RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Divide numerator and denominator by $\cos \alpha$ to get an equation in $\tan \alpha$, then use ${\sin }^2\alpha =\frac{{\tan }^2\alpha }{1+{\tan }^2\alpha }$.

1. $\frac{\tan \alpha +1}{\tan \alpha -1}=3\Rightarrow \tan \alpha +1=3\tan \alpha -3$
2. $2\tan \alpha =4\Rightarrow \tan \alpha =2$
3. ${\sin }^2\alpha =\frac{{\tan }^2\alpha }{1+{\tan }^2\alpha }=\frac{4}{1+4}=\frac{4}{5}$

Trick: By componendo–dividendo, $\frac{\sin \alpha }{\cos \alpha }=\frac{3+1}{3-1}=2$ directly.

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

Concept: If the literacy rate of the whole population equals the literacy rate of males, the female literacy rate must be the same (weighted average).

1. Females $=329000-209000=120000$
2. Total literates $=44\%\times 329000=144760$; male literates $=44\%\times 209000=91960$
3. Female literates $=144760-91960=52800$, so rate $=\frac{52800}{120000}\times 100=44\%$

Trick: Average of two groups = one group’s value ⇒ the other group has the same value.

Answer: (C) $44\%$

Concept: Take the distance MN as 1 unit. Speed = fraction of distance per hour; when trains move towards each other, their speeds add.

1. A takes 7 h, B takes 5 h ⇒ speeds $\frac{1}{7}$ and $\frac{1}{5}$ per hour.
2. By 10 a.m. A covers $\frac{2}{7}$; remaining $\frac{5}{7}$.
3. Time to meet $=\frac{5/7}{\frac{1}{7}+\frac{1}{5}}=\frac{5}{7}\times \frac{35}{12}=\frac{25}{12}\text{h}=2\text{h}5\text{min}$
4. 10:00 a.m. + 2 h 5 min = 12:05 p.m.

Trick: Answer must lie between 10 a.m. and 3 p.m. – only 12:05 p.m. and 11:14 a.m. qualify.

Answer: (B) 12:05 p.m.

Concept: Profit is shared in the ratio of investment (same time period).

1. Total parts $=61+43+90=194$
2. 1 part $=\frac{2910}{194}=15$
3. C − B $=(90-43)\times 15=47\times 15=705$

Trick: Find the difference of ratio terms first, then multiply once by the value of one part.

Answer: (B) ₹705

Concept: Angle sum of a triangle is ${180}^{\circ }$; with $BC\parallel AD$, alternate angles are equal; equal sides face equal angles.

1. In $△ABC$: $\angle BCA={180}^{\circ }-{20}^{\circ }-{131}^{\circ }={29}^{\circ }$
2. $BC\parallel AD$ ⇒ $\angle CAD=\angle BCA={29}^{\circ }$ (alternate angles)
3. $AC=CD$ ⇒ $\angle CDA=\angle CAD={29}^{\circ }$
4. $\angle ACD={180}^{\circ }-{29}^{\circ }-{29}^{\circ }={122}^{\circ }$

Trick: Isosceles triangle: vertex angle $={180}^{\circ }-2\times \text{base angle}$.

Answer: (C) ${122}^{\circ }$

Concept: Compound amount $A=P{(1+\frac{R}{100})}^n$.

1. $\frac{2601}{2500}={\left(1.02\right)}^n$
2. $\frac{2601}{2500}={\left(\frac{51}{50}\right)}^2$ since ${51}^2=2601$, ${50}^2=2500$
3. So $n=2$ years.

Trick: Spot perfect squares: $2601={51}^2$, $2500={50}^2$.

Answer: (A) 2 years

Concept: Follow BODMAS: division and multiplication before addition and subtraction; mind the signs.

1. $(-60)÷(-10)=6$
2. $(-3)\times 8=-24$
3. $(-9)-6+(-24)=-39$

Trick: Negative ÷ negative = positive; then just add all the negatives.

Answer: (C) $-39$

Concept: Total = average × number of items; split the total in the given ratio.

1. Total price $=3\times 16050=48150$
2. Ratio parts $=3+5+7=15$; 1 part $=\frac{48150}{15}=3210$
3. Cheapest $=3\times 3210=9630$

Trick: Average of 3 items = 5 parts (middle term of 3:5:7), so 1 part $=\frac{16050}{5}=3210$. Don’t stop at 3210 – that is one part only.

Answer: (B) 9,630

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

1. $2\times \frac{22}{7}\times r\times 52=1800$
2. $r=\frac{1800\times 7}{44\times 52}=\frac{12600}{2288}$
3. $r\approx 5.507\approx 5.51\text{cm}$

Trick: $\frac{1800}{2\pi \times 52}\approx \frac{1800}{327}\approx 5.5$ – pick the option near 5.5.

Answer: (B) 5.51

Concept: Check each option: the pair must add to 80 and have LCM 558.

1. All pairs add to 80, so test the LCM.
2. $18=2\times 3^2$, $62=2\times 31$
3. LCM $=2\times 3^2\times 31=558$ ✓

Trick: $558=2\times 9\times 31$ – look for an option containing a multiple of 31: only 62.

Answer: (B) 18, 62

Concept: Rule for 7, 11, 13: make groups of three digits from the right and take the alternating sum; if it is divisible by 13, so is the number.

1. The number is odd ⇒ not divisible by 12 or 18.
2. Groups: 788 | 973 | 289 ⇒ $289-973+788=104$
3. $104=13\times 8$ ⇒ divisible by 13 (but $104÷7$ is not exact, so not by 7)

Trick: $7\times 11\times 13=1001$ – the three-digit alternating-sum test works for all three.

Answer: (C) 13

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

1. $SI=\frac{400\times 8\times 4.5}{100}$
2. $=4\times 8\times 4.5=144$

Trick: Yearly interest $=8\%$ of 400 $=32$; $32\times 4.5=144$.

Answer: (C) 144

Concept: Profit = Selling price of the unbroken items − Total cost price.

1. Unbroken tube lights $=56-20=36$
2. SP $=36\times 4.60=165.60$
3. Profit $=165.60-56=109.60$

Trick: Broken items add to cost but earn nothing – always subtract them before computing SP.

Answer: (C) ₹109.60

Concept: Volume of a sphere $=\frac{4}{3}\pi r^3$, radius = half the diameter.

1. $r=\frac{10}{2}=5\text{m}$
2. $V=\frac{4}{3}\times 3.141\times 125$
3. $V=\frac{1570.5}{3}=523.5{\text{m}}^3$

Trick: With $d$ given: $V=\frac{\pi d^3}{6}=\frac{3.141\times 1000}{6}=523.5$.

Answer: (B) 523.5 cubic metre

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

1. $\frac{1}{C}=\frac{1}{3}-\frac{1}{16}-\frac{1}{11}$
2. LCM 528: $\frac{176-33-48}{528}=\frac{95}{528}$
3. C alone $=\frac{528}{95}=5\frac{53}{95}$ days

Trick: C must take more than 3 days but be fastest of the three (A, B are slow), so the answer is just above 5.

Answer: (C) $5\frac{53}{95}$ days

Concept: Expand each bracket and combine like terms.

1. $4(\frac{3}{4}{x}^2-26x+17)=3x^2-104x+68$
2. $-3(x^2+4x-14)=-3x^2-12x+42$
3. Sum: $(3-3)x^2+(-104-12)x+(68+42)=-116x+110$

Trick: The $x^2$ terms cancel; check the constant: $68+42=110$ (positive).

Answer: (A) $-116x+110$

Concept: Write present ages from the past condition, then form the future condition.

1. Let son = $S$, father = $F$ now. $F-7=2(S-7)+13\Rightarrow F=2S+6$
2. After $n$ years: $F+n=2(S+n)$
3. $2S+6+n=2S+2n\Rightarrow n=6$

Trick: Father is "2 × son + 6" now; the extra 6 years is exactly the time needed for "twice".

Answer: (D) 6

Concept: Number to add = divisor − remainder.

1. $57846÷58$: $58\times 997=57826$, remainder $=20$
2. Add $58-20=38$
3. $57846+38=57884=58\times 998$ ✓

Trick: Smallest number to ADD = divisor − remainder; to SUBTRACT = remainder.

Answer: (A) 38

Concept: For a fixed distance, speed is inversely proportional to time.

1. Distance $=95\times 10=950\text{km}$
2. Required speed $=\frac{950}{5}=190\text{km/h}$

Trick: Time halved ⇒ speed doubled: $2\times 95=190$.

Answer: (A) 190 km/hr

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

1. $(7-3)=4$, $2\times 4=8$
2. $\{30+8\}=38$
3. $836÷38=22$

Trick: Check: $38\times 22=836$.

Answer: (C) 22