RRB NTPC Graduate CBT 1 2024 PYQs

Concept: If a number increases by $r\%$, new value $=\text{number}\times \frac{100+r}{100}$.

1. $57\frac{1}{2}\%=\frac{115}{2}\%$, so new value $=x\times \frac{315}{200}$.
2. $x\times \frac{315}{200}=378$
3. $x=\frac{378\times 200}{315}=240$

Trick: Check: $240+57.5\%\text{ of }240=240+138=378$ ✔

Answer: (A) $240$

Concept: The altitude to the base of an isosceles triangle bisects the base; $\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$.

1. Half base $=\frac{18}{2}=9\text{cm}$
2. $h=\sqrt{{41}^2-9^2}=\sqrt{1681-81}=\sqrt{1600}=40\text{cm}$
3. $\text{Area}=\frac{1}{2}\times 18\times 40=360{\text{cm}}^2$

Trick: (9, 40, 41) is a Pythagorean triplet. Or use $\frac{b}{4}\sqrt{4a^2-b^2}=\frac{18}{4}\times 80=360$.

Answer: (C) $360$

Concept: Least number to be added $=\text{divisor}-\text{remainder}$.

1. $22938÷7$: $7\times 3276=22932$, remainder $=6$
2. Number to add $=7-6=1$
3. $22938+1=22939=7\times 3277$ ✔

Trick: Divide by 7, find remainder; add (7 − remainder).

Answer: (C) $1$

Concept: Volume of shell $=\frac{4}{3}\pi (R^3-r^3)$ and $\text{Mass}=\text{Density}\times \text{Volume}$.

1. With $r=0$, $R=3$: $V=\frac{4}{3}\times \frac{22}{7}\times 27=\frac{792}{7}{\text{cm}}^3$
2. $\text{Mass}=7\times \frac{792}{7}=792\text{g}$
3. $792\text{g}=0.792\text{kg}$

Trick: The density 7 cancels the 7 of $\pi =\frac{22}{7}$: mass $=\frac{4}{3}\times 22\times 27=792$ g.

Answer: (C) $0.792$

Concept: A quadratic $a{x}^2+bx+c=0$ has equal roots when the discriminant $D=b^2-4ac=0$.

1. Here $a=1$, $b=4$, $c=K$.
2. $4^2-4\times 1\times K=0$
3. $16-4K=0\Rightarrow K=4$

Trick: Equal roots → perfect square: $x^2+4x+4=(x+2{)}^2$, so $K={\left(\frac{b}{2}\right)}^2=4$.

Answer: (C) $4$

Concept: $\text{Overall average}=\frac{\text{Total sum}}{\text{Total count}}$.

1. Sum of 15 boys $=15\times 160=2400$
2. Remaining $50-15=35$ boys: $35\times 168=5880$
3. Average $=\frac{2400+5880}{50}=\frac{8280}{50}=165.6\text{cm}$

Trick: Deviation method: $160+\frac{35\times 8}{50}=160+5.6=165.6$.

Answer: (B) $165.6$

Concept: If A is $r\%$ more than B, then B is $\frac{r}{100+r}\times 100\%$ less than A.

1. Let B = 100, then A = 112.
2. B is less than A by 12.
3. $\frac{12}{112}\times 100=10.71\%$

Trick: $\frac{r}{100+r}$: $\frac{12}{112}=\frac{3}{28}\approx 10.71\%$.

Answer: (B) $10.71\%$

Concept: Arrange the data in order; for $n$ odd, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending: 15, 33, 34, 54, 60, 65, 66, 68, 70, 71, 77
2. $n=11$, median $={\left(\frac{11+1}{2}\right)}^{\text{th}}=6^{\text{th}}$ value
3. $6^{\text{th}}$ value $=65$

Trick: 5 values below and 5 above: 15, 33, 34, 54, 60 | 65 | 66, 68, 70, 71, 77.

Answer: (D) $65$

Concept: At compound interest, if money becomes 2 times in $n$ years, it becomes $2^k$ times in $kn$ years.

1. $P(1+r{)}^{11}=2P$
2. $4P=2^{2}P=P{[(1+r{)}^{11}]}^2=P(1+r{)}^{22}$
3. So time $=2\times 11=22$ years.

Trick: Doubling time multiplies: 2× in 11 → 4× in 22 → 8× in 33.

Answer: (D) $22$

Concept: Write initial money and spending in ratio form; equal remaining amounts give an equation.

1. Money: $17x$ and $14x$; spending ratio $12:8=3:2$, i.e. $3k$ and $2k$.
2. $17x-3k=3392$ and $14x-2k=3392$
3. From the second: $k=7x-1696$; then $17x-21x+5088=3392\Rightarrow x=424$
4. Sohit $=17\times 424=₹7208$

Trick: Check: $k=2968-1696=1272$; Sohit spends $3816$, $7208-3816=3392$ ✔

Answer: (B) ₹7,208

Concept: Each part of the cluster follows its own pattern.

1. First letters: D, E, F, G, H (+1).
2. Second letters: J, K, L, M, N (+1).
3. Numbers: 26, 37, 48, 59, 70 ($+11$).

Trick: Split cluster into 3 columns and solve each separately.

Answer: (B) HN70

Concept: Look at the differences between consecutive terms.

1. $187-14=173$, $173-13=160$, $160-12=148$, $148-11=137$
2. Next: $137-10=127$

Trick: Differences fall by 1 each time: 14, 13, 12, 11, 10.

Answer: (A) 127

Concept: Facing north, right = east end. Fix the person at an end first.

1. No one right of D ⇒ D at position 7. Three between D and S ⇒ S at 3.
2. Two between S and T ⇒ T at 6.
3. U, A adjacent (A right of U) with C third to left of U ⇒ U = 4, A = 5, C = 1; B gets 2.
4. Order: C B S U A T D. Between B and T: S, U, A = 3.

Trick: Start with “no one to the right/left” clue — it fixes an end seat.

Answer: (B) Three

Concept: Find the operations from the given sets and test each option.

1. $10\times 2=20$, $20+10=30$, $30\times 2=60$
2. $2\times 2=4$, $4+10=14$, $14\times 2=28$ ⇒ rule: $\times 2$, $+10$, $\times 2$
3. Option C: $5\times 2=10$, $10+10=20$, $20\times 2=40$ ✔

Trick: Check the middle step (+10) first — it eliminates options quickly.

Answer: (C) 5 - 10 - 20 - 40

Concept: Write the direct codes, then apply every condition that holds.

1. Direct: 9 = C, # = I, @ = F, 2 = Q, 7 = R.
2. Fourth element 2 is a number ⇒ coded γ.
3. First element 9 is a multiple of 3 ⇒ coded β.
4. Fifth element 7 is a number ⇒ condition 3 does not apply. Code: β I F γ R.

Trick: Check each condition separately; two conditions can apply together.

Answer: (D) βIFγR

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

1. LIKE → 5738 and TASK → 2819.
2. Common letter: K; common digit: 8.
3. So K = 8.

Trick: Common letter ↔ common digit (order may be jumbled).

Answer: (A) 8

Concept: Take first and second differences.

1. Differences: 4, 7, 12, 19, ?, ?
2. Second differences: 3, 5, 7, 9, 11 ⇒ next differences 28 and 39.
3. $46+28=74$; check $74+39=113$ ✔

Trick: If first differences are irregular, look for odd-number second differences.

Answer: (D) 74

Concept: Write the number and its ascending form one below the other.

1. Original: 7 4 1 6 3 8 5
2. Ascending: 1 3 4 5 6 7 8
3. Compare place by place: no position matches.

Trick: Just compare digit by digit — no calculation needed.

Answer: (A) None

Concept: Draw the least-overlap diagram and test each conclusion.

1. All tanks are figs and no eagle is a tank ⇒ the figs that are tanks are not eagles ⇒ Some figs are not eagles — (I) follows.
2. Some mugs are eagles, and no eagle is a tank ⇒ those mugs are not tanks ⇒ Some mugs are not tanks — (II) follows.
3. So both conclusions follow.

Trick: “Some A are B” + “No B is C” ⇒ “Some A are not C” always follows.

Answer: (B) Both conclusions (I) and (II) follow.

Concept: Replace letters and follow BODMAS.

1. $(110-32)÷2+9\times 4-21$
2. $=78÷2+36-21$
3. $=39+36-21=54$

Trick: Bracket first, then $÷$ and $\times$, then $+$ and $-$.

Answer: (C) 54