RRB NTPC Undergraduate CBT 1 2024 PYQs

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) $152$

Concept: Divide numerator and denominator by $\cos \alpha$ to get an equation in $\tan \alpha$, then use ${\sin }^2\alpha =\frac{{\tan }^2\alpha }{1+{\tan }^2\alpha }$.

1. $\frac{\tan \alpha +1}{\tan \alpha -1}=3\Rightarrow \tan \alpha +1=3\tan \alpha -3$
2. $2\tan \alpha =4\Rightarrow \tan \alpha =2$
3. ${\sin }^2\alpha =\frac{{\tan }^2\alpha }{1+{\tan }^2\alpha }=\frac{4}{1+4}=\frac{4}{5}$

Trick: By componendo–dividendo, $\frac{\sin \alpha }{\cos \alpha }=\frac{3+1}{3-1}=2$ directly.

Answer: (C) $\frac{4}{5}$

Concept: If the literacy rate of the whole population equals the literacy rate of males, the female literacy rate must be the same (weighted average).

1. Females $=329000-209000=120000$
2. Total literates $=44\%\times 329000=144760$; male literates $=44\%\times 209000=91960$
3. Female literates $=144760-91960=52800$, so rate $=\frac{52800}{120000}\times 100=44\%$

Trick: Average of two groups = one group’s value ⇒ the other group has the same value.

Answer: (C) $44\%$

Concept: Take the distance MN as 1 unit. Speed = fraction of distance per hour; when trains move towards each other, their speeds add.

1. A takes 7 h, B takes 5 h ⇒ speeds $\frac{1}{7}$ and $\frac{1}{5}$ per hour.
2. By 10 a.m. A covers $\frac{2}{7}$; remaining $\frac{5}{7}$.
3. Time to meet $=\frac{5/7}{\frac{1}{7}+\frac{1}{5}}=\frac{5}{7}\times \frac{35}{12}=\frac{25}{12}\text{h}=2\text{h}5\text{min}$
4. 10:00 a.m. + 2 h 5 min = 12:05 p.m.

Trick: Answer must lie between 10 a.m. and 3 p.m. – only 12:05 p.m. and 11:14 a.m. qualify.

Answer: (B) 12:05 p.m.

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

1. Total parts $=61+43+90=194$
2. 1 part $=\frac{2910}{194}=15$
3. C − B $=(90-43)\times 15=47\times 15=705$

Trick: Find the difference of ratio terms first, then multiply once by the value of one part.

Answer: (B) ₹705

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

1. In $△ABC$: $\angle BCA={180}^{\circ }-{20}^{\circ }-{131}^{\circ }={29}^{\circ }$
2. $BC\parallel AD$ ⇒ $\angle CAD=\angle BCA={29}^{\circ }$ (alternate angles)
3. $AC=CD$ ⇒ $\angle CDA=\angle CAD={29}^{\circ }$
4. $\angle ACD={180}^{\circ }-{29}^{\circ }-{29}^{\circ }={122}^{\circ }$

Trick: Isosceles triangle: vertex angle $={180}^{\circ }-2\times \text{base angle}$.

Answer: (C) ${122}^{\circ }$

Concept: Compound amount $A=P{(1+\frac{R}{100})}^n$.

1. $\frac{2601}{2500}={\left(1.02\right)}^n$
2. $\frac{2601}{2500}={\left(\frac{51}{50}\right)}^2$ since ${51}^2=2601$, ${50}^2=2500$
3. So $n=2$ years.

Trick: Spot perfect squares: $2601={51}^2$, $2500={50}^2$.

Answer: (A) 2 years

Concept: Follow BODMAS: division and multiplication before addition and subtraction; mind the signs.

1. $(-60)÷(-10)=6$
2. $(-3)\times 8=-24$
3. $(-9)-6+(-24)=-39$

Trick: Negative ÷ negative = positive; then just add all the negatives.

Answer: (C) $-39$

Concept: Total = average × number of items; split the total in the given ratio.

1. Total price $=3\times 16050=48150$
2. Ratio parts $=3+5+7=15$; 1 part $=\frac{48150}{15}=3210$
3. Cheapest $=3\times 3210=9630$

Trick: Average of 3 items = 5 parts (middle term of 3:5:7), so 1 part $=\frac{16050}{5}=3210$. Don’t stop at 3210 – that is one part only.

Answer: (B) 9,630

Concept: Curved surface area of a cylinder $=2\pi rh$.

1. $2\times \frac{22}{7}\times r\times 52=1800$
2. $r=\frac{1800\times 7}{44\times 52}=\frac{12600}{2288}$
3. $r\approx 5.507\approx 5.51\text{cm}$

Trick: $\frac{1800}{2\pi \times 52}\approx \frac{1800}{327}\approx 5.5$ – pick the option near 5.5.

Answer: (B) 5.51

Concept: Check each option: the pair must add to 80 and have LCM 558.

1. All pairs add to 80, so test the LCM.
2. $18=2\times 3^2$, $62=2\times 31$
3. LCM $=2\times 3^2\times 31=558$ ✓

Trick: $558=2\times 9\times 31$ – look for an option containing a multiple of 31: only 62.

Answer: (B) 18, 62