RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Ladder, wall and ground form a right triangle; the ladder is the hypotenuse.

1. Distance of foot from wall $=x$ (base), angle with ground $={60}^{\circ }$.
2. $\cos {60}^{\circ }=\frac{x}{10}\Rightarrow x=10\times \frac{1}{2}=5\text{m}$.

Trick: At ${60}^{\circ }$ the base is always half the ladder (30-60-90 triangle).

Answer: (C) $5$ metres

Concept: For runners on a circular track, distinct meeting points = sum of their speed ratio (opposite directions) or difference (same direction), ratio in lowest terms.

1. Speeds: Amit $=\frac{2100}{42}=50\text{m/s}$, Sumit $=\frac{2100}{75}=28\text{m/s}$.
2. Ratio $50:28=25:14$ (lowest terms).
3. Opposite directions ⇒ meeting points $=25+14=39$.

Trick: Speed ratio = inverse of time ratio: $75:42=25:14$ directly.

Answer: (B) $39$

Concept: Total = average × count; then share the total in the given ratio.

1. Total $=3\times 15255=$ ₹$45765$.
2. Ratio parts $=3+5+7=15$, 1 part $=\frac{45765}{15}=$ ₹$3051$.
3. Cheapest $=3\times 3051=$ ₹$9153$.

Trick: Cheapest $=\frac{3}{15}\times 3\times 15255=\frac{3}{5}\times 15255$.

Answer: (A) ₹$9153$

Concept: Amount $A=P(1+\frac{RT}{100})$ for simple interest.

1. $\frac{RT}{100}=\frac{4\times 3}{100}=0.12$.
2. $840=P\times 1.12\Rightarrow P=\frac{840}{1.12}=750$.

Trick: 12% of 750 = 90 and $750+90=840$ ✔.

Answer: (A) ₹$750$

Concept: A root satisfies the equation: substitute $x=3$ in both.

1. $9-24+P=0\Rightarrow P=15$.
2. $9+12+Q=0\Rightarrow Q=-21$.
3. $P+Q=15-21=-6$.

Trick: Adding directly: $(9-24)+(9+12)+P+Q=0\Rightarrow P+Q=-6$.

Answer: (B) $-6$

Concept: Heron's formula: $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$, $s=\frac{a+b+c}{2}$.

1. $s=\frac{8+9+13}{2}=15$.
2. $s-a=7,s-b=6,s-c=2$.
3. Area $=\sqrt{15\times 7\times 6\times 2}=\sqrt{1260}\approx 35.50{\text{cm}}^2$.

Trick: ${35}^2=1225$, ${36}^2=1296$; $1260$ is almost midway ⇒ about $35.5$.

Answer: (D) $35.50{\text{cm}}^2$

Concept: Mode = the observation that occurs most often.

1. Count repeats: 78 occurs 5 times; 90 and 66 occur 3 times each; 86 and 83 twice; the rest once.
2. Highest frequency = 5 for 78.

Trick: Tick only the option values in the list – no need to tally every number.

Answer: (C) 78

Concept: In a proportion $a:b::c:d$, $\frac{b}{a}=\frac{d}{c}$.

1. $6:42$ ⇒ multiplier 7, so $x=7.6\times 7=53.2$.
2. $3:6$ ⇒ multiplier 2, so $y=6\times 2=12$.
3. $x:y=53.2:12=532:120=133:30$.

Trick: Multiply by 10 to clear the decimal, then cancel by 4.

Answer: (A) $133:30$

Concept: A works for the full 54 minutes; B fills only the remaining part.

1. Work by A in 54 min $=\frac{54}{72}=\frac{3}{4}$.
2. Remaining $\frac{1}{4}$ is filled by B: time $=\frac{1}{4}\times 76=19$ min.
3. So B is turned off after 19 minutes.

Trick: Pipe that runs throughout – compute its share first; the other pipe covers the rest.

Answer: (C) 19 minutes

Concept: For 2 years: $\text{CI}-\text{SI}=P{\left(\frac{R}{100}\right)}^2$.

1. $967=P\times {\left(\frac{17}{100}\right)}^2=P\times 0.0289$.
2. $P=\frac{967}{0.0289}\approx 33460.2$.
3. Rounded: ₹$33460$.

Trick: $\frac{967\times 10000}{289}$: $289\times 33460=9669940\approx 9670000$.

Answer: (B) ₹$33460$

Concept: If $a:b=b:c$, then $c$ is the third proportional of $a$ and $b$, so $c=\frac{b^2}{a}$.

1. First three even natural numbers: $2+4+6=12$, so $B=12$.
2. $C=\frac{{12}^2}{29}=\frac{144}{29}$.
3. $\frac{144}{29}\approx 4.9655\approx 4.97$.

Trick: Third proportional = (middle term)${}^2$ ÷ first term.

Answer: (D) $4.97$

Concept: A number divisible by both 6 and 31 (co-prime) must be divisible by $6\times 31=186$.

1. $3720÷186=20$ exactly.
2. Check others: 2132, 4916, 2858 are not divisible by 3 (digit sums 8, 20, 23), so not by 6.

Trick: First test divisibility by 6 (even + digit sum divisible by 3): only 3720 (digit sum 12) passes.

Answer: (A) $3720$

Concept: For $a{x}^2+bx+c=0$, discriminant $D=b^2-4ac$. $D<0$ ⇒ no real roots.

1. $a=2,b=4,c=8$.
2. $D=4^2-4\times 2\times 8=16-64=-48$.
3. $D<0$, so the roots are imaginary (no real roots).

Trick: If $a$ and $c$ have the same sign and $b^2<4ac$, roots are never real.

Answer: (D) no real roots

Concept: Pipe A runs for the full 42 minutes; Pipe B must fill whatever A cannot.

1. A in 42 min fills $\frac{42}{84}=\frac{1}{2}$ of the cistern.
2. Remaining $\frac{1}{2}$ is filled by B.
3. Time for B $=\frac{1}{2}\times 36=18$ min.

Trick: 42 is exactly half of 84, so B only needs half of its 36 minutes.

Answer: (A) 18 minutes

Concept: Mode = the value that occurs most often.

1. Count repeats: 74 appears 4 times; 69, 77, 80, 87, 89 appear twice each; others once.
2. Highest frequency = 4 for 74.

Trick: Tick values that repeat while scanning once; only the repeating ones matter.

Answer: (A) 74

Concept: $A=P(1+\frac{RT}{100})$ for simple interest.

1. $RT=4\times 5=20\%$, so $A=1.2P$.
2. $1.2P=720$ ⇒ $P=\frac{720}{1.2}=600$.

Trick: 20% SI ⇒ Amount : Principal = 6 : 5, so P = ₹$720\times \frac{5}{6}$ = ₹600.

Answer: (A) ₹600

Concept: Dividend = Divisor × Quotient + Remainder; HCF by Euclid's method.

1. Let smaller = $S$: $L=5S+178$ and $L-S=2858$.
2. $4S+178=2858$ ⇒ $S=670$, $L=3528$.
3. $\text{HCF}(3528,3828)$: $3828-3528=300$; $3528=11\times 300+228$; $300-228=72$; $228=3\times 72+12$; $72=6\times 12$.
4. HCF = 12.

Trick: HCF of two numbers also divides their difference (300) – check options: 12 divides 300, 3528 and 3828.

Answer: (C) 12

Concept: $\text{SP}=\text{MP}\times \frac{100-d}{100}$ and $\text{CP}=\text{SP}\times \frac{100}{100+g}$.

1. SP = ₹$5960\times \frac{91}{100}$ = ₹5423.60.
2. CP = ₹$5423.6\times \frac{100}{130}$ = ₹4172.

Trick: One line: CP = ₹$\frac{5960\times 91}{130}=\frac{5960\times 7}{10}$ = ₹4172 (since $\frac{91}{130}=\frac{7}{10}$).

Answer: (C) ₹4,172

Concept: In a right triangle, $\tan \theta =\frac{\text{height}}{\text{base}}$.

1. Base $=15\text{m}$, $\theta ={45}^{\circ }$.
2. $\tan {45}^{\circ }=\frac{h}{15}=1$.
3. $h=15\text{m}$.

Trick: At ${45}^{\circ }$ elevation, height = distance from the base.

Answer: (C) $15\text{m}$

Concept: If $x(1+r{)}^{12}=y(1+r{)}^{13}$, then $x=y(1+r)$.

1. Arjun gets $x$, Kartik gets $y$: $x(1.1{)}^{12}=y(1.1{)}^{13}$ ⇒ $x=1.1y$.
2. $x+y=1218$ ⇒ $2.1y=1218$.
3. $y=\frac{1218}{2.1}=580$.

Trick: Shares are in ratio $11:10$ (one extra year at 10%); Kartik = ₹$\frac{10}{21}\times 1218$ = ₹580.

Answer: (B) ₹580