CAT 2019 Quant Questions — Slot 1

The actual Quantitative Ability section from CAT 2019, Slot 1 — 34 questions in original paper order, each with a full worked solution. Tap an option (or type your answer) to check yourself before reading the solution.

34 questions23 MCQ · 11 TITA60 min section0/34 attempted

Q1LogarithmsMCQ

Let x and y be positive real numbers such that log₅(x + y) + log₅(x − y) = 3, and log₂ y − log₂ x = 1 − log₂ 3. Then xy equals

Show solution

Correct answer: A

What we want: the value of $xy$ , given two logarithmic equations — one in base 5 involving $x+y$ and $x-y$ , and one in base 2 relating $y$ to $x$ . The plan: convert each log equation into a plain algebraic relation, then solve.

Step 1 — Simplify the base-5 equation.

We are given: $\log_5(x+y) + \log_5(x-y) = 3$

Since both terms share the same base, apply the product rule $\log_b A + \log_b B = \log_b(AB)$ : $\log_5\big((x+y)(x-y)\big) = 3$

Now convert from logarithmic to exponential form using $\log_b M = c \Rightarrow M = b^c$ : $(x+y)(x-y) = 5^3 = 125$

The left side is a difference of squares: $x^2 - y^2 = 125 \tag{1}$

Step 2 — Simplify the base-2 equation.

We are given: $\log_2 y - \log_2 x = 1 - \log_2 3$

Apply the quotient rule $\log_b A - \log_b B = \log_b\frac{A}{B}$ on the left, and rewrite $1$ as $\log_2 2$ on the right: $\log_2\frac{y}{x} = \log_2 2 - \log_2 3 = \log_2\frac{2}{3}$

💡 Teacher tip: Whenever a logarithmic equation has the same base on both sides, you can "drop the logs" — equal logs (same base) means equal arguments. This is because $\log_b$ is a one-to-one function.

So: $\frac{y}{x} = \frac{2}{3} \quad\Longrightarrow\quad y = \frac{2x}{3} \tag{2}$

Step 3 — Substitute and solve for $x$ and $y$ .

Plug equation (2) into equation (1): $x^2 - \left(\frac{2x}{3}\right)^2 = 125$ $x^2 - \frac{4x^2}{9} = 125$ $\frac{9x^2 - 4x^2}{9} = 125$ $\frac{5x^2}{9} = 125$ $x^2 = 125 \times \frac{9}{5} = 225$

Since $x > 0$ (given), we take the positive root: $x = 15$

Then from equation (2): $y = \frac{2 \times 15}{3} = 10$

Step 4 — Compute $xy$ .

$xy = 15 \times 10 = 150$

Verification — confirm both original equations hold.

Equation	Substituting $x=15,\; y=10$	Result
$\log_5(x+y)+\log_5(x-y)$	$\log_5 25 + \log_5 5 = 2 + 1$	$= 3$ ✅
$\log_2 y - \log_2 x$	$\log_2 10 - \log_2 15 = \log_2\frac{10}{15} = \log_2\frac{2}{3}$	$= 1 - \log_2 3$ ✅

Both equations check out.

Answer: $\boxed{150}$ (Choice A)

💡 Why this works so cleanly: The two equations decouple beautifully — the base-2 equation fixes the ratio $y/x = 2/3$ , while the base-5 equation fixes the scale through $x^2 - y^2 = 125$ . Once the ratio is known, a single substitution resolves everything. This is a common CAT pattern: one equation gives a proportion, the other gives a magnitude.

Practice more Logarithms ↗Ask in forum ↗

Q2Polynomials & ProgressionsMCQ

If a₁, a₂, ...... are in A.P., then 1/(√a₁ + √a₂) + 1/(√a₂ + √a₃) + ....... + 1/(√aₙ + √aₙ₊₁) is equal to

Show solution

Correct answer: A

Think of this as a classic telescoping series: each fraction looks messy, but rationalising it turns it into a difference of two terms — and differences cancel beautifully when summed.

Step 1 — Rationalise a single term.

Take a typical summand, say the $k$ -th one: $\dfrac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}$ .

Multiply the numerator and denominator by the conjugate $\sqrt{a_{k+1}}-\sqrt{a_k}$ :

\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} \cdot \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{\sqrt{a_{k+1}}-\sqrt{a_k}} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{(\sqrt{a_{k+1}})^2-(\sqrt{a_k})^2} = \frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{a_{k+1}-a_k}.

Step 2 — Use the A.P. structure.

Since $a_1, a_2, \dots$ are in an arithmetic progression, the difference $a_{k+1}-a_k$ is the common difference $d$ , which is the same for every $k$ . So each term becomes:

\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}.

Now write out the entire sum $S$ :

S = \frac{\sqrt{a_2}-\sqrt{a_1}}{d} + \frac{\sqrt{a_3}-\sqrt{a_2}}{d} + \cdots + \frac{\sqrt{a_{n+1}}-\sqrt{a_n}}{d}.

Factor out $\dfrac{1}{d}$ :

S = \frac{1}{d}\Big[(\sqrt{a_2}-\sqrt{a_1}) + (\sqrt{a_3}-\sqrt{a_2}) + \cdots + (\sqrt{a_{n+1}}-\sqrt{a_n})\Big].

This is a telescoping sum: $-\sqrt{a_2}$ cancels $+\sqrt{a_2}$ , $-\sqrt{a_3}$ cancels $+\sqrt{a_3}$ , and so on. Everything vanishes except the first and last terms:

S = \frac{\sqrt{a_{n+1}}-\sqrt{a_1}}{d}.

💡 Teacher tip: Whenever you see an A.P. sitting inside a difference of square roots, the move is almost always the same: rationalise, then telescope. The common difference $d$ becomes a clean constant you can factor out.

Step 3 — Remove the $d$ from the denominator.

In an A.P. with common difference $d$ , the $(n+1)$ -th term satisfies:

a_{n+1} = a_1 + nd \quad\Longrightarrow\quad a_{n+1} - a_1 = nd \quad\Longrightarrow\quad d = \frac{a_{n+1}-a_1}{n}.

Substitute this into $S$ :

S = \frac{\sqrt{a_{n+1}}-\sqrt{a_1}}{\dfrac{a_{n+1}-a_1}{n}} = \frac{n\,\big(\sqrt{a_{n+1}}-\sqrt{a_1}\big)}{a_{n+1}-a_1}.

Now recognise the denominator as a difference of squares in disguise:

a_{n+1}-a_1 = \big(\sqrt{a_{n+1}}\big)^2 - \big(\sqrt{a_1}\big)^2 = \big(\sqrt{a_{n+1}}-\sqrt{a_1}\big)\big(\sqrt{a_{n+1}}+\sqrt{a_1}\big).

Cancel the common factor $\big(\sqrt{a_{n+1}}-\sqrt{a_1}\big)$ :

S = \frac{n}{\sqrt{a_{n+1}}+\sqrt{a_1}} = \frac{n}{\sqrt{a_1}+\sqrt{a_{n+1}}}.

Answer: $\boxed{\dfrac{n}{\sqrt{a_1}+\sqrt{a_{n+1}}}}$ (Choice A)

Why this works so cleanly: the A.P. guarantees that every denominator $a_{k+1}-a_k$ is the same constant $d$ . That uniformity is exactly what lets us factor it out and collapse the sum to a single fraction involving only the first and last terms.

Practice more Polynomials & Progressions ↗Ask in forum ↗

Q3Time & WorkTITA

Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

Your answer:

Show solution

Correct answer: 13

Think of this as: we're told how two mixed crews compare in speed, and we need to convert "machine-power" into "man-power" to find the equivalent crew size.

Step 1 — Set up work rates.

Let one man's daily work rate be $a$ (fraction of job per day). Let one machine's daily work rate be $b$ (fraction of job per day).

Step 2 — Translate the time comparison into a rate equation.

Crew 1: 3 men + 8 machines → rate $= 3a + 8b$ . Crew 2: 3 machines + 8 men → rate $= 3b + 8a$ .

We're told Crew 1 finishes in half the time of Crew 2. Since work rate is inversely proportional to time, finishing in half the time means working at twice the rate:

$3a + 8b = 2\,(3b + 8a)$

💡 Teacher tip: "Half the time" always means "double the rate." Time and rate are reciprocals — if one crew takes $\tfrac{1}{2}$ the time, it produces work at $2$ times the speed.

Step 3 — Solve for the man-to-machine rate ratio.

$3a + 8b = 6b + 16a$ $8b - 6b = 16a - 3a$ $2b = 13a$ $b = \frac{13}{2}\,a$

So one machine does as much work per day as $\frac{13}{2}$ men (i.e., 6.5 men).

Step 4 — Use the concrete data: 2 machines finish in 13 days.

Work done by 2 machines in 13 days $= 2b \times 13 = 1$ (one complete job):

$26b = 1 \;\Rightarrow\; b = \frac{1}{26}$

Now back-substitute to find $a$ :

$a = \frac{2b}{13} = \frac{2}{13 \times 26} = \frac{2}{338} = \frac{1}{169}$

Step 5 — Find how many men finish the job in 13 days.

Let $N$ men finish the job in 13 days. Their total work must equal 1:

$N \times a \times 13 = 1$ $N \times \frac{13}{169} = 1$ $N = \frac{169}{13} = 13$

Answer: $\boxed{13}$

Alternative approach — the shortcut (much faster).

From Step 3, we found $b = \frac{13}{2}\,a$ , meaning 1 machine $\equiv$ $\frac{13}{2}$ men in work rate.

Therefore 2 machines $\equiv$ $2 \times \frac{13}{2} = 13$ men.

Since 2 machines finish the job in 13 days, and 2 machines have the same total work rate as 13 men, it follows immediately that 13 men also finish the job in 13 days.

No need to compute individual rates at all — the ratio does everything.

💡 Why this is elegant: The number 13 appears in the given data (days) and again in the answer (men). That's not a coincidence — it's because $2 \times \frac{13}{2} = 13$ collapses the conversion neatly. Whenever a problem gives you a rate ratio and a fixed time, check whether the ratio alone resolves the question before computing absolute rates.

Practice more Time & Work ↗Ask in forum ↗

Q4Time Speed & DistanceMCQ

The wheels of bicycles A and B have radii 30 cm and 40 cm, respectively. While traveling a certain distance, each wheel of A required 5000 more revolutions than each wheel of B. If bicycle B traveled this distance in 45 minutes, then its speed, in km per hour, was

Show solution

Correct answer: C

Think of this as: two wheels of different sizes roll over the same distance. The smaller wheel must spin more times. We're told exactly how many more — and that's enough to pin down the distance, and hence B's speed.

Step 1 — Distance covered in one revolution.

A wheel covers a distance equal to its circumference in one full revolution.

For bicycle A (radius $30$ cm): $\text{Circumference}_A = 2\pi r_A = 2\pi(30) = 60\pi \text{ cm}$

For bicycle B (radius $40$ cm): $\text{Circumference}_B = 2\pi r_B = 2\pi(40) = 80\pi \text{ cm}$

Step 2 — Set up the revolution equation.

Let the common distance travelled be $D$ cm.

Number of revolutions of each wheel of A $= \dfrac{D}{60\pi}$
Number of revolutions of each wheel of B $= \dfrac{D}{80\pi}$

We are told A's wheel makes $5000$ more revolutions than B's: $\frac{D}{60\pi} - \frac{D}{80\pi} = 5000$

Step 3 — Solve for $D$ .

Factor out $\dfrac{D}{\pi}$ : $\frac{D}{\pi}\left(\frac{1}{60} - \frac{1}{80}\right) = 5000$

Compute the bracket with a common denominator of $480$ : $\frac{1}{60} - \frac{1}{80} = \frac{80 - 60}{4800} = \frac{20}{4800} = \frac{1}{240}$

So: $\frac{D}{240\pi} = 5000 \implies D = 5000 \times 240\pi = 1{,}200{,}000\pi \text{ cm}$

Step 4 — Convert the distance to kilometres.

$1{,}200{,}000\pi \text{ cm} = 12{,}000\pi \text{ m} = 12\pi \text{ km}$

💡 Teacher tip: The "5000 more revolutions" clause is doing the heavy lifting here. Without an actual revolution count for either wheel, the difference in revolutions — combined with the difference in circumferences — is what uniquely determines the distance.

Step 5 — Compute B's speed.

Bicycle B covers $12\pi$ km in $45$ minutes. Converting time to hours: $45 \text{ min} = \frac{45}{60} \text{ h} = \frac{3}{4} \text{ h}$

$\text{Speed of B} = \frac{\text{Distance}}{\text{Time}} = \frac{12\pi}{3/4} = 12\pi \times \frac{4}{3} = 16\pi \text{ km/h}$

Answer: $\boxed{16\pi \text{ km/h}}$ (Choice C)

Closing insight: Notice the distance came out to a clean $12\pi$ km — the $5000$ extra revolutions and the $240\pi$ circumference-gap combined to give a round number. Whenever a problem gives a difference of revolutions rather than absolute counts, expect the distance to be recoverable from that difference alone.

Practice more Time Speed & Distance ↗Ask in forum ↗

Q5Set TheoryMCQ

A club has 256 members of whom 144 can play football, 123 can play tennis, and 132 can play cricket. Moreover, 58 members can play both football and tennis, 25 can play both cricket and tennis, while 63 can play both football and cricket. If every member can play at least one game, then the number of members who can play only tennis is

Show solution

Correct answer: D

Think of this as: we have three overlapping circles (football, tennis, cricket) covering all 256 members. We're told the sizes of each circle and each pairwise overlap, but not the triple overlap. First find the triple overlap, then peel away everything that isn't purely tennis.

Step 1 — Set up notation.

Let:

$F$ = set of football players, $|F| = 144$
$T$ = set of tennis players, $|T| = 123$
$C$ = set of cricket players, $|C| = 132$
$|F \cap T| = 58$ , $\;|T \cap C| = 25$ , $\;|F \cap C| = 63$
$x = |F \cap T \cap C|$ (the unknown triple overlap)

Since every member plays at least one game, $|F \cup T \cup C| = 256$ .

Step 2 — Apply the three-set inclusion-exclusion principle.

$|F \cup T \cup C| = |F| + |T| + |C| - |F \cap T| - |T \cap C| - |F \cap C| + |F \cap T \cap C|$

Substituting the known values:

$256 = 144 + 123 + 132 - 58 - 25 - 63 + x$

$256 = 399 - 146 + x$

$256 = 253 + x$

$x = 3$

So exactly 3 members play all three sports.

💡 Teacher tip: The pairwise numbers ( $58, 25, 63$ ) each include the triple-overlap members. Inclusion-exclusion subtracts the three pairwise intersections, which removes the triple members three times — but they were added three times in the singles, so we add $x$ back once to balance.

Step 3 — Find the number who play only tennis.

"Only tennis" means: in $T$ , but not in $F$ and not in $C$ .

Start with all tennis players ( $123$ ), then remove those who also play football ( $58$ ) and those who also play cricket ( $25$ ):

$\text{Only } T = |T| - |F \cap T| - |T \cap C| + x$

💡 Teacher tip: Why add $x$ back? The 3 triple-sport players are counted inside both $|F \cap T|$ and $|T \cap C|$ , so they get subtracted twice. We add $x$ once to undo the double subtraction.

$\text{Only } T = 123 - 58 - 25 + 3 = 43$

Step 4 — Verify with a quick Venn-diagram sanity check.

Region	Calculation	Count
Only Football	$144 - 58 - 63 + 3$	$26$
Only Tennis	$123 - 58 - 25 + 3$	$43$
Only Cricket	$132 - 63 - 25 + 3$	$47$
Football ∩ Tennis only	$58 - 3$	$55$
Tennis ∩ Cricket only	$25 - 3$	$22$
Football ∩ Cricket only	$63 - 3$	$60$
All three	$x$	$3$
Total	$26+43+47+55+22+60+3$	$256$ ✓

The regions sum to exactly $256$ , confirming our answer.

Answer: $\boxed{43}$ (Choice D)

Closing insight: The triple overlap ( $x=3$ ) is small here, but it still matters — without adding it back, "only tennis" would come out to $40$ instead of $43$ . Whenever a question asks for an "only" region, always account for the triple intersection explicitly.

Practice more Set Theory ↗Ask in forum ↗

Q6Time Speed & DistanceMCQ

Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is

Show solution

Correct answer: A

Think of this as: the second car has one hour less to cover the same distance, so it must be faster. We want the largest possible speed gap — which happens when the first car takes the least allowed time.

What we want: the largest possible percentage by which the second car's speed exceeds the first's, given they cover the same distance and arrive together — but the second car starts an hour later.

Step 1 — Translate the timing into an equation. Let the common distance be $D$ . Let the first car travel for $t_1$ hours and the second for $t_2$ hours.

The first car starts at 10:00 am, the second at 11:00 am, yet both arrive together. So the second car is on the road for exactly one hour less:

$t_2 = t_1 - 1.$

Step 2 — Write the two speeds. Since distance $=$ speed $\times$ time,

$v_1 = \frac{D}{t_1}, \qquad v_2 = \frac{D}{t_2} = \frac{D}{t_1-1}.$

Step 3 — Build the percentage-excess expression. The percentage by which $v_2$ exceeds $v_1$ is

$\frac{v_2 - v_1}{v_1} \times 100 = \left(\frac{v_2}{v_1} - 1\right) \times 100.$

Now $\dfrac{v_2}{v_1} = \dfrac{D/(t_1-1)}{D/t_1} = \dfrac{t_1}{t_1-1}$ , so

$\text{excess } \% = \left(\frac{t_1}{t_1-1} - 1\right) \times 100 = \frac{1}{t_1-1} \times 100.$

💡 Teacher tip: The distance $D$ cancels out completely — the answer depends only on the times. This is why the problem never tells you the actual distance.

Step 4 — Maximise the expression. The percentage $\dfrac{100}{t_1-1}$ gets larger as $t_1 - 1$ gets smaller, i.e. as $t_1$ gets smaller. We are told the first car travelled at least 6 hours, so $t_1 \ge 6$ . The smallest allowed value is therefore

$t_1 = 6.$

💡 Teacher tip: The phrase "at least 6 hours" is the key signal — it sets a floor on $t_1$ , and to maximise a "one-over-something" expression we push that something down to its floor.

Step 5 — Substitute and compute.

$\text{excess } \% = \frac{100}{6 - 1} = \frac{100}{5} = 20.$

Quick check with concrete numbers. Take $D = 30$ km (chosen so the speeds come out clean). Then $t_1 = 6$ h gives $v_1 = 5$ km/h, and $t_2 = 5$ h gives $v_2 = 6$ km/h. The excess is $\dfrac{6-5}{5} \times 100 = 20\%$ . ✓

Quantity	First car	Second car
Distance	30 km	30 km
Time	6 h	5 h
Speed	5 km/h	6 km/h
Excess	—	$\frac{6-5}{5}\times 100 = 20\%$

Answer: $\boxed{20}$ (Option A)

💡 Why this is clean: the percentage excess collapses to the single fraction $\frac{100}{t_1-1}$ , so the entire question reduces to finding the smallest legal value of $t_1$ .

Practice more Time Speed & Distance ↗Ask in forum ↗

Q7MensurationMCQ

If the rectangular faces of a brick have their diagonals in the ratio 3 : 2√3 : √15, then the ratio of the length of the shortest edge of the brick to that of its longest edge is

Show solution

Correct answer: B

Think of this as: a brick has three edges and three face diagonals. If we know the ratio of the three face diagonals, can we back out the ratio of the edges themselves?

Step 1 — Set up the face diagonals.

Let the brick have edges $a, b, c$ . The three rectangular faces have diagonals:

$\sqrt{a^2+b^2},\quad \sqrt{b^2+c^2},\quad \sqrt{a^2+c^2}$

These are given in the ratio $3: 2\sqrt{3}: \sqrt{15}$ .

Step 2 — Square the ratio to remove the square roots.

$3^2: (2\sqrt{3})^2: (\sqrt{15})^2 \;=\; 9: 12: 15$

So the three quantities $a^2+b^2$ , $b^2+c^2$ , $a^2+c^2$ take the values $9, 12, 15$ in some order.

💡 Teacher tip: Squaring a ratio of square roots converts it into a ratio of plain integers — always do this first when face diagonals are involved, since each face diagonal squared is just a sum of two squared edges.

Step 3 — Use the sum to find each squared edge.

Add all three face-diagonal-squared values:

$(a^2+b^2)+(b^2+c^2)+(a^2+c^2) = 9+12+15 = 36$

$\Rightarrow\quad 2(a^2+b^2+c^2) = 36 \quad\Rightarrow\quad a^2+b^2+c^2 = 18$

Now each squared edge is the total $18$ minus the squared diagonal of the opposite face:

Squared edge	Calculation	Value
One edge squared	$18 - 9 = 9$	$9$
Another edge squared	$18 - 12 = 6$	$6$
Third edge squared	$18 - 15 = 3$	$3$

So the three edges are $\sqrt{9}=3$ , $\sqrt{6}$ , and $\sqrt{3}$ .

💡 Teacher tip: The assignment of which diagonal goes with which pair of edges doesn't matter — we always recover the same set $\{3, 6, 9\}$ of squared edges, just with different labels.

Step 4 — Identify shortest and longest.

Shortest edge $= \sqrt{3}$
Longest edge $= 3$

$\text{Ratio} = \sqrt{3}: 3 = 1: \sqrt{3}$

Answer: $\boxed{1: \sqrt{3}}$ (Choice B)

Alternative approach — order the edges first.

Assume $a \le b \le c$ . Then the smallest face diagonal uses the two smallest edges and the largest uses the two largest:

$a^2+b^2 = 9,\quad a^2+c^2 = 12,\quad b^2+c^2 = 15$

Subtract the first from the second: $c^2 - b^2 = 3$ .

Subtract the second from the third: $b^2 - a^2 = 3$ .

So $b^2 = a^2+3$ and $c^2 = a^2+6$ . Substituting into $a^2+b^2=9$ :

$a^2 + (a^2+3) = 9 \;\Rightarrow\; 2a^2 = 6 \;\Rightarrow\; a^2 = 3$

This gives $a=\sqrt{3}$ , $b=\sqrt{6}$ , $c=3$ , and the ratio $\sqrt{3}:3 = 1:\sqrt{3}$ . Same result, reached by subtracting equations instead of summing them.

Closing insight: The three squared face diagonals $9, 12, 15$ and the total $18$ conspire to give squared edges $3, 6, 9$ — an arithmetic progression. That symmetry is what makes the shortest-to-longest ratio come out to the clean value $1:\sqrt{3}$ .

Practice more Mensuration ↗Ask in forum ↗

Q8Functions & GraphsTITA

For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive integer such that 8f(m + 1) − f(m) = 2, then m equals

Your answer:

Show solution

Correct answer: 10

Think of this as: the function $f$ uses one rule on even inputs and a different rule on odd inputs. Since $m$ and $m+1$ are always of opposite parity, the equation $8f(m+1)-f(m)=2$ will mix both rules — so we just need to test which parity of $m$ makes the algebra work out.

The two rules, restated:

$f(n)=\begin{cases} n(n+1), & n\text{ even}\\[4pt] n+3, & n\text{ odd}\end{cases}$

Since consecutive integers have opposite parity, exactly one of $f(m)$ and $f(m+1)$ uses the even-rule and the other uses the odd-rule.

Case 1 — $m$ is even (so $m+1$ is odd).

Here $f(m)$ uses the even-rule and $f(m+1)$ uses the odd-rule:

$f(m)=m(m+1), \qquad f(m+1)=(m+1)+3=m+4.$

Substitute into $8f(m+1)-f(m)=2$ :

$8(m+4)-m(m+1)=2.$

Expand and simplify:

$8m+32-m^2-m=2$

$-m^2+7m+30=0$

$m^2-7m-30=0.$

Factor the quadratic:

$(m-10)(m+3)=0 \;\Rightarrow\; m=10 \text{ or } m=-3.$

Since $m$ must be a positive integer, $m=10$ . Note $10$ is indeed even ✓, so our case assumption holds.

Case 2 — $m$ is odd (so $m+1$ is even).

Here $f(m)$ uses the odd-rule and $f(m+1)$ uses the even-rule:

$f(m)=m+3, \qquad f(m+1)=(m+1)(m+2).$

Substitute:

$8(m+1)(m+2)-(m+3)=2.$

Expand:

$8(m^2+3m+2)-m-3=2$

$8m^2+24m+16-m-3=2$

$8m^2+23m+11=0.$

The discriminant is $23^2-4(8)(11)=529-352=177$ , which is not a perfect square, so this case gives no integer solution.

Verification of $m=10$ :

Quantity	Rule used	Value
$f(10)$	even: $n(n+1)$	$10\times 11=110$
$f(11)$	odd: $n+3$	$11+3=14$
$8f(11)-f(10)$	—	$8(14)-110=112-110=2$ ✓

The equation is satisfied.

Answer: $\boxed{10}$

💡 Teacher tip: Whenever a piecewise function depends on parity and the argument shifts by 1, the two pieces always interact. Try the parity that makes the algebra cleanest first; if it yields a valid integer, you're done — but a quick check of the other case (or its discriminant) confirms uniqueness.

Practice more Functions & Graphs ↗Ask in forum ↗

Q9Permutation & CombinationTITA

With rectangular axes of coordinates, the number of paths from (1, 1) to (8, 10) via (4, 6), where each step from any point (x, y) is either to (x, y+1) or to (x+1, y), is

Your answer:

Show solution

Correct answer: 3920

Think of this as: we need to count monotone (right/up only) lattice paths that must pass through a specific checkpoint. The key idea is to split the journey at that checkpoint into two independent legs.

Step 1 — Split at the via-point $(4,6)$ .

Every valid path goes from $(1,1)$ to $(4,6)$ , and then continues from $(4,6)$ to $(8,10)$ . These two legs are independent: any path for the first leg can be combined with any path for the second leg.

So: $\text{Total paths} = \text{Paths}\,(1,1)\to(4,6) \times \text{Paths}\,(4,6)\to(8,10).$

💡 Teacher tip: Whenever a path must pass through a specific point, split there. The multiplication principle applies because the choice made in the first leg doesn't restrict the choice in the second leg.

Step 2 — Count paths on the first leg: $(1,1)\to(4,6)$ .

To go from $(1,1)$ to $(4,6)$ using only right-steps $(x+1,y)$ and up-steps $(x,y+1)$ :

Right-steps needed: $4 - 1 = 3$
Up-steps needed: $6 - 1 = 5$
Total steps: $3 + 5 = 8$

A path is completely determined by choosing which 3 of the 8 steps are right-steps (the rest are automatically up-steps):

$\binom{8}{3} = \frac{8!}{3!\,5!} = 56.$

Step 3 — Count paths on the second leg: $(4,6)\to(8,10)$ .

Right-steps needed: $8 - 4 = 4$
Up-steps needed: $10 - 6 = 4$
Total steps: $4 + 4 = 8$

Choose which 4 of the 8 steps are right-steps:

$\binom{8}{4} = \frac{8!}{4!\,4!} = 70.$

Step 4 — Multiply the two legs.

$56 \times 70 = 3920.$

Quick verification of the arithmetic:

Leg	Route	Right	Up	Total	Choose	Value
1	$(1,1)\to(4,6)$	3	5	8	$\binom{8}{3}$	56
2	$(4,6)\to(8,10)$	4	4	8	$\binom{8}{4}$	70
Combined	—	—	—	—	$56\times 70$	3920

Answer: $\boxed{3920}$

💡 Closing insight: The general formula for monotone lattice paths from $(x_1,y_1)$ to $(x_2,y_2)$ is $\binom{(x_2-x_1)+(y_2-y_1)}{x_2-x_1}$ . When a via-point is imposed, just evaluate this formula for each leg and multiply — the via-point acts as a natural factorisation of the counting problem.

Practice more Permutation & Combination ↗Ask in forum ↗

Q10MensurationMCQ

Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is

Show solution

Correct answer: A

Think of this as: we're slicing off three identical small equilateral triangles from the corners of a larger equilateral triangle, and the leftover shape is a regular hexagon. We need to figure out how much area survives.

Step 1 — Visualise the cut.

When you cut off each corner of an equilateral triangle with a straight line, the piece removed is itself a small equilateral triangle. After all three corners are cut, the remaining shape has six sides:

3 sides come from the cuts themselves (one per corner removed).
3 sides are the middle segments left on the original triangle's edges.

For this leftover shape to be a regular hexagon, all six sides must be equal.

Step 2 — Set up side lengths.

Let the original equilateral triangle $T$ have side length $1$ .

Suppose each small corner triangle has side length $a$ . Then on each original edge, two segments of length $a$ are removed (one at each end), leaving a middle segment of length:

$1 - 2a$

For a regular hexagon, the cut sides ( $a$ ) must equal the middle segments ( $1 - 2a$ ):

$a = 1 - 2a$

$3a = 1$

$a = \frac{1}{3}$

💡 Teacher tip: The key insight is that a regular hexagon needs all six sides equal. Three of those sides are the cut edges, and three are the leftover middle portions of the original triangle's sides. Setting them equal pins down exactly how big each cut must be.

Step 3 — Compute the areas.

The area of an equilateral triangle with side $s$ is $\dfrac{\sqrt{3}}{4}\,s^2$ .

Original triangle (side $1$ ):

$\text{Area}(T) = \frac{\sqrt{3}}{4} \times 1^2 = \frac{\sqrt{3}}{4}$

Each removed corner triangle (side $\frac{1}{3}$ ):

$\text{Area of one corner} = \frac{\sqrt{3}}{4}\left(\frac{1}{3}\right)^2 = \frac{\sqrt{3}}{4} \times \frac{1}{9} = \frac{\sqrt{3}}{36}$

Three corners removed:

$3 \times \frac{\sqrt{3}}{36} = \frac{\sqrt{3}}{12}$

Hexagon area (original minus the three corners):

$\text{Area}(H) = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{12} = \frac{3\sqrt{3}}{12} - \frac{\sqrt{3}}{12} = \frac{2\sqrt{3}}{12} = \frac{\sqrt{3}}{6}$

Step 4 — Find the ratio.

$\frac{\text{Area}(H)}{\text{Area}(T)} = \frac{\sqrt{3}/6}{\sqrt{3}/4} = \frac{1/6}{1/4} = \frac{4}{6} = \frac{2}{3}$

Quick verification: Each small triangle has $\frac{1}{9}$ the area of the original (since area scales with the square of the side ratio $\frac{1}{3}$ ). Three of them remove $\frac{3}{9} = \frac{1}{3}$ of the total area, leaving $\frac{2}{3}$ . ✓

Answer: $\boxed{2: 3}$ (Choice A)

💡 Closing insight: The ratio $\frac{2}{3}$ holds regardless of the original triangle's size — it depends only on the shape of the construction, not on any particular side length. That's why we could safely set the side to $1$ without loss of generality.

Practice more Mensuration ↗Ask in forum ↗

Q11CirclesMCQ

AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to

Show solution

Correct answer: C

Think of this as: a diameter gives us two free right angles, so both triangles become straightforward Pythagoras applications — one to find $AP$ , the other to find $QB$ .

Step 1 — Spot the right angles (Thales' theorem).

Since $AB$ is a diameter, any point on the circle subtends a right angle with $AB$ . Therefore: $\angle APB = 90^\circ \quad\text{and}\quad \angle AQB = 90^\circ.$

So $\triangle APB$ is right-angled at $P$ , and $\triangle AQB$ is right-angled at $Q$ .

💡 Teacher tip: Whenever you see a diameter in a circle problem, immediately look for points on the circle — Thales' theorem converts them into right-angled triangles for free. This is the single most useful circle fact in CAT geometry.

Step 2 — Find $AP$ using Pythagoras in $\triangle APB$ .

The radius is $5$ cm, so the diameter is: $AB = 2 \times 5 = 10 \text{ cm}.$

Given $PB = 6$ cm, apply Pythagoras: $AP^2 + PB^2 = AB^2$ $AP^2 + 6^2 = 10^2$ $AP^2 + 36 = 100$ $AP^2 = 64 \;\Rightarrow\; AP = 8 \text{ cm}.$

Step 3 — Find $AQ$ from the given ratio.

We are told $AP = 2 \cdot AQ$ : $8 = 2 \cdot AQ \;\Rightarrow\; AQ = 4 \text{ cm}.$

Step 4 — Find $QB$ using Pythagoras in $\triangle AQB$ .

$AQ^2 + QB^2 = AB^2$ $4^2 + QB^2 = 10^2$ $16 + QB^2 = 100$ $QB^2 = 84$ $QB = \sqrt{84} = 2\sqrt{21} \approx 9.165 \text{ cm}.$

Step 5 — Match to the nearest option.

$9.165$ is closest to $9.1$ .

Answer: $\boxed{9.1 \text{ cm}}$ (Choice C)

💡 Why this works so directly: the diameter hands us two right angles for free, so each triangle reduces to a one-line Pythagoras. The only piece of reasoning needed beyond that is using the ratio $AP = 2\cdot AQ$ to transfer information from one triangle to the other.

Practice more Circles ↗Ask in forum ↗

Q12Simple & Compound InterestTITA

A person invested a total amount of Rs 15 lakh. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2 : 1, earning annual interest at the rates of 4% and 3%, respectively. If the total annual interest income is Rs 76000 then the amount (in Rs lakh) invested in the fixed deposit was

Your answer:

Show solution

Correct answer: 9

Think of this as: one chunk earns a fixed 6%, the rest earns a blended rate (since it's split between 4% and 3%), and the overall interest is known. Find the chunk sizes.

Step 1 — Set up the deposits.

Let the fixed deposit (FD) hold $x$ lakh. The remaining $(15 - x)$ lakh is split in the ratio $2:1$ between the 4% and 3% deposits:

Deposit	Amount (Rs lakh)	Rate
Fixed deposit	$x$	$6\%$
4% deposit	$\tfrac{2}{3}(15-x)$	$4\%$
3% deposit	$\tfrac{1}{3}(15-x)$	$3\%$

Step 2 — Find the blended rate on the non-FD portion.

Since the non-FD money is split $2:1$ , the effective rate on that entire $(15-x)$ lakh is the weighted average:

$\frac{2}{3}\times 4\% + \frac{1}{3}\times 3\% = \frac{8}{3}\% + \frac{3}{3}\% = \frac{11}{3}\%.$

💡 Teacher tip: Instead of computing interest from each of the three deposits separately, collapse the two non-FD deposits into one "bucket" earning $\tfrac{11}{3}\%$ . This turns a three-part problem into a clean two-part equation.

Step 3 — Write the interest equation.

Total interest $=$ Rs $76000 = 0.76$ lakh. So:

$0.06\,x + \frac{0.11}{3}(15 - x) = 0.76.$

Multiply through by $3$ to clear the denominator:

$0.18\,x + 0.11(15 - x) = 2.28.$

Expand:

$0.18\,x + 1.65 - 0.11\,x = 2.28.$

Combine like terms:

$0.07\,x = 0.63 \quad\Longrightarrow\quad x = \frac{0.63}{0.07} = 9.$

Step 4 — Verify.

Deposit	Amount (lakh)	Rate	Interest (lakh)
FD	$9$	$6\%$	$0.54$
4% deposit	$\tfrac{2}{3}\times 6 = 4$	$4\%$	$0.16$
3% deposit	$\tfrac{1}{3}\times 6 = 2$	$3\%$	$0.06$
Total	15		0.76

$0.76$ lakh $=$ Rs $76000$ ✓ — matches the given total interest.

Answer: $\boxed{9}$

Alternative approach — Alligation (faster under exam time).

The FD earns $6\%$ ; the non-FD bucket earns $\tfrac{11}{3}\%$ . The overall rate on the full Rs 15 lakh is:

$\frac{0.76}{15}\times 100 = \frac{76}{15}\%.$

Now apply alligation between the FD rate and the non-FD rate:

$\text{FD} \;\left(\frac{90}{15}\%\right) \qquad \text{Non-FD}\;\left(\frac{55}{15}\%\right)$

$\text{Mixture} = \frac{76}{15}\%$

Ratio of FD to Non-FD $= \left(\frac{76}{15}-\frac{55}{15}\right): \left(\frac{90}{15}-\frac{76}{15}\right) = \frac{21}{15}: \frac{14}{15} = 21: 14 = 3: 2.$

So FD $= \tfrac{3}{5}\times 15 = 9$ lakh — same answer, no algebraic equation needed.

💡 Why this works so cleanly: whenever a total is split into parts with different rates and the overall average is known, alligation gives the split ratio directly — no variable, no equation to solve.

Practice more Simple & Compound Interest ↗Ask in forum ↗

Q13Time & WorkMCQ

At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?

Show solution

Correct answer: C

Think of this as: two people work at their normal speeds and finish in 12 days; if one slows down and the other speeds up, they finish in 9 days. We need to back out each person's individual speed.

Step 1 — Set up rate equations.

Let A's usual daily work rate be $a$ (fraction of task per day) and B's usual daily work rate be $b$ .

At usual efficiencies, together they finish in 12 days: $a + b = \frac{1}{12} \quad \cdots (1)$

In the modified scenario, A works at half her usual rate ( $\frac{a}{2}$ ) and B works at triple his usual rate ( $3b$ ), and together they finish in 9 days: $\frac{a}{2} + 3b = \frac{1}{9} \quad \cdots (2)$

💡 Teacher tip: In Time & Work problems, always express each person's contribution as a rate (work per day), not days. Rates add cleanly; days don't.

Step 2 — Solve the system.

From equation (1), express $a$ in terms of $b$ : $a = \frac{1}{12} - b$

Substitute into equation (2): $\frac{1}{2}\left(\frac{1}{12} - b\right) + 3b = \frac{1}{9}$

Expand the first term: $\frac{1}{24} - \frac{b}{2} + 3b = \frac{1}{9}$

Combine the $b$ terms ( $-\frac{b}{2} + 3b = \frac{5b}{2}$ ): $\frac{1}{24} + \frac{5b}{2} = \frac{1}{9}$

Isolate $\frac{5b}{2}$ : $\frac{5b}{2} = \frac{1}{9} - \frac{1}{24}$

Find a common denominator (72): $\frac{5b}{2} = \frac{8}{72} - \frac{3}{72} = \frac{5}{72}$

Solve for $b$ : $b = \frac{2}{5} \times \frac{5}{72} = \frac{1}{36}$

Step 3 — Find A's rate.

Plug $b$ back into equation (1): $a = \frac{1}{12} - \frac{1}{36} = \frac{3}{36} - \frac{1}{36} = \frac{2}{36} = \frac{1}{18}$

So A completes $\frac{1}{18}$ of the task per day at her usual efficiency.

Step 4 — Convert rate to days.

If A does $\frac{1}{18}$ of the task each day, she needs $\frac{1}{a} = 18$ days to finish the whole task alone.

Answer: $\boxed{18}$ (Choice C)

Alternative approach — LCM (units) method.

Take the LCM of 12 and 9, which is 36. Let the total task = 36 units.

At usual rates: $a + b = \frac{36}{12} = 3$ units/day.
Modified rates: $\frac{a}{2} + 3b = \frac{36}{9} = 4$ units/day.

From the first equation, $a = 3 - b$ . Substitute: $\frac{3 - b}{2} + 3b = 4 \implies \frac{3}{2} - \frac{b}{2} + 3b = 4 \implies \frac{5b}{2} = \frac{5}{2} \implies b = 1$

Then $a = 3 - 1 = 2$ units/day. A alone needs $\frac{36}{2} = 18$ days.

💡 The LCM method avoids fractions entirely and is faster under exam conditions — both methods give the same answer.

Practice more Time & Work ↗Ask in forum ↗

Q14CirclesMCQ

In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is

Show solution

Correct answer: D

Think of this as: two chords cross inside a circle, and we know one chord's full length plus pieces of the other. The intersecting-chords theorem links the four segments — from there it's algebra.

Step 1 — Find the other piece of the diameter.

The radius is 11 cm, so the diameter is: $CD = 2 \times 11 = 22\text{ cm}.$

We're told $CE = 7$ cm. Since $E$ lies on $CD$ between $C$ and $D$ : $ED = CD - CE = 22 - 7 = 15\text{ cm}.$

Step 2 — Apply the intersecting-chords theorem.

When two chords $AB$ and $CD$ meet at an interior point $E$ , the theorem says: $AE \cdot EB = CE \cdot ED.$

Substituting: $AE \cdot EB = 7 \times 15 = 105.$

We also know the full chord length $AB = 20.5$ cm, and since $E$ lies on $AB$ : $AE + EB = 20.5.$

So we have two equations:

$AE + EB = 20.5$
$AE \cdot EB = 105$

Step 3 — Find $|BE - AE|$ without solving for each piece.

💡 Teacher tip: Whenever you know the sum and product of two numbers, you can get the difference directly via the identity $(a-b)^2 = (a+b)^2 - 4ab$ — no need to solve a quadratic.

Using $(BE - AE)^2 = (AE + EB)^2 - 4\,AE\cdot EB$ :

$(BE - AE)^2 = (20.5)^2 - 4(105) = 420.25 - 420 = 0.25.$

Taking the positive square root (since we want a length difference): $|BE - AE| = \sqrt{0.25} = 0.5\text{ cm}.$

Quick check. Solving $x^2 - 20.5x + 105 = 0$ gives $x = \frac{20.5 \pm 0.5}{2}$ , i.e. $AE = 10$ and $EB = 10.5$ (or vice versa). Their product is $10 \times 10.5 = 105$ ✓ and their difference is $0.5$ ✓.

Answer: $\boxed{0.5}$ (Choice D)

💡 Closing insight: The numbers are engineered so that $20.5^2 = 420.25$ sits just $0.25$ above $4 \times 105 = 420$ — a tiny gap that produces the clean answer $0.5$ . Whenever a chord-length-difference question gives you the sum and product of segments, reach for $(a-b)^2 = (a+b)^2 - 4ab$ first.

Practice more Circles ↗Ask in forum ↗

Q15Profit Loss & DiscountMCQ

On selling a pen at 5% loss and a book at 15% gain, Karim gains Rs. 7. If he sells the pen at 5% gain and the book at 10% gain, he gains Rs. 13. What is the cost price of the book in Rupees?

Show solution

Correct answer: C

Think of this as: two unknown cost prices, two scenarios with known total gains — a classic simultaneous equations problem dressed up in profit-loss language.

Step 1 — Define variables.

Let the cost price of the pen be $p$ rupees and the cost price of the book be $b$ rupees.

Step 2 — Translate each scenario into an equation.

Recall: a gain of $x\%$ on cost price $C$ means profit $= \frac{x}{100} \times C$ , and a loss of $x\%$ means profit $= -\frac{x}{100} \times C$ .

Scenario 1: Pen at $5\%$ loss, book at $15\%$ gain, total gain $=$ Rs. $7$ .

$-0.05\,p + 0.15\,b = 7 \quad \cdots (1)$

Scenario 2: Pen at $5\%$ gain, book at $10\%$ gain, total gain $=$ Rs. $13$ .

$0.05\,p + 0.10\,b = 13 \quad \cdots (2)$

Step 3 — Eliminate $p$ by adding the two equations.

Notice the pen's coefficient is $-0.05$ in equation (1) and $+0.05$ in equation (2). Adding them cancels $p$ cleanly:

$(-0.05\,p + 0.05\,p) + (0.15\,b + 0.10\,b) = 7 + 13$

$0.25\,b = 20$

$b = \frac{20}{0.25} = 80$

💡 Teacher tip: When one item's profit/loss percentage flips sign between scenarios (here the pen goes from $-5\%$ to $+5\%$ ), adding the equations often eliminates that variable in one step — no substitution needed.

Step 4 — Verify the answer is consistent.

With $b = 80$ , from equation (2): $0.05\,p + 0.10(80) = 13 \Rightarrow 0.05\,p + 8 = 13 \Rightarrow p = 100$ .

Check in equation (1): $-0.05(100) + 0.15(80) = -5 + 12 = 7$ ✓

Both scenarios check out.

Answer: $\boxed{80}$ (Choice C)

The key move was spotting that the pen's percentage flips from $-5\%$ to $+5\%$ , making elimination by addition the fastest route — the book's cost price falls out directly without ever solving for the pen first.

Practice more Profit Loss & Discount ↗Ask in forum ↗

Q16Surds & IndicesMCQ

If m and n are integers such that (√2 $)^{19}$ · $3^4$ · $4^2$ · $9^m$ · $8^n$ = $3^n$ · 1 $6^m$ · ⁴√64, then m is

Show solution

Correct answer: C

Think of this as: every base in the equation is secretly a power of $2$ or $3$ . If we rewrite both sides using only $2$ and $3$ as bases, the exponents on each prime must match — giving us two equations in two unknowns.

Step 1 — Break every term into powers of $2$ and $3$ .

Term	Rewrite	Prime-power form
$(\sqrt{2})^{19}$	$(2^{1/2})^{19}$	$2^{19/2}$
$3^4$	—	$3^4$
$4^2$	$(2^2)^2$	$2^4$
$9^m$	$(3^2)^m$	$3^{2m}$
$8^n$	$(2^3)^n$	$2^{3n}$
$3^n$	—	$3^n$
$16^m$	$(2^4)^m$	$2^{4m}$
$\sqrt[4]{64}$	$(2^6)^{1/4}$	$2^{6/4} = 2^{3/2}$

💡 Teacher tip: Whenever you see mixed bases like $4, 8, 9, 16, 64$ in an indices problem, your first move is to express everything in terms of the underlying primes. Here every base reduces to $2$ or $3$ .

Step 2 — Assemble each side.

$\text{LHS} = 2^{19/2}\cdot 3^4 \cdot 2^4 \cdot 3^{2m}\cdot 2^{3n} = 2^{\,19/2\, +\, 4\, +\, 3n}\cdot 3^{\,4\, +\, 2m}$

$\text{RHS} = 3^n \cdot 2^{4m}\cdot 2^{3/2} = 2^{\,4m\, +\, 3/2}\cdot 3^n$

Step 3 — Match the exponents on each prime.

Since the two sides are equal and $2$ and $3$ are distinct primes, the exponent of $2$ on the LHS must equal the exponent of $2$ on the RHS, and likewise for $3$ .

Powers of $3$ : $\;4 + 2m = n$ ……… (1)
Powers of $2$ : $\;\dfrac{19}{2} + 4 + 3n = 4m + \dfrac{3}{2}$

Simplify the powers-of- $2$ equation:

$\frac{19}{2} + 4 + 3n = 4m + \frac{3}{2}$

$\frac{19}{2} + \frac{8}{2} + 3n = 4m + \frac{3}{2}$

$\frac{27}{2} + 3n = 4m + \frac{3}{2}$

$\frac{27}{2} - \frac{3}{2} + 3n = 4m$

$12 + 3n = 4m \quad \text{……… (2)}$

Step 4 — Solve the system.

From (1): $n = 4 + 2m$ . Substitute into (2):

$12 + 3(4 + 2m) = 4m$

$12 + 12 + 6m = 4m$

$24 + 6m = 4m$

$24 = -2m$

$m = -12$

Step 5 — Verify.

With $m = -12$ : $\;n = 4 + 2(-12) = -20$ .

Check equation (2): $\;12 + 3(-20) = 12 - 60 = -48$ , and $4(-12) = -48$ . ✓

Equation	LHS value	RHS value	Match?
(1) $4 + 2m = n$	$4 + 2(-12) = -20$	$n = -20$	✅
(2) $12 + 3n = 4m$	$12 + 3(-20) = -48$	$4(-12) = -48$	✅

Answer: $\boxed{m = -12}$ (Choice C)

💡 Closing insight: The key technique here is prime-base decomposition — once every term is expressed as a power of $2$ or $3$ , the problem reduces to a simple two-equation system. This is the go-to strategy for any indices equation with composite bases.

Practice more Surds & Indices ↗Ask in forum ↗

Q17Quadratic & Linear EquationsTITA

The number of solutions to the equation |x|(6x² + 1) = 5x² is

Your answer:

Show solution

Correct answer: 5

We need the number of real solutions of $|x|(6x^2+1)=5x^2$ .

The key obstacle is the $|x|$ — it behaves differently for positive and negative $x$ . But since $x^2 = |x|^2$ for every real $x$ , we can rewrite the entire equation in terms of $|x|$ and avoid casework.

Step 1 — Check $x=0$ separately.

Plugging in: LHS $= |0|(6\cdot 0+1)=0$ , RHS $= 5\cdot 0 = 0$ . ✓

So $x=0$ is one solution.

Step 2 — For $x\ne 0$ , substitute $t = |x|$ .

Since $x\ne 0$ , we have $t=|x|>0$ . Using $x^2 = |x|^2 = t^2$ , the equation becomes: $t(6t^2+1)=5t^2$ $6t^3+t=5t^2$ $6t^3-5t^2+t=0$ $t(6t^2-5t+1)=0$

Since $t>0$ , the factor $t=0$ is discarded. We solve the quadratic: $6t^2-5t+1=0$ $t=\frac{5\pm\sqrt{25-24}}{12}=\frac{5\pm 1}{12}$ $\Rightarrow t=\frac{6}{12}=\frac{1}{2}\quad\text{or}\quad t=\frac{4}{12}=\frac{1}{3}$

Both roots are positive, so both are valid.

💡 Teacher tip: Whenever an equation mixes $|x|$ with $x^2$ , try $t=|x|$ . Since $x^2=|x|^2$ , the equation often collapses into a clean polynomial in $t$ — no need to split into $x>0$ and $x<0$ cases.

Step 3 — Convert each $t$ back to $x$ -values.

Each positive value of $t=|x|$ corresponds to two real values of $x$ : $x=+t$ and $x=-t$ . The value $t=0$ gives only $x=0$ .

| $t=|x|$ | Corresponding $x$ -values | Count | |:---:|:---:|:---:| | $0$ | $0$ | $1$ | | $\tfrac{1}{2}$ | $+\tfrac{1}{2},\;-\tfrac{1}{2}$ | $2$ | | $\tfrac{1}{3}$ | $+\tfrac{1}{3},\;-\tfrac{1}{3}$ | $2$ |

Step 4 — Add them up.

$1+2+2=5$

Answer: $\boxed{5}$

💡 Why this works so cleanly: the substitution $t=|x|$ turns a piecewise absolute-value equation into a single cubic. The cubic factors as $t(6t^2-5t+1)$ , giving three non-negative $t$ -values — and each one maps to either one $x$ (if $t=0$ ) or two $x$ -values (if $t>0$ ). That symmetry is what produces the odd count of $5$ .

Practice more Quadratic & Linear Equations ↗Ask in forum ↗

Q18PercentagesMCQ

The income of Amala is 20% more than that of Bimala and 20% less than that of Kamala. If Kamala's income goes down by 4% and Bimala's goes up by 10%, then the percentage by which Kamala's income would exceed Bimala's is nearest to

Show solution

Correct answer: A

Think of this as: three people's incomes are linked by percentage statements, then two of them change. We need the final gap as a percentage.

Step 1 — Anchor on Amala and pick a smart number.

Both percentage statements involve Amala, so she is our anchor. The two multipliers are $1.2$ and $0.8$ , whose product is $0.96$ . A clean choice is to set Amala's income to a multiple of both. Let's try:

$A = 120$

💡 Teacher tip: "20% more" and "20% less" are not symmetric operations — $1.2$ and $0.8$ do not cancel. Always anchor on the common quantity and express the others relative to it. Picking a concrete number (instead of algebra) makes the arithmetic transparent.

Step 2 — Find Bimala's and Kamala's original incomes.

Amala is 20% more than Bimala: $A = 1.2\,B$ , so $B = \dfrac{120}{1.2} = 100$ .
Amala is 20% less than Kamala: $A = 0.8\,K$ , so $K = \dfrac{120}{0.8} = 150$ .

Person	Original income
Amala	$120$
Bimala	$100$
Kamala	$150$

Step 3 — Apply the changes.

Kamala's income drops by $4\%$ : $K_{\text{new}} = 150 \times 0.96 = 144$ .
Bimala's income rises by $10\%$ : $B_{\text{new}} = 100 \times 1.10 = 110$ .

Step 4 — Compute the percentage by which Kamala now exceeds Bimala.

The percentage excess is measured relative to Bimala's (new) income:

$\frac{K_{\text{new}} - B_{\text{new}}}{B_{\text{new}}} \times 100 = \frac{144 - 110}{110} \times 100 = \frac{34}{110} \times 100 = \frac{3400}{110} \approx 30.91\%$

Rounding to the nearest integer: $\approx 31\%$ .

Answer: $\boxed{31}$ (Choice A)

💡 Why the smart-number method wins here: the multipliers $1.2$ and $0.8$ both divide $120$ cleanly, so every step stays in whole numbers — no fractions, no rounding until the very end. Whenever a chain of percentage statements shares a common reference person, set that person to a well-chosen number and the rest falls out.

Alternative approach — Pure algebra.

Let Amala's income be $A$ .

$B = \dfrac{A}{1.2} = \dfrac{5A}{6}$ , and $K = \dfrac{A}{0.8} = \dfrac{5A}{4}$ .
After changes: $K_{\text{new}} = 0.96 \times \dfrac{5A}{4} = 1.20\,A$ , and $B_{\text{new}} = 1.10 \times \dfrac{5A}{6} = \dfrac{11A}{12}$ .
Percentage excess $= \dfrac{1.20 - 11/12}{11/12} \times 100 = \dfrac{17/60}{55/60} \times 100 = \dfrac{17}{55} \times 100 \approx 30.9\%$ .

Same answer, but the smart-number route is faster and less error-prone under exam pressure.

Practice more Percentages ↗Ask in forum ↗

Q19Coordinate GeometryTITA

Let S be the set of all points (x, y) in the x-y plane such that |x| + |y| ≤ 2 and |x| ≥ 1. Then, the area, in square units, of the region represented by S equals

Your answer:

Show solution

Correct answer: 2

We need the area of the region where both conditions hold: $|x| + |y| \le 2$ and $|x| \ge 1$ .

Think of it as: take the diamond $|x|+|y|\le 2$ , then slice off its middle column — we only keep the two side pieces where $x$ is at least 1 unit away from the $y$ -axis.

Step 1 — Picture the diamond.

The inequality $|x| + |y| \le 2$ describes a square rotated $45^\circ$ (a diamond) with vertices at $(\pm 2, 0)$ and $(0, \pm 2)$ . Its diagonals have length $4$ , so its total area is $\frac{1}{2} \times 4 \times 4 = 8$ .

Step 2 — Apply the second condition $|x| \ge 1$ .

This keeps only the two side strips: $x \ge 1$ (right strip) or $x \le -1$ (left strip). The central strip $-1 < x < 1$ is removed.

By symmetry, the right and left strips have equal area, so we compute the right strip and double it.

Step 3 — Find the vertical extent in the right strip.

For $1 \le x \le 2$ , the diamond boundary gives $|y| \le 2 - x$ . So at a fixed $x$ , $y$ ranges from $-(2-x)$ to $+(2-x)$ , a vertical length of $2(2 - x)$ .

Step 4 — Integrate.

$\text{Area (right strip)} = \int_1^2 2(2 - x)\, dx = 2\left[2x - \frac{x^2}{2}\right]_1^2$

$= 2\left[\left(4 - 2\right) - \left(2 - \frac{1}{2}\right)\right] = 2\left[2 - 1.5\right] = 2 \times 0.5 = 1.$

Step 5 — Double for the left strip.

$\text{Total area} = 2 \times 1 = 2.$

Answer: $\boxed{2}$

Alternative approach — pure geometry (faster).

In the right strip ( $1 \le x \le 2$ ), the diamond's boundary is the line from $(1, 1)$ down to $(2, 0)$ and from $(1, -1)$ up to $(2, 0)$ . These three points — $(1, 1)$ , $(1, -1)$ , $(2, 0)$ — form a triangle with:

base $= 2$ (from $y = -1$ to $y = 1$ at $x = 1$ ),
height $= 1$ (from $x = 1$ to $x = 2$ ).

$\text{Area (right triangle)} = \frac{1}{2} \times 2 \times 1 = 1.$

Doubling for the left strip: $\text{Total} = 2 \times 1 = 2$ . ✓

💡 Teacher tip: Whenever a region is carved out of a symmetric shape by a condition like $|x| \ge 1$ , exploit the symmetry — solve one side and double. Also, before reaching for calculus, check if the slice is a simple triangle or trapezoid; the geometric route is often faster on CAT.

Practice more Coordinate Geometry ↗Ask in forum ↗

Q20Polynomials & ProgressionsTITA

If a₁ + a₂ + a₃ + .... + aₙ = 3( $2^{(n+1)}$ − 2), for every n ≥ 1, then a₁₁ equals

Your answer:

Show solution

Correct answer: 6144

We're given the partial sum $S_n = a_1 + a_2 + \cdots + a_n = 3(2^{n+1} - 2)$ for every $n \ge 1$ , and we need the individual term $a_{11}$ .

💡 Teacher tip: Whenever you know the running total $S_n$ but need a single term $a_n$ , remember the bridge between them: each term is the difference of two consecutive partial sums. This is the single most useful identity in sequence problems.

Step 1 — Connect terms to partial sums.

By definition, $S_n = a_1 + a_2 + \cdots + a_n$ and $S_{n-1} = a_1 + a_2 + \cdots + a_{n-1}$ . Subtracting cancels all the shared terms:

$a_n = S_n - S_{n-1} \qquad (n \ge 2).$

So to get $a_{11}$ , we evaluate $S_{11}$ and $S_{10}$ and subtract.

Step 2 — Write out both partial sums.

Using $S_n = 3(2^{n+1} - 2)$ :

$S_{11} = 3(2^{12} - 2), \qquad S_{10} = 3(2^{11} - 2).$

Step 3 — Subtract.

$a_{11} = S_{11} - S_{10} = 3(2^{12} - 2) - 3(2^{11} - 2).$

Factor out the 3:

$a_{11} = 3\bigl[(2^{12} - 2) - (2^{11} - 2)\bigr] = 3(2^{12} - 2^{11}).$

Notice the $-2$ and $+2$ cancel — the constants disappear because both partial sums share the same offset.

Step 4 — Simplify the power of 2.

$2^{12} - 2^{11} = 2^{11}(2 - 1) = 2^{11}.$

So:

$a_{11} = 3 \times 2^{11}.$

Step 5 — Compute the value.

$2^{11} = 2048$ , therefore:

$a_{11} = 3 \times 2048 = 6144.$

Answer: $\boxed{6144}$

💡 Closing insight: The sequence $a_n = 3 \cdot 2^{n}$ (for $n \ge 2$ ) is a geometric progression. The partial-sum formula $3(2^{n+1} - 2)$ is exactly what you'd get summing $3 \cdot 2^1 + 3 \cdot 2^2 + \cdots + 3 \cdot 2^n$ — so the $a_n = S_n - S_{n-1}$ trick is really just "un-summing" a geometric series term by term.

Practice more Polynomials & Progressions ↗Ask in forum ↗

Q21Functions & GraphsTITA

Consider a function f satisfying f(x + y) = f(x) f(y) where x, y are positive integers, and f(1) = 2. If f(a + 1) + f(a + 2) + ... + f(a + n) = 16( $2^n$ − 1) then a is equal to

Your answer:

Show solution

Correct answer: 3

We need to find the value of $a$ given a functional equation and a sum condition.

What we want: the value of $a$ such that $f(a+1)+f(a+2)+\cdots+f(a+n)=16(2^n-1)$ .

Step 1 — Crack the functional equation.

We're told $f(x+y)=f(x)\,f(y)$ for positive integers, and $f(1)=2$ .

This is the classic exponential functional equation. Let's pin down $f$ on every positive integer by building up from $f(1)$ :

$f(2)=f(1+1)=f(1)\cdot f(1)=2\cdot 2=4=2^2$
$f(3)=f(2+1)=f(2)\cdot f(1)=4\cdot 2=8=2^3$
$f(4)=f(3+1)=f(3)\cdot f(1)=8\cdot 2=16=2^4$

The pattern is clear. By induction, if $f(n)=2^n$ , then $f(n+1)=f(n)\cdot f(1)=2^n\cdot 2=2^{n+1}$ .

💡 Teacher tip: Whenever you see $f(x+y)=f(x)\,f(y)$ , think exponential. The function is completely determined by a single value — here $f(1)=2$ — because every input can be built by repeatedly adding $1$ .

So for every positive integer $n$ : $f(n)=2^n.$

Step 2 — Rewrite the sum using $f(n)=2^n$ .

The given sum is: $f(a+1)+f(a+2)+\cdots+f(a+n)=2^{a+1}+2^{a+2}+\cdots+2^{a+n}.$

Factor out $2^{a+1}$ from every term: $=2^{a+1}\bigl(1+2+2^2+\cdots+2^{n-1}\bigr).$

The bracket is a geometric series with first term $1$ , ratio $2$ , and $n$ terms: $1+2+2^2+\cdots+2^{n-1}=\frac{2^n-1}{2-1}=2^n-1.$

So the entire sum collapses to: $2^{a+1}(2^n-1).$

Step 3 — Match against the given total.

We're told the sum equals $16(2^n-1)$ : $2^{a+1}(2^n-1)=16(2^n-1).$

Since $n$ is a positive integer, $2^n-1\neq 0$ , so we can safely divide both sides by $2^n-1$ : $2^{a+1}=16.$

Writing $16$ as a power of $2$ : $2^{a+1}=2^4.$

Equating exponents: $a+1=4\quad\Rightarrow\quad a=3.$

Quick check. With $a=3$ , the sum is $f(4)+f(5)+\cdots+f(3+n)=2^4+2^5+\cdots+2^{3+n}=2^4(2^n-1)=16(2^n-1)$ . ✅ Matches exactly.

Answer: $\boxed{3}$

💡 Why this works so cleanly: the functional equation forces $f$ to be exponential, and the sum of consecutive powers of $2$ factors into a single power times $(2^n-1)$ . That shared factor $(2^n-1)$ cancels on both sides, leaving a simple equation in $a$ alone.

Practice more Functions & Graphs ↗Ask in forum ↗

Q22Quadratic & Linear EquationsMCQ

The product of the distinct roots of |x² − x − 6| = x + 2 is

Show solution

Correct answer: A

Think of this as: an absolute value equation always splits into two sub-equations, but each sub-equation only counts where its sign assumption actually holds — so we solve, then filter.

What we want: the product of the distinct roots of $\left|x^2 - x - 6\right| = x + 2$ .

Step 1 — Sign constraint on the right-hand side.

The left side is an absolute value, so it is always $\ge 0$ . That forces the right side to be non-negative too: $x + 2 \ge 0 \quad\Longrightarrow\quad x \ge -2.$ Any root we eventually find must satisfy $x \ge -2$ .

Step 2 — Factor the expression inside the absolute value.

$x^2 - x - 6 = (x - 3)(x + 2).$

This quadratic changes sign at its roots $x = -2$ and $x = 3$ :

$x^2 - x - 6 \ge 0$ when $x \le -2$ or $x \ge 3$ ,
$x^2 - x - 6 < 0$ when $-2 < x < 3$ .

These two regions give us the two cases for removing the absolute value.

Case 1 — Where $x^2 - x - 6 \ge 0$ (i.e. $x \le -2$ or $x \ge 3$ ).

Here the absolute value does nothing, so the equation becomes: $x^2 - x - 6 = x + 2$ $x^2 - 2x - 8 = 0$ $(x - 4)(x + 2) = 0$ $x = 4 \quad\text{or}\quad x = -2.$

Now check each against the case condition ( $x \le -2$ or $x \ge 3$ ) and the Step 1 condition ( $x \ge -2$ ):

$x = 4$ : satisfies $x \ge 3$ . ✓
$x = -2$ : satisfies $x \le -2$ (boundary) and $x \ge -2$ . ✓

Both are valid.

Case 2 — Where $x^2 - x - 6 < 0$ (i.e. $-2 < x < 3$ ).

Here the expression inside is negative, so the absolute value flips its sign: $-(x^2 - x - 6) = x + 2$ $-x^2 + x + 6 = x + 2$ $-x^2 + 4 = 0$ $x^2 = 4$ $x = 2 \quad\text{or}\quad x = -2.$

Check against the case condition $-2 < x < 3$ :

$x = 2$ : lies in $(-2, 3)$ . ✓
$x = -2$ : does not lie in the open interval $(-2, 3)$ . ✗

So only $x = 2$ is valid from this case.

💡 Teacher tip: The boundary $x = -2$ is subtle. It belongs to Case 1 (where the expression is $\ge 0$ ), not Case 2 (which requires the strict inequality $< 0$ ). This is why $x = -2$ appears in both cases but is only counted once — and only from Case 1.

Step 3 — Collect distinct roots and multiply.

The distinct valid roots are: $x = 4, \quad x = -2, \quad x = 2.$

Root	Came from	Valid?
$4$	Case 1 ( $x \ge 3$ )	✓
$-2$	Case 1 ( $x \le -2$ , boundary)	✓
$2$	Case 2 ( $-2 < x < 3$ )	✓
$-2$	Case 2	✗ rejected (outside open interval)

Their product: $4 \times (-2) \times 2 = -16.$

Answer: $\boxed{-16}$ (Choice A)

💡 Why this works: The absolute value created two quadratics, but each was only "active" on part of the number line. The key discipline is solving and then filtering — every candidate root must satisfy both the sign condition for its case and the original requirement $x \ge -2$ . Here, $x = -2$ is the boundary that sits right at the seam between the two cases, which is exactly the kind of value exam questions love to test.

Practice more Quadratic & Linear Equations ↗Ask in forum ↗

Q23Mixtures & AlligationsMCQ

A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is

Show solution

Correct answer: A

Think of this as: we know the density of each pure liquid, and we know the density of the mixture. From those three densities, we can back out the volume fraction of liquid 1.

Step 1 — Convert everything to consistent units (grams per litre).

Liquid 1: $1$ L weighs $1$ kg $= 1000$ g, so its density is $1000$ g/L.
Liquid 2: $1$ L weighs $800$ g, so its density is $800$ g/L.

Step 2 — Find the density of the mixture.

Half a litre of the mixture weighs $480$ g. Scaling up to a full litre:

$\text{Weight of 1 L of mixture} = 480 \times 2 = 960 \text{ g}.$

So the mixture's density is $960$ g/L.

Step 3 — Let the volume fraction of liquid 1 be $f$ .

In 1 L of mixture, liquid 1 contributes $f$ litres and liquid 2 contributes $(1-f)$ litres. The total weight is the sum of the two contributions:

$1000f + 800(1-f) = 960.$

Step 4 — Solve for $f$ .

$1000f + 800 - 800f = 960$

$200f + 800 = 960$

$200f = 160$

$f = \frac{160}{200} = 0.8.$

So liquid 1 is $0.8 \times 100\% = 80\%$ of the mixture by volume.

Quick check: $80\%$ of a litre is $0.8$ L of liquid 1 (weighing $800$ g) and $20\%$ is $0.2$ L of liquid 2 (weighing $160$ g). Total $= 800 + 160 = 960$ g per litre, so half a litre $= 480$ g. ✓

Answer: $\boxed{80}$ (Choice A)

Alternative approach — Alligation.

Since we are mixing two liquids of known densities to get a mixture of known density, alligation on the densities works directly:

	Liquid 1	Mixture	Liquid 2
Density (g/L)	$1000$	$960$	$800$

By alligation, the volume ratio of liquid 1 to liquid 2 is:

$\frac{\text{Liquid 1}}{\text{Liquid 2}} = \frac{960 - 800}{1000 - 960} = \frac{160}{40} = \frac{4}{1}.$

So liquid 1 is $\frac{4}{4+1} = \frac{4}{5} = 80\%$ of the mixture by volume — confirming the answer.

💡 Teacher tip: Alligation is fastest here because the question is literally a two-ingredient weighted average of densities. The algebraic method is the same idea written out; alligation just skips the equation setup.

Closing insight: The mixture's density ( $960$ ) sits $4$ times closer to liquid 1's density ( $1000$ ) than to liquid 2's ( $800$ ), which is exactly why liquid 1 dominates at $80\%$ .

Practice more Mixtures & Alligations ↗Ask in forum ↗

Q24PercentagesTITA

In a class, 60% of the students are girls and the rest are boys. There are 30 more girls than boys. If 68% of the students, including 30 boys, pass an examination, the percentage of the girls who do not pass is

Your answer:

Show solution

Correct answer: 20

We need to find what percentage of the girls did not pass the exam. Let's build up the numbers step by step.

Step 1 — Find the total number of students.

We're told 60% are girls, so the remaining 40% are boys. The gap between girls and boys is:

$60\% - 40\% = 20\% \text{ of the class.}$

This 20% gap equals 30 students (since there are 30 more girls than boys):

$0.20 \times T = 30 \;\Rightarrow\; T = 150.$

So the class has 150 students total.

💡 Teacher tip: Whenever a percentage difference and an actual count are both given, anchor on the total first. The difference in percentages (20%) maps directly to the difference in count (30), giving you the total in one step.

Step 2 — Split into girls and boys.

Girls: $60\% \times 150 = 90$
Boys: $40\% \times 150 = 60$

Quick check: $90 - 60 = 30$ ✓ (matches the given difference).

Step 3 — Find how many students pass.

68% of all 150 students pass:

$0.68 \times 150 = 102 \text{ students pass.}$

Step 4 — Break the passers down by gender.

We're told 30 boys pass. So the remaining passers are girls:

$102 - 30 = 72 \text{ girls pass.}$

Step 5 — Find girls who do NOT pass.

Out of 90 girls, 72 pass, so:

$90 - 72 = 18 \text{ girls do not pass.}$

Step 6 — Express as a percentage of girls.

The question asks for the percentage of girls (not of the whole class) who do not pass:

$\frac{18}{90} \times 100 = 20\%.$

Answer: $\boxed{20}$

💡 Closing insight: The key trap here is the base of the percentage. The question asks for the percentage of girls who fail — so the denominator is 90 (girls), not 150 (all students). Always confirm which group the percentage is measured against.

Practice more Percentages ↗Ask in forum ↗

Q25Maxima & MinimaMCQ

The number of the real roots of the equation 2cos(x(x + 1)) = $2^x$ + $2^{(−x)}$ is

Show solution

Correct answer: B

We need to find how many real values of $x$ satisfy: $2\cos(x(x+1)) = 2^x + 2^{-x}.$

Think of this as: the left side is trapped above by $2$ , the right side is trapped below by $2$ — so the only hope is when both sides hit exactly $2$ .

Step 1 — Find the minimum of the right-hand side.

Apply the AM–GM inequality to $2^x$ and $2^{-x}$ : $2^x + 2^{-x} \geq 2\sqrt{2^x \cdot 2^{-x}} = 2\sqrt{1} = 2.$

Equality holds when $2^x = 2^{-x}$ , i.e. $x = 0$ .

So the RHS is always at least $2$ , and equals $2$ only at $x = 0$ .

Step 2 — Find the maximum of the left-hand side.

For any real argument, $\cos(\cdot) \leq 1$ , so: $2\cos(x(x+1)) \leq 2.$

The LHS is always at most $2$ .

Step 3 — Squeeze both sides.

Putting the two bounds together: $\underbrace{2\cos(x(x+1))}_{\leq\, 2} = \underbrace{2^x + 2^{-x}}_{\geq\, 2}.$

The only way a quantity that is $\leq 2$ can equal a quantity that is $\geq 2$ is if both are exactly $2$ .

This forces two conditions simultaneously:

RHS $= 2 \Rightarrow x = 0$ (from Step 1).
LHS $= 2 \Rightarrow \cos(x(x+1)) = 1$ .

Step 4 — Verify $x = 0$ satisfies both.

At $x = 0$ :

LHS: $2\cos(0 \cdot 1) = 2\cos(0) = 2 \times 1 = 2$ ✓
RHS: $2^0 + 2^{-0} = 1 + 1 = 2$ ✓

For any $x \neq 0$ , the RHS is strictly greater than $2$ while the LHS is at most $2$ , so equality is impossible.

Answer: $\boxed{1}$ (Choice B)

💡 Why this works: The equation is a classic "squeeze" — one side bounded above, the other bounded below, meeting at a single point. No calculus or graphing needed; the bounds do all the work.

Practice more Maxima & Minima ↗Ask in forum ↗

Q26Time Speed & DistanceTITA

In a race of three horses, the first beat the second by 11 metres and the third by 90 metres. If the second beat the third by 80 metres, what was the length, in metres, of the racecourse?

Your answer:

Show solution

Correct answer: 880

Think of this as: we know three pairwise winning margins in a race, but they don't simply add up — we need to translate each margin into a speed ratio and then make the ratios consistent.

What we want: the length of the racecourse, $L$ .

Step 1 — Translate each "beat by" statement into a speed ratio.

When two horses run for the same amount of time, the distances they cover are in the same ratio as their speeds. This is the core idea: a winning margin of $x$ metres means the loser was $x$ metres short when the winner crossed the finish line.

Let the three horses have constant speeds $v_1, v_2, v_3$ .

When Horse 1 finishes (covers $L$ metres):

Horse 2 has covered $L - 11$ metres (beaten by 11 m).
Horse 3 has covered $L - 90$ metres (beaten by 90 m).

So the speed ratios are: $\frac{v_2}{v_1} = \frac{L-11}{L}, \qquad \frac{v_3}{v_1} = \frac{L-90}{L}.$

When Horse 2 finishes (covers $L$ metres):

Horse 3 has covered $L - 80$ metres (beaten by 80 m).

So: $\frac{v_3}{v_2} = \frac{L-80}{L}.$

💡 Teacher tip: Notice that $11 + 80 = 91 \neq 90$ . The margins are not additive — Horse 3 is 90 m behind Horse 1 when Horse 1 finishes, but 80 m behind Horse 2 when Horse 2 finishes. These are measurements taken at different moments, so we must use speed ratios, not simple addition.

Step 2 — Get a second expression for $v_3 / v_2$ .

We already have $v_3/v_1$ and $v_2/v_1$ from Step 1. Dividing the first by the second:

$\frac{v_3}{v_2} = \frac{v_3 / v_1}{v_2 / v_1} = \frac{\, (L-90)/L \,}{\, (L-11)/L \,} = \frac{L-90}{L-11}.$

Step 3 — Equate the two expressions for $v_3 / v_2$ and solve.

From Steps 1 and 2:

$\frac{L-80}{L} = \frac{L-90}{L-11}.$

Cross-multiply:

$(L-80)(L-11) = L(L-90).$

Expand both sides:

$L^2 - 91L + 880 = L^2 - 90L.$

The $L^2$ terms cancel:

$-91L + 880 = -90L \implies 880 = L.$

Step 4 — Verify the answer is consistent with all three margins.

With $L = 880$ :

Event	Horse 1 position	Horse 2 position	Horse 3 position	Margin check
Horse 1 finishes	880 m	$880 - 11 = 869$ m	$880 - 90 = 790$ m	1 beats 2 by 11 ✓, 1 beats 3 by 90 ✓
Horse 2 finishes	—	880 m	$880 \times \frac{790}{869} = 800$ m	2 beats 3 by $880 - 800 = 80$ ✓

All three margins check out.

Answer: $\boxed{880}$

💡 Why this works: The key insight is that every "beat by $x$ metres" gives a ratio of distances, hence a ratio of speeds. With three horses we get three ratios, and any two of them must produce the third — that consistency condition pins down $L$ uniquely. Whenever you see multiple pairwise margins in a race, reach for speed ratios, not addition.

Practice more Time Speed & Distance ↗Ask in forum ↗

Q27Simple & Compound InterestMCQ

Amala, Bina, and Gouri invest money in the ratio 3 : 4 : 5 in fixed deposits having respective annual interest rates in the ratio 6 : 5 : 4. What is their total interest income (in Rs) after a year, if Bina's interest income exceeds Amala's by Rs 250?

Show solution

Correct answer: D

Think of this as: we know the ratio of the three interest incomes and we know the actual rupee gap between two of them. Combine the two and the total falls out.

Step 1 — Write interest in terms of principal and rate.

For a one-year fixed deposit, interest $= \text{Principal} \times \text{Rate}$ . So the interest income of each person is proportional to the product of her principal ratio-part and her rate ratio-part.

Let the principals be in the ratio $3:4:5$ and the rates be in the ratio $6:5:4$ . Then:

Person	Principal part	Rate part	Interest $\propto$ Principal $\times$ Rate
Amala	$3$	$6$	$3 \times 6 = 18$
Bina	$4$	$5$	$4 \times 5 = 20$
Gouri	$5$	$4$	$5 \times 4 = 20$

So the interest incomes are in the ratio $\text{Amala: Bina: Gouri} = 18: 20: 20.$

💡 Teacher tip: Whenever a question gives ratios for two multiplying quantities (here principal and rate), the ratio of the product is just the product of the ratios — multiply the corresponding parts directly.

Step 2 — Use the known rupee gap to value one unit.

We are told Bina's interest exceeds Amala's by Rs $250$ . From the ratio, $\text{Bina} - \text{Amala} = 20 - 18 = 2 \text{ units}.$ So $2$ units $= 250$ rupees, which gives $1 \text{ unit} = 125 \text{ rupees}.$

Step 3 — Compute the total interest income.

The total number of ratio units is $18 + 20 + 20 = 58.$ Multiplying by the value of one unit: $\text{Total interest} = 58 \times 125 = 7250 \text{ rupees}.$

A quick arithmetic check: $58 \times 125 = 58 \times \frac{1000}{8} = \frac{58000}{8} = 7250$ .

Answer: $\boxed{7250}$ (Choice D)

💡 Why this works so cleanly: the actual principals and rates are never needed — only their products matter, and a single known gap pins down the scale of the entire ratio.

Practice more Simple & Compound Interest ↗Ask in forum ↗

Q28LogarithmsMCQ

If (5.55 $)^x$ = (0.555 $)^y$ = 1000, then the value of (1/x − 1/y) is

Show solution

Correct answer: A

Think of this as: two different bases raised to different powers both equal $1000$ . We need $\frac{1}{x} - \frac{1}{y}$ , so the goal is to express $\frac{1}{x}$ and $\frac{1}{y}$ individually, then subtract.

Step 1 — Take $\log_{10}$ of each equation.

Since $(5.55)^x = 1000 = 10^3$ :

$x \log 5.55 = 3 \quad\Rightarrow\quad \frac{1}{x} = \frac{\log 5.55}{3}.$

Since $(0.555)^y = 1000 = 10^3$ :

$y \log 0.555 = 3 \quad\Rightarrow\quad \frac{1}{y} = \frac{\log 0.555}{3}.$

Step 2 — Relate the two logarithms.

Both numbers share the digits $555$ ; they differ only by a power of $10$ :

$5.55 = \frac{555}{100} = \frac{555}{10^2}, \qquad 0.555 = \frac{555}{1000} = \frac{555}{10^3}.$

Taking logs:

$\log 5.55 = \log 555 - 2, \qquad \log 0.555 = \log 555 - 3.$

💡 Teacher tip: The shared digit pattern $5.55$ and $0.555$ is the clue — both are $555$ scaled by a power of $10$ , so the $\log 555$ terms will cancel exactly when we subtract.

Step 3 — Subtract.

$\frac{1}{x} - \frac{1}{y} = \frac{\log 5.55}{3} - \frac{\log 0.555}{3} = \frac{(\log 555 - 2) - (\log 555 - 3)}{3} = \frac{1}{3}.$

The $\log 555$ terms cancel, leaving just $\frac{-2-(-3)}{3} = \frac{1}{3}$ .

Answer: $\boxed{\dfrac{1}{3}}$ (Choice A)

Alternative approach (faster — using the quotient directly).

Instead of introducing $\log 555$ , subtract the two expressions for $\frac{1}{x}$ and $\frac{1}{y}$ and use the quotient rule of logarithms:

$\frac{1}{x} - \frac{1}{y} = \frac{\log 5.55 - \log 0.555}{3} = \frac{1}{3}\,\log\!\left(\frac{5.55}{0.555}\right).$

Now $\frac{5.55}{0.555} = 10$ , so $\log 10 = 1$ :

$\frac{1}{x} - \frac{1}{y} = \frac{1}{3} \times 1 = \frac{1}{3}.$

This avoids writing $\log 555$ at all — the ratio of the two bases is exactly $10$ , which is the base of the logarithm. That is the elegant core of the problem: the answer is simply $\frac{1}{3}\log_{10}(10) = \frac{1}{3}$ .

Practice more Logarithms ↗Ask in forum ↗

Q29Polynomials & ProgressionsMCQ

The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157 : 3, then the sum of the two numbers is

Show solution

Correct answer: C

Framing: We know the product of two positive numbers and a ratio involving their cubes. The goal is to find their sum. The strategy: translate the cube ratio into something involving only the sum $S = x+y$ and difference $d = x-y$ , then use the product to pin down $S$ .

Step 1 — Set up variables and translate the ratio.

Let the two numbers be $x$ and $y$ (with $x > y > 0$ ). We're given:

$xy = 616$
$\dfrac{x^3 - y^3}{(x - y)^3} = \dfrac{157}{3}$

Factor the numerator using the difference of cubes identity: $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

So the ratio becomes: $\frac{(x-y)(x^2 + xy + y^2)}{(x-y)^3} = \frac{x^2 + xy + y^2}{(x-y)^2} = \frac{x^2 + xy + y^2}{x^2 - 2xy + y^2} = \frac{157}{3}$

Step 2 — Switch to sum and difference.

Let $S = x + y$ and $d = x - y$ . Then:

$xy = \dfrac{S^2 - d^2}{4}$
$x^2 + y^2 = \dfrac{S^2 + d^2}{2}$

Now express the numerator and denominator of our ratio:

Numerator: $x^2 + xy + y^2 = (x^2 + y^2) + xy = \dfrac{S^2 + d^2}{2} + \dfrac{S^2 - d^2}{4} = \dfrac{3S^2 + d^2}{4}$
Denominator: $x^2 - 2xy + y^2 = (x - y)^2 = d^2$

So the ratio becomes: $\frac{3S^2 + d^2}{4d^2} = \frac{157}{3}$

Step 3 — Solve for the relationship between $S$ and $d$ .

Cross-multiply: $3(3S^2 + d^2) = 628\,d^2$ $9S^2 + 3d^2 = 628\,d^2$ $9S^2 = 625\,d^2$ $3S = 25d \quad \Longrightarrow \quad d = \frac{3S}{25}$

(We take the positive root since $x > y > 0$ means both $S > 0$ and $d > 0$ .)

Step 4 — Use the product to find $S$ .

Since $xy = 616$ : $\frac{S^2 - d^2}{4} = 616 \quad \Longrightarrow \quad S^2 - d^2 = 2464$

Substitute $d = \dfrac{3S}{25}$ : $S^2 - \frac{9S^2}{625} = 2464$ $\frac{625S^2 - 9S^2}{625} = 2464$ $\frac{616\,S^2}{625} = 2464$ $S^2 = \frac{2464 \times 625}{616} = 4 \times 625 = 2500$ $S = 50$

💡 Teacher tip: Notice how $616$ appears both in the product and in the coefficient $\frac{616}{625}$ — that's not a coincidence. The $616$ from $625 - 9$ cancels neatly with the $616$ from $4 \times 616 = 2464$ , leaving $S^2 = 4 \times 625 = 2500$ .

Step 5 — Verify.

With $S = 50$ and $d = \frac{3 \times 50}{25} = 6$ :

$x = \frac{S + d}{2} = 28$ , $\quad y = \frac{S - d}{2} = 22$

Check	Computed value	Expected	Status
$xy$	$28 \times 22 = 616$	$616$	✅
$\dfrac{x^3 - y^3}{(x-y)^3}$	$\dfrac{21952 - 10648}{216} = \dfrac{11304}{216} = \dfrac{157}{3}$	$\dfrac{157}{3}$	✅

Answer: $\boxed{50}$ (Choice C)

Alternative approach — Direct algebra (faster).

From Step 1, we had: $\frac{x^2 + xy + y^2}{x^2 - 2xy + y^2} = \frac{157}{3}$

Let $p = x^2 + y^2$ and substitute $xy = 616$ : $\frac{p + 616}{p - 1232} = \frac{157}{3}$

Cross-multiply: $3(p + 616) = 157(p - 1232)$ $3p + 1848 = 157p - 193424$ $154p = 195272$ $p = 1268$

So $x^2 + y^2 = 1268$ . Now use the identity $(x+y)^2 = (x^2 + y^2) + 2xy$ : $(x + y)^2 = 1268 + 2(616) = 1268 + 1232 = 2500$ $x + y = 50$

This approach is faster under exam time because it avoids introducing $S$ and $d$ entirely — it works directly with $x^2 + y^2$ and $xy$ , which is all we need to reach $(x+y)^2$ .

💡 Why this works so cleanly: The ratio $\frac{x^3 - y^3}{(x-y)^3}$ simplifies to $\frac{x^2 + xy + y^2}{(x-y)^2}$ , which depends only on $x^2 + y^2$ and $xy$ . Since $xy$ is already given, one equation pins down $x^2 + y^2$ , and then $(x+y)^2 = (x^2 + y^2) + 2xy$ delivers the sum directly.

Practice more Polynomials & Progressions ↗Ask in forum ↗

Q30Polynomials & ProgressionsMCQ

If the population of a town is p in the beginning of any year then it becomes 3 + 2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be

Show solution

Correct answer: D

Framing: Each year, the population is obtained from the previous year's by doubling and adding 3. The "+3" is a nuisance — our first job is to absorb it so we're left with a clean doubling pattern.

Step 1 — Write the recurrence formally.

Let $p_n$ be the population at the beginning of year $n$ . We are given:

$p_{n+1} = 3 + 2p_n, \qquad p_{2019} = 1000.$

Step 2 — Remove the constant by shifting the variable.

The "+3" stops this from being a simple geometric sequence. Define a shifted variable:

$q_n = p_n + 3.$

Now substitute into the recurrence:

$q_{n+1} = p_{n+1} + 3 = (3 + 2p_n) + 3 = 2p_n + 6 = 2(p_n + 3) = 2\,q_n.$

So $q_n$ is a geometric sequence with common ratio 2 — it simply doubles every year.

💡 Teacher tip: For any recurrence of the form $p_{n+1} = c + rp_n$ , the substitution $q_n = p_n + \dfrac{c}{r-1}$ eliminates the constant and gives a pure geometric sequence $q_{n+1} = r \cdot q_n$ . Here $c = 3$ and $r = 2$ , so the shift is $\dfrac{3}{2-1} = 3$ . This shift is actually the fixed point of the recurrence — the value where $p = 3 + 2p$ would hold.

Step 3 — Compute the starting value of $q$ .

$q_{2019} = p_{2019} + 3 = 1000 + 3 = 1003.$

Step 4 — Count the number of steps.

From the beginning of 2019 to the beginning of 2034, the recurrence is applied:

$2034 - 2019 = 15 \text{ times.}$

Step 5 — Apply the geometric-sequence formula.

Since $q_n$ doubles each step:

$q_{2034} = q_{2019} \times 2^{15} = 1003 \times 2^{15}.$

Step 6 — Convert back to population $p$ .

Since $q_n = p_n + 3$ , we have $p_n = q_n - 3$ :

$p_{2034} = q_{2034} - 3 = 1003 \times 2^{15} - 3.$

Verification — check the first few terms directly:

Year $n$	$p_n$ (direct computation)	$q_n = p_n + 3$	$1003 \times 2^{\,n-2019}$	Match?
2019	$1000$	$1003$	$1003 \times 2^0 = 1003$	✅
2020	$3 + 2(1000) = 2003$	$2006$	$1003 \times 2^1 = 2006$	✅
2021	$3 + 2(2003) = 4009$	$4012$	$1003 \times 2^2 = 4012$	✅
2022	$3 + 2(4009) = 8021$	$8024$	$1003 \times 2^3 = 8024$	✅

The pattern is confirmed. After 15 doublings, $p_{2034} + 3 = 1003 \times 2^{15}$ , so $p_{2034} = 1003 \times 2^{15} - 3$ .

Answer: $\boxed{(1003) \times 2^{15} - 3}$ (Choice D)

Alternative approach — Spot the pattern by computing terms.

If the substitution trick doesn't occur to you immediately, compute a few terms and look for structure:

Year	$p_n$	$p_n + 3$
2019	1000	1003
2020	2003	2006
2021	4009	4012
2022	8021	8024

Notice that $p_n + 3$ doubles each year: $1003,\; 2006,\; 4012,\; 8024,\; \ldots$ — that is, $1003 \times 2^0,\; 1003 \times 2^1,\; 1003 \times 2^2,\; 1003 \times 2^3,\; \ldots$

After 15 doublings (from 2019 to 2034):

$p_{2034} + 3 = 1003 \times 2^{15} \implies p_{2034} = 1003 \times 2^{15} - 3.$

Same result, discovered by pattern recognition. The substitution method is more reliable under exam pressure since it doesn't depend on spotting the pattern.

Why the wrong options fail:

Option A $(1003)^{15}+6$ : confuses multiplying by $2^{15}$ with raising to the 15th power.
Option B $(997)^{15}-3$ : uses $1000-3=997$ (wrong shift direction) and again exponentiates instead of doubling.
Option C $(997)\times 2^{14}+3$ : wrong shift ( $997$ instead of $1003$ ), wrong number of steps ( $14$ instead of $15$ ), and wrong sign on the trailing constant.

Closing insight: The recurring theme in CAT recurrence problems is linearisation — any affine recurrence $p_{n+1}=c+rp_n$ can be turned into a pure geometric one by measuring from its fixed point $\frac{c}{r-1}$ . Once you internalise this, the "+3" in the problem becomes a signal, not an obstacle.

Practice more Polynomials & Progressions ↗Ask in forum ↗

Q31PercentagesMCQ

Meena scores 40% in an examination and after review, even though her score is increased by 50%, she fails by 35 marks. If her post-review score is increased by 20%, she will have 7 marks more than the passing score. The percentage score needed for passing the examination is

Show solution

Correct answer: C

Think of this as: Meena's score gets boosted twice in two different scenarios, and each scenario is compared to the same passing mark. Two equations, two unknowns — the maximum marks and the passing mark.

Step 1 — Set up variables.

Let the maximum marks of the examination be $M$ , and the passing mark be $P$ .

Meena's initial score is $40\%$ of $M$ : $\text{Initial score} = 0.4M$

Step 2 — Translate the first scenario into an equation.

Her score is increased by $50\%$ after review. A $50\%$ increase means multiplying by $1.5$ : $\text{Reviewed score} = 0.4M \times 1.5 = 0.6M$

Even after this increase, she fails by 35 marks — meaning her reviewed score is 35 less than the passing mark: $0.6M = P - 35 \quad \cdots (1)$

Step 3 — Translate the second scenario into an equation.

Now the reviewed score ( $0.6M$ ) is increased by a further $20\%$ . A $20\%$ increase means multiplying by $1.2$ : $\text{Further increased score} = 0.6M \times 1.2 = 0.72M$

This time she has 7 marks more than the passing mark: $0.72M = P + 7 \quad \cdots (2)$

💡 Teacher tip: Notice the structure — both equations link the same two unknowns, $M$ and $P$ . The difference between the two scenarios is entirely captured by the gap between $0.72M$ and $0.6M$ , which equals $35 + 7 = 42$ marks.

Step 4 — Solve the system.

Subtract equation $(1)$ from equation $(2)$ : $0.72M - 0.6M = (P + 7) - (P - 35)$ $0.12M = 42$ $M = \frac{42}{0.12} = 350$

So the maximum marks are 350.

Now substitute back into equation $(1)$ to find $P$ : $0.6 \times 350 = P - 35$ $210 = P - 35$ $P = 245$

Step 5 — Verify with the second equation (quick check).

$0.72 \times 350 = 252, \quad P + 7 = 245 + 7 = 252 \quad \checkmark$

Both equations are satisfied.

Step 6 — Find the passing percentage.

$\text{Passing percentage} = \frac{P}{M} \times 100 = \frac{245}{350} \times 100 = 70\%$

Answer: $\boxed{70}$ (Choice C)

💡 Closing insight: The key to this problem is recognising that both scenarios share the same passing mark $P$ . Once you write both equations, subtracting them eliminates $P$ immediately and isolates $M$ — a classic "two conditions, two unknowns" setup where elimination does the heavy lifting.

Practice more Percentages ↗Ask in forum ↗

Q32Averages & Weighted AveragesMCQ

Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is

Show solution

Correct answer: B

Think of this as: we have two conditions linking the total of all 22 scores to individual scores of Ramesh and Gautam. If we can pin down the total, Gautam's score falls out directly.

What we want: Gautam's score, $G$ .

Let the sum of all 22 scores be $T$ .

Step 1 — Translate each condition into an equation.

Condition 1: The average score of the 21 students other than Gautam is $62$ .

Removing Gautam's score $G$ from the total leaves $T - G$ , spread over 21 students:

$\frac{T - G}{21} = 62 \quad\Rightarrow\quad T - G = 62 \times 21 = 1302. \tag{1}$

Condition 2: The average of all 22 students is one more than the average of the 21 students other than Ramesh (whose score is $82.5$ ).

$\frac{T}{22} = \frac{T - 82.5}{21} + 1. \tag{2}$

💡 Teacher tip: Equation (1) already tells us $G = T - 1302$ . So our only job is to find $T$ from equation (2), and then Gautam's score is a single subtraction away.

Step 2 — Solve equation (2) for the total $T$ .

Multiply both sides by $462 = 22 \times 21$ to clear denominators:

$21T = 22(T - 82.5) + 462.$

Expand the right side:

$21T = 22T - 1815 + 462.$

$21T = 22T - 1353.$

$T = 1353.$

So the sum of all 22 scores is 1353.

Step 3 — Find Gautam's score.

From equation (1):

$G = T - 1302 = 1353 - 1302 = 51.$

Quick verification.

Check	Computation	Result
Average of 21 excluding Gautam	$(1353 - 51)/21 = 1302/21$	$62$ ✓
Average of all 22	$1353/22$	$61.5$
Average of 21 excluding Ramesh	$(1353 - 82.5)/21 = 1270.5/21$	$60.5$
Difference	$61.5 - 60.5$	$1$ ✓

Both conditions hold.

Answer: $\boxed{51}$ (Choice B)

💡 Closing insight: The key move was anchoring on the total sum $T$ rather than on individual averages. Once both conditions are written in terms of $T$ , the problem reduces to one linear equation — and Gautam's score is just $T$ minus the sum of the other 21.

Practice more Averages & Weighted Averages ↗Ask in forum ↗

Q33Coordinate GeometryTITA

Let T be the triangle formed by the straight line 3x + 5y - 45 = 0 and the coordinate axes. Let the circumcircle of T have radius of length L, measured in the same unit as the coordinate axes. Then, the integer closest to L is

Your answer:

Show solution

Correct answer: 9

We need the circumradius of the triangle cut off by the line $3x + 5y - 45 = 0$ and the two coordinate axes. Let's build the triangle first, then use a key property of right triangles.

Step 1 — Find the three vertices.

The triangle is bounded by the given line, the $x$ -axis, and the $y$ -axis. Its vertices are the three pairwise intersection points.

Origin: the axes meet at $(0, 0)$ .
$x$ -intercept: set $y = 0$ in $3x + 5y - 45 = 0$ : $3x = 45 \Rightarrow x = 15 \Rightarrow (15, 0).$
$y$ -intercept: set $x = 0$ : $5y = 45 \Rightarrow y = 9 \Rightarrow (0, 9).$

So the triangle has vertices $O(0,0)$ , $A(15,0)$ , and $B(0,9)$ .

Step 2 — Recognise the right angle.

The legs $OA$ and $OB$ lie along the $x$ -axis and $y$ -axis respectively. Since the axes are perpendicular, the angle at the origin is $90^\circ$ .

💡 Teacher tip: Any triangle with one vertex at the origin and the other two on the coordinate axes is automatically right-angled at the origin. This is the single observation that makes the problem easy.

Step 3 — Use the right-triangle circumradius property.

For a right-angled triangle, the circumcircle has the hypotenuse as its diameter. (The midpoint of the hypotenuse is equidistant from all three vertices.) Therefore: $L = \frac{\text{hypotenuse}}{2}.$

The hypotenuse is $AB$ : $AB = \sqrt{15^2 + 9^2} = \sqrt{225 + 81} = \sqrt{306}.$

So: $L = \frac{\sqrt{306}}{2}.$

Step 4 — Compute the numerical value.

We know $17^2 = 289$ and $17.5^2 = 306.25$ , so $\sqrt{306}$ is just under $17.5$ : $\sqrt{306} \approx 17.49 \Rightarrow L \approx \frac{17.49}{2} \approx 8.746.$

Step 5 — Pick the closest integer.

$8.746$ lies between $8$ and $9$ , and is closer to $9$ (distance $\approx 0.25$ ) than to $8$ (distance $\approx 0.75$ ).

Candidate	Distance from $8.746$	Closest?
$8$	$0.746$	✗
$9$	$0.254$	✓

Answer: $\boxed{9}$

💡 Why no full circle work is needed: because the triangle is right-angled, its circumradius is simply half the hypotenuse — no perpendicular bisector construction or general circumradius formula required.

Practice more Coordinate Geometry ↗Ask in forum ↗

Q34Time Speed & DistanceMCQ

One can use three different transports which move at 10, 20, and 30 kmph, respectively to reach from A to B. Amal took each mode of transport for 1/3 of his total journey time, while Bimal took each mode of transport for 1/3 of the total distance. The percentage by which Bimal's travel time exceeds Amal's travel time is nearest to

Show solution

Correct answer: A

Think of this as: two people cover the same route with the same three vehicles, but one splits his time equally across vehicles while the other splits his distance equally. We need to compare their total travel times.

Let the total distance from A to B be $D$ km, and the three speeds be $10$ , $20$ , and $30$ kmph.

Step 1 — Amal: equal time in each mode

Amal spends one-third of his total time on each transport. Let his total time be $3t$ , so he spends $t$ hours on each mode.

The distance covered in each mode is $\text{speed} \times \text{time}$ :

Mode	Speed (kmph)	Time (h)	Distance (km)
1	10	$t$	$10t$
2	20	$t$	$20t$
3	30	$t$	$30t$

So the total distance is:

$D = 10t + 20t + 30t = 60t \implies t = \frac{D}{60}$

Therefore Amal's total travel time is:

$T_A = 3t = \frac{3D}{60} = \frac{D}{20}$

Step 2 — Bimal: equal distance in each mode

Bimal covers one-third of the total distance on each transport, i.e., $D/3$ km per mode. Time for each leg is $\frac{\text{distance}}{\text{speed}}$ :

Mode	Speed (kmph)	Distance (km)	Time (h)
1	10	$D/3$	$\frac{D/3}{10} = \frac{D}{30}$
2	20	$D/3$	$\frac{D/3}{20} = \frac{D}{60}$
3	30	$D/3$	$\frac{D/3}{30} = \frac{D}{90}$

Bimal's total time:

$T_B = \frac{D}{30} + \frac{D}{60} + \frac{D}{90} = D\left(\frac{1}{30} + \frac{1}{60} + \frac{1}{90}\right)$

The LCM of $30, 60, 90$ is $180$ :

$T_B = D\left(\frac{6 + 3 + 2}{180}\right) = D \cdot \frac{11}{180} = \frac{11D}{180}$

Step 3 — Compare the two times

Let's put both times over a common denominator:

$T_A = \frac{D}{20} = \frac{9D}{180}, \qquad T_B = \frac{11D}{180}$

So Bimal takes $\frac{11D}{180}$ and Amal takes $\frac{9D}{180}$ — Bimal is clearly slower.

Step 4 — Percentage by which Bimal's time exceeds Amal's

$\text{Percentage excess} = \frac{T_B - T_A}{T_A} \times 100 = \frac{\frac{11D}{180} - \frac{9D}{180}}{\frac{9D}{180}} \times 100$

$= \frac{\frac{2D}{180}}{\frac{9D}{180}} \times 100 = \frac{2}{9} \times 100 \approx 22.22\%$

This rounds to 22%.

💡 Teacher tip: When time is split equally, the fastest mode covers the most distance — so the fast mode "pulls its weight." When distance is split equally, you're forced to spend a long time crawling on the slow mode. That's why equal-distance splitting is always slower than equal-time splitting when speeds differ.

Answer: 22 (Choice A)

💡 Why the numbers work out cleanly: Amal's effective speed is the arithmetic mean of the three speeds ( $\frac{10+20+30}{3} = 20$ kmph), while Bimal's is the harmonic mean ( $\frac{3}{\frac{1}{10}+\frac{1}{20}+\frac{1}{30}} = \frac{180}{11} \approx 16.36$ kmph). The harmonic mean is always less than or equal to the arithmetic mean — and the gap here gives us the $\frac{2}{9} \approx 22\%$ excess.

Practice more Time Speed & Distance ↗Ask in forum ↗

Practice more questions like these ↗Ask a doubt in the forum ↗