RRB NTPC Undergraduate CBT 1 2024 PYQs

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

1. Total distance $=282+120+496=898\text{km}$.
2. Times: $\frac{282}{47}=6$, $\frac{120}{20}=6$, $\frac{496}{62}=8$ hours; total $=20$ h.
3. Average speed $=\frac{898}{20}=44.9\text{km/h}$.

Trick: Each distance is an exact multiple of its speed – divide first, then add the hours.

Answer: (B) $44.9$

Concept: Volume of a sphere $=\frac{4}{3}\pi r^3$.

1. Diameter $=30\text{cm}$, so $r=15\text{cm}$ and $r^3=3375$.
2. $V=\frac{4}{3}\times \frac{22}{7}\times 3375=\frac{99000}{7}\approx 14142.9{\text{cm}}^3$.
3. The nearest option is $14130{\text{cm}}^3$ (this exact value comes from $\pi =3.14$: $\frac{4}{3}\times 3.14\times 3375=14130$).

Trick: $\frac{4}{3}\pi r^3=4\pi \times \frac{r^3}{3}$: $\frac{3375}{3}=1125$, then $4\times 1125=4500$, $4500\pi$.

Answer: (A) $14,130{\text{cm}}^3$

Concept: Parallel lines give equal alternate angles; equal sides of a triangle face equal angles.

1. In $△ABC$: $\angle ACB={180}^{\circ }-{54}^{\circ }-{60}^{\circ }={66}^{\circ }$.
2. $BC\parallel AD$, so $\angle CAD=\angle ACB={66}^{\circ }$ (alternate angles).
3. $AC=CD\Rightarrow \angle CDA=\angle CAD={66}^{\circ }$.
4. $\angle ACD={180}^{\circ }-{66}^{\circ }-{66}^{\circ }={48}^{\circ }$.

Trick: Once you get ${66}^{\circ }$, the answer is simply ${180}^{\circ }-2\times {66}^{\circ }$.

Answer: (B) ${48}^{\circ }$

Concept: A number is divisible by 5 if its last digit is 0 or 5.

1. $740705$ ends in $5$ → divisible by $5$.
2. It is odd, so not divisible by $4$ or $12$.
3. Digit sum $=7+4+0+7+0+5=23$, not a multiple of $9$ (or $3$).

Trick: Check the last digit first – it settles 2, 4, 5 and 10 at a glance.

Answer: (C) $5$

Concept: When time is the same, profit is shared in the ratio of investments.

1. Ratio $1350:1470:1380=135:147:138=45:49:46$ (sum $140$).
2. Isha's share $=\frac{46}{140}\times 1330=46\times 9.5=437$.

Trick: Divide total profit by the ratio sum first: $\frac{1330}{140}=9.5$ per part.

Answer: (C) ₹$437$

Concept: Combined average $=\frac{\text{total runs}}{\text{total matches}}$.

1. First 23 matches: $23\times 41=943$ runs.
2. Next 10 matches: $10\times 14=140$ runs.
3. Average $=\frac{943+140}{33}=\frac{1083}{33}\approx 32.82$.

Trick: Answer must lie between $14$ and $41$, closer to $41$ because 23 matches carry more weight.

Answer: (D) $32.82$

Concept: $SI=P\times r\times t$, with rate and time in the same unit (here per month).

1. $P=1200$, $r=4\%$ per month, $t=3$ months.
2. $SI=1200\times \frac{4}{100}\times 3=144$.

Trick: $4\%$ per month for 3 months $=12\%$; $12\%$ of $1200=144$.

Answer: (D) ₹$144$

Concept: If $p\%$ of $A$ = $k$ times $B$, then $\frac{A}{B}=\frac{k}{p/100}$.

1. $\frac{39}{100}A=\frac{8}{9}B$.
2. $\frac{A}{B}=\frac{8}{9}\times \frac{100}{39}=\frac{800}{351}$.

Trick: Cross-multiply: $A:B=(8\times 100):(9\times 39)$.

Answer: (B) $800:351$

Concept: $\tan \theta =\frac{\text{height}}{\text{base}}$; use $\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$, $\tan {60}^{\circ }=\sqrt{3}$.

1. Roof: $h_1=30\tan {30}^{\circ }=\frac{30}{\sqrt{3}}=10\sqrt{3}\text{m}$.
2. Top of pole: $h_2=30\tan {60}^{\circ }=30\sqrt{3}\text{m}$.
3. Flagstaff $=h_2-h_1=20\sqrt{3}\text{m}$.

Trick: With ${30}^{\circ }$ and ${60}^{\circ }$ from the same point, the heights are in ratio $1:3$, so pole $=\frac{2}{3}$ of $30\sqrt{3}$.

Answer: (B) $20\sqrt{3}$

Concept: BODMAS: division and multiplication before addition and subtraction.

1. $12÷6=2$ and $2\times 4=8$.
2. $16+2-8=10$.

Trick: Never add $16+12$ first – solve $÷$ and $\times$ before $+$, $-$.

Answer: (D) $10$

Concept: Selling price $=\text{MP}\times \frac{100-d}{100}$.

1. Discount $=25\%$ of $800=200$.
2. SP $=800-200=600$.

Trick: $25\%$ off means pay $\frac{3}{4}$: $\frac{3}{4}\times 800=600$.

Answer: (B) ₹$600$

Concept: One day's work $=\frac{1}{\text{days}}$; work done together = sum of rates × days.

1. $A+B$ per day $=\frac{1}{13}+\frac{1}{24}=\frac{24+13}{312}=\frac{37}{312}$.
2. In 5 days: $\frac{185}{312}$.
3. Remaining $=1-\frac{185}{312}=\frac{127}{312}$.

Trick: LCM $13\times 24=312$ units: A does 24, B does 13 per day → $5\times 37=185$, left $127$.

Answer: (C) $\frac{127}{312}$

Concept: Successive discounts are multiplied: price $\times (1-d_1)(1-d_2)(1-d_3)$.

1. After $20\%$: $10000\times 0.8=8000$.
2. After $10\%$: $8000\times 0.9=7200$.
3. After $5\%$: $7200\times 0.95=6840$.

Trick: Never add discounts ($35\%$ would give ₹$6500$, a trap).

Answer: (D) ₹$6,840$

Concept: Literates among women = total literates − literate men.

1. Total literates $=22\%$ of $359000=78980$.
2. Literate men $=22\%$ of $179000=39380$.
3. Women $=180000$; literate women $=39600$; rate $=\frac{39600}{180000}\times 100=22\%$.

Trick: If the whole and one part both have $22\%$, the other part must also be $22\%$.

Answer: (C) $22\%$

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

1. Diameter $=217\text{m}$.
2. Circumference $=\pi d=\frac{22}{7}\times 217=22\times 31=682\text{m}$.

Trick: $217=7\times 31$, so $\frac{22}{7}$ cancels neatly.

Answer: (C) $682$

Concept: Rate of a pipe $=\frac{\text{part filled}}{\text{time}}$; an emptying pipe has a negative rate.

1. P: $\frac{3}{4}$ in 15 h → full in $20$ h. Q: $\frac{5}{6}$ in 25 h → full in $30$ h.
2. In 2 h, P+Q fill $2(\frac{1}{20}+\frac{1}{30})=\frac{1}{6}$.
3. R empties $\frac{1}{6}$ in 5 h → R empties full tank in $30$ h.
4. P+Q−R: $\frac{1}{20}+\frac{1}{30}-\frac{1}{30}=\frac{1}{20}$ → $20$ hours.

Trick: Q and R have equal rates and cancel, so only P works: $20$ hours.

Answer: (D) 20 hours

Concept: Form two linear equations and solve.

1. $5s+f=79$ and $2.5f+s=94$.
2. From the first, $f=79-5s$; so $2.5(79-5s)+s=94\Rightarrow 197.5-11.5s=94$.
3. $s=9$, $f=79-45=34$.

Trick: Option check: $f=34\Rightarrow s=9$; $2.5\times 34+9=94$ ✓.

Answer: (A) $34$

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

1. Total $=3\times 15990=47970$.
2. Cheapest $=\frac{3}{15}\times 47970=9594$.

Trick: $\frac{3}{15}\times 3\times \text{avg}=\frac{3}{5}\times 15990=9594$.

Answer: (B) $9,594$

Concept: $SI=\frac{Prt}{100}$; amount $=P+SI$.

1. $P\times \frac{7.7}{100}\times \frac{20}{7}=1188\Rightarrow 0.22P=1188\Rightarrow P=5400$.
2. New SI $=5400\times \frac{8.5}{100}\times \frac{41}{9}=51\times 41=2091$.
3. Amount $=5400+2091=7491$.

Trick: $7.7\times \frac{20}{7}=22\%$ total interest – so $P=\frac{1188}{0.22}$.

Answer: (A) ₹$7,491$

Concept: Distance = speed × time (convert minutes to hours).

1. $5.4\text{min}=\frac{5.4}{60}=0.09\text{h}$.
2. Length $=42\times 0.09=3.78\text{km}$.

Trick: km/h × minutes ÷ 60: $\frac{42\times 5.4}{60}=0.7\times 5.4$.

Answer: (D) $3.78\text{km}$