RRB NTPC Graduate CBT 1 2024 PYQs

Concept: If $a:b=b:c$, then $c$ is the third proportional of $a$ and $b$, so $c=\frac{b^2}{a}$.

1. First three even natural numbers: $2+4+6=12$, so $B=12$.
2. $C=\frac{{12}^2}{29}=\frac{144}{29}$.
3. $\frac{144}{29}\approx 4.9655\approx 4.97$.

Trick: Third proportional = (middle term)${}^2$ ÷ first term.

Answer: (D) $4.97$

Concept: A number divisible by both 6 and 31 (co-prime) must be divisible by $6\times 31=186$.

1. $3720÷186=20$ exactly.
2. Check others: 2132, 4916, 2858 are not divisible by 3 (digit sums 8, 20, 23), so not by 6.

Trick: First test divisibility by 6 (even + digit sum divisible by 3): only 3720 (digit sum 12) passes.

Answer: (A) $3720$

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

1. $a=2,b=4,c=8$.
2. $D=4^2-4\times 2\times 8=16-64=-48$.
3. $D<0$, so the roots are imaginary (no real roots).

Trick: If $a$ and $c$ have the same sign and $b^2<4ac$, roots are never real.

Answer: (D) no real roots

Concept: Pipe A runs for the full 42 minutes; Pipe B must fill whatever A cannot.

1. A in 42 min fills $\frac{42}{84}=\frac{1}{2}$ of the cistern.
2. Remaining $\frac{1}{2}$ is filled by B.
3. Time for B $=\frac{1}{2}\times 36=18$ min.

Trick: 42 is exactly half of 84, so B only needs half of its 36 minutes.

Answer: (A) 18 minutes

Concept: Mode = the value that occurs most often.

1. Count repeats: 74 appears 4 times; 69, 77, 80, 87, 89 appear twice each; others once.
2. Highest frequency = 4 for 74.

Trick: Tick values that repeat while scanning once; only the repeating ones matter.

Answer: (A) 74

Concept: $A=P(1+\frac{RT}{100})$ for simple interest.

1. $RT=4\times 5=20\%$, so $A=1.2P$.
2. $1.2P=720$ ⇒ $P=\frac{720}{1.2}=600$.

Trick: 20% SI ⇒ Amount : Principal = 6 : 5, so P = ₹$720\times \frac{5}{6}$ = ₹600.

Answer: (A) ₹600

Concept: Dividend = Divisor × Quotient + Remainder; HCF by Euclid's method.

1. Let smaller = $S$: $L=5S+178$ and $L-S=2858$.
2. $4S+178=2858$ ⇒ $S=670$, $L=3528$.
3. $\text{HCF}(3528,3828)$: $3828-3528=300$; $3528=11\times 300+228$; $300-228=72$; $228=3\times 72+12$; $72=6\times 12$.
4. HCF = 12.

Trick: HCF of two numbers also divides their difference (300) – check options: 12 divides 300, 3528 and 3828.

Answer: (C) 12

Concept: $\text{SP}=\text{MP}\times \frac{100-d}{100}$ and $\text{CP}=\text{SP}\times \frac{100}{100+g}$.

1. SP = ₹$5960\times \frac{91}{100}$ = ₹5423.60.
2. CP = ₹$5423.6\times \frac{100}{130}$ = ₹4172.

Trick: One line: CP = ₹$\frac{5960\times 91}{130}=\frac{5960\times 7}{10}$ = ₹4172 (since $\frac{91}{130}=\frac{7}{10}$).

Answer: (C) ₹4,172

Concept: In a right triangle, $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. Base $=15\text{m}$, $\theta ={45}^{\circ }$.
2. $\tan {45}^{\circ }=\frac{h}{15}=1$.
3. $h=15\text{m}$.

Trick: At ${45}^{\circ }$ elevation, height = distance from the base.

Answer: (C) $15\text{m}$

Concept: If $x(1+r{)}^{12}=y(1+r{)}^{13}$, then $x=y(1+r)$.

1. Arjun gets $x$, Kartik gets $y$: $x(1.1{)}^{12}=y(1.1{)}^{13}$ ⇒ $x=1.1y$.
2. $x+y=1218$ ⇒ $2.1y=1218$.
3. $y=\frac{1218}{2.1}=580$.

Trick: Shares are in ratio $11:10$ (one extra year at 10%); Kartik = ₹$\frac{10}{21}\times 1218$ = ₹580.

Answer: (B) ₹580

Concept: Empirical formula: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $61.6=3M-2\times 47.4=3M-94.8$.
2. $3M=156.4$.
3. $M=\frac{156.4}{3}=52.13\approx 52.1$.

Trick: $\text{Median}=\frac{\text{Mode}+2\text{Mean}}{3}=\frac{61.6+94.8}{3}$.

Answer: (B) $52.1$

Concept: Angle sum of a quadrilateral is ${360}^{\circ }$. If two adjacent angles add to ${180}^{\circ }$, the sides opposite their common side are parallel.

1. $2+8+3+7=20$ parts ⇒ 1 part $={18}^{\circ }$: $\angle P={36}^{\circ }$, $\angle Q={144}^{\circ }$, $\angle R={54}^{\circ }$, $\angle S={126}^{\circ }$.
2. $\angle P+\angle Q={180}^{\circ }$ ⇒ $PS\parallel QR$.
3. $\angle Q+\angle R={198}^{\circ }\ne {180}^{\circ }$ ⇒ $PQ$ not parallel to $SR$.
4. Exactly one pair of parallel sides ⇒ trapezium.

Trick: Ratio terms $2+8=3+7=10$ (half of 20) ⇒ co-interior pair ⇒ trapezium; all angles unequal rules out rhombus/square.

Answer: (A) Trapezium

Concept: After paying 40% rent, the remaining 60% of salary covers the other expenses and savings.

1. Other expenses + savings = ₹$442+9115+946$ = ₹10,503.
2. $60\%$ of salary = ₹10,503.
3. Salary = ₹$\frac{10503\times 100}{60}$ = ₹17,505.

Trick: Salary must be a multiple of 5 such that $0.6\times$ salary = 10503 – only 17,505 works ($17505\times 0.6=10503$).

Answer: (C) ₹17,505

Concept: $V=\frac{1}{3}\pi r^{2}h$, $l=\sqrt{r^2+h^2}$, $\text{TSA}=\pi r(r+l)$.

1. $\frac{1}{3}\times \frac{22}{7}\times r^2\times 21=1078$ ⇒ $22r^2=1078$ ⇒ $r^2=49$, $r=7$.
2. $l=\sqrt{49+441}=\sqrt{490}=7\sqrt{10}$.
3. $\text{TSA}=\frac{22}{7}\times 7\times (7+7\sqrt{10})=22\times 7(1+\sqrt{10})=154(1+\sqrt{10}){\text{cm}}^2$.

Trick: With $h=21$, $\frac{1}{3}\pi h=22$, so $r^2=\frac{1078}{22}=49$ instantly.

Answer: (C) $154(1+\sqrt{10}){\text{cm}}^2$

Concept: Take total work = LCM of the days and find each person's daily work.

1. Total work = LCM(2, 52) = 52 units; A = 26 units/day, B = 1 unit/day.
2. B in 26 days: $26\times 1=26$ units; remaining $52-26=26$ units.
3. A's time $=\frac{26}{26}=1$ day.

Trick: B did half the work (26 of 52 days), so A needs half of 2 days = 1 day.

Answer: (C) 1

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

1. $R^3-r^3=729-512=217$.
2. Volume $=\frac{4}{3}\times \frac{22}{7}\times 217=\frac{2728}{3}{\text{cm}}^3$.
3. Mass $=21\times \frac{2728}{3}=19096\text{g}=19.096\text{kg}$.

Trick: $217=7\times 31$ cancels the 7 of $\pi$ and $21$ cancels the 3 → $4\times 22\times 31\times 7=19096$.

Answer: (B) 19.096

Concept: Two consecutive integers are $x$ and $x+1$.

1. Product: $x(x+1)=960$.
2. $x^2+x=960\Rightarrow x^2+x-960=0$.
3. Only the form of the equation is asked, so no need to solve it.

Trick: Consecutive → $x(x+1)$; move the constant to the left with a minus sign.

Answer: (B) $x^2+x-960=0$

Concept: Secant–power theorem: for external point P, $PA\times PB=P{O}^2-r^2$.

1. $PB=PA+AB=8+7=15\text{cm}$.
2. $PA\times PB=8\times 15=120$.
3. ${17}^2-r^2=120\Rightarrow r^2=289-120=169\Rightarrow r=13\text{cm}$.

Trick: $PA·PB$ = (tangent length)${}^2$ = $d^2-r^2$.

Answer: (A) 13 cm

Concept: Follow BODMAS: brackets (with 'of') first, then division.

1. $\frac{9}{9}\text{ of }\frac{9}{8}=1\times \frac{9}{8}=\frac{9}{8}$.
2. $4\times \frac{9}{8}=\frac{9}{2}$.
3. $34÷\frac{9}{2}=34\times \frac{2}{9}=\frac{68}{9}$.

Trick: $\frac{9}{9}=1$ – spot such simplifications at a glance.

Answer: (A) $\frac{68}{9}$

Concept: Sum of numbers = average $\times$ count.

1. Sum $=23\times 3=69$.
2. Third number $=69-(11+25)=69-36=33$.

Trick: Deviation method: $11$ is $-12$, $25$ is $+2$ from 23 → third $=23+10=33$.

Answer: (D) 33