RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: For a fixed SP of ₹1, the number of articles sold is inversely proportional to $\text{SP}$ per article, and $\text{SP}=\text{CP}\times \frac{100\pm \text{rate}}{100}$.

1. At 40% loss, ₹1 = 60% of CP of 6 bands.
2. For 20% gain, ₹1 = 120% of CP of $n$ bands.
3. So $60\times 6=120\times n$.
4. $n=\frac{360}{120}=3$.

Trick: $n=6\times \frac{100-40}{100+20}=6\times \frac{60}{120}=3$.

Answer: (C) $3$

Concept: Increasing a number by $r\%$ multiplies it by $(1+\frac{r}{100})$.

1. Increase by $100\%$ means the number is doubled: $2x$.
2. $2x=2780$.
3. $x=\frac{2780}{2}=1390$.

Trick: 100% increase = double, so just halve 2780.

Answer: (A) $1390$

Concept: For consecutive numbers $n(n+1)=P$, $n$ is close to $\sqrt{P}$.

1. $\sqrt{210}\approx 14.5$, so try $14$ and $15$.
2. $14\times 15=210$. ✓
3. The greater number is $15$.

Trick: Product ends in 0 and $\sqrt{210}\approx 14.5$, so $14\times 15$.

Answer: (C) $15$

Concept: Power of a power: $(a^m{)}^n=a^{mn}$; if $a^p=a^q$ (with $a\ne 1$), then $p=q$.

1. $x^z=({149}^{0.59}{)}^z={149}^{0.59z}$.
2. $y^6=({149}^{0.94}{)}^6={149}^{5.64}$.
3. $0.59z=5.64\Rightarrow z=\frac{5.64}{0.59}\approx 9.56$.

Trick: Base is the same, so ignore 149 and compare exponents: $z=\frac{6\times 0.94}{0.59}$.

Answer: (C) $9.56$

Concept: Sum of first $n$ even numbers $=n(n+1)$, so their average $=n+1$.

1. $n=151$.
2. Average $=\frac{151\times 152}{151}=152$.

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

Answer: (A) $152$

Concept: With two candidates, margin $=(\text{winner}\%-\text{loser}\%)$ of valid votes.

1. Let registered voters $=R$. Votes polled $=0.8R$.
2. Valid votes $=95\%$ of $0.8R=0.76R$.
3. Margin $=(75\%-25\%)$ of valid $=0.5\times 0.76R=0.38R$.
4. $0.38R=1805\Rightarrow R=\frac{1805}{0.38}=4750$.

Trick: Answer must make $0.38R$ a whole number; only $4750$ works ($0.38\times 4750=1805$).

Answer: (A) $4750$

Concept: Work rate of a pipe $=\frac{1}{\text{time}}$; for pipes together, add the rates.

1. Combined rate $=\frac{1}{16}+\frac{1}{25}+\frac{1}{12}$.
2. LCM of 16, 25, 12 is 1200: $\frac{75+48+100}{1200}=\frac{223}{1200}$.
3. Time $=\frac{1200}{223}=5\frac{85}{223}$ hours (since $223\times 5=1115$, remainder 85).

Trick: Take tank $=1200$ units (LCM): rates 75, 48, 100 units/h; time $=\frac{1200}{223}$ h.

Answer: (A) $5\frac{85}{223}$ hours

Concept: Mean proportional of $a$ and $b$ is $\sqrt{ab}$ (since $a:x=x:b$).

1. $x=\sqrt{25\times 16}$.
2. $=\sqrt{400}=20$.

Trick: Mean proportional = geometric mean: $\sqrt{25}\times \sqrt{16}=5\times 4=20$.

Answer: (B) $20$

Concept: Trade discount is allowed on the list price: $\text{Discount}\%=\frac{\text{LP}-\text{Price paid}}{\text{LP}}\times 100$.

1. List price ₹3,600; retailer pays ₹2,800.
2. Discount $=3600-2800=800$.
3. $\text{Discount}\%=\frac{800}{3600}\times 100=\frac{200}{9}\approx 22.2\%$.

Trick: $\frac{8}{36}=\frac{2}{9}$ and $\frac{1}{9}=11.1\%$, so $\frac{2}{9}=22.2\%$.

Answer: (D) $22.2\%$

Concept: $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$; doubling the speed halves the time.

1. Let Kevin take $t$ h. Salaj (speed $s$): $\frac{182}{s}=t+4$.
2. At $2s$: $\frac{182}{2s}=t-9$.
3. Difference of the two times: $\frac{182}{s}-\frac{182}{2s}=13\Rightarrow \frac{91}{s}=13$.
4. $s=7\text{km/h}$.

Trick: Half of Salaj's time $=4+9=13$ h, so his time $=26$ h and speed $=\frac{182}{26}=7$.

Answer: (A) $7\text{km/hr}$

Concept: Profit is shared in the ratio of (capital $\times$ time).

1. Anubhav + Gaurav $=7155$ and Anubhav $-$ Gaurav $=1431$ ⇒ Anubhav $=4293$, Gaurav $=2862$.
2. Ratio $=4293:2862=3:2$.
3. Let Gaurav invest $G$: $\frac{(G+55000)\times 2}{G\times 5}=\frac{3}{2}$ ⇒ $4G+220000=15G$ ⇒ $G=20000$.
4. Anubhav $=20000+55000=75000$.

Trick: Sum and difference of shares give each share directly: $\frac{7155\pm 1431}{2}$.

Answer: (B) ₹75,000

Concept: Make the decimal places equal, find the LCM of the whole numbers, then put the decimal back.

1. $2.7=\frac{2700}{1000}$, $0.096=\frac{96}{1000}$.
2. $2700=2^2\times 3^3\times 5^2$, $96=2^5\times 3$.
3. $\text{LCM}=2^5\times 3^3\times 5^2=21600$.
4. So LCM $=\frac{21600}{1000}=21.6$.

Trick: LCM of decimals must be a multiple of the larger number: $21.6=8\times 2.7$ and $21.6=225\times 0.096$.

Answer: (C) $21.6$

Concept: Product of two numbers $=\text{HCF}\times \text{LCM}$.

1. $82\times x=2\times 492=984$.
2. $x=\frac{984}{82}=12$.
3. Check: $\text{HCF}(82,12)=2$ ✓.

Trick: Other number $=\frac{\text{HCF}\times \text{LCM}}{\text{given number}}$.

Answer: (D) $12$

Concept: Simplify the bracket first, then expand and combine like terms.

1. $\frac{4}{2}=2$, so the first part is $2(2x^2-21x+15)=4x^2-42x+30$.
2. $-4(x^2+5x-15)=-4x^2-20x+60$.
3. Add: $4x^2-4x^2-42x-20x+30+60=-62x+90$.

Trick: Here the $x^2$ terms cancel; just track the signs of $x$ and constant terms.

Answer: (C) $-62x+90$

Concept: Express both selling prices in terms of cost price (CP).

1. Let actual SP $=S$. At $\frac{2}{5}S$ there is a $24\%$ loss: $\frac{2}{5}S=0.76\text{CP}$.
2. $S=\frac{0.76\times 5}{2}\text{CP}=1.9\text{CP}$.
3. New SP $=0.88\times 1.9\text{CP}=1.672\text{CP}$.
4. Profit $=67.2\%$.

Trick: Take CP $=100$: $\frac{2}{5}S=76$ ⇒ $S=190$ ⇒ $88\%$ of $190=167.2$.

Answer: (A) $67.2\%$

Concept: For the same distance, time is inversely proportional to speed.

1. Distance $=77\times 12=924\text{km}$.
2. Meghna's time $=\frac{924}{33}=28$ hours.

Trick: $t_2=t_1\times \frac{s_1}{s_2}=12\times \frac{77}{33}=12\times \frac{7}{3}=28$.

Answer: (A) 28 hours

Concept: Profit is divided in the ratio of investment.

1. Total parts $=86+60+24=170$.
2. 1 part $=\frac{4590}{170}=27$.
3. Y $-$ Z $=(60-24)\times 27=36\times 27=972$.

Trick: Find only the difference of parts and multiply by the value of one part.

Answer: (D) ₹972

Concept: $24=3\times 8$ (co-prime), so the number must be divisible by both 3 and 8.

1. Odd numbers (A) are out; for 8 check the last three digits.
2. B: $390$, D: $978$ are not divisible by 8; C: $904=8\times 113$ ✓.
3. C digit sum $=6+8+3+0+9+0+4=30$, divisible by 3 ✓.

Trick: Test 8 first (last three digits) – it removes most options quickly.

Answer: (C) 6830904

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

1. $3400=\text{MP}\times 0.85$.
2. $\text{MP}=\frac{3400}{0.85}=4000$.

Trick: $85\%\to 3400$ ⇒ $1\%\to 40$ ⇒ $100\%\to 4000$.

Answer: (C) ₹4,000

Concept: $\sin \theta =\frac{1}{2}$ ⇒ $\theta ={30}^{\circ }$; use ${\text{cosec}}^2\theta$ and ${\cot }^2\theta$ values.

1. $\text{cosec}\theta =2$ ⇒ ${\text{cosec}}^2\theta =4$.
2. $\cot \theta =\sqrt{3}$ ⇒ ${\cot }^2\theta =3$.
3. $13\times 4-3\times 3-15=52-9-15=28$.

Trick: Write $13{\text{cosec}}^2\theta -3{\cot }^2\theta =10{\text{cosec}}^2\theta +3$ (since ${\text{cosec}}^2\theta -{\cot }^2\theta =1$): $40+3-15=28$.

Answer: (D) $28$