Q&A: practice questions with answers

Concept: Track the actual money: total investment is ₹200 (the second purchase is made from the first sale's money).

1. Sale 1: $200\times \frac{70}{100}=140$ (loss 30%).
2. Second article bought for ₹140, sold at 75% gain: $140\times \frac{175}{100}=245$.
3. Overall gain $=245-200=45$ on ₹200 $\Rightarrow \frac{45}{200}\times 100=22.5\%$.

Trick: Successive % formula: $-30+75+\frac{(-30)(75)}{100}=45-22.5=22.5\%$.

Answer: (C) 22.5%

Concept: Successive increases multiply: $P(1+\frac{r_1}{100})(1+\frac{r_2}{100})$.

1. After year 1: $2,90,000\times 1.1=3,19,000$.
2. After year 2: $3,19,000\times 1.3=4,14,700$.

Trick: Net % $=10+30+\frac{10\times 30}{100}=43\%$; $2,90,000\times 1.43=4,14,700$.

Answer: (C) 4,14,700

Concept: If every observation changes by the same amount $k$, the average changes by exactly $k$.

1. Each of the 7 numbers decreases by 7, so the total decreases by $7\times 7=49$.
2. New average $=\frac{47\times 7-49}{7}=47-7=40$.

Trick: Add/subtract to all → average shifts by the same; multiply all → average multiplies by the same.

Answer: (B) 40

Concept: Order of operations: Division and Multiplication first (left to right), then Addition and Subtraction.

1. $18÷6=3$ and $3\times 4=12$.
2. $16+3-12=7$.

Trick: Never add $16+18$ first – that is the common trap.

Answer: (D) 7

Concept: $\frac{2\tan \theta }{1+{\tan }^2\theta }=\sin 2\theta$.

1. With $\theta ={30}^{\circ }$: value $=\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$.
2. Direct check: $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$, so $\frac{2/\sqrt{3}}{1+\frac{1}{3}}=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\frac{\sqrt{3}}{2}$.
3. $\frac{\sqrt{3}}{2}=\cos {30}^{\circ }$.

Trick: $\sin {60}^{\circ }=\cos {30}^{\circ }$ (complementary angles) – the options are in cos, so convert.

Answer: (B) $\cos {30}^{\circ }$

Concept: $\frac{9}{12}={\left(\frac{12}{9}\right)}^{-1}$; make all bases equal, then equate powers.

1. LHS $={\left(\frac{12}{9}\right)}^{19}{\left(\frac{12}{9}\right)}^{-5}{\left(\frac{12}{9}\right)}^8={\left(\frac{12}{9}\right)}^{22}$.
2. RHS $={\left(\frac{9}{12}\right)}^{6m+11}={\left(\frac{12}{9}\right)}^{-(6m+11)}$.
3. $22=-(6m+11)\Rightarrow 6m=-33\Rightarrow m=-\frac{33}{6}$.

Trick: Flipping a fraction flips the sign of its exponent – convert everything to one base before adding powers.

Answer: (B) $-\frac{33}{6}$

Concept: $\text{Overall average}=\frac{\sum (\text{members}\times \text{average})}{\text{total members}}$.

1. Totals: $42\times 335=14070$, $31\times 103=3193$, $32\times 271=8672$.
2. Sum $=14070+3193+8672=25935$; members $=42+31+32=105$.
3. Average $=\frac{25935}{105}=247$.

Trick: $105\times 247=105\times 250-105\times 3=26250-315=25935$ – verify by multiplying back.

Answer: (B) 247

Concept: For similar triangles, $\frac{{\text{Area}}_1}{{\text{Area}}_2}={\left(\frac{{\text{side}}_1}{{\text{side}}_2}\right)}^2$, and corresponding (longest) sides pair up.

1. ${\left(\frac{AC}{PR}\right)}^2=\frac{432}{588}=\frac{36}{49}$.
2. $\frac{AC}{PR}=\frac{6}{7}\Rightarrow AC=56\times \frac{6}{7}=48\text{cm}$.

Trick: Divide both areas by 12 to see the perfect squares 36 and 49 at once.

Answer: (A) 48 cm

Concept: Sum of interior angles of an $n$-sided polygon $=(n-2)\times {180}^{\circ }$.

1. $n=6$: $(6-2)\times {180}^{\circ }=4\times {180}^{\circ }={720}^{\circ }$.
2. Each angle of a regular hexagon $=\frac{{720}^{\circ }}{6}={120}^{\circ }$.

Yaad rakhein: Triangle 180°, quadrilateral 360°, pentagon 540°, hexagon 720° – add 180° per extra side.

Answer: (B) ${720}^{\circ }$

Concept: Take CP = 100; "150% above CP" means MP = 250; required SP = 120.

1. Discount $=250-120=130$.
2. Discount% $=\frac{130}{250}\times 100=52\%$.

Trick: "150% above" = 2.5 times (not 1.5 times) – read "above" carefully.

Answer: (B) 52%

Concept: Curved surface area of a cylinder $=2\pi rh$.

1. $2\times \frac{22}{7}\times r\times 60=1300$.
2. $r=\frac{1300\times 7}{2\times 22\times 60}=\frac{9100}{2640}\approx 3.447$.
3. Correct to two decimals: $r\approx 3.45\text{cm}$.

Trick: $\frac{9100}{2640}=\frac{455}{132}$; $132\times 3.45=455.4$, so the answer is just under 3.45 → rounds to 3.45 (not 3.48).

Answer: (A) 3.45

Concept: Do $÷$ and $\times$ before $+$ and $-$.

1. $5.6÷7=0.8$ and $1.6\times 0.5=0.8$.
2. $25-0.8+0.8=25$.

Trick: The two middle terms are equal and opposite – they cancel, leaving 25.

Answer: (D) 25

Concept: Compare cost price and selling price of one article; the number of articles does not affect the percentage.

1. CP of one article $=\frac{474}{12}=39.5$ rupees.
2. SP of one article = ₹79, so profit $=79-39.5=39.5$ rupees.
3. Profit$\%=\frac{39.5}{39.5}\times 100=100\%$.

Trick: $79\times 12=948=2\times 474$ – SP per dozen is exactly double the CP per dozen, so profit is 100%.

Answer: (B) 100%

Concept: Monthly saving = salary − expenses; count only the months in which the saving is actually kept.

1. Monthly saving $=18000-5000-3000=10000$ rupees.
2. In the birthday month the whole saving is spent, so savings are kept for only 11 months.
3. Yearly savings $=11\times 10000=110000$ rupees.

Trick: Read the 'birthday' condition carefully – it removes exactly one month; 12 × 10000 = 120000 is the trap.

Answer: (B) 110000

Concept: If HCF = $h$, write the numbers as $ha$ and $hb$ with $a,b$ co-prime; then LCM $=hab$.

1. Given LCM $=21\times$ (first number) $\Rightarrow hab=21ha\Rightarrow b=21$.
2. So the second number $=21h$, where $h$ is a two-digit prime.
3. Smallest two-digit prime $h=11$, so least second number $=21\times 11=231$.

Trick: LCM ÷ first number = the co-prime factor of the second number; multiply it by the HCF.

Answer: (C) 231

Concept: Take successive differences; if they form a simple pattern, extend it.

1. $19-15=4,27-19=8,39-27=12,55-39=16$.
2. Differences are $4,8,12,16$ – multiples of 4 – so the next difference is $20$.
3. $55+20=75$.

Trick: Differences increasing by a constant = second-level AP; just add the next multiple.

Answer: (A) 75

Concept: Successive differences often form multiples of a fixed number.

1. $40-33=7,54-40=14,75-54=21,103-75=28$.
2. Differences $7,14,21,28$ are multiples of 7, so the next is $35$.
3. $103+35=138$.

Trick: First difference 7 and it keeps adding 7 – table of 7.

Answer: (D) 138

Concept: The product of two consecutive numbers lies just below the square of the larger one.

1. Let the numbers be $n$ and $n+1$: $n(n+1)=380$.
2. ${19}^2=361$ and ${20}^2=400$, so try $19\times 20=380$ ✓.
3. Greater number $=20$.

Trick: $\sqrt{380}\approx 19.5$ – the two consecutive numbers sit on either side of it: 19 and 20.

Answer: (A) 20

Concept: When $a$ and $b$ are reciprocals of each other, work with $a+b$ and $ab$ instead of the individual values.

1. $a=\frac{3+\sqrt{5}}{3-\sqrt{5}}$ and $b=\frac{3-\sqrt{5}}{3+\sqrt{5}}$, so $ab=1$.
2. $a+b=\frac{(3+\sqrt{5}{)}^2+(3-\sqrt{5}{)}^2}{(3-\sqrt{5})(3+\sqrt{5})}=\frac{2(9+5)}{9-5}=\frac{28}{4}=7$.
3. $a^2+b^2+3ab=(a+b{)}^2+ab=49+1=50$.

Trick: $a^2+b^2+3ab=(a+b{)}^2+ab$ – never rationalise separately when $ab=1$.

Answer: (B) 50

Concept: If two sums become equal after $n$ and $n+1$ years at the same CI rate, the sums are in the ratio $(1+r):1$.

1. Let Mahesh get $M$, Vijay $V$. $M(1.1{)}^{12}=V(1.1{)}^{13}\Rightarrow M=1.1V$.
2. $M:V=11:10$, total $21$ parts $=1407\Rightarrow$ 1 part $=67$.
3. Vijay's share $=10\times 67=670$ rupees.

Trick: One year's difference at 10% → ratio $11:10$; just divide the total in that ratio.

Answer: (C) 670