RRB ALP CBT 1 2025 PYQs

Concept: Change in average = (error in total) ÷ number of students.

1. Error in total $=96-32=64$
2. $\frac{64}{n}=\frac{4}{3}$
3. $n=48$

Short trick: $n=$ error ÷ change $=64\times \frac{3}{4}=48$.

Answer: (B) 48

Concept: $SI=\frac{P\times R\times T}{100}$.

1. $480=\frac{P\times 5\times 3}{100}$
2. $P=\frac{48000}{15}=3200$

Short trick: Total rate = 5 × 3 = 15% → 15% = 480 → 1% = 32 → 100% = 3200.

Answer: (C) 3200

Concept: Successive change: multiply by $1.15$ then $0.80$.

1. $2100\times 1.15=2415$
2. $2415\times 0.8=1932$

Short trick: Net effect $15-20-\frac{15\times 20}{100}=-8\mathrm{\%}$ → $2100\times 0.92=1932$.

Answer: (A) 1932

Concept: ${(x+\frac{1}{4x})}^2=x^2+\frac{1}{16x^2}+\frac{1}{2}$.

1. Divide the given equation by 4: $x^2+\frac{1}{16x^2}=\frac{14}{4}=3.5$
2. ${(x+\frac{1}{4x})}^2=3.5+0.5=4$
3. Positive value $=2$

Short trick: Note $4x^2+\frac{1}{4x^2}=4(x^2+\frac{1}{16x^2})$ – scale once, then add $2\cdot x\cdot \frac{1}{4x}=\frac{1}{2}$.

Answer: (C) 2

Concept: Recognise perfect powers: $2744={14}^3$, $4096=8^4$.

1. $\sqrt[3]{2744}=14$, $\sqrt[4]{4096}=8$
2. $\frac{14}{8}\times \frac{4}{42}\times 84$
3. $=\frac{14}{8}\times 8=14$

Short trick: $\frac{4}{42}\times 84=8$ cancels the 8 in the denominator → answer is just $\sqrt[3]{2744}=14$.

Answer: (A) 14

Concept: $V=\pi r^{2}h$ → new/old $=(0.81{)}^2\times 2.29$.

1. ${0.81}^2=0.6561$
2. $0.6561\times 2.29=1.5025$
3. Increase ≈ 50.25% ≈ 50%

Short trick: Radius effect: $-19-19+3.61=-34.39\mathrm{\%}$; with +129%: $-34.39+129-\frac{34.39\times 129}{100}\approx 50.2\mathrm{\%}$.

Answer: (B) 50%

Concept: $\sum x_i=n\overline{x}$; split the sum term by term.

1. $\sum x_i=10\times 20=200$
2. $\sum \frac{3x_i-4}{5}=\frac{3\times 200-4\times 10}{5}$
3. $=\frac{560}{5}=112$

Short trick: Transform the mean: new mean $=\frac{3\times 20-4}{5}=11.2$; × 10 terms = 112.

Answer: (D) 112

Concept: SP = MP × (1 − d₁)(1 − d₂); without discount SP = MP = 1.25 × CP.

1. $MP\times 0.40\times 0.69=324.3$ → $MP\times 0.276=324.3$
2. $MP=1175$
3. $CP=\frac{1175}{1.25}=940$

Short trick: $CP=\frac{324.3}{0.276\times 1.25}=\frac{324.3}{0.345}=940$.

Answer: (C) 940

Concept: Rates add; find P + Q from the second phase, then R = (P + Q + R) − (P + Q).

1. In 2 h all three fill $\frac{2}{6}=\frac{1}{3}$
2. P + Q fill remaining $\frac{2}{3}$ in 7 h → $P+Q=\frac{2}{21}$ per h
3. $R=\frac{1}{6}-\frac{2}{21}=\frac{7-4}{42}=\frac{1}{14}$ → 14 hours

Short trick: Tank = 42 units: P+Q+R = 7/h; after 2 h, 28 left in 7 h → P+Q = 4/h → R = 3/h → 42 ÷ 3 = 14 h.

Answer: (B) 14

Concept: Interest in year 2 = $P\times \frac{r}{100}\times (1+\frac{r}{100})$.

1. $P\times 0.1\times 1.1=0.11P$
2. $0.11P=8063$
3. $P=73300$

Short trick: 2nd-year CI at 10% = 11% of P → $8063÷0.11=73300$.

Answer: (A) ₹73300

Concept: For any interior point of an equilateral triangle, sum of perpendiculars to the sides = height.

1. $h=18+30+39=87$ cm
2. Area $=\frac{h^2}{\sqrt{3}}=\frac{7569}{\sqrt{3}}=2523\sqrt{3}$ → $m=2523$, $k=3$
3. $m-k^7=2523-2187=336$

Short trick: Area of equilateral triangle from height: $\frac{h^2}{\sqrt{3}}=\frac{h^2}{3}\sqrt{3}$ → $m=\frac{{87}^2}{3}=29\times 87=2523$.

Answer: (D) 336

Concept: Mode = the value that occurs most often.

1. 41 occurs 4 times
2. 54 occurs 3 times; 55 and 49 twice
3. Mode = 41

Short trick: Tally only the option values: 54 (3), 49 (2), 55 (2), 41 (4).

Answer: (D) 41

Concept: Average speed = total distance ÷ total time.

1. Times: $\frac{160}{80}=2$, $\frac{160}{120}=\frac{4}{3}$, $\frac{160}{80}=2$ h
2. Total time $=\frac{16}{3}$ h, distance 480 km
3. Average $=480\times \frac{3}{16}=90$ km/h

Short trick: Equal distances → harmonic mean: $\frac{3}{\frac{1}{80}+\frac{1}{120}+\frac{1}{80}}=\frac{3\times 240}{3+2+3}=90$.

Answer: (A) 90 km/hr

Concept: Equal % up and down always give a net loss of $\frac{x^2}{100}\mathrm{\%}$.

1. Let CP = 100 → MP = 110
2. SP = 110 × 0.9 = 99
3. Loss = 1%

Short trick: $\frac{{10}^2}{100}=1\mathrm{\%}$ loss.

Answer: (A) 1% loss

Concept: At the same time, height : shadow is the same for all objects.

1. $\frac{12}{8}=1.5$
2. Pole + flag $=1.5\times 15=22.5$ m
3. Pole $=22.5-3=19.5$ m

Short trick: Shadow 15 is $\frac{15}{8}$ times 8 → height $12\times \frac{15}{8}=22.5$; minus 3 m flag.

Answer: (C) 19.5 m

Concept: Bring all sine terms to one side and divide by $\cos y$.

1. $\sqrt{9}=3$, so $6\sin y+\cos y=3\sin y$.
2. $3\sin y=-\cos y$.
3. Divide by $3\cos y$: $\tan y=-\frac{1}{3}$.

Short trick: In $a\sin y+b\cos y=0$, $\tan y=-\frac{b}{a}$ directly: here $a=6-3=3,b=1$.

Answer: (A) $-\frac{1}{3}$

Concept: Write $\tan ,\cot$ in $\sin ,\cos$ and use $1-{\cos }^{2}A=(1-\cos A)(1+\cos A)$.

1. First term $=\frac{{\sin }^{2}A}{\cos A(1-\cos A)}=\frac{(1-\cos A)(1+\cos A)}{\cos A(1-\cos A)}=\frac{1+\cos A}{\cos A}=\sec A+1$.
2. Second term $=\frac{{\cos }^{2}A}{\sin A(1-\sin A)}=\frac{1+\sin A}{\sin A}=\mathrm{cosec}A+1$.
3. Difference $=\sec A-\mathrm{cosec}A$.

Short trick: Put $A={60}^{\circ }$: value $=(2+1)-(\frac{2}{\sqrt{3}}+1)\approx 0.845$, and $\sec {60}^{\circ }-\mathrm{cosec}{60}^{\circ }=2-1.155=0.845$ ✓ (B, C give negative values; D gives 0.366).

Answer: (A) $\sec A-\mathrm{cosec}A$

Concept: Unengraved area = total surface area of cube − area of all spots.

1. TSA $=6a^2=6\times {50}^2=15000{\text{cm}}^2$.
2. Number of spots $=1+2+3+4+5+6=21$.
3. Spot area $=21\times \pi \times 5^2=525\pi \approx 1649.3{\text{cm}}^2$.
4. Required $=15000-1649.3\approx 13351{\text{cm}}^2$.

Short trick: Spots $=\frac{6\times 7}{2}=21$; $525\pi \approx 1650$ → $15000-1650\approx 13350$ → only 13351 is close.

Answer: (B) 13351

Concept: Change in total = change in average × number of students.

1. Total decreased by $59-23=36$.
2. $n\times \frac{9}{7}=36$.
3. $n=36\times \frac{7}{9}=28$.

Short trick: $n=\frac{\text{error}}{\text{change in average}}$; and $n$ must be a multiple of 7 (denominator) → only 28 or 35; $28\times \frac{9}{7}=36$ ✓.

Answer: (B) 28

Concept: Mode = the value that occurs most often.

1. Frequencies of option values: 64 → 4 times, 81 → 3, 82 → 3, 89 → 2.
2. Highest frequency is 4 for 64.

Short trick: Only count the four option values – no need to tally all 25 numbers.

Answer: (B) 64