RRB NTPC Graduate CBT 1 2024 PYQs

Concept: $SP=CP\times (1+\text{markup})\times (1-\text{discount})$.

1. $SP=CP\times 1.64\times 0.5=0.82CP$.
2. $0.82CP=59860$.
3. $CP=\frac{59860}{0.82}=73000$.

Trick: Only the round option 73,000 gives $0.82\times 73000=59860$.

Answer: (D) 73,000

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

1. $(a-b{)}^2=83-12=71$; since $a>b$, $a-b=\sqrt{71}$.
2. $(a+b{)}^2=83+12=95$, so $a+b=\sqrt{95}$ (here $a,b$ are positive since $ab>0$).
3. $\frac{a-b}{a+b}=\frac{\sqrt{71}}{\sqrt{95}}=\sqrt{\frac{71}{95}}$.

Trick: ${\left(\frac{a-b}{a+b}\right)}^2=\frac{a^2+b^2-2ab}{a^2+b^2+2ab}$ — then take the root.

Answer: (B) $\sqrt{\frac{71}{95}}$

Concept: Sum = average × count.

1. Total of 8 numbers $=8\times 15=120$.
2. Total of 5 numbers $=5\times 18=90$.
3. Remaining 3: $\frac{120-90}{3}=10$.

Trick: Deviation method: 5 numbers are $+3$ each $=+15$, so other 3 are $-5$ each → $15-5=10$.

Answer: (D) 10

Concept: TSA of cube $=6a^2$, volume $=a^3$.

1. $6a^2=216\Rightarrow a^2=36\Rightarrow a=6$ cm.
2. Volume $=6^3=216{\text{cm}}^3$.

Trick: For a cube with side 6, TSA and volume are numerically equal (216).

Answer: (C) $216{\text{cm}}^3$

Concept: Write everything in powers of 2 and 7.

1. $8^2=64$, $7^3=343$; $\sqrt{64}=8$, $\sqrt{2401}=49$.
2. $\frac{64\times 343}{8\times 49}=\frac{64}{8}\times \frac{343}{49}$.
3. $=8\times 7=56$.

Trick: $2401=7^4$, so $\sqrt{2401}=7^2$; then $\frac{8^2}{8}\times \frac{7^3}{7^2}=8\times 7$.

Answer: (B) 56

Concept: In a quadrilateral, if bisectors of $\angle C$ and $\angle D$ meet at O, then $\angle DOC=\frac{\angle A+\angle B}{2}$.

1. $\angle C+\angle D={360}^{\circ }-({53}^{\circ }+{46}^{\circ })={261}^{\circ }$.
2. In $△DOC$: $\angle DOC={180}^{\circ }-\frac{{261}^{\circ }}{2}={180}^{\circ }-{130.5}^{\circ }$.
3. $\angle DOC={49.5}^{\circ }$.

Trick: Directly: $\frac{{53}^{\circ }+{46}^{\circ }}{2}={49.5}^{\circ }$.

Answer: (C) ${49.5}^{\circ }$

Concept: If girls : boys = 1 : 5, then boys = 5 × girls.

1. $\frac{\text{Girls}}{\text{Boys}}=\frac{1}{5}$
2. $\text{Boys}=5\times 201=1005$

Trick: 1 part = 201, so 5 parts = 1005.

Answer: (C) 1005

Concept: Median = middle value after arranging data in order; for $n$ odd it is the ${\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending order: 11, 12, 38, 38, 49, 53, 55, 60, 70, 93, 98.
2. $n=11$, so median = ${\left(\frac{11+1}{2}\right)}^{\text{th}}=6^{\text{th}}$ value = 53.

Trick: Always sort first – the unsorted middle (98) is a trap.

Answer: (B) 53

Concept: Production ∝ (manpower × hours). If one factor falls, the other must rise to keep the product constant.

1. Manpower effect: production becomes $80\%$, i.e. factor $=\frac{4}{5}$.
2. To restore, hours must be multiplied by $\frac{5}{4}$.
3. Increase $=\frac{5}{4}-1=\frac{1}{4}=25\%$.

Trick: Decrease $x\%$ → required increase $=\frac{x}{100-x}\times 100=\frac{20}{80}\times 100=25\%$.

Answer: (C) $25\%$

Concept: Centroid of a triangle (meeting point of medians) $G=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$.

1. $x=\frac{1+7+10}{3}=\frac{18}{3}=6$
2. $y=\frac{4+8+12}{3}=\frac{24}{3}=8$
3. $G=(6,8)$

Trick: Centroid = simple average of the three x's and the three y's.

Answer: (B) $(6,8)$

Concept: Ladder, wall and ground form a right triangle; the ladder is the hypotenuse, height is opposite to the 30° angle.

1. $\sin {30}^{\circ }=\frac{h}{8}$
2. $h=8\times \frac{1}{2}=4\text{m}$

Trick: In a 30°-60°-90° triangle, side opposite 30° = half of the hypotenuse.

Answer: (A) $4\text{m}$

Concept: Volume of a hemisphere $=\frac{2}{3}\pi r^3$.

1. $r^3={24}^3=13824$
2. $V=\frac{2}{3}\pi \times 13824=2\pi \times 4608=9216\pi {\text{cm}}^3$

Trick: Divide by 3 first: $\frac{13824}{3}=4608$, then double.

Answer: (C) $9216\pi$

Concept: Opposite directions: relative speed = sum of speeds; distance covered while crossing = sum of the lengths.

1. Relative speed $=99+72=171\text{km/h}=171\times \frac{5}{18}=47.5\text{m/s}$.
2. $x+1.5x=47.5\times 46=2185\Rightarrow x=874\text{m}$; length of B $=1.5\times 874=1311\text{m}$.
3. Speed of B $=72\times \frac{5}{18}=20\text{m/s}$; in 82 s it covers $20\times 82=1640\text{m}$.
4. Bridge $=1640-1311=329\text{m}$.

Trick: Train crossing a bridge covers (train + bridge) length.

Answer: (D) 329

Concept: $(a-b{)}^2=a^2-2ab+b^2$ and $(a+b{)}^2=a^2+2ab+b^2$.

1. $(5z-9y{)}^2=25z^2-90yz+81y^2$
2. $(9z+5y{)}^2=81z^2+90yz+25y^2$
3. Sum $=106z^2+106y^2$ (the $yz$ terms cancel).
4. $106z^2+106y^2-81z^2=25z^2+106y^2$

Trick: $(ax-by{)}^2+(bx+ay{)}^2=(a^2+b^2)(x^2+y^2)=106(z^2+y^2)$.

Answer: (C) $25z^2+106y^2$

Concept: If numbers are in the ratio $a:b:c$ with HCF of the ratio terms 1, the numbers are $ak,bk,ck$; HCF $=k$ and LCM $=\text{LCM}(a,b,c)\times k$.

1. Numbers $=8k,2k,k$. $\text{LCM}(8,2,1)=8$, so LCM $=8k$.
2. $8k=7552\Rightarrow k=944$, so HCF $=944$.
3. $944$ is not among the options (502, 472, 486, 456), so no option is correct.
4. If the ratio had been $16:2:1$, LCM $=16k=7552$ gives $k=472$, option (B).

Note: For a ratio like $8:2:1$, the largest number is itself the LCM. 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: For $a{x}^2+bx+c=0$: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

1. $a=9,b=-9,c=1$
2. $D=b^2-4ac=81-36=45$
3. $x=\frac{9\pm \sqrt{45}}{2\times 9}=\frac{9\pm \sqrt{45}}{18}$

Trick: Check: sum of roots $=\frac{18}{18}=1=-\frac{b}{a}$ ✓; options with denominator 9 or 4.5 fail this.

Answer: (A) $\frac{9+\sqrt{45}}{18}$ & $\frac{9-\sqrt{45}}{18}$

Concept: A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

1. In 32043p88, the last two digits are 88.
2. $88÷4=22$ ✓ – divisible for every value of $p$.
3. So $p$ can be any digit 0–9; maximum $p=9$.

Trick: If $p$ is not among the last two digits, it is free – take the largest digit 9.

Answer: (D) 9

Concept: Average speed $=\frac{\text{Total distance}}{\text{Total time}}$ (never the average of speeds).

1. Times: $\frac{204}{68}=3\text{h}$, $\frac{424}{53}=8\text{h}$, $\frac{366}{61}=6\text{h}$; total $=17\text{h}$.
2. Distance $=204+424+366=994\text{km}$.
3. Average speed $=\frac{994}{17}\approx 58.47\text{km/h}$.

Trick: All divisions are exact (3, 8, 6) – a sign you are on the right track.

Answer: (C) 58.47

Concept: With 30% gain, SP $=\frac{130}{100}\times$ CP, so profit $=\frac{30}{130}\times$ SP.

1. CP = ₹ $\frac{2938\times 100}{130}=2260$
2. Profit = ₹ $2938-2260=678$

Trick: Profit $=\frac{3}{13}\times 2938=3\times 226=678$.

Answer: (C) 678

Concept: Successive discounts multiply the remaining fractions: net price $=P(1-d_1)(1-d_2)(1-d_3)$.

1. Remaining fraction $=0.71\times 0.72\times 0.87$.
2. $0.71\times 0.72=0.5112$; $0.5112\times 0.87=0.444744$.
3. Single discount $=(1-0.444744)\times 100=55.5256\%\approx 55.53\%$.

Trick: Two at a time: $29\%,28\%\to 29+28-\frac{29\times 28}{100}=48.88\%$; then with $13\%$: $48.88+13-\frac{48.88\times 13}{100}=55.53\%$.

Answer: (C) $55.53\%$