RRB NTPC Graduate CBT 2 2024 PYQs

Concept: Weighted average $=\frac{\sum w_i{x}_i}{\sum w_i}$.

1. $\sum w=29+87+15+26+18=175$
2. $\sum wx=493+1044+345+1170+5418=8470$
3. Weighted average $=\frac{8470}{175}=48.4$

Trick: The heavy value 301 (weight 18) pulls the average far above the simple mean of most numbers – expect an answer near 48.

Answer: (D) 48.4

Concept: Evaluate powers and roots first, then add/subtract.

1. $6^2=36$
2. $\sqrt{2^2}=2$, $5\sqrt{25}=5\times 5=25$
3. $36+2-25-11=2$

Trick: $\sqrt{a^2}=a$ for positive $a$ – no calculation needed.

Answer: (D) 2

Concept: The sum of the interior angles of a quadrilateral is ${360}^{\circ }$.

1. Total parts $=2+3+3+4=12$
2. 1 part $=\frac{{360}^{\circ }}{12}={30}^{\circ }$
3. $\angle x=3\times {30}^{\circ }={90}^{\circ }$

Trick: For a quadrilateral, angle $=\frac{\text{its share}}{\text{total share}}\times {360}^{\circ }$.

Answer: (B) ${90}^{\circ }$

Concept: Combined average $=\frac{n_1{a}_1+n_2{a}_2}{n_1+n_2}$.

1. Total marks $=30\times 70+20\times 80=2100+1600=3700$
2. Total students $=50$
3. Average $=\frac{3700}{50}=74$

Trick: Deviation method: $70+\frac{20\times 10}{50}=70+4=74$.

Answer: (B) 74

Concept: Number of bricks $=\frac{\text{volume occupied by bricks}}{\text{volume of one brick}}$.

1. Wall volume $=81\times 29\times 0.8=1879.2{\text{m}}^3$
2. Bricks occupy $60\%$: $0.6\times 1879.2=1127.52{\text{m}}^3$
3. One brick $=0.72\times 0.5\times 0.25=0.09{\text{m}}^3$
4. Bricks $=\frac{1127.52}{0.09}=12528$

Trick: Convert brick sizes to metres first; only $100-40=60\%$ of the wall is brick.

Answer: (C) 12528

Concept: Convert each ratio to a decimal (or compare with $\frac{1}{2}$) and pick the largest.

1. $\frac{13}{40}=0.325$, $\frac{19}{43}\approx 0.442$ – both less than $\frac{1}{2}$.
2. $\frac{23}{45}\approx 0.511$ and $\frac{20}{38}\approx 0.526$ – both more than $\frac{1}{2}$.
3. Cross-multiply: $20\times 45=900>23\times 38=874$, so $\frac{20}{38}>\frac{23}{45}$.

Trick: First drop ratios below $\frac{1}{2}$, then cross-multiply the remaining two.

Answer: (D) 20 : 38

Concept: Net change for $+a\%$ then $-b\%$: $a-b-\frac{ab}{100}$.

1. $a=15$, $b=8$.
2. Net $=15-8-\frac{15\times 8}{100}=7-1.2=5.8\%$.
3. Check: $100\times 1.15\times 0.92=105.8$ → increase $5.8\%$.

Trick: Take price ₹100: $115\to 115-9.2=105.8$.

Answer: (D) $5.8\%$

Concept: The gap between the two cabs stays fixed; the man (moving towards them) covers this gap at relative speed (cab + man).

1. Gap $=57\times \frac{12}{60}=11.4\text{km}$.
2. Man meets them $\frac{10}{60}$ h apart: $(57+v)\times \frac{10}{60}=11.4$.
3. $57+v=68.4$ → $v=11.4\text{km/h}$.

Trick: Shortcut: $v=\frac{57\times (12-10)}{10}=11.4$.

Answer: (D) $11.4$

Concept: Multiply each term of the ratio by its own growth factor.

1. First: $3\times 1.10=3.3$.
2. Second: $4\times 1.20=4.8$.
3. New ratio $=3.3:4.8=33:48=11:16$.

Trick: Use $3\times 110:4\times 120=330:480=11:16$.

Answer: (D) 11 : 16

Concept: For 2 years, CI $=P[{(1+\frac{r}{100})}^2-1]$; effective 2-year rate $=2r+\frac{r^2}{100}$.

1. At $8\%$: $16+0.64=16.64\%$ → CI $=25000\times 0.1664=4160$.
2. At $6\%$: $12+0.36=12.36\%$ → CI $=25000\times 0.1236=3090$.
3. Difference $=4160-3090=1070$.

Trick: Difference of rates $=16.64-12.36=4.28\%$; $4.28\%$ of $25000=1070$.

Answer: (A) ₹1,070

Concept: For an odd number of consecutive integers, the average is the middle number.

1. Middle (3rd) integer $=12$.
2. Numbers: 10, 11, 12, 13, 14.
3. Largest $=12+2=14$.

Trick: Largest $=\text{average}+\frac{n-1}{2}=12+2$.

Answer: (D) 14

Concept: Volume is conserved when a solid is melted and recast. Flat surface area = area of circular bases.

1. Cylinder: $r=7$, $h=20$ → $V=\pi \times 49\times 20=980\pi$; two bases $=2\times 49\pi =98\pi$.
2. Cones get $\frac{3}{7}\times 980\pi =420\pi$ and $\frac{4}{7}\times 980\pi =560\pi$.
3. $\frac{1}{3}\pi r^2\times 35=420\pi$ → $r^2=36$; $\frac{1}{3}\pi r^2\times 35=560\pi$ → $r^2=48$. Bases $=84\pi$.
4. Change $=\frac{98\pi -84\pi }{98\pi }\times 100=\frac{100}{7}=14\frac{2}{7}\%$ decrease.

Trick: Keep $\pi$ as it is – it cancels in the percentage.

Answer: (D) $14\frac{2}{7}\%$ decrease

Concept: Time = distance ÷ speed; total distance for a round trip $=2d$.

1. $\frac{d}{125}+\frac{d}{25}=25$.
2. $\frac{d+5d}{125}=25$ → $6d=3125$ → $d=\frac{3125}{6}$.
3. Total distance $=2d=\frac{3125}{3}=1041\frac{2}{3}\text{km}$.

Trick: Don't stop at one-way distance $520\frac{5}{6}$ – the question asks for the whole trip.

Answer: (D) $1041\frac{2}{3}$

Concept: Express all numbers in terms of the smallest one.

1. Let third $=x$, second $=2x$, first $=4x$.
2. Sum $=7x=3\times 49=147$ → $x=21$.
3. Largest (first) $=4x=84$.

Trick: Parts 4 : 2 : 1 = 7 parts = 147 → 1 part = 21.

Answer: (C) 84

Concept: Form two equations from the two price conditions and solve.

1. Let pen $=p$, notebook $=n$: $4p+5n=130$.
2. New prices $1.25p$ and $0.8n$: $2(1.25p)+4(0.8n)=82$ → $2.5p+3.2n=82$.
3. Multiply the first equation by $0.625$: $2.5p+3.125n=81.25$. Subtract: $0.075n=0.75$ → $n=10$, $p=20$.
4. $p+2n=20+20=40$.

Trick: Check: $4\times 20+5\times 10=130$ ✓, $2\times 25+4\times 8=82$ ✓.

Answer: (D) ₹40

Concept: If $x$ varies inversely as $(y^3-1)$, then $x(y^3-1)=k$ (constant).

1. At $y=4$: $y^3-1=63$ → $k=6\times 63=378$.
2. At $y=8$: $y^3-1=511$.
3. $x=\frac{378}{511}=\frac{54}{73}$ (divide by 7).

Trick: $378=7\times 54$ and $511=7\times 73$.

Answer: (D) $\frac{54}{73}$

Concept: Find the cost price of the mixture from the selling price and profit, then use alligation.

1. CP of mixture $=\frac{91}{1.30}=70$ per kg.
2. Ghee: $70-56=14$; Oil: $112-70=42$.
3. Ghee : Oil $=14:42=1:3$.

Trick: Alligation always uses the mean cost price, never the selling price.

Answer: (D) 1 : 3

Concept: Value after $n$ years $=P{(1-\frac{r}{100})}^n$.

1. $P=100000$, $r=10$, $n=2$.
2. Value $=100000\times 0.9\times 0.9=100000\times 0.81$.
3. $=81000$.

Trick: Two successive 10% falls = $19\%$ fall ($10+10-1$).

Answer: (B) ₹81,000

Concept: $39=3\times 13$, so the number must be divisible by both 3 and 13.

1. 625092: digit sum $=6+2+5+0+9+2=24$ → divisible by 3.
2. $625092÷13=48084$ exactly ($13\times 48084=625092$) → divisible by 13.
3. Hence divisible by 39: $625092=39\times 16028$. Others fail (e.g. 625649 has digit sum 32, not divisible by 3).

Trick: Use the digit-sum test for 3 to cut options quickly, then divide only the survivors by 13.

Answer: (D) 625092

Concept: If two numbers are in ratio $a:b$ (co-prime) with HCF $h$, the numbers are $ha$ and $hb$, and LCM $=h\times a\times b$.

1. Numbers $=5\times 5=25$ and $5\times 4=20$.
2. LCM $=5\times 5\times 4=100$.
3. Check: $\text{HCF}\times \text{LCM}=5\times 100=500=25\times 20$ ✓.

Trick: LCM = HCF × product of ratio terms.

Answer: (C) 100