CAT 2019 Quant Questions — Slot 2

Two ants start from the same point P on a circular track and move in opposite directions. We know where they meet, how fast A finishes the lap, and we need to find when B finishes. The key is to use the meeting point to extract the speed ratio, then use A's finishing time to calibrate actual speeds.

Step 1 — Extract the speed ratio from the meeting point.

Since A and B move in opposite directions on a closed circle, their first meeting happens when their combined distances equal one full lap. At 10:00 am:

• A has covered 60% of the track (clockwise).
• B has covered 40% of the track (anti-clockwise).

They traveled for the same amount of time, so their speeds are in the ratio of their distances: $v_A: v_B = 60: 40 = 3: 2$

💡 Teacher tip: On a circular track with two bodies moving in opposite directions, the first meeting always splits the track into two parts whose ratio equals the speed ratio. This is the fastest way to get relative speeds.

Step 2 — Use A's return time to find A's actual speed.

At the meeting point (10:00 am), A has already covered 60% going clockwise. To return to P, A must continue clockwise and cover the remaining 40% of the track.

We're told A reaches P at 10:12 am, i.e., 12 minutes after the meeting.

So A's speed: $v_A = \frac{40\% \text{ of track}}{12 \text{ min}}$

Step 3 — Find B's remaining distance and time.

At the meeting point, B has covered 40% going anti-clockwise. To return to P, B must continue anti-clockwise and cover the remaining 60% of the track.

Now compare B's remaining task to A's remaining task:

• Distance factor: B must cover 60% while A covered 40%, so B's distance is $\frac{60}{40} = \frac{3}{2}$ times A's distance.
• Speed factor: Since $v_A: v_B = 3: 2$, B's speed is $\frac{2}{3}$ of A's. Slower speed means more time, so B takes $\frac{3}{2}$ times as long per unit distance.

Putting it together: $T_B = 12 \times \frac{60}{40} \times \frac{3}{2} = 12 \times \frac{3}{2} \times \frac{3}{2} = 12 \times \frac{9}{4} = 27 \text{ minutes}$

Step 4 — Convert to clock time.

B's 27-minute return trip starts at 10:00 am: $10{:}00 \text{ am} + 27 \text{ min} = 10{:}27 \text{ am}$

Answer: $\boxed{10{:}27 \text{ am}}$ (Choice D)

💡 Why this works so cleanly: The meeting point gives us the speed ratio for free, and A's finishing time gives us the actual speed. Once both are known, B's time is just a ratio-and-proportion calculation — no need to find the track length or the actual starting time.

Think of this as: first compute Amal's total interest from two separate investments (one annual compounding, one semi-annual), then find the principal that earns the same interest at simple interest.

Step 1 — Amal's first investment: Rs 12,000 at 8% compounded annually, 1 year.

For a single year, annual compounding behaves just like simple interest — there's only one compounding cycle. So the interest is:

$\text{Interest}_1 = 12000 \times \frac{8}{100} = 12000 \times 0.08 = \text{Rs } 960.$

Step 2 — Amal's second investment: Rs 10,000 at 6% compounded semi-annually, 1 year.

💡 Teacher tip: "Compounded semi-annually" means interest is applied twice a year. So the annual rate gets halved into a half-yearly rate, and one year gives two compounding periods.

• Half-yearly rate $= \frac{6\%}{2} = 3\%$
• Number of half-year periods in 1 year $= 2$

The maturity amount is:

$A = 10000 \times \left(1 + \frac{3}{100}\right)^2 = 10000 \times (1.03)^2.$

Compute $(1.03)^2$:

$(1.03)^2 = 1 + 2(0.03) + (0.03)^2 = 1 + 0.06 + 0.0009 = 1.0609.$

So:

$A = 10000 \times 1.0609 = \text{Rs } 10{,}609.$

The interest earned is the amount minus the principal:

$\text{Interest}_2 = 10609 - 10000 = \text{Rs } 609.$

Step 3 — Amal's total interest.

$\text{Amal's total interest} = 960 + 609 = \text{Rs } 1{,}569.$

Step 4 — Set Bimal's simple interest equal to Amal's interest.

Bimal invests a principal $P$ at $7.5\%$ simple interest for 1 year. Simple interest for one year is:

$\text{SI} = P \times \frac{7.5}{100} = 0.075\,P.$

Since the interests are equal:

$0.075\,P = 1569.$

Step 5 — Solve for $P$.

$P = \frac{1569}{0.075} = \frac{1569 \times 1000}{75} = \frac{1{,}569{,}000}{75}.$

Simplify the fraction — divide numerator and denominator by 3:

$P = \frac{523{,}000}{25}.$

Now divide by 25 (multiply by 4 and divide by 100):

$P = \frac{523{,}000 \times 4}{100} = \frac{2{,}092{,}000}{100} = 20{,}920.$

Quick check:

Both interests match at Rs 1,569.

Answer: $\boxed{20920}$

💡 Closing insight: The key trap here is the semi-annual investment — a 6% annual rate does not mean 6% per half-year. Always split the annual rate by the number of compounding periods per year, and multiply the number of years by that same factor to get the total number of periods.

Think of this as: we know the shape of the salary distribution in two years (via ratios), and one person's actual % change. That's enough to pin down everyone else's % change.

Step 1 — Set up the 2010 salaries.

The 2010 ratio Ramesh: Ganesh: Rajesh = $6:5:7$. Let the common multiplier be $k$, so:

Step 2 — Use Ramesh's 25% increase to find his 2015 salary.

Ramesh's salary rose by 25%, so his 2015 salary is: $6k \times 1.25 = 7.5k$

💡 Teacher tip: A ratio only gives us relative sizes. To connect 2010 and 2015, we need one actual link — and Ramesh's 25% increase is exactly that link. Anchor everything on Ramesh.

Step 3 — Match Ramesh's 2015 salary to the 2015 ratio.

The 2015 ratio is Ramesh: Ganesh: Rajesh = $3:4:3$. Let the 2015 multiplier be $m$.

Ramesh's 2015 share is $3m$, and we already know it equals $7.5k$: $3m = 7.5k \;\Rightarrow\; m = 2.5k$

Step 4 — Find Rajesh's 2015 salary.

Rajesh's share in the 2015 ratio is also $3$: $\text{Rajesh's 2015 salary} = 3m = 3 \times 2.5k = 7.5k$

Step 5 — Compute Rajesh's percentage increase.

Rajesh went from $7k$ (2010) to $7.5k$ (2015): $\text{Percentage increase} = \frac{7.5k - 7k}{7k} \times 100 = \frac{0.5k}{7k} \times 100 = \frac{50}{7} \approx 7.14\%$

Step 6 — Pick the closest option.

$7.14\%$ is closest to $7\%$.

Answer: $\boxed{7\%}$ (Choice B)

💡 Why this works: The two ratios use different multipliers ($k$ for 2010, $m$ for 2015). Ramesh's known % change is the bridge that ties $m$ to $k$. Once that bridge is built, every other person's 2015 salary — and hence their % change — is determined.

Think of this as: the total job is split into three shares — Anil's, Sunil's, and Bimal's. We know each person's daily rate, and we know Bimal's total share is 10%. The only unknown is the total time $T$, which controls how long Anil and Sunil each worked.

Step 1 — Write down each person's daily rate.

• Anil finishes the job alone in 20 days, so his rate is $\frac{1}{20}$ of the job per day.
• Sunil finishes it alone in 40 days, so his rate is $\frac{1}{40}$ of the job per day.
• Bimal's daily rate is not given — but we don't actually need it, because we're told his total contribution directly.

Step 2 — Translate the timeline into work-days.

Let the total time to finish the job be $T$ days.

• Anil starts on day 1 and works until the end, so he works all $T$ days.
• Sunil joins after 3 days, so he works $T - 3$ days.
• Bimal joins later still, but instead of tracking when he joined, we use the given fact: Bimal completed $10\% = \frac{1}{10}$ of the job in total.

💡 Teacher tip: When a problem gives you someone's total contribution as a fraction of the job, you can skip figuring out their daily rate and join-day entirely. Just plug their total share straight into the work equation.

Step 3 — Build the work equation.

The three shares must add up to one whole job:

$\underbrace{\frac{T}{20}}_{\text{Anil}} + \underbrace{\frac{T-3}{40}}_{\text{Sunil}} + \underbrace{\frac{1}{10}}_{\text{Bimal}} = 1.$

Step 4 — Solve for $T$.

Multiply every term by 40 (the LCM of 20, 40, and 10):

$2T + (T - 3) + 4 = 40.$

Simplify:

$3T + 1 = 40 \implies 3T = 39 \implies T = 13.$

Step 5 — Verify the shares add up.

Everything checks out — the job is fully accounted for.

Answer: $\boxed{13}$ (Choice B)

💡 Closing insight: The beauty of this problem is that Bimal's join-day and daily rate are both irrelevant — his total contribution is handed to you. Whenever a problem states a worker's share of the output rather than their speed, anchor the equation on that share directly.

Think of this as: two equal circles sit on a common tangent line, and a smaller circle nestles between them on the same line, touching both. We need its radius.

Step 1 — Set up a clean coordinate picture.

Let the two given circles have radius $R = 4$. Since they touch externally, the distance between their centres is $4 + 4 = 8$.

Place their centres at $A = (0, 4), \quad B = (8, 4).$ Both circles rest on the line $y = 0$, which is the common tangent to all three circles.

Step 2 — Locate the third circle.

Let the third circle have radius $r$.

• Since it is also tangent to the line $y = 0$, its centre must be at height $r$ above that line.
• It touches each of the two big circles externally, so the distance from its centre to each big centre is $R + r = 4 + r$.

By symmetry, the third circle sits exactly midway between the two equal circles, so its centre $C$ lies on the perpendicular bisector of $AB$: $C = (4, r).$

Step 3 — Use the tangency condition to solve for $r$.

The distance $CA$ must equal $4 + r$: $CA^2 = (4 - 0)^2 + (r - 4)^2 = 16 + (r - 4)^2.$

Set this equal to $(4 + r)^2$: $16 + (r - 4)^2 = (4 + r)^2.$

Expand both sides: $16 + r^2 - 8r + 16 = 16 + 8r + r^2,$ $32 - 8r = 16 + 8r.$

Collect terms: $16 = 16r \quad\Rightarrow\quad r = 1.$

💡 Teacher tip: The key move is translating each geometric fact ("tangent to the line", "externally tangent to a big circle") into a precise algebraic statement about the centre's coordinates. Once the picture is coded into coordinates, the radius falls out of one equation.

Step 4 — Quick sanity check.

With $r = 1$, the centre is $C = (4, 1)$. $CA = \sqrt{4^2 + (1-4)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 = 4 + 1. \;\checkmark$ The distance equals $R + r$, confirming external tangency.

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

Alternative approach — using the tangent-point distance formula.

A useful fact: if two circles of radii $r_1$ and $r_2$ both touch a common line and touch each other externally, the distance between their points of contact with the line is $2\sqrt{r_1 r_2}.$

• For the two big circles ($r_1 = r_2 = 4$): the contact points are $2\sqrt{16} = 8$ apart — consistent with centres $8$ apart. ✓
• For the third circle ($r$) and one big circle ($4$): the contact points are $2\sqrt{4r} = 4\sqrt{r}$ apart.

By symmetry the third circle's contact point is the midpoint, so it is $4$ units from each big circle's contact point: $4\sqrt{r} = 4 \quad\Rightarrow\quad \sqrt{r} = 1 \quad\Rightarrow\quad r = 1.$

This confirms the same answer with less algebra.

💡 Why this is elegant: the answer $r = 1$ is independent of the specific value $R = 4$ in the sense that the same method works for any $R$ — the smaller circle's radius always comes out to $R/4$ when it nests symmetrically between two equal circles on a common tangent.

We need the condition on $A$ so that $x^2 - 4x - \log_2 A = 0$ has real and distinct roots.

Step 1 — Recall the discriminant test. For a quadratic $ax^2 + bx + c = 0$, the roots are real and distinct when the discriminant is strictly positive: $\Delta = b^2 - 4ac > 0.$ Here $a = 1$, $b = -4$, and $c = -\log_2 A$.

Step 2 — Compute the discriminant. $\Delta = (-4)^2 - 4(1)(-\log_2 A) = 16 + 4\log_2 A.$ For real, distinct roots: $16 + 4\log_2 A > 0.$

Step 3 — Solve the inequality. $4\log_2 A > -4 \implies \log_2 A > -1.$

💡 Teacher tip: When an inequality involves a logarithm, always check whether the log's base is greater than or less than $1$ — that tells you whether the inequality direction stays the same or flips. Since $\log_2$ has base $2 > 1$, the function is increasing, so the direction is preserved.

Step 4 — Convert to a condition on $A$. Because $\log_2$ is increasing: $\log_2 A > -1 \implies A > 2^{-1} = \frac{1}{2}.$

Wait — let me recheck Step 3 carefully. Dividing $16 + 4\log_2 A > 0$ by $4$: $4 + \log_2 A > 0 \implies \log_2 A > -4.$ So the correct conversion is: $A > 2^{-4} = \frac{1}{16}.$

Step 5 — Check the domain of the logarithm. $\log_2 A$ is defined only when $A > 0$. Since $\frac{1}{16} > 0$, the condition $A > \frac{1}{16}$ already guarantees the logarithm is defined. No extra restriction is needed.

Verification at the boundary:

The boundary $A = \frac{1}{16}$ gives a repeated root, so we need strict inequality $A > \frac{1}{16}$.

Answer: $\boxed{A > \dfrac{1}{16}}$ (Choice A)

💡 Closing insight: The key trap here is the arithmetic of dividing by $4$ — a careless $16/4$ can become $1$ instead of $4$, leading to $\frac{1}{2}$ or $\frac{1}{8}$. Always isolate the logarithm term cleanly before exponentiating.

Think of this as: two medians cross at right angles — that crossing point (the centroid) hands us a right triangle whose area we can compute instantly, and the whole triangle is just three copies of it.

Step 1 — Locate the centroid and split each median.

The medians $AD = 12$ and $BE = 9$ meet at the centroid $G$. A fundamental property of the centroid is that it divides every median in the ratio $2:1$, with the longer part toward the vertex.

So: $AG = \tfrac{2}{3}\times 12 = 8, \qquad BG = \tfrac{2}{3}\times 9 = 6.$

Step 2 — Use the perpendicularity to get a right triangle.

We are told the two medians are perpendicular, so $\angle AGB = 90^\circ$. Triangle $AGB$ is therefore a right triangle with legs $AG = 8$ and $BG = 6$: $\text{Area}(\triangle AGB) = \tfrac12 \times AG \times BG = \tfrac12 \times 8 \times 6 = 24.$

💡 Teacher tip: The centroid-to-vertex segments $GA$, $GB$, $GC$ split $\triangle ABC$ into three triangles of equal area: $\triangle AGB$, $\triangle BGC$, $\triangle CGA$. This is because each has the same height-from-centroid setup and equal bases are not required — the equality comes from the median property (each median bisects the opposite side, forcing all six small sub-triangles to have equal area, hence each of the three centroid-to-vertex triangles contains two of them).

Step 3 — Scale up to the full triangle.

Since $\triangle AGB$ is one of three equal-area pieces: $\text{Area}(\triangle ABC) = 3 \times \text{Area}(\triangle AGB) = 3 \times 24 = 72.$

Quick verification via a known formula. When two medians of lengths $m_1$ and $m_2$ are perpendicular, the area of the triangle is $\frac{2}{3}\,m_1 m_2$: $\text{Area} = \tfrac{2}{3}\times 12 \times 9 = \tfrac{2}{3}\times 108 = 72. \quad \checkmark$

Answer: $\boxed{72}$ (Choice C)

💡 Why this works so cleanly: the perpendicular medians give us a right triangle for free, and the centroid's $2:1$ split plus the equal-area partition turn one small right triangle into exactly one-third of the whole — no need to find any side of $\triangle ABC$.

Think of this as a salt-gram tracking problem: instead of juggling percentages through three transfers, we track exactly how many grams of salt each vessel holds at every step. Since strength is defined as grams of salt per 100 ml, this keeps everything concrete.

Initial setup (each vessel has 500 ml):

Using the definition that a $p\%$ solution has $p$ grams of salt per 100 ml, each 500 ml vessel holds $5p$ grams of salt.

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Step 1 — Transfer 100 ml from A to B

100 ml is $\frac{1}{5}$ of A's 500 ml, so it carries $\frac{1}{5}$ of A's salt: $\text{Salt transferred} = \frac{100}{500} \times 50 = 10 \text{ g}$

B's new strength: $\frac{120}{600} \times 100 = 20\%$.

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Step 2 — Transfer 100 ml from B to C

B is now at 20%, so 100 ml carries $20\% \times 100 = 20$ g of salt.

C's new strength: $\frac{180}{600} \times 100 = 30\%$.

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Step 3 — Transfer 100 ml from C to A

C is now at 30%, so 100 ml carries $30\% \times 100 = 30$ g of salt.

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Step 4 — Final strength of A

A now has 70 g of salt in 500 ml of solution: $\text{Strength of A} = \frac{70}{500} \times 100 = 14\%$

💡 Teacher tip: Notice that after all three transfers, every vessel is back to 500 ml — each vessel gives away 100 ml once and receives 100 ml once. Only the salt content has shifted. Tracking grams of salt (not percentages) is the cleanest way through any multi-step mixture problem.

Answer: $\boxed{14}$ (Choice D)

Closing insight: The three transfers form a cycle (A→B→C→A), so each vessel ends at its original volume. The net effect is that A lost 10 g to B but gained 30 g from C — a net gain of 20 g — turning its original 50 g into 70 g, which is exactly $14\%$ of 500 ml.

Think of this as: we know the total books in 2010 and 2015, and we know the percentage increase for each category. We need to work backwards to split the 2010 total into fiction and non-fiction, then apply the fiction increase.

Step 1 — Set up variables for 2010.

Let the number of fiction books in 2010 be $F$, and non-fiction books be $N$.

We're told the 2010 total is 11500:

$F + N = 11500 \quad \cdots (1)$

Step 2 — Express the 2015 total.

Fiction grows by 10%, so 2015 fiction $= 1.10\,F$.

Non-fiction grows by 12%, so 2015 non-fiction $= 1.12\,N$.

The 2015 total is 12760:

$1.10\,F + 1.12\,N = 12760 \quad \cdots (2)$

Step 3 — Solve the system.

From equation (1), substitute $N = 11500 - F$ into equation (2):

$1.10\,F + 1.12\,(11500 - F) = 12760$

$1.10\,F + 12880 - 1.12\,F = 12760$

$-0.02\,F = 12760 - 12880$

$-0.02\,F = -120$

$F = \frac{-120}{-0.02} = 6000$

So there were 6000 fiction books in 2010.

Step 4 — Find fiction books in 2015.

Apply the 10% increase:

$\text{Fiction in 2015} = 1.10 \times 6000 = 6600$

💡 Teacher tip: The key move is realising you have two unknowns ($F$ and $N$) and two equations — one from the 2010 total, one from the 2015 total. Once you spot that, it's a straightforward simultaneous-equations problem.

Quick check: If $F = 6000$, then $N = 5500$. Verify the 2015 total: $1.10 \times 6000 + 1.12 \times 5500 = 6600 + 6160 = 12760$ ✓

Answer: $\boxed{6600}$ (Choice B)

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Alternative approach — "What if everything grew by 10%?"

Suppose both categories had grown by 10%. Then the 2015 total would have been:

$11500 \times 1.10 = 12650$

But the actual 2015 total is 12760, which is 110 more than this hypothetical. Where does the extra 110 come from? It's the additional 2% increase that non-fiction got (12% instead of 10%):

$0.02 \times N = 110 \implies N = 5500$

So $F = 11500 - 5500 = 6000$, and fiction in 2015 $= 1.10 \times 6000 = 6600$.

💡 This method is faster under exam time — it avoids solving simultaneous equations by anchoring on a single "what if" scenario and isolating the difference.

We're told the quadratic $x^2 + bx + c = 0$ has roots $4a$ and $3a$ (with $a$ an integer), and we need to find which option could equal $b^2 + c$.

Step 1 — Express $b$ and $c$ in terms of $a$ using Vieta's formulas.

For a quadratic $x^2 + bx + c = 0$ with roots $r_1$ and $r_2$:

• Sum of roots: $r_1 + r_2 = -b$
• Product of roots: $r_1 \cdot r_2 = c$

Substituting $r_1 = 4a$ and $r_2 = 3a$:

$-b = 4a + 3a = 7a \;\Rightarrow\; b = -7a$

$c = (4a)(3a) = 12a^2$

Step 2 — Build the expression $b^2 + c$.

$b^2 + c = (-7a)^2 + 12a^2 = 49a^2 + 12a^2 = 61a^2$

💡 Teacher tip: Since $a$ is an integer, $a^2$ must be a perfect square. So $b^2 + c = 61a^2$ means the correct option must be $61$ times a perfect square. That's our filter.

Step 3 — Test each option by dividing by $61$.

Step 4 — Verify the valid case.

If $61a^2 = 549$, then $a^2 = 9$, so $a = \pm 3$ — which is indeed an integer. ✓

Answer: $\boxed{549}$ (Choice D)

💡 Closing insight: Whenever a quadratic's roots are given as multiples of a single variable, Vieta's formulas collapse both $b$ and $c$ into expressions of that one variable — turning the problem into a simple divisibility/perfect-square check.

We need to find how many pairs of positive integers $(m, n)$ satisfy $m^2 + 105 = n^2$.

Think of this as: we're looking for two perfect squares that are exactly 105 apart — and we want to count how many such pairs exist with both values positive.

Step 1 — Rearrange into a difference of squares.

Move $m^2$ to the other side: $n^2 - m^2 = 105$

Now apply the difference-of-squares factorisation: $(n - m)(n + m) = 105$

💡 Teacher tip: Whenever you see $n^2 - m^2 = \text{constant}$, always factor it as $(n-m)(n+m)$. This converts a Diophantine equation into a factorisation problem — which is far easier to handle.

Step 2 — Establish the properties of the two factors.

Since $n^2 = m^2 + 105 > m^2$ and both are positive, we know $n > m > 0$. This tells us:

• Both factors are positive: $n - m > 0$ and $n + m > 0$.
• Both factors have the same parity: $(n - m)$ and $(n + m)$ are either both even or both odd. (Adding or subtracting the same two integers always gives results of the same parity — check: if $m, n$ are both odd, both factors are even; if one is odd and one even, both factors are odd; if both even, both factors are even.)

Since $105$ is odd, both factors must be odd. This means every factorisation of 105 into two positive factors is automatically valid — no factor pair gets rejected on parity grounds.

Step 3 — List all factor pairs of 105.

First, factorise: $105 = 3 \times 5 \times 7$.

The divisors of 105 are: $1, 3, 5, 7, 15, 21, 35, 105$.

Since $n - m < n + m$ (because $m > 0$), we assign the smaller factor to $n - m$ and the larger to $n + m$. This gives four ordered factor pairs:

Step 4 — Recover $m$ and $n$ from each pair.

From the system $n - m = a$ and $n + m = b$: $n = \frac{a + b}{2}, \qquad m = \frac{b - a}{2}$

Since both $a$ and $b$ are odd, $a + b$ and $b - a$ are both even, so $m$ and $n$ are always integers. And since $a < b$, we get $m > 0$ automatically.

Step 5 — Verify each pair.

All four pairs check out.

Step 6 — Count.

There are exactly 4 valid pairs of positive integers.

Answer: $\boxed{4}$

💡 Why this works so cleanly: The number 105 is odd, which forces both factors to be odd — so every factor pair survives. If 105 were even, some factor pairs would have mismatched parity and would be rejected, reducing the count. The number of solutions equals the number of ways to write 105 as a product of two positive factors of the same parity — and for an odd number, that's just half the number of divisors, i.e. $\frac{8}{2} = 4$.

Think of this as: we know how the side counts of two regular polygons relate, and how their interior angles relate — that's enough to pin down both polygons, and then we just read off the angle of a third one.

Step 1 — Recall the interior-angle formula.

For a regular polygon with $n$ sides, each interior angle is:

$\text{Interior angle} = \frac{(n-2)\times 180^\circ}{n} = 180^\circ - \frac{360^\circ}{n}$

The second form — $180 - \frac{360}{n}$ — is often handier because it separates the "straight line" part from the "turn" part.

Step 2 — Write the angles for A and B.

Polygon $A$ has $a$ sides, so each interior angle is: $\alpha = 180 - \frac{360}{a}$

Polygon $B$ has $b = 2a$ sides, so each interior angle is: $\beta = 180 - \frac{360}{2a} = 180 - \frac{180}{a}$

Step 3 — Apply the angle relationship $\beta = \frac{3}{2}\,\alpha$.

Substitute the expressions from Step 2:

$180 - \frac{180}{a} = \frac{3}{2}\left(180 - \frac{360}{a}\right)$

Expand the right side:

$180 - \frac{180}{a} = 270 - \frac{540}{a}$

Now gather the $\frac{1}{a}$ terms on one side and the constants on the other:

$\frac{540}{a} - \frac{180}{a} = 270 - 180$

$\frac{360}{a} = 90$

$a = 4$

💡 Teacher tip: Whenever you see $\frac{360}{n}$ in a polygon problem, think exterior angle. The equation $\frac{360}{a} = 90$ is literally saying: the exterior angle of polygon $A$ is $90^\circ$ — which immediately tells you $A$ is a square.

Step 4 — Identify the polygons.

• $a = 4$: polygon $A$ is a square (interior angle $90^\circ$).
• $b = 2a = 8$: polygon $B$ is a regular octagon (interior angle $135^\circ$).
• Check: $\frac{3}{2} \times 90 = 135$ ✓

So a regular polygon with $a + b = 4 + 8 = 12$ sides is a regular dodecagon.

Step 5 — Compute the interior angle of the 12-gon.

$\frac{(12-2)\times 180}{12} = \frac{10 \times 180}{12} = \frac{1800}{12} = 150^\circ$

Answer: $\boxed{150}$

💡 Closing insight: The condition $\beta = \frac{3}{2}\alpha$ combined with $b = 2a$ forces $A$ to be a square — the simplest regular polygon. Once $a = 4$ falls out, everything else is just arithmetic. The key habit here is reaching for the $180 - \frac{360}{n}$ form of the interior-angle formula, which makes the algebra far cleaner than working with $\frac{(n-2)\,180}{n}$ directly.

We're given a sum of terms that all look like $2n + \text{(odd number)}$, and we need to work backwards to find $n$, then compute $1 + 2 + \cdots + n$.

Step 1 — Recognise the structure.

The terms $2n+1,\; 2n+3,\; 2n+5,\; \ldots,\; 2n+47$ form an arithmetic progression with:

• First term $a = 2n+1$
• Common difference $d = 2$
• Last term $l = 2n+47$

Step 2 — Count the number of terms.

Using the AP term-count formula: $\text{Number of terms} = \frac{l - a}{d} + 1 = \frac{(2n+47) - (2n+1)}{2} + 1 = \frac{46}{2} + 1 = 23 + 1 = 24.$

So there are 24 terms in the sum.

💡 Teacher tip: Notice that $n$ cancels out when computing the number of terms — the count depends only on the odd numbers $1, 3, 5, \ldots, 47$, not on $n$.

Step 3 — Apply the AP sum formula.

$\text{Sum} = \frac{\text{number of terms}}{2} \times (\text{first term} + \text{last term})$

$5280 = \frac{24}{2}\bigl((2n+1) + (2n+47)\bigr)$

$5280 = 12\,(4n + 48)$

$5280 = 48n + 576$

Step 4 — Solve for $n$.

$48n = 5280 - 576 = 4704$

$n = \frac{4704}{48} = 98.$

Step 5 — Compute $1 + 2 + 3 + \cdots + n$.

Using the triangular number formula: $1 + 2 + \cdots + n = \frac{n(n+1)}{2} = \frac{98 \times 99}{2} = 49 \times 99.$

$49 \times 99 = 49 \times (100 - 1) = 4900 - 49 = 4851.$

Answer: $\boxed{4851}$

💡 Closing insight: The key move was spotting that the given sum is an AP with common difference $2$. Once you count the terms correctly (24, independent of $n$), the rest is straightforward algebra. Always check whether a sequence of expressions in $n$ hides a simple AP or GP underneath.

We need to find how many terms appear in both sequences simultaneously.

Step 1 — Identify each sequence as an arithmetic progression.

• Sequence 1: $15, 19, 23, 27, \ldots, 415$ — first term $15$, common difference $d_1 = 4$.
• Sequence 2: $14, 19, 24, 29, \ldots, 464$ — first term $14$, common difference $d_2 = 5$.

Step 2 — Find the first common term.

Writing out the first few terms of each:

The first term common to both is $\mathbf{19}$.

Step 3 — Determine the pattern of common terms.

💡 Teacher tip: If a number $t$ is in both APs, then $t + \text{lcm}(d_1, d_2)$ is also in both — because adding a multiple of $4$ keeps you in Sequence 1, and adding a multiple of $5$ keeps you in Sequence 2. The smallest such step is $\text{lcm}(4,5)$.

$\text{lcm}(4, 5) = 20$

So the common terms themselves form an AP:

$19,\; 39,\; 59,\; 79,\; \ldots$

with first term $a = 19$ and common difference $D = 20$.

Step 4 — Find the upper bound for common terms.

A common term must appear in both sequences, so it cannot exceed the smaller of the two last terms:

$\min(415,\; 464) = 415$

We need the largest $k$ such that the $k$-th common term $\le 415$:

$19 + 20(k-1) \le 415$ $20(k-1) \le 396$ $k - 1 \le \frac{396}{20} = 19.8$ $k \le 20.8$

So $k = 20$.

Step 5 — Verify the boundary.

The 20th common term ($399$) lies within both sequences; the 21st ($419$) exceeds Sequence 1's last term ($415$), so it is excluded.

Answer: $\boxed{20}$ (Choice B)

💡 Closing insight: The common terms of two APs always form a new AP whose common difference is the LCM of the two original differences. Once you spot that, the problem reduces to a single counting step — find the first common term, then count how many steps of size $\text{lcm}(d_1, d_2)$ fit before hitting the smaller last term.

We're given an alternating-sign sum that always equals $n$, and we need the plain (non-alternating) sum of $a_{51}$ through $a_{1023}$. The key is to first pin down exactly what each $a_k$ is.

Step 1 — Extract the individual terms.

Define the given alternating sum as: $S_n = a_1 - a_2 + a_3 - a_4 + \cdots + (-1)^{n-1}a_n = n.$

The last term of $S_n$ is $(-1)^{n-1}a_n$, and the last term of $S_{n-1}$ is $(-1)^{n-2}a_{n-1}$. So subtracting: $S_n - S_{n-1} = (-1)^{n-1}a_n.$

But from the given condition, $S_n - S_{n-1} = n - (n-1) = 1$. Therefore: $(-1)^{n-1}a_n = 1 \quad\Longrightarrow\quad a_n = (-1)^{n-1}.$

💡 Teacher tip: This is the standard "telescoping difference" trick for sequence problems: if you know a running sum $S_n$ for every $n$, then $a_n = S_n - S_{n-1}$ (with the sign adjustment here). You don't need to guess the pattern — you can derive it in one line.

Step 2 — Spell out the pattern.

Quick sanity check with the first few: $a_1=1,\; a_2=-1,\; a_3=1,\; a_4=-1$, giving $S_1=1,\; S_2=1-(-1)=2,\; S_3=2+1=3,\; S_4=3-(-1)=4$. ✓

**Step 3 — Count odd vs even indices in $[51, 1023]$.

Both endpoints $51$ and $1023$ are odd, so the range starts and ends on an odd index.

• Odd indices $51, 53, \ldots, 1023$: count $= \dfrac{1023 - 51}{2} + 1 = \dfrac{972}{2} + 1 = 487$, each contributing $+1$.
• Even indices $52, 54, \ldots, 1022$: count $= \dfrac{1022 - 52}{2} + 1 = \dfrac{970}{2} + 1 = 486$, each contributing $-1$.

Check: $487 + 486 = 973 = 1023 - 51 + 1$. ✓

Step 4 — Compute the sum. $a_{51} + a_{52} + \cdots + a_{1023} = 487(1) + 486(-1) = 487 - 486 = 1.$

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

💡 Why it's 1 and not 0: Whenever a range of consecutive integers starts and ends on the same parity, the two counts (odd vs even) differ by exactly one. Here both ends are odd, so odd indices win by one — giving a sum of $+1$. Had both ends been even, the sum would have been $-1$.

We need to count the factors of $2^4 \times 3^5 \times 10^4$ that are perfect squares and greater than 1.

Think of this as: first express the number in clean prime-factor form, then count how many of its factors have only even exponents (the hallmark of a perfect square), and finally remove the trivial factor $1$.

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Step 1 — Prime-factorise the number.

The key move is to break $10^4$ into primes before combining:

$10^4 = (2 \times 5)^4 = 2^4 \times 5^4.$

Now combine everything:

$2^4 \times 3^5 \times 10^4 = 2^4 \times 3^5 \times 2^4 \times 5^4 = 2^8 \times 3^5 \times 5^4.$

So the number is $2^8 \cdot 3^5 \cdot 5^4$.

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Step 2 — When is a factor a perfect square?

Any factor of $2^8 \cdot 3^5 \cdot 5^4$ looks like $2^a \cdot 3^b \cdot 5^c$, where

$0 \le a \le 8,\quad 0 \le b \le 5,\quad 0 \le c \le 4.$

💡 Teacher tip: A number is a perfect square if and only if every prime in its factorisation has an even exponent. So we need $a$, $b$, and $c$ to each be even — not the factor itself to "look square."

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Step 3 — Count the even-exponent choices for each prime.

Note carefully: for $3$, the exponent can go up to $5$, but among even numbers $\le 5$ we only get $\{0, 2, 4\}$ — $6$ would exceed the limit.

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Step 4 — Multiply the independent choices.

Since the exponents for different primes are chosen independently:

$\text{Total perfect-square factors} = 5 \times 3 \times 3 = 45.$

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Step 5 — Exclude the trivial factor $1$.

The combination $a = 0,\, b = 0,\, c = 0$ gives the factor $2^0 \cdot 3^0 \cdot 5^0 = 1$, which is technically a perfect square but the question asks for squares greater than 1.

$45 - 1 = 44.$

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Answer: $\boxed{44}$

💡 Closing insight: The subtraction of $1$ at the end is the classic "off-by-one" in factor-counting problems. Whenever a question says greater than 1 (or non-trivial), build the full count first, then remove the all-zero-exponent case — it is always exactly one factor.

Think of this as: we know how long the cyclist takes, and we can use the "45 motorcycles by 11 am" clue to figure out how fast the motorcycles are. Once we know that, we just re-run the clock with a faster cyclist.

Step 1 — Cyclist's normal trip.

The cyclist leaves A at 10:00 am and reaches B at 11:00 am, so the A-to-B journey takes him 60 minutes.

Step 2 — Find each motorcycle's travel time.

Motorcycles leave A starting at 10:01 am, one every minute — so they depart at minutes $k = 1, 2, 3, \ldots$ after 10:00 am. Since they all travel at the same speed over the same distance, each motorcycle has the same travel time $T$ minutes.

Motorcycle $k$ (leaving at minute $k$) arrives at B at minute $k + T$.

It reaches B by 11:00 am (minute 60) if and only if: $k + T \le 60 \quad\Longrightarrow\quad k \le 60 - T$

We are told that 45 motorcycles reach B by 11:00 am. That means motorcycles $k = 1, 2, \ldots, 45$ all qualify, but motorcycle $k = 46$ does not. So: $60 - T = 45 \quad\Longrightarrow\quad T = 15 \text{ minutes}$

💡 Teacher tip: The 45th motorcycle leaves at 10:45 am and just barely makes it to B by 11:00 am. That single observation tells you the travel time: $11{:}00 - 10{:}45 = 15$ minutes. No algebra needed — just read off the boundary.

So each motorcycle takes 15 minutes to go from A to B, which means a motorcycle is $4\times$ as fast as the cyclist ($60/15 = 4$).

Step 3 — Doubled-speed cyclist.

If the cyclist doubles his speed, his travel time is halved: $\text{New cyclist time} = \frac{60}{2} = 30 \text{ minutes}$

He now reaches B at 10:30 am (minute 30).

Step 4 — Count motorcycles arriving by 10:30 am.

Motorcycle $k$ arrives at minute $k + 15$. It has reached B by 10:30 am (minute 30) if: $k + 15 \le 30 \quad\Longrightarrow\quad k \le 15$

So motorcycles $k = 1, 2, \ldots, 15$ — that is, the ones leaving at 10:01 through 10:15 — all arrive by 10:30 am.

Verification of the boundary:

The count is exactly 15.

Answer: $\boxed{15}$ (Choice C)

💡 Why this works so cleanly: The motorcycle's 15-minute travel time and the doubled-speed cyclist's 30-minute travel time share a simple relationship — $30 - 15 = 15$. The answer is just the number of departure minutes that fit in that gap. Whenever a problem gives you "$N$ vehicles arrive by time $X$," find the travel time from the boundary vehicle first, then re-apply it to the new scenario.