RRB NTPC Graduate CBT 1 2024 PYQs

Concept: For $a{x}^2+bx+c=0$, real and equal roots ⇔ $D=b^2-4ac=0$.

1. Here $a=2$, $b=-8$, $c=K$.
2. $D=64-8K=0$ ⇒ $K=8$.
3. $K^2-5=64-5=59$.

Trick: Equal roots: $c=\frac{b^2}{4a}=\frac{64}{8}=8$ directly.

Answer: (B) 59

Concept: Mode is the value that occurs the greatest number of times.

1. Count: 41 occurs 3 times.
2. 47, 48, 53, 54 occur twice each; all others once.
3. So mode = 41.

Trick: Sort the data quickly or tally – the highest tally wins.

Answer: (A) 41

Concept: Brackets first, then division/multiplication, then addition.

1. $(5-9)=-4$ and $\frac{123}{41}=3$.
2. $10\times (-4)=-40$.
3. $-40+3=-37$.

Trick: $41\times 3=123$ – check easy divisions first.

Answer: (D) −37

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

1. SP $=3930\times 0.6=₹2358$.
2. $1.2\text{CP}=2358$.
3. CP $=\frac{2358}{1.2}=₹1965$.

Trick: CP $=\text{MP}\times \frac{60}{120}=\frac{3930}{2}=1965$.

Answer: (B) ₹1,965

Concept: Net filling rate = filling rate − emptying rate.

1. Net per minute $=\frac{1}{18}-\frac{1}{19}=\frac{19-18}{342}=\frac{1}{342}$.
2. Full tank: 342 minutes.
3. One-third: $\frac{342}{3}=114$ minutes.

Trick: For $\frac{1}{a}-\frac{1}{b}$ with $b-a=1$, time $=a\times b$.

Answer: (A) 114

Concept: Required number = LCM of the divisors + the amount diminished.

1. $\text{LCM}(13,20,10,23)$: 10 divides 20, so $=13\times 20\times 23=5980$.
2. Number $=5980+9=5989$.

Trick: Subtract 9 from each option and test divisibility by 13 and 23 – only 5989 works.

Answer: (A) 5989

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

1. Times: $\frac{445}{89}=5$ h, $\frac{495}{55}=9$ h, $\frac{250}{50}=5$ h → total $19$ h.
2. Distance $=445+495+250=1190\text{km}$.
3. Average $=\frac{1190}{19}=62.63\text{km/h}$.

Trick: Never average the speeds directly; each leg's time here is a whole number.

Answer: (A) 62.63

Concept: CSA $=2\pi rh$, TSA $=2\pi r(r+h)$.

1. $2\pi rh=\frac{1}{2}\times 2\pi r(r+h)$.
2. $2h=r+h$ ⇒ $r=h$.
3. $r=102\text{cm}$.

Trick: CSA = half TSA ⇔ the two circular ends equal the CSA ⇔ r = h.

Answer: (D) 102

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

1. Average $=\frac{300}{6}=50$.

Trick: Divide the total by the count.

Answer: (D) 50

Concept: An increase from one year to another is found from the two values: increase $=$ later value $-$ earlier value, and percentage increase $=\frac{\text{increase}}{\text{earlier value}}\times 100$.

1. The question asks for the change in urban infrastructure investment from FY 2019 to FY 2023.
2. The paper gives no table, chart or figure for these investments, so neither the FY 2019 value nor the FY 2023 value is known.
3. Without the data, 5%, 10%, 8% or 3% can only be a guess; the question cannot be solved as printed.

Note: In a data-interpretation question, first locate the data set; if it is missing, the question has no definite answer. In the exam, every candidate gets full marks (bonus) for such a question.

Answer: Bonus – no option is correct as printed; full marks to all.

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

1. Ascending: 41, 44, 45, 55, 58, 67, 69, 71, 78, 88, 95.
2. $n=11$, so median = $\frac{11+1}{2}=6^{\text{th}}$ value.
3. $6^{\text{th}}$ value = 67.

Trick: Strike one number from each end alternately; the one left in the middle is the median.

Answer: (C) 67

Concept: Mean $=\frac{\sum fx}{\sum f}$.

1. $\sum fx=19\times 22+31\times 35+54\times 64+69\times 34+95\times 57$
2. $=418+1085+3456+2346+5415=12720$
3. $\sum f=22+35+64+34+57=212$
4. Mean $=\frac{12720}{212}=60$

Trick: Check unit digits: $12720$ ends in 0 and $212\times 60=12720$ – exact.

Answer: (A) 60

Concept: For 2 years, $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$.

1. $407=P{\left(\frac{10}{100}\right)}^2=\frac{P}{100}$
2. $P=407\times 100=40700$

Trick: At 10% for 2 years, difference = 1% of the principal.

Answer: (A) ₹40,700

Concept: Production $\propto$ (manpower) $\times$ (hours). To keep production constant, hours must rise in inverse ratio.

1. Manpower falls 10%: factor $\frac{9}{10}$.
2. Hours must become $\frac{10}{9}$ times.
3. Increase $=\frac{10-9}{9}\times 100=\frac{100}{9}\%=11\frac{1}{9}\%$.

Trick: A fall of $\frac{1}{10}$ needs a rise of $\frac{1}{9}$ (denominator minus numerator).

Answer: (C) $11\frac{1}{9}\%$

Concept: Base area $=\pi r^2$; curved surface area $=\pi rl$.

1. $\pi r^2=3136\pi \Rightarrow r^2=3136\Rightarrow r=56\text{cm}$
2. $\text{CSA}=\pi \times 56\times 65$
3. $=3640\pi {\text{cm}}^2$

Trick: $56\times 65=56\times 60+56\times 5=3360+280=3640$.

Answer: (A) $3640\pi$

Concept: When a transversal cuts two parallel lines, angles in matching positions at the two intersections are called corresponding angles.

1. In the figure, ∠1 and ∠2 occupy the same relative position (upper-right) at each intersection.
2. Corresponding angles axiom: if the lines are parallel, each pair of corresponding angles is equal.
3. Co-interior angles (not corresponding) are supplementary.

Trick: Corresponding & alternate angles → equal; co-interior → sum ${180}^{\circ }$.

Answer: (C) equal

Concept: Successive increases multiply: $P\times (1+\frac{r_1}{100})(1+\frac{r_2}{100})$.

1. $100000\times \frac{105}{100}=105000$
2. $105000\times \frac{120}{100}=126000$

Trick: Net % $=5+20+\frac{5\times 20}{100}=26\%$ → 1,26,000.

Answer: (A) 1,26,000

Concept: Let brother = $B$, sister = $S$ and form two equations.

1. $3B=5S+25$ and $5S=B-5$.
2. Substitute: $3B=(B-5)+25\Rightarrow 2B=20\Rightarrow B=10$.
3. $5S=5\Rightarrow S=1$; difference $=10-1=9$.

Trick: Substitute the whole term $5S$ directly – no need to find $S$ first.

Answer: (D) 9

Concept: Sum = average $\times$ count.

1. Sum of 9 numbers $=9\times 17=153$.
2. Sum of 7 numbers $=7\times 17=119$.
3. Remaining two: $153-119=34$, average $=\frac{34}{2}=17$.

Trick: If the average of a part equals the overall average, the rest also has the same average.

Answer: (A) 17

Concept: Area of a regular hexagon of side $s$ $=\frac{3\sqrt{3}}{2}{s}^2$.

1. Square side $=\sqrt{2304}=48$, perimeter $=4\times 48=192\text{cm}$.
2. Hexagon side $s=\frac{192}{6}=32\text{cm}$.
3. Area $=\frac{3\sqrt{3}}{2}\times {32}^2=\frac{3\sqrt{3}}{2}\times 1024=1536\sqrt{3}{\text{cm}}^2$.

Trick: Hexagon = 6 equilateral triangles: $6\times \frac{\sqrt{3}}{4}{s}^2$.

Answer: (A) $1536\sqrt{3}$