RRB NTPC Graduate CBT 1 2024 PYQs

Concept: B’s one-day work $=\frac{1}{18}-\frac{1}{24}$.

1. $\frac{1}{18}-\frac{1}{24}=\frac{4-3}{72}=\frac{1}{72}$ → B alone takes $72$ days.
2. One-fourth work $=\frac{72}{4}=18$ days.

Trick: Total work $=\text{LCM}(18,24)=72$ units; A+B $=4$, A $=3$, so B $=1$ unit/day.

Answer: (A) 18

Concept: Downstream speed $=b+s$, upstream speed $=b-s$; time $=\frac{\text{distance}}{\text{speed}}$.

1. Downstream $=17+16=33\text{km/h}$; upstream $=17-16=1\text{km/h}$.
2. Time $=\frac{26.4}{33}+\frac{26.4}{1}=0.8+26.4=27.2$ hours.

Trick: When $b-s$ is tiny, the upstream trip dominates – the answer must exceed $26.4$.

Answer: (D) 27.2

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

1. $3\cot A=2\Rightarrow \cot A=\frac{2}{3}$, $\tan A=\frac{3}{2}$.
2. $\sqrt{\frac{3{\cot }^{2}A}{{\tan }^{2}A}}=\sqrt{3{\cot }^{4}A}=\sqrt{3}{\cot }^{2}A$.
3. $=\sqrt{3}\times \frac{4}{9}=\frac{4\sqrt{3}}{9}$.

Trick: $\frac{{\cot }^{2}A}{{\tan }^{2}A}={\cot }^{4}A$ – the root then just gives ${\cot }^{2}A$.

Answer: (C) $\frac{4\sqrt{3}}{9}$

Concept: Effective price $=MP\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})(1-\frac{d_3}{100})$.

1. $0.92\times 0.83\times 0.83=0.92\times 0.6889$.
2. $=0.633788\approx 63.38\%$ of marked price.

Trick: Discounts cannot simply be added ($42\%$ would give $58\%$) – always multiply the remaining fractions.

Answer: (C) $63.38\%$

Concept: Let a chair cost $c$ and a table $t$; solve the two equations.

1. $15c+3t=7800$ … (i), $14c+6t=9200$ … (ii).
2. $2\times$(i) $-$ (ii): $30c-14c=15600-9200\Rightarrow 16c=6400\Rightarrow c=400$.
3. $3t=7800-6000=1800\Rightarrow t=600$.
4. $17c+18t=6800+10800=17600$.

Trick: Doubling (i) makes the table terms equal ($6t$), so subtract directly.

Answer: (C) ₹17,600

Concept: Expenditure $=(100-\text{saving}\%)$ of income.

1. Expenditure $=80\%$ of $7915$.
2. $=\frac{4}{5}\times 7915=4\times 1583=6332$.

Trick: $80\%=\frac{4}{5}$: divide by 5, multiply by 4.

Answer: (B) ₹6,332

Concept: Speed in still water $=\frac{\text{downstream}+\text{upstream}}{2}$.

1. Downstream speed $=\frac{91}{13}=7\text{km/h}$.
2. Upstream speed $=\frac{39}{13}=3\text{km/h}$.
3. Still-water speed $=\frac{7+3}{2}=5\text{km/h}$.

Trick: Stream speed $=\frac{7-3}{2}=2$ – useful if asked next.

Answer: (A) 5

Concept: Integer roots need the discriminant $D=b^2-4ac$ to be a perfect square.

1. $D=(a-1{)}^2-4(4a+7)=a^2-18a-27=(a-9{)}^2-108$.
2. Let $D=k^2$: $(a-9-k)(a-9+k)=108$; both factors even → $(2,54)$ or $(6,18)$.
3. $(6,18)$: $a-9=12\Rightarrow a=21$; $(2,54)$: $a=37$. Least odd positive $a=21$.
4. Check: $x^2-20x+91=(x-7)(x-13)$ ✔.

Trick: Option test: put each option in $(a-9{)}^2-108$; only $a=21$ gives $144-108=36$, a perfect square.

Answer: (C) 21

Concept: Rotating a rectangle about its length makes length $=h$ and breadth $=r$; $\frac{V}{CSA}=\frac{\pi r^{2}h}{2\pi rh}=\frac{r}{2}$.

1. $\frac{r}{2}=\frac{4851}{462}=10.5\Rightarrow r=21\text{cm}$.
2. $2\times \frac{22}{7}\times 21\times h=462\Rightarrow 132h=462\Rightarrow h=3.5\text{cm}$.
3. Perimeter $=2(21+3.5)=49\text{cm}$.

Trick: Divide volume by CSA to get $\frac{r}{2}$ at once.

Answer: (B) 49

Concept: The perpendicular from the centre bisects a chord: $r^2=(\text{half chord}{)}^2+(\text{distance}{)}^2$.

1. Let AB be at distance $d$ from O; CD (same side) is at $d+4$.
2. $r^2=7^2+d^2=3^2+(d+4{)}^2\Rightarrow 49=25+8d\Rightarrow d=3$.
3. $r^2=49+9=58\Rightarrow r=\sqrt{58}$; diameter $=2\sqrt{58}$.

Trick: Longer chord is nearer the centre, so on the same side the shorter chord is at $d+4$.

Answer: (B) $2\sqrt{58}$

Concept: If numbers are $x,11x,19x$ with $1,11,19$ pairwise co-prime, then HCF $=x$ and LCM $=1\times 11\times 19\times x$.

1. LCM $=209x$.
2. $209x=8778\Rightarrow x=\frac{8778}{209}=42$.
3. So HCF $=42$ (numbers 42, 462, 798).

Trick: For co-prime ratio terms, HCF $=\frac{\text{LCM}}{\text{product of ratio terms}}$.

Answer: (A) 42

Concept: Mode is the observation that occurs most often.

1. Frequencies: 27 occurs 4 times; 20, 24, 32 occur twice each; the rest once.
2. Highest frequency is 4, for 27.

Trick: Just tally repeats — no calculation needed.

Answer: (C) 27

Concept: For equal distances at speeds $a,b,c$: average speed $=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$.

1. $\frac{1}{31}+\frac{1}{4}+\frac{1}{34}\approx 0.03226+0.25+0.02941=0.31167$.
2. Average speed $=\frac{3}{0.31167}\approx 9.63$.
3. Rounded to one decimal: $9.6\text{km/h}$.

Trick: The slow 4 km/h stretch dominates: time on it alone is $\frac{1}{12}$ of the journey per km-unit, so the answer must be a little under $3\times 4=12$ — only 9.6 fits sensibly.

Answer: (D) $9.6$

Concept: If $DE\parallel AC$, then $△BDE~△BAC$ and the ratio of areas equals the square of the ratio of corresponding sides.

1. $BA=BD+AD=10+17=27$.
2. $\frac{\text{ar}(BDE)}{\text{ar}(BAC)}={\left(\frac{10}{27}\right)}^2=\frac{100}{729}$.
3. Trapezium $ADEC=729-100=629$ parts.
4. Required ratio $=100:629$.

Trick: Square the side ratio, then subtract: ${10}^2:({27}^2-{10}^2)$.

Answer: (D) $100:629$

Concept: Amount $A=P(1+\frac{RT}{100})$ and $SI=\frac{PRT}{100}$.

1. 3 years at 20% → SI $=60\%$ of P, so $A=1.6P=8640$.
2. $P=\frac{8640}{1.6}=5400$.
3. SI for 2 years $=5400\times \frac{40}{100}=2160$.

Trick: $160\%\to 8640$, so $40\%\to \frac{8640}{4}=2160$.

Answer: (B) ₹2,160

Concept: Total cost = number of dolls × rate per doll.

1. Total $=(4x-3y)(x+y)$.
2. $=4x^2+4xy-3xy-3y^2$.
3. $=4x^2+xy-3y^2$.

Trick: Check with $x=y=1$: dolls $=1$, rate ₹2, total ₹2; only $4+1-3=2$ matches.

Answer: (B) ₹$(4x^2+xy-3y^2)$

Concept: $\text{CP}=\frac{\text{SP}\times 100}{100-\text{Loss}\%}$ and new $\text{SP}=\text{CP}\times \frac{100+\text{Gain}\%}{100}$.

1. $\text{CP}=\frac{225\times 100}{90}=250$.
2. For 24% gain: $\text{SP}=250\times 1.24=310$.

Trick: Directly: $225\times \frac{124}{90}=310$.

Answer: (A) ₹310

Concept: Volume of a sphere $V=\frac{4}{3}\pi r^3$.

1. $r=\frac{4.2}{2}=2.1\text{cm}$, $r^3=9.261$.
2. $V=\frac{4}{3}\times \frac{22}{7}\times 9.261=\frac{88}{21}\times 9.261$.
3. $=88\times 0.441=38.808\approx 38.8{\text{cm}}^3$.

Trick: $2.1=3\times 0.7$, so $\frac{22}{7}$ cancels neatly: $\frac{88}{21}\times 9.261=88\times 0.441$.

Answer: (C) $38.8$

Concept: Average of groups $=\sum (\text{share}\times \text{average})$.

1. Children $=100-20-30=50\%$.
2. Average $=0.2\times 50+0.3\times 40+0.5\times 40$.
3. $=10+12+20=42\text{kg}$.

Trick: 80% weigh 40 kg; men add $0.2\times 10=2$ extra → $40+2=42$.

Answer: (C) 42

Concept: Solve innermost brackets first, then division and multiplication from left to right.

1. $18-\frac{12}{2}=18-6=12$.
2. $2÷2\times 12=1\times 12=12$.
3. $(24\times 6)\times 12=144\times 12=1728$.

Trick: $1728={12}^3$ — a familiar cube.

Answer: (D) 1728