RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Take total work = LCM of days; work in 2-day cycles and finish the remainder.

1. Work $=100$ units; A $=5$/day, B $=4$/day; one cycle (2 days) $=9$ units.
2. 11 cycles $=22$ days $\to 99$ units; 1 unit left.
3. Day 23 is A's: $\frac{1}{5}$ day. Total $=22\frac{1}{5}$ days.

Trick: Stop the cycles just before the work exceeds 100, then check whose turn it is.

Answer: (A) $22\frac{1}{5}$ days

Concept: $a^3+b^3=(a+b)(a^2-ab+b^2)$, so $\frac{a^3+b^3}{a^2-ab+b^2}=a+b$.

1. With $a=525$, $b=321$: the fraction $=525+321=846$.
2. $846\times \frac{1}{96-36}=\frac{846}{60}=14.1$.

Trick: Whenever you see $a^2-ab+b^2$ below $a^3+b^3$, just write $a+b$.

Answer: (D) 14.1

Concept: Rise in level $=\frac{\text{volume of submerged solid}}{\text{base area of container}}$.

1. Cube: $4a^2=324\Rightarrow a=9$ cm, volume $=729{\text{ cm}}^3$.
2. Container: $4\times s\times 18=864\Rightarrow s=12$ cm, base area $=144{\text{ cm}}^2$.
3. Rise $=\frac{729}{144}\approx 5.06\approx 5$ cm.

Trick: LSA of cube $=4a^2$; LSA of square prism $=4sh$ – use them to get $a$ and $s$ fast.

Answer: (D) 5 cm

Concept: Write the numbers as $3x,5x,7x$.

1. Largest + smallest $=7x+3x=10x=60\Rightarrow x=6$.
2. Sum $=3x+5x+7x=15x=90$.

Trick: Sum of all $=\frac{15}{10}\times 60=90$ – scale directly from ratio parts.

Answer: (A) 90

Concept: $\sin 2A=2\sin A\cos A$, $\cos 3A=4{\cos }^{3}A-3\cos A$.

1. $\cot A=\frac{3}{4}\Rightarrow \sin A=\frac{4}{5},\cos A=\frac{3}{5}$ (3-4-5 triangle).
2. $\sin 2A=2\times \frac{4}{5}\times \frac{3}{5}=\frac{24}{25}=\frac{120}{125}$.
3. $\cos 3A=4\times \frac{27}{125}-\frac{9}{5}=\frac{108-225}{125}=-\frac{117}{125}$.
4. $\sin 2A-\cos 3A=\frac{120+117}{125}=\frac{237}{125}$.

Trick: Denominator 125 hints at ${\cos }^{3}A$ with $\cos A=\frac{3}{5}$.

Answer: (C) $\frac{237}{125}$

Concept: A number is divisible by 4 if its last two digits form a number divisible by 4.

1. Last two digits of 54468A0 are "A0".
2. A0 divisible by 4 → A = 0, 2, 4, 6, 8 (00, 20, 40, 60, 80).
3. Minimum value of A $=0$.

Trick: Only the last two digits matter for 4; A = 0 gives 00, which is divisible by 4.

Answer: (C) 0

Concept: Division and multiplication first, then addition/subtraction from left to right.

1. $(-56)÷(-14)=4$; $(-3)\times 8=-24$.
2. Expression $=-9-4+(-24)=-37$.

Trick: Sign rule: $(-)÷(-)=+$, $(-)\times (+)=-$.

Answer: (D) $-37$

Concept: $\text{Overall average}=\frac{\sum (n_i\times {\overline{x}}_i)}{\sum n_i}$.

1. Total spent $=12\times 184+73\times 234+80\times 390=2208+17082+31200=50490$.
2. Total members $=12+73+80=165$.
3. Average $=\frac{50490}{165}=306$.

Trick: Check the last digit: $50490÷165$ must end in 6 since $165\times 6=990$.

Answer: (B) 306

Concept: For $a{x}^2+bx+c=0$, product of roots $=\frac{c}{a}$.

1. $2x(2x+1)=1\Rightarrow 4x^2+2x-1=0$.
2. Product of roots $=\frac{-1}{4}$.
3. If one root is $\alpha$, the other $=\frac{-1/4}{\alpha }=-\frac{1}{4\alpha }$.

Trick: When one root is given as $\alpha$, the other is simply $\frac{c}{a\alpha }$.

Answer: (C) $-\frac{1}{4\alpha }$

Concept: Winning margin = difference of the two candidates' shares of the valid votes.

1. Let registered voters $=N$. Votes polled $=84\%$ of $N=0.84N$.
2. Valid votes $=80\%$ of $0.84N=0.672N$.
3. Winner $58\%$, loser $42\%$ → margin $=16\%$ of valid votes $=0.16\times 0.672N=0.10752N$.
4. $0.10752N=336\Rightarrow N=\frac{336}{0.10752}=3125$.

Trick: Margin $=(58-42)\%=16\%$ of valid votes. Chain the percentages: $N\times 0.84\times 0.8\times 0.16=336$.

Answer: (A) 3125

Concept: If HCF of two numbers is $h$, write them as $X=ha$, $Y=hb$ with $a,b$ co-prime.

1. $X=13a$, $Y=13b$ ⇒ $169ab=6760\Rightarrow ab=40$.
2. Co-prime factor pairs of 40: $(40,1)$ and $(8,5)$. Pairs $(20,2)$ and $(10,4)$ are not co-prime.
3. With $X>Y$: $(520,13)$ and $(104,65)$ → 2 pairs.

Trick: Number of co-prime pairs of $n$ = $2^{k-1}$ where $k$ = number of distinct prime factors. $40=2^3\times 5$, $k=2$ → $2$ pairs.

Answer: (B) 2

Concept: Check the differences between consecutive terms; they often form a known pattern (squares, cubes, primes).

1. $20-19=1=1^3$, $28-20=8=2^3$, $55-28=27=3^3$, $119-55=64=4^3$.
2. Next difference $=5^3=125$.
3. $?=119+125=244$.

Trick: Differences 1, 8, 27, 64 → cubes. Memorise cubes up to 15.

Answer: (C) 244

Concept: Combine chained ratios by making the common term equal.

1. $A:B=2:3$, $B:C=4:5$, $C:D=6:7$.
2. $A:B:C:D=(2\times 4\times 6):(3\times 4\times 6):(3\times 5\times 6):(3\times 5\times 7)=48:72:90:105$.
3. Sum of parts $=315$. One part $=\frac{7770}{315}=24\frac{2}{3}$.
4. Share of A $=48\times \frac{7770}{315}=\frac{7770\times 16}{105}=74\times 16=1184$.

Trick: For $a:b,c:d,e:f$ → $A=ace$, $B=bce$, $C=bde$, $D=bdf$.

Answer: (B) ₹1184

Concept: On a circular track, A overtakes B each time A gains one full lap: time $=\frac{\text{track length}}{\text{relative speed}}$.

1. 1st overtake: relative speed $=225-180=45$ m/min → $t_1=\frac{900}{45}=20$ min.
2. Speeds drop by 5%, so relative speed also drops by 5%: $45\times 0.95=42.75$ → $t_2=\frac{900}{42.75}\approx 21.05$ min.
3. Again $\times 0.95$: $40.6125$ → $t_3=\frac{900}{40.6125}\approx 22.16$ min.
4. Total $=20+21.05+22.16=63.21$ min.

Trick: Each stage's time is the previous one $÷0.95$: $20(1+\frac{1}{0.95}+\frac{1}{{0.95}^2})\approx 63.21$.

Answer: (D) 63.21 min

Concept: Average of the first $n$ natural numbers $=\frac{n+1}{2}$ (average of an AP = mean of first and last term).

1. Here $n=63$.
2. Average $=\frac{63+1}{2}=\frac{64}{2}=32$.

Trick: For consecutive numbers, average = middle term. The middle of 1…63 is the 32nd term = 32.

Answer: (C) 32

Concept: Successive increases multiply: final $=P(1+\frac{a}{100})(1+\frac{b}{100})$.

1. After 1st year: $2,90,000\times 1.10=3,19,000$.
2. After 2nd year: $3,19,000\times 1.20=3,82,800$.

Trick: Net change $=a+b+\frac{ab}{100}=10+20+2=32\%$; $2,90,000\times 1.32=3,82,800$.

Answer: (C) 3,82,800

Concept: With parallel lines, corresponding angles are equal, co-interior angles add to ${180}^{\circ }$ and vertically opposite angles are equal.

1. $t_1\parallel t_2$ cut by line $u$: $\angle C$ is corresponding to $\angle A$ → $\angle C={92}^{\circ }$.
2. $u\parallel v$ cut by $t_2$: $\angle B$ and $\angle C$ are co-interior → $\angle B={180}^{\circ }-{92}^{\circ }={88}^{\circ }$.
3. $\angle D$ is vertically opposite to $\angle B$ → $\angle D={88}^{\circ }$.

Trick: Two pairs of parallel lines form a parallelogram; $\angle A$ and $\angle D$ are at opposite corners but $\angle D$ is the exterior angle there, so $\angle D={180}^{\circ }-\angle A$.

Answer: (A) ${88}^{\circ }$

Concept: With a false weight, the dealer charges for 1 kg but hands over only $w$ kg; his real cost is for $w$ kg.

1. Let cost price $=$ ₹100 per kg. Professed 25% loss → selling price of "1 kg" $=75$.
2. Actual goods given $=w$ kg, costing $100w$. Gain 70% → $75=1.7\times 100w$.
3. $w=\frac{75}{170}=0.4412$ kg $=441.2$ g.

Trick: $\frac{\text{True weight}}{\text{False weight}}=\frac{100+\text{gain}\%}{100-\text{loss}\%}$ → $\frac{1000}{w}=\frac{170}{75}$, $w=441.2$ g.

Answer: (B) 441.2 gm

Concept: Add the per-hour work of filling taps and subtract that of the emptying tap.

1. Net work per hour $=\frac{1}{4}+\frac{1}{27}-\frac{1}{27}=\frac{1}{4}$.
2. So the full cistern fills in 4 hours.
3. One-fourth of the cistern $=\frac{1}{4}\times 4=1$ hour.

Trick: The 27-hour filler and the 27-hour emptier cancel each other – only the 4-hour tap matters.

Answer: (C) 1

Concept: For a discrete frequency distribution, median is the value of $x$ whose cumulative frequency first reaches or exceeds $\frac{N}{2}$.

1. $N=8+10+11+16+20+25+15+9+6=120$, so $\frac{N}{2}=60$.
2. Cumulative frequencies: 8, 18, 29, 45, 65, 90, 105, 114, 120.
3. The first cumulative frequency $\ge 60$ is 65, at $x=5$ → median $=5$.

Trick: Both the 60th and 61st observations lie in the $x=5$ group (46th to 65th), so median = 5 without any averaging.

Answer: (D) 5