RRB NTPC Graduate CBT 1 2024 PYQs

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

Concept: First swap the signs as told, then solve using BODMAS (÷, × before +, −).

1. Swap: $35+112÷84\times 48-75$
2. $112÷84=\frac{4}{3}$, and $\frac{4}{3}\times 48=64$
3. $35+64-75=24$

Trick: Rewrite the whole expression with the new signs first – mistakes happen when swapping in your head.

Answer: (A) $24$

Concept: Place the fixed clues first (L, H), then the "only K below M" clue.

1. L = floor 5; H is above L, so H = 6.
2. Only K lives below M → K = 1, M = 2.
3. F is on an even floor; 2 and 6 are taken → F = 4. Remaining: G = 3.
4. Below G: K and M = two persons.

Trick: "Only X below Y" always means X is on floor 1 and Y on floor 2.

Answer: (A) Two

Concept: Change every digit by the rule (odd $+2$, even $-1$), then count digits that repeat.

1. Number: 7 5 2 4 6 1 8 3
2. New: 9 7 1 3 5 3 7 5
3. Repeated digits: 7 (twice), 3 (twice), 5 (twice) → three digits.

Trick: Write the new digit directly under each old digit – no digit gets skipped.

Answer: (A) Three

Concept: Check the differences between consecutive terms.

1. Differences: $8,10,12,14$ – they rise by $2$.
2. Next difference $=16$
3. $1456+16=1472$

Trick: When differences are consecutive even numbers, the next difference is just $+2$ more.

Answer: (D) $1472$

Concept: Facing north, left and right are the solver's own left and right. Number seats 1–7 from the left.

1. Only two people left of L → L = 3. Three between L and E → E = 7.
2. G second to the right of L → G = 5. Two between G and M → M = 2.
3. Seats 1, 4, 6 remain for N, D, F with N < D < F → N = 1, D = 4, F = 6.
4. Right of D: G, F, E = three.

Trick: People to the right of seat $k$ in a row of 7 $=7-k=7-4=3$.

Answer: (C) Three

Concept: Find the position change of each letter (A = 1 … Z = 26).

1. ICK → KEF: I $+2$ = K, C $+2$ = E, K $-5$ = F.
2. HDW → JFR: H $+2$ = J, D $+2$ = F, W $-5$ = R. Pattern: $+2,+2,-5$.
3. VJF → XLA: V $+2$ = X, J $+2$ = L, F $-5$ = A ✓
4. RFM → TIH ($+3$ in middle), KLR → MNN ($-4$), GGP → IJK ($+3$) fail.

Trick: Check only the middle letter first ($+2$) – it removes A and D at once.

Answer: (B) VJF : XLA

Concept: Find one rule linking first, second and third numbers that works for both sets.

1. $(7,80,22)$: $2\times 7+3\times 22=14+66=80$ ✓
2. $(43,143,19)$: $2\times 43+3\times 19=86+57=143$ ✓ → Rule: $2a+3c=b$
3. $(13,89,21)$: $26+63=89$ ✓
4. Others: $50+45=95\ne 91$; $22+45=67\ne 63$; $98+45=143\ne 139$.

Trick: The middle number is largest → it is built from the other two; try $2a+3c$, $a+2c$, $3a+c$ quickly.

Answer: (C) $(13,89,21)$

Concept: Remove all numbers, rewrite only the symbols, then count from the right.

1. Symbols left (left → right): @ $ # % ≺ ⋟ ^ &
2. From right: & (1), ^ (2), ⋟ (3), ≺ (4), % (5), # (6).

Trick: 8 symbols remain, so 6th from right = 3rd from left (8 − 6 + 1 = 3) = #.

Answer: (C) #

Concept: Track the facing direction after each left turn (W → S → E → N → W) and add net displacements.

1. West 15, left → South 11, left → East 19, left → North 16, left → West 4.
2. East–West: $-15+19-4=0$ → P is directly in line with A.
3. North–South: $16-11=5$ km north of A.
4. To return: 5 km towards south.

Trick: Net E–W $=19-15-4=0$, net N–S $=16-11=5$ – no need for Pythagoras.

Answer: (B) 5 km to the south

Concept: Identify the highest and lowest numbers first, then pick the asked digits.

1. Highest = 815 → second digit = 1.
2. Lowest = 147 → third digit = 7.
3. $7-1=6$

Trick: Read the order carefully: "A subtracted from B" means $B-A$.

Answer: (A) $6$