RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Take capacity $=\text{LCM}(35,15,6)=210$ 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: $\frac{210}{35}=6$ units/min; inlet 2: $\frac{210}{15}=14$ units/min; together $=20$ units/min.
2. All three pipes: $\frac{210}{6}=35$ units/min.
3. Outlet rate $=20-35=-15$ 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 6 min; the time asked has no valid value and none of 14, 5, 8, 17 min is correct.
5. If the third pipe were also a filling pipe, it alone would fill the tank in $\frac{210}{15}=14$ min.

Note: Quick check: two inlets alone take $\frac{210}{20}$ min, which is more than 6 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: 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. $\frac{4}{2}=\frac{6}{3}=2$, so the second equation is $2\times$ the first.
2. Then $(k+1)=2\times 5=10$.
3. $k=9$.

Trick: Check: $2(2x^2-5x+3)=4x^2-10x+6$ → $k+1=10$.

Answer: (A) 9

Concept: A regular hexagon is made of 6 equilateral triangles, so Area $=6\times \frac{\sqrt{3}}{4}{a}^2=\frac{3\sqrt{3}}{2}{a}^2$.

1. $a=16\sqrt{2}$ → $a^2=256\times 2=512$.
2. Area $=\frac{3\sqrt{3}}{2}\times 512=3\sqrt{3}\times 256$.
3. $=768\sqrt{3}{\text{cm}}^2$.

Trick: Options differ only by digit order (768, 786, 867, 876) — compute $3\times 256=768$ carefully.

Answer: (D) $768\sqrt{3}{\text{cm}}^2$

Concept: Opposite directions → relative speed = sum of speeds; crossing distance = sum of the two lengths.

1. Relative speed $=32+28=60\text{km/h}=60\times \frac{5}{18}=\frac{50}{3}\text{m/s}$.
2. $A+B=\frac{50}{3}\times 33=550$ m; $A+1.5A=2.5A=550$ → $A=220$ m, $B=330$ m.
3. Train B on the bridge: $28\times \frac{5}{18}\times 54=420$ m $=330+\text{bridge}$.
4. Bridge $=420-330=90$ m.

Trick: $x\text{km/h}\times t\text{s}=\frac{5xt}{18}$ m: $\frac{5\times 28\times 54}{18}=420$.

Answer: (D) 90

Concept: Write both sides with the same base 12 and compare powers.

1. $x^z={\left({12}^{0.36}\right)}^z={12}^{0.36z}$.
2. $y^4={\left({12}^{0.64}\right)}^4={12}^{2.56}$.
3. $0.36z=2.56$ → $z=\frac{256}{36}=\frac{64}{9}\approx 7.11$.

Trick: Same base → equate exponents directly.

Answer: (A) $7.11$

Concept: Equal areas: $\pi r^2=l\times b$; then circumference $=2\pi r$.

1. Area of rectangle $=11\times 14=154{\text{m}}^2$.
2. $\frac{22}{7}{r}^2=154$ → $r^2=49$ → $r=7$ m.
3. Circumference $=2\times \frac{22}{7}\times 7=44$ m.

Trick: Area 154 with $\pi =\frac{22}{7}$ → $r=7$ (a very common pair to remember).

Answer: (B) 44

Concept: Distance from the point to the top of the hill is the hypotenuse of the right triangle; use $\sin \theta =\frac{\text{perpendicular}}{\text{hypotenuse}}$.

1. Height $=171\text{m}$, angle $={60}^{\circ }$.
2. $\sin {60}^{\circ }=\frac{171}{d}\Rightarrow \frac{\sqrt{3}}{2}=\frac{171}{d}$
3. $d=\frac{342}{\sqrt{3}}=\frac{342\sqrt{3}}{3}=114\sqrt{3}\text{m}$

Trick: With ${60}^{\circ }$, hypotenuse $=\frac{2}{\sqrt{3}}\times$ height; $171=57\times 3$, so $d=2\times 57\sqrt{3}$.

Answer: (D) $114\sqrt{3}\text{m}$

Concept: $X\propto A^2$ and $A\propto \frac{1}{Y}$ together give $X\propto \frac{1}{Y^2}$, i.e. $X{Y}^2=k$.

1. $k=39\times {20}^2=39\times 400=15600$
2. When $Y=2$: $X=\frac{15600}{2^2}=\frac{15600}{4}=3900$

Trick: $Y$ becomes $\frac{1}{10}$ times, so $X$ becomes ${10}^2=100$ times: $39\times 100=3900$.

Answer: (D) $3900$

Concept: Write present ages in one variable, then form the future equation.

1. 3 years ago: son $=x$, father $=2x+30$.
2. Now: son $=x+3$, father $=2x+33$.
3. After $n$ years: $2x+33+n=2(x+3+n)$
4. $33+n=6+2n\Rightarrow n=27$

Trick: The $x$ cancels, so you can test with any son age, e.g. son 0 then: father 30 → now 33 and 3 → $33+n=2(3+n)$ gives 27.

Answer: (D) $27$

Concept: Rotating a right triangle about one leg makes a cone: that leg is the height, the other leg is the radius, the hypotenuse is the slant height.

1. $h=56$, $r=42$, $l=70\text{cm}$ (check: ${42}^2+{56}^2={70}^2$).
2. $\text{TSA}=\pi r(l+r)=\pi \times 42\times (70+42)$
3. $=\pi \times 42\times 112=4704\pi {\text{cm}}^2$

Trick: Sides are $14\times (3,4,5)$; the side you rotate about is always the height.

Answer: (B) $4704\pi$

Concept: Income = Savings + Expenditure.

1. Savings $=50\%$, so expenditure $=50\%$ of income.
2. $50\%$ of income = ₹320 → income $=320\times 2=$ ₹640.

Trick: If savings and expenditure are equal, income is just double the expenditure.

Answer: (B) $640$

Concept: Selling price after discount $=\text{MP}\times (100-d)\%$; with no discount, SP = MP.

1. $\text{MP}\times \frac{24}{100}=3330\Rightarrow \text{MP}=\frac{3330\times 100}{24}=13875$
2. With no discount, gain $50\%$: $\text{MP}=1.5\times \text{CP}$
3. $\text{CP}=\frac{13875}{1.5}=9250$

Trick: $\text{CP}=\frac{3330}{0.24\times 1.5}=\frac{3330}{0.36}=9250$.

Answer: (D) $9,250$

Concept: The perpendicular from the centre bisects the chord; use Pythagoras in the half-chord triangle.

1. $OB=6$, $OM=4.5$, $MB=x$.
2. $x^2=6^2-{4.5}^2=36-20.25=15.75=\frac{63}{4}$
3. $x=\frac{\sqrt{63}}{2}=\frac{3\sqrt{7}}{2}$
4. $AB=2x=3\sqrt{7}\text{cm}$

Trick: $AB=2\sqrt{r^2-d^2}=\sqrt{4(36-20.25)}=\sqrt{63}=3\sqrt{7}$.

Answer: (C) $3\sqrt{7}\text{cm}$

Concept: Average of first $n$ natural numbers $=\frac{n+1}{2}$.

1. Sum $=\frac{12\times 13}{2}=78$
2. Average $=\frac{78}{12}=6.5$

Trick: Average of consecutive numbers = (first + last) ÷ 2 $=\frac{1+12}{2}=6.5$.

Answer: (B) $6.5$

Concept: $\text{SI}=\frac{P\times R\times T}{100}$, with rate and time in the same unit (here per month).

1. $P=900$, $R=6\%$ per month, $T=12$ months.
2. $\text{SI}=\frac{900\times 6\times 12}{100}=648$

Trick: $6\%$ per month = $72\%$ per year; $72\%$ of 900 $=648$.

Answer: (B) $648$

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

1. Loss $30\%$: $\text{CP}\times 0.7=112\Rightarrow \text{CP}=160$
2. Gain $15\%$: $\text{SP}=160\times 1.15=184$

Trick: $\text{New SP}=112\times \frac{115}{70}=16\times \frac{115}{10}=184$.

Answer: (A) $184$

Concept: Time $=\frac{\text{Distance}}{\text{Speed}}$; one-way distance $d$ is covered once downstream and once upstream.

1. $\frac{d}{41}+\frac{d}{20.5}=13$
2. $\frac{d}{41}+\frac{2d}{41}=\frac{3d}{41}=13\Rightarrow d=\frac{533}{3}\text{km}$
3. Total distance $=2d=\frac{1066}{3}\text{km}$

Trick: $\frac{533}{3}$ is only one way — the question asks total distance, so double it.

Answer: (D) $\frac{1066}{3}\text{km}$

Concept: Sum $=$ Average $\times$ Count.

1. Sum $=124\times 8=992$
2. Least number $=\frac{1}{16}\times 992=62$

Trick: $\frac{8\times 124}{16}=\frac{124}{2}=62$ — half the average.

Answer: (C) $62$

Concept: Third proportional of $a$ and $b$: $a:b=b:c\Rightarrow c=\frac{b^2}{a}$.

1. $B=2+4+6=12$
2. $C=\frac{{12}^2}{57}=\frac{144}{57}$
3. $=2.526\dots \approx 2.53$

Trick: $\frac{144}{57}=\frac{48}{19}$; $19\times 2.5=47.5$ so the answer is just above 2.5.

Answer: (A) $2.53$

Concept: Work per day adds up; A's rate = B's rate + C's rate.

1. $A+B=\frac{1}{7}$, $C=\frac{1}{9}$, $A=B+\frac{1}{9}$
2. $2B+\frac{1}{9}=\frac{1}{7}\Rightarrow 2B=\frac{9-7}{63}=\frac{2}{63}$
3. $B=\frac{1}{63}$ → B alone takes 63 days.

Trick: Total work $=63$ units: $A+B=9$, $C=7$, $A=B+7\Rightarrow 2B=2$, $B=1$ unit/day → 63 days.

Answer: (B) $63$