RRB NTPC Undergraduate CBT 2 2024 PYQs

Concept: $\text{Average}=\frac{\text{Sum}}{\text{Number of terms}}$

1. $\text{Average}=\frac{612}{10}$
2. $=61.2$

Trick: Dividing by 10 just shifts the decimal one place left.

Answer: (D) $61.2$

Concept: Write everything as powers of 3: $27=3^3$, $9=3^2$, ${\left(\frac{1}{3}\right)}^{-5}=3^5$.

1. $3^5-{\left(\frac{1}{3}\right)}^{-5}=243-243=0$.
2. Numerator: $3^{36}+3^{26}=3^{26}(3^{10}+1)$.
3. Denominator: $3^{18}+3^{28}=3^{18}(1+3^{10})$.
4. $\sqrt{\frac{3^{26}}{3^{18}}}=\sqrt{3^8}=3^4=81$.

Trick: Take out the smaller power as common – the brackets $(3^{10}+1)$ cancel.

Answer: (D) 81

Concept: $\text{HCF}\times \text{LCM}=\text{Product of the two numbers}$

1. $49\times 490=245\times x$
2. $x=\frac{49\times 490}{245}=49\times 2=98$
3. Check: $98=49\times 2$, $245=49\times 5$ → HCF 49, LCM $49\times 10=490$ ✔.

Trick: $\frac{490}{245}=2$, so other number $=49\times 2$.

Answer: (C) 98

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

1. $P=\frac{1560\times 100}{20\times 4}$
2. $=\frac{156000}{80}=1950$

Trick: Total rate = $20\%\times 4=80\%$; 80% of P = 1560 → $P=1560\times \frac{5}{4}=1950$.

Answer: (C) ₹1,950

Concept: Take CP = 100 and compare MP with the required SP.

1. $\text{MP}=100+90=190$.
2. Gain 71% → $\text{SP}=171$.
3. $\text{Discount}=190-171=19$.
4. $\text{Discount}\%=\frac{19}{190}\times 100=10\%$.

Trick: $\frac{100+71}{100+90}=\frac{171}{190}=0.9$ → 10% off.

Answer: (D) $10\%$

Concept: $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$ for each part, then add.

1. $\frac{180}{45}=4\text{h}$
2. $\frac{180}{60}=3\text{h}$
3. $\frac{180}{90}=2\text{h}$
4. Total $=4+3+2=9\text{h}$.

Trick: Never use the average of speeds; add the times of each leg.

Answer: (B) 9 hours

Concept: Alligation: Cheaper : Dearer = (Dearer − Mean) : (Mean − Cheaper).

1. $150-145.5=4.5$
2. $145.5-130=15.5$
3. Ratio $=4.5:15.5=9:31$.

Trick: The mean (145.5) is close to 150, so more of the dearer variety → smaller first term.

Answer: (C) 9 : 31

Concept: Make the common term ($\beta$) equal in both ratios using LCM.

1. LCM of 22 and 12 $=132$.
2. $\alpha :\beta =15:22=90:132$ (× 6).
3. $\beta :\gamma =12:13=132:143$ (× 11).
4. $\alpha :\beta :\gamma =90:132:143$.

Trick: $\alpha :\beta :\gamma =(15\times 12):(22\times 12):(22\times 13)=180:264:286=90:132:143$.

Answer: (C) 90 : 132 : 143

Concept: $\text{CP}=\frac{\text{SP}}{1-\text{loss}\%}$; then compare the new SP with CP.

1. 95% of CP = 850 → $\text{CP}=\frac{850}{0.95}$.
2. $\frac{\text{New SP}}{\text{CP}}=\frac{1530\times 0.95}{850}=\frac{1453.5}{850}=1.71$.
3. So gain $=71\%$, i.e. $x=71$.

Trick: $\frac{1530}{850}=1.8$; $1.8\times 95\%=171\%$ → gain 71%.

Answer: (B) 71

Concept: Take distance MN = D. Speed of A $=\frac{D}{8}$ (7:20 a.m.–3:20 p.m.), speed of B $=\frac{D}{7}$ (9:20 a.m.–4:20 p.m.).

1. By 9:20 a.m. A has run 2 h: $\frac{2D}{8}=\frac{D}{4}$; remaining $\frac{3D}{4}$.
2. Relative speed $=\frac{D}{8}+\frac{D}{7}=\frac{15D}{56}$.
3. Time $=\frac{3D}{4}÷\frac{15D}{56}=\frac{3\times 56}{4\times 15}=2.8\text{h}=2\text{h}48\text{min}$.
4. 9:20 a.m. + 2 h 48 min = 12:08 p.m.

Trick: Take D = LCM(8, 7) = 56 km: speeds 7 and 8, gap after 2 h = 42 km, $\frac{42}{15}=2.8\text{h}$.

Answer: (A) 12:08 p.m.

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

1. $\text{SP}=6500\times \frac{85}{100}$
2. $=65\times 85=5525$

Trick: 15% of 6500 = 975; $6500-975=5525$.

Answer: (C) ₹5525

Concept: Successive discounts multiply: $\text{SP}=\text{MP}\times (1-d_1)(1-d_2)(1-d_3)$.

1. $960\times 0.75=720$
2. $720\times 0.75=540$
3. $540\times 0.85=459$

Trick: $960\times \frac{3}{4}\times \frac{3}{4}=540$; 15% of 540 = 81 → $540-81=459$.

Answer: (A) 459

Concept: "A is what % of B" $=\frac{A}{B}\times 100$.

1. $\frac{75}{54}\times 100$
2. $=\frac{7500}{54}\approx 138.89\%$

Trick: A (75) > B (54), so the answer must exceed 100% – only 101.07% and 138.89%; $\frac{75}{54}\approx 1.39$.

Answer: (D) $138.89\%$

Concept: Water per second = (cross-section area of pipe) × speed. Time = tank volume ÷ water per second.

1. $r=10.5\text{cm}=0.105\text{m}$; area $=\frac{22}{7}\times {0.105}^2=0.03465{\text{m}}^2$.
2. Flow $=0.03465\times 8=0.2772{\text{m}}^3\text{/s}$.
3. Tank volume $=12\times 9\times 4.41=476.28{\text{m}}^3$.
4. Time $=\frac{476.28}{0.2772}\approx 1718.2\text{s}=\frac{1718.2}{60}\approx 28.6\text{min}$.

Trick: Convert cm to m first; per-minute flow $=0.2772\times 60=16.632{\text{m}}^3$, and $\frac{476.28}{16.632}\approx 28.6$.

Answer: (A) 28.6 minutes

Concept: With different yearly rates, $A=P(1+\frac{r_1}{100})(1+\frac{r_2}{100})$. Here $r_1=10\%$, $r_2=10+3=13\%$.

1. After year 1: $57000\times 1.10=62700$.
2. After year 2: $62700\times 1.13=70851$.
3. $\text{CI}=70851-57000=13851$.

Trick: Effective rate $=10+13+\frac{10\times 13}{100}=24.3\%$; $24.3\%$ of 57000 = 13851.

Answer: (D) ₹13,851