RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Discount = discount rate × list price.

1. Discount $=25\%\text{ of }3000=\frac{25}{100}\times 3000$.
2. $=$ ₹750.

Trick: 25% = one-fourth: $3000÷4=750$.

Answer: (C) ₹750

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

1. Aman's profit $=11600-2900=$ ₹8700; ratio Aman : Pawan $=8700:2900=3:1$.
2. Investments: Aman $=\frac{3}{4}\times 62800=$ ₹47100, Pawan $=$ ₹15700.
3. Difference $=47100-15700=$ ₹31400.

Trick: Difference $=\frac{3-1}{4}\times 62800=\frac{1}{2}\times 62800=$ ₹31400.

Answer: (C) ₹31,400

Concept: Product $n(n+1)$ is close to $n^2$; take $\sqrt{306}\approx 17.5$.

1. Try $17\times 18=306$ ✓.
2. The two numbers are 17 and 18; greater is 18.

Trick: Last digit 6 → numbers end in 2&3 or 7&8; near $\sqrt{306}$ gives 17, 18.

Answer: (B) $18$

Concept: Curved surface area of a cylinder $=2\pi rh$.

1. $2\times \frac{22}{7}\times r\times 87=7500$.
2. $r=\frac{7500\times 7}{44\times 87}=\frac{52500}{3828}$.
3. $r\approx 13.71\text{cm}$.

Trick: Quick estimate: $2\pi h\approx 547$, and $7500÷547\approx 13.7$.

Answer: (B) $13.71$

Concept: Share = (its ratio part ÷ total parts) × amount.

1. Total parts $=2+11+17=30$; 1 part $=\frac{976}{30}$.
2. Difference = $17-2=15$ parts.
3. $15\times \frac{976}{30}=\frac{976}{2}=$ ₹488.

Trick: 15 parts out of 30 is exactly half of ₹976.

Answer: (A) $488$

Concept: Divisible by 5 if the last digit is 0 or 5; by 3 if the digit sum is divisible by 3.

1. Last digit is 5 → divisible by 5 ✓.
2. Digit sum $=8+6+9+6+7+1+1+9+5=52$, not divisible by 3 → not by 15 or 12.
3. It is odd, so not divisible by 12 either.

Trick: Check the easiest rule first – the last digit.

Answer: (A) $5$

Concept: Total = average × count; split total in the given ratio.

1. Total price $=3\times 16035=$ ₹48105.
2. Total parts $=3+5+7=15$; 1 part $=\frac{48105}{15}=$ ₹3207.
3. Cheapest $=3\times 3207=$ ₹9621.

Trick: Average of the ratio terms is 5 parts = ₹16035, so 1 part = ₹3207 directly.

Answer: (D) $9,621$

Concept: $a^m\times b^m\times c^m=(abc{)}^m$ and $a^{-m}=\frac{1}{a^m}$.

1. ${14}^{-23}\times {25}^{-23}\times {15}^{-23}=(14\times 25\times 15{)}^{-23}$.
2. $14\times 25\times 15=5250$.
3. Value $={5250}^{-23}=\frac{1}{{5250}^{23}}$.

Trick: $14\times 15=210$, $210\times 25=5250$.

Answer: (D) $\frac{1}{{5250}^{23}}$

Concept: Relate the selling price (SP) to the cost price (CP) using the loss condition.

1. $\frac{2}{5}\text{SP}=0.76\text{CP}$ (loss 24%).
2. $\text{SP}=\frac{0.76\times 5}{2}\text{CP}=1.9\text{CP}$.
3. New price $=0.81\times 1.9\text{CP}=1.539\text{CP}$ → profit $53.9\%$.

Trick: Take CP = 100: $\frac{2}{5}\text{SP}=76$ → SP = 190; $81\%\text{ of }190=153.9$.

Answer: (D) 53.9%

Concept: Third proportional of $a$ and $b$ is $x$ where $a:b=b:x$, i.e. $x=\frac{b^2}{a}$.

1. $x=\frac{{52}^2}{26}=\frac{2704}{26}$.
2. $x=104$.

Trick: $52=2\times 26$, so $x=2\times 52=104$ (ratio doubles each time).

Answer: (B) $104$

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

1. Let Anmol’s speed $=a$ and Nikhil’s time $=t$. Then $\frac{162}{a}=t+3$ and $\frac{162}{2a}=\frac{81}{a}=t-6$.
2. Subtract: $\frac{162}{a}-\frac{81}{a}=(t+3)-(t-6)=9$
3. $\frac{81}{a}=9\Rightarrow a=9\text{km/h}$
4. Check: Anmol takes $18$ h, Nikhil $15$ h; at double speed Anmol takes $9$ h $=15-6$ ✓

Trick: The gap $3-(-6)=9$ h equals half of Anmol’s time, so his time $=18$ h and speed $=\frac{162}{18}=9$.

Answer: (A) $9\text{km/hr}$

Concept: Use $(a^m{)}^n=a^{mn}$ and compare powers of the same base.

1. $x^z=({17}^{0.24}{)}^z={17}^{0.24z}$
2. $y^4=({17}^{0.51}{)}^4={17}^{2.04}$
3. $0.24z=2.04\Rightarrow z=\frac{2.04}{0.24}=8.5$

Trick: Same base → just equate exponents: $z=\frac{4\times 0.51}{0.24}=\frac{204}{24}=8.5$.

Answer: (A) $8.5$

Concept: Median = middle value after arranging data in order; for odd $n$ it is the ${\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending order: 15, 24, 30, 39, 40, 57, 78, 80, 82, 82, 97 ($n=11$).
2. Position $=\frac{11+1}{2}=6$
3. 6th value $=57$

Trick: Always sort first – the unsorted middle value (57 here by chance) can mislead.

Answer: (A) $57$

Concept: For a cube of edge $a$: diagonal $=\sqrt{3}a$, total surface area $=6a^2$.

1. $\sqrt{3}a=15\Rightarrow a=\frac{15}{\sqrt{3}}=5\sqrt{3}$
2. $a^2=75$
3. TSA $=6\times 75=450{\text{cm}}^2$

Trick: $\text{TSA}=2d^2$ directly: $2\times {15}^2=450$.

Answer: (C) $450\text{sq.cm}$

Concept: Discount $=\text{Discount}\%\times \text{Marked price}$.

1. Discount $=\frac{12}{100}\times 4300$
2. $=12\times 43=516$

Trick: 1% of ₹4,300 = ₹43; so 12% = $12\times 43=$ ₹516.

Answer: (D) ₹516

Concept: Successive changes multiply: $P_3=P\times (1+\frac{r_1}{100})(1+\frac{r_2}{100})(1-\frac{r_3}{100})$.

1. Factors: $+25\%\to \frac{5}{4}$, $+60\%\to \frac{8}{5}$, $-50\%\to \frac{1}{2}$
2. Net factor $=\frac{5}{4}\times \frac{8}{5}\times \frac{1}{2}=1$
3. Population $=16362\times 1=16362$

Trick: If the net multiplier is 1, the answer is the starting value – no calculation needed.

Answer: (B) $16,362$

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

1. CP of 1 pencil $=\frac{180}{12}=$ ₹15
2. SP of 1 pencil = ₹18, profit = ₹3
3. $\text{Profit}\%=\frac{3}{15}\times 100=20\%$

Trick: Compare per dozen: SP $=18\times 12=216$ vs CP $180$ → $\frac{36}{180}=20\%$.

Answer: (A) $20$

Concept: Value of 1 part $=\frac{\text{Total}}{\text{Sum of ratio terms}}$.

1. Sum of terms $=19+11+18=48$
2. 1 part $=\frac{552}{48}=11.5$
3. Difference $=(19-11)\times 11.5=8\times 11.5=92$

Trick: Find only the difference of terms (8 parts) – no need to compute each share.

Answer: (C) $92$

Concept: Use ${\sin }^2\theta =1-{\cos }^2\theta$ and ${\sec }^2\theta -{\tan }^2\theta =1$.

1. $\frac{{\sin }^2\theta }{1+\cos \theta }=\frac{(1-\cos \theta )(1+\cos \theta )}{1+\cos \theta }=1-\cos \theta$
2. $\cot \theta -\csc \theta =\frac{\cos \theta -1}{\sin \theta }$, so $\frac{\sin \theta }{\cot \theta -\csc \theta }=\frac{{\sin }^2\theta }{\cos \theta -1}=-(1+\cos \theta )$
3. $X=(1-\cos \theta )+(1+\cos \theta )=2\Rightarrow X-1=1$
4. Among the options only ${\sec }^2\theta -{\tan }^2\theta =1$.

Trick: Put $\theta ={90}^{\circ }$: $X=\frac{1}{1}-\frac{1}{0-1}=2$, so $X-1=1$; check options at $\theta ={45}^{\circ }$: only ${\sec }^2-{\tan }^2=2-1=1$.

Answer: (C) ${\sec }^2\theta -{\tan }^2\theta$

Concept: $a^0=1$ and $a^{-n}=\frac{1}{a^n}$.

1. $3^0+3^{-2}+3^{-1}=1+\frac{1}{9}+\frac{1}{3}$
2. Multiply each term by 27: $27+3+9$
3. $=39$

Trick: Multiplying by $27=3^3$ just raises each power by 3: $3^3+3^1+3^2=27+3+9$.

Answer: (A) $39$