RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Simplify the inner (complex) fraction first, then add/subtract using the LCM of denominators.

1. $1+\frac{5}{6}=\frac{11}{6}$, so $\frac{1}{\frac{11}{6}}=\frac{6}{11}$.
2. Expression $=\frac{3}{4}+\frac{6}{11}-\frac{3}{5}$; LCM of 4, 11, 5 $=220$.
3. $=\frac{165+120-132}{220}=\frac{153}{220}$.

Yaad rakhein: Denominator must be a multiple of 4, 11 and 5 → only 220 fits.

Answer: (B) $\frac{153}{220}$

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

1. $T=3\frac{5}{9}=\frac{32}{9}$ years: $2808=\frac{P\times 2.7\times 32}{900}=0.096P\Rightarrow P=29250$.
2. New SI $=\frac{29250\times 9.2\times 17}{300}=15249$ (as $5\frac{2}{3}=\frac{17}{3}$).
3. Amount $=29250+15249=44499$.

Yaad rakhein: Rate × time: $2.7\times \frac{32}{9}=9.6\%$ and $9.2\times \frac{17}{3}=52.13\%$; so amount $=\frac{2808}{9.6}\times (100+52.13)$ ≈ ₹44,499.

Answer: (A) ₹44,499

Concept: Volume of a sphere $V=\frac{4}{3}\pi r^3$, where $r=\frac{d}{2}$.

1. $r=\frac{16.8}{2}=8.4\text{cm}$, $r^3=592.704$.
2. $V=\frac{4}{3}\times \frac{22}{7}\times 592.704=\frac{88}{21}\times 592.704$.
3. $\frac{592.704}{21}=28.224$; $28.224\times 88=2483.712\approx 2483.71{\text{cm}}^3$.

Yaad rakhein: $8.4=21\times 0.4$, so divide by 21 first – the calculation stays easy.

Answer: (D) $2483.71$

Concept: Mode is the value that occurs most often in the data.

1. Count each value: 11 → 2 times, 12 → 3 times, 13, 14, 15, 16, 17 → once each.
2. 12 has the highest frequency (3).

Yaad rakhein: Arrange in order (11, 11, 12, 12, 12, 13 …) – the longest run is the mode.

Answer: (C) 12

Concept: Use known perfect squares; for decimals, count decimal places in pairs.

1. $\sqrt{400}=20$.
2. $\sqrt{0.0289}=0.17$ (since ${17}^2=289$, four decimal places → two).
3. $\sqrt{7.84}=2.8$ (since ${28}^2=784$).
4. $20+0.17-2.8=17.37$.

Yaad rakhein: Decimal places in the root = half the decimal places in the number.

Answer: (B) $17.37$

Concept: Simplify the bracket, expand using the distributive law, then combine like terms.

1. $\frac{6}{3}=2$, so the first part is $3(2x^2-29x+14)=6x^2-87x+42$.
2. $-6(x^2+5x-13)=-6x^2-30x+78$.
3. Add: $(6x^2-6x^2)+(-87x-30x)+(42+78)=-117x+120$.

Yaad rakhein: $x^2$ terms cancel ($6-6=0$); constant $42+78=120$ is positive → only option (D) fits.

Answer: (D) $-117x+120$

Concept: Sum of interior angles $=(n-2)\times {180}^{\circ }$; sum of exterior angles of any polygon $={360}^{\circ }$; diagonals $=\frac{n(n-3)}{2}$.

1. $(n-2)\times 180=4\times 360=1440$.
2. $n-2=8\Rightarrow n=10$ (decagon).
3. Diagonals $=\frac{10\times 7}{2}=35$.

Yaad rakhein: Interior sum $=k\times$ exterior sum $\Rightarrow n=2k+2$; here $k=4$, $n=10$.

Answer: (C) $35$

Concept: If SP of $a$ articles = CP of $b$ articles, then $\text{loss}\%=\frac{a-b}{a}\times 100$ (when $a>b$).

1. Let CP of one book = ₹1. CP of 50 books = ₹50; SP of 50 books = CP of 15 = ₹15.
2. Loss $=50-15=35$.
3. $\text{Loss}\%=\frac{35}{50}\times 100=70\%$.

Yaad rakhein: More books sold for the money of fewer → loss; $\frac{50-15}{50}=70\%$.

Answer: (B) 70% loss

Concept: Mean proportional of $a$ and $b$ is $x$ such that $a:x=x:b$, i.e. $x=\sqrt{ab}$.

1. $x=\sqrt{81\times 36}$.
2. $=\sqrt{81}\times \sqrt{36}=9\times 6=54$.

Yaad rakhein: Both are perfect squares → multiply their roots directly.

Answer: (C) $54$

Concept: Convert each part to a number, then subtract.

1. $65\%\text{ of }75=\frac{65\times 75}{100}=48.75$.
2. $\frac{3}{5}\text{ of }35=21$.
3. Difference $=48.75-21=27.75$.

Yaad rakhein: $65\%$ of 75 $=75\times 0.65$: $75\times 0.6=45$, $75\times 0.05=3.75$ → $48.75$.

Answer: (A) $27.75$

Concept: Literate females = total literate − literate males.

1. Females $=360000-231000=129000$.
2. Total literate $=66\%\times 360000=237600$.
3. Literate males $=66\%\times 231000=152460$.
4. Literate females $=237600-152460=85140$; $\frac{85140}{129000}\times 100=66\%$.

Trick: If the whole group and one part both have $66\%$, the other part must also have $66\%$ — no calculation needed.

Answer: (B) $66\%$

Concept: For a fixed SP (₹1), number of articles sold is proportional to $\frac{1}{\text{SP per article}}$, and SP per article $\propto (100\pm \text{gain/loss}\%)$.

1. At $30\%$ loss, SP $=70\%$ of CP; at $5\%$ gain, SP $=105\%$ of CP.
2. Number $\times$ (SP %) stays constant for ₹1: $6\times 70=n\times 105$.
3. $n=\frac{420}{105}=4$.

Trick: $n=6\times \frac{100-30}{100+5}=6\times \frac{70}{105}=4$.

Answer: (B) 4

Concept: A number is divisible by 3 if its digit sum is divisible by 3.

1. Digit sum of 563607: $5+6+3+6+0+7=27$, divisible by 3 → divisible by 3.
2. Last digit 7 is odd → not divisible by 8; not ending in 0 → not by 10.
3. $13\times 43354=563602$, remainder 5 → not by 13.

Trick: Odd last digit instantly removes 8 and 10; check digit sum for 3.

Answer: (C) 3

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

1. $(-60)÷(-10)=6$, so $-(-60)÷(-10)=-6$.
2. $(-4)\times 8=-32$.
3. $-9-6-32=-47$.

Trick: Negative ÷ negative = positive; then subtract it.

Answer: (D) $-47$

Concept: Effective discount $=\frac{\text{free items}}{\text{total items received}}\times 100$.

1. Pay for 3, receive $3+1=4$ trousers.
2. Discount $=\frac{1}{4}\times 100=25\%$.

Trick: Buy $n$ get 1 free → discount $=\frac{100}{n+1}\%$.

Answer: (D) $25\%$

Concept: Same time, equal SI → principals are in inverse ratio of rates.

1. $P_1:P_2=9:5$ (inverse of $5\%:9\%$).
2. Sum at $5\%=\frac{9}{14}\times 20300=9\times 1450=13050$.
3. Check: $13050\times 5\%\times 7=4567.5$ and $7250\times 9\%\times 7=4567.5$.

Trick: Equal SI → $P\propto \frac{1}{R}$.

Answer: (B) ₹13,050

Concept: Form an equation from the past condition, then use the future condition.

1. 3 years ago: $F-3=2(S-3)+40\Rightarrow F=2S+37$.
2. After $n$ years: $F+n=2(S+n)\Rightarrow n=F-2S$.
3. $n=37$ years.

Trick: When "twice" is asked, $n=F-2S$ directly.

Answer: (A) 37

Concept: For fixed distance, speed $\propto \frac{1}{\text{time}}$.

1. Distance $=84\times 6=504\text{km}$.
2. Speed $=\frac{504}{12}=42\text{km/h}$.

Trick: Time doubles → speed halves: $\frac{84}{2}=42$.

Answer: (B) $42\text{km/h}$

Concept: Same time period → profit shared in ratio of investments.

1. $1020:1740:1840=51:87:92$ (divide by 20); sum $=230$.
2. Siddharth $=\frac{92}{230}\times 1050=420$.

Trick: $\frac{1050}{230}\times 92$: $\frac{1050}{230}=\frac{105}{23}$, $92=4\times 23$ → $105\times 4=420$.

Answer: (C) ₹420

Concept: ${\sec }^2\varphi {\cot }^2\varphi ={\csc }^2\varphi$, ${\csc }^2\varphi -{\cot }^2\varphi =1$, ${\sin }^2\varphi +{\cos }^2\varphi =1$.

1. ${\sec }^2\varphi {\cot }^2\varphi =\frac{1}{{\cos }^2\varphi }·\frac{{\cos }^2\varphi }{{\sin }^2\varphi }={\csc }^2\varphi$.
2. ${\csc }^2\varphi -\frac{1}{{\tan }^2\varphi }={\csc }^2\varphi -{\cot }^2\varphi =1$.
3. $2(\frac{1}{{\csc }^2\varphi }+\frac{1}{{\sec }^2\varphi })=2({\sin }^2\varphi +{\cos }^2\varphi )=2$.
4. Total $=1+2=3$.

Trick: Put $\varphi ={45}^{\circ }$: $2\times 1-1+2(\frac{1}{2}+\frac{1}{2})=3$.

Answer: (B) 3