RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Write both selling prices in terms of CP $x$ and equate the ₹294 gap.

1. Original SP $=1.15x$
2. New CP $=0.75x$, new SP $=1.30\times 0.75x=0.975x$
3. $1.15x-0.975x=294\Rightarrow 0.175x=294$
4. $x=\frac{294}{0.175}=1680$

Trick: Gap in SP $=17.5\%$ of CP $=294$ → CP $=\frac{294\times 100}{17.5}$.

Answer: (D) ₹1,680

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

1. Times: $\frac{90}{45}=2$, $\frac{120}{60}=2$, $\frac{150}{75}=2$ hours
2. Total distance $=360\text{km}$, total time $=6\text{h}$
3. Average $=\frac{360}{6}=60\text{km/h}$

Trick: Equal times on each part → average speed = simple mean $\frac{45+60+75}{3}=60$.

Answer: (D) $60\text{km/h}$

Concept: Take tank $=\text{LCM}(12,15,20)=60$ units; filling pipes add, emptying pipe subtracts.

1. Rates: A $=5$, B $=4$, C $=-3$ units/h
2. First 2 h (A+B): $9\times 2=18$ units
3. Remaining $42$ units at $5+4-3=6$ units/h → $7$ h
4. Total $=2+7=9$ hours

Trick: LCM-unit method avoids fractions completely.

Answer: (D) 9 hours

Concept: $\text{SP}=\text{CP}\times \frac{100\pm \%}{100}$. Find CP from the loss, then apply the profit.

1. $\text{CP}=\frac{1786.40}{0.88}=2030$
2. Required SP $=2030\times 1.14=2314.20$

Trick: Directly: new SP $=1786.40\times \frac{114}{88}$.

Answer: (D) ₹2,314.20

Concept: Net work per hour $=$ sum of filling rates $-$ emptying rate.

1. Net rate $=\frac{1}{4}+\frac{1}{8}-\frac{1}{8}=\frac{1}{4}$ per hour
2. Time for $\frac{1}{4}$ tank $=\frac{1/4}{1/4}=1$ hour

Trick: The 8-hour filler and 8-hour emptier cancel each other — only the 4-hour tap works.

Answer: (A) 1

Concept: When coefficients are swapped, add the two equations.

1. $(5c+3t)+(3c+5t)=12500+15500$
2. $8c+8t=28000\Rightarrow c+t=3500$
3. $2c+2t=7000$

Trick: Add equations → $c+t$ directly; no need to find each price.

Answer: (A) ₹7,000

Concept: When solids are melted and recast, the volume stays the same.

1. Volume of one cube $=2^3=8{\text{cm}}^3$
2. Cuboid volume $=16\times 8=128{\text{cm}}^3$
3. Check: $2b\times b\times b=2b^3=128\Rightarrow b=4$, so $8\times 4\times 4$

Trick: The shape details are not needed for the volume — just add the volumes.

Answer: (B) $128{\text{cm}}^3$

Concept: Form two equations: one for the present sum, one for 2 years ago.

1. $P+Q=46$
2. $P-2=2(Q-2)\Rightarrow P=2Q-2$
3. $3Q-2=46\Rightarrow Q=16$, $P=30$

Trick: Check options: $P=30$, $Q=16$ → 2 years ago $28$ and $14$ ✓.

Answer: (B) 30 years

Concept: Mean proportional of $a$ and $b$ $=\sqrt{ab}$.

1. Side $=\sqrt{16\times 25}=\sqrt{400}=20\text{m}$
2. Check: $16:20=20:25$ ✓

Trick: $\sqrt{16}\times \sqrt{25}=4\times 5$.

Answer: (D) 20 m

Concept: Convert time to minutes and find pages per minute.

1. 1 h 20 min $=80$ min; rate $=\frac{2400}{80}=30$ pages/min
2. In 45 min: $30\times 45=1350$

Trick: $2400\times \frac{45}{80}=2400\times \frac{9}{16}=1350$.

Answer: (B) 1350

Concept: Numbers in ratio $9:7:5$ are $9k,7k,5k$ with HCF $k$ (ratio terms are co-prime).

1. $\text{LCM}=9\times 7\times 5\times k=315k$
2. $315k=2205\Rightarrow k=7$

Trick: HCF $=\frac{\text{LCM}}{\text{LCM of ratio terms}}=\frac{2205}{315}$.

Answer: (A) 7

Concept: Add whole parts and fractional parts separately.

1. Whole parts: $55\times 4=220$
2. Fractions: $\frac{1+3+5+7}{8}=\frac{16}{8}=2$
3. Total $=220+2=222$

Trick: Sum of first $n$ odd numbers $=n^2$: $1+3+5+7=16$.

Answer: (B) 222

Concept: $(a-b{)}^2=(a+b{)}^2-4ab$.

1. $(a-b{)}^2={32}^2-4\times 220=1024-880=144$
2. $a-b=12$ (numbers $22$ and $10$)

Trick: Find two numbers with sum 32, product 220: $22\times 10$ → difference 12.

Answer: (B) 12

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

1. $P=\frac{1440\times 100}{16\times 2}=\frac{144000}{32}=4500$

Trick: Total rate $=16\times 2=32\%$; $32\%$ of $P=1440$.

Answer: (C) 4,500

Concept: Mean $=\frac{\sum fx}{\sum f}$.

1. $\sum f=2+3+1+4+f+4=14+f$
2. $\sum fx=20+60+30+160+50f+240=510+50f$
3. $\frac{510+50f}{14+f}=40\Rightarrow 510+50f=560+40f$
4. $10f=50\Rightarrow f=5$

Trick: Deviation method: $\sum f(x-40)=0$ → $-60-60-10+0+10f+80=0$ → $f=5$.

Answer: (C) 5

Concept: $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. $\tan {60}^{\circ }=\frac{18}{d}\Rightarrow \sqrt{3}=\frac{18}{d}$
2. $d=\frac{18}{\sqrt{3}}=\frac{18\sqrt{3}}{3}=6\sqrt{3}\text{m}$

Trick: At ${60}^{\circ }$, base $=\frac{\text{height}}{\sqrt{3}}$; at ${30}^{\circ }$, base $=\text{height}\times \sqrt{3}$.

Answer: (A) $6\sqrt{3}$

Concept: Surface area of a sphere $=4\pi r^2$.

1. Sum $=4\pi (2^2+3^2+6^2)=4\pi \times 49$
2. $=4\times \frac{22}{7}\times 49=616{\text{cm}}^2$

Trick: $r^2$ sum $=49$ cancels the 7 in $\pi =\frac{22}{7}$ neatly.

Answer: (A) $616{\text{cm}}^2$

Concept: Convert each scheme into an effective % discount on 5 toys; least beneficial = smallest discount.

1. (A) $1-0.76\times 0.67=1-0.5092\Rightarrow 49.08\%$
2. (B) $1-0.61\times 0.63\times 0.77=1-0.2959\Rightarrow 70.4\%$
3. (C) $\frac{2\times 24+3\times 12}{5}=\frac{84}{5}=16.8\%$
4. (D) pays for 5, gets 7 → $\frac{2}{7}\approx 28.6\%$
5. Smallest discount is scheme C

Trick: Successive discounts multiply the remaining fractions, they never simply add.

Answer: (B) C

Concept: Selling price after successive discounts $=\text{MP}\times (1-d_1)(1-d_2)$.

1. $3000\times 0.9=2700$
2. $2700\times 0.8=2160$

Trick: Single equivalent discount $=10+20-\frac{10\times 20}{100}=28\%$; $72\%$ of $3000=2160$.

Answer: (D) ₹2,160

Concept: $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$, using the class with the highest frequency.

1. Modal class $50-55$: $l=50$, $f_1=13$, $f_0=4$, $f_2=5$, $h=5$
2. $\text{Mode}=50+\frac{13-4}{26-4-5}\times 5=50+\frac{45}{17}$
3. $=50+2.647=52.65$

Trick: Mode must lie inside the modal class (50–55) — only 51.34 and 52.65 qualify; formula picks 52.65.

Answer: (C) 52.65