Elementary Mathematics 2019 (II)

Set C

Questions 1–20 of 100

1Statistics • Mean

ModerateIdeal time: about 2.5 min

Read the following frequency distribution for two series of observations and answer the two items that follow :

Class interval	Frequency Series-I	Frequency Series-II
10-20	20	4
20-30	15	8
30-40	10	4
40-50	$x$	$2 x$
50-60	$y$	$y$
Total	100	100

What is the mean of frequency distribution of Series-I ?

33.6

35.6

37.6

39.6

Show answer & solution

Option C

37.6

Approach

Use the two column totals to pin down $x$ and $y$ , then take the class marks as the values in the direct mean formula.

Mean of a frequency distribution

Step-by-step solution

1
$20 + 15 + 10 + x + y = 100 \Rightarrow x + y = 55$
Why: The Series-I frequencies must add up to the stated total of $100$ .
Mean of a frequency distribution
2
$4 + 8 + 4 + 2 x + y = 100 \Rightarrow 2 x + y = 84$
Why: The same total applies to Series-II, and it involves the same two unknowns, so the pair can be solved together.
Turn a worded condition into a linear equation in one unknown
3
$(2 x + y) - (x + y) = 84 - 55 \Rightarrow x = 29$ , and then $y = 26$
Why: Subtracting the first equation from the second removes $y$ in one move.
4
Class marks $15, 25, 35, 45, 55$ against frequencies $20, 15, 10, 29, 26$
Why: For a grouped distribution the mid-point of each class stands in for every observation in it.
5
$x ˉ = \frac{20 ( 15 ) + 15 ( 25 ) + 10 ( 35 ) + 29 ( 45 ) + 26 ( 55 )}{100} = \frac{300 + 375 + 350 + 1305 + 1430}{100} = \frac{3760}{100} = 37.6$
Why: Direct mean formula: the sum of frequency times class mark, divided by the total frequency.

Shortcut

Only Series-II pins down $x$ : $2 x + y = 84$ with $x + y = 55$ gives $x = 29$ at once. Then note $20 (15) + 15 (25) + 10 (35) = 1025$ and $29 (45) + 26 (55) = 2735$ , so the mean is $\frac{3760}{100}$ .

Common mistakes

Option A

No route reaches $33.6$ : with $x + y = 55$ the weighted sum cannot fall below $3500$ .

Option B

No route reaches $35.6$ either; the four options are simply spaced $2$ apart.

Option D

Mis-addition of the five products, most easily $1305 + 1430$ carried as $1630$ , gives $3960$ and so $39.6$ .

Skills used

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

2Statistics • Mode

ModerateIdeal time: about 2 min

Same directions as Q1

What is the mode of the frequency distribution of Series-II ?

26

36

46

56

Show answer & solution

Option C

46

Approach

Recover $x$ and $y$ from the two column totals, identify the modal class of Series-II, then interpolate inside it.

Turn a worded condition into a linear equation in one unknown

Step-by-step solution

1
$x + y = 55$ from Series-I and $2 x + y = 84$ from Series-II, so $x = 29$ and $y = 26$
Why: Both totals are $100$ , which is what makes the two unknowns determinate.
Turn a worded condition into a linear equation in one unknown
2
Series-II frequencies: $4, 8, 4, 2 x = 58, y = 26$
Why: Substituting the values back gives the actual distribution to be worked with.
Mode: the most frequent value, and it need not be unique
3
Modal class is $40$ - $50$ , since $58$ is the greatest frequency
Why: The classes all have width $10$ , so the largest frequency alone fixes the modal class.
4
$Mode = 40 + \frac{58 - 4}{2 ( 58 ) - 4 - 26} \times 10 = 40 + \frac{54}{86} \times 10 = 40 + \frac{270}{43} \approx 46.28$
Why: Inside the modal class the mode is placed by the interpolation formula, leaning towards whichever neighbour has the smaller frequency.

Shortcut

Only $2 x + y = 84$ and $x + y = 55$ are needed; $x = 29$ makes $2 x = 58$ the dominant frequency, and the mode must then lie in $40$ - $50$ — which already singles out $46$ .

Common mistakes

Option A

Taking the modal class from the three printed numbers $4, 8, 4$ alone gives $20$ - $30$ , whose mid-point is close to $26$ .

Option B

Nothing reaches $36$ : it is the mid-point of $30$ - $40$ , a class with the smallest frequency.

Option D

Nothing reaches $56$ : the last class has frequency $26$ , well short of $58$ .

Skills used

Turn a worded condition into a linear equation in one unknown Mode: the most frequent value, and it need not be unique

3Statistics • Data Interpretation • Table

ModerateIdeal time: about 1.5 min

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

Let the distribution of number of scooters of companies $X$ and $Y$ sold by $5$ showrooms ( $A$ , $B$ , $C$ , $D$ and $E$ ) in a certain year be denoted by $S 1$ and the distribution of number of scooters of only company $X$ sold by the five showrooms in the same year be denoted by $S 2$ .

Showroom	$A$	$B$	$C$	$D$	$E$	Total number of scooters sold
$S 1$ (in $%$ )	19	21	15	33	12	6400
$S 2$ (in $%$ )	24	18	20	30	8	3000

Number of scooters of company $Y$ sold by showroom $E$ is what per cent of the number of scooters of both companies sold by showroom $C$ ?

52

54

55

56

Show answer & solution

Option C

55

Approach

$S 2$ counts company $X$ only and $S 1$ counts both companies, so subtract to get company $Y$ showroom by showroom.

x %

of a quantity

Step-by-step solution

1
Showroom $E$ , both companies: $12%$ of $6400 = 768$
Why: $S 1$ percentages act on its own total of $6400$ .
$x %$ of a quantity
2
Showroom $E$ , company $X$ : $8%$ of $3000 = 240$
Why: $S 2$ percentages act on the company- $X$ total of $3000$ , a different base.
3
Showroom $E$ , company $Y$ : $768 - 240 = 528$
Why: Company $Y$ is what is left of both companies once company $X$ is removed.
4
Showroom $C$ , both companies: $15%$ of $6400 = 960$
Why: This is the base the question asks the percentage against.
5
$\frac{528}{960} \times 100 = 55$
Why: "What per cent of" puts the second quantity in the denominator.
Percentage change as a multiplier

Shortcut

$12%$ of $6400$ is $768$ and $8%$ of $3000$ is $240$ , so $Y$ at $E$ is $528$ ; $15%$ of $6400$ is $960$ , and $\frac{528}{960} = \frac{11}{20}$ .

Common mistakes

Option A

Nothing reaches $52$ ; it would need a numerator of $499.2$ .

Option B

A slip in $768 - 240$ (reading $768 - 250$ ) gives $518$ and about $54%$ .

Option D

Nothing reaches $56$ ; it would need a numerator of $537.6$ .

Skills used

x %

of a quantity Percentage change as a multiplier

4Statistics • Data Interpretation • Table

ModerateIdeal time: about 1.5 min

Same directions as Q3

Number of scooters of both the companies sold by showroom $B$ is what per cent more than the number of scooters of company $X$ sold by showroom $A$ ?

78 \frac{2}{3}

83 \frac{1}{3}

86 \frac{2}{3}

88 \frac{1}{3}

Show answer & solution

Option C

86 \frac{2}{3}

Approach

Read both companies at $B$ off $S 1$ and company $X$ at $A$ off $S 2$ , then measure the excess against the second figure.

x %

of a quantity

Step-by-step solution

1
Both companies at $B$ : $21%$ of $6400 = 1344$
Why: $S 1$ is the distribution of both companies, on a total of $6400$ .
$x %$ of a quantity
2
Company $X$ at $A$ : $24%$ of $3000 = 720$
Why: $S 2$ is company $X$ alone, on a total of $3000$ .
3
Excess $= 1344 - 720 = 624$
Why: "Per cent more than" is measured on the difference, not on the larger figure.
Percentage change as a multiplier
4
$\frac{624}{720} \times 100 = \frac{260}{3} = 86 \frac{2}{3}$
Why: The base of "more than" is the quantity that follows those words, here the $720$ .

Shortcut

$1344$ against $720$ is a ratio of $\frac{28}{15}$ , so the excess is $\frac{13}{15} = 86 \frac{2}{3} %$ .

Common mistakes

Option A

Nothing reaches $78 \frac{2}{3}$ ; it would need a base of about $793$ .

Option B

Nothing reaches $83 \frac{1}{3}$ ; it would need a base of about $749$ .

Option D

A slip in $21%$ of $6400$ (as $1358$ ) raises the excess to $638$ and the answer to about $88 \frac{1}{3} %$ .

Skills used

x %

of a quantity Percentage change as a multiplier

5Statistics • Data Interpretation • Table

ModerateIdeal time: about 2 min

Same directions as Q3

What is the average number of scooters of company $Y$ sold by the showrooms $A$ , $C$ and $E$ ?

461 \frac{1}{3}

431 \frac{1}{3}

426 \frac{1}{3}

416 \frac{1}{3}

Show answer & solution

Option A

461 \frac{1}{3}

Approach

Get company $Y$ at each of $A$ , $C$ and $E$ as ( $S 1$ figure) $-$ ( $S 2$ figure), then average the three.

x %

of a quantity

Step-by-step solution

1
At $A$ : $19%$ of $6400 - 24%$ of $3000 = 1216 - 720 = 496$
Why: Company $Y$ at a showroom is both companies there less company $X$ there.
$x %$ of a quantity
2
At $C$ : $15%$ of $6400 - 20%$ of $3000 = 960 - 600 = 360$
Why: Same subtraction, with each percentage on its own total.
3
At $E$ : $12%$ of $6400 - 8%$ of $3000 = 768 - 240 = 528$
Why: Same subtraction again.
4
Average $= \frac{496 + 360 + 528}{3} = \frac{1384}{3} = 461 \frac{1}{3}$
Why: The average of three figures is their sum over $3$ .
Mean of a frequency distribution

Shortcut

Company $Y$ at $A$ , $C$ , $E$ is $(19 - 24 \cdot \frac{30}{64})$ -style bookkeeping; faster to read $1216 - 720$ , $960 - 600$ , $768 - 240$ straight off and add to $1384$ .

Common mistakes

Option B

A slip of $90$ somewhere in the three subtractions (for instance $20%$ of $3000$ taken as $690$ ) drops the sum to $1294$ and the average to $431 \frac{1}{3}$ .

Option C

Nothing reaches $426 \frac{1}{3}$ ; it needs a sum of $1279$ , which no combination of the row figures gives.

Option D

Nothing reaches $416 \frac{1}{3}$ either; it needs a sum of $1249$ .

Skills used

x %

of a quantity Mean of a frequency distribution

6Statistics • Data Interpretation • Table

ModerateIdeal time: about 1.5 min

Same directions as Q3

What is the difference between the number of scooters of both companies sold by showroom $A$ and total number of scooters of company $X$ sold by showrooms $B$ and $E$ together ?

416

426

432

436

Show answer & solution

Option D

436

Approach

Take both companies at $A$ from $S 1$ , company $X$ at $B$ and $E$ from $S 2$ , then subtract.

x %

of a quantity

Step-by-step solution

1
Both companies at $A$ : $19%$ of $6400 = 1216$
Why: The first quantity is a $S 1$ figure, so its base is $6400$ .
$x %$ of a quantity
2
Company $X$ at $B$ and $E$ : $(18 + 8) %$ of $3000 = 26%$ of $3000 = 780$
Why: Both are $S 2$ figures on the same base, so the percentages may be added before multiplying.
3
Difference $= 1216 - 780 = 436$
Why: This is the quantity asked for.
Percentage change as a multiplier

Shortcut

Add the $S 2$ percentages first: $26%$ of $3000 = 780$ against $19%$ of $6400 = 1216$ , difference $436$ .

Common mistakes

Option A

Nothing reaches $416$ ; it would need the second quantity to be $800$ .

Option B

Nothing reaches $426$ ; it would need $790$ .

Option C

A slip in $19 \times 64$ (taken as $1212$ ) gives $432$ .

Skills used

x %

of a quantity Percentage change as a multiplier

7Statistics • Data Interpretation • Table

EasyIdeal time: about 55 s

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

The data shows that Indian roads are turning deadlier over the years.

Year	2014	2015	2016	2017
Number of bikers killed	40957	46070	52750	48746
Number of pedestrians killed	12330	13894	15746	20457
Number of cyclists killed	4037	3125	2585	3559

What was the average number of pedestrians killed per day in the year $2017$ ?

51

53

54

56

Show answer & solution

Option D

56

Approach

Divide the year- $2017$ pedestrian total by the number of days in a year.

Unitary method: find one unit, then scale

Step-by-step solution

1
Pedestrians killed in $2017$ $= 20457$
Why: Read the $2017$ entry of the pedestrian row.
Unitary method: find one unit, then scale
2
$\frac{20457}{365} = 56.05$
Why: A daily average is the yearly total shared equally over the $365$ days.
3
$56.05 \approx 56$
Why: The count asked for is a whole number of people per day.

Shortcut

$365 \times 56 = 20440$ , just under $20457$ , so the answer is $56$ without finishing the division.

Common mistakes

Option A

Dividing by $400$ instead of $365$ as a rough estimate gives $51.1$ .

Option B

Nothing reaches $53$ ; $365 \times 53 = 19345$ .

Option C

Nothing reaches $54$ ; $365 \times 54 = 19710$ .

Skills used

Unitary method: find one unit, then scale

8Statistics • Data Interpretation • Table

EasyIdeal time: about 1 min

Same directions as Q7

What is the approximate percentage change in the pedestrians' fatalities during the period 2014-17 ?

66%

68%

71%

76%

Show answer & solution

Option A

66%

Approach

Measure the $2014$ -to- $2017$ rise in pedestrian deaths against the $2014$ figure.

Percentage change as a multiplier

Step-by-step solution

1
$2014$ : $12330$ ; $2017$ : $20457$
Why: The period $2014$ - $17$ runs from the first column to the last.
Percentage change as a multiplier
2
Rise $= 20457 - 12330 = 8127$
Why: Percentage change works on the change, not on the new value.
3
$\frac{8127}{12330} \times 100 \approx 65.9%$
Why: The base of a percentage change is always the starting value.
4
$65.9% \approx 66%$
Why: The question says "approximate".

Shortcut

$\frac{2}{3}$ of $12330$ is $8220$ against an actual rise of $8127$ , so the answer is just below $66 \frac{2}{3} %$ : $66%$ .

Common mistakes

Option B

Rounding $\frac{8127}{12330}$ up from a part-finished division gives about $68%$ .

Option C

Nothing reaches $71%$ ; that would need a rise of about $8754$ .

Option D

Nothing reaches $76%$ ; that would need a rise of about $9371$ .

Skills used

Percentage change as a multiplier

9Statistics • Data Interpretation • Table

EasyIdeal time: about 55 s

Same directions as Q7

What is the average number of bikers killed daily in road accidents in the year $2017$ ?

163

152

147

134

Show answer & solution

Option D

134

Approach

Divide the year- $2017$ biker total by $365$ .

Unitary method: find one unit, then scale

Step-by-step solution

1
Bikers killed in $2017$ $= 48746$
Why: Read the $2017$ entry of the biker row, not the larger $2016$ figure.
Unitary method: find one unit, then scale
2
$\frac{48746}{365} = 133.55$
Why: A daily average is the yearly total shared over $365$ days.
3
$133.55 \approx 134$
Why: The nearest whole number of deaths per day.

Shortcut

$365 \times 133 = 48545$ and $365 \times 134 = 48910$ straddle $48746$ , so the answer is $134$ .

Common mistakes

Option A

Dividing by $300$ as a rough estimate gives $162.5$ , which rounds to $163$ .

Option B

Nothing reaches $152$ ; $365 \times 152 = 55480$ .

Option C

Taking the $2016$ figure $52750$ by mistake gives $144.5$ , and rounding that upward lands near $147$ .

Skills used

Unitary method: find one unit, then scale

10Statistics • Data Interpretation • Table

EasyIdeal time: about 45 s

Same directions as Q7

What is the average number of cyclists killed daily in road accidents in $2017$ ?

10

12

19

21

Show answer & solution

Option A

10

Approach

Divide the year- $2017$ cyclist total by $365$ .

Unitary method: find one unit, then scale

Step-by-step solution

1
Cyclists killed in $2017$ $= 3559$
Why: Read the $2017$ entry of the cyclist row.
Unitary method: find one unit, then scale
2
$\frac{3559}{365} = 9.75$
Why: A daily average is the yearly total shared over $365$ days.
3
$9.75 \approx 10$
Why: The nearest whole number per day.

Shortcut

$365 \times 10 = 3650$ , just above $3559$ , so the answer is $10$ .

Common mistakes

Option B

Dividing by $300$ as a rough estimate gives $11.9$ , which rounds to $12$ .

Option C

Nothing reaches $19$ ; that would need about $6935$ deaths.

Option D

Nothing reaches $21$ ; that would need about $7665$ .

Skills used

Unitary method: find one unit, then scale

11Algebra • Surds • Simplification

HardIdeal time: about 2.5 min

Let $a$ and $b$ be two positive real numbers such that $a a + b b = 32$ and $a b + b a = 31$ . What is the value of $\frac{5 ( a + b )}{7}$ ?

5

7

9

DCannot be determined

Show answer & solution

Option C

9

Approach

Put $p = a$ and $q = b$ ; both givens then become symmetric expressions in $p$ and $q$ that the cube-of-a-sum identity ties together.

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

Step-by-step solution

1
$a a = p^{3}$ , $b b = q^{3}$ , so $p^{3} + q^{3} = 32$
Why: $a a = a^{3/2} = (a^{1/2})^{3}$ , so the first given is a sum of cubes in the new letters.
Power of a power, and the move to a common base
2
$a b + b a = p^{2} q + p q^{2} = pq (p + q) = 31$
Why: The second given factorises completely, leaving only the two symmetric quantities $p + q$ and $pq$ .
3
$(p + q)^{3} = p^{3} + q^{3} + 3 pq (p + q) = 32 + 3 (31) = 125$ , so $p + q = 5$
Why: The grouped form of the cube of a sum is exactly a combination of the two givens — this is why the substitution was worth making.
Cube of a sum and of a difference
4
$pq (p + q) = 31$ with $p + q = 5$ gives $pq = \frac{31}{5}$
Why: The second given now yields the product.
5
$a + b = p^{2} + q^{2} = (p + q)^{2} - 2 pq = 25 - \frac{62}{5} = \frac{63}{5}$
Why: The square-of-a-sum identity converts sum and product into the sum of squares, which is what $a + b$ is.
Square of a sum and of a difference
6
$\frac{5 ( a + b )}{7} = \frac{5}{7} \cdot \frac{63}{5} = \frac{63}{7} = 9$
Why: The factor $\frac{5}{7}$ is built to cancel the $\frac{63}{5}$ , which is the hint that $\frac{63}{5}$ is the intended value of $a + b$ .

Shortcut

$32 + 3 (31) = 125 = 5^{3}$ gives $a + b = 5$ immediately; then $ab = \frac{31}{5}$ and $a + b = 25 - \frac{62}{5} = \frac{63}{5}$ , so the expression is $9$ .

Common mistakes

Option A

Stopping at $p + q = a + b = 5$ and reporting that instead of evaluating the expression asked for.

Option B

Adding the two givens gives $(p + q) (p^{2} + q^{2}) = 63$ ; dividing $63$ by $9$ rather than by $p + q = 5$ returns $7$ .

Option D

Judging two equations in two unknowns carrying surds to be unsolvable, without trying the substitution.

Skills used

Power of a power, and the move to a common base Cube of a sum and of a difference Square of a sum and of a difference

12Algebra • Expressions • Algebraic Identities

HardIdeal time: about 2.5 min

If $x = \frac{1 + 3}{2}$ and $y = x^{3}$ , then $y$ satisfies which one of the following equations ?

8 y^{2} - 20 y - 1 = 0

8 y^{2} + 20 y - 1 = 0

8 y^{2} + 20 y + 1 = 0

8 y^{2} - 20 y + 1 = 0

Show answer & solution

Option A

8 y^{2} - 20 y - 1 = 0

Approach

Turn $x$ into a quadratic with integer coefficients, use it to reduce $x^{3}$ to a linear expression in $x$ , then eliminate $x$ .

Square both sides to clear a radical

Step-by-step solution

1
$x = \frac{1 + 3}{2} \Rightarrow 2 x - 1 = 3 \Rightarrow 4 x^{2} - 4 x + 1 = 3 \Rightarrow 2 x^{2} - 2 x - 1 = 0$
Why: Isolating the surd and squaring clears it, giving the quadratic $x$ satisfies.
Square both sides to clear a radical
2
$x^{2} = x + \frac{1}{2}$
Why: Rearranged, the quadratic is a rule for replacing any $x^{2}$ that appears.
3
$y = x^{3} = x \cdot x^{2} = x (x + \frac{1}{2}) = x^{2} + \frac{x}{2} = (x + \frac{1}{2}) + \frac{x}{2} = \frac{3 x + 1}{2}$
Why: Applying the replacement rule twice drops the cube to a linear expression — this is cheaper than expanding $(\frac{1 + 3}{2})^{3}$ .
4
$x = \frac{2 y - 1}{3}$
Why: Invert the linear relation so that $x$ can be eliminated.
Turn a worded condition into a linear equation in one unknown
5
$2 (\frac{2 y - 1}{3})^{2} - 2 (\frac{2 y - 1}{3}) - 1 = 0 \Rightarrow 2 (2 y - 1)^{2} - 6 (2 y - 1) - 9 = 0$
Why: Substituting into the quadratic $x$ satisfies and multiplying through by $9$ clears the denominators.
Clear the denominators of an equation
6
$2 (4 y^{2} - 4 y + 1) - 12 y + 6 - 9 = 8 y^{2} - 20 y - 1 = 0$
Why: Expanding the square and collecting terms gives the equation in $y$ .
Square of a sum and of a difference

Shortcut

From $2 x^{2} = 2 x + 1$ , $y = x^{3} = \frac{3 x + 1}{2}$ , so $x = \frac{2 y - 1}{3}$ ; substituting into $2 x^{2} - 2 x - 1 = 0$ gives $8 y^{2} - 20 y - 1 = 0$ .

Common mistakes

Option B

Dropping the sign when moving $6 (2 y - 1)$ across produces $+ 20 y$ instead of $- 20 y$ .

Option C

Both the middle-term sign and the constant sign flipped, as happens when $- 6 (2 y - 1) - 9$ is expanded as $- 12 y - 6 + 9$ .

Option D

Expanding $2 (2 y - 1)^{2} - 6 (2 y - 1) - 9$ with the constants as $2 + 6 - 9 = - 1$ taken as $+ 1$ leaves $8 y^{2} - 20 y + 1$ .

Skills used

Square both sides to clear a radical Turn a worded condition into a linear equation in one unknown Clear the denominators of an equation Square of a sum and of a difference

13Arithmetic • Number System • LCM and HCF

EasyIdeal time: about 50 s

HCF of two numbers is $12$ . Which one of the following can never be their LCM ?

80

60

36

24

Show answer & solution

Option A

80

Approach

The HCF divides both numbers, so it divides their LCM; test each option for divisibility by $12$ .

Coprime means no shared prime factor

Step-by-step solution

1
$HCF = 12$ means the numbers are $12 m$ and $12 n$ with $HCF (m, n) = 1$
Why: Pulling the HCF out leaves two coprime cofactors — that is what makes $12$ the greatest common divisor and not merely a common one.
Coprime means no shared prime factor
2
$12 m \cdot 12 n = 12 \times LCM \Rightarrow LCM = 12 mn$
Why: The HCF-LCM product identity turns the LCM into $12$ times an integer, so the LCM must be a multiple of $12$ .
HCF $\times$ LCM = product of the two numbers
3
$80 = 12 \times 6 + 8$ , so $12 ∤ 80$ ; but $60 = 12 \times 5$ , $36 = 12 \times 3$ and $24 = 12 \times 2$
Why: Only one option fails the divisibility test, and it is the answer.
Prime factorisation

Shortcut

The LCM is always a multiple of the HCF; $80$ is the only option not divisible by $12$ .

Common mistakes

Option B

Picked without applying the rule; $12$ and $60$ do have LCM $60$ .

Option C

Picked without applying the rule; $12$ and $36$ do have LCM $36$ .

Option D

Believing the LCM must exceed the HCF by more than a factor of $2$ ; but $12$ and $24$ have HCF $12$ and LCM $24$ .

Skills used

Coprime means no shared prime factor HCF

\times

LCM = product of the two numbers Prime factorisation

14Arithmetic • Number System • Divisibility

HardIdeal time: about 2.5 min

Consider the following statements :

1: Unit digit in $1 7^{174}$ is $7$ .

2: Difference of the squares of any two odd numbers is always divisible by $8$ .

3: Adding $1$ to the product of two consecutive odd numbers makes it a perfect square.

Which of the above statements are correct ?

A1, 2 and 3B1 and 2 onlyC2 and 3 onlyD1 and 3 only

Show answer & solution

Option C

2 and 3 only

Approach

Settle the three claims separately: a unit-digit cycle for the first, the general odd form for the second, and the square-of-a-sum identity for the third.

Unit digit of a power from its cycle

Step-by-step solution

1
Unit digits of $7^{1}, 7^{2}, 7^{3}, 7^{4}$ are $7, 9, 3, 1$ , a cycle of length $4$
Why: The unit digit of $1 7^{174}$ depends only on the unit digit $7$ of the base.
Unit digit of a power from its cycle
2
$174 = 4 \times 43 + 2$ , so the unit digit of $1 7^{174}$ is the second in the cycle, namely $9$
Why: A remainder of $2$ points at the second entry, not the first. Statement 1 is false.
3
$(2 m + 1)^{2} - (2 n + 1)^{2} = 4 (m^{2} + m) - 4 (n^{2} + n) = 4 [m (m + 1) - n (n + 1)]$
Why: Writing both odd numbers in the form $2 m + 1$ turns the claim into an identity.
Write an odd number as $2 m + 1$
4
$m (m + 1)$ and $n (n + 1)$ are each even, so their difference is even and the whole expression is $4 \times (even)$ , a multiple of $8$
Why: A product of two consecutive integers is always even, which supplies the extra factor $2$ . Statement 2 is true.
Difference of two squares
5
Two consecutive odd numbers are $n$ and $n + 2$ ; $n (n + 2) + 1 = n^{2} + 2 n + 1 = (n + 1)^{2}$
Why: The square-of-a-sum identity shows the result is a perfect square for every such pair. Statement 3 is true.
Square of a sum and of a difference

Shortcut

Statement 1 fails because $174$ leaves remainder $2$ on division by $4$ , giving unit digit $9$ ; both the other statements are standard identities. Only C has statement 1 out.

Common mistakes

Option A

Reading the remainder $174 \div 4 = 43$ remainder $2$ as pointing at the first entry of the cycle makes the unit digit $7$ and statement 1 look true.

Option B

The same cycle-position slip on statement 1, plus rejecting statement 3 after testing non-consecutive odd numbers such as $3$ and $7$ , where $21 + 1 = 22$ is not a square.

Option D

The same cycle-position slip on statement 1, plus rejecting statement 2 for want of the $m (m + 1)$ parity argument that supplies the second factor of $2$ .

Skills used

Unit digit of a power from its cycle Write an odd number as

2 m + 1

Difference of two squares Square of a sum and of a difference

15Arithmetic • Interest • Compound Interest

ModerateIdeal time: about 2 min

The rate of interest on two different schemes is the same and it is $20%$ . But in one of the schemes, the interest is compounded half yearly and in the other the interest is compounded annually. Equal amounts are invested in the schemes. If the difference of the returns after $2$ years is Rs. $482$ , then what is the principal amount in each scheme ?

ARs.

10, 000

BRs.

16, 000

CRs.

20, 000

DRs.

24, 000

Show answer & solution

Option C

Rs.

20, 000

Approach

Write both returns with the compound amount formula — four half-yearly periods at $10%$ against two yearly periods at $20%$ — and set their difference to Rs. $482$ .

Compound interest amount

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

Step-by-step solution

1
Half-yearly: rate $\frac{20}{2} = 10%$ per period, $n = 4$ periods, amount $P (1.1)^{4}$
Why: Half-yearly compounding halves the rate and doubles the number of periods.
Compound interest amount $A = P (1 + \frac{r}{100})^{n}$
2
$(1.1)^{4} = 1.4641$
Why: $(1.1)^{2} = 1.21$ and $1.2 1^{2} = 1.4641$ — squaring twice is quicker than four multiplications.
Percentage change as a multiplier
3
Annual: $n = 2$ periods at $20%$ , amount $P (1.2)^{2} = 1.44 P$
Why: The same principal, compounded once a year for two years.
4
$1.4641 P - 1.44 P = 0.0241 P = 482$
Why: Equal principals make the difference of the returns a single multiple of $P$ ; the difference of the amounts equals the difference of the interests.
5
$P = \frac{482}{0.0241} = \frac{4820000}{241} = 20000$
Why: $241 \times 20 = 4820$ , so the division is exact.

Shortcut

$(1.1)^{4} - (1.2)^{2} = 1.4641 - 1.44 = 0.0241$ , and $\frac{482}{0.0241} = 20000$ .

Common mistakes

Option A

Doubling the factor to $0.0482$ — as happens if the two-year difference is taken as twice a one-year one — gives $P = 10000$ .

Option B

Rounding $(1.1)^{4}$ to $1.47$ makes the factor $0.03$ and the principal about $16000$ .

Option D

Rounding $(1.1)^{4}$ to $1.46$ makes the factor $0.02$ and the principal $24100$ , read off as $24000$ .

Skills used

Compound interest amount

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

Percentage change as a multiplier

16Algebra • Expressions • Factorisation

ModerateIdeal time: about 1.5 min

For what value of $k$ can the expression $x^{3} + k x^{2} - 7 x + 6$ be resolved into three linear factors ?

0

1

2

3

Show answer & solution

Option A

0

Approach

Three linear factors means three roots; test the divisors of the constant term $6$ with the factor theorem and read off the $k$ each demands.

Factor and remainder theorem

Step-by-step solution

1
$p (x) = x^{3} + k x^{2} - 7 x + 6$ ; $p (1) = 1 + k - 7 + 6 = k$
Why: The factor theorem says $(x - 1)$ divides $p (x)$ exactly when $p (1) = 0$ , and $x = 1$ is the cheapest divisor of $6$ to test.
Factor and remainder theorem
2
$p (1) = 0 \Rightarrow k = 0$
Why: So $k = 0$ is the only value of $k$ for which $x = 1$ is a root.
3
With $k = 0$ : $x^{3} - 7 x + 6 = (x - 1) (x^{2} + x - 6)$
Why: Dividing out the known factor leaves a quadratic, which is what has to split for three linear factors.
4
$x^{2} + x - 6 = (x + 3) (x - 2)$ , so $x^{3} - 7 x + 6 = (x - 1) (x + 3) (x - 2)$
Why: Splitting the middle term with $3$ and $- 2$ completes the factorisation into three linear factors.
Factorise a quadratic by splitting the middle term

Shortcut

$p (1) = k$ , so $x = 1$ is a root only when $k = 0$ ; and $x^{3} - 7 x + 6 = (x - 1) (x - 2) (x + 3)$ confirms it.

Common mistakes

Option B

Setting $p (1) = 1 + k - 7 + 6$ equal to $1$ instead of $0$ gives $k = 1$ .

Option C

Guessed; with $k = 2$ the cubic has no rational root, so it cannot split into three linear factors.

Option D

Guessed; with $k = 3$ the cubic likewise has no rational root.

Skills used

Factor and remainder theorem Factorise a quadratic by splitting the middle term

17Arithmetic • Number System • LCM and HCF

ModerateIdeal time: about 1.5 min

$X$ , $Y$ and $Z$ start at same point and same time in the same direction to run around a circular stadium. $X$ completes a round in $252$ seconds, $Y$ in $308$ seconds and $Z$ in $198$ seconds. After what time will they meet again at the starting point ?

26

minutes

18

secondsB

42

minutes

36

secondsC

45

minutesD

46

minutes

12

seconds

Show answer & solution

Option D

46

minutes

12

seconds

Approach

All three are back at the start together after any common multiple of their lap times, so take the LCM of $252$ , $308$ and $198$ .

Prime factorisation

Step-by-step solution

1
$X$ is at the start after multiples of $252$ s, $Y$ after multiples of $308$ s, $Z$ after multiples of $198$ s
Why: A runner returns to the starting point only at whole numbers of laps, so the meeting times are the common multiples.
Prime factorisation
2
$252 = 2^{2} \cdot 3^{2} \cdot 7$ , $308 = 2^{2} \cdot 7 \cdot 11$ , $198 = 2 \cdot 3^{2} \cdot 11$
Why: Prime factorisation is what makes the LCM readable off the three numbers at once.
3
$LCM = 2^{2} \cdot 3^{2} \cdot 7 \cdot 11 = 4 \times 9 \times 77 = 2772$ s
Why: Take each prime to the highest power in which it appears anywhere.
4
$2772 = 46 \times 60 + 12$ , i.e. $46$ minutes $12$ seconds
Why: The options are in minutes and seconds, so convert.
km/hour to m/second

Shortcut

$252 = 2^{2} \cdot 3^{2} \cdot 7$ , $308 = 2^{2} \cdot 7 \cdot 11$ , $198 = 2 \cdot 3^{2} \cdot 11$ , so the LCM is $4 \cdot 9 \cdot 7 \cdot 11 = 2772$ s $= 46$ min $12$ s.

Common mistakes

Option A

Nothing reaches $1578$ s; it is not even a multiple of $252$ .

Option B

Nothing reaches $2556$ s; $2556 \div 252$ is not a whole number.

Option C

$2700$ s is a round $45$ minutes but is not a multiple of any of the three lap times.

Skills used

Prime factorisation km/hour to m/second

18Arithmetic • Number System • LCM and HCF

EasyIdeal time: about 55 s

What is the LCM of $\frac{1}{3}, \frac{5}{6}, \frac{2}{9}, \frac{4}{27}$ ?

\frac{5}{18}

\frac{1}{27}

\frac{10}{27}

\frac{20}{3}

Show answer & solution

Option D

\frac{20}{3}

Approach

Apply the rule for the LCM of fractions: LCM of the numerators over HCF of the denominators, with every fraction first in lowest terms.

Comparing two fractions

Step-by-step solution

1
$\frac{1}{3}, \frac{5}{6}, \frac{2}{9}, \frac{4}{27}$ are already in lowest terms
Why: The rule only works on reduced fractions; a common factor left in place corrupts both the numerator LCM and the denominator HCF.
Comparing two fractions
2
$LCM (1, 5, 2, 4) = 20$
Why: $4$ carries the $2$ s and $5$ is coprime to it, so the LCM is $4 \times 5$ .
Prime factorisation
3
$HCF (3, 6, 9, 27) = 3$
Why: $3$ divides all four denominators and nothing larger does, since $6$ does not divide $9$ .
4
$LCM = \frac{20}{3}$
Why: Numerator LCM over denominator HCF.

Shortcut

$\frac{LCM ( 1 , 5 , 2 , 4 )}{HCF ( 3 , 6 , 9 , 27 )} = \frac{20}{3}$ . A sanity filter: the LCM of a set of fractions cannot be smaller than the largest of them, which rules out three options immediately.

Common mistakes

Option A

Nothing reaches $\frac{5}{18}$ ; it is smaller than $\frac{1}{3}$ , so it cannot be a common multiple at all.

Option B

$\frac{1}{27}$ is smaller than every one of the four fractions, so it cannot be their LCM.

Option C

Using the LCM of the denominators instead of their HCF gives $\frac{20}{54} = \frac{10}{27}$ .

Skills used

Comparing two fractions Prime factorisation

19Algebra • Quadratic Equations • Roots

ModerateIdeal time: about 1.5 min

If the equations $x^{2} + 5 x + 6 = 0$ and $x^{2} + k x + 1 = 0$ have a common root, then what is the value of $k$ ?

- \frac{5}{2}

- \frac{10}{3}

\frac{5}{2}

\frac{10}{3}

\frac{5}{2}

- \frac{10}{3}

- \frac{5}{2}

\frac{10}{3}

Show answer & solution

Option B

\frac{5}{2}

\frac{10}{3}

Approach

Factorise the first equation to get its two roots, then make each in turn satisfy the second and solve for $k$ .

Factorise a quadratic by splitting the middle term

Step-by-step solution

1
$x^{2} + 5 x + 6 = (x + 2) (x + 3) = 0$ , so $x = - 2$ or $x = - 3$
Why: Only these two values can be the common root, since the shared root must satisfy the first equation too.
Factorise a quadratic by splitting the middle term
2
A common root is one of $- 2$ and $- 3$
Why: The zero-product property is what reduces an equation to a short list of candidates.
A product is zero only when a factor is zero
3
$x = - 2$ : $(- 2)^{2} + k (- 2) + 1 = 0 \Rightarrow 5 - 2 k = 0 \Rightarrow k = \frac{5}{2}$
Why: Substituting the candidate into the second equation leaves a linear equation in $k$ . Mind the sign: $k (- 2) = - 2 k$ .
Turn a worded condition into a linear equation in one unknown
4
$x = - 3$ : $(- 3)^{2} + k (- 3) + 1 = 0 \Rightarrow 10 - 3 k = 0 \Rightarrow k = \frac{10}{3}$
Why: The second candidate gives the second admissible value of $k$ .

Shortcut

Roots of the first are $- 2, - 3$ ; substituting, $k = \frac{x ^{2} + 1}{- x}$ gives $\frac{5}{2}$ and $\frac{10}{3}$ . Both are positive, which alone settles the option.

Common mistakes

Option A

Taking the roots of $x^{2} + 5 x + 6 = 0$ as $+ 2$ and $+ 3$ instead of $- 2$ and $- 3$ gives $k = - \frac{5}{2}$ and $- \frac{10}{3}$ .

Option C

The sign handled correctly for $x = - 2$ but not for $x = - 3$ , where $k (- 3)$ is read as $+ 3 k$ .

Option D

The mirror image of C: the sign lost on $x = - 2$ and kept on $x = - 3$ .

Skills used

Factorise a quadratic by splitting the middle term A product is zero only when a factor is zero Turn a worded condition into a linear equation in one unknown

20Arithmetic • Interest • Simple Interest

EasyIdeal time: about 1.5 min

$A$ lent Rs. $25000$ to $B$ and at the same time lent some amount to $C$ at same $7%$ simple interest. After $4$ years $A$ received Rs. $11200$ as interest from $B$ and $C$ . How much did $A$ lend to $C$ ?

ARs.

20000

BRs.

25000

CRs.

15000

DRs.

10000

Show answer & solution

Option C

Rs.

15000

Approach

Work out the interest $B$ owes with the simple-interest formula, subtract it from the total received, and run the formula backwards on the remainder.

Simple interest

= \frac{P R T}{100}

Step-by-step solution

1
From $B$ : $SI = \frac{25000 \times 7 \times 4}{100} = Rs. 7000$
Why: Simple interest on a fixed principal at a fixed rate, for $4$ years.
Simple interest $= \frac{P R T}{100}$
2
From $C$ : $11200 - 7000 = Rs. 4200$
Why: The Rs. $11200$ is the combined interest, so what is left after $B$ 's share is $C$ 's.
3
$\frac{P \times 7 \times 4}{100} = 4200 \Rightarrow 0.28 P = 4200$
Why: Same rate and same period for $C$ , so the same formula applies with the principal unknown.
4
$P = \frac{4200}{0.28} = \frac{420000}{28} = 15000$
Why: Divide out to recover the principal.

Shortcut

Four years at $7%$ is $28%$ of the principal. Total interest Rs. $11200$ is $28%$ of the total lent, so the total lent is Rs. $40000$ ; subtract $B$ 's Rs. $25000$ to get Rs. $15000$ .

Common mistakes

Option A

Using $3$ years for $C$ instead of $4$ gives $\frac{4200}{0.21} = 20000$ .

Option B

Assuming "some amount" means the same amount, Rs. $25000$ — but that would make the total interest Rs. $14000$ .

Option D

Using $6$ years for $C$ , or halving the remaining interest to Rs. $2100$ , gives $10000$ .

Skills used

Simple interest

= \frac{P R T}{100}