RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Upstream speed $=B-S$, downstream speed $=B+S$.

1. Upstream speed $=\frac{33.2}{83/60}=\frac{33.2\times 60}{83}=24\text{km/h}$.
2. $B:S=8:5$ → $B-S=3$ parts $=24$ → 1 part $=8$: $B=64$, $S=40$, downstream $=104\text{km/h}$.
3. Time $=\frac{47.1}{24}+\frac{55.9}{104}=1.9625+0.5375=2.5$ hours.

Trick: $33.2÷83=0.4\text{km/min}=24\text{km/h}$.

Answer: (B) 2.5

Concept: Take CP $=100$; then MP and SP follow directly from the percentages.

1. CP $=100$ → MP $=126$ (26% above CP).
2. Profit 13% → SP $=113$.
3. Discount $=\frac{126-113}{126}\times 100=\frac{1300}{126}\approx 10.32\%$.

Trick: The ₹568 is not needed – percentages alone decide the discount.

Answer: (B) 10.32%

Concept: $(x+y+z{)}^2=x^2+y^2+z^2+2xy+2yz+2zx$.

1. Squares: $49a^2+36b^2+16c^2$.
2. $2(7a)(6b)=84ab$, $2(6b)(4c)=48bc$, $2(7a)(4c)=56ac$.
3. Sum: $49a^2+36b^2+16c^2+84ab+48bc+56ac$.

Trick: Only check the cross terms: $84ab$, $48bc$, $56ac$ – only option (C) has all three.

Answer: (C) $49a^2+36b^2+16c^2+84ab+48bc+56ac$

Concept: Use one-hour work (rates) and subtract.

1. $C=\frac{1}{27}-\frac{1}{63}=\frac{7-3}{189}=\frac{4}{189}$
2. $B=\frac{1}{42}-\frac{4}{189}=\frac{9-8}{378}=\frac{1}{378}$
3. B alone takes 378 hours.

Trick: LCM method: total work 378 units → A = 6, A+C = 14, B+C = 9 → C = 8, B = 1 → 378 h.

Answer: (D) 378

Concept: $a:b::c:d\Rightarrow ad=bc$; mean proportional: $y^2=$ product of extremes.

1. $(29+x)(8+x)=(31+x)(7+x)$ → $232+37x=217+38x$ → $x=15$.
2. $5x=75$, $2x-3=27$.
3. $y^2=75\times 27=2025\Rightarrow y=45$ ($y>0$).

Trick: $75\times 27=25\times 81$, so $y=5\times 9=45$.

Answer: (C) 45

Concept: Gain $=\text{SP}-\text{CP}$, Loss $=\text{CP}-\text{SP}$. Here gain $=1.25\times$ loss.

1. $8845-x=1.25(x-1987)$
2. $8845+2483.75=2.25x\Rightarrow x=\frac{11328.75}{2.25}=5035$
3. SP for 20% gain $=5035\times \frac{120}{100}=6042$

Trick: Gain : Loss $=5:4$, so CP divides the gap $8845-1987=6858$ in ratio $5:4$: $\text{CP}=1987+\frac{4}{9}\times 6858=5035$.

Answer: (A) ₹6,042

Concept: $a^m\times a^n=a^{m+n}$ and ${\left(\frac{2}{11}\right)}^n={\left(\frac{11}{2}\right)}^{-n}$.

1. LHS $={\left(\frac{11}{2}\right)}^{12}\times {\left(\frac{11}{2}\right)}^{-2}\times {\left(\frac{11}{2}\right)}^{16}={\left(\frac{11}{2}\right)}^{26}$
2. $={\left(\frac{2}{11}\right)}^{-26}$
3. So $9m+9=-26\Rightarrow 9m=-35\Rightarrow m=-\frac{35}{9}$

Trick: Flip the base → change the sign of the power.

Answer: (B) $-\frac{35}{9}$

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

1. Sum $=40+49+56+63+37=245$
2. Average $=\frac{245}{5}=49$ kg

Trick: Take 49 as base: deviations $-9,0,+7,+14,-12$ sum to 0 → average 49.

Answer: (A) 49

Concept: Two quadratics have both roots common when their coefficients are proportional: $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.

1. $\frac{2}{k}=\frac{8}{4k}=\frac{6}{3}$
2. $\frac{2}{k}=2\Rightarrow k=1$

Trick: Divide the first by 2: $x^2+4x+3=0$; it must match $k{x}^2+4kx+3$ → $k=1$.

Answer: (B) 1

Concept: $\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$; get each sine from a right triangle.

1. $\cos \alpha =\frac{3}{5}\Rightarrow \sin \alpha =\frac{4}{5}$ (triplet 3, 4, 5)
2. $\cos \beta =\frac{15}{17}\Rightarrow \sin \beta =\frac{8}{17}$ (triplet 8, 15, 17)
3. $\sin (\alpha +\beta )=\frac{4}{5}\times \frac{15}{17}+\frac{3}{5}\times \frac{8}{17}=\frac{60+24}{85}=\frac{84}{85}$

Trick: Know triplets (3, 4, 5), (8, 15, 17), (5, 12, 13), (7, 24, 25).

Answer: (B) $\frac{84}{85}$

Concept: Take capacity $=\text{LCM}(44,84,28)=924$ units. A pipe that empties the tank lowers the combined rate, so with the outlet open the tank must fill slower than with the two inlets alone.

1. Inlet 1: $\frac{924}{44}=21$ units/min; inlet 2: $\frac{924}{84}=11$ units/min; together $=32$ units/min.
2. All three pipes: $\frac{924}{28}=33$ units/min.
3. Outlet rate $=32-33=-1$ units/min. The negative value means the third pipe would be adding water, not draining it.
4. So an emptying pipe cannot give a filling time of 28 min; the time asked has no valid value and none of 916, 921, 920, 924 min is correct.
5. If the third pipe were also a filling pipe, it alone would fill the tank in $\frac{924}{1}=924$ min.

Note: Quick check: two inlets alone take $\frac{924}{32}$ min, which is more than 28 min, so the third pipe cannot be an outlet. 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: Speed $=\frac{\text{Distance}}{\text{Time}}$ for the remaining part.

1. Distance covered $=\frac{2}{3}\times 222=148$ km in $\frac{3}{7}\times 14=6$ h
2. Remaining: $222-148=74$ km in $14-6=8$ h
3. Speed $=\frac{74}{8}=9.25$ km/h

Trick: Remaining distance $=\frac{1}{3}$, remaining time $=\frac{4}{7}$ of total.

Answer: (C) 9.25

Concept: A number divisible by both 27 and 19 (co-prime) is divisible by $27\times 19=513$.

1. Digit sum must be a multiple of 9: 35691 → 24 ✗, 33488 → 26 ✗, 34371 → 18 ✓, 34962 → 24 ✗
2. Check: $34371÷513=67$ exactly ✓

Trick: Use the digit-sum test for 9 to eliminate quickly, then divide only the survivor.

Answer: (C) 34371

Concept: If $a:x::x:b$, then $x^2=ab$ (x is the mean proportional).

1. $A^2=10\times 40=400$
2. $A=\sqrt{400}=20$ (positive value)

Trick: Mean proportional $=\sqrt{ab}$.

Answer: (D) 20

Concept: Follow BODMAS: bracket first, then multiplication, then addition.

1. $70+42=112$
2. $\frac{93}{28}\times 112=93\times 4=372$
3. $372+82=454$

Trick: Look for cancellation: $112=28\times 4$.

Answer: (A) 454

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

1. At 25% loss, SP $=75\%$ of CP $=112$
2. Required SP $=150\%$ of CP $=112\times \frac{150}{75}=224$

Trick: Ratio method: $\frac{\text{new SP}}{\text{old SP}}=\frac{150}{75}=2$ → $112\times 2$.

Answer: (B) ₹224

Concept: Linear pair on the transversal + angle sum of triangle EGF.

1. $\angle BEF={180}^{\circ }-\angle PEB={180}^{\circ }-{42}^{\circ }={138}^{\circ }$
2. $\angle GEF=\angle BEF-\angle BEG={138}^{\circ }-{21}^{\circ }={117}^{\circ }$
3. In $△EGF$: $\angle EGF={180}^{\circ }-{117}^{\circ }-{34}^{\circ }={29}^{\circ }$

Trick: Shortcut: $\angle EGF=\angle PEB+\angle BEG-\angle GFE={42}^{\circ }+{21}^{\circ }-{34}^{\circ }={29}^{\circ }$.

Answer: (A) ${29}^{\circ }$

Concept: Expenditure $=$ Income $-$ Savings.

1. Incomes $4k$, $3k$; expenditures $4k-15000$ and $3k-9000$.
2. $\frac{4k-15000}{3k-9000}=\frac{16}{17}\Rightarrow 68k-255000=48k-144000\Rightarrow k=5550$
3. Expenditures: $22200-15000=7200$ and $16650-9000=7650$
4. Total $=7200+7650=14850$

Trick: Check: $7200:7650=16:17$ ✓.

Answer: (A) ₹14,850

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

1. $\text{SI}=\frac{350\times 4\times 2.5}{100}=\frac{3500}{100}=35$

Trick: $4\%\times 2.5=10\%$ of 350 $=35$.

Answer: (D) ₹35

Concept: Greatest number leaving the same remainder = HCF of the differences.

1. Differences: $2636-1272=1364$, $4209-2636=1573$
2. HCF(1364, 1573): $1573-1364=209$; $1364=6\times 209+110$; $209=110+99$; $110=99+11$; $99=9\times 11$ → $x=11$
3. $1272=11\times 115+7$ → $y=7$
4. $3x-14y=33-98=-65$

Trick: Same remainder → use differences, never the numbers themselves.

Answer: (B) $-65$