RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Profit is shared in the ratio of investment (same time period).

1. Total parts $=45+37+71=153$.
2. Difference Z − Y $=71-37=34$ parts.
3. Required difference $=\frac{34}{153}\times 2790=\frac{2}{9}\times 2790=620$.

Trick: $\frac{34}{153}$ simplifies to $\frac{2}{9}$ — simplify the fraction before multiplying.

Answer: (B) ₹620

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

1. Put $x+\frac{1}{x}=15$.
2. ${15}^3=3375$ and $3\times 15=45$.
3. $x^3+\frac{1}{x^3}=3375-45=3330$.

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

Answer: (B) $3330$

Concept: Going back one year means dividing by the growth factor: $25\%=\frac{1}{4}$, so factor $=\frac{5}{4}$.

1. 2 years ago: $160000\times \frac{4}{5}\times \frac{4}{5}=102400$.
2. 3 years ago: $102400\times \frac{4}{5}=81920$.
3. Difference $=102400-81920=20480$.

Trick: Difference $=\frac{1}{4}$ of the 3-years-ago value $=\frac{81920}{4}=20480$, or $\frac{1}{5}$ of the 2-years-ago value.

Answer: (B) 20,480

Concept: Take CP $=100$; then MP $=131$ and SP $=109$.

1. Discount $=131-109=22$ on MP $131$.
2. Discount $\%=\frac{22}{131}\times 100\approx 16.79\%$.
3. (MP ₹744 is not needed — the percentage does not depend on it.)

Trick: Discount $\%=\frac{\text{MP}\%-\text{SP}\%}{\text{MP}\%}\times 100=\frac{131-109}{131}\times 100$.

Answer: (D) $16.79\%$

Concept: Being 1% empty means 51 L fills $99\%$ of the tank.

1. $99\%$ of capacity $=51$ L.
2. Capacity $=\frac{51\times 100}{99}=51.5151\dots \approx 51.52$ L.
3. So 51.52 L of petrol in all is needed to fill the empty tank completely.

Trick: $x\%$ empty ⇒ given amount $=(100-x)\%$ of capacity.

Answer: (C) $51.52$

Concept: The greatest number giving the same remainder is the HCF of the differences of the numbers.

1. Differences: $2815-1063=1752$, $3451-2815=636$.
2. $\text{HCF}(1752,636)$: $1752=2\times 636+480$, $636=480+156$, $480=3\times 156+12$, $156=13\times 12$ → $x=12$.
3. $1063=12\times 88+7$ → $y=7$.
4. $3x-14y=36-98=-62$.

Trick: Same remainder → HCF of differences; then remainder from the smallest number.

Answer: (C) $-62$

Concept: Mean $=\frac{\text{sum of observations}}{\text{number of observations}}$.

1. Sum $=69+60+39+58+55+31+84+64+80=540$.
2. Number of observations $=9$.
3. Mean $=\frac{540}{9}=60$.

Trick: Pair numbers making round sums: $69+31=100$, $60+80=140$, $39+64=103$, $58+84=142$, $+55$ → $540$.

Answer: (D) 60

Concept: Do division first, then addition and subtraction from left to right.

1. $\frac{49}{7}=7$.
2. $47+7+65-27$.
3. $=119-27=92$.

Trick: Add all positives ($47+7+65=119$), then subtract $27$.

Answer: (A) 92

Concept: Brackets first, then multiplication, then addition/subtraction.

1. $11+4=15$.
2. $15\times 15=225$.
3. $139-225+52=191-225=-34$.

Trick: Collect positives ($139+52=191$), then subtract $225$.

Answer: (C) $-34$

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

1. $40.1=3\text{Median}-2\times 33.9=3\text{Median}-67.8$.
2. $3\text{Median}=107.9$.
3. Median $=\frac{107.9}{3}=35.966\dots \approx 35.97$.

Trick: Median $=\frac{\text{Mode}+2\text{Mean}}{3}$ — it always lies between mean and mode, closer to the mean.

Answer: (C) $35.97$

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) $4704\pi$