RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Apply BODMAS – brackets first, then "of", then $÷$, $\times$, $+$, $-$.

1. $\frac{16}{36}÷\frac{16}{4}=\frac{4}{9}\times \frac{4}{16}=\frac{1}{9}$
2. $\frac{2}{7}\times \frac{14}{6}+\frac{5}{12}=\frac{2}{3}+\frac{5}{12}=\frac{13}{12}$, so first part $=\frac{1}{9}÷\frac{13}{12}=\frac{1}{9}\times \frac{12}{13}=\frac{4}{39}$
3. $\frac{39}{12}\text{ of }\frac{12}{7}=\frac{39}{7}$, so $\frac{4}{7}÷\frac{39}{7}=\frac{4}{7}\times \frac{7}{39}=\frac{4}{39}$
4. Total $=\frac{4}{39}+\frac{4}{39}=\frac{8}{39}$

Trick: "of" is done before $÷$. Cancel common factors early – both parts come out as $\frac{4}{39}$, so the answer is simply double.

Answer: (B) $\frac{8}{39}$

Concept: Two ratios are equal if they reduce to the same simplest form.

1. $35:25=7:5$, $45:20=9:4$, $60:45=4:3$
2. $120:80$ – divide both by 40: $3:2$ ✓

Trick: Check $\frac{a}{b}=1.5$ quickly: $\frac{120}{80}=1.5$, others give $1.4,2.25,1.33$.

Answer: (D) 120:80

Concept: Effective CI rate for 2 years at $r\%$ $=2r+\frac{r^2}{100}$.

1. At $4\%$: $8+0.16=8.16\%$ of the sum $=$ ₹765
2. Sum $y=\frac{765\times 100}{8.16}=$ ₹9,375
3. Amount $=9375+765=$ ₹10,140

Trick: Amount = Principal + CI, so just add 765 to the principal – no need to recompute $(1.04{)}^2$.

Answer: (C) ₹10,140

Concept: Discounted price $=\text{MP}\times (1-d\%)$; apply the extra offer on this price.

1. Price after $18\%$ discount $=2800\times 0.82=$ ₹2,296
2. 5 pairs: $5\times 2296=$ ₹11,480
3. 2 pairs at half: $2\times 1148=$ ₹2,296
4. Total $=11480+2296=$ ₹13,776

Trick: 2 pairs at $50\%$ off = 1 pair at full discounted price, so total $=6\times 2296$.

Answer: (A) ₹13,776

Concept: If numbers are $3k,7k,2k$ (co-prime ratio terms), HCF $=k$ and LCM $=3\times 7\times 2\times k=42k$.

1. $42k=8820\Rightarrow k=210$
2. HCF $=k=210$

Trick: For co-prime ratio terms, $\text{HCF}=\frac{\text{LCM}}{\text{product of ratio terms}}$.

Answer: (C) 210

Concept: Take total work = LCM of times and work in units/hour.

1. LCM(10,20,12) $=60$ units. A+B $=6$, B+C $=3$, C+A $=5$ units/h
2. Adding: $2(A+B+C)=14\Rightarrow A+B+C=7$
3. B $=7-(C+A)=7-5=2$ units/h
4. Time for B $=\frac{60}{2}=30$ hours

Trick: Single pipe = (sum of all three pairs)$÷2$ − (pair without it).

Answer: (D) 30 hours

Concept: Surface area of a sphere $=4\pi r^2$.

1. $r=\frac{10}{2}=5$ m
2. Area $=4\times 3.14\times 25=314{\text{m}}^2$
3. Cost $=314\times 5=$ ₹1,570

Trick: $4\pi r^2=\pi d^2$; with $d=10$, area $=100\pi =314$ directly.

Answer: (C) ₹1570

Concept: New member = new total − old total.

1. Old total $=6\times 24000=1,44,000$
2. New total $=7\times 25500=1,78,500$
3. New employee $=178500-144000=$ ₹34,500

Trick: New income = new average + (rise in average × old count) $=25500+1500\times 6=34500$.

Answer: (D) ₹34,500

Concept: Proportion of science students $=\frac{\text{science}}{\text{total}}$; compare the fractions.

1. A: $\frac{5}{13}\approx 0.385$
2. B: $\frac{7}{18}\approx 0.389$
3. C: $\frac{3}{8}=0.375$
4. Highest is Batch B.

Trick: Compare $\frac{5}{13}$ and $\frac{7}{18}$ by cross-multiplying: $5\times 18=90<7\times 13=91$, so $\frac{7}{18}$ is bigger.

Answer: (A) Batch B

Concept: Third proportional to $a$ and $b$ is $\frac{b^2}{a}$.

1. $m=\frac{{20}^2}{4}=100$
2. $n=\frac{{50}^2}{100}=25$

Trick: $a:b::b:c\Rightarrow c=\frac{b^2}{a}$ – the middle term is squared.

Answer: (D) 25

Concept: Angles opposite equal sides are equal; angle sum of a triangle $={180}^{\circ }$.

1. $XY=XZ\Rightarrow \angle Z=\angle Y$
2. $\angle Y+\angle Z={180}^{\circ }-{40}^{\circ }={140}^{\circ }$
3. $\angle Z=\frac{{140}^{\circ }}{2}={70}^{\circ }$

Trick: Base angle of an isosceles triangle $=\frac{{180}^{\circ }-\text{vertex angle}}{2}$.

Answer: (B) ${70}^{\circ }$

Concept: Mean $=\frac{\sum fx}{\sum f}$.

1. $\frac{10\times 5+15p}{5+p}=12.5$
2. $50+15p=62.5+12.5p\Rightarrow 2.5p=12.5$
3. $p=5$

Trick: Mean 12.5 is exactly midway between 10 and 15, so the frequencies must be equal: $p=5$.

Answer: (B) 5

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$ (never the average of speeds).

1. Times: $\frac{100}{50}=2$ h, $\frac{150}{60}=2.5$ h, $\frac{50}{40}=1.25$ h
2. Total: 300 km in $5.75$ h
3. Average speed $=\frac{300}{5.75}=52.17\text{km/h}$

Trick: $\frac{300}{5.75}=\frac{1200}{23}\approx 52.17$ – only option B lies between 50 and 60 close to this value.

Answer: (B) 52.17 km/h

Concept: Successive changes multiply: $P\times (1+10\%)\times (1-15\%)$.

1. $P\times 1.1\times 0.85=0.935P$
2. $0.935P=187000\Rightarrow P=\frac{187000}{0.935}=2,00,000$

Trick: Net change $=10-15-\frac{10\times 15}{100}=-6.5\%$, so $93.5\%$ of $P=187000$.

Answer: (A) 2,00,000

Concept: A number is divisible by 11 if (sum of digits at odd places) − (sum at even places) is 0 or a multiple of 11.

1. 3762: $(3+6)-(7+2)=0$ ✓ divisible
2. 6184: $(6+8)-(1+4)=9$ ✗; 4915: $(4+1)-(9+5)=-9$ ✗; 7393: $(7+9)-(3+3)=10$ ✗

Trick: $3762=11\times 342$ – alternate-digit difference zero means divisible.

Answer: (A) 3,762

Concept: Dividend $=\text{Divisor}\times \text{Quotient}+\text{Remainder}$.

1. Divisor $=5\times 56=280$
2. Quotient $=\frac{280}{14}=20$
3. Dividend $=280\times 20+56=5656$

Trick: Find the divisor from the remainder first – it links both given relations.

Answer: (B) 5656

Concept: Net change for $+a\%$ then $-b\%$ $=a-b-\frac{ab}{100}$.

1. Net $=12-10-\frac{120}{100}=+0.8\%$
2. $0.8\%$ of salary $=1584\Rightarrow$ salary $=\frac{1584\times 100}{0.8}=$ ₹1,98,000

Trick: $1.12\times 0.9=1.008$; the extra $0.008$ part equals ₹1,584, so salary $=1584÷0.008$.

Answer: (D) 1,98,000

Concept: Speed $=\frac{\text{Distance}}{\text{Time}}$.

1. Original speed $=\frac{300}{5}=60\text{km/h}$
2. New speed $=\frac{300}{6}=50\text{km/h}$
3. Change $=60-50=10\text{km/h}$ decrease

Trick: More time for the same distance always means lower speed – so only "decrease" options qualify.

Answer: (D) 10 km/h decrease

Concept: Multiply inside each bracket, then divide.

1. $0.25\times 1.6=0.4$
2. $0.04\times 0.5=0.02$
3. $\frac{0.4}{0.02}=\frac{40}{2}=20$

Trick: Remove decimals: $\frac{25\times 16}{4\times 5}\times \frac{{10}^{-3}}{{10}^{-4}}=20\times 1=20$… or simply note $\frac{0.25}{0.5}=0.5$ and $\frac{1.6}{0.04}=40$, product 20.

Answer: (A) 20

Concept: Total work = LCM of days; efficiency = work per day.

1. LCM(12,39) $=156$ units; A $=13$, B $=4$ units/day
2. 4 days together: $4\times 17=68$ units; remaining $=88$
3. B alone: $\frac{88}{4}=22$ days

Trick: In 4 days A does $\frac{4}{12}=\frac{1}{3}$ of the work; B must do the rest alone, so B's remaining time $=39\times (1-\frac{1}{3}-\frac{4}{39})=22$.

Answer: (D) 22