RRB NTPC Undergraduate CBT 2 2024 PYQs

Concept: Simplify a nested root from the innermost bracket outward.

1. $\sqrt{4}=2$, so $7+2=9$ and $\sqrt{9}=3$.
2. $97+3=100$, $\sqrt{100}=10$.
3. $15+10=25$, $\sqrt{25}=5$.
4. $31+5=36$, $\sqrt{36}=6$.

Trick: Each step gives a perfect square – if a step does not, recheck the arithmetic.

Answer: (D) $6$

Concept: Filling rates are added and the emptying rate is subtracted: $\frac{1}{A}+\frac{1}{B}-\frac{1}{C}=\frac{1}{T}$.

1. $\frac{1}{99}+\frac{1}{90}=\frac{10+11}{990}=\frac{7}{330}$.
2. $\frac{1}{C}=\frac{7}{330}-\frac{1}{50}=\frac{35-33}{1650}=\frac{2}{1650}=\frac{1}{825}$.
3. So the outlet empties the full cistern in $825$ min.

Trick: Use LCM units: total $=4950$ units; rates $50+55=105$; together $99$/min, so outlet $=6$/min; $4950÷6=825$.

Answer: (B) $825$ min

Concept: For a round trip of distance $d$ each way: $\frac{d}{\text{downstream}}+\frac{d}{\text{upstream}}=\text{total time}$.

1. $\frac{d}{115}+\frac{d}{23}=25\Rightarrow \frac{d+5d}{115}=25$.
2. $6d=2875\Rightarrow d=\frac{2875}{6}=479\frac{1}{6}\text{km}$.
3. Total distance $=2d=\frac{2875}{3}=958\frac{1}{3}\text{km}$.

Trick: Downstream speed is $5\times$ upstream, so time splits $1:5$: upstream time $=\frac{5}{6}\times 25$ h; $d=23\times \frac{125}{6}$.

Answer: (B) $958\frac{1}{3}\text{km}$

Concept: Solve the innermost bracket first, then multiply, then add.

1. $26-14+6=18$.
2. $-9\times 18=-162$.
3. $30+(-162)=-132$.

Trick: A positive number plus a bigger negative number: subtract and keep the minus sign.

Answer: (C) $-132$

Concept: Take total work as the LCM of the given days; then work $=$ rate $\times$ time.

1. Total work $=\text{LCM}(14,42)=42$ units.
2. A+B $=3$ units/day; A $=1$ unit/day $\Rightarrow$ B $=2$ units/day.
3. One-third work $=14$ units $\Rightarrow$ time $=\frac{14}{2}=7$ days.

Trick: B alone takes $21$ days for full work, so $\frac{1}{3}$ of it takes $\frac{21}{3}=7$ days.

Answer: (D) $7$

Concept: Let present ages be $A$ (Aman) and $C$ (Chaman); form two equations.

1. $A-3=4(C-3)\Rightarrow A=4C-9$.
2. $A+3=2(C+3)\Rightarrow A=2C+3$.
3. $4C-9=2C+3\Rightarrow C=6$, so $A=15$.

Trick: Check options: $A=15$ gives $12=4\times 3$ and $18=2\times 9$ ✓.

Answer: (D) $15$

Concept: Expand each bracket and combine like terms.

1. $x(5x-5)=5x^2-5x$.
2. $7(x^2-2)=7x^2-14$.
3. Sum: $5x^2-5x+7x^2-14+15=12x^2-5x+1$.

Trick: The constant term decides quickly: $-14+15=+1$, and $x^2$ coefficient $5+7=12$ is positive.

Answer: (A) $12x^2-5x+1$

Concept: $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ and ${\left(\frac{a}{b}\right)}^{-n}={\left(\frac{b}{a}\right)}^n$.

1. $\sqrt{\frac{1764}{2025}}=\frac{42}{45}=\frac{14}{15}$ and $\sqrt{\frac{625}{441}}=\frac{25}{21}$.
2. Product $=\frac{14}{15}\times \frac{25}{21}=\frac{10}{9}$.
3. ${\left(\frac{3}{4}\right)}^{-2}=\frac{16}{9}$.
4. $\frac{10}{9}-\frac{16}{9}=-\frac{6}{9}=-\frac{2}{3}$.

Trick: Know squares up to $50$: ${42}^2=1764$, ${45}^2=2025$, ${21}^2=441$.

Answer: (C) $-\frac{2}{3}$

Concept: Monthly saving $=$ income $-$ expenses; one month's saving is spent, so only $11$ months' savings remain.

1. Expenses $=5500+2500=₹8000$.
2. Monthly saving $=16000-8000=₹8000$.
3. Yearly saving $=8000\times 11=₹88000$.

Trick: Subtract the birthday month: $12-1=11$ months.

Answer: (A) ₹88,000

Concept: Weighted average $=\sum (\text{weight}\times \text{value})$ when weights are in $\%$ form.

1. Women: $0.55\times 85=46.75$.
2. Men: $0.45\times 95=42.75$.
3. Average $=46.75+42.75=89.5$ years.

Trick: Start at $85$ and add $45\%$ of the gap $10$: $85+4.5=89.5$.

Answer: (B) $89.5$

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