RRB NTPC Graduate CBT 1 2024 PYQs

Concept: A common root of both equations must satisfy the second equation.

1. $x^2-5x+6=0\Rightarrow (x-2)(x-3)=0$, roots $2,3$
2. If $x=2$: $8-2k+6=0\Rightarrow k=7$
3. If $x=3$: $18-3k+6=0\Rightarrow k=8$ (not an option)
4. So $k=7$

Trick: Factorise the easy equation first, then substitute its roots.

Answer: (C) $7$

Concept: Use $(a-b)(a+b)=a^2-b^2$ with numbers around 95.

1. $86\times 104=(95-9)(95+9)$
2. $={95}^2-9^2=9025-81=8944$

Trick: Or $86\times 104=8600+344=8944$; last digit $6\times 4=24$ → ends in 4.

Answer: (A) $8944$

Concept: Write the two time conditions and add them to eliminate Sameer's time.

1. Let Abhay's speed $=a$, Sameer's $=s$. $\frac{162}{a}-\frac{162}{s}=6$
2. Doubled speed: $\frac{162}{s}-\frac{162}{2a}=3$
3. Adding: $\frac{162}{a}-\frac{81}{a}=9\Rightarrow \frac{81}{a}=9$
4. $a=9\text{km/h}$ (Sameer: 12 h; Abhay: 18 h ✓)

Trick: Option check: $a=9$ → 18 h; doubled → 9 h; Sameer $18-6=12$ h and $12-9=3$ ✓.

Answer: (D) $9\text{km/h}$

Concept: In a trapezium, the segment joining the midpoints of the diagonals is parallel to the bases and equals half their difference.

1. $AB\parallel DC$, $AB=30$, $DC=86$
2. $EF=\frac{DC-AB}{2}=\frac{86-30}{2}$
3. $EF=28\text{cm}$ (BC and AD are not needed)

Trick: Midpoints of sides → half the sum; midpoints of diagonals → half the difference.

Answer: (A) $28$

Concept: Take tank capacity $=$ LCM of the times and add the hourly units.

1. Capacity $=\text{LCM}(5,6,12)=60$ units
2. m $=12$, n $=10$, p $=5$ units/hour
3. Together $=27$ units/hour
4. Time $=\frac{60}{27}=\frac{20}{9}=2\frac{2}{9}$ hours

Trick: $\frac{1}{5}+\frac{1}{6}+\frac{1}{12}=\frac{12+10+5}{60}=\frac{27}{60}$, invert it.

Answer: (D) $2\frac{2}{9}$ hours

Concept: Value of each coin group $=$ number of coins $\times$ coin value.

1. Let coins be $18k,12k,10k$
2. Value: $0.50\times 18k+0.25\times 12k+0.10\times 10k=9k+3k+k=13k$
3. $13k=130\Rightarrow k=10$
4. 10 paise coins $=10\times 10=100$

Trick: Value ratio $9:3:1$ → ₹130 splits as ₹90 : ₹30 : ₹10; ₹10 in 10-paise coins $=100$ coins.

Answer: (C) $100$

Concept: Sum $=$ average $\times$ count; remove the discarded numbers and divide by the new count.

1. Sum of 50 numbers $=50\times 34=1700$
2. After removing 34 and 22: $1700-56=1644$
3. New average $=\frac{1644}{48}=34.25$

Trick: Removing 34 (= average) changes nothing; removing 22 (12 below) raises average by $\frac{12}{48}=0.25$.

Answer: (C) $34.25$

Concept: Use $1+{\tan }^{2}A={\sec }^{2}A$ and $1+{\cot }^{2}A={\csc }^{2}A$.

1. Numerator $=3(1+{\tan }^{2}A)\cot A=3{\sec }^{2}A\cot A$
2. Denominator $={\csc }^{2}A\sec A$
3. Ratio $=\frac{3\sec A\cot A}{{\csc }^{2}A}=3·\frac{1}{\cos A}·\frac{\cos A}{\sin A}·{\sin }^{2}A$
4. $=3\sin A$

Trick: Check with $A={45}^{\circ }$: $\frac{6·1}{2\sqrt{2}}=\frac{3}{\sqrt{2}}=3\sin {45}^{\circ }$ ✓.

Answer: (C) $3\sin A$

Concept: Correct sum $=$ wrong sum $-$ wrong value $+$ correct value.

1. Wrong sum $=10\times 40=400$
2. Correct sum $=400-31+50=419$
3. Correct average $=\frac{419}{10}=41.9$

Trick: Average changes by $\frac{50-31}{10}=1.9$ → $40+1.9=41.9$.

Answer: (A) $41.9$

Concept: If $\frac{x}{11}=\frac{275}{x}$, cross-multiply to get $x^2$.

1. $x^2=11\times 275=3025$
2. $x=\sqrt{3025}=55$ (positive)

Trick: $3025$ ends in 25 → square of a number ending in 5: ${55}^2=3025$.

Answer: (A) $55$

Concept: Work backwards: $\text{CP}=\frac{\text{SP}}{1\pm r}$ at each sale.

1. Bharath sold at 44% loss: his CP $=\frac{4480}{0.56}=8000$.
2. Bharath's CP = Anand's SP at 25% profit: Anand's total cost $=\frac{8000}{1.25}=6400$.
3. Total cost includes repair: original cost $=6400-169=6231$.

Trick: Undo each step in reverse order: $÷0.56$, then $÷1.25$, then $-169$.

Answer: (B) ₹6,231

Concept: $\text{Sum}=\text{Average}\times \text{Count}$.

1. Sum of 5 numbers $=56\times 5=280$.
2. Sum of four given $=35+42+32+26=135$.
3. Fifth number $=280-135=145$.

Trick: Deviation method: deviations from 56 are $-21,-14,-24,-30$ (total $-89$), so fifth $=56+89=145$.

Answer: (A) 145

Concept: Empirical relation: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. Given $\text{Mode}-\text{Median}=20.8$.
2. From the formula, $\text{Mode}-\text{Median}=2\text{Median}-2\text{Mean}=2(\text{Median}-\text{Mean})$.
3. So $\text{Median}-\text{Mean}=\frac{20.8}{2}=10.4$.

Trick: $\text{Mode}-\text{Median}=2(\text{Median}-\text{Mean})$ – just halve the given gap.

Answer: (D) $10.4$

Concept: Surface area of a cuboid $=2(lb+bh+hl)$ and diagonal $=\sqrt{l^2+b^2+h^2}$. The surface area is always more than $2lb$.

1. $2lb=2\times 70\times 96=13440{\text{cm}}^2$, which is already far more than the given $1488{\text{cm}}^2$.
2. Solving anyway: $2(6720+96h+70h)=1488\Rightarrow 6720+166h=744\Rightarrow 166h=-5976\Rightarrow h=-36$.
3. A length cannot be negative, so no cuboid with these measurements exists and the diagonal cannot be found; none of the options is correct.
4. If the height had been $36\text{cm}$ (surface area $25392{\text{cm}}^2$), the diagonal would be $\sqrt{4900+9216+1296}=\sqrt{15412}$, option (D).

Note: Quick check: total surface area $>2lb$ always; test this before solving. 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: $\text{SP}=\text{CP}\times (1-\text{loss}\%)$, $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. At 40% of $x$, loss is 50%: $0.4x=0.5\text{CP}$.
2. $\text{CP}=0.8x$.
3. Selling at $x$: profit $\%=\frac{x-0.8x}{0.8x}\times 100=25\%$.

Trick: Take $x=100$: SP $=40$ at 50% loss → CP $=80$; selling at 100 gives $\frac{20}{80}=25\%$.

Answer: (A) $25\%$

Concept: $(a+b{)}^2-(a-b{)}^2=4ab$.

1. $(a+b{)}^2={32}^2=1024$.
2. $4ab=1024-372=652$.
3. $ab=\frac{652}{4}=163$.

Trick: Remember $4ab=(a+b{)}^2-(a-b{)}^2$ – no need to find $a$ and $b$ separately.

Answer: (B) 163

Concept: For any point P inside rectangle ABCD: $P{A}^2+P{C}^2=P{B}^2+P{D}^2$ (opposite corners pair up).

1. $P{A}^2+P{C}^2={94}^2+{67}^2=8836+4489=13325$.
2. $P{D}^2=13325-P{B}^2=13325-{61}^2=13325-3721=9604$.
3. $PD=\sqrt{9604}=98\text{cm}$.

Trick: Diagonally opposite corners go together: A with C, B with D.

Answer: (C) $98$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$.

1. Amount $=3P$, so SI $=3P-P=2P$.
2. $2P=\frac{P\times R\times 8}{100}$.
3. $R=\frac{200}{8}=25\%$.

Trick: Becomes $n$ times in $T$ years: $R=\frac{(n-1)\times 100}{T}=\frac{2\times 100}{8}=25$.

Answer: (C) $25$

Concept: Total investment $=\frac{\text{Profit}}{14\%}$; split it in the investment ratio.

1. Total investment $=\frac{36400\times 100}{14}=260000$.
2. Ratio $5:6:13$, total parts $=24$.
3. Naveen $=\frac{6}{24}\times 260000=65000$.

Trick: $\frac{6}{24}=\frac{1}{4}$ – Naveen has exactly one-fourth of the total.

Answer: (A) ₹65,000

Concept: $P_n=P{(1+\frac{r}{100})}^n$.

1. 1956 to 1976 = two decades, 20% each.
2. $P=207300\times 1.2\times 1.2=207300\times 1.44$.
3. $=298512$.

Trick: Two successive 20% rises = $20+20+\frac{20\times 20}{100}=44\%$ net increase.

Answer: (B) 2,98,512