RRB ALP CBT 1 2025 PYQs

Concept: Mode = the value that occurs most often.

1. 75 occurs 4 times
2. 61 occurs 3 times; 77 and 69 occur 2 times each
3. Mode = 75

Short trick: Tally only the options: 75 (4), 61 (3), 77 (2), 69 (2) – no need to sort all 25 values.

Answer: (A) 75

Concept: Required % = (first quantity ÷ second quantity) × 100.

1. Let number = 100: 4% = 4, 20% = 20
2. $\frac{4}{20}\times 100=20\mathrm{\%}$

Short trick: The number cancels: just divide the percentages, $\frac{4}{20}=\frac{1}{5}=20\mathrm{\%}$.

Answer: (D) 20%

Concept: If a sum of squares is 0, each square is 0.

1. $a^2-6a+b^2-24b+c^2-12c+189=0$
2. $(a-3{)}^2+(b-12{)}^2+(c-6{)}^2=9+144+36-189=0$
3. $a=3,b=12,c=6$
4. $ab+bc+ca=36+72+18=126$, $\sqrt{126}=3\sqrt{14}$

Short trick: Halve the coefficients on the right: $6a,24b,12c$ → $a=3,b=12,c=6$; check $3^2+{12}^2+6^2=189$ ✓.

Answer: (D) $3\sqrt{14}$

Concept: In $\mathrm{△}ABC\sim \mathrm{△}EDF$, A↔E, B↔D, C↔F, so $\frac{AB}{ED}=\frac{BC}{DF}=\frac{AC}{EF}$.

1. $\frac{AB}{ED}=\frac{5}{12.5}=\frac{2}{5}$
2. $BC=\frac{2}{5}\times 15=6$ cm
3. $EF=7.5\times \frac{5}{2}=18.75$ cm
4. Sum $=6+18.75=24.75$ cm

Short trick: Match letters by position (ABC ↔ EDF), never by alphabet. Ratio 2 : 5 → small side ×2/5, big side ×5/2.

Answer: (A) 24.75 cm

Concept: Work backwards from the last selling price.

1. Bharath's CP $=\frac{8415}{0.85}=9900$
2. This is Anand's SP at 25% profit ⇒ Anand's total cost $=\frac{9900}{1.25}=7920$
3. (Purchase price = 7920 − 63 = 7857; the cost just before selling includes repair = 7920.)

Short trick: $8415\times \frac{20}{17}\times \frac{4}{5}=8415\times \frac{16}{17}=495\times 16=7920$.

Answer: (C) 7920

Concept: Profit is shared in the ratio of investments (same time period).

1. Ratio $70000:32000:46000=35:16:23$ (sum 74)
2. Amount to share $=40\mathrm{\%}$ of 52540 $=21016$
3. Vishal $=\frac{23}{74}\times 21016=23\times 284=6532$

Short trick: 21016 ÷ 74 = 284 exactly; so the answer must be a multiple of 23 → 6532 = 23 × 284 (6532 is the only option divisible by 23).

Answer: (B) 6,532

Concept: Find the cost of one item, then multiply.

1. 1 book $=\frac{960}{12}=80$; 1 pen $=\frac{320}{8}=40$
2. 15 books $=1200$; 10 pens $=400$
3. Total $=1600$

Short trick: 15 books = $960\times \frac{15}{12}=1200$, 10 pens = $320\times \frac{10}{8}=400$ → 1600.

Answer: (B) 1600

Concept: Empirical relation: Mode = 3 Median − 2 Mean.

1. Let mode = $x$, mean = $4x$
2. $x=3(63)-2(4x)=189-8x$
3. $9x=189\Rightarrow x=21$; mean $=84$

Short trick: Use $3\text{Median}=\text{Mode}+2\text{Mean}$: $189=x+8x=9x$ → mean $=4\times 21=84$.

Answer: (D) 84

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

1. Change in total $=92-44=48$
2. $\frac{48}{n}=\frac{4}{3}$
3. $n=\frac{48\times 3}{4}=36$

Short trick: $n=48÷\frac{4}{3}=48\times \frac{3}{4}=36$ in one line.

Answer: (C) 36

Concept: Rate ∝ (diameter)${}^2$.

1. Rates ∝ $2^2:3^2:4^2=4:9:16$, total 29
2. Smallest pipe (4 units) takes 232 min ⇒ capacity $=4\times 232=928$ units
3. All together: $\frac{928}{29}=32$ min

Short trick: Time ∝ 1/rate: $232\times \frac{4}{29}=8\times 4=32$ (232 = 29 × 8).

Answer: (C) 32

Concept: Alligation: cheaper : dearer = (dearer − mean) : (mean − cheaper).

1. A = ₹245/7 = ₹35 per kg; B = ₹300/5 = ₹60 per kg
2. A : B = (60 − 50) : (50 − 35) = 10 : 15
3. = 2 : 3

Short trick: Distances from ₹50: A is 15 below, B is 10 above; quantities go inversely → A : B = 10 : 15 = 2 : 3.

Answer: (B) 2 : 3

Concept: Average speed = total distance ÷ total time.

1. Distance $=59\times 3.5+68\times 1=206.5+68=274.5$ km
2. Time $=4.5$ h
3. Average $=\frac{274.5}{4.5}=61$ km/h

Short trick: Time-weighted: $59+\frac{9\times 1}{4.5}=59+2=61$.

Answer: (A) 61

Concept: Dividend = Divisor × Quotient + Remainder.

1. Divisor $=4\times 22=88$
2. Quotient $=\frac{88}{2}=44$
3. Dividend $=88\times 44+22=3872+22=3894$

Short trick: Dividend − remainder must be divisible by the divisor: only $3894-22=3872=88\times 44$ ✓.

Answer: (A) 3894

Concept: Successive increases multiply: new = old × (1 + a%) × (1 + b%).

1. $1600\times 1.1=1760$
2. $1760\times 1.2=2112$

Short trick: Net increase $=10+20+\frac{10\times 20}{100}=32\mathrm{\%}$; $1600\times 1.32=2112$.

Answer: (D) 2112

Concept: AB ∥ OC ∥ DE (arrows). Co-interior angles add up to ${180}^{\circ }$; angles round a point add up to ${360}^{\circ }$.

1. AB ∥ OC with transversal AO: $\mathrm{\angle }BAO+\mathrm{\angle }AOC={180}^{\circ }$ ⇒ $\mathrm{\angle }AOC={145}^{\circ }$
2. Around O: $\mathrm{\angle }AOC+\mathrm{\angle }AOD+x={360}^{\circ }$
3. $x={360}^{\circ }-{145}^{\circ }-{64}^{\circ }={151}^{\circ }$

Short trick: $x={180}^{\circ }+{35}^{\circ }-{64}^{\circ }={151}^{\circ }$ (the "180 + top angle − middle angle" shortcut for this Z-shape).

Answer: (D) ${151}^{\circ }$

Concept: Net rate = filling rate − emptying rate.

1. Net part per hour $=\frac{1}{5}-\frac{1}{9}=\frac{4}{45}$
2. Time $=\frac{45}{4}=11.25$ h
3. $0.25$ h $=15$ min ⇒ 11 h 15 min

Short trick: Time $=\frac{ab}{b-a}=\frac{5\times 9}{9-5}=\frac{45}{4}$ h = 11 h 15 min.

Answer: (B) 11 hours 15 minutes

Concept: Overall profit % = weighted average of the parts.

1. Parts: 50%, then 30% of remaining 50% = 15%, rest = 35%
2. $x=0.50(40)+0.15(46)-0.35(24)$
3. $=20+6.9-8.4=18.5$

Short trick: Take 100 items of CP ₹1: profit $=20+6.9-8.4=18.5$ on ₹100 → 18.5%.

Answer: (A) 18.5

Concept: Volume ∝ $r^3$, surface area ∝ $r^2$; paint ∝ surface area.

1. Each small sphere has volume ${\left(\frac{1}{10}\right)}^3=\frac{1}{1000}$ of the big one.
2. Usable volume $=95\mathrm{\%}$ ⇒ number of small spheres $=0.95\times 1000=950$
3. Each small sphere has area $\frac{1}{100}$ of the big one ⇒ total area $=\frac{950}{100}=9.5$ times
4. Paint $=9.5\times 14=133$ litres

Short trick: Total area factor $=0.95\times k$ where $k=10$ (radius ratio): $14\times 0.95\times 10=133$.

Answer: (C) 133 liters

Concept: Equal final amounts: $R(1.08{)}^{10}=B(1.08{)}^{11}$.

1. Rakesh: 10 years to 23; Baban: 11 years
2. $R=1.08B\Rightarrow R:B=108:100=27:25$
3. $R=\frac{27}{52}\times 24700=27\times 475=12825$

Short trick: The older child needs more money: ratio = $(1+\frac{8}{100}{)}^1$ for the 1-year gap = 27 : 25 → $24700÷52=475$; Rakesh $=27\times 475$.

Answer: (A) 12825

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

1. $480=\frac{P\times 12\times 2}{100}$
2. $P=\frac{48000}{24}=2000$

Short trick: Total rate $=12\times 2=24\mathrm{\%}$; 24% of P = 480 → P = 2000.

Answer: (C) 2000