RRB NTPC Graduate CBT 1 2024 PYQs

Concept: $\text{CP}=\frac{\text{MP}}{1.25}$ when MP is 25% above CP; $\text{SP}=\text{MP}\times (1-\text{discount})$.

1. $\text{CP}=\frac{7725}{1.25}=6180$
2. $\text{SP}=7725\times 0.96=7416$
3. $\text{Profit}=7416-6180=1236$; $\frac{1236}{6180}\times 100=20\%$

Trick: Take CP = 100 → MP = 125 → SP = $125\times 0.96=120$ → profit $20\%$ directly.

Answer: (C) $20\%$

Concept: Gap between the two cabs stays fixed. The man (moving towards them) closes this gap at relative speed $(74+v)$.

1. Gap $=74\times \frac{28}{60}\text{km}$
2. Man covers it in 10 min: $(74+v)\times \frac{10}{60}=74\times \frac{28}{60}$
3. $740+10v=2072\Rightarrow v=133.2\text{km/h}$

Trick: $v=\frac{74(28-10)}{10}=\frac{74\times 18}{10}=133.2$.

Answer: (B) $133.2$

Concept: Monthly saving = income − expenses; in the birthday month the saving becomes zero.

1. Monthly saving $=10000-6000-2500=1500$
2. Savings occur in 11 months only.
3. Yearly savings $=1500\times 11=16500$

Trick: One month lost → multiply by 11, not 12.

Answer: (B) ₹16,500

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

1. Net per minute $=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$
2. Time = 20 minutes.

Trick: Time $=\frac{ab}{b-a}=\frac{10\times 20}{20-10}=20$.

Answer: (D) 20

Concept: Make the common term (y) equal to combine the ratios.

1. $x:y=2:5=8:20$, $y:z=4:7=20:35$ → $x:y:z=8:20:35$
2. Total parts = 63; one part $=rac1512063=240$
3. $x=1920,y=4800,z=8400$

Trick: Multiply crosswise: $x:y:z=2\times 4:5\times 4:5\times 7$.

Answer: (D) 1,920, 4,800 and 8,400

Concept: If 40% is spent, 60% is saved.

1. $60\%$ of income $=18000$
2. Income $=\frac{18000\times 100}{60}=30000$

Trick: $1\%=300$ → $100\%=30000$.

Answer: (B) ₹30,000

Concept: Covered $=\frac{4}{3}$ × remaining → covered : remaining = 4 : 3.

1. Covered $=\frac{4}{7}\times 37=\frac{148}{7}\text{km}$
2. Time $=1\text{h}5\text{min}=\frac{13}{12}\text{h}$
3. Speed $=\frac{148}{7}\times \frac{12}{13}=\frac{1776}{91}\approx 19.5\text{km/h}$

Trick: "a is $\frac{p}{q}$ of b" → split total in $p:q$.

Answer: (B) $19.5$

Concept: $a:b::c:x$ means $\frac{a}{b}=\frac{c}{x}$, so $x=\frac{b\times c}{a}$.

1. $x=\frac{17.4\times 14}{3.5}$
2. $\frac{14}{3.5}=4$ → $x=17.4\times 4=69.6$

Trick: 14 is 4 times 3.5, so x is 4 times 17.4.

Answer: (C) $69.6$

Concept: Range $=y-x$; sum of marks $=\text{mean}\times n$.

1. Sum $=64\times 12=768$; known 10 marks add to $648$ → $x+y=120$
2. $y-x=70$
3. $x=25,y=95$ (25 below 28 and 95 above 83 ✓)

Trick: Sum and difference of x, y → add and halve.

Answer: (C) $(25,95)$

Concept: "At 70% profit" means the price is marked 70% above cost; a discount is then taken on that price.

1. Pen 1: $630\times 1.70=1071$; after 9% discount $1071\times 0.91=974.61$ → profit $344.61$
2. Pen 2: profit $=722\times 0.71=512.62$
3. Total $=344.61+512.62=857.23$

Trick: Pen 1 profit directly: $630\times (1.7\times 0.91-1)=630\times 0.547$.

Answer: (B) ₹857.23

Concept: Use $a^2-b^2=(a+b)(a-b)$.

1. $a+b=3x+4$, $a-b=x+2$
2. $(3x+4)(x+2)=3x^2+6x+4x+8=3x^2+10x+8$

Trick: Check with $x=0$: $9-1=8$ ✓.

Answer: (A) $3x^2+10x+8$

Concept: Solve the fraction first, then add/subtract.

1. $\frac{72}{6}=12$
2. $70+12+29-30=81$

Trick: Group: $70-30=40$, $40+12+29=81$.

Answer: (B) 81

Concept: Curved surface area of a cone $=\pi rl$ and $l^2=r^2+h^2$.

1. $r=\frac{100}{2}=50$; $\pi \times 50\times l=6500\pi \Rightarrow l=130$
2. $h=\sqrt{{130}^2-{50}^2}=\sqrt{16900-2500}=\sqrt{14400}=120$

Trick: 50, 120, 130 = 10 × (5, 12, 13) triplet.

Answer: (B) 120

Concept: Find litres per hour of each pipe and add.

1. A: $\frac{3700}{9}$, B: $\frac{3700}{6}$ litres/h
2. Together $=3700(\frac{1}{9}+\frac{1}{6})=3700\times \frac{5}{18}=\frac{9250}{9}$ litres/h
3. Time $=\frac{6400\times 9}{9250}=\frac{57600}{9250}=\frac{1152}{185}$ hours

Trick: Together they fill 3700 L in $\frac{18}{5}$ h → 6400 L in $\frac{18}{5}\times \frac{6400}{3700}$ h.

Answer: (A) $\frac{1152}{185}$ hours

Concept: For 2 years: $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$.

1. $47100\times {\left(\frac{9}{100}\right)}^2=47100\times 0.0081$
2. $=381.51$

Trick: 2-year difference = interest on the first year's interest.

Answer: (C) ₹381.51

Concept: New mean = old mean + $\frac{\text{total increase}}{n}$.

1. Total increase $=10\times 12\frac{1}{2}+10\times 15=125+150=275$
2. Increase in mean $=\frac{275}{25}=11$
3. New mean $=110+11=121$

Trick: Spread only the extra amount over all 25 numbers.

Answer: (B) 121

Concept: The largest container that divides both volumes exactly = HCF.

1. $156=2^2\times 3\times 13$
2. $256=2^8$
3. HCF $=2^2=4$ litres

Trick: "Maximum size that fits exactly" → HCF.

Answer: (B) 4

Concept: Follow BODMAS: innermost bracket first, then division, then addition.

1. Innermost bracket: $1-6+5=0$.
2. Square bracket: $6\times 0=0$.
3. Division: $0÷12=0$.
4. Addition: $194+0=194$.

Trick: Spot that $(1-6+5)=0$ at once – then the whole bracket term vanishes and the answer is just 194.

Answer: (B) 194

Concept: Curved surface area of a cone $=\pi rl$ and $l^2=r^2+h^2$.

1. Radius $r=\frac{144}{2}=72\text{cm}$.
2. $\pi \times 72\times l=5400\pi \Rightarrow l=\frac{5400}{72}=75\text{cm}$.
3. $h=\sqrt{{75}^2-{72}^2}=\sqrt{(75-72)(75+72)}=\sqrt{3\times 147}=\sqrt{441}=21\text{cm}$.

Trick: Use $a^2-b^2=(a-b)(a+b)$ to avoid squaring big numbers; $(72,75)=3\times (24,25)$ so $h=3\times 7=21$.

Answer: (B) 21

Concept: Mode $=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$, where $l$ is the lower limit of the modal class.

1. Mode 140 is the upper end of class 135–140, so take this as the modal class: $l=135$, $h=5$, $f_1=33$, $f_0=30$, $f_2=x$.
2. $140=135+\frac{33-30}{66-30-x}\times 5$.
3. $5=\frac{15}{36-x}\Rightarrow 36-x=3\Rightarrow x=33$.

Trick: Mode $=l+h$ happens only when $f_1-f_0=2f_1-f_0-f_2$, i.e. $f_2=f_1$, so $x=33$ directly.

Answer: (B) 33