M.4 MPT 2024 — Questions and Worked Solutions

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Source: FPSC MPT 2024 — no official key. Questions 1, 2 and 4 are fully re-derived below. Questions 3 and 5 are only partly present in the extraction, so the method is shown and the printed answer is quoted as extracted.

Q1 · Exponents

Question: Evaluate $4^{\frac12}\times4^{2}-4^{0}$.

Options: (A) 24 (B) 28 (C) 31 (D) None of these

✅ Answer: (C) 31

Solution:

$$4^{\frac12}\times4^{2}-4^{0}=2\times16-1=31$$

Concept: $4^{\frac12}=\sqrt4=2$, $4^{2}=16$, $4^{0}=1$. Repeated in 2026 with the same numbers.

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Q2 · Exponent equation

Question: $27^{k}=9$. Find $k$.

Options: (A) 1/3 (B) 2/3 (C) 4/3 (D) None of these

✅ Answer: (B) 2/3

Solution:

$$27^{k}=9\;\Rightarrow\;(3^{3})^{k}=3^{2}\;\Rightarrow\;3k=2\;\Rightarrow\;k=\frac23$$

Concept: write both sides as powers of the same base, then equate the exponents.

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Q3 · Food supply (cows and days)

Question: A stock of food lasts a herd for a number of days; if the herd is reduced by 10 cows, the food lasts 10 more days.

Options: (A) 150 (B) 160 (C) 100 (D) None of these

✅ Answer as extracted: (B) 160

Method: let $N$ cows eat the stock in $d$ days. Total food stays constant.

$$\begin{aligned} Nd&=(N-10)(d+10)\\ Nd&=Nd+10N-10d-100\\ N-d&=10 \end{aligned}$$

Check with the answer: $N=160,\ d=150$ gives $160\times150=24\,000=150\times160$ ✅.

🔎 The extraction omits the second condition that fixes $N$; use the relation $N-d=10$ plus that condition.

Concept: "man-days" (or animal-days) are constant: $M_1D_1=M_2D_2$.

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Q4 · Train crossing a station

Question: A 99 m train passes a 231 m station in 11 s. Find the speed in km/h.

Options: (A) 108 (B) 75.6 (C) 30 (D) None of these

✅ Answer: (A) 108 km/h

Solution:

$$\begin{aligned} v&=\frac{99+231}{11}=\frac{330}{11}=30\ \text{m/s}\\ v&=30\times\frac{18}{5}=108\ \text{km/h} \end{aligned}$$

Concept: distance covered = train length + station length. Option (C) 30 is the unconverted m/s trap.

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Q5 · Sharing money among A, B, C

Question: A, B and C share a total of 112 under given conditions; find the required share.

Options: (A) 34 (B) 44 (C) 50 (D) None of these

✅ Answer as extracted: (C) 50

Method: express the three shares in one variable and set their sum to 112.

$$x+(x+a)+(x+b)=112\;\Rightarrow\;3x=112-a-b$$

🔎 Conditions are not in the extraction; the same item returned in 2026 with answer 50.

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Q96–100 · Quadratic and algebra word problems

Pattern: form a quadratic from the story, factorise, reject impossible roots.

$$ax^{2}+bx+c=0\;\Rightarrow\;x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$

Typical story: product of two consecutive numbers, age products, area of a rectangle with a given perimeter.

Rule: a length, age or count cannot be negative — discard negative roots.

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Q101–149 · Mixed block (maths part)

• Ratio and proportion: $a:b$ sharing, adding to terms, time and work ratios.
• Ages: equations built from "$N$ years ago / later".
• Work rate: $\frac1T=\frac1a+\frac1b$; men-days.
• Geometry: perimeter, area, angle sums.
• Probability: $P(E)=\frac{\text{favourable}}{\text{total}}$.
• Sequences: differences, ratios, alternating patterns.
• Logic-word problems: see the Analytical / Reasoning page for the non-maths items in this range.