RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Convert averages to totals; the common member is counted twice in the pair totals.

1. $A+B+C=3\times 45=135$
2. $A+B=2\times 39=78$, $B+C=2\times 48=96$
3. $B=(A+B)+(B+C)-(A+B+C)=78+96-135=39$

Trick: Common member $=$ sum of the two pair totals $-$ group total.

Answer: (B) $39$

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

1. $R=\frac{700\times 100}{1400\times 5}$
2. $=\frac{70000}{7000}=10\%$

Trick: SI is half the principal in 5 years → $50\%$ in 5 years → $10\%$ per year.

Answer: (A) $10\%$

Concept: Number to add $=$ divisor $-$ remainder.

1. $67\times 207=13869$, so $13874÷67$ leaves remainder $13874-13869=5$
2. Number to add $=67-5=62$
3. Check: $13874+62=13936=67\times 208$ ✓

Trick: $67\times 200=13400$; $13874-13400=474$; $474=67\times 7+5$ → remainder $5$.

Answer: (D) $62$

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

1. First part: $23\times 4.5=103.5\text{km}$
2. New speed $=23+11=34\text{km/h}$; second part $=34\times 1=34\text{km}$
3. Total distance $=137.5\text{km}$, total time $=5.5\text{h}$
4. Average $=\frac{137.5}{5.5}=25\text{km/h}$

Trick: Never average the two speeds directly; always divide total distance by total time.

Answer: (B) $25$

Concept: Difference of ratio parts corresponds to the given money difference.

1. Ratio A : B : C : D $=10:4:8:5$
2. C − B $=8-4=4$ parts $=$ ₹24, so 1 part $=$ ₹6
3. A $=10\times 6=$ ₹60

Trick: Value of 1 part = difference ÷ difference of parts.

Answer: (C) ₹60

Concept: Mode is the value that occurs most often in the data.

1. Count repeats: 43 appears 4 times.
2. 53 appears 3 times; 42 and 54 appear 2 times each; others once.
3. Highest frequency is 4, for 43.

Trick: Sort mentally or tally only the repeated numbers – no need to count singles.

Answer: (D) $43$

Concept: Covered : remaining $=5:4$, so the covered part is $\frac{5}{9}$ of the total.

1. Distance covered $=\frac{5}{9}\times 39=\frac{65}{3}\approx 21.67\text{km}$
2. Time $=5\text{h}5\text{min}=5\frac{5}{60}=\frac{61}{12}\text{h}$
3. Speed $=\frac{65}{3}÷\frac{61}{12}=\frac{65\times 12}{3\times 61}=\frac{260}{61}\approx 4.26$
4. Rounded: $4.3\text{km/h}$

Trick: 'a/b of remaining' ⇒ covered $=\frac{a}{a+b}$ of total.

Answer: (B) $4.3$

Concept: $\text{TSA}=2\pi r(r+h)$, $V=\pi r^{2}h$

1. $2\times \frac{22}{7}\times 7\times (7+h)=396$
2. $44(7+h)=396\Rightarrow 7+h=9\Rightarrow h=2\text{cm}$
3. $V=\frac{22}{7}\times 7^2\times 2=22\times 7\times 2=308{\text{cm}}^3$

Trick: With $r=7$, $2\pi r=44$ – divide TSA by 44 to get $r+h$ instantly.

Answer: (B) $308$

Concept: Find both values separately and subtract.

1. $60\%\text{ of }40=\frac{60}{100}\times 40=24$
2. $\frac{3}{5}\text{ of }15=9$
3. Difference $=24-9=15$

Trick: $60\%=\frac{3}{5}$, so the answer is $\frac{3}{5}\times (40-15)=15$.

Answer: (B) $15$

Concept: Angle sum of a quadrilateral $={360}^{\circ }$; the angle between bisectors of two adjacent angles $=\frac{1}{2}(\text{sum of the other two angles})$.

1. $\angle S+\angle T={360}^{\circ }-({52}^{\circ }+{44}^{\circ })={264}^{\circ }$
2. In $△TES$: $\frac{\angle S}{2}+\frac{\angle T}{2}={132}^{\circ }$
3. $\angle TES={180}^{\circ }-{132}^{\circ }={48}^{\circ }$

Trick: $\angle TES=\frac{\angle Q+\angle R}{2}=\frac{{96}^{\circ }}{2}={48}^{\circ }$.

Answer: (A) ${48}^{\circ }$

Concept: $\text{SP}=\text{CP}\times \frac{100\pm \text{gain/loss}\%}{100}$

1. At 10% loss: $\text{CP}=\frac{203700}{0.9}$
2. Required SP $=\frac{203700}{0.9}\times 1.68=203700\times \frac{168}{90}$
3. $=203700\times \frac{28}{15}=13580\times 28=380240$

Trick: Directly scale SP: $\text{new SP}=\text{old SP}\times \frac{168}{90}$.

Answer: (B) ₹3,80,240

Concept: $(a\pm b{)}^2=a^2+b^2\pm 2ab$

1. $(a-b{)}^2=132-72=60$
2. $(a+b{)}^2=132+72=204$
3. $\frac{a-b}{a+b}=\sqrt{\frac{60}{204}}=\sqrt{\frac{5}{17}}$ (positive since $a>b$)

Trick: Squared ratio $=\frac{a^2+b^2-2ab}{a^2+b^2+2ab}$ – no need to find a and b.

Answer: (A) $\sqrt{\frac{5}{17}}$

Concept: Successive changes multiply: $\text{Final price}={\text{CP}}_A\times \frac{100-60}{100}\times \frac{100+25}{100}$

1. $62616={\text{CP}}_A\times 0.4\times 1.25={\text{CP}}_A\times 0.5$
2. ${\text{CP}}_A=62616\times 2=125232$

Trick: $\frac{2}{5}\times \frac{5}{4}=\frac{1}{2}$ – C's price is exactly half of A's cost.

Answer: (D) ₹1,25,232

Concept: After 50% rent, the remaining 50% covers all other spending and savings.

1. Remaining amount $=780+9800+110=10690$
2. $50\%$ of salary $=10690$
3. Salary $=10690\times 2=21380$

Trick: Remaining % × salary = sum of the leftover items.

Answer: (D) ₹21,380

Concept: Form two equations with Anjali's age $A$ and Kamal's age $K$.

1. $7A=6K+14$ … (1)
2. $7K=7A-7\Rightarrow K=A-1$ … (2)
3. Put (2) in (1): $7A=6A-6+14\Rightarrow A=8$, $K=7$
4. Sum $=8+7=15$

Trick: Check: $7\times 8=56=6\times 7+14$ ✔.

Answer: (B) $15$

Concept: $\frac{x+1}{x}=1+\frac{1}{x}$, so the given condition fixes $\frac{1}{x}$ directly; also $\frac{x^2+1}{x^2}=1+\frac{1}{x^2}$.

1. $\frac{x+1}{x}=33\Rightarrow 1+\frac{1}{x}=33\Rightarrow \frac{1}{x}=32$.
2. $\frac{x^2+1}{x^2}=1+{\left(\frac{1}{x}\right)}^2=1+{32}^2=1+1024=1025$.
3. $1025$ is not among the options (1091, 1083, 1087, 1089), so no option is correct.
4. If the condition had been $x+\frac{1}{x}=33$, then $x^2+\frac{1}{x^2}={33}^2-2=1087$, option (C).

Note: Read the fraction carefully: $\frac{x+1}{x}$ is not the same as $x+\frac{1}{x}$. 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: If $N$ leaves remainders that are each 10 less than the divisors, then $N+10$ is divisible by all divisors.

1. $27-17=10$, $19-9=10$, $15-5=10$ ⇒ $N+10$ is a multiple of $\text{LCM}(27,19,15)=2565$
2. $N=2565k-10$: $k=1\Rightarrow 2555$ (not divisible by 80), $k=2\Rightarrow 5120$
3. $5120÷80=64$ ✔ ⇒ $N=5120$, which lies between 5110 and 5130.

Trick: Constant (divisor − remainder) ⇒ $N=\text{LCM}\times k-\text{that constant}$.

Answer: (B) 5110 and 5130

Concept: Net rate = filling rate − emptying rate.

1. Net rate $=\frac{1}{15}-\frac{1}{40}=\frac{8-3}{120}=\frac{1}{24}$ tank per hour
2. Full tank takes 24 h, so $\frac{1}{3}$ tank takes $\frac{24}{3}=8$ h

Trick: LCM method: tank $=120$ units; filler $+8$, emptier $-3$ ⇒ net 5/h; $\frac{1}{3}$ tank $=40$ units ⇒ 8 h.

Answer: (A) $8$

Concept: Take LCM and use $1-{\sin }^2\theta ={\cos }^2\theta$.

1. $\frac{\cos \theta (1+\sin \theta )+\cos \theta (1-\sin \theta )}{1-{\sin }^2\theta }=\frac{2\cos \theta }{{\cos }^2\theta }=\frac{2}{\cos \theta }$
2. $\frac{2}{\cos \theta }=4\Rightarrow \cos \theta =\frac{1}{2}$
3. $\theta ={60}^{\circ }$

Trick: The expression always simplifies to $2\sec \theta$; so $\sec \theta =2\Rightarrow \theta ={60}^{\circ }$.

Answer: (C) ${60}^{\circ }$

Concept: Let total time be $t$ minutes; D works alone for $\frac{t}{2}$, C and D together for $\frac{t}{2}$.

1. Rates: C $=\frac{1}{60}$, D $=\frac{1}{40}$, together $=\frac{1}{24}$ per minute
2. $\frac{t}{2}(\frac{1}{40}+\frac{1}{24})=1\Rightarrow \frac{t}{2}\times \frac{8}{120}=1$
3. $\frac{t}{30}=1\Rightarrow t=30$ minutes

Trick: Units: tank $=120$; D $=3$/min, C+D $=5$/min ⇒ $3x+5x=120$, $x=15$ ⇒ total $30$ min.

Answer: (B) 30 minutes