RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Equivalent ratios are obtained by multiplying both terms by the same number.

1. $6:4=6\times 10:4\times 10=60:40$.
2. Others: $66:56=33:28$, $61:41$, $62:46=31:23$ – none equals $3:2$.

Trick: Reduce each option: only $60:40=3:2=6:4$.

Answer: (A) 60 : 40

Concept: For a pure recurring decimal, $0.\overline{abc}=\frac{abc}{999}$.

1. Let $x=0.583583\dots$; then $1000x=583.583583\dots$
2. $1000x-x=583\Rightarrow x=\frac{583}{999}$.
3. $583=11\times 53$ and $999=27\times 37$ share no factor, so it is already in lowest terms.

Trick: As many 9s in the denominator as there are digits under the bar.

Answer: (A) $\frac{583}{999}$

Concept: Compounded half-yearly: rate is halved and time (number of periods) is doubled.

1. Rate per half-year $=\frac{2\%}{2}=1\%$; periods $=2$.
2. $\text{CI}=P[{(1+\frac{1}{100})}^2-1]=660000\times 0.0201$.
3. $660000\times 0.0201=13266$.

Trick: Two periods at $1\%$ → effective rate $1+1+\frac{1\times 1}{100}=2.01\%$; $2.01\%$ of ₹6,60,000 = ₹13,266.

Answer: (C) ₹13,266

Concept: Multiply by 111 using $111=100+10+1$.

1. $82\times 100=8200$, $82\times 10=820$, $82\times 1=82$.
2. $8200+820+82=9102$.

Trick: For a 2-digit $ab\times 111$: write $a,(a+b),(a+b),b$ with carry → 8, 10, 10, 2 → 9102.

Answer: (C) 9102

Concept: Make Q common in both ratios to combine them.

1. $P:Q=15:4=45:12$ and $Q:R=3:1=12:4$.
2. $P:Q:R=45:12:4$; total parts $=61$.
3. One part $=\frac{12200}{61}=200$; R's share $=4\times 200=800$.

Trick: Total 12,200 is divisible by 61 – a quick check that the ratio is right.

Answer: (B) 800

Concept: $\text{SI}=\frac{P\times R\times T}{100}$.

1. Amount $=3P$, so $\text{SI}=2P$.
2. $2P=\frac{P\times R\times 10}{100}\Rightarrow R=20\%$.

Trick: To become $n$ times in $T$ years: $R=\frac{(n-1)\times 100}{T}=\frac{2\times 100}{10}=20\%$.

Answer: (A) $20\%$

Concept: Write the condition as an equation and find $\frac{a}{b}$.

1. $\frac{20}{100}a=\frac{3}{4}b\Rightarrow \frac{a}{5}=\frac{3b}{4}$.
2. $\frac{a}{b}=\frac{3}{4}\times 5=\frac{15}{4}$.
3. Ratio $a:b=15:4$.

Trick: If $x\%$ of A = $y$ of B, then $A:B=y:\frac{x}{100}$ → $\frac{3}{4}:\frac{1}{5}=15:4$.

Answer: (A) 15 : 4

Concept: $(a+b{)}^2-(a-b{)}^2=4ab$.

1. $(a+b{)}^2={70}^2=4900$.
2. $4ab=4900-208=4692$.
3. $ab=\frac{4692}{4}=1173$.

Trick: $ab=\frac{(a+b{)}^2-(a-b{)}^2}{4}$ – one line.

Answer: (A) 1173

Concept: In an isosceles triangle the two base angles are equal, and the angles add up to ${180}^{\circ }$.

1. Let each base angle be $x$: ${44}^{\circ }+2x={180}^{\circ }$.
2. $2x={136}^{\circ }\Rightarrow x={68}^{\circ }$.

Trick: Base angle $={90}^{\circ }-\frac{\text{vertex}}{2}={90}^{\circ }-{22}^{\circ }={68}^{\circ }$.

Answer: (C) ${68}^{\circ }$ and ${68}^{\circ }$

Concept: Extra selling price corresponds to the extra gain percentage on the cost price.

1. Gain rises from $39\%$ to $64\%$ → difference $=25\%$ of CP.
2. $25\%$ of CP $=₹558$.
3. CP $=558\times 4=₹2232$.

Trick: Difference in gain % × CP = difference in SP.

Answer: (B) 2,232

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

1. $\text{SP}=54000\times \frac{175}{100}=94500$ (₹)
2. After 10% discount, SP = 90% of MP: $\text{MP}=\frac{94500\times 100}{90}=105000$ (₹)

Trick: One line: $\text{MP}=54000\times \frac{175}{90}=600\times 175=105000$.

Answer: (C) ₹$1,05,000$

Concept: Sum = average × number. Adding the two pair-sums counts B twice.

1. $A+B+C=45\times 3=135$
2. $A+B=37\times 2=74$, $B+C=47\times 2=94$
3. $B=(A+B)+(B+C)-(A+B+C)=74+94-135=33$

Trick: Middle person in both pairs: $B=74+94-135$.

Answer: (D) $33$ kg

Concept: In an isosceles triangle the two base angles are equal and the angle sum is ${180}^{\circ }$.

1. Let each base angle be $x$: $x+x+{124}^{\circ }={180}^{\circ }$
2. $2x={56}^{\circ }\Rightarrow x={28}^{\circ }$

Trick: Base angle $=\frac{{180}^{\circ }-\text{vertex}}{2}=\frac{{56}^{\circ }}{2}$; options with unequal angles are rejected at sight.

Answer: (B) ${28}^{\circ }$ and ${28}^{\circ }$

Concept: Both sat the same exam, so maximum marks are the same: $\text{Max}=\frac{\text{marks}}{\text{percent}}\times 100$.

1. $26\%$ of Max $=624\Rightarrow \text{Max}=\frac{624\times 100}{26}=2400$
2. Sudha: $\frac{720}{2400}\times 100=30\%$

Trick: Percent is proportional to marks: $26\times \frac{720}{624}=26\times \frac{15}{13}=30\%$.

Answer: (A) $30\%$

Concept: Half-yearly compounding: rate is halved and time is doubled.

1. Rate $=\frac{6}{2}=3\%$ per half-year, $n=2$
2. $\text{CI}=820000[(1.03{)}^2-1]=820000\times 0.0609$
3. $=49938$ (₹)

Trick: Effective rate for two periods of 3%: $3+3+\frac{3\times 3}{100}=6.09\%$ → $6.09\%$ of 820000.

Answer: (B) ₹$49,938$

Concept: Convert mixed fractions to improper fractions, then use BODMAS (multiply before adding).

1. $2\frac{2}{9}=\frac{20}{9}$, $2\frac{2}{3}=\frac{8}{3}$, $8\frac{3}{4}=8.75$
2. $\frac{20}{9}\times \frac{8}{3}\times 567=\frac{160}{27}\times 567=160\times 21=3360$
3. $3360+8.75=3368.75$

Trick: $567=27\times 21$ cancels the denominator 27; answer must end in .75 → only (C).

Answer: (C) $3368.75$

Concept: Work per hour: pump fills $\frac{1}{4}$; with the leak the net filling is $\frac{1}{9\frac{1}{3}}=\frac{3}{28}$.

1. Leak per hour $=\frac{1}{4}-\frac{3}{28}=\frac{7-3}{28}=\frac{4}{28}=\frac{1}{7}$
2. So the leak alone empties the full tank in $7$ hours.

Trick: Shortcut: $\text{Leak time}=\frac{a\times b}{b-a}=\frac{4\times \frac{28}{3}}{\frac{28}{3}-4}=\frac{112}{16}=7$.

Answer: (D) $7$ hours

Concept: Volume of a hollow sphere $=\frac{4}{3}\pi (R^3-r^3)$; mass = density × volume.

1. $R^3-r^3=125-8=117{\text{cm}}^3$
2. $V=\frac{4}{3}\times \frac{22}{7}\times 117=\frac{10296}{21}\approx 490.29{\text{cm}}^3$
3. Mass $=28\times \frac{10296}{21}=\frac{4}{3}\times 10296=13728\text{g}=13.728\text{kg}$

Trick: 28 and the 21 in the denominator cancel to $\frac{4}{3}$: mass $=\frac{4}{3}\times \frac{4}{3}\times 22\times 117$ g.

Answer: (A) $13.728$ kg

Concept: For $x^2-x-6=0$: $\alpha +\beta =1$, $\alpha \beta =-6$. Use ${\alpha }^4+{\beta }^4=({\alpha }^2+{\beta }^2{)}^2-2(\alpha \beta {)}^2$.

1. ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =1+12=13$
2. ${\alpha }^4+{\beta }^4={13}^2-2\times 36=169-72=97$

Trick: Roots are $3$ and $-2$: $3^4+(-2{)}^4=81+16=97$.

Answer: (D) $97$

Concept: Mode is the observation that occurs most often.

1. Frequencies: 24 → 4 times; 25, 27, 34 → 2 times each; all others once.
2. Highest frequency is for 24.

Trick: Just tally the repeats – no calculation needed.

Answer: (C) $24$