RRB NTPC Graduate CBT 1 2024 PYQs

Concept: SP at loss $=\text{CP}\times \frac{100-\text{loss}\%}{100}$.

1. $\text{CP}=\frac{950\times 100}{95}=1000$.
2. Gain $=1360-1000=360$.
3. $x=\frac{360}{1000}\times 100=36\%$.

Trick: 5% loss on ₹950 → CP is a round ₹1000; gain is then read directly.

Answer: (C) 36

Concept: Form two equations and eliminate one variable.

1. $A+4B=34$ … (i); $3A+7B=97$ … (ii).
2. (i) $\times 3$: $3A+12B=102$; subtract (ii): $5B=5\Rightarrow B=1$.
3. $A=34-4=30$; $A+B=31$.

Trick: Multiply the equation with coefficient 1 to match the other's $A$.

Answer: (B) 31

Concept: Successive increases multiply: $P\times (1+\frac{r_1}{100})(1+\frac{r_2}{100})$.

1. After year 1: $110000\times \frac{105}{100}=115500$.
2. After year 2: $115500\times \frac{125}{100}=144375$.

Trick: $25\%=\frac{1}{4}$: add a quarter, $115500+28875=144375$.

Answer: (A) 1,44,375

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

1. Times: $\frac{576}{16}=36$, $\frac{630}{7}=90$, $\frac{315}{9}=35$ → total $161\text{h}$.
2. Distance $=576+630+315=1521\text{km}$.
3. $\frac{1521}{161}=9\frac{72}{161}\text{km/h}$ (since $161\times 9=1449$).

Trick: Never average the speeds; compute each time separately.

Answer: (C) $9\frac{72}{161}$

Concept: SP $=\text{MP}\times \frac{100-d}{100}$; without discount SP = MP.

1. $\text{MP}=\frac{5180}{0.20}=25900$.
2. Selling at MP gives 48% gain: $\text{CP}\times 1.48=25900$.
3. $\text{CP}=\frac{25900}{1.48}=17500$.

Trick: $1.48=\frac{37}{25}$: $25900\times \frac{25}{37}=700\times 25=17500$.

Answer: (D) 17,500

Concept: Mean proportional of $a$ and $b$ is $\sqrt{ab}$.

1. $(18+8\sqrt{2})(15-2\sqrt{2})=270-36\sqrt{2}+120\sqrt{2}-16\times 2$.
2. $=270-32+84\sqrt{2}=238+84\sqrt{2}$.
3. Mean proportional $=\sqrt{238+84\sqrt{2}}$.

Trick: All options share 238; only the $\sqrt{2}$ term matters: $120-36=84$.

Answer: (A) $\sqrt{238+84\sqrt{2}}$

Concept: For fixed distance, time $\propto \frac{1}{\text{speed}}$.

1. Speed ratio new : usual $=5:7$, so time ratio $=7:5$.
2. Extra time $=7-5=2$ parts $=20$ minutes → 1 part $=10$ minutes.
3. Usual time $=5$ parts $=50$ minutes.

Trick: Speed $\frac{a}{b}$ of normal and late by $t$ → usual time $=\frac{at}{b-a}=\frac{5\times 20}{2}$.

Answer: (C) 50 minutes

Concept: Write everything as powers of 146 and compare exponents.

1. $x^z={146}^{0.93z}$ and $y^7={146}^{0.22\times 7}={146}^{1.54}$.
2. $0.93z=1.54\Rightarrow z=\frac{1.54}{0.93}\approx 1.66$.

Trick: Same base → equate exponents.

Answer: (C) 1.66

Concept: Total = average $\times$ count; difference of totals gives the excluded value.

1. Total of 7 subjects $=75\times 7=525$.
2. Total of 6 subjects $=70\times 6=420$.
3. Science $=525-420=105$.

Trick: Science $=75+6\times (75-70)=105$.

Answer: (A) 105

Concept: Convert each ratio to a decimal (or compare with $\frac{1}{2}$).

1. $\frac{33}{51}\approx 0.647$, $\frac{37}{64}\approx 0.578$.
2. $\frac{32}{86}\approx 0.372$, $\frac{26}{59}\approx 0.441$.
3. Greatest: $33:51$.

Trick: $\frac{32}{86}$ and $\frac{26}{59}$ are below $\frac{1}{2}$; between the other two, $33:51$ has larger numerator and smaller denominator.

Answer: (A) $33:51$

Concept: A letter common to two words gets the digit common to both codes.

1. PACE → 2468 and COST → 1327.
2. Common letter: C. Common digit: 2.
3. So C = 2.

Trick: Match common letters with common digits – no need to decode the full word.

Answer: (D) 2

Concept: Replace each sign as instructed, then apply BODMAS.

1. $55-8÷3+15\times 5$ becomes $55\times 8+3-15÷5$.
2. $=440+3-3$.
3. $=440$.

Trick: Write the new expression fully before calculating – most errors come from mixing old and new signs.

Answer: (A) $440$

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

1. Numbers: 236, 758, 230, 563, 263.
2. Highest = 758 (first digit 7); lowest = 230 (first digit 2).
3. $7-2=5$.

Trick: Compare only first digits to find the extremes; for ties (236 vs 230) look at the next digit.

Answer: (A) 5

Concept: Find the letter shift from the complete pair and apply the same shift.

1. RLT → %: try $-1$ on each letter: R→Q, L→K, T→S ⇒ % = QKS.
2. So # → NHP also by $-1$: # must be O I Q (O−1=N, I−1=H, Q−1=P).
3. # = OIQ, % = QKS.

Trick: Check the option whose % gives the same shift as # – only option D is consistent.

Answer: (D) # = OIQ, % = QKS

Concept: Facing the centre, "left" means clockwise and "right" means anticlockwise.

1. S is between C and D; A is third to the left of C.
2. T and D have two people between them (from left of T); B is fourth to the right of R.
3. Only arrangement (clockwise): A, B, D, S, C, R, T.
4. From the left of S going to B: C, R, T, A ⇒ 4 people.

Trick: Start with fixed blocks (C–S–D) and place the others around them.

Answer: (D) Four

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

1. Numbers: 144, 743, 290, 258, 799.
2. Highest = 799 → second digit 9. Lowest = 144 → third digit 4.
3. $4-9=-5$.

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

Answer: (A) $-5$

Concept: Look at successive differences.

1. $402-2=400$, $400-4=396$, $396-6=390$, $390-8=382$.
2. Differences increase by 2: next is $-10$.
3. $382-10=372$.

Trick: Differences 2, 4, 6, 8, 10 – consecutive even numbers.

Answer: (C) 372

Concept: Check each condition on the digits 4 8 7 1 9 and apply only the ones that fit; the codes for 1 and 9 (J and S) are read from the options.

1. Condition 1: 1st digit 4 and 2nd digit 8 are both even → both coded as A. ✔
2. Condition 2: 1st digit 4 is even → does not apply.
3. Condition 3: 3rd digit 7 and 4th digit 1 are odd → does not apply.
4. Remaining digits use the table: 7 → Z, 1 → J, 9 → S.

Trick: Test the conditions first; usually only one fits – here the 'both even' start immediately gives A A, so only option (C) survives.

Answer: (C) A A Z J S

Concept: Track each position of the clusters separately (A = 1 … Z = 26, cycling after Z).

1. 1st letters: Z, G, N, U → each $+7$ ⇒ U + 7 = B.
2. 2nd letters: Y, F, M, T → each $+7$ ⇒ T + 7 = A.
3. 3rd letters: B, I, P, W → each $+7$ ⇒ W + 7 = D.
4. Next term: BAD.

Trick: Inside each term, 2nd letter = 1st − 1 and 3rd = 1st + 2 (Z Y B, G F I). U + 7 = B ⇒ B A D.

Answer: (C) BAD

Concept: Facing the centre, left = clockwise, right = anticlockwise.

1. P is immediate right of S, and P is second to the left of T ⇒ (clockwise) P, S, _, _, _, T.
2. U is third to the left of S ⇒ U sits just before T; V is next to U.
3. Two people between T and Q from the left of Q ⇒ Q sits right after S; V then sits between Q and U; R fills the last seat.
4. Clockwise: P, S, Q, V, U, T, R. From the left of R to Q: P, S ⇒ 2.

Trick: Fix S–P first because two statements link them with T.

Answer: (B) 2