Q&A: practice questions with answers

Concept: Follow BODMAS: division and multiplication first (left to right), then addition and subtraction.

1. $20÷4=5$ and $2\times 4=8$.
2. $16+5-8=13$.

Trick: Never add first – '16 + 20' before dividing gives a wrong answer.

Answer: (A) 13

Concept: Convert each ratio to a decimal (or compare with a benchmark such as $\frac{1}{2}$).

1. $3:17=\frac{3}{17}\approx 0.176$; $13:67=\frac{13}{67}\approx 0.194$.
2. $10:37=\frac{10}{37}\approx 0.270$; $5:9=\frac{5}{9}\approx 0.556$.
3. Greatest is $5:9$.

Trick: Only $\frac{5}{9}$ is more than $\frac{1}{2}$ (numerator more than half the denominator) – spot it without any division.

Answer: (B) 5 : 9

Concept: In a right triangle with one acute angle ${45}^{\circ }$, the other is also ${45}^{\circ }$, so the two legs are equal and hypotenuse $=\text{leg}\times \sqrt{2}$.

1. $\angle C={90}^{\circ }$, $\angle A={45}^{\circ }\Rightarrow \angle B={45}^{\circ }$, so $BC=AC=7$.
2. $AB=\sqrt{7^2+7^2}=\sqrt{98}=7\sqrt{2}$ units.
3. Check with trigonometry: $\cos {45}^{\circ }=\frac{AC}{AB}\Rightarrow AB=\frac{7}{1/\sqrt{2}}=7\sqrt{2}$.

Trick: 45-45-90 triangle sides are in the ratio $1:1:\sqrt{2}$; 30-60-90 are $1:\sqrt{3}:2$.

Answer: (C) $7\sqrt{2}$ units

Concept: Work with prime powers: $240=2^4\times 3\times 5$. Each HCF fixes the minimum power common to two numbers.

1. gcd$(a,b)=8$ and gcd$(b,c)=12$ → $b$ is a multiple of $24$; gcd$(c,a)=4$ → $c=2^2\times 3\times \dots$ and $a$ has no factor 3.
2. LCM needs $2^4$ (in $a$ or $b$) and a factor $5$ (in any one). Smallest bases: $a=8,b=24,c=12$.
3. Options: $a=16,b=24,c=60$ → $100$; $a=40,b=48,c=12$ → $100$; $a=8,b=48,c=60$ → $116$; $a=80,b=24,c=12$ → $116$.
4. Check $(16,24,60)$: gcds $8,12,4$ and LCM $240$ ✓. Minimum sum $=100$.

Trick: Put the 'expensive' extra factors ($2^4$ and $5$) on different numbers and compare – $16+24+60$ and $40+48+12$ both give 100.

Answer: (A) 100

Concept: With options given, substitute the triple that satisfies all three equations; otherwise eliminate one variable at a time.

1. Try $(3,-2,1)$: $3-2+1=2$ ✓.
2. $6(3)-4(-2)+5(1)=18+8+5=31$ ✓.
3. $5(3)+2(-2)+2(1)=15-4+2=13$ ✓ – all three satisfied.
4. Option (A) fails the first equation ($3-2-1=0$), (D) fails ($3+2+1=6$), (B) fails ($-3-2+1=-4$).

Trick: The first equation $x+y+z=2$ alone rejects A, B and D – check the simplest equation first.

Answer: (C) (3, −2, 1)

Concept: For similar triangles, ratio of areas = square of the ratio of corresponding sides.

1. Side ratio $=16:14$.
2. Area ratio $={16}^2:{14}^2=256:196$ (which simplifies to $64:49$).

Trick: Sides → square for area, cube for volume; never take the same ratio for area.

Answer: (A) 256 : 196

Concept: Overall average $=\frac{\text{total amount}}{\text{total members}}$ – weight each group average by its size.

1. Totals: $12\times 160=1920$, $55\times 109=5995$, $89\times 181=16109$.
2. Sum $=1920+5995+16109=24024$; members $=12+55+89=156$.
3. Average $=\frac{24024}{156}=154$.

Trick: Never average the averages $(\frac{160+109+181}{3}=150)$ – group sizes differ.

Answer: (D) 154

Concept: If SP of $m$ articles = CP of $n$ articles, then $\text{Profit}\%=\frac{n-m}{m}\times 100$.

1. Let CP of each piece $=$ ₹1. Then SP of 48 pieces $=$ ₹$x$ (the CP of $x$ pieces).
2. Profit $25\%$ → SP of 48 pieces $=48\times \frac{125}{100}=60$.
3. So $x=60$.

Trick: $\frac{x-48}{48}=\frac{1}{4}\Rightarrow x-48=12\Rightarrow x=60$; with profit, $x>48$ – reject 40 instantly.

Answer: (A) 60

Concept: Adding the three pair-sums gives twice the total $x+y+z$.

1. Let $x+y=6k,y+z=7k,z+x=8k$.
2. Adding: $2(x+y+z)=21k\Rightarrow 28=21k\Rightarrow k=\frac{4}{3}$.
3. $z=(x+y+z)-(x+y)=14-6k=14-8=6$.

Trick: Simpler: the pair-sum that leaves out $z$ is $x+y=\frac{6}{21}\times 28=8$, so $z=14-8=6$.

Answer: (B) 6

Concept: Number of coins $=\frac{\text{Volume of cuboid}}{\text{Volume of one coin}}$ (volume is conserved on melting).

1. Circumference $2\pi r=5.5\Rightarrow r=\frac{5.5\times 7}{2\times 22}=\frac{7}{8}$ cm; thickness $h=2$ mm $=\frac{1}{5}$ cm.
2. Coin volume $=\pi r^{2}h=\frac{22}{7}\times \frac{49}{64}\times \frac{1}{5}=\frac{77}{160}$ cm³.
3. Cuboid volume $=14\times 11\times 8=1232$ cm³.
4. Coins $=1232÷\frac{77}{160}=\frac{1232\times 160}{77}=16\times 160=2560$.

Trick: Convert mm to cm before anything else; $\frac{1232}{77}=16$ because $77\times 16=1232$.

Answer: (A) 2560

Concept: If $a+b+c=0$, then $a^3+b^3+c^3=3abc$.

1. $55+(-30)+(-25)=0$, so the identity applies.
2. $3abc=3\times 55\times (-30)\times (-25)=3\times 55\times 750=123750$.
3. Direct check: $166375-27000-15625=123750$.

Trick: Whenever three numbers add to zero, skip the cubes – just compute $3abc$.

Answer: (D) 123750

Concept: Under compound interest, if money doubles in $n$ years, it becomes $2^k$ times in $k\times n$ years.

1. Doubles in 7 years: $(1+r{)}^7=2$.
2. Eight times $=2^3={[(1+r{)}^7]}^3=(1+r{)}^{21}$.
3. So the time is $21$ years.

Trick: CI: double in $n$ → 4× in $2n$, 8× in $3n$. (In SI it would be $7\times 7=49$ years – a different rule.)

Answer: (C) 21

Concept: Under simple interest, if a sum becomes $n$ times itself, the interest earned is $(n-1)$ times the principal, so $R=\frac{(n-1)\times 100}{T}$.

1. Sum becomes 4 times, so $SI=3P$.
2. $SI=\frac{P\times R\times T}{100}\Rightarrow 3P=\frac{P\times R\times 25}{100}$.
3. $R=\frac{300}{25}=12\%$ per annum.

Trick: "$n$ times in $T$ years" $\Rightarrow R=\frac{100(n-1)}{T}$. Here $\frac{100\times 3}{25}=12$.

Answer: (C) 12%

Concept: Profit is shared in the ratio of (capital × time). When capital changes mid-year, add the products for each period.

1. X: $9\times 6+\frac{9}{2}\times 6=54+27=81$; Y: $4\times 12=48$; Z: $8\times 12=96$. Ratio $81:48:96$, total 225 parts.
2. Difference of X and Y $=81-48=33$ parts.
3. $33\times \frac{33000}{225}=33\times 146.67=$ ₹4840.

Trick: Z's ₹23,000 is only a distractor; the profit split needs only the ratio. $\frac{33000}{225}=146\frac{2}{3}$, times 33 gives 4840.

Answer: (C) ₹4840

Concept: Sum of interior angles $=(n-2)\times {180}^{\circ }$; each exterior angle of a regular polygon $=\frac{{360}^{\circ }}{n}$; interior $+$ exterior $={180}^{\circ }$.

1. $(n-2)\times 180=1620\Rightarrow n-2=9\Rightarrow n=11$.
2. Interior angle $=\frac{1620}{11}$, exterior angle $=\frac{360}{11}$.
3. Difference $=\frac{1620-360}{11}=\frac{1260}{11}=114\frac{6}{11}$ degrees.

Trick: Difference $={180}^{\circ }-2\times \text{exterior}=180-\frac{720}{11}=\frac{1260}{11}$.

Answer: (A) $114\frac{6}{11}$

Concept: Express MP in terms of CP from the loss condition, then compute the new SP.

1. $\frac{3}{11}MP=0.85CP\Rightarrow MP=\frac{0.85\times 11}{3}CP=\frac{9.35}{3}CP$.
2. New SP $=0.54\times MP=0.54\times \frac{9.35}{3}CP=1.683CP$.
3. Profit $=68.3\%$.

Trick: Let MP $=1100$. Then $\frac{3}{11}\times 1100=300=0.85CP\Rightarrow CP=352.94$; SP $=594$; profit $=\frac{594-352.94}{352.94}\times 100\approx 68.3\%$.

Answer: (B) 68.3%

Concept: Mode $=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$, where $l$ = lower limit of the modal class, $f_1$ its frequency, $f_0$, $f_2$ the frequencies before and after, $h$ class width.

1. Highest frequency 39 → modal class 60–70; $l=60$, $h=10$, $f_1=39$, $f_0=28$, $f_2=10$.
2. Mode $=60+\frac{39-28}{78-28-10}\times 10=60+\frac{11}{40}\times 10$.
3. $=60+2.75=62.75$.

Trick: The mode always lies inside the modal class (60–70); only options A and B qualify, and $\frac{110}{40}=2.75$ settles it.

Answer: (B) 62.75

Concept: If $x=a-b$ with $ab=1$, then $x^3=a^3-b^3-3ab(a-b)=a^3-b^3-3x$, so $x^3+3x=a^3-b^3$.

1. Here $a=7^{2/3}$, $b=7^{-2/3}$, and $ab=7^0=1$.
2. $a^3=7^2=49$, $b^3=7^{-2}=\frac{1}{49}$.
3. $x^3+3x=49-\frac{1}{49}=\frac{2401-1}{49}=\frac{2400}{49}$.

Trick: Whenever $x=t-\frac{1}{t}$, remember $x^3+3x=t^3-\frac{1}{t^3}$ (and for $x=t+\frac{1}{t}$, $x^3-3x=t^3+\frac{1}{t^3}$).

Answer: (B) $\frac{2400}{49}$

Concept: A number is divisible by 3 when the sum of its digits is divisible by 3.

1. Digit sum of 230A7807 $=2+3+0+A+7+8+0+7=27+A$.
2. 27 is already a multiple of 3, so $A$ must be a multiple of 3: $A=0,3,6,9$.
3. Maximum single digit value $=9$.

Trick: Drop digits that already sum to a multiple of 3 ($2+7$, $3$, $8+7$) → only $A$ remains, so $A\in \{0,3,6,9\}$.

Answer: (C) 9

Concept: Cutting a cylinder parallel to its bases does not change the curved surface; it only adds new circular faces. TSA of cylinder $=2\pi r^2+2\pi rh$.

1. 5 pieces have 10 flat faces of area $\pi r^2$ each, total $10\pi r^2$; their total curved surface is still $2\pi rh$.
2. Given $2\pi rh=10\pi r^2$, so the original TSA $=2\pi r^2+10\pi r^2=12\pi r^2$.
3. One piece: $S=2\pi r^2+\frac{2\pi rh}{5}=2\pi r^2+2\pi r^2=4\pi r^2$. Hence TSA $=\frac{12\pi r^2}{4\pi r^2}S=3S$.

Trick: Per piece, flat = curved = $2\pi r^2$, so $S=4\pi r^2$; original TSA $=2\pi r^2$ (ends) $+10\pi r^2$ (curved) $=12\pi r^2=3S$.

Answer: (D) 3S