RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Equal areas: $l\times b=\pi r^2$, then perimeter of circle $=2\pi r$.

1. Area of rectangle $=55\times 70=3850{\text{m}}^2$.
2. $\frac{22}{7}{r}^2=3850\Rightarrow r^2=3850\times \frac{7}{22}=1225\Rightarrow r=35\text{m}$.
3. Circumference $=2\times \frac{22}{7}\times 35=220\text{m}$.

Trick: $3850=22\times 175=22\times 7\times 25$, so $r^2=7^2\times 5^2$ and $r=35$ at a glance.

Answer: (C) 220

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

1. Here $22+(-18)+(-4)=0$.
2. So the sum $=3\times 22\times (-18)\times (-4)=3\times 22\times 72$.
3. $=66\times 72=4752$.

Trick: Whenever the three bases add to zero, never cube them – just multiply $3abc$.

Answer: (C) 4752

Concept: Form two linear equations from the two time points.

1. Let present ages be $S$ (Shyam) and $R$ (Rajat). 4 years ago: $S-4=7(R-4)\Rightarrow S=7R-24$.
2. After 8 years: $S+8=3(R+8)\Rightarrow S=3R+16$.
3. $7R-24=3R+16\Rightarrow 4R=40,R=10$, so $S=7\times 10-24=46$.

Trick: Check: 4 years ago 42 and 6 (ratio 7:1); after 8 years 54 and 18 (ratio 3:1). Both fit.

Answer: (A) 46 years

Concept: Same distance both ways: average speed $=\frac{2xy}{x+y}$.

1. $x=32,y=24$.
2. Average $=\frac{2\times 32\times 24}{32+24}=\frac{1536}{56}$.
3. $=27.428\dots \approx 27.43\text{km/h}$.

Trick: The 48 km distance is a distractor – for equal distances it cancels out.

Answer: (D) 27.43 km/hr

Concept: Length of train $=$ relative speed $\times$ time; relative speed (same direction) $=v-u$.

1. Let train speed be $v$ km/h. $L=(v-5)\times \frac{5}{18}\times 18=5(v-5)$.
2. Also $L=(v-7)\times \frac{5}{18}\times 20=\frac{50(v-7)}{9}$.
3. $45(v-5)=50(v-7)\Rightarrow 5v=125,v=25$; $L=5\times 20=100\text{m}$.

Trick: Convert km/h to m/s with $\times \frac{5}{18}$ first; the 18 s cancels neatly.

Answer: (D) 100

Concept: Two numbers are co-prime when their HCF is 1.

1. 44 and 99: HCF 11. 42 and 56: HCF 14. 27 and 36: HCF 9.
2. 25 $=5^2$ and 36 $=2^2\times 3^2$ share no prime factor, HCF $=1$.

Trick: Two consecutive numbers, or two perfect squares of different primes, are always co-prime.

Answer: (B) 25 and 36

Concept: Take total work = LCM(10, 15, 20) = 60 units.

1. A $=+6$, B $=+4$, C $=-3$ units/hour.
2. All three for 4 hours: $(6+4-3)\times 4=28$ units. Remaining $=32$ units.
3. A + B $=10$ units/hour $\Rightarrow \frac{32}{10}=3.2$ hours. Total $=4+3.2=7.2$ hours.

Trick: With LCM units the fractions vanish – work only with whole numbers till the last step.

Answer: (B) 7.2 hours

Concept: $\%\text{ decrease}=\frac{\text{decrease}}{\text{original}}\times 100$.

1. Decrease $=500-450=50$.
2. $\frac{50}{500}\times 100=10\%$.

Trick: Always divide by the original (earlier) value, not the new one.

Answer: (B) 10%

Concept: In a right triangle, horizontal distance $=\frac{\text{height}}{\tan \theta }$; angle of depression = angle of elevation from the ship.

1. Nearer ship (60°): $BC=\frac{80}{\tan {60}^{\circ }}=\frac{80}{\sqrt{3}}$.
2. Farther ship (30°): $BD=\frac{80}{\tan {30}^{\circ }}=80\sqrt{3}$.
3. $CD=80\sqrt{3}-\frac{80}{\sqrt{3}}=\frac{80\times 2}{\sqrt{3}}=\frac{160}{1.732}\approx 92.38\text{m}$.

Trick: For angles 30° and 60° the gap is always $\frac{2h}{\sqrt{3}}$ – here $\frac{2\times 80}{\sqrt{3}}$.

Answer: (D) 92.38 meters

Concept: Compare what the customer pays with the cost of what he actually receives.

1. Let CP $=$ ₹100 per kg. MP $=124$, SP after 18% discount $=124\times 0.82=101.68$.
2. He gives only 800 g, whose cost $=80$.
3. Profit $=\frac{101.68-80}{80}\times 100=\frac{21.68}{80}\times 100=27.1\%$.

Trick: Price multiplier $1.24\times 0.82=1.0168$; weight multiplier $\frac{1000}{800}=1.25$; overall $1.0168\times 1.25=1.271\Rightarrow 27.1\%$.

Answer: (D) 27.1%

Concept: Write every number as a product of primes and cancel powers.

1. Numerator: $3^{-5}\times 2^{15}\times 5^9\times (2^2·3^2{)}^3=2^{21}\times 5^9\times 3^{6-5}=2^{21}\times 5^9\times 3$.
2. Denominator: $(2^2{5}^2{)}^4\times 2^8\times 2^3{3}^3\times 2^2=2^{21}\times 5^8\times 3^3$.
3. Ratio $=\frac{5^9\times 3}{5^8\times 3^3}=\frac{5}{9}$.

Trick: Count the powers of 2 first – they are equal (21 each), so only 3s and 5s survive.

Answer: (B) $\frac{5}{9}$

Concept: Mean proportional between $a$ and $b$ is $\sqrt{ab}$.

1. $\sqrt{75}=5\sqrt{3}$, so the second number is $10-5\sqrt{3}=5(2-\sqrt{3})$.
2. Product $=(2+\sqrt{3})\times 5(2-\sqrt{3})=5(4-3)=5$.
3. Mean proportional $=\sqrt{5}$.

Trick: Spot the conjugate pair $(2+\sqrt{3})(2-\sqrt{3})=1$ – the product collapses to 5 instantly.

Answer: (C) $\sqrt{5}$

Concept: A number is divisible by 9 if its digit sum is a multiple of 9.

1. Digit sum of 7KM93 $=7+K+M+9+3=19+K+M$.
2. Need $19+K+M=27$ or $36\Rightarrow K+M=8$ or $17$.
3. Minimise $4K^2+2M$: take $K=0,M=8\Rightarrow 0+16=16$ (any $K\ge 1$ gives more).

Trick: $K^2$ carries the big coefficient, so push $K$ to 0 and load everything on $M$.

Answer: (A) 16

Concept: No overall profit/loss $\Rightarrow$ total CP = total SP.

1. Total SP $=2\times 2973=5946=$ total CP.
2. CP of first (50% gain) $=\frac{2973}{1.5}=1982$.
3. CP of second $=5946-1982=3964$; loss $=3964-2973=991$; $\frac{991}{3964}\times 100=25\%$.

Trick: Gain $\frac{1}{2}$ on CP₁ means CP₁ $=\frac{2}{3}$SP; the other must then lose the same rupee amount: $\frac{1}{3}$SP on CP₂ $=\frac{4}{3}$SP $\Rightarrow 25\%$.

Answer: (D) 25%

Concept: $\sqrt[3]{4096}=16$ (since ${16}^3=4096$) and $\sqrt[4]{10000}=10$.

1. $\frac{16}{10}\times \frac{5}{48}\times 186$.
2. $\frac{16\times 5}{10\times 48}=\frac{80}{480}=\frac{1}{6}$.
3. $\frac{186}{6}=31$.

Trick: Remember $4096=2^{12}$, so its cube root is $2^4=16$ and its square root is 64.

Answer: (D) 31

Concept: Profit is shared in the ratio of (capital $\times$ time).

1. X : Y : Z $=60000\times 12:80000\times 6:120000\times 4=720:480:480=3:2:2$.
2. Z's share $=\frac{2}{7}\times 77000=22000$.

Trick: Divide all capitals by 20,000 first (3, 4, 6) then multiply by months: 36 : 24 : 24.

Answer: (A) ₹22,000

Concept: Apply multipliers one after another: $-30\%\to \times 0.7$, $+15\%\to \times 1.15$.

1. $800\times 0.7=560$.
2. $560\times 1.15=644$.

Trick: Net change $=-30+15-\frac{30\times 15}{100}=-19.5\%$; $800\times 0.805=644$.

Answer: (C) ₹644

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

1. $R=\frac{3600\times 100}{15000\times 4}$.
2. $=\frac{360000}{60000}=6\%$.

Trick: SI per year $=900$; $900$ is $6\%$ of $15000$.

Answer: (D) 6%

Concept: SP $=$ MP $\times (1-d_1)(1-d_2)$.

1. After 20%: $1250\times 0.8=1000$.
2. $1000\to 900$ is a drop of $100$, i.e. $\frac{100}{1000}=10\%$.

Trick: Second discount is always calculated on the already-reduced price, never on the MP.

Answer: (A) 10%

Concept: Net rate = sum of filling rates − emptying rate.

1. Net rate $=\frac{1}{4}+\frac{1}{22}-\frac{1}{22}=\frac{1}{4}$ cistern per hour.
2. Time for $\frac{1}{4}$ cistern $=\frac{1/4}{1/4}=1$ hour.

Trick: The 22-hour filler and 22-hour emptier cancel exactly – only the 4-hour tap matters.

Answer: (C) 1