RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: For the same distance, time is inversely proportional to speed: $T_1{S}_1=T_2{S}_2$.

1. Distance $=138\times 8=1104\text{km}$.
2. Malaika's time $=\frac{1104}{92}=12$ hours.

Trick: Speed ratio $138:92=3:2$, so time ratio $=2:3$. $8\times \frac{3}{2}=12$ hours.

Answer: (B) 12 hours

Concept: With a false weight, the customer pays for 1 kg but gets less. Effective $\text{SP}:\text{CP}=\frac{1+\text{profit}}{\text{actual weight}}$.

1. Let marked weight be 1 kg costing ₹100; actual weight given $=0.92$ kg, costing ₹92.
2. He charges $100\times 1.47=147$.
3. Profit $\%=\frac{147-92}{92}\times 100=\frac{55}{92}\times 100\approx 59.8\%\approx 60\%$.

Trick: $\frac{1.47}{0.92}=1.598$ → about $60\%$ profit.

Answer: (A) $60\%$

Concept: Convert both statements into linear equations and solve.

1. Let Mary $=M$, Diya $=D$. $3M=5D+20$ and $2D=M-2\Rightarrow M=2D+2$.
2. $3(2D+2)=5D+20\Rightarrow 6D+6=5D+20\Rightarrow D=14$.
3. $M=2(14)+2=30$. Difference $=30-14=16$ years.

Trick: Check options: difference 16 → $M=D+16$ and $M=2D+2$ give $D=14$, $M=30$; $3\times 30=90=5\times 14+20$ ✓.

Answer: (C) 16

Concept: Successive discounts multiply the remaining fractions: price left $=(1-d_1)(1-d_2)(1-d_3)$.

1. Remaining $=0.80\times 0.75\times 0.85$.
2. $0.80\times 0.75=0.60$; $0.60\times 0.85=0.51$.
3. Single discount $=1-0.51=0.49=49\%$.

Trick: Take ₹100: $100\to 80\to 60\to 51$. Discount $=49\%$.

Answer: (A) $49\%$

Concept: Test each number for divisibility by primes up to $\sqrt{345}\approx 18.6$ (2, 3, 5, 7, 11, 13, 17).

1. Even numbers and those ending in 5 are out: 332, 334, …, 335, 345.
2. 333 and 339 are divisible by 3; $341=11\times 31$; $343=7^3$.
3. Left: 331 and 337 — both prime. Count $=2$.

Yaad rakhein: 341 looks prime but $=11\times 31$ (alternating digit sum $3-4+1=0$).

Answer: (D) 2

Concept: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$, both per eraser.

1. CP of one eraser $=\frac{150}{12}=12.5$ (₹).
2. SP $=18$, profit $=5.5$.
3. Profit $\%=\frac{5.5}{12.5}\times 100=44\%$.

Trick: Per dozen: SP $=18\times 12=216$ vs CP 150 → $\frac{66}{150}=44\%$.

Answer: (C) 44

Concept: Make decimal places equal, find the LCM of the whole numbers, then put back the decimal places.

1. $15.4=15.400$ and $0.006$ → whole numbers 15400 and 6 (3 decimal places).
2. $15400=2^3\times 5^2\times 7\times 11$, $6=2\times 3$ → LCM $=15400\times 3=46200$.
3. Put back 3 decimal places: $46.200=46.2$.

Trick: LCM must be a multiple of 15.4: $15.4\times 3=46.2$, and $46.2÷0.006=7700$ ✓.

Answer: (C) 46.2

Concept: Divide by $\cos \theta$ to get $\tan \theta$; use ${\sec }^2\theta =1+{\tan }^2\theta$.

1. $\frac{\tan \theta +1}{\tan \theta -4}=11\Rightarrow \tan \theta +1=11\tan \theta -44\Rightarrow \tan \theta =\frac{45}{10}=\frac{9}{2}$.
2. Expression $=13{\tan }^2\theta +3-3{\sec }^2\theta =13{\tan }^2\theta +3-3-3{\tan }^2\theta =10{\tan }^2\theta$.
3. $=10\times \frac{81}{4}=\frac{405}{2}$.

Trick: $\frac{3}{{\cos }^2\theta }=3{\sec }^2\theta$; the constants 3 and $-3$ cancel, leaving $10{\tan }^2\theta$.

Answer: (C) $\frac{405}{2}$

Concept: The largest circle inside a rectangle has diameter equal to the shorter side.

1. Shorter side $=14\text{m}$ → diameter $=14$, radius $=7\text{m}$.
2. Circumference $=\pi d=\frac{22}{7}\times 14=44\text{m}$.

Yaad rakhein: The longer side (116 m) does not matter — the circle is limited by the width.

Answer: (B) 44

Concept: $A=P(1+\frac{RT}{100})$ for simple interest.

1. Total interest rate $=4\times 5=20\%$.
2. $P\times 1.20=2400\Rightarrow P=\frac{2400}{1.2}=2000$.

Trick: Amount $=120\%$ of principal → $\frac{2400}{120}\times 100=2000$.

Answer: (C) ₹2,000

Concept: $a^m÷a^n=a^{m-n}$ and $a^m\times a^n=a^{m+n}$ (work left to right).

1. ${62}^3÷{62}^2={62}^1$.
2. ${62}^1\times {62}^{-13}={62}^{1-13}={62}^{-12}$.

Trick: Just add/subtract powers: $3-2-13=-12$.

Answer: (D) ${62}^{-12}$

Concept: Apply each discount on the reduced price; effective discount $=a+b-\frac{ab}{100}$.

1. After $32\%$: $4000\times 0.68=2720$.
2. After $22\%$: $2720\times 0.78=2121.6$.
3. Effective discount $=32+22-\frac{32\times 22}{100}=54-7.04=46.96\%$.

Trick: Effective discount is always less than the plain sum ($54\%$), so (A) is a trap.

Answer: (D) ₹2,121.6 ; $46.96\%$

Concept: Combined average $=\frac{\text{total runs}}{\text{total matches}}$.

1. Runs in 23 matches $=23\times 41=943$.
2. Runs in 10 matches $=10\times 16=160$.
3. Average $=\frac{943+160}{33}=\frac{1103}{33}=33.42$.

Trick: Deviation method: $41-\frac{10\times 25}{33}=41-7.58=33.42$.

Answer: (B) 33.42

Concept: Time $=\frac{\text{Distance}}{\text{Speed}}$; doubling speed halves the time.

1. Let Pranay's time $=t_p=\frac{150}{p}$, Aman's time $=t_a$. Then $t_p=t_a+2$.
2. With double speed: $\frac{t_p}{2}=t_a-3$.
3. Subtract: $t_p-\frac{t_p}{2}=5\Rightarrow t_p=10$ h.
4. $p=\frac{150}{10}=15\text{km/hr}$.

Trick: Halving the time saves $2+3=5$ h, so original time $=10$ h.

Answer: (C) 15 km/hr

Concept: For decimals, take the root of the digits and halve the decimal places.

1. $\sqrt{169}=13$.
2. $\sqrt{0.0144}=0.12$ (since ${12}^2=144$).
3. $\sqrt{5.76}=2.4$ (since ${24}^2=576$).
4. $13+0.12-2.4=10.72$.

Trick: 4 decimal places → root has 2; 2 decimal places → root has 1.

Answer: (A) 10.72

Concept: Count the day of withdrawal but not the day of deposit; 2024 is a leap year (366 days).

1. Days: February $26$–$29$ = 4, March = 31, April = 26 → total 61 days.
2. Time $=\frac{61}{366}=\frac{1}{6}$ year.
3. SI $=\frac{4000\times 6.75\times 1}{100\times 6}=\frac{270}{6}=45$.

Yaad rakhein: In a leap year, 61 days $=\frac{1}{6}$ year exactly.

Answer: (C) ₹45

Concept: For the same time period, profit is shared in the ratio of investments.

1. Total investment $=1050+1320+1730=4100$.
2. Deepa's share $=\frac{1730}{4100}\times 1230$.
3. $\frac{1230}{4100}=0.3$, so share $=1730\times 0.3=519$.

Trick: Profit is $30\%$ of total capital, so each gets $30\%$ of own investment.

Answer: (B) ₹519

Concept: One-day work of C $=$ (A+B+C) one-day work $-$ A's $-$ B's.

1. A $=\frac{1}{22}$, B $=\frac{1}{11}=\frac{2}{22}$ → A+B $=\frac{3}{22}$.
2. C $=\frac{1}{5}-\frac{3}{22}=\frac{22-15}{110}=\frac{7}{110}$.
3. C alone $=\frac{110}{7}=15\frac{5}{7}$ days.

Trick: Total work = LCM$(22,11,5)=110$ units: A = 5, B = 10, all = 22 → C $=7$ units/day → $\frac{110}{7}$ days.

Answer: (A) $15\frac{5}{7}$ days

Concept: Expand each bracket, then combine like terms.

1. $3(\frac{7}{3}{x}^2-27x+14)=7x^2-81x+42$.
2. $-7(x^2+8x-11)=-7x^2-56x+77$.
3. Add: $(7x^2-7x^2)+(-81x-56x)+(42+77)=-137x+119$.

Trick: $x^2$ terms cancel; constant $42+77=119$ is positive, $x$ term negative → only (C).

Answer: (C) $-137x+119$

Concept: Arrange data in order; for $n$ odd, median is the ${\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending: 17, 20, 21, 27, 35, 41, 42, 49, 68, 76, 95.
2. $n=11$ → $\frac{11+1}{2}=6^{\text{th}}$ value.
3. 6th value $=41$.

Yaad rakhein: Always sort first — 42 is the middle of the unsorted list, a trap.

Answer: (C) 41