RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Profit is shared in the ratio of investments (same time).

1. Total ratio $=3+53+24=80$ parts; 1 part $=\frac{1280}{80}=16$
2. Difference of Q and R $=53-24=29$ parts
3. $29\times 16=464$

Trick: Find the difference in parts first, multiply only once.

Answer: (B) ₹464

Concept: $\text{SP}=\text{CP}\times \frac{100+P\%}{100}$ and $\text{MP}=\text{SP}\times \frac{100}{100-D\%}$.

1. $\text{SP}=576\times \frac{115}{100}=662.4$
2. $\text{MP}=662.4\times \frac{100}{96}=690$

Trick: Directly: $\text{MP}=576\times \frac{115}{96}=6\times 115=690$ (since $576=96\times 6$).

Answer: (B) ₹690

Concept: If the whole and one part have the same rate, the other part must have that rate too.

1. Females $=373000-227000=146000$
2. Literate total $=0.69\times 373000$; literate males $=0.69\times 227000$
3. Literate females $=0.69\times (373000-227000)=0.69\times 146000$ → $69\%$

Trick: Overall 69% and males 69% ⇒ females also 69% – no calculation needed.

Answer: (B) 69%

Concept: Form two equations in pen price $x$ and pencil price $y$ and solve.

1. $18x+4y=200$ … (i)
2. $19(x-1)+3(y+3)=151\Rightarrow 19x+3y=161$ … (ii)
3. $4\times$(ii): $76x+12y=644$; $3\times$(i): $54x+12y=600$ → $22x=44$, $x=2$, $y=41$
4. $12x+10y=24+410=434$

Trick: Make the $y$-coefficients equal (12) and subtract – one step gives $x$.

Answer: (B) ₹434

Concept: If $a\%$ of A $=$ $b\%$ of B, then $A:B=b:a$.

1. $\frac{77}{100}A=\frac{3}{4}B=\frac{75}{100}B$
2. $77A=75B\Rightarrow A:B=75:77$

Trick: Write $\frac{3}{4}$ as 75% and simply swap: $75:77$.

Answer: (A) $75:77$

Concept: Solve innermost bracket first, then multiplication, then addition, then division.

1. $(2-3)=-1$
2. $7\times (-1)=-7$; $17-7=10$
3. $290÷10=29$

Trick: Order: ( ) → { } → [ ].

Answer: (D) 29

Concept: Use ${\sec }^{2}A-{\tan }^{2}A=1$ and the standard result $\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\sec A+\tan A=\frac{1+\sin A}{\cos A}$.

1. Denominator $=\tan A+\sec A-1$.
2. So $\frac{\tan A-\sec A+1}{\tan A+\sec A-1}=\frac{\cos A}{1+\sin A}$.
3. Expression $=(1+\sin A)\sec A\times \frac{\cos A}{1+\sin A}=\sec A\cos A=1$

Trick: Test $A={45}^{\circ }$: numerator $=(1+\frac{1}{\sqrt{2}})(2-\sqrt{2})\sqrt{2}=\sqrt{2}$ and denominator $=\sqrt{2}$, so the value is 1.

Answer: (B) 1

Concept: Net rate $=$ sum of filling rates − emptying rate (in tank per minute).

1. $\frac{1}{70}+\frac{1}{30}-\frac{1}{x}=\frac{1}{28}$
2. LCM 420: $\frac{6+14}{420}-\frac{1}{x}=\frac{15}{420}$
3. $\frac{1}{x}=\frac{5}{420}=\frac{1}{84}$ → $x=84$ min

Trick: Take capacity $=420$ units: fillers $6+14=20$/min, net $15$/min ⇒ leak $=5$/min ⇒ $\frac{420}{5}=84$ min.

Answer: (B) 84 min

Concept: Angle sum of a triangle is ${180}^{\circ }$; with parallel lines, alternate angles are equal; equal sides face equal angles.

1. In $△ABC$: $\angle BCA={180}^{\circ }-{40}^{\circ }-{120}^{\circ }={20}^{\circ }$
2. $BC\parallel AD$ ⇒ $\angle CAD=\angle BCA={20}^{\circ }$ (alternate angles)
3. $AC=CD$ ⇒ $\angle CDA=\angle CAD={20}^{\circ }$
4. $\angle ACD={180}^{\circ }-{20}^{\circ }-{20}^{\circ }={140}^{\circ }$

Trick: Isosceles triangle with base angle $\theta$ has apex angle ${180}^{\circ }-2\theta$.

Answer: (C) ${140}^{\circ }$

Concept: Let present ages be father $F$, son $S$; translate each statement into an equation.

1. 5 years ago: $F-5=2(S-5)+14\Rightarrow F=2S+9$
2. After $n$ years: $F+n=2(S+n)$
3. $2S+9+n=2S+2n\Rightarrow n=9$

Trick: Father is $9$ years more than twice the son now; that extra 9 disappears after exactly 9 years.

Answer: (A) 9

Concept: A prime has no divisor other than 1 and itself. To test $n$, check primes up to $\sqrt{n}$; here $\sqrt{485}\approx 22$, so test 2, 3, 5, 7, 11, 13, 17, 19.

1. Even numbers 476, 478, 480, 482, 484 and 485 (ends in 5) are out.
2. $477=3\times 159$, $483=3\times 161$ (digit sums 18 and 15).
3. $481=13\times 37$.
4. 479 is not divisible by 2, 3, 5, 7, 11, 13, 17 or 19, so it is prime.

Trick: Strike out evens and multiples of 3 and 5 first; only 479 and 481 remain to test.

Answer: (B) $1$

Concept: Trade discount is a percentage reduction on the list price of the goods bought; a scheme discount given as free units does not change it.

1. List price $=100\times 50=5000$ (₹).
2. Trade discount $=10\%\text{ of }5000=\frac{10}{100}\times 5000=500$ (₹).
3. The 5% free units are extra goods, so they are not part of this amount.

Trick: $10\%$ of any amount = move the decimal one place left: $5000\to 500$.

Answer: (A) ₹$500$

Concept: Work = (number of workers) $\times$ (days). Take total work as a convenient number of units.

1. Work $=5\times 18=90$ man-days $=9\times 5=45$ woman-days, so 1 man does 1 unit/day and 1 woman does 2 units/day (total 90 units).
2. 3 men + 3 women $=3\times 1+3\times 2=9$ units/day.
3. Time $=\frac{90}{9}=10$ days.

Trick: $1\text{W}=\frac{90}{45}=2\text{M}$, so $3\text{M}+3\text{W}=9\text{M}$; $\frac{5\times 18}{9}=10$ days.

Answer: (B) $10$ days

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

1. $65\%\text{ of }65=\frac{65\times 65}{100}=\frac{4225}{100}=42.25$
2. $\frac{4}{5}\text{ of }15=\frac{4\times 15}{5}=12$
3. Difference $=42.25-12=30.25$

Trick: ${65}^2=4225$ (ends in 5: $6\times 7=42$, then 25), so $65\%$ of $65=42.25$.

Answer: (D) $30.25$

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

1. After a $13\%$ discount, $\text{SP}=87\%$ of MP.
2. $\frac{87}{100}\times \text{MP}=348$
3. $\text{MP}=\frac{348\times 100}{87}=4\times 100=400$ (₹).

Trick: $348=87\times 4$, so MP $=4\times 100=400$.

Answer: (D) ₹$400$

Concept: If $a\%$ of $x$ equals a fraction of $y$, write the equation and find $\frac{x}{y}$.

1. $\frac{36}{100}x=\frac{4}{7}y$
2. $\frac{x}{y}=\frac{4}{7}\times \frac{100}{36}=\frac{400}{252}$
3. Divide by 4: $\frac{x}{y}=\frac{100}{63}$, i.e. $100:63$.

Trick: $\frac{x}{y}=\frac{\text{coefficient of }y}{\text{coefficient of }x}=\frac{4/7}{9/25}=\frac{100}{63}$.

Answer: (B) $100:63$

Concept: For the same time period, profit is shared in the ratio of investments.

1. Kushal's profit $=12000-5700=6300$, so ratio $K:B=6300:5700=21:19$.
2. Total parts $=40$ ↔ ₹37,100, so 1 part $=\frac{37100}{40}=927.5$.
3. Difference $=21-19=2$ parts $=2\times 927.5=1855$ (₹).

Trick: Difference in investment $=\frac{6300-5700}{12000}\times 37100=\frac{37100}{20}=1855$.

Answer: (A) ₹$1,855$

Concept: With a false weight, the trader gives less goods (cost) but charges the full price: $\text{SP}:\text{CP}=(100+\text{claimed profit}):(100-\text{weight shortfall})$.

1. Let the goods marked 100 cost ₹100. He charges ₹129 (profit $29\%$).
2. He actually gives only $100-13=87$ units, costing ₹87.
3. $\text{Profit}\%=\frac{129-87}{87}\times 100=\frac{42}{87}\times 100\approx 48.28\%\approx 48\%$

Trick: $\frac{129}{87}=1.4828$ → profit $\approx 48\%$.

Answer: (C) $48\%$

Concept: Two consecutive natural numbers are $n$ and $n+1$.

1. $n(n+1)=6\Rightarrow n^2+n-6=0$
2. $(n+3)(n-2)=0\Rightarrow n=2$ (natural number).
3. Numbers are 2 and 3; the greater is 3.

Trick: Check options: $2\times 3=6$ directly.

Answer: (C) $3$

Concept: Sum of the first $n$ odd natural numbers $=n^2$, so their average $=\frac{n^2}{n}=n$.

1. Sum of first 152 odd numbers $={152}^2$.
2. Average $=\frac{{152}^2}{152}=152$.
3. Check: average $=\frac{\text{first}+\text{last}}{2}=\frac{1+303}{2}=152$.

Yaad rakhein: Average of first $n$ odd numbers $=n$; of first $n$ even numbers $=n+1$.

Answer: (C) $152$