RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Take CP $=100$. Then $\text{MP}=122$ and $\text{SP}=107$; $\text{Discount}\%=\frac{\text{MP}-\text{SP}}{\text{MP}}\times 100$.

1. $\text{MP}=100+22=122$, $\text{SP}=100+7=107$.
2. Discount $=122-107=15$.
3. $\text{Discount}\%=\frac{15}{122}\times 100\approx 12.295\%\approx 12.30\%$.
4. The actual price ₹668 is not needed.

Trick: Discount% $=\frac{m-p}{100+m}\times 100=\frac{22-7}{122}\times 100$.

Answer: (B) $12.30\%$

Concept: Net filling rate = filling rate − emptying rate.

1. Net rate $=\frac{1}{18}-\frac{1}{24}=\frac{4-3}{72}=\frac{1}{72}$ tank per hour.
2. Full tank takes $72$ hours.
3. $\frac{3}{4}$ tank takes $\frac{3}{4}\times 72=54$ hours.

Trick: LCM method: capacity $=72$ units; filler $+4$, emptier $-3$ → net $1$ unit/h; $\frac{3}{4}\times 72=54$ units → 54 h.

Answer: (C) 54

Concept: $a^3+b^3+c^3=(a+b+c{)}^3-3(a+b+c)(ab+bc+ca)+3abc$.

1. Given $a+b+c=5$, $ab+bc+ca=-2$, $abc=3$.
2. $(a+b+c{)}^3=125$.
3. $-3(5)(-2)=+30$ and $3abc=9$.
4. $a^3+b^3+c^3=125+30+9=164$.

Trick: Alternatively, $a^2+b^2+c^2=25+4=29$, then $a^3+b^3+c^3-3abc=5(29+2)=155$, so the sum $=155+9=164$.

Answer: (A) $164$

Concept: Upstream speed $=b-s$, downstream speed $=b+s$; $b=\frac{\text{down}+\text{up}}{2}$.

1. Upstream speed $=\frac{14}{14}=1\text{km/h}$.
2. Downstream speed $=\frac{171}{9}=19\text{km/h}$.
3. Speed in still water $=\frac{19+1}{2}=10\text{km/h}$.
4. (Stream speed $=\frac{19-1}{2}=9\text{km/h}$.)

Trick: Boat = average of the two speeds; stream = half of their difference.

Answer: (D) 10 km/hr

Concept: Net filling rate = tap rate − leak rate.

1. Net rate $=\frac{1}{8}-\frac{1}{10}=\frac{5-4}{40}=\frac{1}{40}$ cistern per hour.
2. Time to fill $=40$ hours.

Trick: For one filler and one leak: time $=\frac{xy}{y-x}=\frac{8\times 10}{10-8}=40$.

Answer: (C) 40 hours

Concept: If numbers are $2k$, $k$, $5k$ (2, 1, 5 co-prime), HCF $=k$ and LCM $=k\times \text{LCM}(2,1,5)=10k$.

1. $10k=5380$.
2. $k=538$.
3. HCF $=538$.

Trick: HCF $=\frac{\text{LCM}}{\text{LCM of ratio terms}}=\frac{5380}{10}$.

Answer: (A) 538

Concept: Let a bag cost $b$ and a pen cost $p$ (in ₹). Form two equations.

1. $10b+10p=1310\Rightarrow b+p=131$.
2. $2b+11p=505\Rightarrow 2(131-p)+11p=505\Rightarrow 9p=243\Rightarrow p=27$.
3. $b=131-27=104$.
4. $8b-8p=8(104-27)=8\times 77=616$.

Trick: Only $b-p$ is needed: $2(b+p)=262$, so $9p=505-262=243$.

Answer: (C) ₹616

Concept: Sum = average × count.

1. Sum of 36 numbers $=36\times 45=1620$.
2. After removing 37 and 53: $1620-90=1530$.
3. Average of 34 numbers $=\frac{1530}{34}=45$.

Trick: The removed pair has average $\frac{37+53}{2}=45$, equal to the overall average, so the average stays 45.

Answer: (A) 45

Concept: Empirical formula: $\text{Mode}=3\text{Median}-2\text{Mean}$, i.e. $\text{Mode}-\text{Mean}=3(\text{Median}-\text{Mean})$.

1. $37.8=3(\text{Median}-\text{Mean})\Rightarrow \text{Median}-\text{Mean}=12.6$.
2. $\text{Mode}-\text{Median}=(\text{Mode}-\text{Mean})-(\text{Median}-\text{Mean})$.
3. $=37.8-12.6=25.2$.

Trick: Mode − Median $=\frac{2}{3}(\text{Mode}-\text{Mean})=\frac{2}{3}\times 37.8=25.2$.

Answer: (D) 25.2

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

1. $(13+8\sqrt{5})(10-2\sqrt{5})=130-26\sqrt{5}+80\sqrt{5}-16\times 5$.
2. $=130-80+54\sqrt{5}=50+54\sqrt{5}$.
3. Mean proportional $=\sqrt{50+54\sqrt{5}}$.

Trick: Watch the rational part: $130-80=50$ (not 54), which rules out option (C).

Answer: (D) $\sqrt{50+54\sqrt{5}}$

Concept: Two quadratics with both roots common are proportional (one is a constant multiple of the other).

1. $x^2+3x+2=(x+1)(x+2)$, roots $-1$ and $-2$.
2. A quadratic with the same two roots is $a(x+1)(x+2)=a{x}^2+3ax+2a$.
3. So $b=3a$, $c=2a$: $a{x}^2+bx+c$ must be a multiple of $x^2+3x+2$, i.e. $\frac{a}{1}=\frac{b}{3}=\frac{c}{2}$.
4. Check (A): $a+3b+2c=a+9a+4a=14a\ne 0$.

Trick: Both roots common ⇒ $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.

Answer: (D) $a{x}^2+bx+c$ must be a multiple of $x^2+3x+2$

Concept: A number is divisible by 9 when the sum of its digits is a multiple of 9.

1. Digit sum of 31425p99 $=3+1+4+2+5+9+9+p=33+p$.
2. Next multiples of 9 after 33: 36, 45 …
3. $33+p=36\Rightarrow p=3$; $33+p=45$ needs $p=12$, not a digit.
4. So maximum $p=3$.

Trick: Drop 9s: $3+1+4+2+5=15\to 6$; need $6+p=9$, so $p=3$ (or 12, invalid).

Answer: (A) 3

Concept: Volume of a cube $=a^3$.

1. ${42}^3=42\times 42\times 42$.
2. $42\times 42=1764$.
3. $1764\times 42=74088{\text{m}}^3$.

Trick: Unit digit of ${42}^3$ is $2^3=8$ → only 74,088 ends in 8.

Answer: (D) 74,088

Concept: $\text{SP}=\text{CP}\times \frac{100+\text{gain}\%}{100}$

1. $\text{CP}=3417\times \frac{100}{134}=2550$
2. $\text{Profit}=3417-2550=867$

Trick: Profit $=\text{SP}\times \frac{34}{134}=3417\times \frac{17}{67}=51\times 17=867$.

Answer: (B) ₹867

Concept: For 3 years: $\text{CI}-\text{SI}=P\times \frac{r^2(300+r)}{{100}^3}$

1. $186=P\times \frac{100\times 310}{1000000}=0.031P$
2. $P=\frac{186}{0.031}=6000$

Trick: At 10% for 3 years, CI − SI $=3.1\%$ of $P$, so $P=\frac{186}{3.1}\times 100=6000$.

Answer: (B) ₹6,000

Concept: Solve innermost bracket first; inside it, multiply before subtracting.

1. $(30-30\times 3)=30-90=-60$
2. $\{30-(-60)\}=90$
3. $[30-90]=-60$
4. $30-(-60)=90$

Trick: Each pair of minus signs flips: the value alternates −60, 90, −60, 90.

Answer: (A) 90

Concept: $\text{Average price}=\frac{\text{Total cost}}{\text{Total books}}$

1. $51\times 1120=57120$
2. $48\times 900=43200$
3. $\text{Average}=\frac{57120+43200}{99}=\frac{100320}{99}=1013.33$

Trick: Average $=900+\frac{51\times 220}{99}=900+113.33=1013.33$.

Answer: (C) ₹1,013.33

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

1. $\frac{(18xy{)}^2}{6x^2}=\frac{324x^2{y}^2}{6x^2}$
2. $=54y^2$

Trick: $x^2$ cancels; just compute $\frac{{18}^2}{6}=54$.

Answer: (C) $54y^2$

Concept: In congruent triangles ABC and XYZ, corresponding angles are equal: $\angle X=\angle A,\angle Z=\angle C$.

1. $2k+4k+4k={180}^{\circ }\Rightarrow k={18}^{\circ }$
2. $\angle A={36}^{\circ },\angle B=\angle C={72}^{\circ }$
3. $\angle X+\angle Z={36}^{\circ }+{72}^{\circ }={108}^{\circ }$

Trick: Sum of the required parts $=\frac{2+4}{10}\times {180}^{\circ }={108}^{\circ }$.

Answer: (C) ${108}^{\circ }$

Concept: $a^m\times a^n=a^{m+n}$; write every term as a power of 7.

1. Numerator $=7^2\times 7^3=7^5$
2. $\sqrt{49}=7,\sqrt{2401}=49=7^2$ → Denominator $=7^3$
3. $\frac{7^5}{7^3}=7^2=49$

Trick: $2401=7^4$; its square root is $7^2$.

Answer: (B) 49