RRB NTPC Graduate CBT 1 2025 PYQs

Concept: $\text{Overall average}=\frac{\sum (\text{members}\times \text{average})}{\text{total members}}$.

1. Totals: $42\times 335=14070$, $31\times 103=3193$, $32\times 271=8672$.
2. Sum $=14070+3193+8672=25935$; members $=42+31+32=105$.
3. Average $=\frac{25935}{105}=247$.

Trick: $105\times 247=105\times 250-105\times 3=26250-315=25935$ – verify by multiplying back.

Answer: (B) 247

Concept: For similar triangles, $\frac{{\text{Area}}_1}{{\text{Area}}_2}={\left(\frac{{\text{side}}_1}{{\text{side}}_2}\right)}^2$, and corresponding (longest) sides pair up.

1. ${\left(\frac{AC}{PR}\right)}^2=\frac{432}{588}=\frac{36}{49}$.
2. $\frac{AC}{PR}=\frac{6}{7}\Rightarrow AC=56\times \frac{6}{7}=48\text{cm}$.

Trick: Divide both areas by 12 to see the perfect squares 36 and 49 at once.

Answer: (A) 48 cm

Concept: Sum of interior angles of an $n$-sided polygon $=(n-2)\times {180}^{\circ }$.

1. $n=6$: $(6-2)\times {180}^{\circ }=4\times {180}^{\circ }={720}^{\circ }$.
2. Each angle of a regular hexagon $=\frac{{720}^{\circ }}{6}={120}^{\circ }$.

Yaad rakhein: Triangle 180°, quadrilateral 360°, pentagon 540°, hexagon 720° – add 180° per extra side.

Answer: (B) ${720}^{\circ }$

Concept: Take CP = 100; "150% above CP" means MP = 250; required SP = 120.

1. Discount $=250-120=130$.
2. Discount% $=\frac{130}{250}\times 100=52\%$.

Trick: "150% above" = 2.5 times (not 1.5 times) – read "above" carefully.

Answer: (B) 52%

Concept: Curved surface area of a cylinder $=2\pi rh$.

1. $2\times \frac{22}{7}\times r\times 60=1300$.
2. $r=\frac{1300\times 7}{2\times 22\times 60}=\frac{9100}{2640}\approx 3.447$.
3. Correct to two decimals: $r\approx 3.45\text{cm}$.

Trick: $\frac{9100}{2640}=\frac{455}{132}$; $132\times 3.45=455.4$, so the answer is just under 3.45 → rounds to 3.45 (not 3.48).

Answer: (A) 3.45

Concept: Do $÷$ and $\times$ before $+$ and $-$.

1. $5.6÷7=0.8$ and $1.6\times 0.5=0.8$.
2. $25-0.8+0.8=25$.

Trick: The two middle terms are equal and opposite – they cancel, leaving 25.

Answer: (D) 25

Concept: Compare cost price and selling price of one article; the number of articles does not affect the percentage.

1. CP of one article $=\frac{474}{12}=39.5$ rupees.
2. SP of one article = ₹79, so profit $=79-39.5=39.5$ rupees.
3. Profit$\%=\frac{39.5}{39.5}\times 100=100\%$.

Trick: $79\times 12=948=2\times 474$ – SP per dozen is exactly double the CP per dozen, so profit is 100%.

Answer: (B) 100%

Concept: Monthly saving = salary − expenses; count only the months in which the saving is actually kept.

1. Monthly saving $=18000-5000-3000=10000$ rupees.
2. In the birthday month the whole saving is spent, so savings are kept for only 11 months.
3. Yearly savings $=11\times 10000=110000$ rupees.

Trick: Read the 'birthday' condition carefully – it removes exactly one month; 12 × 10000 = 120000 is the trap.

Answer: (B) 110000

Concept: If HCF = $h$, write the numbers as $ha$ and $hb$ with $a,b$ co-prime; then LCM $=hab$.

1. Given LCM $=21\times$ (first number) $\Rightarrow hab=21ha\Rightarrow b=21$.
2. So the second number $=21h$, where $h$ is a two-digit prime.
3. Smallest two-digit prime $h=11$, so least second number $=21\times 11=231$.

Trick: LCM ÷ first number = the co-prime factor of the second number; multiply it by the HCF.

Answer: (C) 231

Concept: Take successive differences; if they form a simple pattern, extend it.

1. $19-15=4,27-19=8,39-27=12,55-39=16$.
2. Differences are $4,8,12,16$ – multiples of 4 – so the next difference is $20$.
3. $55+20=75$.

Trick: Differences increasing by a constant = second-level AP; just add the next multiple.

Answer: (A) 75

Concept: Successive differences often form multiples of a fixed number.

1. $40-33=7,54-40=14,75-54=21,103-75=28$.
2. Differences $7,14,21,28$ are multiples of 7, so the next is $35$.
3. $103+35=138$.

Trick: First difference 7 and it keeps adding 7 – table of 7.

Answer: (D) 138

Concept: The product of two consecutive numbers lies just below the square of the larger one.

1. Let the numbers be $n$ and $n+1$: $n(n+1)=380$.
2. ${19}^2=361$ and ${20}^2=400$, so try $19\times 20=380$ ✓.
3. Greater number $=20$.

Trick: $\sqrt{380}\approx 19.5$ – the two consecutive numbers sit on either side of it: 19 and 20.

Answer: (A) 20

Concept: When $a$ and $b$ are reciprocals of each other, work with $a+b$ and $ab$ instead of the individual values.

1. $a=\frac{3+\sqrt{5}}{3-\sqrt{5}}$ and $b=\frac{3-\sqrt{5}}{3+\sqrt{5}}$, so $ab=1$.
2. $a+b=\frac{(3+\sqrt{5}{)}^2+(3-\sqrt{5}{)}^2}{(3-\sqrt{5})(3+\sqrt{5})}=\frac{2(9+5)}{9-5}=\frac{28}{4}=7$.
3. $a^2+b^2+3ab=(a+b{)}^2+ab=49+1=50$.

Trick: $a^2+b^2+3ab=(a+b{)}^2+ab$ – never rationalise separately when $ab=1$.

Answer: (B) 50

Concept: If two sums become equal after $n$ and $n+1$ years at the same CI rate, the sums are in the ratio $(1+r):1$.

1. Let Mahesh get $M$, Vijay $V$. $M(1.1{)}^{12}=V(1.1{)}^{13}\Rightarrow M=1.1V$.
2. $M:V=11:10$, total $21$ parts $=1407\Rightarrow$ 1 part $=67$.
3. Vijay's share $=10\times 67=670$ rupees.

Trick: One year's difference at 10% → ratio $11:10$; just divide the total in that ratio.

Answer: (C) 670

Concept: Convert each condition on the fraction into a linear equation in $a$ and $b$, then solve.

1. $\frac{a-2}{b-2}=\frac{1}{4}\Rightarrow 4a-8=b-2\Rightarrow b=4a-6$.
2. $\frac{a+17}{b+8}=\frac{7}{7}=1\Rightarrow a+17=b+8\Rightarrow b=a+9$.
3. $4a-6=a+9\Rightarrow a=5,b=14$; so $4a-b=20-14=6$.

Trick: $\frac{7}{7}$ simply means numerator = denominator – that gives $b=a+9$ instantly.

Answer: (C) 6

Concept: Mean proportional between $a$ and $b$ is $\sqrt{ab}$ (since $a:x=x:b$).

1. $x=\sqrt{7.2\times 20}=\sqrt{144}$.
2. $x=12$.

Trick: $7.2\times 20=72\times 2=144$ – spot the perfect square.

Answer: (D) 12

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$; assume a convenient total distance.

1. Let the distance be $1000$ km (LCM-friendly): parts $200$ km, $100$ km, $700$ km.
2. Times: $\frac{200}{8}=25$ h, $\frac{100}{25}=4$ h, $\frac{700}{20}=35$ h; total $=64$ h.
3. Average speed $=\frac{1000}{64}=15.625\text{km/h}$.

Trick: Never average the speeds; pick a distance divisible by 5, 10, 8, 25, 20 (1000 works) and add times.

Answer: (C) 15.625 km/h

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

1. $840=\frac{8400\times 5\times T}{100}=420T$.
2. $T=\frac{840}{420}=2$ years.

Trick: SI is 10% of the principal and the rate is 5% per year → $10÷5=2$ years.

Answer: (A) 2

Concept: Volume of water delivered by a pipe = (cross-section area) × (speed) × (time); equate it to the volume of the cylindrical layer formed in the tank.

1. Pipe radius $=1$ cm, speed $=4$ m/s $=400$ cm/s, time $=24\times 60=1440$ s.
2. Water delivered $=\pi \times 1^2\times 400\times 1440$ cm³.
3. Rise $h$: $\pi \times {80}^2\times h=\pi \times 400\times 1440\Rightarrow h=\frac{576000}{6400}=90$ cm.

Trick: $h=\frac{r_{\text{pipe}}^2\times \text{speed}\times \text{time}}{R_{\text{tank}}^2}$ – $\pi$ cancels, just keep all units in cm and s.

Answer: (B) 90 cm

Concept: One-day work of B = one-day work of (A+B) − one-day work of A.

1. Total work $=44$ units (LCM of 22, 44). A+B do $2$ units/day, A does $1$ unit/day.
2. B does $2-1=1$ unit/day → whole work in $44$ days.
3. Three-fourth of the work: $\frac{3}{4}\times 44=33$ days.

Trick: A alone takes exactly double the pair's time, so B is as fast as A → B also needs 44 days; then take ¾.

Answer: (D) 33