RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Write the condition as an equation and form the ratio.

1. $\frac{56}{100}a=\frac{5}{9}b\Rightarrow \frac{a}{b}=\frac{5}{9}\times \frac{100}{56}$
2. $=\frac{500}{504}=\frac{125}{126}$
3. So $a:b=125:126$

Trick: $a\times 56\%=b\times \frac{5}{9}\Rightarrow a:b=\frac{5}{9}:\frac{56}{100}$ – cross-multiply.

Answer: (D) $125:126$

Concept: BODMAS: Brackets, then Multiplication, then Addition/Subtraction from left to right.

1. Bracket: $5+2=7$
2. Multiply: $4\times 7=28$
3. $81-28+54=107$

Trick: Never subtract before multiplying: $81-4$ first would give a wrong value.

Answer: (C) $107$

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

1. Total distance $=60+80+100=240\text{km}$
2. Total time $=3$ h
3. Average speed $=\frac{240}{3}=80\text{km/h}$

Trick: Equal time intervals → average speed = simple average of the speeds.

Answer: (D) $80\text{km/h}$

Concept: Find the value of one ratio part from the given difference.

1. C − B $=11-7=4$ parts $=196$ (₹)
2. 1 part $=\frac{196}{4}=49$ (₹)
3. A $=14\times 49=686$ (₹)

Trick: A is $\frac{14}{4}$ of the difference: $196\times 3.5=686$.

Answer: (B) ₹$686$

Concept: Convert a ratio of fractions into whole numbers by multiplying each term by the LCM of the denominators.

1. Ratio $=\frac{17}{2}:\frac{14}{6}:\frac{2}{5}=\frac{17}{2}:\frac{7}{3}:\frac{2}{5}$.
2. LCM of 2, 3, 5 is 30: $\frac{17}{2}\times 30=255,\frac{7}{3}\times 30=70,\frac{2}{5}\times 30=12$.
3. Sum $=255+70+12=337$ parts; one part $=\frac{674}{337}=2$.
4. Third monkey $=12\times 2=24$.

Trick: 674 is exactly $2\times 337$, so each part is 2 bananas.

Answer: (A) 24

Concept: If divisor − remainder is the same for all divisors, the number = k × LCM − (that common difference).

1. $32-4=28,41-13=28,48-20=28$, so $N+28$ is divisible by 32, 41 and 48.
2. $\text{LCM}(32,41,48)=96\times 41=3936$, so $N=3936k-28$.
3. $k=1$: $3908$ (not divisible by 53). $k=2$: $7872-28=7844=53\times 148$ ✓.
4. $N=7844$, which lies between 7834 and 7854.

Trick: Common 'shortfall' 28 → subtract it from multiples of the LCM, then test the extra condition.

Answer: (D) 7834 and 7854

Concept: Use $\sec A-\tan A=\frac{1-\sin A}{\cos A}$ and $1-{\sin }^{2}A={\cos }^{2}A$.

1. $\sec A(1+\sin A)(\sec A-\tan A)=\frac{1}{\cos A}\times (1+\sin A)\times \frac{1-\sin A}{\cos A}$.
2. $=\frac{1-{\sin }^{2}A}{{\cos }^{2}A}=\frac{{\cos }^{2}A}{{\cos }^{2}A}=1$.
3. Expression $=7\times 1-4=3$.

Trick: Put $A=0^{\circ }$: $7\times 1\times 1\times 1-4=3$.

Answer: (B) 3

Concept: Volume of cuboid $=l\times b\times h$.

1. $7524=33\times 4\times h=132h$.
2. $h=\frac{7524}{132}=57\text{cm}$.

Trick: Unit digit check: $132\times 57$ ends in 4 ($2\times 7=14$) ✓.

Answer: (B) 57

Concept: When a transversal cuts two parallel lines, corresponding angles are equal and angles on a straight line add up to ${180}^{\circ }$.

1. $\angle QND={50}^{\circ }$ (given).
2. $\angle QMB$ and $\angle QND$ are corresponding angles ($AB\parallel CD$), so $\angle QMB={50}^{\circ }$.
3. $\angle PMB$ and $\angle QMB$ form a linear pair: $\angle PMB={180}^{\circ }-{50}^{\circ }={130}^{\circ }$.

Trick: The angle at N is acute (${50}^{\circ }$); $\angle PMB$ opens on the other side of the transversal, so it is obtuse $={180}^{\circ }-{50}^{\circ }$.

Answer: (C) ${130}^{\circ }$

Concept: Successive selling: final price = original cost × product of the multipliers.

1. Loss $60\%$ → multiplier $0.4$; profit $25\%$ → multiplier $1.25$.
2. C's price $=x\times 0.4\times 1.25=0.5x$.
3. $0.5x=42206\Rightarrow x=84412$.

Trick: $0.4\times 1.25=\frac{1}{2}$, so A's cost is just double C's price.

Answer: (A) ₹84,412

Concept: Net filling rate = filling rate − emptying rate.

1. Net rate $=\frac{1}{27}-\frac{1}{28}=\frac{28-27}{756}=\frac{1}{756}$ tank per hour.
2. Full tank takes 756 hours.
3. One-fourth tank $=\frac{756}{4}=189$ hours.

Trick: For consecutive numbers $n,n+1$: net time $=n(n+1)$; here $27\times 28=756$.

Answer: (D) 189

Concept: Successive percentage increases are applied one after another as multipliers.

1. After year 1: $120000\times 1.10=132000$.
2. After year 2: $132000\times 1.30=171600$.

Trick: Net change $=10+30+\frac{10\times 30}{100}=43\%$; $120000\times 1.43=171600$.

Answer: (C) 1,71,600

Concept: If profit % at one SP equals loss % at another SP, the cost price is the average of the two SPs.

1. $\frac{863-CP}{CP}=\frac{CP-433}{CP}\Rightarrow CP=\frac{863+433}{2}=648$.
2. SP for $25\%$ profit $=648\times \frac{125}{100}=810$.

Trick: Equal profit% and loss% → $CP=\frac{S{P}_1+S{P}_2}{2}$.

Answer: (B) ₹810

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

1. $\text{Mode}=3\times 100-2\times 76$.
2. $=300-152=148$.

Trick: Remember '3 Median minus 2 Mean'.

Answer: (A) 148

Concept: $\frac{x+1}{x}=1+\frac{1}{x}$, so the given condition fixes $\frac{1}{x}$ directly; also $\frac{x^2+1}{x^2}=1+\frac{1}{x^2}$.

1. $\frac{x+1}{x}=11\Rightarrow 1+\frac{1}{x}=11\Rightarrow \frac{1}{x}=10$.
2. $\frac{x^2+1}{x^2}=1+{\left(\frac{1}{x}\right)}^2=1+{10}^2=1+100=101$.
3. $101$ is not among the options (119, 129, 114, 112), so no option is correct.
4. If the condition had been $x+\frac{1}{x}=11$, then $x^2+\frac{1}{x^2}={11}^2-2=119$, option (A).

Note: Read the fraction carefully: $\frac{x+1}{x}$ is not the same as $x+\frac{1}{x}$. 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: Form two linear equations from the statements and solve.

1. $9A+8B=29$ and $2B-2A=20\Rightarrow B=A+10$.
2. $9A+8(A+10)=29\Rightarrow 17A=-51\Rightarrow A=-3,B=7$.
3. Present sum $A+B=4$.
4. Sum after 6 years $=4+6+6=16$ years.

Trick: Sum of two ages after n years = present sum + 2n.

Answer: (C) 16

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

1. $R=\frac{500\times 100}{1250\times 10}$.
2. $=\frac{50000}{12500}=4\%$.

Trick: Interest is $\frac{500}{1250}=40\%$ in 10 years → $4\%$ per year.

Answer: (B) $4\%$

Concept: Overtaking in the same direction: time = (sum of lengths) ÷ (difference of speeds).

1. Relative speed $=110-100=10\text{km/h}=10\times \frac{5}{18}=\frac{25}{9}\text{m/s}$.
2. Distance $=100+500=600\text{m}$.
3. Time $=600\times \frac{9}{25}=216\text{s}=\frac{216}{60}=3.6$ minutes.

Trick: In km and hours: $\frac{0.6}{10}\text{h}=0.06\times 60=3.6$ min.

Answer: (C) 3.6

Concept: Express both conditions in terms of Burhan's time and subtract.

1. Let Anirudh's speed $=a$, Burhan's time $=t$. Then $\frac{90}{a}=t+3$ and $\frac{90}{2a}=t-2$.
2. Subtract: $\frac{90}{a}-\frac{90}{2a}=5\Rightarrow \frac{45}{a}=5$.
3. $a=9\text{km/h}$ (check: 10 h normally, 5 h at double speed; Burhan 7 h ✓).

Trick: Doubling speed halves the time: half of Anirudh's time = 3 + 2 = 5 h, so time = 10 h, speed = 9 km/h.

Answer: (C) 9 km/hr

Concept: ${(x-\frac{1}{x})}^2=x^2+\frac{1}{x^2}-2$ and $x^3-\frac{1}{x^3}={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})$.

1. ${(x-\frac{1}{x})}^2=83-2=81\Rightarrow x-\frac{1}{x}=9$.
2. $x^3-\frac{1}{x^3}=9^3+3\times 9=729+27=756$.

Trick: If $x-\frac{1}{x}=k$, then $x^3-\frac{1}{x^3}=k^3+3k$.

Answer: (D) 756