RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: When parts of the stock are sold at different profit/loss rates, the overall profit % is the weighted average of the rates (weights = fraction of goods, since cost per unit is the same).

1. Fractions: first $50\%$, then $40\%$ of the remaining $50\%$ = $20\%$, the rest = $30\%$.
2. Overall profit $=0.5\times 21+0.2\times 36-0.3\times 29=10.5+7.2-8.7=9$.

Trick: Think of total cost as ₹100: profit = $10.5+7.2-8.7=9$ rupees = $9\%$ directly.

Answer: (C) 9

Concept: A number leaving the same remainder $r$ on division by several numbers = (multiple of their LCM) $+r$.

1. $\text{LCM}(6,14,7)=42$.
2. Largest 4-digit multiple of 42: $9999=42\times 238+3$, so $42\times 238=9996$. But $9996+5=10001$ is 5-digit, so take $42\times 237=9954$.
3. Required number $=9954+5=9959$.

Trick: Always check that adding the remainder does not push the number past 9999 – that is exactly the trap here.

Answer: (B) 9959

Concept: SP $=$ MP $\times$ (product of discount multipliers); compare SP with CP.

1. Let CP $=100$, MP $=140$.
2. SP $=140\times 0.82\times 0.95=140\times 0.779=109.06$.
3. Profit $=9.06\%$.

Trick: Combined discount of 18% and 5% $=18+5-\frac{18\times 5}{100}=22.1\%$; then $140\times 0.779$.

Answer: (D) 9.06%

Concept: Successive profits multiply: final price $=\text{CP}\times (1+p_1)(1+p_2)$.

1. First sale: $200\times 1.10=220$.
2. Second sale: $220\times 1.15=253$.

Trick: Net % $=10+15+\frac{10\times 15}{100}=26.5\%$ → $200\times 1.265=253$.

Answer: (A) ₹253

Concept: Divisible by 8 ⇔ last three digits divisible by 8; divisible by 9 ⇔ digit sum divisible by 9.

1. Last three digits 512 $=8\times 64$ ✓ for any $y$.
2. Digit sum $=6+y+5+1+2=14+y$; must be 18 → $y=4$.

Trick: The 8-test does not involve $y$ here, so only the 9-test decides.

Answer: (C) 4

Concept: $\frac{1}{{\cot }^{2}A}={\tan }^{2}A$, $\frac{1}{{\cos }^{2}A}={\sec }^{2}A$ and ${\sec }^{2}A-{\tan }^{2}A=1$.

1. Expression $=15+12{\tan }^{2}A-12{\sec }^{2}A$.
2. $=15-12({\sec }^{2}A-{\tan }^{2}A)=15-12=3$.

Yaad rakhein: $1+{\tan }^{2}A={\sec }^{2}A$ – whenever ${\tan }^2$ and ${\sec }^2$ appear with equal coefficients, they collapse to a constant.

Answer: (B) 3

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

1. $(a-b{)}^2={35}^2-4\times 286=1225-1144=81$.
2. $a-b=9$ (numbers are 22 and 13).

Trick: Factor 286 $=2\times 11\times 13$ → pair $22\times 13$ sums to 35, difference 9.

Answer: (B) 9

Concept: Sum $=$ Average $\times$ count.

1. Total $=15\times 5=75$.
2. $x=75-(7+11+15+23)=75-56=19$.

Trick: Deviations from 15: $-8,-4,0,+8$ sum to $-4$, so $x=15+4=19$.

Answer: (A) 19

Concept: If every observation changes by the same amount $k$, the average changes by exactly $k$.

1. Each of the 8 numbers decreases by 4.
2. New average $=45-4=41$.

Trick: No need of the count 8 – it is a distractor; average simply shifts by 4.

Answer: (B) 41

Concept: The diagonal of a square bisects its right angles, so it makes ${45}^{\circ }$ with each side. Extending the diagonal does not change the line, so the angle with the side-line stays ${45}^{\circ }$ (acute angle).

1. In square ABCD, diagonal AC makes $\angle ACB={45}^{\circ }$.
2. Extension CE is the same straight line; the acute angle between CE and side BC (produced) is also ${45}^{\circ }$.
3. The length 6 units does not matter – angles are independent of it.

Yaad rakhein: Diagonal of a square ↔ ${45}^{\circ }$ with every side; diagonal of a rectangle ↔ not ${45}^{\circ }$ in general.

Answer: (B) 45 degrees

Concept: Form equations for present and future ages; both ages increase by the same number of years.

1. Let daughter $=d$, mother $=76-d$.
2. After 4 years: $80-d=3(d+4)\Rightarrow 80-d=3d+12\Rightarrow d=17$.
3. Mother $=76-17=59$.

Trick: After 4 years total $=84$ in ratio $3:1$ → mother $=63$, daughter $=21$; subtract 4 → 59 and 17.

Answer: (D) 59

Concept: Let the quantities be $5x$ and $7x$; the total stays the same after transfer.

1. $\frac{5x+12}{7x-12}=\frac{7}{5}\Rightarrow 25x+60=49x-84$.
2. $24x=144\Rightarrow x=6$.
3. Second bag originally $=7\times 6=42$.

Trick: Total $=12$ parts both times; the ratio flips $5:7\to 7:5$, so 12 marbles $=2$ parts → 1 part $=6$.

Answer: (B) 42

Concept: First find the maximum marks from the given percentage condition.

1. $16\%$ of max $=448+64=512\Rightarrow \text{max}=\frac{512\times 100}{16}=3200$.
2. Friend: $\frac{672}{3200}\times 100=21\%$.

Trick: $672=512+160$; $512=16\%$ and $160=5\%$ of 3200 → $21\%$.

Answer: (D) 21%

Concept: Take total work = LCM of days; work per day = efficiency.

1. Total work $=\text{LCM}(20,25)=100$ units; A $=5$/day, B $=4$/day.
2. 5 days together: $9\times 5=45$ units done; $55$ remain.
3. A alone: $\frac{55}{5}=11$ days.

Trick: Remaining fraction $=1-5(\frac{1}{20}+\frac{1}{25})=\frac{11}{20}$ → A needs $\frac{11}{20}\times 20=11$ days.

Answer: (B) 11 days

Concept: Perimeter $=2(l+b)$.

1. $l=2b+4$, so $2(2b+4+b)=56\Rightarrow 3b+4=28$.
2. $b=8$ cm (and $l=20$ cm).

Trick: $l+b=28$; check option (D): $b=8\Rightarrow l=20$, sum 28 ✓.

Answer: (D) 8 cm

Concept: $12.5\%=\frac{1}{8}$, so amount multiplies by $\frac{9}{8}$ each year.

1. Amount $=8000\times \frac{9}{8}\times \frac{9}{8}=125\times 81=10125$.
2. CI $=10125-8000=2125$.

Trick: Effective 2-year rate $=12.5+12.5+\frac{12.5\times 12.5}{100}=26.5625\%$ → $8000\times 0.265625=2125$.

Answer: (D) ₹2,125

Concept: For 2 years, effective CI rate $=2r+\frac{r^2}{100}$.

1. At $5\%$: $10+0.25=10.25\%$.
2. $P\times \frac{10.25}{100}=102.50\Rightarrow P=1000$.

Yaad rakhein: 5% for 2 years $=10.25\%$; 10% for 2 years $=21\%$ – memorise these.

Answer: (A) ₹1,000

Concept: Melting preserves volume; TSA of cuboid $=2(lb+bh+hl)$.

1. Cuboid volume $=\frac{V}{2}$, so $V-\frac{V}{2}=250\Rightarrow V=500$, cuboid $=250{\text{cm}}^3$.
2. Dimensions $2k,k,k$: $2k^3=250\Rightarrow k=5$ → $10\times 5\times 5$.
3. TSA $=2(50+25+50)=250{\text{cm}}^2$.

Trick: For a $2k\times k\times k$ cuboid, TSA $=10k^2$ and volume $=2k^3$.

Answer: (D) $250{\text{cm}}^2$

Concept: Convert every scheme into an effective discount on 5 toys (MP ₹100 each, total ₹500).

1. (A) $0.95\times 0.70=0.665$ → pays ₹332.5 (discount $33.5\%$).
2. (B) $0.82\times 0.85\times 0.95\approx 0.662$ → pays ≈ ₹331 (discount ≈ $33.8\%$).
3. (C) only $11\%$ on all 5 toys → pays ₹445; the 47% part never applies.
4. (D) 2 free with 5 → 7 toys for ₹500, effective discount $\frac{2}{7}\approx 28.6\%$ per toy.

Trick: Scheme C's big 47% starts only from the 6th toy – for exactly 5 toys it is the weakest offer.

Answer: (C) C

Concept: Brackets first, then "of", then ÷ and ×, then + and −.

1. $\frac{14}{24}÷\frac{14}{9}=\frac{9}{24}=\frac{3}{8}$.
2. $\frac{3}{7}\times \frac{14}{9}+\frac{5}{8}=\frac{2}{3}+\frac{5}{8}=\frac{31}{24}$; so $\frac{3}{8}÷\frac{31}{24}=\frac{9}{31}$.
3. $\frac{31}{24}\text{ of }\frac{24}{10}=\frac{31}{10}$; $\frac{9}{10}÷\frac{31}{10}=\frac{9}{31}$.
4. Sum $=\frac{9}{31}+\frac{9}{31}=\frac{18}{31}$.

Trick: Both halves give $\frac{9}{31}$ – spotting the repeated 31 saves time.

Answer: (B) $\frac{18}{31}$