Q&A: practice questions with answers

Concept: Use the angle-sum property of a triangle and the linear pair at E.

1. In right $△BDC$: $\angle C={180}^{\circ }-{90}^{\circ }-{63}^{\circ }={27}^{\circ }$.
2. In $△AEC$: $\angle AEC={180}^{\circ }-{76}^{\circ }-{27}^{\circ }={77}^{\circ }$.
3. $\angle AEB$ and $\angle AEC$ form a linear pair: $\angle AEB={180}^{\circ }-{77}^{\circ }={103}^{\circ }$.

Trick: Exterior angle: $\angle AEB=\angle CAE+\angle C={76}^{\circ }+{27}^{\circ }={103}^{\circ }$.

Answer: (B) ${103}^{\circ }$

Concept: Extra expenses (decoration, repair, transport) are added to the cost price.

1. Effective CP = ₹$461+15=476$; SP = ₹$918$.
2. Profit = ₹$918-476=442$.
3. $\text{Profit}\%=\frac{442}{476}\times 100=\frac{13}{14}\times 100=\frac{650}{7}\%$.

Trick: $442=34\times 13$ and $476=34\times 14$ – cancel 34 first.

Answer: (C) $\frac{650}{7}\%$ profit

Concept: $\text{SI}=\frac{P\times R\times T}{100}$; first find the principal, then the new amount.

1. $R\times T=3.6\times \frac{41}{9}=16.4\%$, so $0.164P=5453\Rightarrow P=33250$.
2. New $R\times T=2.7\times \frac{56}{9}=16.8\%$; SI $=0.168\times 33250=5586$.
3. Amount $=33250+5586=38836$.

Trick: Total SI % = rate × time; Amount $=P\times \frac{100+16.8}{100}$.

Answer: (D) ₹38,836

Concept: $\text{SI}=\frac{P\times R\times T}{100}$.

1. $P=400$, $R=8\%$, $T=5.5$ years.
2. $\text{SI}=\frac{400\times 8\times 5.5}{100}=4\times 44=176$.

Trick: 1 year SI $=32$; $32\times 5.5=176$.

Answer: (A) ₹176

Concept: For two numbers: $\text{HCF}\times \text{LCM}=\text{product of the numbers}$.

1. $2\times 410=82\times x$
2. $x=\frac{820}{82}=10$.
3. Check: $\text{HCF}(82,10)=2$, $\text{LCM}=410$ ✓.

Trick: Other number $=\frac{\text{HCF}\times \text{LCM}}{\text{given number}}$.

Answer: (C) $10$

Concept: Savings = Income − Expenditure; apply each % change to its own base.

1. Savings $=28\%$ of ₹34,000 $=9520$; expenditure $=34000-9520=24480$.
2. New income $=34000\times 1.11=37740$.
3. New expenditure $=24480\times 1.30=31824$.
4. New savings $=37740-31824=5916$; change $=5916-9520=-3604$.

Trick: Never apply % changes to savings directly – first split income into savings and expenditure.

Answer: (A) decrease by ₹3,604

Concept: Take each square root separately, then add/subtract.

1. $\sqrt{169}=13$
2. $\sqrt{0.04}=0.2$ and $\sqrt{6.76}=2.6$
3. $13+0.2-2.6=10.6$

Trick: For decimals, root of digits then halve the decimal places: $\sqrt{676}=26\Rightarrow \sqrt{6.76}=2.6$.

Answer: (C) $10.6$

Concept: Express the actual SP in terms of CP using the loss condition.

1. $\frac{3}{8}\text{SP}=0.87\text{CP}$ (loss $13\%$).
2. $\text{SP}=\frac{8\times 0.87}{3}\text{CP}=2.32\text{CP}$.
3. $70\%$ of SP $=0.7\times 2.32\text{CP}=1.624\text{CP}$.
4. Profit $=62.4\%$.

Trick: Take CP $=100$: SP $=\frac{87\times 8}{3}=232$, $0.7\times 232=162.4$ → profit $62.4\%$.

Answer: (D) $62.4\%$

Concept: Check for a Pythagorean triple before using Heron's formula.

1. ${65}^2+{72}^2=4225+5184=9409={97}^2$ → right triangle.
2. Legs are $65$ and $72$.
3. Area $=\frac{1}{2}\times 65\times 72=2340{\text{cm}}^2$.
4. (Heron check: $s=117$, $\sqrt{117\times 45\times 20\times 52}=2340$.)

Trick: 65-72-97 is a Pythagorean triple – remember it.

Answer: (C) $2340$

Concept: Take CP $=100$ and find SP; then the % change needed to reach CP.

1. MP $=100\times 1.9=190$.
2. SP $=190\times 0.4=76$ → loss.
3. For no profit no loss SP must be $100$: increase $=24$.
4. % change $=\frac{24}{76}\times 100\approx 31.6\%$ increase.

Trick: % change is always on the current SP ($76$), not on CP.

Answer: (C) $31.6\%$ increase

Concept: Sum of first $n$ even numbers $=n(n+1)$, so their average $=n+1$.

1. Here $n=157$.
2. Sum $=157\times 158$.
3. Average $=\frac{157\times 158}{157}=158$.

Yaad rakhein: Average of first $n$ even numbers $=n+1$; of first $n$ odd numbers $=n$.

Answer: (D) $158$

Concept: Product of two numbers $=\text{HCF}\times \text{LCM}$.

1. $168\times x=28\times 840$
2. $x=\frac{23520}{168}=140$
3. Check: $\text{HCF}(168,140)=28$ ✓

Trick: $\frac{840}{168}=5$, so other number $=28\times 5=140$.

Answer: (A) $140$

Concept: Each ratio unit $=\frac{\text{Total}}{\text{sum of ratio terms}}$.

1. Sum of terms $=5+14+17=36$.
2. 1 unit $=\frac{432}{36}=12$.
3. Difference $=(17-5)\times 12=144$.

Trick: Work directly with the difference of ratio terms ($12$ units).

Answer: (A) $144$

Concept: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. Discount $=750-660=90$.
2. Rate $=\frac{90}{750}\times 100=12\%$.

Trick: Discount % is always on marked price.

Answer: (C) $12\%$

Concept: Mean proportional of $a$ and $b$ $=\sqrt{ab}$.

1. $15\times 240=3600$
2. $\sqrt{3600}=60$
3. Check: $15:60=60:240$ ✓

Yaad rakhein: Mean proportional $=\sqrt{ab}$; third proportional $=\frac{b^2}{a}$.

Answer: (A) $60$

Concept: ${\sec }^2\theta -{\tan }^2\theta =1\Rightarrow (\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1$.

1. $P=\sec \theta +\tan \theta \Rightarrow \frac{1}{P}=\sec \theta -\tan \theta$.
2. $\frac{P^2+1}{2P}=\frac{1}{2}(P+\frac{1}{P})$.
3. $=\frac{1}{2}(2\sec \theta )=\sec \theta$.

Trick: Put $\theta =0^{\circ }$: $P=1$, value $=\frac{2}{2}=1=\sec 0^{\circ }$ – but $\tan 0^{\circ }=0$, so try $\theta ={60}^{\circ }$ to separate: matches $\sec \theta$.

Answer: (D) $\sec \theta$

Concept: Consecutive numbers $n(n+1)$ – their product lies just above $n^2$.

1. $\sqrt{272}\approx 16.5$, so try $16$ and $17$.
2. $16\times 17=272$ ✓
3. Greater number $=17$.

Trick: Unit digit 2 ($6\times 7$) confirms $16\times 17$.

Answer: (D) $17$

Concept: Add the hourly filling rates: $\frac{1}{17}+\frac{1}{27}+\frac{1}{12}$.

1. LCM $(17,27,12)=1836$ (tank capacity units).
2. Rates: $108+68+153=329$ units/hour.
3. Time $=\frac{1836}{329}=5\frac{191}{329}$ hours.

Trick: Time must be less than the fastest pipe alone ($12$ h) – only $5\frac{191}{329}$ fits.

Answer: (C) $5\frac{191}{329}$ hours

Concept: Expand brackets, then combine like terms.

1. $x(3x-6)=3x^2-6x$
2. $4(x^2-5)=4x^2-20$
3. Total $=7x^2-6x-20+15=7x^2-6x-5$

Trick: Constant term: $-20+15=-5$ rules out options with $+5$.

Answer: (D) $7x^2-6x-5$

Concept: Find each pipe's full time, then the net rate.

1. P: $\frac{5}{9}$ in $25$ h → full in $45$ h; Q: $\frac{3}{7}$ in $27$ h → full in $63$ h.
2. In 3 h: $\frac{3}{45}+\frac{3}{63}=\frac{1}{15}+\frac{1}{21}=\frac{4}{35}$ of tank.
3. R empties $\frac{4}{35}$ in $4$ h → empties full tank in $35$ h.
4. Net: $\frac{1}{45}+\frac{1}{63}-\frac{1}{35}=\frac{7+5-9}{315}=\frac{1}{105}$ → $105$ h.

Trick: Full time $=\text{time}÷\text{fraction}$: $25÷\frac{5}{9}=45$.

Answer: (C) $105$ hours