Q&A: practice questions with answers

Concept: For two numbers, test the options: sum must be 57 and LCM must be 812.

1. $812=2^2\times 7\times 29=28\times 29$.
2. 28 and 29 are consecutive, so co-prime: $\text{LCM}(28,29)=28\times 29=812$.
3. Sum $=28+29=57$ ✓. Check others: $21+36=57$ but $\text{LCM}=252$; $33+24=57$ but $\text{LCM}=264$; $26+31=57$ but $\text{LCM}=806$.

Trick: Consecutive numbers are co-prime, so LCM = product; $\sqrt{812}\approx 28.5$ points straight to 28 and 29.

Answer: (C) 28, 29

Concept: Greatest number dividing $a,b,c$ leaving remainders $r_1,r_2,r_3$ = HCF of $(a-r_1),(b-r_2),(c-r_3)$.

1. $131-8=123,91-9=82,215-10=205$.
2. $123=3\times 41,82=2\times 41,205=5\times 41$.
3. $\text{HCF}=41$.

Trick: Subtract remainders, then check options: only 41 divides 123, 82 and 205.

Answer: (D) 41

Concept: Let cost price $=C$ and actual selling price $=S$. Write each condition as an equation in $C$.

1. $\frac{2}{5}S=C(1-0.26)=0.74C$.
2. $S=\frac{5\times 0.74}{2}C=1.85C$.
3. New SP $=0.84S=0.84\times 1.85C=1.554C$.
4. $\text{Profit}\%=(1.554-1)\times 100=55.4\%$.

Trick: Take $C=100$: $\frac{2}{5}S=74\Rightarrow S=185$; $84\%$ of $185=155.4$ → profit $55.4\%$.

Answer: (C) $55.4\%$

Concept: For equal distances covered at speeds $x$ and $y$, average speed $=\frac{2xy}{x+y}$.

1. Here $x=55$, $y=66\text{km/h}$.
2. Average $=\frac{2\times 55\times 66}{55+66}=\frac{7260}{121}$.
3. $=60\text{km/h}$.

Trick: $55:66=5:6$ → average $=\frac{2\times 5\times 6}{11}\times 11=60$.

Answer: (B) $60$

Concept: Sum $=$ average $\times$ number of terms.

1. Sum of first two $=2\times 80=160$; third $=101$.
2. Sum of last two $=655-160-101=394$.
3. Average of last two $=\frac{394}{2}=197$.

Trick: Subtract known parts from total, then divide by count left.

Answer: (D) $197$

Concept: If $a^m=a^n$ (with $a>0,a\ne 1$), then $m=n$; also $(a^m{)}^n=a^{mn}$.

1. $x^z=({58}^{0.5}{)}^z={58}^{0.5z}$.
2. $y^{10}=({58}^{0.79}{)}^{10}={58}^{7.9}$.
3. $0.5z=7.9\Rightarrow z=15.8$.

Trick: Just compare powers: $z=\frac{0.79\times 10}{0.5}=15.8$.

Answer: (D) $15.8$

Concept: Overhead (decoration) cost is added to the cost price. $\text{Loss}\%=\frac{\text{Loss}}{\text{CP}}\times 100$.

1. CP $=740+16=$ ₹$756$; SP $=$ ₹$558$ → loss.
2. Loss $=756-558=$ ₹$198$.
3. $\text{Loss}\%=\frac{198}{756}\times 100=\frac{11}{42}\times 100=\frac{550}{21}\%\approx 26.19\%$.

Trick: SP below CP → only “loss” options; $\frac{198}{756}$ simplifies by 18 to $\frac{11}{42}$.

Answer: (D) $\frac{550}{21}\%$ loss

Concept: Increasing a number by $80\%$ multiplies it by $1.8$.

1. $1.8N=2790$.
2. $N=\frac{2790}{1.8}=1550$.

Trick: Check: $1550+80\%$ of $1550=1550+1240=2790$ ✓.

Answer: (C) $1550$

Concept: Trade discount is deducted from the list price: Net price $=\text{LP}\times \frac{100-d}{100}$.

1. Discount $=30\%$ of $900=$ ₹$270$.
2. Net price $=900-270=$ ₹$630$.

Trick: $30\%$ off → pay $70\%$: $0.7\times 900=630$.

Answer: (D) ₹$630$

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

1. $2\times \frac{22}{7}\times r\times 72=3300$.
2. $r=\frac{3300\times 7}{2\times 22\times 72}=\frac{23100}{3168}$.
3. $r=7.2916\dots \approx 7.29\text{cm}$.

Trick: $\frac{3300}{44}=75$, so $r=\frac{75\times 7}{72}=\frac{525}{72}=7.29$.

Answer: (B) $7.29$

Concept: A number is divisible by 5 if its last digit is 0 or 5; by 3 if its digit sum is divisible by 3.

1. Last digit of 4,81,355 is 5 → divisible by 5.
2. Digit sum $=4+8+1+3+5+5=26$, not divisible by 3 → not by 3, 6 or 15.
3. Odd number → not divisible by 6 either.

Trick: Check the last digit first – it settles 5 and 2 instantly.

Answer: (C) 5

Concept: Each exterior angle of a regular $n$-gon $=\frac{{360}^{\circ }}{n}$; interior angle $={180}^{\circ }-$ exterior angle.

1. $n=30$, exterior angle $=\frac{{360}^{\circ }}{30}={12}^{\circ }$.
2. Interior angle $={180}^{\circ }-{12}^{\circ }={168}^{\circ }$.
3. Check: $\frac{(n-2)\times {180}^{\circ }}{n}=\frac{28\times {180}^{\circ }}{30}={168}^{\circ }$.

Yaad rakhein: ${12}^{\circ }$ is the exterior angle – a trap option; always subtract from ${180}^{\circ }$.

Answer: (A) ${168}^{\circ }$

Concept: Price after successive discounts $=\text{MP}\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})$.

1. After 12% discount: ₹$1400\times 0.88=1232$.
2. Second factor $=\frac{1133.44}{1232}=0.92$.
3. So second discount $=1-0.92=0.08=8\%$.

Trick: Divide final price by price after first discount; $0.92$ means 8% off.

Answer: (B) 8%

Concept: Use $\sin ({90}^{\circ }-x)=\cos x$, $\cos ({90}^{\circ }-x)=\sin x$, $\cos ({180}^{\circ }-A)=-\cos A$, $\sin ({180}^{\circ }-A)=\sin A$ and $\cos (A-B)=\cos A\cos B+\sin A\sin B$.

1. Expression $=\cos x(-\cos (x-y))+\sin x\sin (y-x)$.
2. $\sin (y-x)=-\sin (x-y)$, so it becomes $-[\cos x\cos (x-y)+\sin x\sin (x-y)]$.
3. $=-\cos (x-(x-y))=-\cos y$.

Trick: Put $x=y=0$: value $=1\times (-1)+0=-1=-\cos 0$. Only $-\cos y$ fits.

Answer: (C) $-\cos y$

Concept: Third proportional of $a$ and $b$ is $x$ where $a:b=b:x$, i.e. $x=\frac{b^2}{a}$.

1. $x=\frac{{27}^2}{81}=\frac{729}{81}$.
2. $x=9$.

Yaad rakhein: Check: $81:27=3:1$ and $27:9=3:1$ ✓.

Answer: (B) 9

Concept: Successive % changes multiply: $P_3=P\times (1+\frac{r_1}{100})(1+\frac{r_2}{100})(1-\frac{r_3}{100})$.

1. Factors: $1.25\times 1.6\times 0.5$.
2. $1.25\times 1.6=2$, and $2\times 0.5=1$.
3. So population $=12014\times 1=12,014$.

Trick: In fractions: $\frac{5}{4}\times \frac{8}{5}\times \frac{1}{2}=1$ – no change at all.

Answer: (A) 12,014

Concept: BODMAS: division and multiplication before addition/subtraction; mind the signs.

1. $(-60)÷(-10)=6$.
2. $(-3)\times 6=-18$.
3. $-9-6+(-18)=-33$.

Yaad rakhein: Negative ÷ negative = positive; negative × positive = negative.

Answer: (B) $-33$

Concept: For the same time period, profit is shared in the ratio of investments.

1. Ritik's profit $=16500-3600=12900$, ratio $=12900:3600=43:12$.
2. Sanjay's investment $=79200\times \frac{12}{55}=17280$; Ritik's $=79200-17280=61920$.
3. Difference $=61920-17280=44640$.

Trick: Difference $=79200\times \frac{43-12}{55}=1440\times 31=44640$.

Answer: (B) ₹44,640

Concept: Total = average × number of items; a common member appears in both pair sums.

1. $V+M+H=3\times 45=135$ (H = Mohammad).
2. $V+M=86$, $M+H=100$; adding: $V+2M+H=186$.
3. $M=186-135=51\text{kg}$.

Trick: Mohan $=$ (sum of both pair totals) $-$ (three-person total).

Answer: (C) 51

Concept: Same direction: relative speed = difference; opposite: sum. Crossing distance = sum of lengths.

1. Overtaking man: relative speed $=\frac{195}{13}=15\text{m/s}=54\text{km/h}$, so train speed $=54+5=59\text{km/h}$.
2. Opposite train: $59+76=135\text{km/h}=135\times \frac{5}{18}=37.5\text{m/s}$.
3. Time $=\frac{195+255}{37.5}=\frac{450}{37.5}=12\text{s}$.

Yaad rakhein: $\text{m/s}\to \text{km/h}$: multiply by $\frac{18}{5}$.

Answer: (D) 12