RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Area of trapezium $=\frac{1}{2}\times (\text{sum of parallel sides})\times \text{height}$.

1. Sum of bases $=8+12=20\text{cm}$.
2. Area $=\frac{1}{2}\times 20\times 5=50{\text{cm}}^2$

Trick: Average base × height $=10\times 5=50$.

Answer: (B) $50{\text{cm}}^2$

Concept: Alligation: ratio of quantities $=$ (opposite differences from the mean).

1. Managers : Supervisors $=(44400-42000):(48000-44400)=2400:3600=2:3$.
2. Supervisors $=50\times \frac{3}{5}=30$.
3. Check: $\frac{20\times 48000+30\times 42000}{50}=44400$ ✔

Trick: Mean is closer to 42000, so supervisors are more than half → 30.

Answer: (D) $30$

Concept: BODMAS – brackets first, then 'of', then division/multiplication, then addition.

1. $\frac{19}{18}÷\frac{19}{6}=\frac{19}{18}\times \frac{6}{19}=\frac{1}{3}$
2. $\frac{5}{18}\times \frac{16}{15}+\frac{4}{6}=\frac{8}{27}+\frac{18}{27}=\frac{26}{27}$
3. $\frac{1}{3}÷\frac{26}{27}=\frac{9}{26}$; and $\frac{24}{25}\text{ of }\frac{25}{8}=3$, so $\frac{6}{8}÷3=\frac{1}{4}$.
4. $\frac{9}{26}+\frac{1}{4}=\frac{18+13}{52}=\frac{31}{52}$

Trick: 'of' is solved before $÷$ – the most common trap in such questions.

Answer: (C) $\frac{31}{52}$

Concept: $SP=MP\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})$.

1. After 15%: $8000\times 0.85=6800$.
2. After 10%: $6800\times 0.9=6120$.

Trick: Net discount $=15+10-\frac{15\times 10}{100}=23.5\%$ → $8000\times 0.765=6120$.

Answer: (A) ₹$6,120$

Concept: ${\text{cosec}}^{2}A=1+{\cot }^{2}A$.

1. $2\cot A=4\Rightarrow \cot A=2$.
2. ${\text{cosec}}^{2}A=1+4=5$.
3. $\frac{5+1}{5-1}=\frac{6}{4}=1.5$

Trick: Expression $=\frac{2+{\cot }^{2}A}{{\cot }^{2}A}=\frac{2+4}{4}=1.5$ directly.

Answer: (A) $1.5$

Concept: Third proportional to $a$ and $b$ is $\frac{b^2}{a}$ (since $a:b=b:n$).

1. $n=\frac{{5.6}^2}{0.8}=\frac{31.36}{0.8}$
2. $n=39.2$

Trick: $\frac{5.6}{0.8}=7$, so $n=5.6\times 7=39.2$.

Answer: (A) $39.2$

Concept: BODMAS with 'of' solved before division.

1. $3\text{ of }2=6$; $24÷6=4$.
2. $4\times 4=16$; $18+16=34$.
3. $34-10=24$

Trick: If you do $24÷3\times 2$ instead of using 'of', you get 72 – a trap option.

Answer: (C) $24$

Concept: Find 1 part from the known height.

1. Shyam $=7$ parts $=140\Rightarrow 1$ part $=20\text{cm}$.
2. Manish $=8\times 20=160\text{cm}$.
3. Manish is taller by $160-140=20\text{cm}$.

Trick: Difference is 1 part $=\frac{140}{7}=20$, and the bigger ratio term (8) belongs to Manish.

Answer: (B) Manish is taller by 20 cm

Concept: Take total work $=$ LCM of days; efficiency $=\frac{\text{work}}{\text{days}}$.

1. Work $=\text{LCM}(12,28)=84$; A $=7$/day, B $=3$/day.
2. In 3 days together: $3\times 10=30$ units.
3. Remaining $=54$ units; B alone: $\frac{54}{3}=18$ days.

Trick: Fractions: $3(\frac{1}{12}+\frac{1}{28})=\frac{5}{14}$; rest $\frac{9}{14}\times 28=18$.

Answer: (C) $18$

Concept: Net change $=a+b+\frac{ab}{100}$ (use $-$ sign for decrease).

1. $a=+20$, $b=-25$.
2. Net $=20-25-\frac{20\times 25}{100}=-5-5=-10\%$

Trick: $100\to 120\to 90$ → 10% decrease.

Answer: (C) $10\%$ decrease

Concept: Average speed $=\frac{\text{Total distance}}{\text{Total time}}$ (not the average of speeds).

1. Times: $\frac{120}{40}=3$, $\frac{180}{60}=3$, $\frac{100}{50}=2$ hours.
2. Total distance $=400\text{km}$, total time $=8\text{h}$.
3. Average speed $=\frac{400}{8}=50\text{km/h}$

Trick: Never take $\frac{40+60+50}{3}$ blindly – here it also gives 50 only by coincidence.

Answer: (A) $50\text{km/h}$

Concept: Petrol used is directly proportional to distance.

1. Mileage $=\frac{120}{8}=15\text{km}$ per litre.
2. Petrol for 225 km $=\frac{225}{15}=15$ litres.

Trick: $\frac{225}{120}=\frac{15}{8}$ → $8\times \frac{15}{8}=15$.

Answer: (C) $15$ liters

Concept: Total work $=$ LCM of times; efficiency $=\frac{\text{work}}{\text{time}}$.

1. Work $=\text{LCM}(12,18,24)=72$; A $=6$, B $=4$, C $=3$ units/h.
2. 4 hours together: $4\times 13=52$ units.
3. Remaining $=20$ units; B + C $=7$ units/h.
4. Time $=\frac{20}{7}=2\frac{6}{7}$ hours.

Trick: $\frac{20}{7}$ lies between 2 and 3 → only (A) or (B); $20=2\times 7+6$ → $2\frac{6}{7}$.

Answer: (A) $2\frac{6}{7}$ hours

Concept: Heron's formula: Area $=\sqrt{s(s-a)(s-b)(s-c)}$, $s=\frac{a+b+c}{2}$.

1. Sides $=5k+6k+7k=90\Rightarrow k=5$ → $25,30,35\text{cm}$.
2. $s=45$; $s-a=20$, $s-b=15$, $s-c=10$.
3. Area $=\sqrt{45\times 20\times 15\times 10}=\sqrt{135000}=\sqrt{22500\times 6}=150\sqrt{6}$

Trick: For ratio $5:6:7$, area $=6\sqrt{6}{k}^2=6\sqrt{6}\times 25=150\sqrt{6}$.

Answer: (D) $150\sqrt{6}{\text{cm}}^2$

Concept: Mode is the value that occurs most often.

1. Frequencies: 12 → 2, 15 → 2, 18 → 4, 20 → 1, 22 → 3, 25 → 1, 28 → 1.
2. 18 occurs the maximum (4) times.

Trick: Only compare the repeated values – 18 (4 times) beats 22 (3 times).

Answer: (B) $18$

Concept: Equivalent single change $=a+b+\frac{ab}{100}$.

1. $a=b=15$.
2. $15+15+\frac{15\times 15}{100}=30+2.25=32.25\%$

Trick: ${1.15}^2=1.3225$ → $32.25\%$ increase.

Answer: (C) $32.25\%$

Concept: Under CI, if money becomes $x$ times in $n$ years, it becomes $x^k$ times in $kn$ years.

1. $3$ times in 6 years.
2. $27=3^3$ → time $=3\times 6=18$ years.

Trick: CI multiplies: $\times 3$ every 6 years → $1\to 3\to 9\to 27$ (3 steps = 18 years).

Answer: (D) $18$ years

Concept: If numbers are in ratio $a:b$ (co-prime) with HCF $h$, numbers are $ha$, $hb$ and LCM $=h\times a\times b$.

1. Numbers $=6\times 3=18$ and $6\times 8=48$.
2. LCM $=6\times 3\times 8=144$

Trick: LCM must be a multiple of 6 and 8 (i.e. of 24) → only 144 fits.

Answer: (D) $144$

Concept: Number $=10a+b$, reversed $=10b+a$.

1. $4(a+b)=10b+a\Rightarrow 3a=6b\Rightarrow a=2b$.
2. $a-b=3\Rightarrow b=3$, $a=6$.
3. Number $=63$; check: $4\times 9=36$ ✔

Trick: Test options: only 63 gives $4\times (6+3)=36$ = its reverse.

Answer: (A) $63$

Concept: $\text{Profit}\%=\frac{SP-CP}{CP}\times 100$.

1. CP per kg $=\frac{1000}{20}=50$.
2. Profit per kg $=55-50=5$.
3. $\text{Profit}\%=\frac{5}{50}\times 100=10\%$

Trick: Total SP $=20\times 55=1100$ → ₹100 profit on ₹1000 = 10%.

Answer: (B) $10\%$