RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Mean proportional of $a$ and $b$ is $x$ where $a:x=x:b$, so $x=\sqrt{ab}$.

1. $x=\sqrt{16\times 256}$
2. $=\sqrt{4096}=64$
3. Check: $16:64=1:4$ and $64:256=1:4$ ✓

Trick: $\sqrt{16}\times \sqrt{256}=4\times 16=64$ – no need to multiply first.

Answer: (A) $64$

Concept: Mode is the observation that occurs the maximum number of times.

1. Arrange: $10,11,12,12,12,14,14,15,16,19$
2. Frequencies: $12\to 3$ times, $14\to 2$ times, all others once.
3. Highest frequency is for $12$.

Trick: Just tally – no formula needed for ungrouped data.

Answer: (D) $12$

Concept: Total surface area of a cuboid $=2(lb+bh+hl)$.

1. $l=18,b=12,h=9$
2. $lb=216,bh=108,hl=162$
3. Sum $=486$; TSA $=2\times 486=972{\text{m}}^2$

Trick: Unit digit: $2(6+8+2)=2\times 16\to$ ends in $2$; only $972$ fits.

Answer: (D) $972{\text{m}}^2$

Concept: Find each value and subtract: $x\%$ of $N=\frac{x}{100}\times N$.

1. $65\%$ of $65=\frac{65\times 65}{100}=\frac{4225}{100}=42.25$
2. $\frac{3}{5}$ of $25=15$
3. Difference $=42.25-15=27.25$

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

Answer: (C) $27.25$

Concept: Diagonals of a rectangle are equal and bisect each other, so $OP=OQ=OR=OS$.

1. $\angle POS=\angle QOR={36}^{\circ }$ (vertically opposite angles).
2. In $△OPS$, $OP=OS$, so $\angle OPS=\angle OSP$.
3. $\angle OPS=\frac{{180}^{\circ }-{36}^{\circ }}{2}={72}^{\circ }$

Trick: Base angle of the isosceles triangle $={90}^{\circ }-\frac{{36}^{\circ }}{2}={72}^{\circ }$.

Answer: (B) ${72}^{\circ }$

Concept: Work back from the margin: margin $=$ (winner $\%$ − loser $\%$) of valid votes.

1. Let registered voters $=N$. Votes polled $=0.9N$; valid $=82\%$ of $0.9N=0.738N$.
2. Winner $51\%$, loser $49\%$ → margin $=2\%$ of valid $=0.01476N$.
3. $0.01476N=1845\Rightarrow N=\frac{1845}{0.01476}=125000$

Trick: One line: $N\times 0.9\times 0.82\times 0.02=1845$.

Answer: (D) $125000$

Concept: BODMAS: do division and multiplication before addition and subtraction; take care of signs.

1. $(-60)÷(-6)=10$
2. $(-4)\times 6=-24$
3. $(-9)-10+(-24)=-9-10-24=-43$

Trick: Negative ÷ negative = positive; then the minus before it flips it back.

Answer: (A) $-43$

Concept: Divisibility by 13 (also 7, 11): split the number into 3-digit groups from the right; the alternating sum must be divisible by 13.

1. Not by $16$/$20$: number is odd. Groups: $716|710|501$
2. $501-710+716=507=13\times 39$ ✓ → divisible by $13$.
3. Indeed $716710501=13\times 55131577$; for $19$ the remainder is non-zero.

Trick: Odd number → kill 16 and 20 at once; test 13 by the 3-digit alternating rule.

Answer: (D) $13$

Concept: Simplify the inner complex fraction first, then take LCM.

1. $1+\frac{3}{8}=\frac{11}{8}$, so $\frac{1}{\frac{11}{8}}=\frac{8}{11}$
2. Expression $=\frac{5}{2}+\frac{8}{11}-\frac{3}{5}$; LCM $=110$
3. $=\frac{275+80-66}{110}=\frac{289}{110}$

Trick: LCM of $2,11,5$ is $110$ – only option C has denominator $110$.

Answer: (C) $\frac{289}{110}$

Concept: The broken part, the standing part and the ground form a right triangle; standing part $h$ is opposite the ${30}^{\circ }$ angle and the broken part $18-h$ is the hypotenuse.

1. $\sin {30}^{\circ }=\frac{h}{18-h}=\frac{1}{2}$
2. $2h=18-h\Rightarrow 3h=18$
3. $h=6\text{ft}$ (broken part $=12\text{ft}$)

Trick: At ${30}^{\circ }$, standing part : broken part $=1:2$, so standing $=\frac{18}{3}=6$.

Answer: (B) $6\text{ft}$

Concept: Convert everybody into one-day work using total work = workers × days.

1. Work of 1 man in 1 day $=\frac{1}{5\times 16}=\frac{1}{80}$; work of 1 woman in 1 day $=\frac{1}{3\times 15}=\frac{1}{45}$.
2. 9 women per day $=\frac{9}{45}=\frac{1}{5}$; 4 men per day $=\frac{4}{80}=\frac{1}{20}$.
3. Together $=\frac{1}{5}+\frac{1}{20}=\frac{5}{20}=\frac{1}{4}$ of the work per day.
4. Time $=4$ days.

Trick: 9 women = 3 × (3 women) finish in $15÷3=5$ days; 4 men finish in $\frac{80}{4}=20$ days; together $\frac{5\times 20}{25}=4$ days.

Answer: (B) 4 days

Concept: Profit = money received from the unbroken mirrors − total cost.

1. Mirrors left $=92-35=57$.
2. Selling price $=57\times 4.40=$ ₹250.80.
3. Profit $=250.80-92=$ ₹158.80.

Trick: Broken items still count in the cost – only the remaining items are sold.

Answer: (A) ₹158.80

Concept: Two successive discounts $a\%$ and $b\%$ equal a single discount of $(a+b-\frac{ab}{100})\%$.

1. $a=8,b=5$.
2. Single discount $=8+5-\frac{8\times 5}{100}=13-0.4=12.6\%$.

Trick: Check with ₹100: after 8% → 92, after 5% → $92\times 0.95=87.4$, so total discount $=12.6\%$.

Answer: (B) $12.6\%$

Concept: If SP of $x$ articles = CP of $y$ articles, then $\text{Profit/Loss}\%=\frac{y-x}{x}\times 100$.

1. SP of 22 books = CP of 11 books.
2. $\%=\frac{11-22}{22}\times 100=-50\%$.
3. Negative sign → loss of $50\%$.

Trick: SP of more books = CP of fewer books → always a loss. Here SP of one book is half its CP.

Answer: (D) 50% loss

Concept: Pipes working together: add the parts filled per hour.

1. In 1 hour: $\frac{1}{12}+\frac{1}{30}+\frac{1}{18}$.
2. LCM $=180$: $\frac{15+6+10}{180}=\frac{31}{180}$.
3. Time $=\frac{180}{31}=5\frac{25}{31}$ hours.

Trick: Take tank $=180$ units (LCM): A, B, C fill 15, 6, 10 units/h → $\frac{180}{31}$ h.

Answer: (D) $5\frac{25}{31}$ hours

Concept: Write present ages as $F$ and $S$ and form equations.

1. 7 years ago: $F-7=2(S-7)+20\Rightarrow F=2S+13$.
2. After $x$ years: $F+x=2(S+x)\Rightarrow x=F-2S$.
3. $x=13$ years.

Trick: In “father = 2 × son” problems, $x=F-2S$; just read it from the equation.

Answer: (A) 13

Concept: Cost price = marked price after successive discounts + extra expenses; $\text{Profit}\%=\frac{\text{Profit}}{\text{CP}}\times 100$.

1. Purchase price $=8000\times 0.80\times 0.75=$ ₹4,800.
2. Total CP $=4800+300=$ ₹5,100.
3. Profit $=8000-5100=$ ₹2,900.
4. $\text{Profit}\%=\frac{2900}{5100}\times 100=\frac{2900}{51}=56\frac{44}{51}\%$.

Trick: $20\%$ and $25\%$ successive = $40\%$ single discount, so purchase price $=0.6\times 8000$.

Answer: (C) $56\frac{44}{51}\%$

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

1. Rate is per month and time is in months: $R=4,T=8$.
2. $\text{SI}=\frac{1200\times 4\times 8}{100}=$ ₹384.
3. (Had the rate been $4\%$ per annum, the interest for 8 months would be $\frac{1200\times 4\times 8}{1200}=$ ₹32.)

Trick: Monthly rate × number of months gives the total percentage: $4\times 8=32\%$ of ₹1,200 $=$ ₹384.

Answer: (D) ₹384

Concept: Profit is shared in the ratio of investments.

1. Total parts $=16+85+54=155$.
2. 1 part $=\frac{3750}{155}=\frac{750}{31}$.
3. Difference Y − Z $=85-54=31$ parts $=31\times \frac{750}{31}=$ ₹750.

Trick: Difference of 31 parts out of 155 = $\frac{1}{5}$ of the profit $=\frac{3750}{5}$.

Answer: (B) ₹750

Concept: Eliminate options with divisibility rules, then divide.

1. 769501 is odd → not divisible by 96 or 90; it does not end in 0 or 5 → not divisible by 105.
2. Check 97: $97\times 7933=769501$.
3. So 97 divides it exactly.

Trick: An odd number can never be divided by an even number (96, 90).

Answer: (D) 97