RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Write each ratio as a decimal (or fraction) and compare.

1. $46:75=\frac{46}{75}\approx 0.613$
2. $34:77\approx 0.442$, $39:89\approx 0.438$
3. $36:69\approx 0.522$
4. Largest value is $0.613$, i.e. $46:75$.

Trick: Only $46:75$ has the first term more than half of the second ($46>37.5$); compare others with $\frac{1}{2}$ to eliminate fast.

Answer: (A) $46:75$

Concept: Use ${\sin }^2\theta +{\cos }^2\theta =1$.

1. $3{\sin }^2\theta +7{\cos }^2\theta =3({\sin }^2\theta +{\cos }^2\theta )+4{\cos }^2\theta =3+4{\cos }^2\theta$
2. $3+4{\cos }^2\theta =4\Rightarrow {\cos }^2\theta =\frac{1}{4}$
3. $\cos \theta =\frac{1}{2}$ (acute angle) $\Rightarrow \theta ={60}^{\circ }$

Trick: Split the bigger coefficient: $7=3+4$, so the identity removes ${\sin }^2\theta$ at once.

Answer: (B) ${60}^{\circ }$

Concept: If SP of $x$ articles = CP of $y$ articles, then $\text{Gain}\%=\frac{y-x}{x}\times 100$.

1. SP of 20 books = CP of 24 books, so SP is more than CP: it is a gain.
2. $\text{Gain}\%=\frac{24-20}{20}\times 100$
3. $=\frac{4}{20}\times 100=20\%$

Trick: Divide the difference by the number of articles sold (20), not by 24 – dividing by 4 or 24 gives the trap options.

Answer: (B) $20\%$ gain

Concept: Add one-day works: $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{\text{together}}$.

1. A + B in one day: $\frac{1}{22}+\frac{1}{11}=\frac{3}{22}$
2. A + B + C in one day: $\frac{1}{4}$
3. C in one day: $\frac{1}{4}-\frac{3}{22}=\frac{11-6}{44}=\frac{5}{44}$
4. C alone: $\frac{44}{5}=8\frac{4}{5}$ days

Trick: LCM method: total work $=44$ units; A $=2$, B $=4$, all $=11$ units/day, so C $=5$ units/day $\Rightarrow \frac{44}{5}$ days.

Answer: (C) $8\frac{4}{5}$ days

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

1. Diameter $d=168\text{m}$ (the shorter side)
2. Circumference $=\pi d=\frac{22}{7}\times 168$
3. $=22\times 24=528\text{m}$

Trick: The circle can never be wider than the smaller side – the 249 m side is extra data.

Answer: (B) $528$

Concept: Angle sum of a triangle is ${180}^{\circ }$; an exterior angle equals the sum of the two opposite interior angles.

1. In right $△BDC$: $\angle C={180}^{\circ }-{90}^{\circ }-{76}^{\circ }={14}^{\circ }$
2. In $△AEC$: $\angle AEC={180}^{\circ }-{74}^{\circ }-{14}^{\circ }={92}^{\circ }$
3. B, E, C are collinear: $\angle AEB={180}^{\circ }-{92}^{\circ }={88}^{\circ }$

Trick: Directly: $\angle AEB$ is exterior to $△AEC$, so $\angle AEB={74}^{\circ }+{14}^{\circ }={88}^{\circ }$.

Answer: (A) ${88}^{\circ }$

Concept: Form two linear equations from the two statements and solve.

1. $4P-Q=41\Rightarrow Q=4P-41$
2. After 10 years: $3(Q+10)=7(P+10)-33\Rightarrow 3Q=7P+7$
3. $3(4P-41)=7P+7\Rightarrow 5P=130\Rightarrow P=26$
4. $Q=4\times 26-41=63$

Trick: Check options: $Q=63\Rightarrow P=26$; then $3\times 73=219$ and $7\times 36-33=219$ ✔.

Answer: (B) $63$

Concept: In 'buy 4 get 2 free', the customer pays only for 4 out of every 6 items.

1. Six shirts = 4 paid + 2 free
2. Amount $=4\times 1250=5000$, i.e. ₹5,000

Trick: Effective discount $=\frac{2}{6}=33\frac{1}{3}\%$; $\frac{2}{3}$ of ₹7,500 = ₹5,000.

Answer: (B) ₹5,000

Concept: Find each pipe's full-tank time, then use filled part = rate × time.

1. P fills $\frac{7}{9}$ in 14 h $\Rightarrow$ full tank in $14\times \frac{9}{7}=18$ h; Q fills $\frac{5}{9}$ in 20 h $\Rightarrow$ 36 h
2. In 2 h, P + Q fill $2(\frac{1}{18}+\frac{1}{36})=\frac{1}{6}$ tank
3. R empties $\frac{1}{6}$ tank in 6 h $\Rightarrow$ full tank in 36 h
4. Together: $\frac{1}{18}+\frac{1}{36}-\frac{1}{36}=\frac{1}{18}\Rightarrow 18$ h

Trick: Q's filling rate and R's emptying rate are equal (36 h each), so they cancel – only P works: 18 h.

Answer: (B) 18 hours

Concept: After a $70\%$ increase, the number becomes $170\%$ of itself.

1. $1.7x=3230$
2. $x=\frac{3230}{1.7}=1900$

Trick: $3230÷17=190$, so $x=190\times 10=1900$.

Answer: (A) $1900$

Concept: $\sqrt{\text{decimal}}$: take the root of the digits and halve the number of decimal places.

1. $\sqrt{144}=12$
2. $\sqrt{0.0225}=0.15$ (since ${15}^2=225$)
3. $\sqrt{6.25}=2.5$
4. $12+0.15-2.5=9.65$

Trick: 4 decimal places under the root give 2 decimal places in the answer.

Answer: (C) $9.65$

Concept: Let the numbers be $n$ and $n+2$.

1. $n^2+(n+2{)}^2=1684\Rightarrow 2n^2+4n-1680=0$
2. $n^2+2n-840=0\Rightarrow (n+30)(n-28)=0\Rightarrow n=28$
3. Numbers: 28 and 30; sum $=58$

Trick: $\frac{1684}{2}=842\approx {29}^2$, so the numbers sit around 29: 28 and 30 ($784+900=1684$).

Answer: (D) $58$

Concept: Equal SI for equal time means principal × rate is the same for both parts.

1. Let the sum at $14\%$ be $x$: $x\times 14=(21300-x)\times 11$
2. $25x=234300\Rightarrow x=9372$

Trick: Principals are in inverse ratio of rates, $11:14$: $\frac{11}{25}\times 21300=9372$.

Answer: (A) ₹9,372

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

1. Sum of first $n$ even numbers $=n(n+1)$
2. Average $=\frac{n(n+1)}{n}=n+1=161+1=162$

Yaad rakhein: First $n$ even numbers: average $n+1$; first $n$ odd numbers: average $n$.

Answer: (D) $162$

Concept: Profit share is proportional to capital × time.

1. M + N $=6915$, M − N $=461\Rightarrow$ M $=3688$, N $=3227$; ratio $=8:7$
2. Let N's capital $=x$, M's $=x+38000$: $\frac{2(x+38000)}{3x}=\frac{8}{7}$
3. $14x+532000=24x\Rightarrow x=53200$
4. M's capital $=53200+38000=91200$

Trick: $3688=8\times 461$ and $3227=7\times 461$ – spotting the common factor gives $8:7$ instantly.

Answer: (C) ₹91,200

Concept: Expand the brackets and combine like terms.

1. $x(2x-8)=2x^2-8x$
2. $5(x^2-3)=5x^2-15$
3. Total: $2x^2-8x+5x^2-15+12=7x^2-8x-3$

Trick: Constant term check: $-15+12=-3$ removes options (A) and (D); $x^2$ coefficient $+7$ removes (B).

Answer: (C) $7x^2-8x-3$

Concept: $SI=\frac{P\times R\times T}{100}$, with $T$ in years $=\frac{\text{days}}{366}$ in a leap year (count the day of deposit out, day of withdrawal in).

1. 2024 is a leap year: Feb has 29 days. Days: Feb 3 (27, 28, 29) + Mar 31 + Apr 27 $=61$
2. $SI=\frac{4000\times 6.75\times 61}{100\times 366}$
3. $=\frac{270\times 61}{366}=\frac{16470}{366}=45$

Trick: $366=6\times 61$, so $\frac{61}{366}=\frac{1}{6}$ year: $SI=\frac{270}{6}=45$.

Answer: (D) ₹45

Concept: Product of two numbers $=\text{HCF}\times \text{LCM}$.

1. Other number $=\frac{83\times 498}{166}$
2. $=\frac{498}{2}=249$

Trick: Both numbers must be multiples of the HCF 83: $249=3\times 83$ ✔, while 230 is not.

Answer: (D) $249$

Concept: $\text{Average speed}=\frac{\text{total distance}}{\text{total time}}$.

1. Times: $\frac{200}{25}=8$ h, $\frac{210}{21}=10$ h, $\frac{204}{51}=4$ h; total $22$ h
2. Distance $=200+210+204=614\text{km}$
3. Average speed $=\frac{614}{22}\approx 27.91\text{km/h}$

Trick: Never average the three speeds directly – that gives 32.33, which is wrong.

Answer: (C) $27.91$

Concept: Test divisibility of odd numbers by primes up to $\sqrt{476}\approx 21.8$.

1. Even numbers 468, 470, 472, 474, 476 are not prime; 475 ends in 5
2. $469=7\times 67$, $471=3\times 157$, $473=11\times 43$
3. No prime lies in this range.

Trick: Check 3 (digit sum), 7 and 11 first – they catch most composite numbers.

Answer: (A) $0$