RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Angle sum of any polygon $=(n-2)\times {180}^{\circ }$; a trapezium has $n=4$.

1. A diagonal splits the trapezium into two triangles.
2. Each triangle has angle sum ${180}^{\circ }$ → total $=2\times {180}^{\circ }={360}^{\circ }$.
3. Equivalently $(4-2)\times {180}^{\circ }={360}^{\circ }$.

Trick: Every quadrilateral – square, kite, trapezium, any shape – has interior angles summing to ${360}^{\circ }$.

Answer: (C) ${360}^{\circ }$

Concept: On a circular track, runners moving in opposite directions first meet when together they cover one full lap; relative speed = sum of speeds.

1. Relative speed $=6+4=10\text{m/s}$.
2. Time for first meeting $=\frac{400}{10}=40$ s.

Trick: Opposite directions → add speeds; same direction → subtract speeds, and divide the track length by it.

Answer: (B) 40 seconds

Concept: Under simple interest, if a sum becomes $n$ times, then $\text{SI}=(n-1)P$ and $R=\frac{(n-1)\times 100}{T}$.

1. Five times ⇒ $\text{SI}=4P$ in 11 years.
2. $R=\frac{4\times 100}{11}=\frac{400}{11}=36\frac{4}{11}\%$ per annum.

Trick: "$n$ times in $T$ years" ⇒ $R\times T=(n-1)\times 100$.

Answer: (D) $36\frac{4}{11}\%$

Concept: Eliminate one variable by subtracting equations that share the same coefficients.

1. (1) − (2): $3z=7\Rightarrow z=\frac{7}{3}$.
2. From (2): $x+2y=7-\frac{7}{3}=\frac{14}{3}$; from (3): $2x+y=2-\frac{14}{3}=-\frac{8}{3}$.
3. $2(x+2y)-(2x+y)=3y=\frac{28}{3}+\frac{8}{3}=12\Rightarrow y=4$.
4. $x=\frac{14}{3}-8=-\frac{10}{3}$.

Trick: Equations (1) and (2) differ only in $z$ – subtract them first and $z$ falls out instantly.

Answer: (D) $x=-\frac{10}{3},y=4,z=\frac{7}{3}$

Concept: Net work per hour = sum of filling rates − emptying rate; time for a part = part ÷ net rate.

1. Net rate $=\frac{1}{4}+\frac{1}{32}-\frac{1}{4}=\frac{1}{32}$ cistern/hour.
2. Full cistern takes 32 h, so one-fourth takes $\frac{32}{4}=8$ h.

Trick: The 4-hour inlet and the 4-hour outlet cancel exactly – only the 32-hour tap matters.

Answer: (B) 8

Concept: Melting and recasting keeps the volume same: $\pi r^{2}h=\frac{1}{3}\pi r^{2}H$.

1. Same radius cancels on both sides ⇒ $h=\frac{H}{3}$.
2. $H=3h=3\times 10=30$ cm.

Trick: Cone with the same base and volume as a cylinder is always 3 times as tall – no need to use $\pi =\frac{22}{7}$ at all.

Answer: (B) 30 cm

Concept: Time is directly proportional to the number of pages.

1. Time per page $=\frac{40}{2}=20$ minutes.
2. For 82 pages: $82\times 20=1640$ minutes.

Trick: $82\times 20$ = $82\times 2$ with a zero = 1640.

Answer: (D) 1640

Concept: Overall average $=\frac{\text{total amount}}{\text{total members}}$.

1. Totals: $52\times 420=21840$, $22\times 437=9614$, $36\times 101=3636$.
2. Sum $=21840+9614+3636=35090$; members $=52+22+36=110$.
3. Average $=\frac{35090}{110}=319$.

Trick: $35090÷110$: drop a zero → $3509÷11=319$.

Answer: (B) 319

Concept: Let upstream speed $=u$ km/h and downstream speed $=d$ km/h; time $=\frac{\text{distance}}{\text{speed}}$. Boat speed $=\frac{u+d}{2}$.

1. $\frac{20}{u}+\frac{44}{d}=8$ and $\frac{15}{u}+\frac{22}{d}=5$.
2. Double the second: $\frac{30}{u}+\frac{44}{d}=10$; subtract the first ⇒ $\frac{10}{u}=2\Rightarrow u=5$.
3. $\frac{44}{d}=8-4=4\Rightarrow d=11$. Boat speed $=\frac{5+11}{2}=8$ km/h.
4. $8\times \frac{5}{18}=\frac{20}{9}=2\frac{2}{9}$ m/s.

Trick: Convert metres to km first (20, 44, 15, 22); km/h → m/s multiply by $\frac{5}{18}$.

Answer: (D) $2\frac{2}{9}$ m/s

Concept: Diagonals of a rhombus bisect each other at right angles, so half-diagonals and a side form a right triangle; Area $=\frac{1}{2}{d}_1{d}_2$.

1. Half of the given diagonal $=15$ cm.
2. Other half-diagonal $=\sqrt{{17}^2-{15}^2}=\sqrt{289-225}=\sqrt{64}=8$ cm ⇒ $d_2=16$ cm.
3. Area $=\frac{1}{2}\times 30\times 16=240{\text{cm}}^2$.

Trick: 8-15-17 is a Pythagorean triplet – spot it and skip the square root.

Answer: (D) $240{\text{cm}}^2$

Concept: Compare money received (on 104 articles at 90% of MP) with cost of articles given away (120 articles).

1. Let CP of one article $=1$ and MP $=m$. Customer pays $104\times 0.9m=93.6m$.
2. Articles given $=104+16=120$, cost $=120$. Profit 17% ⇒ $93.6m=1.17\times 120=140.4$.
3. $m=\frac{140.4}{93.6}=1.5$ ⇒ MP is 50% above CP.

Trick: Use the chain $\frac{\text{MP}}{\text{CP}}=\frac{120}{104}\times \frac{1.17}{0.9}=\frac{15}{13}\times 1.3=1.5$.

Answer: (C) 50%

Concept: Average $=\frac{\text{sum}}{\text{number of terms}}$.

1. Average $=\frac{540}{12}=45$.

Trick: $12\times 45=540$ – check by multiplying back.

Answer: (D) 45

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

1. $50=3M-2\times 55\Rightarrow 3M=50+110=160$.
2. $M=\frac{160}{3}=53.33$.

Trick: Median always lies between mean and mode, one-third of the way from the mean: $55-\frac{5}{3}=53.33$.

Answer: (A) 53.33

Concept: $(a^m{)}^n=a^{mn}$; equal bases ⇒ equal exponents.

1. $(5^4{)}^{105}=5^{420}$, so $3x+4=420$.
2. $3x=416\Rightarrow x=\frac{416}{3}=138.67$.
3. Integral part $=138$.

Trick: $416÷3$: $3\times 138=414$, remainder 2 ⇒ integer part 138.

Answer: (D) 138

Concept: When CP of $m$ articles = SP of $n$ articles, $\text{profit}\%=\frac{m-n}{n}\times 100$.

1. CP of 48 books = SP of 36 books. Let CP of each book be ₹1, so SP of 36 books = ₹48.
2. SP of one book $=\frac{48}{36}=\frac{4}{3}$, profit per book $=\frac{1}{3}$.
3. Profit$\%=\frac{1/3}{1}\times 100=33.33\%$.

Trick: $\frac{48-36}{36}\times 100=\frac{12}{36}\times 100=33.33\%$ – divide by the SP count.

Answer: (C) $33.33\%$

Concept: Butter needed is directly proportional to the number of waffles.

1. Butter for 1 waffle $=\frac{680}{4}=170\text{g}$.
2. Butter for 17 waffles $=170\times 17=2890\text{g}$.

Trick: $170\times 17=170\times 10+170\times 7=1700+1190=2890$.

Answer: (C) $2890\text{g}$

Concept: Things repeating every $a,b,c$ minutes coincide every $\text{LCM}(a,b,c)$ minutes.

1. $\text{LCM}(12,15,20)=60$ minutes, so all three sprays activate together every 1 hour.
2. First together at 9:45 AM; the sixth time (including the first) is after $5\times 60=300$ minutes = 5 hours.
3. 9:45 AM + 5 h = 2:45 PM.

Trick: $n$th time together = start + $(n-1)\times$LCM; don't add 6 intervals.

Answer: (B) 2:45 PM

Concept: When a hemisphere is scooped out of a face, the flat circle ($\pi r^2$) is removed and the curved surface ($2\pi r^2$) is added – net gain $\pi r^2$ per face.

1. Cube surface $=6a^2=6\times {14}^2=1176{\text{cm}}^2$.
2. Each of the 6 faces: $-\pi r^2+2\pi r^2=+\pi r^2=\frac{22}{7}\times 49=154{\text{cm}}^2$.
3. Total $=1176+6\times 154=1176+924=2100{\text{cm}}^2$.

Trick: Remaining area $=6a^2+6\pi r^2$ (one hemisphere per face).

Answer: (B) $2100$

Concept: The perpendicular from the centre bisects the chord; radius, half-chord and distance form a right triangle: $r^2=d^2+{\left(\frac{c}{2}\right)}^2$.

1. Half chord $=\frac{18}{2}=9\text{cm}$, distance $d=12\text{cm}$.
2. $r=\sqrt{{12}^2+9^2}=\sqrt{144+81}=\sqrt{225}=15\text{cm}$.
3. Diameter $=2r=30\text{cm}$.

Trick: 9-12-15 is the 3-4-5 triple $\times 3$ – no calculation needed.

Answer: (C) $30\text{cm}$

Concept: For equal time, profit is shared in the ratio of investments.

1. Ratio $=1230:1710:1010=123:171:101$ (dividing by 10); sum $=395$.
2. Shanaya's share $=\frac{101}{395}\times 1580$.
3. $1580÷395=4$, so share $=101\times 4=₹404$.

Trick: Total profit 1580 is exactly $4\times 395$ – spot such multiples to avoid long division.

Answer: (B) ₹404