RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Mode = the observation that occurs most often.

1. Count frequencies: 32 occurs 5 times, 27 occurs 3 times, 31 occurs 2 times; others once.
2. Highest frequency is 5, for 32.

Trick: Tally only the repeated values — that is enough to spot the mode.

Answer: (B) 32

Concept: $\text{SP}=\text{CP}+\text{Profit}$; cost price stays the same.

1. Let CP $=C$, profit $=P$, so SP $=C+P$.
2. New: $2(C+P)=C+7P\Rightarrow C=5P$.
3. Profit% $=\frac{P}{5P}\times 100=20\%$.

Trick: Doubling SP adds $C+P$ to profit; this equals $6P$, so $C=5P$ → $20\%$.

Answer: (B) $20$

Concept: Write numbers as $7x$ and $11x$.

1. $\frac{7x+10}{11x-20}=\frac{2}{3}$
2. $21x+30=22x-40\Rightarrow x=70$.
3. Sum $=18x=18\times 70=1260$.

Trick: Sum must be a multiple of $7+11=18$: only 1260 ($18\times 70$) qualifies.

Answer: (C) 1260

Concept: The value of $a+b$ is given directly.

1. $a+b=33$
2. $(a+b{)}^2={33}^2=1089$.
3. The value $a-b=37$ is not needed.

Trick: Do not over-solve — read what is asked; extra data is a distractor.

Answer: (C) $1089$

Concept: Work done = rate × time; find what is left for Jitesh.

1. Rates: Jitesh $\frac{1}{5}$, Kamal $\frac{1}{18}$; together $\frac{1}{5}+\frac{1}{18}=\frac{23}{90}$ per day.
2. In 3 days: $3\times \frac{23}{90}=\frac{23}{30}$; remaining $=\frac{7}{30}$.
3. Jitesh's time $=\frac{7}{30}\times 5=\frac{7}{6}$ days.

Trick: LCM method: total work 90 units; rates 18 and 5; 3 days → 69; left 21; $\frac{21}{18}=\frac{7}{6}$.

Answer: (C) $\frac{7}{6}$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$; count days excluding the first day.

1. Days: Feb $28-3=25$, March 31, April 17 → total 73 days $=\frac{73}{365}=\frac{1}{5}$ year.
2. $\text{SI}=\frac{2000\times 6\times \frac{1}{5}}{100}=24$.

Trick: 73 days $=\frac{1}{5}$ year — a favourite exam number.

Answer: (B) ₹24

Concept: SP at loss $=\text{CP}\times (100-\text{L}\%)÷100$.

1. $\text{CP}=\frac{544}{0.64}=850$.
2. SP for 24% gain $=850\times 1.24=1054$.

Trick: Direct ratio: $544\times \frac{124}{64}=1054$.

Answer: (A) ₹1,054

Concept: B's rate = (A+B)'s rate − A's rate.

1. $\frac{1}{14}-\frac{1}{63}=\frac{9-2}{126}=\frac{7}{126}=\frac{1}{18}$ → B alone takes 18 days.
2. Half the work: $\frac{18}{2}=9$ days.

Trick: LCM 126 units: A+B = 9/day, A = 2/day, B = 7/day → 63 units (half) in 9 days.

Answer: (A) 9

Concept: Diagonals of a rhombus bisect at right angles; $\text{Area}=\frac{1}{2}{d}_1{d}_2$.

1. Side $=\frac{102}{4}=25.5$ cm. Let diagonals be $8k,15k$; half-diagonals $4k,7.5k$.
2. $(4k{)}^2+(7.5k{)}^2={25.5}^2\Rightarrow 8.5k=25.5\Rightarrow k=3$.
3. Diagonals 24 and 45; Area $=\frac{1}{2}\times 24\times 45=540$.

Trick: 8 : 15 : 17 is a Pythagorean triple, so side $=8.5k$ directly.

Answer: (A) $540$

Concept: $\text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}$.

1. Times: $\frac{80}{40}=2$ h and $\frac{45}{15}=3$ h → total 5 h.
2. Total distance $=125$ km; average $=\frac{125}{5}=25\text{km/h}$.

Trick: Never average the two speeds directly when distances differ.

Answer: (A) $25$

Concept: Real and equal roots ⇔ discriminant $b^2-4ac=0$.

1. Here $a=k$, $b=-4$, $c=1$.
2. $16-4k=0\Rightarrow k=4$.

Trick: Equal roots → perfect square: $4x^2-4x+1=(2x-1{)}^2$.

Answer: (B) 4

Concept: Equivalent ratios: multiply both terms by the same number.

1. $7\times 10=70$, $9\times 10=90$ → $70:90=7:9$.
2. $67:93$, $67:91$, $72:95$ do not reduce to $7:9$.

Trick: Cross-check: $70\times 9=90\times 7=630$.

Answer: (D) 70 : 90

Concept: Half-yearly: rate $=\frac{7.4}{2}=3.7\%$ per half-year.

1. Jan deposit earns 2 half-years: $8022\times {1.037}^2=8022\times 1.075369$.
2. July deposit earns 1 half-year: $8022\times 1.037$.
3. Interest $=8022\times (1.075369+1.037-2)=8022\times 0.112369\approx 901.42$.

Trick: Interest factor $=({1.037}^2-1)+(1.037-1)=0.075369+0.037$.

Answer: (B) ₹901.42

Concept: $\frac{1}{{\sec }^2\theta }={\cos }^2\theta$ and $1+{\cot }^2\theta ={\csc }^2\theta$.

1. $\frac{4}{{\sec }^2\theta }=4{\cos }^2\theta$; $\frac{3}{1+{\cot }^2\theta }=\frac{3}{{\csc }^2\theta }=3{\sin }^2\theta$.
2. Sum $=4{\cos }^2\theta +3{\sin }^2\theta +{\sin }^2\theta =4({\cos }^2\theta +{\sin }^2\theta )=4$.

Trick: Put $\theta =0^{\circ }$: $4+0+0=4$.

Answer: (B) 4

Concept: $p^3+m^3=(p+m{)}^3-3pm(p+m)$.

1. $(p+m{)}^3={12}^3=1728$.
2. $360=1728-3pm\times 12=1728-36pm$.
3. $36pm=1728-360=1368$.
4. $pm=\frac{1368}{36}=38$.

Trick: $pm=\frac{(p+m{)}^3-(p^3+m^3)}{3(p+m)}=\frac{1728-360}{36}$.

Answer: (C) $38$

Concept: Share of a total in a ratio = (its parts ÷ total parts) × total.

1. Total parts $=11+19+5+7=42$.
2. One part $=\frac{2289}{42}=54.5$.
3. First + third $=(11+5)\times 54.5=16\times 54.5=872$.

Trick: $\frac{16}{42}\times 2289=\frac{8}{21}\times 2289=8\times 109=872$.

Answer: (A) $872$

Concept: BODMAS – solve the innermost brackets first, then ÷ and × from left to right.

1. $28÷\frac{3}{9}=28\times 3=84$.
2. $6÷2\times \frac{18-2}{6}=3\times \frac{16}{6}=8$.
3. $84\times 8=672$.
4. $8\times 6÷3=16$, so $672÷16=42$.

Trick: $\frac{3}{9}=\frac{1}{3}$, so dividing by it means multiplying by 3.

Answer: (C) $42$

Concept: Average price = total cost ÷ total number of candles.

1. Pack A: $5\times 4=$ ₹20; Pack B: $5\times 6=$ ₹30.
2. Total = ₹50 for 10 candles.
3. Average $=\frac{50}{10}=$ ₹5.

Trick: Equal quantities ⇒ simple average of prices: $\frac{4+6}{2}=5$.

Answer: (B) ₹5

Concept: Sum = average × number of persons.

1. Amit + Arjun + Anya $=3\times 30=90$ years.
2. Amit + Anya $=2\times 35=70$ years.
3. Arjun $=90-70=20$ years.
4. After 2 years: $20+2=22$ years.

Trick: Don't stop at 20 – the question asks for age after 2 years.

Answer: (C) 22 years

Concept: A product stays the same if the decimal shifts multiply to 1 overall.

1. Option D: $73.62=\frac{7362}{100}$ (factor $÷100$).
2. $7268=726.8\times 10$ (factor $\times 10$).
3. $7.09=0.709\times 10$ (factor $\times 10$).
4. Net factor $=\frac{10\times 10}{100}=1$ ⇒ equal value.

Trick: Count total decimal places: original $0+1+3=4$; option D $2+0+2=4$ with the same digits.

Answer: (D) $73.62\times 7268\times 7.09$