RRB NTPC Undergraduate CBT 1 2024 PYQs

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) $39$

Concept: Volume of a cylinder $V=\pi r^{2}h$, so the new volume factor = (radius factor)${}^2\times$(height factor).

1. Radius factor $=1-0.19=0.81$; height factor $=1+1.02=2.02$.
2. New volume factor $={0.81}^2\times 2.02=0.6561\times 2.02=1.3253$.
3. Increase $=32.53\%\approx 33\%$.

Trick: Successive change: $-19,-19,+102$ → $-19-19+\frac{361}{100}=-34.39$; then $-34.39+102-\frac{34.39\times 102}{100}\approx 32.5\%$.

Answer: (B) $33\%$

Concept: Two successive discounts $a\%$ and $b\%$ equal a single discount of $(a+b-\frac{ab}{100})\%$.

1. Equivalent single discount $=20+10-\frac{20\times 10}{100}=28\%$.
2. Single discount offered is also $28\%$.
3. Price paid in both cases $=80000\times 0.72=$ ₹57,600, so difference = ₹0.

Trick: Multiply the "pay" factors: $0.8\times 0.9=0.72$ → same as $28\%$ off.

Answer: (A) Both the discounts are the same; ₹0

Concept: Convert each ratio to a decimal (or compare with $\frac{1}{2}$) and pick the largest.

1. $\frac{38}{83}\approx 0.458$, $\frac{35}{77}\approx 0.455$ (both below $\frac{1}{2}$).
2. $\frac{47}{67}\approx 0.701$.
3. $\frac{50}{62}\approx 0.806$ – the largest.

Trick: The ratio whose antecedent is closest to its consequent is largest here: $50$ vs $62$.

Answer: (D) $50:62$

Concept: Take total work = LCM of the days; efficiency = work ÷ days.

1. Total work $=\text{LCM}(12,21)=84$ units.
2. A $=7$ units/day, B $=4$ units/day; together $11$ units/day.
3. In 3 days: $33$ units done; left $=84-33=51$ units.
4. Fraction left $=\frac{51}{84}=\frac{17}{28}$.

Trick: Work done $=3(\frac{1}{12}+\frac{1}{21})=\frac{11}{28}$, so left $=1-\frac{11}{28}$.

Answer: (C) $\frac{17}{28}$

Concept: Average of the first $n$ even numbers $=n+1$.

1. First 166 even numbers: $2,4,\dots ,332$.
2. Average $=\frac{2+332}{2}=167$.
3. Formula check: $n+1=166+1=167$.

Yaad rakhein: First $n$ even → $n+1$; first $n$ odd → $n$; first $n$ natural → $\frac{n+1}{2}$.

Answer: (B) $167$

Concept: Evaluate powers and roots first (BODMAS), then add/subtract.

1. $2^2=4$ and $\sqrt{8^2}=8$.
2. $3\sqrt{100}=3\times 10=30$.
3. Value $=4+8-30-19=-37$.

Yaad rakhein: $\sqrt{a^2}=a$ for $a\ge 0$.

Answer: (B) $-37$

Concept: If the numbers are $ha$ and $hb$ with HCF $h$, then sum $=h(a+b)$ and LCM $=hab$.

1. $84=2^2\times 3\times 7$ and $245=5\times 7^2$; common factor $h=7$.
2. $a+b=12$, $ab=35$ → $a=7,b=5$.
3. Numbers $=49$ and $35$; check: $49+35=84$, LCM $=245$.

Trick: Test options: only $49,35$ gives sum $84$ and LCM $245$.

Answer: (A) $49,35$

Concept: Expand each bracket and combine like terms.

1. $4(\frac{5}{4}{x}^2-25x+19)=5x^2-100x+76$.
2. $-5(x^2+5x-14)=-5x^2-25x+70$.
3. Sum $=(5x^2-5x^2)+(-100x-25x)+(76+70)=-125x+146$.

Trick: The $x^2$ terms cancel; just track $x$: $-100-25=-125$ and constants $76+70=146$.

Answer: (B) $-125x+146$

Concept: If $4A=9B=3C=k$, then $A:B:C=\frac{1}{4}:\frac{1}{9}:\frac{1}{3}$.

1. Multiply by $\text{LCM}(4,9,3)=36$: $A:B:C=9:4:12$.
2. Total parts $=9+4+12=25$.
3. A $=\frac{9}{25}\times 87400=$ ₹31,464.

Yaad rakhein: When $xA=yB=zC$, shares are in ratio $\frac{1}{x}:\frac{1}{y}:\frac{1}{z}$.

Answer: (A) ₹31,464

Concept: For equal distances at speeds $x$ and $y$: average speed $=\frac{2xy}{x+y}$.

1. $x=24$, $y=40\text{km/h}$.
2. Average $=\frac{2\times 24\times 40}{24+40}=\frac{1920}{64}$.
3. $=30\text{km/h}$.

Trick: Average speed is the harmonic mean, never the simple mean $32$.

Answer: (D) $30$