RRB NTPC Graduate CBT 1 2024 PYQs

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

Concept: Form one equation for the total cost using the total number of items.

1. Let sarees $=s$, so suits $=8-s$.
2. $1500s+1000(8-s)=10000$.
3. $500s+8000=10000\Rightarrow s=4$; suits $=8-4=4$.

Trick: If all 8 were suits: ₹8,000. Extra ₹2,000, and each saree costs ₹500 more, so sarees $=\frac{2000}{500}=4$.

Answer: (A) 4 sarees and 4 suits

Concept: The angle subtended by an arc (chord) at the centre is double the angle it subtends at any point on the remaining part of the circle.

1. Let the angle at the circumference be $\theta$.
2. Then the angle at the centre $=2\theta$.
3. Ratio $=2\theta :\theta =2:1$.

Trick: Centre angle is always double – e.g. $\angle ACB={40}^{\circ }\Rightarrow \angle AOB={80}^{\circ }$.

Answer: (B) $2:1$

Concept: With the same base, multiply by adding powers; and ${\left(\frac{a}{b}\right)}^n={\left(\frac{b}{a}\right)}^{-n}$.

1. Write ${\left(\frac{4}{29}\right)}^2={\left(\frac{29}{4}\right)}^{-2}$.
2. LHS $={\left(\frac{29}{4}\right)}^{3-2+11}={\left(\frac{29}{4}\right)}^{12}={\left(\frac{4}{29}\right)}^{-12}$.
3. So $9m+7=-12\Rightarrow 9m=-19\Rightarrow m=-\frac{19}{9}$.

Trick: Flipping the base flips the sign of the power – convert everything to one base first.

Answer: (A) $-\frac{19}{9}$

Concept: For $a{x}^2+bx+c=0$, discriminant $D=b^2-4ac$. $D>0$: two real roots; $D=0$: equal roots; $D<0$: no real roots.

1. Here $a=1$, $b=-2$, $c=13$.
2. $D=(-2{)}^2-4\times 1\times 13=4-52=-48$.
3. $D<0$, so there are no real roots.

Trick: If $c$ is positive and $b^2<4ac$ (here $4<52$), the roots are imaginary at once.

Answer: (B) $-48$, No real roots

Concept: Weighted average $=\frac{\sum (\text{share}\times \text{average})}{\sum \text{share}}$.

1. Children $=100\%-20\%-50\%=30\%$.
2. Average $=0.2\times 45+0.5\times 60+0.3\times 20$.
3. $=9+30+6=45\text{kg}$.

Trick: Take 10 people: 2 men, 5 women, 3 children → total $90+300+60=450$, average $=45$.

Answer: (B) 45

Concept: $a:b::c:d\Rightarrow ad=bc$; and if $p:y::y:q$ then $y^2=pq$.

1. $(26+x)(34+x)=(40+x)(22+x)$.
2. $884+60x+x^2=880+62x+x^2\Rightarrow 2x=4\Rightarrow x=2$.
3. $4x=8$, $7x-6=8$; so $y^2=8\times 8=64$.
4. $y>0\Rightarrow y=8$.

Trick: Check options: $x=2$ gives $28:42=24:36$, both $=2:3$ ✓.

Answer: (B) 8

Concept: In the right triangle formed by the tower and the ground, $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. $\tan {30}^{\circ }=\frac{h}{48}$.
2. $\frac{1}{\sqrt{3}}=\frac{h}{48}\Rightarrow h=\frac{48}{\sqrt{3}}$.
3. $h=\frac{48\sqrt{3}}{3}=16\sqrt{3}\text{m}$.

Trick: For ${30}^{\circ }$: height $=\frac{\text{base}}{\sqrt{3}}$; dividing by $\sqrt{3}$ is the same as $\frac{\text{base}}{3}\times \sqrt{3}$, i.e. $16\sqrt{3}$.

Answer: (D) $16\sqrt{3}\text{m}$

Concept: Angles on a straight line add to ${180}^{\circ }$, and the angles of a triangle add to ${180}^{\circ }$.

1. $\angle PEB$ and $\angle BEF$ are a linear pair: $\angle BEF={180}^{\circ }-{56}^{\circ }={124}^{\circ }$.
2. G lies inside $\angle BEF$: $\angle GEF={124}^{\circ }-{32}^{\circ }={92}^{\circ }$.
3. In $△EGF$: $\angle EGF={180}^{\circ }-{92}^{\circ }-{47}^{\circ }={41}^{\circ }$.

Trick: Shortcut: $\angle EGF={180}^{\circ }-({180}^{\circ }-{56}^{\circ }-{32}^{\circ })-{47}^{\circ }={56}^{\circ }+{32}^{\circ }-{47}^{\circ }={41}^{\circ }$.

Answer: (D) ${41}^{\circ }$

Concept: $\text{CP}=\frac{\text{SP}}{1+\text{gain}\%}$; number bought for ₹1 $=\frac{1}{\text{CP of one}}$.

1. SP of 1 chocolate $=₹\frac{1}{5}$.
2. CP of 1 chocolate $=\frac{1}{5}÷1.4=\frac{1}{7}$ rupee.
3. So for ₹1 he bought 7 chocolates.

Trick: With gain, he buys fewer per rupee than he sells: $5\times 1.4=7$.

Answer: (D) 7