RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Time $=\frac{\text{Distance}}{\text{Speed}}$. Let Abhay's speed be $a$ and Sameer's time be $t$.

1. $\frac{156}{a}=t+5$ … (1)
2. With double speed: $\frac{156}{2a}=\frac{78}{a}=t-8$ … (2)
3. (1) − (2): $\frac{78}{a}=13\Rightarrow a=6\text{km/h}$.
4. Check: Abhay 26 h, Sameer 21 h, doubled speed 13 h $=21-8$ ✓.

Trick: Doubling speed halves the time: half of Abhay's time is saved, and that saving $=5+8=13$ h, so full time $=26$ h → $156÷26=6$.

Answer: (C) 6 km/hr

Concept: BODMAS: do division and multiplication before addition and subtraction; watch the signs.

1. $(-60)÷(-10)=6$.
2. $(-2)\times 6=-12$.
3. $(-9)-6+(-12)=-9-6-12=-27$.

Trick: Negative ÷ negative = positive; then simply add all negatives: $9+6+12=27$.

Answer: (A) −27

Concept: Write both numbers in terms of one variable and use the product.

1. Second $=x$, first $=5x$.
2. $5x^2=10125\Rightarrow x^2=2025\Rightarrow x=45$.
3. Sum $=x+5x=6x=270$.

Trick: $2025={45}^2$ is worth remembering (also a year!).

Answer: (D) 270

Concept: CI of the 2nd year = interest on (P + 1st year interest) $=P\times \frac{R}{100}\times (1+\frac{R}{100})$.

1. 2nd-year interest $=P\times 0.1\times 1.1=0.11P$.
2. $0.11P=5214\Rightarrow P=\frac{5214}{0.11}=47400$.

Trick: At 10%, 1st year interest $=10\%$ of P, 2nd year $=11\%$ of P. So $P=5214\times \frac{100}{11}$.

Answer: (D) ₹47,400

Concept: With a false weight the trader gives only 85% of the goods but charges the price of 100%, plus 3% profit on that price.

1. Cost of goods actually given $=0.85$ (per unit of true price).
2. Money received $=1.03$.
3. Profit $\%=\frac{1.03-0.85}{0.85}\times 100=\frac{18}{85}\times 100\approx 21.18\%\approx 21\%$.

Trick: Profit $\%=\frac{(100+p)-(100-w)}{100-w}\times 100=\frac{103-85}{85}\times 100$.

Answer: (C) $21\%$

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

1. Brijesh's share $=\frac{2200}{17600}=\frac{1}{8}$ of profit, so Kushal's $=\frac{7}{8}$.
2. Investment ratio K : B $=7:1$; total $=63300$, so 1 part $=\frac{63300}{8}=7912.5$.
3. Difference $=7-1=6$ parts $=6\times 7912.5=47475$.

Trick: Difference $=\frac{7-1}{8}\times 63300=\frac{3}{4}\times 63300=47475$.

Answer: (C) ₹47,475

Concept: Pipes working together: add their one-hour work.

1. $\frac{1}{13}+\frac{1}{29}+\frac{1}{10}=\frac{290+130+377}{3770}=\frac{797}{3770}$.
2. Time $=\frac{3770}{797}$ hours.
3. $797\times 4=3188$, remainder $3770-3188=582$, so time $=4\frac{582}{797}$ hours.

Trick: Estimate first: C alone takes 10 h, so together it must be well below 10 h; and $\frac{3770}{797}\approx 4.7$, so only $4\frac{582}{797}$ fits.

Answer: (C) $4\frac{582}{797}$ hours

Concept: Overall average $=\frac{\text{total runs}}{\text{total matches}}$.

1. Runs in 23 matches $=23\times 42=966$.
2. Runs in next 10 $=10\times 14=140$.
3. Average $=\frac{966+140}{33}=\frac{1106}{33}\approx 33.52$.

Trick: Deviation method: $42-\frac{10\times (42-14)}{33}=42-\frac{280}{33}=42-8.48=33.52$.

Answer: (A) 33.52

Concept: Savings = Income − Expenditure; apply each percentage change to its own base.

1. Savings $=26\%$ of 60500 $=15730$; expenditure $=60500-15730=44770$.
2. New income $=1.17\times 60500=70785$; new expenditure $=1.5\times 44770=67155$.
3. New savings $=70785-67155=3630$.
4. Change $=3630-15730=-12100$ → decrease of ₹12,100.

Trick: Change in savings = change in income − change in expenditure $=10285-22385=-12100$.

Answer: (C) decrease by ₹12,100

Concept: Profit share ∝ investment × time.

1. Sharma: $4000\times 6=24000$; Gupta: $8000\times 8=64000$; Sinha: $6000\times 8=48000$.
2. Ratio $=24:64:48=3:8:6$ (total 17).
3. Gupta's share $=\frac{8}{17}\times 34000=16000$.

Trick: Cancel the 1000s and common factors early: $(4\times 6):(8\times 8):(6\times 8)=3:8:6$.

Answer: (C) ₹16,000

Concept: Selling price after discount $=\text{MP}\times \frac{100-\text{Discount}\%}{100}$.

1. After a $28\%$ discount, SP $=72\%$ of MP.
2. $72\%$ of MP $=$ ₹$3240$
3. MP $=\frac{3240\times 100}{72}=45\times 100=$ ₹$4500$

Trick: Check: $10\%$ of 4500 = 450; $28\%$ = 1260; $4500-1260=3240$ ✓.

Answer: (D) ₹4,500

Concept: A number is divisible by 11 if (sum of digits at odd places) $-$ (sum of digits at even places) is $0$ or a multiple of 11.

1. Odd places: $N+4+6=N+10$; even places: $8+M+9=M+17$.
2. Difference: $(N+10)-(M+17)=N-M-7$.
3. $N-M-7=0\Rightarrow N-M=7$, i.e. $M-N=-7$ (e.g. $N=9,M=2$: 984269 $=11\times 89479$).
4. $N-M-7=-11$ gives $M-N=4$, not an option; $+11$ is impossible for digits.

Trick: Write the digits with alternate $+$ and $-$ signs: $N-8+4-M+6-9=N-M-7$.

Answer: (D) $M-N=-7$

Concept: Take tank capacity $=$ LCM of times; each pipe's efficiency $=\frac{\text{capacity}}{\text{time}}$.

1. LCM$(12,22,8)=264$ units.
2. A $=\frac{264}{12}=22$, B $=\frac{264}{22}=12$, C $=\frac{264}{8}=33$ units/h.
3. Together $=22+12+33=67$ units/h.
4. Time $=\frac{264}{67}=3\frac{63}{67}$ hours.

Trick: All options end in $\frac{63}{67}$; only the whole part matters: $67\times 3=201<264<268$, so it is 3.

Answer: (B) $3\frac{63}{67}$ hours

Concept: Same direction: relative speed $=$ difference; opposite direction: relative speed $=$ sum. $1\text{km/h}=\frac{5}{18}\text{m/s}$.

1. Overtaking the man: relative speed $=\frac{220}{44}=5\text{m/s}=5\times \frac{18}{5}=18\text{km/h}$.
2. Train speed $=18+10=28\text{km/h}$.
3. Opposite train: relative speed $=28+79=107\text{km/h}=\frac{107\times 5}{18}\text{m/s}$; distance $=220+315=535\text{m}$.
4. Time $=\frac{535\times 18}{107\times 5}=\frac{5\times 18}{5}=18\text{s}$.

Trick: $535=5\times 107$ – the numbers are chosen to cancel neatly.

Answer: (D) 18

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

1. $\frac{24389}{15625}={(1+\frac{R}{100})}^3$
2. $15625={25}^3$ and $24389={29}^3$, so ${\left(\frac{29}{25}\right)}^3={(1+\frac{R}{100})}^3$.
3. $1+\frac{R}{100}=\frac{29}{25}\Rightarrow \frac{R}{100}=\frac{4}{25}\Rightarrow R=16\%$.

Trick: Know cubes up to 30: ${25}^3=15625$, ${29}^3=24389$. Rate $=\frac{29-25}{25}\times 100=16\%$.

Answer: (B) $16\%$

Concept: BODMAS: solve innermost brackets first, then multiplication/division, then addition/subtraction.

1. $(3-7)=-4$
2. $6\times (-4)=-24$
3. $\{26-24\}=2$
4. $868÷2=434$

Trick: Watch the negative sign: $6\times (-4)$ is $-24$, not $+24$.

Answer: (D) 434

Concept: $\text{SI}=\frac{P\times R\times T}{100}$. Equal SI for the same time $\Rightarrow$ principal $\propto \frac{1}{\text{rate}}$.

1. Let $x$ at $3\%$ and $y$ at $12\%$: $\frac{x\times 3\times 7}{100}=\frac{y\times 12\times 7}{100}\Rightarrow 3x=12y$.
2. $x:y=4:1$ (5 parts in total).
3. 1 part $=\frac{21700}{5}=4340$, so $x=4\times 4340=17360$.

Trick: Equal interest: ratio of sums $=$ inverse ratio of rates $=12:3=4:1$.

Answer: (A) 17,360

Concept: If a number leaves the same remainder with $a$, $b$, $c$, it divides their differences. Greatest such number $=$ HCF of the differences.

1. Differences: $183-127=56$, $211-183=28$, $211-127=84$.
2. HCF$(56,28,84)=28$.
3. Check: $127=4\times 28+15$, $183=6\times 28+15$, $211=7\times 28+15$ – remainder 15 each time.

Trick: HCF of $|a-b|$ and $|b-c|$ is enough: HCF$(56,28)=28$.

Answer: (B) 28

Concept: Use $\tan A=\frac{\sin A}{\cos A}$ and ${\sin }^{2}A=1-{\cos }^{2}A=(1-\cos A)(1+\cos A)$.

1. $\frac{\sin A\tan A}{1-\cos A}=\frac{{\sin }^{2}A}{\cos A(1-\cos A)}$
2. $=\frac{(1-\cos A)(1+\cos A)}{\cos A(1-\cos A)}=\frac{1+\cos A}{\cos A}$
3. $=\sec A+1=1+\sec A$

Trick: Put $A={60}^{\circ }$: $\frac{\frac{\sqrt{3}}{2}\times \sqrt{3}}{\frac{1}{2}}=3$, and $1+\sec {60}^{\circ }=1+2=3$ ✓.

Answer: (A) $1+\sec A$

Concept: Product of consecutive numbers $n(n+1)$ lies close to $n^2$, so look near $\sqrt{182}$.

1. $\sqrt{182}\approx 13.5$, so try 13 and 14.
2. $13\times 14=182$ ✓
3. The greater number is 14.

Trick: Unit digit 2 comes from $3\times 4$ – so the numbers end in 3 and 4: 13, 14.

Answer: (A) 14