RRB NTPC Undergraduate CBT 1 2025 PYQs

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

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

1. $SI=\frac{8000\times 7.5\times 3}{100}=1800$
2. Amount $=8000+1800=$ ₹9,800

Trick: $7.5\%\times 3=22.5\%$ of 8000 $=1800$ in one step.

Answer: (A) ₹9800

Concept: $SP=CP\times (1+\text{profit}\%)$. Let marked price $=M$.

1. $CP=0.8M$, $SP=M+10$
2. $M+10=1.4\times 0.8M=1.12M$
3. $0.12M=10\Rightarrow M=\frac{10}{0.12}=83.33$

Trick: $1.4\times 0.8=1.12$, so the ₹10 above MP is exactly $12\%$ of MP.

Answer: (A) ₹83.33

Concept: If $a+b+c=0$, then $a^3+b^3+c^3=3abc$.

1. $23+(-14)+(-9)=0$
2. Sum $=3\times 23\times (-14)\times (-9)=3\times 23\times 126$
3. $=69\times 126=8694$

Trick: Whenever three cubes are added and the bases sum to zero, skip cubing – just use $3abc$.

Answer: (B) 8694

Concept: Translate words into an equation: $A=2B+5$.

1. $B=15\Rightarrow 2B=30$
2. $A=30+5=35$ years

Trick: "5 more than twice" = double first, then add 5.

Answer: (A) 35

Concept: Overall $\%=\frac{\text{net gain}}{\text{total CP}}\times 100$.

1. One-third: CP ₹800, profit $20\%=+160$
2. One-fourth: CP ₹600, loss $10\%=-60$
3. Rest at cost: $0$. Net gain $=100$
4. $\frac{100}{2400}\times 100=\frac{25}{6}=4\frac{1}{6}\%$ profit

Trick: Weighted profit $=\frac{1}{3}(20)+\frac{1}{4}(-10)+0=\frac{20}{3}-\frac{5}{2}=\frac{25}{6}\%$ – no need to use ₹2400 at all.

Answer: (A) $4\frac{1}{6}\%$ profit

Concept: Compare the effective price paid per toy (take MP = ₹100 each).

1. (A) $100\times 0.79\times 0.92=$ ₹72.68 per toy
2. (B) $100\times 0.62\times 0.58\times 0.70\approx$ ₹25.17 per toy
3. (C) $8\times 83+2\times 80=824$ for 10, i.e. ₹82.40 per toy
4. (D) Pay for 10, get 17: $\frac{1000}{17}\approx$ ₹58.82 per toy
5. Highest effective price = scheme C → least beneficial.

Trick: Successive discounts always beat the plain sum; a scheme with only $17$–$20\%$ off can never compete with $21\%+8\%$ or three big discounts.

Answer: (C) C

Concept: Symmetric equations – add them to get $m+o$ directly.

1. $8m+5o=164$, $5m+8o=148$
2. Adding: $13(m+o)=312\Rightarrow m+o=24$
3. $3(m+o)=72$

Trick: When coefficients are swapped, adding gives the sum and subtracting gives the difference – no need to solve for each item.

Answer: (C) ₹72

Concept: In the right triangle, $\cos \theta =\frac{\text{base}}{\text{hypotenuse}}$.

1. $\cos {60}^{\circ }=\frac{15}{L}$
2. $\frac{1}{2}=\frac{15}{L}\Rightarrow L=30$ m

Trick: In a ${30}^{\circ }$-${60}^{\circ }$-${90}^{\circ }$ triangle the hypotenuse is twice the side opposite ${30}^{\circ }$ (the base here).

Answer: (B) 30 m

Concept: $a^m\times a^n=a^{m+n}$, $(a^m{)}^n=a^{mn}$, $\frac{a^m}{a^n}=a^{m-n}$.

1. Numerator: $5^{3x}\times 5^{2(x-1)}=5^{3x+2x-2}=5^{5x-2}$
2. Divide: $5^{(5x-2)-(4x-1)}=5^{x-1}$

Trick: Put $x=1$: expression $=\frac{5^3\times 1}{5^3}=1=5^0$; only $5^{x-1}$ gives $5^0$.

Answer: (A) $5^{x-1}$

Concept: Surface area $=4\pi r^2$, so the ratio of areas $={\left(\frac{r_1}{r_2}\right)}^2$.

1. $\frac{r_A}{r_B}=\frac{6}{3}=2$
2. Area ratio $=2^2=4:1$

Trick: Double the radius → area $\times 4$, volume $\times 8$.

Answer: (A) 4:1