RRB NTPC Graduate CBT 1 2024 PYQs

Concept: If $X\propto \frac{1}{Y}$ and $Y\propto \frac{1}{Z}$, then $X\propto Z$, i.e. $X=kZ$.

1. $X=\frac{k_1}{Y}$ and $Y=\frac{k_2}{Z}$ give $X=\frac{k_1}{k_2}Z=kZ$.
2. $\frac{1}{2}=k\times \frac{1}{74}\Rightarrow k=37$
3. When $Z=7$: $X=37\times 7=259$

Trick: Two inverse variations make a direct variation: $\frac{X_2}{X_1}=\frac{Z_2}{Z_1}$, so $X=\frac{1}{2}\times 7\times 74=259$.

Answer: (A) $259$

Concept: $\text{SP}=\text{MP}\times (1-d_1)(1-d_2)$ and $\text{SP}=\text{CP}\times (1+\text{gain})$.

1. Required SP $=3300\times 1.33=4389$
2. Net factor after discounts $=0.75\times 0.70=0.525$
3. $\text{MP}=\frac{4389}{0.525}=8360$

Trick: Successive 25% and 30% = single discount of $25+30-\frac{25\times 30}{100}=47.5\%$.

Answer: (B) ₹8,360

Concept: Every factor of $x$ must divide the LCM, and the LCM must still be reached.

1. $44=2^2\times 11$, $86=2\times 43$, $7568=2^4\times 11\times 43$
2. $\text{LCM}(44,86)=2^2\times 11\times 43=1892$; so $x$ must supply $2^4$ and have no other primes.
3. $176=2^4\times 11$ ✔; 145, 165 contain 5, and 101 is prime – they do not divide 7568.

Trick: Quick test: the option must divide 7568 – only $7568÷176=43$ works.

Answer: (D) $176$

Concept: At a loss of $l\%$, $\text{SP}=\text{CP}\times \frac{100-l}{100}$.

1. $344=\text{CP}\times \frac{43}{100}\Rightarrow \text{CP}=800$
2. For 71% profit: $\text{SP}=800\times \frac{171}{100}=1368$

Trick: Direct ratio: $\text{new SP}=344\times \frac{171}{43}=8\times 171=1368$.

Answer: (A) $1,368$

Concept: $\cos A\cos B-\sin A\sin B=\cos (A+B)$

1. $\cos {42}^{\circ }\cos {18}^{\circ }-\sin {42}^{\circ }\sin {18}^{\circ }=\cos ({42}^{\circ }+{18}^{\circ })$
2. $=\cos {60}^{\circ }=\frac{1}{2}$

Trick: 'cos cos minus sin sin' → cos of the sum; angles here add to ${60}^{\circ }$.

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

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

1. $\text{SI}=\frac{2200\times 7\times 2}{100}$
2. $=22\times 14=308$

Trick: 1 year interest = 154; two years = 308.

Answer: (D) ₹308

Concept: Powers first, then ÷ and × from left to right, then + and −.

1. $3^2=9$; $91÷13\times 9=7\times 9=63$
2. $18\times 14÷2=252÷2=126$
3. $z=63-126+11=-52$

Trick: Do × and ÷ strictly left to right – never multiply before an earlier division.

Answer: (D) $-52$

Concept: Total = average × number of items.

1. Total of 5 students $=5\times 60=300$
2. Total of 6 students $=6\times 62=372$
3. New student $=372-300=72\text{kg}$

Trick: New member = new average + (old count × rise) $=62+5\times 2=72$.

Answer: (D) $72\text{kg}$

Concept: For an odd number $n$ of observations, median = ${\left(\frac{n+1}{2}\right)}^{\text{th}}$ value after arranging in order.

1. Ascending: 15, 19, 26, 46, 53, 72, 80, 82, 86, 87, 91
2. $n=11\Rightarrow \frac{11+1}{2}=6^{\text{th}}$ value
3. 6th value = 72

Trick: Always sort first; the unsorted middle value (91) is a trap.

Answer: (B) $72$

Concept: Sum of exterior angles of any polygon $={360}^{\circ }$; sum of interior angles $=(n-2)\times {180}^{\circ }$.

1. 200% more than 360° means $3\times {360}^{\circ }={1080}^{\circ }$.
2. $(n-2)\times 180=1080\Rightarrow n=8$
3. Each interior angle $=\frac{{1080}^{\circ }}{8}={135}^{\circ }$

Trick: Each exterior angle $=\frac{{360}^{\circ }}{8}={45}^{\circ }$, so interior $={180}^{\circ }-{45}^{\circ }={135}^{\circ }$.

Answer: (D) $135$

Concept: For 2 years, $\text{CI}=P[{(1+\frac{R}{100})}^2-1]$ and $\text{SI}=\frac{P\times R\times T}{100}$.

1. At $5\%$: $(1.05{)}^2-1=0.1025$
2. $0.1025P=102.5\Rightarrow P=1000$
3. $\text{SI}=\frac{1000\times 5\times 2}{100}=100$

Trick: 2-year CI at $R\%$ = SI $+$ interest on interest: $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$. Here $\text{CI}=\text{SI}(1+\frac{R}{200})$ → $\text{SI}=\frac{102.5}{1.025}=100$.

Answer: (C) ₹$100$

Concept: $(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2$.

1. Take $a=4x$, $b=3y$: $3a^{2}b=3(16x^2)(3y)=144x^{2}y$ and $3a{b}^2=3(4x)(9y^2)=108x{y}^2$ ✓
2. So the expression $=(4x+3y{)}^3$
3. $4(5)+3(-5)=20-15=5$, and $5^3=125$

Trick: Cube roots of the first and second terms ($4x$, $3y$) tell you the bracket instantly.

Answer: (B) $125$

Concept: Solve brackets inside-out, then ÷ and ×, then subtraction; track the signs carefully.

1. $(-31-58-19)=-108$; $\{7+(-2)(-6)\}=19$; $[9\times 19]=171$
2. $(-98)\times (-108)=10584$
3. $10584÷171=\frac{10584}{171}=\frac{1176}{19}$
4. $72-\frac{1176}{19}=\frac{1368-1176}{19}=\frac{192}{19}$

Trick: All options have denominator $19$; cancel the $9$ first: $\frac{10584}{9}=1176$.

Answer: (A) $\frac{192}{19}$

Concept: For a cubic $a{x}^3+b{x}^2+cx+d=0$, product of roots $=-\frac{d}{a}$.

1. Here $a=k$, $d=k$.
2. Product $=-\frac{k}{k}=-1$

Trick: Cubic signs: sum $=-\frac{b}{a}$, pair-sum $=+\frac{c}{a}$, product $=-\frac{d}{a}$ (signs alternate $-,+,-$).

Answer: (B) $-1$

Concept: $\text{Average}=\frac{\text{Sum of values}}{\text{Number of values}}$

1. Sum $=250+300+275+325+350=1500$
2. Average $=\frac{1500}{5}=300$

Trick: Take $300$ as base: deviations $-50,0,-25,+25,+50$ sum to $0$ → average $=300$.

Answer: (C) ₹$300$

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

1. $\text{SP}=1301\times \frac{141}{100}=1834.41$
2. $\text{MP}\times \frac{47}{100}=1834.41$
3. $\text{MP}=\frac{1834.41\times 100}{47}=3903$

Trick: $\text{MP}=\text{CP}\times \frac{141}{47}=1301\times 3=3903$, since $141=3\times 47$.

Answer: (D) ₹$3903$

Concept: Write each ratio as a fraction/decimal and compare.

1. $\frac{9}{11}\approx 0.82$, $\frac{10}{27}\approx 0.37$, $\frac{31}{38}\approx 0.82$
2. $\frac{11}{11}=1$
3. Every other ratio has antecedent smaller than consequent, so each is less than $1$.

Trick: A ratio equal to $1$ beats every ratio whose first term is smaller than its second.

Answer: (C) $11:11$

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

1. Highest frequency $37$ → modal class $50$–$60$: $l=50$, $h=10$, $f_1=37$, $f_0=34$, $f_2=32$.
2. $\text{Mode}=50+\frac{37-34}{74-34-32}\times 10=50+\frac{3}{8}\times 10$
3. $=50+3.75=53.75$

Trick: The mode always lies inside the modal class; among the options only $53.75$ lies in $50$–$60$.

Answer: (B) $53.75$

Concept: $l^2=r^2+h^2$ and $V=\frac{1}{3}\pi r^{2}h$.

1. $l=2r\Rightarrow h=\sqrt{4r^2-r^2}=\sqrt{3}r$
2. $V=\frac{1}{3}\pi r^2·\sqrt{3}r=\frac{\sqrt{3}}{3}\pi r^3=401\pi$
3. $r^3=\frac{3\times 401}{\sqrt{3}}=401\sqrt{3}$

Trick: $\frac{3}{\sqrt{3}}=\sqrt{3}$, so $r^3=401\sqrt{3}$ directly.

Answer: (A) $401\sqrt{3}$

Concept: Sum of interior angles $=(n-2)\times {180}^{\circ }$; number of diagonals $=\frac{n(n-3)}{2}$.

1. $(n-2)\times 180=1260\Rightarrow n-2=7\Rightarrow n=9$
2. Diagonals $=\frac{9\times 6}{2}=27$

Trick: $\frac{1260}{180}=7$ → add $2$ → $9$ sides (nonagon).

Answer: (B) $27$