$ and $40$ .
denotes the non-negative root
3
$b^{2} - a^{2} = (5 k)^{2} - k^{2} = 24 k^{2} = 24 \times 64 = 1536$
Why: Everything is homogeneous of degree $2$ , so the difference of the squares is just $24 k^{2}$ .
Difference of two squares

Shortcut

$b^{2} - a^{2} = (b - a) (b + a) = 32 \times 48 = 1536$ .

Common mistakes

Option A

Squared the difference, $(40 - 8)^{2} = 3 2^{2} = 1024$ , instead of taking the difference of the squares.

Option B

$1256$ is neither $b^{2} - a^{2}$ nor any other symmetric combination of $8$ and $40$ .

Option D

$1640$ is $1600 + 40$ , mixing a square with a number rather than subtracting the squares.

Skills used

Introduce a common factor

k

for a ratio

denotes the non-negative root Difference of two squares

50Arithmetic • Mixtures and Alligation

ModerateIdeal time: about 1.5 min

$25$ kg of alloy X is mixed with $125$ kg of alloy Y. If the amount of lead and tin in the alloy X is in the ratio $1 : 2$ and the amount of lead and tin in the alloy Y is in the ratio $2 : 3$ , then what is the ratio of lead to tin in the mixture?

1 : 2

2 : 3

3 : 5

7 : 11

Show answer & solution

Option D

7 : 11

Approach

Split each alloy into its lead and tin masses, add the leads and add the tins, then form the ratio.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$X : lead = k, tin = 2 k, 3 k = 25 \Rightarrow lead = \frac{25}{3}, tin = \frac{50}{3}$
Why: A $1 : 2$ split of $25$ kg puts one part in three into lead.
Introduce a common factor $k$ for a ratio
2
$Y : lead = 2 m, tin = 3 m, 5 m = 125 \Rightarrow lead = 50, tin = 75$
Why: A $2 : 5$ share of $125$ kg is lead and a $3 : 5$ share is tin; the two alloys need separate letters because their ratios are unrelated.
3
$lead = \frac{25}{3} + 50 = \frac{175}{3}, tin = \frac{50}{3} + 75 = \frac{275}{3}$
Why: Mixing adds the masses of each metal; the alloys themselves do not interact.
4
$\frac{175/3}{275/3} = \frac{175}{275} = \frac{7}{11}$
Why: The common $\frac{1}{3}$ cancels, and dividing both by $25$ puts the ratio in lowest terms.

Common mistakes

Option A

Reported alloy X's own lead-to-tin ratio, ignoring the mixing.

Option B

Reported alloy Y's own lead-to-tin ratio, ignoring the mixing.

Option C

Added the ratio terms instead of the masses: $(1 + 2) : (2 + 3) = 3 : 5$ , which ignores the $25$ kg and $125$ kg weights.

Skills used

Introduce a common factor

k

for a ratio

51Statistics • Mean

Very easyIdeal time: about 30 s

The mean of $5$ numbers is $15$ . If one more number is included, the mean of the $6$ numbers becomes $17$ . What is the included number?

24

25

26

27

Show answer & solution

Option D

27

Approach

Turn each mean back into a total; the extra number is the difference between the two totals.

Mean of a frequency distribution

Step-by-step solution

1
$\sum_{5} = 5 \times 15 = 75$
Why: The mean times the count recovers the sum.
Mean of a frequency distribution
2
$\sum_{6} = 6 \times 17 = 102$
Why: Same relation applied to the enlarged set.
3
$102 - 75 = 27$
Why: The six numbers are the original five plus one more, so the new number is the whole of the increase in the total.

Shortcut

The new number is $17 + 5 \times (17 - 15) = 17 + 10 = 27$ : it must cover its own share plus the lift it gives the other five.

Common mistakes

Option A

$24$ would lift the mean only to $16.5$ .

Option B

$25$ would lift the mean only to $16. 6$ .

Option C

Slipped in the subtraction $102 - 75$ , reading it as $26$ .

Skills used

Mean of a frequency distribution

52Statistics • Mean • Weighted Mean

EasyIdeal time: about 55 s

The mean marks obtained by $300$ students in a subject are $60$ . The mean of top $100$ students was found to be $80$ and the mean of last $100$ students was found to be $50$ . The mean marks of the remaining $100$ students are

70

65

60

50

Show answer & solution

Option D

50

Approach

Convert all three means into totals and take what is left for the middle hundred.

Mean of a frequency distribution

Step-by-step solution

1
$total for 300 = 300 \times 60 = 18000$
Why: The overall mean times the number of students gives the grand total of marks.
Mean of a frequency distribution
2
$100 \times 80 = 8000, 100 \times 50 = 5000$
Why: The same conversion for the top and bottom groups of $100$ .
3
$18000 - 8000 - 5000 = 5000$
Why: The three groups of $100$ partition the $300$ students, so the remaining total is the grand total less the other two.
4
$\frac{5000}{100} = 50$
Why: Dividing the remaining total by its own count of $100$ gives that group's mean.

Common mistakes

Option A

$70$ would push the overall total to $20000$ and the overall mean to $66. 6$ .

Option B

Averaged the two given means, $\frac{80 + 50}{2} = 65$ , instead of using the overall total.

Option C

Assumed the middle group must sit at the overall mean of $60$ , which the totals contradict.

Skills used

Mean of a frequency distribution

53Statistics • Mean

ModerateIdeal time: about 2 min

Consider the following distribution:

Class	Frequency
$0 - 20$	$17$
$20 - 40$	$28$
$40 - 60$	$32$
$60 - 80$	$f$
$80 - 100$	$19$

If the mean of the above distribution is $50$ , what is the value of $f$ ?

24

34

56

96

Show answer & solution

Option A

24

Approach

Use class midpoints in the grouped-mean formula, set it equal to $50$ , and solve the resulting linear equation for $f$ .

Mean of a frequency distribution

Step-by-step solution

1
$x_{i} = 10, 30, 50, 70, 90$
Why: For grouped data the representative value of a class is its midpoint, here the average of the two class limits.
Mean of a frequency distribution
2
$\sum f_{i} x_{i} = 170 + 840 + 1600 + 70 f + 1710 = 4320 + 70 f$
Why: Each midpoint is weighted by its frequency: $10 (17) + 30 (28) + 50 (32) + 70 f + 90 (19)$ .
3
$\sum f_{i} = 17 + 28 + 32 + f + 19 = 96 + f$
Why: The divisor is the total number of observations, not the number of classes.
4
$\frac{4320 + 70 f}{96 + f} = 50 \Rightarrow 4320 + 70 f = 4800 + 50 f \Rightarrow 20 f = 480 \Rightarrow f = 24$
Why: Cross-multiplying turns the mean condition into a linear equation in $f$ .
Turn a worded condition into a linear equation in one unknown

Common mistakes

Option B

Arithmetic slip in the weighted total (for instance $50 \times 32 = 1500$ ), leaving $20 f = 680$ .

Option C

$f = 56$ gives a mean of $\frac{8240}{152} \approx 54.2$ , not $50$ .

Option D

Reported $96$ , the total of the four known frequencies, rather than solving for $f$ .

Skills used

Mean of a frequency distribution Turn a worded condition into a linear equation in one unknown

54Statistics • Data Interpretation • Pie Chart

ModerateIdeal time: about 1.5 min

In a pie diagram, there are four slices with angles $15 0^{\circ}$ , $9 0^{\circ}$ , $6 0^{\circ}$ and $6 0^{\circ}$ . A new pie diagram is formed by deleting one of the slices having angle $6 0^{\circ}$ in the given pie diagram. In the new pie diagram

AThe largest slice has angle

15 0^{\circ}

BThe smallest slice has angle

7 0^{\circ}

CThe largest slice has angle

18 0^{\circ}

DThe smallest slice has angle

9 0^{\circ}

Show answer & solution

Option C

The largest slice has angle

18 0^{\circ}

Approach

Convert the surviving slices back to shares of the reduced total, then re-issue the full $36 0^{\circ}$ among them.

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

Step-by-step solution

1
$150 + 90 + 60 + 60 = 36 0^{\circ}$
Why: Confirms the four angles are a complete pie before anything is removed.
A pie chart sector's angle is proportional to its share
2
$remaining share total = 150 + 90 + 60 = 300$
Why: Deleting one $6 0^{\circ}$ slice removes that category's data; the three survivors keep their sizes relative to one another.
3
$\frac{150}{300} \times 360 = 18 0^{\circ}, \frac{90}{300} \times 360 = 10 8^{\circ}, \frac{60}{300} \times 360 = 7 2^{\circ}$
Why: A new pie must again fill $36 0^{\circ}$ , so every surviving share is scaled by $\frac{360}{300} = 1.2$ .
4
$largest = 18 0^{\circ}, smallest = 7 2^{\circ}$
Why: Reading off the new angles settles all four options at once.

Common mistakes

Option A

Deleted the slice but left the other three angles untouched, so the largest still reads $15 0^{\circ}$ .

Option B

Rescaled but slipped on the smallest: $\frac{60}{300} \times 360 = 7 2^{\circ}$ , not $7 0^{\circ}$ .

Option D

Confused the smallest new angle with the unchanged $9 0^{\circ}$ slice, which in fact becomes $10 8^{\circ}$ .

Skills used

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

55Statistics • Mode

EasyIdeal time: about 40 s

In an asymmetrical distribution, if the mean and median of the distribution are $270$ and $220$ respectively, then the mode of the data is

120

220

280

370

Show answer & solution

Option A

120

Approach

Apply the empirical relation between mean, median and mode for a moderately asymmetrical distribution.

Empirical relation between mean, median and mode

Step-by-step solution

1
$Mode = 3 Median - 2 Mean$
Why: For an asymmetrical distribution the three averages lie roughly on a line with the median dividing mean-to-mode in the ratio $1 : 2$ , which rearranges to this empirical relation.
Empirical relation between mean, median and mode
2
$Mode = 3 (220) - 2 (270) = 660 - 540 = 120$
Why: Substituting the given median $220$ and mean $270$ .

Common mistakes

Option B

Took the mode to equal the median, which is only true of a symmetrical distribution.

Option C

$280$ matches no rearrangement of the empirical relation.

Option D

Swapped mean and median in the formula: $3 (270) - 2 (220) = 370$ .

Skills used

Empirical relation between mean, median and mode

56Statistics • Mean

HardIdeal time: about 2 min

Let $a$ , $b$ , $c$ , $d$ , $e$ , $f$ , $g$ be consecutive even numbers and $j$ , $k$ , $l$ , $m$ , $n$ be consecutive odd numbers. What is the average of all the numbers?

\frac{3 ( a + n )}{2}

\frac{( 5 l + 7 d )}{4}

\frac{( a + b + m + n )}{4}

DNone of the above

Show answer & solution

Option D

None of the above

Approach

Sum each run of consecutive numbers as its count times its middle term, add the two sums over the twelve numbers, then compare with each option.

An evenly spaced run averages to its middle term

Step-by-step solution

1
$a, a + 2, \dots, a + 12 (seven evens), j, j + 2, \dots, j + 8 (five odds)$
Why: Consecutive even numbers rise by $2$ , and so do consecutive odd numbers, so each block is an arithmetic run of step $2$ .
Write an odd number as $2 m + 1$
2
$a + b + \dots + g = 7 (a + 6) = 7 d, j + k + l + m + n = 5 (j + 4) = 5 l$
Why: An evenly spaced run sums to its count times its middle term: $d$ is the fourth of seven, $l$ the third of five.
An evenly spaced run averages to its middle term
3
$average = \frac{7 d + 5 l}{12}$
Why: There are twelve numbers in all, so the divisor is $12$ , not the number of blocks.
Mean of a frequency distribution
4
$\frac{7 d + 5 l}{12} \neq = \frac{3 ( a + n )}{2}, \frac{5 l + 7 d}{4}, \frac{a + b + m + n}{4}$
Why: The first and third options collapse to multiples of $a + n$ , which cannot capture two independent runs, and the second divides the correct total by $4$ instead of $12$ ; none is the average.

Shortcut

Test with the concrete evens $2..14$ and odds $1..9$ : the average is $6.75$ and no option returns it.

Common mistakes

Option A

Averaged the extreme terms as though all twelve numbers formed one evenly spaced run, then scaled by $3$ .

Option B

Found the correct total $7 d + 5 l$ but divided by $4$ instead of by the twelve numbers.

Option C

Averaged the two end pairs, $\frac{a + b + m + n}{4} = \frac{a + n}{2}$ , which ignores the different lengths of the two runs.

Skills used

An evenly spaced run averages to its middle term Write an odd number as

2 m + 1

Mean of a frequency distribution

57Statistics • Mean • Weighted Mean

ModerateIdeal time: about 1.5 min

An individual purchases three qualities of pencils. The relevant data is given below:

Quality	Price per pencil (in Rs.)	Money spent (in Rs.)
A	$1.00$	$50$
B	$1.50$	$x$
C	$2.00$	$20$

It is known that the average price per pencil is Rs. $1.25$ . What is the value of $x$ ?

ARs.

10

BRs.

30

CRs.

40

DRs.

60

Show answer & solution

Option B

Rs.

30

Approach

Convert each money figure into a pencil count, then set total money over total pencils equal to Rs. $1.25$ .

Mean of a frequency distribution

Step-by-step solution

1
$n_{A} = \frac{50}{1.00} = 50, n_{B} = \frac{x}{1.50} = \frac{2 x}{3}, n_{C} = \frac{20}{2.00} = 10$
Why: Money spent divided by price per pencil gives the number bought — the unitary step the table is designed to hide.
Unitary method: find one unit, then scale
2
$average price = \frac{total money}{total pencils} = \frac{50 + x + 20}{60 + \frac{2 x}{3}}$
Why: An average price per pencil is a weighted mean over pencils, so the divisor must be the pencil count, not the number of qualities.
Mean of a frequency distribution
3
$\frac{70 + x}{60 + \frac{2 x}{3}} = \frac{5}{4} \Rightarrow 4 (70 + x) = 5 (60 + \frac{2 x}{3})$
Why: Rs. $1.25 = \frac{5}{4}$ , and cross-multiplying clears the fraction.
Turn a worded condition into a linear equation in one unknown
4
$280 + 4 x = 300 + \frac{10 x}{3} \Rightarrow \frac{2 x}{3} = 20 \Rightarrow x = 30$
Why: Collecting the $x$ terms leaves a one-step linear equation.

Common mistakes

Option A

Rs. $10$ gives $66 \frac{2}{3}$ pencils for Rs. $80$ , an average of Rs. $1.20$ .

Option C

Rs. $40$ gives $86 \frac{2}{3}$ pencils for Rs. $110$ , an average of about Rs. $1.27$ .

Option D

Rs. $60$ gives $100$ pencils for Rs. $130$ , an average of Rs. $1.30$ .

Skills used

Mean of a frequency distribution Unitary method: find one unit, then scale Turn a worded condition into a linear equation in one unknown

58Statistics • Frequency Distribution

EasyIdeal time: about 40 s

Consider the following frequency distribution:

$x$	Frequency	Cumulative frequency
$1$	$8$	$8$
$2$	$10$	$18$
$3$	$f_{1}$	$29$
$4$	$f_{2}$	$45$

What are the values of $f_{1}$ and $f_{2}$ respectively?

10

and

17

17

and

10

11

and

16

16

and

11

Show answer & solution

Option C

11

and

16

Approach

Each frequency is the jump in the cumulative frequency at that value.

Cumulative frequency

Step-by-step solution

1
$f_{1} = 29 - 18 = 11$
Why: The cumulative frequency at $x = 3$ counts everything up to and including $x = 3$ , so subtracting the running total at $x = 2$ leaves the frequency of $x = 3$ alone.
Cumulative frequency
2
$f_{2} = 45 - 29 = 16$
Why: The same jump taken one row further down.

Common mistakes

Option A

$10$ and $17$ rebuild the cumulative column as $8, 18, 28, 45$ , missing the printed $29$ .

Option B

$17$ and $10$ rebuild it as $8, 18, 35, 45$ , missing the printed $29$ .

Option D

Read the two jumps in the wrong order, reporting $f_{1}$ and $f_{2}$ interchanged.

Skills used

Cumulative frequency

59Trigonometry • Trigonometric Identity • Degree and Radian

EasyIdeal time: about 35 s

If D is the number of degrees and R is the number of radians in an angle $θ$ , then which one of the following is correct?

π D = 180 R

π D = 90 R

π R = 180 D

π R = 90 D

Show answer & solution

Option A

π D = 180 R

Approach

Express the same angle as a fraction of a full turn in each unit and equate the fractions.

Converting between degrees and radians

Step-by-step solution

1
$18 0^{\circ} = π radians$
Why: A straight angle is $180$ degrees and $π$ radians, which is the single fact linking the two units.
Converting between degrees and radians
2
$\frac{D}{180} = \frac{R}{π}$
Why: The angle $θ$ is the same fraction of a straight angle however it is measured.
3
$π D = 180 R$
Why: Cross-multiplying clears both denominators.

Common mistakes

Option B

Used $9 0^{\circ} = π$ radians instead of $18 0^{\circ} = π$ radians.

Option C

Cross-multiplied $\frac{D}{180} = \frac{R}{π}$ with $D$ and $R$ interchanged.

Option D

Both errors at once: the wrong straight angle and the two measures interchanged.

Skills used

Converting between degrees and radians

60Trigonometry • Trigonometric Identity • Maximum and Minimum Values

ModerateIdeal time: about 1 min

What is the minimum value of $9 tan^{2} θ + 4 cot^{2} θ$ ?

6

9

12

13

Show answer & solution

Option C

12

Approach

The two terms have a constant product, so their sum is least when they are equal.

A product with a fixed sum is largest when the factors are equal

Step-by-step solution

1
$(9 tan^{2} θ) (4 cot^{2} θ) = 36 (tan θ cot θ)^{2} = 36$
Why: $tan θ cot θ = 1$ , so however $θ$ moves the product of the two terms is fixed at $36$ .
Reciprocal and quotient relations between the ratios
2
$u + v \geq 2 uv = 236 = 12 (u, v > 0)$
Why: For a fixed product the sum of two positive quantities is smallest when they are equal, and that least sum is twice their geometric mean.
A product with a fixed sum is largest when the factors are equal
3
$9 tan^{2} θ = 4 cot^{2} θ \Rightarrow tan^{4} θ = \frac{4}{9} \Rightarrow tan^{2} θ = \frac{2}{3}$
Why: The bound is only a minimum if it is attained; $tan^{2} θ = \frac{2}{3}$ is a real possibility, so $12$ is achieved.

Common mistakes

Option A

Took $9 \times 4 = 6$ and forgot the factor $2$ in $2 uv$ .

Option B

$9$ is the first coefficient; no route through the expression produces it as a minimum.

Option D

Evaluated at $θ = 4 5^{\circ}$ , getting $9 + 4 = 13$ , and assumed that single case was the minimum.

Skills used

A product with a fixed sum is largest when the factors are equal Reciprocal and quotient relations between the ratios