RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Number of identical pipes and time are inversely proportional: $N_1{T}_1=N_2{T}_2$.

1. $43\times 40=9\times T$
2. $T=\frac{1720}{9}=191.11$ minutes

Trick: More pipes → less time; just multiply and divide: $\frac{43\times 40}{9}$.

Answer: (A) $191.11$

Concept: Before taking a ratio, both quantities must be in the same unit.

1. $7\text{ days}=7\times 24=168\text{ hours}$
2. $76:168$; divide by 4: $19:42$

Trick: Always convert days to hours first: 1 day = 24 h.

Answer: (B) $19:42$

Concept: Simplify nested roots from the innermost root outward.

1. $\sqrt{900}=30$
2. $\sqrt{139+30}=\sqrt{169}=13$
3. $\sqrt{212+13}=\sqrt{225}=15$

Trick: Each step should give a perfect square: 169, then 225.

Answer: (B) 15

Concept: Monthly saving = salary − expenses; the birthday month adds no saving.

1. Monthly saving = $12000-4000-1500=6500$
2. Savings are made in only $11$ months.
3. Yearly saving = $6500\times 11=71500$

Trick: Do not take 12 months – one month's saving is fully spent.

Answer: (C) ₹71,500

Concept: In the right triangle formed, $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. $\tan {45}^{\circ }=\frac{h}{27.6}$
2. $\tan {45}^{\circ }=1\Rightarrow h=27.6\text{m}$

Trick: At ${45}^{\circ }$, height = distance from the foot.

Answer: (B) $27.6\text{m}$

Concept: For $a{x}^2+bx+c=0$, discriminant $D=b^2-4ac$.

1. Here $a=1,b=3,c=k$: $D=9-4k$
2. $9-4k=9\Rightarrow k=0$

Trick: $D=b^2$ already, so $4ac$ must be $0$ → $k=0$.

Answer: (D) 0

Concept: If numbers are in ratio $a:b:c$ (co-prime parts), they are $ax,bx,cx$ with HCF $=x$ and LCM $=\text{LCM}(a,b,c)\times x$.

1. Numbers: $x,9x,7x$
2. $\text{LCM}=63x=6930$
3. $x=\frac{6930}{63}=110$ = HCF

Trick: HCF = $\frac{\text{LCM}}{\text{LCM of ratio terms}}=\frac{6930}{63}$.

Answer: (B) 110

Concept: For a pure recurring decimal, $0.\bar{a}=\frac{a}{9}$.

1. $1.222\dots =1+0.\bar{2}=1+\frac{2}{9}$
2. $=\frac{9+2}{9}=\frac{11}{9}$

Trick: $1.\bar{2}=\frac{12-1}{9}=\frac{11}{9}$ (whole number − non-repeating part, over 9).

Answer: (A) $\frac{11}{9}$

Concept: $\text{Average}=\frac{\text{Sum of observations}}{\text{Number of observations}}$

1. Sum $=5+8+6+7+9=35$
2. Average $=\frac{35}{5}=7\text{km}$

Trick: 5, 6, 7, 8, 9 are consecutive → average = middle value 7.

Answer: (C) $7\text{km}$

Concept: Use $(a-b)(a+b)=a^2-b^2$.

1. $891\times 909=(900-9)(900+9)$
2. $={900}^2-9^2=810000-81=809919$

Trick: Both numbers are equally far (9) from 900 → square of 900 minus square of 9. Last digits: $1\times 9=9$ ends in 9.

Answer: (A) 809919

Concept: $\text{SP}=\text{CP}\times \frac{100\pm \%}{100}$.

1. Loss $78\%$ ⇒ SP $=22\%$ of CP.
2. $\text{CP}=\frac{44\times 100}{22}=$ ₹200.
3. For $78\%$ profit: $\text{SP}=200\times \frac{178}{100}=$ ₹356.

Trick: $\text{New SP}=44\times \frac{178}{22}=2\times 178=356$.

Answer: (A) ₹356

Concept: Half-yearly compounding: rate per period $=\frac{R}{2}\%$, periods $=2\times$ years.

1. 1 year 6 months $=3$ half-years.
2. $\frac{19683}{15625}={\left(\frac{27}{25}\right)}^3$.
3. So $1+\frac{r}{100}=\frac{27}{25}$ ⇒ $r=8\%$ per half-year.
4. Annual rate $=2\times 8=16\%$.

Trick: $19683={27}^3$ and $15625={25}^3$ – spot the cubes.

Answer: (C) $16\%$

Concept: Stoppage time = time with stoppages − time without stoppages.

1. Without stops: $\frac{630}{70}=9$ h.
2. With stops: $\frac{630}{30}=21$ h.
3. Stoppage time $=21-9=12$ h.

Trick: Per hour stoppage $=\frac{70-30}{70}\times 60\approx 34$ min; over 21 h this gives 12 h.

Answer: (B) 12

Concept: Convert each ratio to a decimal (or compare with $\frac{1}{2}$).

1. $\frac{17}{42}\approx 0.405$
2. $\frac{20}{36}=\frac{5}{9}\approx 0.556$
3. $\frac{21}{45}\approx 0.467$
4. $\frac{25}{50}=0.5$

Trick: Only $\frac{20}{36}$ has numerator more than half the denominator, so it alone exceeds $\frac{1}{2}$.

Answer: (B) $20:36$

Concept: Mean proportional of $a$ and $b$ is $\sqrt{ab}$.

1. $ab=(21+16\sqrt{2})(16-7\sqrt{2})$.
2. $=336-147\sqrt{2}+256\sqrt{2}-16\times 7\times 2$.
3. $=336-224+109\sqrt{2}=112+109\sqrt{2}$.
4. Mean proportional $=\sqrt{112+109\sqrt{2}}$.

Trick: Rational part: $336-224=112$; surd part: $256-147=109$.

Answer: (B) $\sqrt{112+109\sqrt{2}}$

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

1. $\text{Mode}-\text{Median}=2\text{Median}-2\text{Mean}$.
2. $45.98=2(\text{Median}-\text{Mean})$.
3. $\text{Median}-\text{Mean}=\frac{45.98}{2}=22.99$.

Trick: Mode − Median is always twice Median − Mean.

Answer: (B) 22.99

Concept: Leak's rate = pump's rate − net rate.

1. Pump: $\frac{1}{3}$ per hour; net: $\frac{1}{21/4}=\frac{4}{21}$ per hour.
2. Leak $=\frac{1}{3}-\frac{4}{21}=\frac{7-4}{21}=\frac{3}{21}=\frac{1}{7}$.
3. Leak empties the tank in 7 hours.

Trick: Leak time $=\frac{ab}{b-a}=\frac{3\times 5.25}{5.25-3}=\frac{15.75}{2.25}=7$.

Answer: (B) 7 hours

Concept: Downstream speed $=b+s$, upstream speed $=b-s$.

1. Downstream $=9+5=14\text{km/h}$; upstream $=9-5=4\text{km/h}$.
2. Time down $=\frac{75.6}{14}=5.4$ h; time up $=\frac{75.6}{4}=18.9$ h.
3. Total $=5.4+18.9=24.3$ h.

Trick: Total time $=\frac{2bd}{b^2-s^2}=\frac{2\times 9\times 75.6}{81-25}=\frac{1360.8}{56}=24.3$.

Answer: (A) 24.3

Concept: $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$.

1. $\frac{0.011025}{0.001225}=\frac{11025}{1225}=9$.
2. $\sqrt{9}=3$.
3. Check: $\sqrt{0.011025}=0.105$, $\sqrt{0.001225}=0.035$, and $\frac{0.105}{0.035}=3$.

Trick: Equal decimal places cancel – divide the plain numbers.

Answer: (C) 3

Concept: $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. $\tan {30}^{\circ }=\frac{50}{PB}$.
2. $\frac{1}{\sqrt{3}}=\frac{50}{PB}$ ⇒ $PB=50\sqrt{3}$ m.

Trick: At ${30}^{\circ }$, base $=\sqrt{3}\times$ height; at ${60}^{\circ }$, base $=\frac{\text{height}}{\sqrt{3}}$.

Answer: (A) $50\sqrt{3}$ m