M.1 Formula Rule Book — All Quant Types as Equations

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How to use: learn each rule as an equation, redo the practice variant on paper, then check the answer. Every MPT quantitative MCQ in 2022–2026 falls into one of the types below.

1️⃣ Percentages — "sold X%, Y left"

Rule

$$\text{Original}=\frac{R}{1-\frac{a}{100}}$$

• $a\%$ sold or spent, $R$ remains.
• Percentage change:

$$\text{Change}\,\%=\frac{\text{New}-\text{Old}}{\text{Old}}\times100$$

• Successive changes (up $a\%$ then down $b\%$):

$$\text{Factor}=\left(1+\frac{a}{100}\right)\left(1-\frac{b}{100}\right)$$

Past paper (2023): 40% sold, 420 left

$$\text{Original}=\frac{420}{1-0.40}=\frac{420}{0.6}=700$$

Practice: 25% of stock sold, 1500 left. Find the original stock.

$$\frac{1500}{0.75}=2000$$

⚡ Shortcut: remaining fraction is $1-\frac{a}{100}$, so divide by it. ⚠️ Trap: a 20% rise followed by a 20% fall is $1.2\times0.8=0.96$, i.e. a 4% fall, not 0%.

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2️⃣ Ratio and proportion

Sharing a total $T$ in ratio $a:b$

$$\text{First part}=\frac{a}{a+b}\times T,\qquad \text{Second part}=\frac{b}{a+b}\times T$$

Adding $x$ to both terms of $a:b$ to get $c:d$

$$\frac{a+x}{b+x}=\frac{c}{d}\;\Rightarrow\;x=\frac{ad-bc}{c-d}$$

If the ratio of TIME is $a:b$ and the first takes $t$

$$t_{2}=t\times\frac{b}{a}$$

Past paper (2025): boys : girls = 9 : 5, total 1050

$$\text{Boys}=\frac{9}{14}\times1050=675$$

Past paper (2025): add $x$ to 7 : 13 to get 2 : 3

$$x=\frac{7\cdot3-13\cdot2}{2-3}=\frac{21-26}{-1}=5$$

Practice: time ratio 4 : 3, the second person takes 60 min.

$$t_{1}=60\times\frac{4}{3}=80\ \text{min}$$

⚠️ Trap: always write which quantity the ratio refers to (time vs speed). Time and speed are inversely proportional.

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3️⃣ Ages

Rules

• Present ages are variables; "$N$ years ago" means $x-N$; "$N$ years later" means $x+N$.
• After $t$ years the sum of two ages rises by $2t$.
• The difference between two ages never changes.

Past paper (2026): Zaid is 6 times Zoya's age; in 7 years the sum is 49

$$\begin{aligned} (6x+7)+(x+7)&=49\\ 7x&=35\\ x&=5 \end{aligned}$$

So Zoya is 5, Zaid is 30 and the age difference is 25.

Past paper (2025): Ali is 40 when Asim is born. When is Ali 5 times Asim's age?

$$40+A=5A\;\Rightarrow\;A=10\;\Rightarrow\;\text{Ali}=50$$

Practice: a father is 4 times his son's age; in 5 years the sum of their ages is 50.

$$(4x+5)+(x+5)=50\;\Rightarrow\;x=8,\ \text{father}=32$$

⚠️ Trap: highest error rate on the page. Write the equation in full; use the constant age difference to check.

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4️⃣ Mixtures

Rules

• Track the total and each component separately.
• Removing a part of a mixture removes components in the same ratio.
• Adding a pure component changes only that component.

Method for "add water to change the ratio" (alcohol : water = $p:q$, add $w$ litres of water, new ratio $m:n$)

$$\frac{pk}{qk+w}=\frac{m}{n}$$

Past paper (2025, numbers as extracted): alcohol : water = 12 : 5, add 14 L water, ratio becomes 4 : 3

$$\frac{12k}{5k+14}=\frac{4}{3}\;\Rightarrow\;36k=20k+56\;\Rightarrow\;k=3.5\;\Rightarrow\;\text{Alcohol}=42\ \text{L}$$

⚠️ The extracted key says 24 L: the numbers as listed give 42 L, so check the booklet wording.

Practice: a 1 L mixture has milk : water = 2 : 3. Remove 200 ml, then add pure milk to make the ratio 1 : 1.

Key idea: when you remove some amount from a well-mixed mixture, you remove both ingredients in the same ratio as the mixture.

Start with 1 L = 1000 ml mixture.

Milk : Water = 2 : 3

Total parts:

$$2+3=5$$

So initially:

$$\text{Milk}=\frac{2}{5}\times1000=400\text{ ml}$$ $$\text{Water}=\frac{3}{5}\times1000=600\text{ ml}$$

Now 200 ml mixture is removed.

Because the mixture is in the ratio (2:3), the removed 200 ml also contains milk and water in the same ratio.

$$\text{Milk removed}=\frac{2}{5}\times200=80\text{ ml}$$ $$\text{Water removed}=\frac{3}{5}\times200=120\text{ ml}$$

So the remaining quantities are:

$$\text{Milk}=400-80=320\text{ ml}$$ $$\text{Water}=600-120=480\text{ ml}$$

Now add pure milk until:

$$\text{Milk : Water}=1:1$$

Water remains:

$$480\text{ ml}$$

For a 1:1 ratio, milk must also be:

$$480\text{ ml}$$

Current milk is:

$$320\text{ ml}$$

Therefore milk to be added:

$$480-320=160\text{ ml}$$

Hence:

$$\boxed{160\text{ ml}}$$ $$\begin{aligned} \text{Milk}&=400,\ \text{Water}=600\\ \text{After removing 200 ml}&:\ \text{Milk}=400-80=320,\ \text{Water}=600-120=480\\ 320+x&=480\;\Rightarrow\;x=160\ \text{ml} \end{aligned}$$

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5️⃣ Exponents and surds

$$a^{m}\cdot a^{n}=a^{m+n},\qquad \frac{a^{m}}{a^{n}}=a^{m-n},\qquad (a^{m})^{n}=a^{mn}$$ $$a^{0}=1,\qquad a^{-n}=\frac{1}{a^{n}},\qquad a^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^{m}$$

Past paper (2022, 2025):

$$x^{\frac12}\cdot x^{-\frac12}=x^{\frac12-\frac12}=x^{0}=1$$

Past paper (2024, 2026):

$$4^{\frac12}\times4^{2}-4^{0}=2\times16-1=31$$ $$27^{k}=9\;\Rightarrow\;3^{3k}=3^{2}\;\Rightarrow\;k=\frac23$$

Square root of a negative number

$$\sqrt{-25}=\sqrt{25}\cdot\sqrt{-1}=5i$$

Practice:

$$9^{\frac12}\times9^{-\frac12}+9^{0}=1+1=2$$

⚠️ Trap: $x^{\frac12}\cdot x^{-\frac12}$ is $1$, not $0$; mis-keyed by FPSC in 2025.

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6️⃣ Standard form

$$N=a\times10^{n},\qquad 1\le a<10$$

Past paper (2023):

$$0.0000738=7.38\times10^{-5}$$

Practice:

$$45\,000\,000=4.5\times10^{7}$$

Rule: moving the decimal point $n$ places right gives $10^{-n}$; moving left gives $10^{+n}$.

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7️⃣ Geometry formulas

$$\text{Circle: }C=2\pi r,\qquad A=\pi r^{2}$$ $$\text{Semicircle: }P=\pi r+2r,\qquad A=\tfrac12\pi r^{2}$$ $$\text{Cylinder: }V=\pi r^{2}h,\qquad \text{TSA}=2\pi r(r+h)$$ $$\text{Cuboid: }V=l\,w\,h,\qquad 1\ \text{m}^{3}=1000\ \text{litres}$$ $$\text{Regular }n\text{-gon: exterior angle}=\frac{360^{\circ}}{n},\qquad \text{interior sum}=(n-2)\times180^{\circ}$$

Past paper (2026): exterior angle 36°

$$n=\frac{360}{36}=10\ (\text{decagon})$$

Past paper (2026): cylinder $r=3.5,\ h=5.5$ with $\pi=\frac{22}{7}$

$$\text{TSA}=2\cdot\frac{22}{7}\cdot3.5\cdot(3.5+5.5)=198\ \text{cm}^{2}$$

Practice: exterior angle 45°

$$n=\frac{360}{45}=8\ (\text{octagon})$$

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8️⃣ Sets and Venn diagrams

$$|A\cup B|=|A|+|B|-|A\cap B|$$ $$\text{Neither}=N-|A\cup B|$$ $$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|A\cap C|+|A\cap B\cap C|$$

Past paper (2026): $N=200$, tennis 90, cricket 108, both 46

$$|A\cup B|=90+108-46=152\;\Rightarrow\;\text{Neither}=200-152=48$$

Practice: a class of 50, cricket 30, football 25, both 10

$$50-(30+25-10)=5$$

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9️⃣ Speed, distance and time

$$v=\frac{d}{t},\qquad \text{km/h}\to\text{m/s}:\ \times\frac{5}{18},\qquad \text{m/s}\to\text{km/h}:\ \times\frac{18}{5}$$ $$\text{Average speed}=\frac{\text{total distance}}{\text{total time}}$$

Train crossing a platform: distance = length of train + length of platform.

Past paper (2024, 2026): 99 m train, 231 m station, 11 s

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

Practice: 120 m train crosses a 280 m bridge in 20 s

$$v=\frac{400}{20}=20\ \text{m/s}=72\ \text{km/h}$$

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🔟 Simple interest

• P = Principal → the original amount of money
• R = Rate → interest rate per year
• T = Time → number of years

$$SI=\frac{P\,R\,T}{100},\qquad A=P+SI$$

Two amounts at two times: the difference of amounts over the difference of times is one year's interest.

$$SI_{\text{1 yr}}=\frac{A_{2}-A_{1}}{t_{2}-t_{1}},\qquad P=A_{1}-t_{1}\cdot SI_{\text{1 yr}}$$

Past paper (2023): Rs. 815 in 3 years and Rs. 854 in 4 years

$$SI_{\text{1 yr}}=39,\qquad P=815-3\times39=698$$

Note: the 2023 key is 737, which fits Rs. 815 in 2 years and Rs. 854 in 3 years: $815-2\times39=737$.

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1️⃣1️⃣ Work and time

$$\text{Rate}=\frac{1}{\text{days}},\qquad \frac{1}{T}=\frac{1}{a}+\frac{1}{b}$$ $$M_{1}D_{1}=M_{2}D_{2}\qquad(\text{same work})$$

Practice: 12 men finish in 15 days. How long for 18 men?

$$D_{2}=\frac{12\times15}{18}=10\ \text{days}$$

Practice: A takes 10 days, B takes 15 days. Together?

$$\frac{1}{T}=\frac1{10}+\frac1{15}=\frac16\;\Rightarrow\;T=6\ \text{days}$$

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1️⃣2️⃣ Algebra and quadratics

$$ax^{2}+bx+c=0\;\Rightarrow\;x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$ $$x_{1}+x_{2}=-\frac{b}{a},\qquad x_{1}x_{2}=\frac{c}{a}$$ $$a^{2}+b^{2}=(a+b)^{2}-2ab,\qquad (a-b)^{2}=(a+b)^{2}-4ab$$

Completing the square

$$x^{2}+bx+c=\left(x+\frac{b}{2}\right)^{2}-\frac{b^{2}}{4}+c$$

Past paper (2026): $a+b=5,\ ab=6$

$$a^{2}+b^{2}=25-12=13$$

Past paper (2026): $y=x^{2}+8x-5$

$$y=(x+4)^{2}-21$$

Discriminant: $b^{2}-4ac$ is a perfect square only when roots are rational; if not a perfect square, no integer roots.

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1️⃣3️⃣ Averages

$$\bar{x}=\frac{\sum x}{n}$$

• For evenly spaced numbers the average equals the middle term.

Practice: 5 children born 2 years apart have average age 18.

$$x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20\;\Rightarrow\;\bar{x}=x+4=18\;\Rightarrow\;x=14$$

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1️⃣4️⃣ Series and patterns

$$\text{AP: }a_{n}=a+(n-1)d,\qquad S_{n}=\frac{n}{2}\left[2a+(n-1)d\right]$$ $$\text{GP: }a_{n}=a\,r^{n-1}$$

• Method: write the differences; if they are not constant, take differences again, test ratios, squares, cubes, alternating patterns.

Past paper (2026): 11, 20, 38, 56, 83 → differences 9, 18, 18, 27 (pattern $9\times1,\ 9\times2,\ 9\times2,\ 9\times3$).

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1️⃣5️⃣ Probability

$$P(E)=\frac{\text{favourable}}{\text{total}},\qquad P(\text{not }E)=1-P(E)$$ $$P(\text{neither }A\text{ nor }B)=1-P(A\cup B)$$

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1️⃣6️⃣ Inequalities

• Multiplying or dividing by a negative number flips the sign.

Past paper (2026): $-7<4n\le8$

$$-\frac{7}{4}<n\le2\;\Rightarrow\;n\in\{-1,0,1,2\}$$

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📋 Quick reference lists

• Squares: $11^{2}=121,\ 12^{2}=144,\ 13^{2}=169,\ 14^{2}=196,\ 15^{2}=225,\ 16^{2}=256,\ 17^{2}=289,\ 18^{2}=324,\ 19^{2}=361,\ 20^{2}=400$
• Cubes: $2^{3}=8,\ 3^{3}=27,\ 4^{3}=64,\ 5^{3}=125,\ 6^{3}=216,\ 7^{3}=343,\ 8^{3}=512,\ 9^{3}=729$
• Fractions to percent: $\frac12=50\%,\ \frac13\approx33.3\%,\ \frac14=25\%,\ \frac15=20\%,\ \frac16\approx16.7\%,\ \frac18=12.5\%,\ \frac1{10}=10\%$
• Values of pi: use $\pi=\frac{22}{7}$ when the radius is a multiple of 7 or 3.5; use $3.14$ otherwise.
• Units: $1\ \text{km}=1000\ \text{m}$, $1\ \text{m}^{3}=1000\ \text{L}$, $1\ \text{hour}=60\ \text{min}$.