RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Volume of shell $=\frac{4}{3}\pi (R^3-r^3)$; mass $=$ density $\times$ volume.

1. $R^3-r^3=4^3-1^3=64-1=63$
2. $V=\frac{4}{3}\times \frac{22}{7}\times 63=264{\text{cm}}^3$
3. Mass $=40\times 264=10560\text{g}$
4. $10560\text{g}=10.56\text{kg}$

Trick: $\frac{22}{7}\times 63=198$, then $198\times \frac{4}{3}=264$ – pick $R^3-r^3$ as a multiple of 7 to cancel fast.

Answer: (D) $10.56$

Concept: Share $=\frac{\text{ratio part}}{\text{sum of parts}}\times \text{total}$.

1. Sum of parts $=2+4+3=9$; 1 part $=\frac{936}{9}=104$
2. Largest share $=4\times 104=416$, smallest $=2\times 104=208$
3. Difference $=(4-2)\times 104=208$

Trick: Difference = (largest part − smallest part) × one part = $2\times 104$.

Answer: (D) $208$

Concept: $x\%$ of $N=\frac{x}{100}\times N$.

1. $65\%$ of $75=\frac{65\times 75}{100}=48.75$
2. $\frac{3}{5}$ of $15=9$
3. Difference $=48.75-9=39.75$

Trick: $65\%$ of $75$ = $75\%$ of $65$ = $\frac{3}{4}\times 65=48.75$.

Answer: (C) $39.75$

Concept: If angles of elevation from the foot of each pole to the top of the other are complementary, then $x^2=h_1\times h_2$.

1. From B to top P of pole A: $\tan \theta =\frac{25}{x}$
2. From A to top Q of pole B: $\tan ({90}^{\circ }-\theta )=\cot \theta =\frac{36}{x}$
3. Multiply: $\tan \theta ·\cot \theta =1=\frac{25\times 36}{x^2}$
4. $x^2=900\Rightarrow x=30\text{m}$

Trick: Distance $=\sqrt{h_1{h}_2}=\sqrt{25\times 36}=5\times 6=30$.

Answer: (B) $30$

Concept: Successive discounts are applied one after another: $\text{SP}=\text{MP}\times \frac{100-d_1}{100}\times \frac{100-d_2}{100}$.

1. After 20%: $5000\times \frac{80}{100}=4000$
2. After 10% on 4000: $4000\times \frac{90}{100}=3600$
3. Final price = ₹3,600

Trick: Net discount $=20+10-\frac{20\times 10}{100}=28\%$; $72\%$ of 5000 = 3600.

Answer: (D) ₹3,600

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

1. $9:30=30:x$
2. $x=\frac{{30}^2}{9}=\frac{900}{9}=100$

Trick: Third proportional $=\frac{(\text{second}{)}^2}{\text{first}}$.

Answer: (D) $100$

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$ (not the average of speeds).

1. Times: $\frac{252}{63}=4\text{h}$, $\frac{150}{30}=5\text{h}$, $\frac{680}{68}=10\text{h}$
2. Total distance $=252+150+680=1082\text{km}$; total time $=19\text{h}$
3. Average speed $=\frac{1082}{19}\approx 56.95\text{km/h}$

Trick: $19\times 57=1083$, so the answer is just below 57 → 56.95.

Answer: (D) $56.95$

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

1. Equivalent of 10% and 40%: $10+40-\frac{10\times 40}{100}=46\%$
2. Single discount 48% vs 46%: gap $=2\%$
3. $2\%$ of $1450=\frac{2\times 1450}{100}=29$

Trick: Find only the % gap and apply it to the price – no need to compute both selling prices (754 and 667).

Answer: (B) ₹29

Concept: Open brackets by distributing, then combine like terms.

1. $4(\frac{3}{4}{x}^2-25x+17)=3x^2-100x+68$
2. $-3(x^2+8x-13)=-3x^2-24x+39$
3. Add: $(3x^2-3x^2)+(-100x-24x)+(68+39)=-124x+107$

Trick: The $x^2$ terms cancel ($4\times \frac{3}{4}=3$), so only the $x$ and constant terms need checking: sign of $x$ is −, constant is +.

Answer: (D) $-124x+107$

Concept: For the same distance, time is inversely proportional to speed: $\frac{t_1}{t_2}=\frac{s_2}{s_1}$.

1. Distance $=42\times 14=588\text{km}$
2. Mira’s time $=\frac{588}{98}=6\text{h}$
3. Check: $\frac{t_M}{14}=\frac{42}{98}=\frac{3}{7}\Rightarrow t_M=6$

Trick: Speed ratio $42:98=3:7$, so time ratio $=7:3$; $14\to 6$.

Answer: (C) 6 hours

Concept: For CI compounded annually, $A=P{(1+\frac{r}{100})}^n$.

1. $\frac{A}{P}=\frac{529}{400}={(1+\frac{r}{100})}^2$
2. $\sqrt{\frac{529}{400}}=\frac{23}{20}=1+\frac{r}{100}$
3. $\frac{r}{100}=\frac{3}{20}\Rightarrow r=15\%$

Trick: 529 and 400 are perfect squares (${23}^2,{20}^2$): rate $=\frac{23-20}{20}\times 100=15\%$.

Answer: (B) $15\%$

Concept: With a false weight, Selling price per real unit rises; $\text{Profit}\%=(\frac{\text{SP}}{\text{CP of goods actually given}}-1)\times 100$.

1. Let CP of 1 kg = ₹100; claimed SP = ₹158.
2. He actually gives only $0.92\text{kg}$, costing ₹92.
3. $\text{Profit}\%=\frac{158-92}{92}\times 100=\frac{6600}{92}\approx 71.74\%$
4. Rounded: $72\%$.

Trick: Overall gain $=\frac{100+\text{gain}}{100-\text{error}}\times 100-100=\frac{158}{92}\times 100-100$.

Answer: (C) $72\%$

Concept: Write both numbers in one variable and use the product.

1. Let second number $=x$, first $=5x$.
2. $5x\times x=80\Rightarrow x^2=16\Rightarrow x=4$
3. Numbers: 20 and 4; sum $=24$.

Trick: Check options: $20\times 4=80$ and $20=5\times 4$ ✓.

Answer: (B) 24

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

1. $15.4=15.400=\frac{15400}{1000}$, $0.006=\frac{6}{1000}$
2. $\text{LCM}(15400,6)$: $15400=2^3\times 5^2\times 7\times 11$, $6=2\times 3$ → LCM $=46200$
3. Put back 3 decimal places: $46.200=46.2$

Trick: LCM of decimals = LCM of numerators ÷ common denominator.

Answer: (B) 46.2

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

1. $P=400,R=16,T=3.5$
2. $\text{SI}=\frac{400\times 16\times 3.5}{100}=4\times 56=224$

Trick: $16\times 3.5=56\%$ of ₹400 = ₹224.

Answer: (C) 224

Concept: Solve innermost bracket first: ( ) → { } → [ ].

1. $(4-9)=-5$
2. $\{12+2\times (-5)\}=12-10=2$
3. $[480÷2]=240$

Trick: Bracket order: small → curly → square.

Answer: (B) 240

Concept: Form equations from present ages $F$ and $S$.

1. 3 years ago: $F-3=2(S-3)+18\Rightarrow F=2S+15$
2. After $n$ years: $F+n=2(S+n)\Rightarrow F=2S+n$
3. Compare: $n=15$ years.

Trick: If $F-a=2(S-a)+k$, then father becomes twice the son after exactly $k-a$ years: $18-3=15$.

Answer: (D) 15

Concept: BODMAS: division and multiplication before addition/subtraction; watch signs.

1. $(-60)÷(-10)=6$
2. $(-4)\times 6=-24$
3. $(-9)-6+(-24)=-39$

Trick: Negative ÷ negative = positive.

Answer: (A) $-39$

Concept: One-day work adds up: $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{\text{together}}$.

1. $\frac{1}{C}=\frac{1}{3}-\frac{1}{19}-\frac{1}{11}$
2. LCM $=627$: $\frac{209-33-57}{627}=\frac{119}{627}$
3. $C=\frac{627}{119}=5\frac{32}{119}$ days

Trick: $627=3\times 19\times 11$; integer part must be just over 5 since $\frac{119}{627}\approx 0.19$.

Answer: (C) $5\frac{32}{119}$ days

Concept: Successive % changes multiply: $P\times (1+r_1)(1+r_2)(1-r_3)$.

1. $12800\times 1.25=16000$
2. $16000\times 1.35=21600$
3. $21600\times 0.95=20520$

Trick: Use fractions: $12800\times \frac{5}{4}\times \frac{27}{20}\times \frac{19}{20}=20520$.

Answer: (A) 20,520