RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Do multiplication before subtraction (BODMAS).

1. $\frac{2}{5}\times \frac{10}{20}=\frac{1}{5}=0.2$.
2. $5.25+2.75=8$; $8-0.6-0.2=7.2$.

Trick: $5.25+2.75$ is a neat 8; the rest is just $0.6+0.2$.

Answer: (B) 7.2

Concept: The altitude to the base of an isosceles triangle bisects the base; use Pythagoras and $a^2-b^2=(a-b)(a+b)$.

1. Equal side $a$, half-base $b$: $2a+2b=48\Rightarrow a+b=24$.
2. $a^2-b^2=(12\sqrt{2}{)}^2=288\Rightarrow a-b=\frac{288}{24}=12$ → $a=18,b=6$.
3. Area $=\frac{1}{2}\times 12\times 12\sqrt{2}=72\sqrt{2}{\text{cm}}^2$.

Trick: $a+b$ and $a^2-b^2$ known → $a-b$ in one division; no need to solve quadratics.

Answer: (D) $72\sqrt{2}{\text{cm}}^2$

Concept: Third proportional to $a$ and $b$ is $c$ with $a:b::b:c$, i.e. $c=\frac{b^2}{a}$.

1. $c=\frac{{12}^2}{6}=\frac{144}{6}=24$.

Yaad rakhein: Third proportional $=\frac{b^2}{a}$; mean proportional $=\sqrt{ab}$ – do not mix them.

Answer: (A) 24

Concept: Fourth proportional to $a,b,c$ is $d=\frac{bc}{a}$.

1. Numbers $2k,5k,7k$: $d=\frac{5k\times 7k}{2k}=\frac{35k}{2}$.
2. $\frac{35k}{2}=140\Rightarrow k=8$.
3. Middle number $=5k=40$ (numbers are 16, 40, 56).

Trick: Check: $16:40=56:140$ (both $=2:5$) ✓.

Answer: (D) 40

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

1. Original speed $=\frac{90}{3}=30\text{km/h}$; new speed $=35\text{km/h}$.
2. Time $=\frac{70}{35}=2$ hours.

Trick: $70=2\times 35$ – the numbers are chosen to cancel cleanly.

Answer: (D) 2 hours

Concept: To undo an increase of $r\%$, reduce by $\frac{r}{100+r}\times 100\%$.

1. $\frac{25}{125}\times 100=20\%$.
2. Check: $125\times 0.8=100$ ✓.

Yaad rakhein: $25\%=\frac{1}{4}$ up → $\frac{1}{5}=20\%$ down; $20\%$ up → $16\frac{2}{3}\%$ down.

Answer: (B) 20%

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

1. Divisor $=4\times 15=60$; quotient $=\frac{60}{5}=12$.
2. Dividend $=60\times 12+15=735$.

Trick: Remainder 15 must be less than the divisor 60 – a quick sanity check.

Answer: (D) 735

Concept: Take capacity = LCM of the times; add filling rates, subtract emptying rate.

1. Capacity $=\text{LCM}(4,10,10)=20$ units; rates $+5,+2,-2$ per hour → net $5$/h.
2. One-fourth $=5$ units → time $=\frac{5}{5}=1$ hour.

Trick: The 10-h filler and 10-h emptier cancel, leaving just the 4-h tap: quarter of 4 h = 1 h.

Answer: (C) 1

Concept: $x^m\times x^n=x^{m+n}$ and $\frac{x^m}{x^n}=x^{m-n}$.

1. $x^2\times x^3=x^5\Rightarrow a=5$.
2. $\frac{x^2}{x^b}=x^1\Rightarrow 2-b=1\Rightarrow b=1$.
3. $a+b=6$.

Yaad rakhein: Multiply → add powers; divide → subtract powers.

Answer: (B) 6

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

1. Part 1: $40$ km at $20$ km/h → $2$ h.
2. Part 2: $3$ h at $40$ km/h → $120$ km.
3. Average $=\frac{40+120}{2+3}=\frac{160}{5}=32\text{km/h}$.

Trick: Time-weighted: $\frac{2\times 20+3\times 40}{5}=32$.

Answer: (A) 32 km/h

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