RRB NTPC Graduate CBT 1 2024 PYQs

Concept: $\text{SP}=\text{MP}\times (1-\text{discount})$ and $\text{CP}=\frac{\text{SP}}{1+\text{gain}}$.

1. $\text{SP}=5180\times 0.69=3574.2$
2. $\text{CP}=\frac{3574.2}{1.05}=3404$

Trick: Directly: $\text{CP}=5180\times \frac{69}{105}=3404$ ; check $3404\times 105=357420$ ✓.

Answer: (C) ₹3,404

Concept: Net filling rate = filling rate − leaking rate.

1. Tap fills $\frac{1}{6}$ per hour; hole empties $\frac{1}{15}$ per hour.
2. Net $=\frac{1}{6}-\frac{1}{15}=\frac{5-2}{30}=\frac{1}{10}$ per hour.
3. Time $=10$ hours.

Trick: Shortcut: $\frac{ab}{b-a}=\frac{6\times 15}{15-6}=\frac{90}{9}=10$.

Answer: (C) 10 hours

Concept: Downstream $=u+v$, upstream $=u-v$, so still-water speed $u=\frac{\text{down}+\text{up}}{2}$.

1. $u=\frac{37+27}{2}=\frac{64}{2}=32\text{km/h}$
2. (Stream speed $=\frac{37-27}{2}=5\text{km/h}$.)

Trick: Still water = average of the two speeds; stream = half the difference.

Answer: (C) 32 km/hr

Concept: Use $(a\pm b{)}^2=a^2\pm 2ab+b^2$; the cross terms cancel.

1. $(3z-9y{)}^2=9z^2-54yz+81y^2$
2. $(9z+3y{)}^2=81z^2+54yz+9y^2$
3. Sum $=90z^2+90y^2$; subtract $81z^2$ → $9z^2+90y^2$.

Trick: $(a-b{)}^2+(c+d{)}^2$ with $ab=cd$ cancels the $yz$ terms – just add squares.

Answer: (B) $9z^2+90y^2$

Concept: Consecutive integers are $x$ and $x+1$.

1. $x(x+1)=247$
2. $x^2+x-247=0$
3. Only the form of the equation is asked, so this is the required equation.

Trick: Product = $c$ moves across as $-c$; the $+x$ comes from $x\times 1$.

Answer: (C) $x^2+x-247=0$

Concept: $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$

1. Mode 165 lies at the end of class 160–165 (highest known frequency 37): $l=160,f_1=37,f_0=25,f_2=x,h=5$.
2. $165=160+\frac{12}{74-25-x}\times 5$ → $5=\frac{60}{49-x}$
3. $49-x=12$ → $x=37$.

Trick: Mode at the upper boundary means $f_1-f_0=2f_1-f_0-f_2$, i.e. $f_2=f_1$ → $x=37$ instantly.

Answer: (A) 37

Concept: Find profit on one bag, then divide the target profit by it.

1. CP per bag $=\frac{41622}{42}=991$
2. SP per bag $=\frac{37345}{35}=1067$
3. Profit per bag $=76$; bags $=\frac{3952}{76}=52$.

Trick: Always reduce to 'per unit' first – then target ÷ unit profit.

Answer: (A) 52

Concept: Income = Expenditure + Savings.

1. Expenditure $=50\%$ of income, so savings $=50\%$ too.
2. Hence expenditure = savings = ₹7,056.

Trick: Equal percentages ⇒ equal amounts – no calculation needed.

Answer: (A) 7,056

Concept: Centroid $G=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$.

1. x-coordinate: $\frac{0+(-2)+p}{3}=3$
2. $p-2=9$ → $p=11$
3. (y check: $\frac{4+0+2}{3}=2$ ✓)

Trick: Sum of x's $=3\times 3=9$; already have $-2$, so $p=11$.

Answer: (D) 11

Concept: Let a pen cost $a$ and a pencil $b$; form two equations.

1. $15a+6b=291$
2. $9(a-9)+3(b+2)=75$ → $9a+3b=150$ → $18a+6b=300$
3. Subtract: $3a=9$ → $a=3$, $b=\frac{291-45}{6}=41$
4. $12a+3b=36+123=159$

Trick: Double the second equation to match the $6b$ term and subtract.

Answer: (D) ₹159

Concept: Net rate = filling − emptying.

1. Net $=\frac{1}{9}-\frac{1}{90}=\frac{10-1}{90}=\frac{1}{10}$ per hour.
2. One tank: 10 hours; twice the volume: $2\times 10=20$ hours.

Trick: Don't stop at 10 – the question asks for double the volume.

Answer: (D) 20

Concept: Third proportional of $a$ and $b$ is $\frac{b^2}{a}$.

1. $\frac{(12xy{)}^2}{2x^2}=\frac{144x^2{y}^2}{2x^2}$
2. $=72y^2$

Trick: $a:b=b:c$ ⇒ $c=\frac{b^2}{a}$.

Answer: (D) $72y^2$

Concept: If bisectors of $\angle R$ and $\angle S$ meet at T, then $\angle STR=\frac{\angle P+\angle Q}{2}$.

1. $\angle STR={180}^{\circ }-\frac{R+S}{2}=\frac{{360}^{\circ }-(R+S)}{2}=\frac{P+Q}{2}$
2. $P+Q=2\times {32.5}^{\circ }={65}^{\circ }$
3. $P:Q=4:9$ → 13 parts $={65}^{\circ }$ → 1 part $=5^{\circ }$
4. $Q-P=5\times 5^{\circ }={25}^{\circ }$

Trick: Angle between bisectors of two adjacent angles = half the sum of the other two angles.

Answer: (C) 25°

Concept: Dividend = Divisor × Quotient + Remainder.

1. $L-S=2134$ and $L=2S+147$ → $S+147=2134$ → $S=1987$, $L=4121$.
2. HCF(4121, 4225): $4225-4121=104$; $4121=104\times 39+65$.
3. $104\to 65\to 39\to 26\to 13$ (Euclid) → HCF $=13$.

Trick: $4225=13\times 325$ and $4121=13\times 317$ – test the options as divisors.

Answer: (D) 13

Concept: $a^2-b^2=(a+b)(a-b)$ and ${\sin }^2\theta +{\cos }^2\theta =1$.

1. ${\sin }^4\theta -{\cos }^4\theta =({\sin }^2\theta +{\cos }^2\theta )({\sin }^2\theta -{\cos }^2\theta )={\sin }^2\theta -{\cos }^2\theta$
2. ${\sin }^2\theta -{\cos }^2\theta ={\sin }^2\theta -(1-{\sin }^2\theta )=2{\sin }^2\theta -1$
3. So $2{\sin }^2\theta -1=\frac{1}{2}$.

Trick: The asked expression equals the given one – no need to find $\theta$.

Answer: (B) $\frac{1}{2}$

Concept: Rolling along the length makes the length the circumference; the width becomes the height.

1. $2\pi r=32$ → $r=\frac{16}{\pi }$, $h=14$
2. $V=\pi r^{2}h=\pi \times \frac{256}{{\pi }^2}\times 14$
3. $=\frac{3584}{\pi }{\text{cm}}^3$

Trick: Options keep $\pi$ in the denominator – don't substitute $\frac{22}{7}$.

Answer: (D) $\frac{3584}{\pi }$

Concept: Translate the words into an equation.

1. $1200\%\text{ of }10=120$
2. $0.2k=120-10=110$
3. $k=\frac{110}{0.2}=550$

Trick: $20\%$ of $k$ = $\frac{k}{5}$, so $k=5\times 110$.

Answer: (B) 550

Concept: $\text{Mean}=\frac{\text{Sum of observations}}{\text{Number of observations}}$

1. Sum $=83+17+62+87+62+96+86+92+90=675$.
2. Number of observations $=9$.
3. Mean $=\frac{675}{9}=75$.

Trick: Pair to make round numbers: $83+17=100$, $62+87+...$ – add quickly.

Answer: (B) $75$

Concept: Compare ratios as fractions; a ratio greater than $\frac{1}{2}$ beats all those below $\frac{1}{2}$.

1. $\frac{38}{77}<\frac{1}{2}$ (since $38\times 2=76<77$).
2. $\frac{34}{51}=\frac{2}{3}>\frac{1}{2}$.
3. $\frac{27}{76}$ and $\frac{31}{72}$ are both $<\frac{1}{2}$.

Trick: Benchmark with $\frac{1}{2}$: only one fraction crosses it.

Answer: (B) $34:51$

Concept: A number is divisible by 8 if its last three digits are divisible by 8.

1. Last three digits: $p08$.
2. Try largest first: $908÷8=113.5$ ✗; $808÷8=101$ ✓.
3. So maximum $p=8$.

Trick: Start from 9 and come down – the first value that works is the maximum.

Answer: (C) $8$