RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$

1. Discount = ₹$500-425=$ ₹$75$.
2. $\text{Discount}\%=\frac{75}{500}\times 100=15\%$.

Trick: Each ₹100 of MP gives ₹$15$ off, since $425=500\times 0.85$.

Answer: (B) $15\%$

Concept: Working together, the rates (part of the tank per hour) add: $\frac{1}{T}=\frac{1}{15}+\frac{1}{27}+\frac{1}{10}$.

1. Take tank = LCM$(15,27,10)=270$ units.
2. Rates: A $=\frac{270}{15}=18$, B $=\frac{270}{27}=10$, C $=\frac{270}{10}=27$ units/h.
3. Together $=18+10+27=55$ units/h.
4. Time $=\frac{270}{55}=\frac{54}{11}=4\frac{10}{11}$ hours.

Trick: Together time must be less than the fastest pipe alone (10 h): A + B together fill $\frac{1}{15}+\frac{1}{27}>\frac{1}{10}$, so total rate $>\frac{2}{10}$ → time $<5$ h → only (C).

Answer: (C) $4\frac{10}{11}$ hours

Concept: With false weights, actual profit = (money received ÷ cost of goods actually given) − 1.

1. Let CP of 1 kg = ₹100. At 20% loss he charges ₹80 for "1 kg".
2. He actually gives 35% less, i.e. $0.65$ kg, costing ₹65.
3. Profit $\%=\frac{80-65}{65}\times 100=\frac{1500}{65}=23.08\%$.

Trick: Profit% $=\frac{(100-\text{loss})}{(100-\text{cheat})}\times 100-100=\frac{80}{65}\times 100-100$.

Answer: (C) $23.08\%$

Concept: If $a^2+b^2=c^2$ the triangle is right-angled and Area $=\frac{1}{2}\times \text{base}\times \text{height}$.

1. $9^2+{12}^2=81+144=225={15}^2$, so the triangle is right-angled with legs 9 cm and 12 cm.
2. Area $=\frac{1}{2}\times 9\times 12=54{\text{cm}}^2$.
3. Check by Heron: $s=18$, Area $=\sqrt{18\times 9\times 6\times 3}=\sqrt{2916}=54$.

Trick: 9, 12, 15 = $3\times (3,4,5)$ triplet → area $=9\times \frac{3\times 4}{2}=54$.

Answer: (B) $54{\text{cm}}^2$

Concept: Simplify the bracket first, expand each term, then combine like terms.

1. $\frac{6}{2}=3$, so the first part is $2(3x^2-22x+15)=6x^2-44x+30$.
2. $-6(x^2+5x-18)=-6x^2-30x+108$.
3. Add: $(6x^2-6x^2)+(-44x-30x)+(30+108)=-74x+138$.

Trick: Watch the sign: $-6\times (-18)=+108$, so the constant is positive.

Answer: (D) $-74x+138$

Concept: Profit = total selling price − total cost price (broken items earn nothing).

1. Bulbs left $=37-19=18$.
2. SP $=18\times 3.90=$ ₹$70.20$.
3. Profit $=70.20-37=$ ₹$33.20$.

Trick: Cost is paid on all 37 bulbs, revenue only on the 18 unbroken ones.

Answer: (B) ₹$33.20$

Concept: Form two equations in the present ages $P$ and $Q$ and solve.

1. $3P-Q=23$ ... (i)
2. After 7 years: $6(Q+7)=6(P+7)-30\Rightarrow 6Q=6P-30\Rightarrow P=Q+5$ ... (ii)
3. Put (ii) in (i): $3(Q+5)-Q=23\Rightarrow 2Q=8\Rightarrow Q=4$.
4. Check: $P=9$, $3\times 9-4=23$ ✓.

Trick: Age difference never changes: $6(P-Q)=30\Rightarrow P-Q=5$ directly.

Answer: (D) $4$

Concept: An exterior angle of a triangle equals the sum of the two interior opposite angles.

1. In right $△BDC$: $\angle C={90}^{\circ }-\angle DBC={90}^{\circ }-{44}^{\circ }={46}^{\circ }$.
2. In $△AEC$, $\angle AEB$ is the exterior angle at E (B, E, C are collinear).
3. $\angle AEB=\angle EAC+\angle ECA={23}^{\circ }+{46}^{\circ }={69}^{\circ }$.

Trick: Perpendicular → complementary: $\angle C={90}^{\circ }-{44}^{\circ }$; then exterior angle = sum of opposite interior angles.

Answer: (B) ${69}^{\circ }$

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

1. Total price $=3\times 5300=$ ₹$15900$.
2. Ratio parts $=3+5+7=15$, so 1 part $=\frac{15900}{15}=$ ₹$1060$.
3. Cheapest $=3\times 1060=$ ₹$3180$.

Trick: Middle ratio term 5 = average of 3, 5, 7, so 5 parts = ₹$5300$ → 1 part = ₹$1060$.

Answer: (B) ₹$3,180$

Concept: Eliminate options with quick rules, then confirm the remaining one by division.

1. $87=3\times 29$. Digit sum $=5+4+1+0+9+6+9+4+3=41$, not divisible by 3 → not divisible by 87.
2. Divide by 97: $541096943÷97=5578319$ exactly (remainder 0).
3. Dividing by 103 leaves remainder 39 and by 107 leaves remainder 83.

Trick: For 97, use $100\equiv 3\left(\mathrm{mod}97\right)$: take 2-digit groups from the right 43, 69, 09, 41, 5 → $43+69\times 3+9\times 9+41\times 27+5\times 81=1843=97\times 19$ ✓.

Answer: (A) $97$

Concept: $SI=\frac{P\times R\times T}{100}$, where R and T must be in the same time unit.

1. Rate is 4% per month and time is 12 months, so $SI=\frac{1200\times 4\times 12}{100}$.
2. $SI=12\times 48=576$, i.e. ₹576.
3. Note: ₹48 would be the interest only if the rate were 4% per annum for one year ($\frac{1200\times 4\times 1}{100}=48$).

Trick: Interest per month $=4\%$ of 1200 $=48$; for 12 months $=48\times 12=576$.

Answer: (D) ₹576

Concept: A prime has no divisor other than 1 and itself; test divisibility by primes up to $\sqrt{n}\approx 20$.

1. Even numbers 394, 396, 398, 400 are out; 395 ends in 5 (divisible by 5).
2. 393: digit sum 15 → divisible by 3. 399: digit sum 21 → divisible by 3 ($399=3\times 133$).
3. 397 and 401 are not divisible by 2, 3, 5, 7, 11, 13, 17 or 19 → both prime.

Trick: Only odd numbers not ending in 5 need checking: 393, 397, 399, 401 → strike multiples of 3.

Answer: (B) 2

Concept: For consecutive numbers $n(n+1)=N$, $n$ is just below $\sqrt{N}$.

1. $\sqrt{462}\approx 21.5$, so try $n=21$.
2. $21\times 22=462$ ✓.
3. Greater number $=22$.

Trick: Unit digit 2 of 462 comes from $1\times 2$ → numbers end in 1 and 2: 21 and 22.

Answer: (B) 22

Concept: $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. Discount $=1500-1200=300$.
2. $\text{Discount}\%=\frac{300}{1500}\times 100=20\%$.

Trick: $\frac{300}{1500}=\frac{1}{5}=20\%$ — discount is always on marked price.

Answer: (A) $20\%$

Concept: Count days excluding the deposit day and including the withdrawal day; 2024 is a leap year (366 days).

1. Feb: 29 − 8 = 21 days; March: 31; April: 9 → total $21+31+9=61$ days.
2. $T=\frac{61}{366}=\frac{1}{6}$ year.
3. $SI=\frac{4000\times 7.5\times 1}{100\times 6}=\frac{300}{6}=50$.

Trick: Yearly interest $=300$; 61 days of a 366-day year $=\frac{1}{6}$ year → $300÷6=50$.

Answer: (A) ₹50

Concept: Find each quantity separately and subtract.

1. $65\%\text{ of }65=\frac{65\times 65}{100}=\frac{4225}{100}=42.25$.
2. $\frac{2}{5}\text{ of }25=10$.
3. Difference $=42.25-10=32.25$.

Trick: ${65}^2=4225$ (numbers ending in 5: $6\times 7=42$, then 25).

Answer: (D) $32.25$

Concept: One person’s one-day work = $\frac{1}{\text{persons}\times \text{days}}$.

1. 1 man’s 1-day work $=\frac{1}{4\times 10}=\frac{1}{40}$; 1 woman’s $=\frac{1}{3\times 16}=\frac{1}{48}$.
2. 6 women + 5 men per day $=\frac{6}{48}+\frac{5}{40}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$.
3. Time $=4$ days.

Trick: Total work $=40$ man-days $=48$ woman-days → 6 W = $\frac{1}{8}$, 5 M = $\frac{1}{8}$ per day.

Answer: (C) 4 days

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

1. Times: $\frac{612}{68}=9$ h, $\frac{312}{39}=8$ h, $\frac{620}{62}=10$ h → total 27 h.
2. Distance $=612+312+620=1544$ km.
3. Average $=\frac{1544}{27}\approx 57.19\text{km/h}$.

Trick: Never average the speeds directly; each part here gives an exact whole-number time.

Answer: (B) $57.19$

Concept: Digit-sum test for 3 (and hence 18, 21); direct division for 13.

1. Digit sum $=7+1+8+7+5+4+0+2+3=37$ → not divisible by 3, so 3, 18 and 21 (all multiples of 3) are ruled out.
2. $718754023÷13=55288771$ exactly.

Trick: If a number fails the 3-test, every option that is a multiple of 3 is eliminated at once.

Answer: (D) 13

Concept: Profit is shared in the ratio of investments.

1. $8:36:76=2:9:19$, total 30 parts.
2. 1 part $=\frac{3030}{30}=101$.
3. C − B $=(19-9)\times 101=1010$.

Trick: Difference of shares = difference of ratio parts × value of one part.

Answer: (B) ₹1,010