Q&A: practice questions with answers

Concept: If $a:b=c:x$, then $x$ is the fourth proportional: $x=\frac{b\times c}{a}$.

1. Here $a=15,b=25,c=12$.
2. $x=\frac{25\times 12}{15}=\frac{300}{15}=20$.
3. Check: $\frac{15}{25}=\frac{3}{5}=\frac{12}{20}$ ✓

Trick: Fourth proportional = (2nd × 3rd) ÷ 1st, i.e. $\frac{bc}{a}$.

Answer: (A) 20

Concept: Read the labels on each line: Company X (lower line) and Company Y (upper line); values are in ₹ crore.

1. January: X $=95$, Y $=100$ → $195$.
2. March: X $=85$, Y $=100$ → $185$.
3. Total $=195+185=380$ crore.

Yaad rakhein: Use the printed data labels, not the y-axis height – the Y line is drawn stacked above X.

Answer: (D) 380 crore

Concept: To add fractions, convert to a common denominator using the LCM of denominators.

1. $\frac{7}{-27}=-\frac{7}{27}$.
2. LCM of 18 and 27 is 54.
3. $\frac{11}{18}-\frac{7}{27}=\frac{33}{54}-\frac{14}{54}=\frac{19}{54}$.

Trick: $\frac{11}{18}\approx 0.61$ is bigger than $\frac{7}{27}\approx 0.26$, so the answer is positive and less than 1 – only $\frac{19}{54}$ fits.

Answer: (C) $\frac{19}{54}$

Concept: Pass marks = marks obtained + marks by which the candidate failed; pass marks are 55% of the total.

1. Pass marks $=120+78=198$.
2. $55\%$ of total $=198$.
3. Total $=\frac{198\times 100}{55}=360$.

Trick: $198÷55=3.6$, so $1\%=3.6$ and $100\%=360$.

Answer: (D) 360

Concept: A sum doubles when SI = Principal. So $R\times T=100$.

1. $\text{SI}=P=\frac{P\times R\times 8}{100}$.
2. $8R=100\Rightarrow R=\frac{100}{8}=12.5\%$.

Trick: Doubling time on SI: $R=\frac{100}{T}$ (here $\frac{100}{8}=12.5\%$).

Answer: (D) 12.5%

Concept: For $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$: if $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, the lines intersect – exactly one solution.

1. $\frac{a_1}{a_2}=\frac{5}{7}$, $\frac{b_1}{b_2}=\frac{-3}{4}$.
2. $\frac{5}{7}\ne \frac{-3}{4}$ → unique solution.
3. (Solving: $x=2,y=1$.)

Yaad rakhein: Unequal ratios → one; all three equal → infinite; first two equal but not the third → none.

Answer: (B) one

Concept: All four sides of a rhombus are equal (side $a$).

1. Sum of squares of sides $=4a^2=1600$.
2. $a^2=400\Rightarrow a=20$ cm.
3. Perimeter $=4a=80$ cm.

Trick: $4a^2=1600$ gives $a=20$ directly; perimeter is just $4a$.

Answer: (A) 80 cm

Concept: Work done = rate × time. One tap fills $\frac{1}{6}$ tank per hour.

1. First half by one tap: $\frac{1}{2}\times 6=3$ h.
2. Now 4 taps: rate $=\frac{4}{6}=\frac{2}{3}$ tank/h.
3. Second half: $\frac{1}{2}÷\frac{2}{3}=\frac{3}{4}$ h $=45$ min.
4. Total $=3$ h $45$ min.

Trick: 4 equal taps do the remaining half in $\frac{3}{4}$ of the time one tap needs ($\frac{3}{4}$ h).

Answer: (D) 3 hours 45 min

Concept: Sum of numbers = average × count.

1. Sum of two numbers $=2M$.
2. One number $=N$.
3. Other $=2M-N$.

Trick: Check with numbers: average 5 of 4 and 6 → $2\times 5-4=6$ ✓

Answer: (D) 2M − N

Concept: If A is 3 times as fast as B, rates are A $=3$ units, B $=1$ unit per minute.

1. Together $=4$ units/min; tank $=4\times 36=144$ units.
2. A alone $=\frac{144}{3}=48$ min.

Trick: Time of A $=\text{together time}\times \frac{3+1}{3}=36\times \frac{4}{3}=48$.

Answer: (C) 48 min

Concept: Write in $\sin$ and $\cos$ and use ${\sin }^{2}A+{\cos }^{2}A=1$.

1. $\tan A+\frac{1}{\tan A}=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}$.
2. $=\frac{{\sin }^{2}A+{\cos }^{2}A}{\sin A\cos A}=\frac{1}{\sin A\cos A}$.
3. $=\text{cosec}A\sec A$.

Yaad rakhein: $\tan A+\cot A=\sec A\text{cosec}A$ – a standard result worth memorising.

Answer: (B) $\text{cosec}A\sec A$

Concept: Share = parts × value of one part.

1. Total parts $=5+7+15=27$.
2. B's 7 parts $=$ ₹1,400 → 1 part $=$ ₹200.
3. Total $=27\times 200=$ ₹5,400.

Trick: Total must be a multiple of 27 – among the options only 5,400 and 8,100 are; 5,400 gives B $=1400$.

Answer: (D) ₹5,400

Concept: Volume of a cylinder $V=\pi r^{2}h$, where $r=\frac{\text{diameter}}{2}$.

1. $r=\frac{10}{2}=5$ cm, $h=10.5$ cm.
2. $V=\frac{22}{7}\times 25\times 10.5$.
3. $=22\times 25\times 1.5=825{\text{cm}}^3$.

Trick: $10.5÷7=1.5$ cancels the 7 neatly.

Answer: (A) 825 ${\text{cm}}^3$

Concept: A sum doubles when SI equals the principal, so $R\times T=100$.

1. $\frac{P\times R\times 16}{100}=P$.
2. $R=\frac{100}{16}=6.25\%$.

Trick: Doubling on SI: $R=\frac{100}{T}$.

Answer: (C) 6.25%

Concept: BODMAS: division and multiplication (left to right) before addition and subtraction.

1. $9÷3=3$, then $3\times 4=12$.
2. $5+9-12=2$.

Trick: Do $÷$ and $\times$ strictly left to right: $9÷3\times 4=12$, not $\frac{9}{12}$.

Answer: (A) 2

Concept: Use $(a-b)(a+b)=a^2-b^2$ repeatedly.

1. $(x-y)(x+y)=x^2-y^2$.
2. $(x^2-y^2)(x^2+y^2)=x^4-y^4$.
3. $(x^4-y^4)(x^4+y^4)=x^8-y^8$.

Trick: Each extra $(x^n+y^n)$ factor doubles the power: $2\to 4\to 8$.

Answer: (B) $x^8-y^8$

Concept: BODMAS: Brackets first, then $÷/\times$ left to right, then $+/-$.

1. Bracket: $5+4=9$.
2. $6\times 9=54$, $54÷3=18$.
3. $3+18-7=14$.

Trick: $6\times 9÷3=6\times 3=18$ – simplify $9÷3$ first if easier; the result is the same.

Answer: (B) 14

Concept: Trains moving in opposite directions: relative speed = sum of speeds; distance = sum of lengths.

1. Distance $=135+155=290$ m.
2. Relative speed $=80+65=145$ km/h $=145\times \frac{5}{18}$ m/s.
3. Time $=\frac{290\times 18}{145\times 5}=\frac{2\times 18}{5}=7.2$ s.

Trick: $290=2\times 145$, so time $=2\times \frac{18}{5}=7.2$ s.

Answer: (D) 7.2 sec

Concept: “Subtract $a$ from $b$” means $b-a$.

1. $\frac{-3}{10}-(-2)=\frac{-3}{10}+2$.
2. $=\frac{-3+20}{10}=\frac{17}{10}$.
3. $=1\frac{7}{10}$.

Trick: Minus a negative becomes plus: $-0.3+2=1.7$.

Answer: (A) $1\frac{7}{10}$

Concept: A number divisible by both 8 and 12 is a multiple of their LCM.

1. $\text{LCM}(8,12)=24$.
2. Multiples of 24: …, 96, 120, …
3. Smallest 3-digit multiple $=24\times 5=120$.

Trick: $100÷24$ leaves remainder 4, so the next multiple is $100+(24-4)=120$.

Answer: (D) 120