RRB NTPC Graduate CBT 1 2025 PYQs

Concept: From $a+\frac{1}{a}$ build higher powers: $a^2+\frac{1}{a^2}=(a+\frac{1}{a}{)}^2-2$, $a^3+\frac{1}{a^3}=(a+\frac{1}{a}{)}^3-3(a+\frac{1}{a})$, and $a^5+\frac{1}{a^5}=(a^2+\frac{1}{a^2})(a^3+\frac{1}{a^3})-(a+\frac{1}{a})$.

1. $a^2+\frac{1}{a^2}=16-2=14$.
2. $a^3+\frac{1}{a^3}=64-12=52$.
3. $a^5+\frac{1}{a^5}=14\times 52-4=728-4=724$.

Trick: Multiply the 2nd and 3rd power sums and subtract the 1st – works for any $a+\frac{1}{a}$.

Answer: (A) 724

Concept: In $a:b::c:d$, $d=\frac{b\times c}{a}$.

1. $x=\frac{28\times 4.1}{4}=7\times 4.1=28.7$.
2. $y=\frac{32\times 3}{4}=24$.
3. $x:y=28.7:24=287:240$.

Trick: $4:28$ means ×7 and $4:32$ means ×8 – just multiply $4.1\times 7$ and $3\times 8$.

Answer: (A) 287 : 240

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

1. Ratio $31:34$, total parts $=65$; difference of parts $=3$.
2. One part $=\frac{44850}{65}=690$.
3. Difference $=3\times 690=$ ₹$2,070$.

Trick: Directly compute $\frac{\text{difference of ratio}}{\text{sum of ratio}}\times \text{profit}=\frac{3}{65}\times 44850$.

Answer: (D) ₹2,070

Concept: The angle in a semicircle is ${90}^{\circ }$, and the angles of a triangle add to ${180}^{\circ }$.

1. In $△ACB$: $\angle ACB={90}^{\circ }$, $\angle CAB={48}^{\circ }$ → $\angle CBA={42}^{\circ }$.
2. In $△ADB$: $\angle ADB={90}^{\circ }$, $\angle ABD={55}^{\circ }$ → $\angle BAD={35}^{\circ }$.
3. C and D are on opposite sides of AB, so $\angle CBD={42}^{\circ }+{55}^{\circ }={97}^{\circ }$ and $\angle CAD={48}^{\circ }+{35}^{\circ }={83}^{\circ }$.
4. Difference $={97}^{\circ }-{83}^{\circ }={14}^{\circ }$.

Trick: $\angle CBD+\angle CAD={180}^{\circ }$ (opposite angles of cyclic quadrilateral ACBD); difference $=(42+55)-(48+35)$.

Answer: (A) ${14}^{\circ }$

Concept: Provisions are fixed, so $\text{men}\times \text{days}$ is constant.

1. Remaining men $=100-55=45$.
2. $100\times 27=45\times d$.
3. $d=\frac{2700}{45}=60$ days.

Trick: Fewer men → more days; $\frac{100}{45}=\frac{20}{9}$, so $27\times \frac{20}{9}=60$.

Answer: (A) 60

Concept: Let the numbers bought be $4k,2k,5k$ and equate total cost.

1. Cost $=4k\times 280+2k\times 180+5k\times 115=1120k+360k+575k=2055k$.
2. $2055k=6165\Rightarrow k=3$.
3. Articles of type Q $=2k=6$.

Trick: Cost of one 'set' (4P+2Q+5R) $=2055$; $6165÷2055=3$ sets.

Answer: (B) 6

Concept: $\text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h$, where $f_1$ is the highest frequency.

1. Modal class $=30\text{–}40$ ($f_1=20$), $l=30$, $h=10$, $f_0=12$, $f_2=10$.
2. $\text{Mode}=30+\frac{20-12}{40-12-10}\times 10=30+\frac{80}{18}$.
3. $=30+4.44=34.44$.

Trick: Mode always lies inside the modal class ($30$–$40$) – only option D qualifies.

Answer: (D) 34.44

Concept: Look at consecutive differences.

1. $65-45=20$, $81-65=16$, $93-81=12$, $101-93=8$.
2. Differences fall by 4 each time: $20,16,12,8$ → next $4$.
3. $101+4=105$.

Trick: Decreasing differences in an arithmetic pattern – just continue the step.

Answer: (A) 105

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

1. Distance 1 $=20\times 90=1800\text{m}$; distance 2 $=55\times 60=3300\text{m}$.
2. Total distance $=5100\text{m}$, total time $=150$ min.
3. Average $=\frac{5100}{150}=34\text{m/min}$.

Trick: Time-weighted mean: $\frac{20\times 3+55\times 2}{5}=\frac{170}{5}=34$.

Answer: (D) 34

Concept: Average speed $=\frac{\text{total distance}}{\text{total time}}$; find each segment's time.

1. Distances $=2:3:1$ of $480$ → $160,240,80$ km.
2. Times: $\frac{160}{48}=\frac{10}{3}$ h, $\frac{240}{72}=\frac{10}{3}$ h, $\frac{80}{96}=\frac{5}{6}$ h.
3. Total time $=\frac{10}{3}+\frac{10}{3}+\frac{5}{6}=\frac{45}{6}=7.5$ h.
4. Average speed $=\frac{480}{7.5}=64\text{km/h}$.

Trick: Work with 1 km units: $\frac{2}{48}+\frac{3}{72}+\frac{1}{96}=\frac{4+4+1}{96}=\frac{9}{96}$ h per 6 km → $\frac{6\times 96}{9}=64$.

Answer: (B) 64 km/h

Concept: Total surface area of a cube $=6a^2$.

1. $a=84$ cm → $a^2=7056$.
2. $6\times 7056=42336{\text{cm}}^2$.

Trick: ${84}^2=(80+4{)}^2=6400+640+16=7056$; unit digit $6\times 6=36$ → ends in 6, only option D.

Answer: (D) $42336{\text{cm}}^2$

Concept: If a sum becomes $n$ times under SI, interest $=(n-1)P$ and $R=\frac{(n-1)\times 100}{T}$.

1. Five times → SI $=4P$ in 5 years.
2. $R=\frac{4\times 100}{5}=80\%$ per annum.

Trick: $R\times T=(n-1)\times 100$ → $R=\frac{400}{5}=80$.

Answer: (C) $80\%$

Concept: If amounts are equal after different periods, the shares are in the inverse ratio of growth factors.

1. $P(1.1{)}^{18}=M(1.1{)}^{19}\Rightarrow P=1.1M$.
2. Piyush : Manoj $=11:10$; total $21$ parts $=1407$ → one part $=67$.
3. Manoj $=10\times 67=$ ₹$670$.

Trick: One extra year at $10\%$ → ratio $11:10$; the big '18 years / 19 years' is only to scare.

Answer: (A) 670

Concept: Margin = (winner % − loser %) of valid votes; chain the percentages back to registered voters.

1. Let registered voters $=N$. Polled $=0.8N$; valid $=90\%$ of that $=0.72N$.
2. Winner $89\%$, loser $11\%$ → margin $=78\%$ of $0.72N=0.5616N$.
3. $0.5616N=351\Rightarrow N=625$.

Trick: $351=0.78\times 450$, and $450=0.72\times 625$.

Answer: (A) 625

Concept: Convert each percentage statement into a multiplier and link the numbers.

1. Let the numbers be $A$, $B$, $C$. $A\times \frac{125}{100}=B$ and $B=C\times \frac{80}{100}$.
2. So $\frac{5A}{4}=\frac{4C}{5}$ → $25A=16C$.
3. $A:C=16:25$.

Trick: $+25\%=\frac{5}{4}$, $-20\%=\frac{4}{5}$; equate $\frac{5}{4}A=\frac{4}{5}C$ → $A:C=\frac{4}{5}·\frac{4}{5}=16:25$.

Answer: (C) 16:25

Concept: A right triangle revolved about one leg forms a cone: that leg is the height, the other leg is the radius, the hypotenuse is the slant height.

1. Sides 3, 4, 5 form a right triangle ($3^2+4^2=5^2$). Revolving about the side 3 cm: $h=3$, $r=4$, $l=5$.
2. $V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times 3.14\times 16\times 3$
3. $=3.14\times 16=50.24{\text{cm}}^3$.

Trick: The axis of rotation is the height; the other leg sweeps the circle (radius). Revolving about 4 cm would give $37.68$ instead.

Answer: (D) 50.24 cubic cm

Concept: Equal roots ⇒ discriminant zero and root $=-\frac{b}{2a}$; the common root satisfies both equations.

1. $x^2-3px+q=0$ has equal roots: $9p^2=4q$ → $q=\frac{9p^2}{4}$, root $\alpha =\frac{3p}{2}$.
2. $\alpha$ satisfies $x^2-2kx+t=0$: $\frac{9p^2}{4}-3pk+t=0$ → $t=3pk-\frac{9p^2}{4}$.
3. $t+q=3pk$ → $(t+q{)}^2=9p^2{k}^2$.

Trick: With equal roots, the common root is simply the repeated root $\frac{3p}{2}$ – plug it straight in.

Answer: (C) $9p^2{k}^2$

Concept: With a false weight, the shopkeeper charges for 1000 g but gives less; his real gain depends on the goods actually given.

1. Let cost be ₹1 per gram. He "sells 1 kg" at a 60% loss, i.e. collects ₹$0.40\times 1000=400$.
2. Real gain is 90%, so the cost of goods actually given $=\frac{400}{1.9}=210.53$ → he gives about $210.5$ g.
3. Check: cost ₹210.5, sold for ₹400 → gain $\frac{189.5}{210.5}\approx 90\%$ ✓.

Trick: $\text{Weight given}=1000\times \frac{100-\text{claimed loss}\%}{100+\text{actual gain}\%}=1000\times \frac{40}{190}$.

Answer: (A) 210.5 gm

Concept: Evaluate roots and "of" first, then multiplication, then subtraction.

1. $15\%\text{ of }80=12$; $\sqrt[3]{216}=6$; $\sqrt{36}=6$.
2. $\frac{3}{6}\times 6\times 6=18$.
3. $y=12-18=-6$.

Trick: "of" means multiply and ranks with brackets – do it before the minus sign.

Answer: (A) $-6$

Concept: In a cyclic quadrilateral the exterior angle at a vertex equals the interior opposite angle.

1. PQ is produced to T, so $\angle RQT$ is the exterior angle at Q.
2. Interior opposite angle of Q is $\angle PSR$ (at S).
3. $\angle PSR=\angle RQT={58}^{\circ }$.
4. Check: $\angle PQR={180}^{\circ }-{58}^{\circ }={122}^{\circ }$ and ${122}^{\circ }+{58}^{\circ }={180}^{\circ }$ ✓.

Trick: Exterior angle = opposite interior angle (because opposite angles sum to ${180}^{\circ }$). Option A (${122}^{\circ }$) is the trap – that is $\angle PQR$.

Answer: (D) ${58}^{\circ }$