RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Combined average $=\frac{n_1{a}_1+n_2{a}_2}{n_1+n_2}$.

1. Total weight $=22\times 40+44\times 37=880+1628=2508$
2. Average $=\frac{2508}{66}=38$ kg

Trick: Ratio $22:44=1:2$, so average $=\frac{40+2\times 37}{3}=\frac{114}{3}=38$.

Answer: (B) $38$

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

1. Distance $=20\times 80+65\times 70=1600+4550=6150$ m
2. Time $=80+70=150$ min
3. Average $=\frac{6150}{150}=41$ m/min

Trick: Never average the two speeds directly ($\frac{20+65}{2}=42.5$ is wrong) – weight them by time.

Answer: (C) $41$

Concept: Successive discounts multiply the remaining fractions.

1. Remaining $=0.76\times 0.82\times 0.80$
2. $0.76\times 0.82=0.6232$; $0.6232\times 0.8=0.49856$
3. Single discount $=1-0.49856=0.50144\approx 50.14\%$

Trick: Two at a time: $a+b-\frac{ab}{100}$: $24,18\to 37.68$; $37.68,20\to 50.144$.

Answer: (D) $50.14\%$

Concept: Time $=\frac{\text{capacity}}{\text{rate}}$. Let filling rate $=x$, emptying rate $=x+10$ (m³/min).

1. $\frac{1800}{x}-\frac{1800}{x+10}=6$
2. $1800\times 10=6x(x+10)\Rightarrow x(x+10)=3000$
3. $x^2+10x-3000=0\Rightarrow (x+60)(x-50)=0\Rightarrow x=50$

Trick: Option check: $\frac{1800}{50}=36$, $\frac{1800}{60}=30$, difference $6$ ✓.

Answer: (C) $50$

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

1. After year 1: $110000\times 1.05=115500$
2. After year 2: $115500\times 1.30=150150$

Trick: Net increase $=5+30+\frac{5\times 30}{100}=36.5\%$; $110000\times 1.365=150150$.

Answer: (A) $1,50,150$

Concept: Let apple $=a$, mango $=m$ (in ₹).

1. $2a+5m=239$
2. Apple at half price: $2\times \frac{a}{2}+4m=184\Rightarrow a+4m=184$
3. Double it: $2a+8m=368$; subtract: $3m=129\Rightarrow m=43$, $a=184-172=12$
4. $8a+8m=8\times 55=440$

Trick: You only need $a+m=55$, then multiply by 8.

Answer: (B) ₹$440$

Concept: Clear fractions by multiplying by the LCM of denominators.

1. $\frac{1}{14}:\frac{1}{2}:\frac{9}{3}=\frac{1}{14}:\frac{1}{2}:3$
2. Multiply by 14: $1:7:42$, total $=50$ parts
3. Third monkey $=\frac{42}{50}\times 600=504$

Trick: $600÷50=12$ per part; $42\times 12=504$.

Answer: (D) $504$

Concept: Convert each ratio to a decimal (or compare with 1).

1. $\frac{18}{37}\approx 0.49$, $\frac{13}{32}\approx 0.41$ – both below $\frac{1}{2}$.
2. $\frac{23}{25}=0.92$, $\frac{25}{26}\approx 0.96$
3. Greatest: $25:26$

Trick: For fractions just below 1, the one with the smaller gap from 1 is bigger: $\frac{1}{26}<\frac{2}{25}$.

Answer: (C) $25:26$

Concept: A cube has 12 equal edges, so its perimeter (total edge length) $=12a$. Use $x^3+y^3=(x+y{)}^3-3xy(x+y)$.

1. $12x+12y=300\Rightarrow x+y=25$; $xy=156$
2. $x^3+y^3={25}^3-3\times 156\times 25$
3. $=15625-11700=3925{\text{cm}}^3$

Trick: $x+y=25$, $xy=156$ gives $x=13$, $y=12$: $2197+1728=3925$.

Answer: (D) $3925{\text{cm}}^3$

Concept: $(1-a)(1+a)=1-a^2$; ${\sec }^{2}A-1={\tan }^{2}A$, ${\csc }^{2}A-1={\cot }^{2}A$.

1. Expression $=\frac{1-{\sec }^{2}A}{1-{\csc }^{2}A}=\frac{-{\tan }^{2}A}{-{\cot }^{2}A}={\tan }^{4}A$
2. $\tan A=\frac{4}{3}$
3. ${\tan }^{4}A=\frac{256}{81}$

Trick: $\frac{{\tan }^{2}A}{{\cot }^{2}A}={\tan }^{4}A$ – no need to find sec or cosec.

Answer: (B) $\frac{256}{81}$

Concept: Compare each letter of the first cluster with the matching letter of the second by its alphabet position.

1. AJK → BEG: A $+1$ = B, J $-5$ = E, K $-4$ = G.
2. JIL → KDH: J $+1$ = K, I $-5$ = D, L $-4$ = H.
3. KKQ → LFM: K $+1$ = L, K $-5$ = F, Q $-4$ = M ✔. (RLO → SIK fails: L $-3$ = I.)

Trick: Pattern $(+1,-5,-4)$; check only the 2nd letter first – it eliminates most options fast.

Answer: (D) KKQ : LFM

Concept: Look for the pattern number–number–number (three consecutive digits); the middle one is counted.

1. Digit runs in the series: "9 4 9", "6 6 5", "5 6 3 8 7 6".
2. "9 4 9" → 4 counted (1). "6 6 5" → middle 6 counted (1).
3. "5 6 3 8 7 6" → 6, 3, 8, 7 counted (4); the end digits of a run never count.
4. Total $=1+1+4=6$.

Trick: In a run of $k$ consecutive digits, $k-2$ digits qualify: $(3-2)+(3-2)+(6-2)=6$.

Answer: (D) 6

Concept: Change each digit separately: even $+2$, odd $-1$.

1. 2 6 3 4 7 5 9 1 → 4 8 2 6 6 4 8 0.
2. Second digit from left $=8$.
3. Third digit from right (from 4 8 2 6 6 4 8 0) $=4$.
4. Sum $=8+4=12$.

Trick: Only transform the two positions asked: 2nd from left is 6 → 8; 3rd from right is 5 → 4.

Answer: (D) 12

Concept: Odd digit $+1$, even digit $-2$, applied to every digit.

1. 1 8 4 7 6 5 3 → 2 6 2 8 4 6 4.
2. Highest digit $=8$, lowest digit $=2$.
3. Sum $=8+2=10$.

Trick: Odd digits become even after $+1$, so the new number has only even digits – quick check.

Answer: (D) 10

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

1. Numbers: 657, 435, 578, 659, 619. Highest $=659$, lowest $=435$.
2. Second digit of 659 $=5$; third digit of 435 $=5$.
3. $5+5=10$.

Trick: Compare first digits first (6 vs 4 or 5), then second digits only among ties.

Answer: (A) 10

Concept: Find the gaps between consecutive letters (alphabet positions, wrap-around after Z).

1. SXBE: S $+5$ X $+4$ B $+3$ E.
2. BGKN: B $+5$ G $+4$ K $+3$ N; FKOR: F $+5$ K $+4$ O $+3$ R.
3. FYUB: F → Y is not $+5$ (it is $+19$), Y $-4$ U, U $+7$ B – pattern broken.

Trick: Check only the first gap $(+5)$: F → Y fails immediately.

Answer: (C) FYUB

Concept: Seven seats 1–7 from left (all face north, so left = our left).

1. Three people left of Q → Q at 4.
2. Only H right of E → E at 6, H at 7.
3. Three between E and J → J at 2.
4. Seats 1, 3, 5 remain for G, I, F with G < I < F → G 1, I 3, F 5. Between G and F: J, I, Q = 3.

Trick: Fix the "only" clues first – they pin exact seats.

Answer: (B) Three

Concept: Find the single operation linking each pair.

1. $78+13=91$; $94+13=107$ → rule: add 13.
2. $86+13=99$ ✔.
3. $82+23=105$, $71+3=74$, $99+15=114$ – do not fit.

Trick: Just subtract: second − first must be 13.

Answer: (B) 86, 99

Concept: The same shift must take # to OWF and VDM to %.

1. Try # = TBK: T $-5$ = O, B $-5$ = W, K $-5$ = F → rule "$-5$ each letter".
2. Apply to VDM: V $-5$ = Q, D $-5$ = Y, M $-5$ = H → % = QYH.
3. Only option (C) has both TBK and QYH.

Trick: Moving 5 back: A–V, B–W, C–X, D–Y, E–Z (wrap-around pairs).

Answer: (C) # = TBK, % = QYH

Concept: Odd digit $+1$, even digit $-2$, then count repeated digits.

1. 1 3 8 5 7 2 4 → 2 4 6 6 8 0 2.
2. 2 appears twice, 6 appears twice; 4, 8, 0 appear once.
3. Digits appearing more than once: 2 and 6 → 2 digits.

Trick: Count distinct digits that repeat, not the number of repetitions.

Answer: (B) 2