RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Compounded half-yearly → rate becomes half and time (periods) becomes double: $A=P{(1+\frac{R/2}{100})}^{2T}$.

1. Rate per half-year $=\frac{10}{2}=5\%$, periods $=1.5\times 2=3$.
2. $A=12000\times {\left(\frac{21}{20}\right)}^3=12000\times \frac{9261}{8000}$.
3. $A=1.5\times 9261=13891.50$

Trick: 3 years of 5% CI = $15.7625\%$ → $12000\times 1.157625=13891.5$.

Answer: (C) ₹$13,891.50$

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

1. After 15% discount: $16000\times 0.85=13600$.
2. $13600\times (1-\frac{N}{100})=12240\Rightarrow 1-\frac{N}{100}=\frac{12240}{13600}=0.9$.
3. $N=10\%$

Trick: Second discount $=\frac{13600-12240}{13600}\times 100=\frac{1360}{13600}\times 100=10\%$.

Answer: (C) $10\%$

Concept: Solve the bracket first (BODMAS): division and multiplication, then addition.

1. $\frac{8.16}{2.4}=\frac{81.6}{24}=3.4$
2. $3.64\times 1.5=5.46$
3. $3.4+5.46=8.86$

Trick: $3.64\times 1.5=3.64+1.82=5.46$ (half of the number added to itself).

Answer: (D) $8.86$

Concept: In a ratio, find the value of 1 part from the known quantity, then multiply by total parts.

1. Passed : Failed $=11:2$; total parts $=13$.
2. Failed $=2$ parts $=44\Rightarrow 1$ part $=22$.
3. Total students $=13\times 22=286$

Trick: Total must be a multiple of 13 → only 286 ($13\times 22$) fits.

Answer: (C) $286$

Concept: $(a-b{)}^2=(a+b{)}^2-4ab$.

1. $(a-b{)}^2={31}^2-4\times 220=961-880=81$
2. $a-b=\sqrt{81}=9$
3. Check: numbers are 20 and 11 ($20+11=31$, $20\times 11=220$).

Trick: Factor 220 into a pair adding to 31 → $20\times 11$ → difference 9.

Answer: (B) $9$

Concept: For a fixed distance, time $\propto \frac{1}{\text{speed}}$.

1. New speed $=\frac{3}{4}$ of original.
2. New time $=\frac{4}{3}$ of original.
3. Increase $=\frac{4-3}{3}\times 100=\frac{100}{3}=33.33\%$

Trick: If speed falls by $x\%$, time rises by $\frac{x}{100-x}\times 100=\frac{25}{75}\times 100=33.33\%$.

Answer: (D) $33.33\%$

Concept: Let the son's present age be $x$; form the equation for the future condition.

1. Rahul $=x$, Mohit $=3x$.
2. After 12 years: $3x+12=2(x+12)\Rightarrow x=12$.
3. Mohit $=36$, sum $=36+12=48$ years.

Trick: Check options: sum must split as $1:3$ → 48 gives 12 and 36; after 12 years 48 and 24 ✔.

Answer: (A) $48$

Concept: Number of items $=\frac{\text{Total cost}}{\text{Price per item}}$; form an equation from the difference in counts.

1. Let mystery novel price $=x$, biography price $=x+40$.
2. $\frac{4800}{x}-\frac{4800}{x+40}=10\Rightarrow 4800\times 40=10x(x+40)$
3. $x^2+40x-19200=0\Rightarrow (x+160)(x-120)=0\Rightarrow x=120$
4. Biography price $=160$ → books $=\frac{4800}{160}=30$ (mystery $=40$ ✔).

Trick: Test options: 30 books → ₹160 each; 40 novels → ₹120 each; difference ₹40 ✔.

Answer: (D) $30$

Concept: Diagonals of a rhombus bisect each other at right angles: ${\text{side}}^2={\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2$.

1. Let diagonals be $3k$ and $4k$; halves $1.5k$ and $2k$.
2. $(1.5k{)}^2+(2k{)}^2=(12.5{)}^2\Rightarrow 6.25k^2=156.25\Rightarrow k=5$
3. Shorter diagonal $=3k=15$ units.

Trick: Half-diagonals and side form a $3:4:5$ triangle → $5$ parts $=12.5$, so half-diagonals are $7.5$ and $10$.

Answer: (D) $15$ units

Concept: Repair cost is part of the cost price: $CP=\frac{SP\times 100}{100+\text{Profit}\%}$.

1. Total CP $=\frac{600\times 100}{125}=480$.
2. Purchase price $=480-50=430$.

Trick: 25% profit → $SP=\frac{5}{4}CP$ → $CP=\frac{4}{5}\times 600=480$; remove repair ₹50.

Answer: (B) ₹$430$

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