Q&A: practice questions with answers

Concept: Fix the marked price as the base, find the cost price from the first sale, then compare the second selling price with that cost price.

1. Let marked price $M=130$ (a number divisible by 13 and 10). First SP $=\frac{4}{13}\times 130=40$.
2. Loss $30\%$ means $\text{SP}=0.7\text{CP}$, so $\text{CP}=\frac{40}{0.7}=\frac{400}{7}$.
3. New SP $=60\%$ of $130=78$. Profit $=78-\frac{400}{7}=\frac{146}{7}$.
4. Profit$\%=\frac{146/7}{400/7}\times 100=\frac{146}{4}=36.5\%$.

Trick: Both sales use the same CP, so $\frac{{\text{SP}}_2}{{\text{SP}}_1}=\frac{100+p}{100-30}$: $\frac{78}{40}=\frac{100+p}{70}\Rightarrow p=36.5$.

Answer: (A) $36.5\%$

Concept: Solve each bracket separately, then combine. Convert decimals to fractions to cancel quickly.

1. $3.6÷\frac{3}{5}=3.6\times \frac{5}{3}=6$.
2. $\frac{7}{3}\times \frac{2.4}{1.4}=\frac{7}{3}\times \frac{12}{7}=4$.
3. $\frac{5}{6}+\frac{14}{12}=\frac{5}{6}+\frac{7}{6}=2$.
4. Total $=6+4-2=8$.

Trick: $\frac{2.4}{1.4}=\frac{24}{14}=\frac{12}{7}$ – the 7 cancels with $\frac{7}{3}$ instantly.

Answer: (A) 8

Concept: If two sums become equal after $n$ and $n+1$ years at the same rate, the first sum is already one year "ahead": $P_1=P_2(1+\frac{r}{100})$.

1. $P(1.08{)}^{18}=G(1.08{)}^{19}\Rightarrow P=1.08G$.
2. $P:G=108:100=27:25$.
3. Gopal $=\frac{25}{52}\times 1404=25\times 27=675$.

Trick: Ratio of shares = $(100+r):100$ in favour of the one who invests for fewer years. Here $27:25$, and $1404÷52=27$.

Answer: (B) 675

Concept: Write $A=12x$, $B=36y$, $C=18z$ and use the HCF/LCM conditions to see what $x,y,z$ must contain.

1. $\text{HCF}(A,C)=6·\text{HCF}(2x,3z)=6$ → $x$ has no factor 3, $z$ is odd and has no factor 3.
2. $\text{LCM}=1080=2^3\times 3^3\times 5$: the $3^3$ can come only from $y$ (so $3\mid y$); the $2^3$ needs an even $x$ or $y$; the 5 must sit somewhere.
3. Smallest choice: $y=3$, $x=2$, $z=5$ → $A=24$, $B=108$, $C=90$. Check: HCFs $12,18,6$ ✓, LCM $1080$ ✓.
4. $A+B+C=24+108+90=222$.

Trick: Put the extra primes on the smallest multipliers: 2 on $12x$ and 5 on $18z$ keeps the sum low (putting 5 on $12x$ gives 246).

Answer: (A) 222

Concept: Factorise the quadratic to get the roots, assign them as per $b>a$, then evaluate.

1. $6x^2+x-2=0\Rightarrow (3x+2)(2x-1)=0$.
2. Roots: $x=-\frac{2}{3}$ and $x=\frac{1}{2}$. Since $b>a$: $a=-\frac{2}{3}$, $b=\frac{1}{2}$.
3. $2a+3b=-\frac{4}{3}+\frac{3}{2}=\frac{-8+9}{6}=\frac{1}{6}$.

Trick: Split the middle term: need two numbers with product $6\times (-2)=-12$ and sum $1$ → $4$ and $-3$.

Answer: (C) $\frac{1}{6}$

Concept: Take present ages as $2x$ and $3x$ and form the equation for the future ratio.

1. $\frac{2x+6}{3x+6}=\frac{8}{9}\Rightarrow 18x+54=24x+48$.
2. $6x=6\Rightarrow x=1$. Present ages: 2 years and 3 years.
3. Difference $=3-2=1$ year.

Trick: Difference of ages never changes: $3x-2x=x$. After 6 years the ratio $8:9$ has difference 1 unit, and $8$ units $=2x+6$ → $x=1$.

Answer: (D) 1

Concept: Mean $=\frac{\sum f_i{x}_i}{\sum f_i}$ with $x_i$ = class mid-points 15, 25, 35, 45, 55.

1. $\sum f_i{x}_i=49(15)+33(25)+24(35)+21(45)+55X=735+825+840+945+55X=3345+55X$.
2. $\sum f_i=127+X$.
3. $\frac{3345+55X}{127+X}=35\Rightarrow 3345+55X=4445+35X\Rightarrow 20X=1100\Rightarrow X=55$.

Trick: Use deviations from the assumed mean 35: $49(-20)+33(-10)+0+21(10)+X(20)=0\Rightarrow -980-330+210+20X=0\Rightarrow X=55$.

Answer: (D) 55

Concept: If A is $x\%$ more than B, then B is $\frac{x}{100+x}\times 100\%$ less than A.

1. Let B $=100$, so A $=226$.
2. B is less than A by $226-100=126$ on a base of 226.
3. $\frac{126}{226}\times 100=55.75\%$ (approx.).

Trick: $\frac{126}{226}=\frac{63}{113}$; since $\frac{63}{113}\approx 0.5575$, pick $55.75\%$. Note the base changes to A, so the answer is always less than $126\%$.

Answer: (B) $55.75\%$

Concept: For any polygon the exterior angles add to ${360}^{\circ }$; the interior angles add to $(n-2)\times {180}^{\circ }$.

1. Given: $360=\frac{2}{3}(n-2)\times 180$.
2. $360=120(n-2)\Rightarrow n-2=3$.
3. $n=5$ – a regular pentagon (interior ${108}^{\circ }$, exterior ${72}^{\circ }$ each).

Trick: Exterior sum : interior sum $=360:(n-2)180=2:(n-2)$. Set $\frac{2}{n-2}=\frac{2}{3}$ → $n=5$.

Answer: (D) 5

Concept: In congruent triangles, corresponding parts are equal (CPCT). The matching of vertices decides which sides correspond.

1. $\angle A=\angle D$ and $AC=DF$ tell us A ↔ D and C ↔ F; hence B ↔ E, i.e. $△ABC\cong △DEF$ in that order.
2. So $AB$ corresponds to $DE$.
3. $AB=DE=5\text{cm}$.

Trick: Write the congruence with vertices in matching order (ABC ↔ DEF); then just read off side pairs letter by letter.

Answer: (C) 5

Concept: Take total work = LCM of the days, find the work done in one 2-day cycle, then handle the leftover.

1. Work $=\text{LCM}(14,19)=266$ units. P does $19$/day, Q does $14$/day.
2. One cycle (P day + Q day) $=33$ units. $8$ cycles $=16$ days → $264$ units; $2$ units remain.
3. Day 17 is P's turn: $2$ units take $\frac{2}{19}$ day.
4. Total $=16\frac{2}{19}$ days.

Trick: Always check who starts the next day after the full cycles – P starts odd days, so the remainder is divided by P's rate (19).

Answer: (B) $16\frac{2}{19}$ days

Concept: Apply the multipliers one after another: new value $=P\times \frac{115}{100}\times \frac{130}{100}$.

1. $2,90,000\times 1.15=3,33,500$.
2. $3,33,500\times 1.30=4,33,550$.

Trick: Net increase $=15+30+\frac{15\times 30}{100}=49.5\%$; $2,90,000\times 1.495=4,33,550$.

Answer: (A) 4,33,550

Concept: Chain the ratios by making the common term equal in consecutive pairs.

1. $A:B=2:3$, $B:C=4:3$ → make B $=12$: $A:B:C=8:12:9$.
2. $C:D=6:7$ → make C $=18$: multiply the first by 2 and the second by 3: $A:B:C:D=16:24:18:21$.

Trick: $A:B:C:D=(2·4·6):(3·4·6):(3·3·6):(3·3·7)=48:72:54:63=16:24:18:21$.

Answer: (B) 16 : 24 : 18 : 21

Concept: When two items are sold at the same SP, one at $x\%$ profit and the other at $x\%$ loss, there is always a loss of $\frac{x^2}{100}\%$.

1. CP of first $=\frac{10000}{0.8}=12500$; CP of second $=\frac{10000}{1.2}\approx 8333.33$.
2. Total CP $\approx 20833.33$, total SP $=20000$ → loss $\approx 833.33$.
3. Loss$\%=\frac{833.33}{20833.33}\times 100=4\%$.

Trick: $\frac{x^2}{100}=\frac{{20}^2}{100}=4\%$ loss – no calculation needed.

Answer: (A) 4% loss

Concept: A head start shortens the distance a runner must cover; finishing later adds to his time. Speed $=\frac{\text{distance run}}{\text{time taken}}$.

1. A runs $1200\text{m}$ at $10\text{m/s}$ → $120\text{s}$.
2. B runs $1200-300=900\text{m}$ in $120+30=150\text{s}$ → $6\text{m/s}$.
3. C runs $1200-600=600\text{m}$ in $120+60=180\text{s}$ → $3.33\approx 3.3\text{m/s}$.

Trick: "A beats B by 30 s" means B is still running 30 s after A finishes – add the seconds to A's time, subtract the head start from the distance.

Answer: (B) $6\text{m/s}$, $3.3\text{m/s}$

Concept: The mean proportional between $a$ and $b$ is $\sqrt{ab}$ (since $a:x=x:b$).

1. $18\times 128=2304$.
2. $\sqrt{2304}=48$ (as ${48}^2=2304$).

Trick: $18\times 128=(2\times 9)(2\times 64)=4\times 9\times 64$ → $\sqrt{}=2\times 3\times 8=48$.

Answer: (D) 48

Concept: Form two equations from "total items" and "total money", then eliminate one variable.

1. Let burgers $=b$, soft drinks $=s$: $b+s=87$ and $15b+5s=785$.
2. From the first, $5b+5s=435$. Subtract: $10b=350\Rightarrow b=35$.
3. $s=87-35=52$.

Trick: Alligation – if all 87 were drinks, money $=435$; extra $350$ comes from burgers at $10$ extra each → $35$ burgers.

Answer: (A) 52

Concept: Provisions are constant: $\text{men}\times \text{days}$ stays the same.

1. Total provision $=100\times 27=2700$ man-days.
2. Remaining men $=100-25=75$.
3. Days $=\frac{2700}{75}=36$.

Trick: Men become $\frac{3}{4}$, so days become $\frac{4}{3}$: $27\times \frac{4}{3}=36$.

Answer: (D) 36

Concept: Average of the first $n$ multiples of $k$ = $k\times \frac{n+1}{2}$.

1. Multiples: $8,16,24,\dots ,104$ (13 terms, an AP).
2. Average of an AP = middle term = 7th term = $8\times 7=56$.
3. Check: sum $=8\times \frac{13\times 14}{2}=728$, $\frac{728}{13}=56$.

Trick: For an odd count of equally spaced numbers the average is simply the middle one.

Answer: (A) 56

Concept: Evaluate the cube root and fourth root first, then cancel.

1. $\sqrt[3]{5832}=18$ (since ${18}^3=5832$) and $\sqrt[4]{256}=4$ (since $4^4=256$).
2. Expression $=\frac{18}{4}\times \frac{2}{54}\times 54=\frac{18}{4}\times 2=9$.

Trick: $\frac{2}{54}\times 54$ cancels to 2 instantly; memorise cubes up to 20 (${18}^3=5832$).

Answer: (A) 9