RRB NTPC Undergraduate CBT 2 2024 PYQs

Concept: A cap is open at the base, so only the curved surface is painted: $\text{CSA}=\pi rl$, with $l=\sqrt{r^2+h^2}$.

1. $r=\frac{40}{2}=20$ cm, $h=21$ cm $\Rightarrow l=\sqrt{400+441}=\sqrt{841}=29$ cm.
2. $\text{CSA}=\frac{22}{7}\times 20\times 29=\frac{12760}{7}\approx 1822.86{\text{cm}}^2$.
3. Cost $=1822.86\times 1.12\approx ₹2041.6$.

Trick: $(20,21,29)$ is a Pythagorean triplet; $112$ paise $=₹1.12$.

Answer: (C) ₹2,041.6

Concept: Use $\tan \theta =\frac{\text{height}}{\text{base}}$ in the two right triangles from P.

1. $\tan {45}^{\circ }=\frac{25}{d}=1\Rightarrow d=25$ m.
2. $\tan {60}^{\circ }=\frac{25+p}{25}=\sqrt{3}\Rightarrow 25+p=25\sqrt{3}$.
3. $p=25\sqrt{3}-25=25(\sqrt{3}-1)$ m.

Trick: With ${45}^{\circ }$ the base equals the house height; then pole $=$ base$\times (\tan {60}^{\circ }-\tan {45}^{\circ })$.

Answer: (B) $25(\sqrt{3}-1)$

Concept: The median of 5 values is the 3rd value after arranging them in order; test the options.

1. $x=3$: values $7,3,15,4,11$.
2. Ordered: $3,4,7,11,15$ → median $=7$ ✓.
3. $x=2$ gives $6,1,10,3,8$ → median $6$; $x=4$ gives median $8$ – not $7$.

Trick: For odd $n$, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ term of the sorted data.

Answer: (A) $3$

Concept: $\text{LCM of fractions}=\frac{\text{LCM of numerators}}{\text{HCF of denominators}}$.

1. LCM$(4,5,8)=40$.
2. HCF$(7,21,63)=7$.
3. LCM $=\frac{40}{7}$.

Trick: LCM of fractions is a multiple of each fraction: $\frac{40}{7}÷\frac{4}{7}=10$ ✓.

Answer: (D) $\frac{40}{7}$

Concept: Supplementary angles add up to ${180}^{\circ }$.

1. Let smaller $=x$, larger $=x+16$.
2. $2x+16=180\Rightarrow x={82}^{\circ }$.
3. Larger $={98}^{\circ }$; check $82+98=180$ ✓.

Trick: Smaller $=\frac{180-\text{difference}}{2}=\frac{164}{2}=82$.

Answer: (D) $82$

Concept: A's one-day work $=$ (A+B)'s one-day work $-$ B's one-day work.

1. $\frac{1}{10}-\frac{1}{15}=\frac{3-2}{30}=\frac{1}{30}$.
2. So A alone takes $30$ days.

Trick: LCM $30$ units: A+B $=3$/day, B $=2$/day → A $=1$/day → $30$ days.

Answer: (C) $30$ days

Concept: $\text{Mean}=\frac{\text{sum of observations}}{\text{number of observations}}$.

1. There are $20$ numbers.
2. Sum (in groups): $12+3+18+17+4=54$; $9+17+19+20+15=80$; $8+17+2+3+16=46$; $11+3+1+0+5=20$.
3. Total $=54+80+46+20=200$; mean $=\frac{200}{20}=10$.

Trick: Add in groups of five to avoid slips.

Answer: (D) $10$

Concept: If B is 50% of (A + C), then B : (A + C) = 1 : 2, so B is $\frac{1}{3}$ of the total.

1. Let $A+C=x$, then $B=\frac{x}{2}$.
2. $x+\frac{x}{2}=9900\Rightarrow x=6600$.
3. $B=\frac{6600}{2}=3300$. (The ratio 4 : 18 of A and C is not even needed.)

Trick: B : (A + C) = 1 : 2 → B = $\frac{1}{3}\times 9900=3300$ directly.

Answer: (D) ₹3,300

Concept: For a square, $\text{Area}=a^2$ and $\text{Perimeter}=4a$.

1. $a^2=1369\Rightarrow a=\sqrt{1369}=37\text{m}$.
2. $\text{Perimeter}=4\times 37=148\text{m}$.

Trick: 1369 ends in 9, so the root ends in 3 or 7; ${37}^2=1369$ fits.

Answer: (B) 148

Concept: A number is divisible by 3 if the sum of its digits is divisible by 3.

1. Known digits: $4+0+8+6+1=19$.
2. $19+(A+B)$ must be a multiple of 3; the next multiple is 21.
3. So minimum $A+B=21-19=2$.

Trick: $19÷3$ leaves remainder 1, so add $3-1=2$.

Answer: (B) 2

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.