RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Area $=l\times b$, Perimeter $=2(l+b)$.

1. Let longer side $=L$, shorter $=L-24$.
2. $L(L-24)=8\times 2(2L-24)=32L-384$
3. $L^2-56L+384=0\Rightarrow (L-48)(L-8)=0$
4. $L=8$ gives a negative side, so $L=48\text{cm}$.

Trick: Check options: $48\times 24=1152$ and $8\times 2(48+24)=1152$ ✓.

Answer: (B) $48$

Concept: Work is fixed, so pipes × time is constant: $N_1{T}_1=N_2{T}_2$.

1. $44\times 38=37\times T$
2. $T=\frac{1672}{37}=45.189\dots$
3. $T\approx 45.19$ minutes

Trick: Fewer pipes ⇒ more time; answer must be a little more than 38 — only 43.91, 45.19, 49.75 qualify; $\frac{44}{37}\times 38$ gives 45.19.

Answer: (D) $45.19$

Concept: List the primes between 20 and 40 and take their mean.

1. Primes: $23,29,31,37$
2. Sum $=120$
3. Average $=\frac{120}{4}=30$

Trick: Pairs 23 + 37 and 29 + 31 both equal 60 ⇒ average 30.

Answer: (C) 30

Concept: $\text{Profit}\%=\frac{SP-CP}{CP}\times 100$; compare per dozen.

1. CP per dozen $=728$
2. SP per dozen $=91\times 12=1092$
3. $\text{Profit}\%=\frac{1092-728}{728}\times 100=\frac{364}{728}\times 100=50\%$

Trick: The number of dozens (59) does not matter — profit % is same per dozen.

Answer: (A) $50\%$

Concept: After successive increases: Present $=x\times \frac{100+a}{100}\times \frac{100+b}{100}$.

1. $x\times 1.02\times 1.20=190400$
2. $x\times 1.224=190400$
3. $x=\frac{190400}{1.224}=155555.56$

Trick: Net increase $=2+20+\frac{2\times 20}{100}=22.4\%$; divide by 1.224.

Answer: (C) ₹1,55,555.56

Concept: Volume of cuboid $=l\times b\times h$; cost $=$ volume × rate.

1. Volume $=6\times 64\times 10=3840{\text{m}}^3$
2. Cost $=3840\times 15=57600$

Trick: Options differ only in last digits — volume × 15 must end in 0, so 57,600.

Answer: (A) ₹57,600

Concept: $\sin ({90}^{\circ }-\theta )=\cos \theta$, $\cos ({90}^{\circ }-\theta )=\sin \theta$ and ${\sin }^2\theta +{\cos }^2\theta =1$.

1. ${\sin }^2{68}^{\circ }={\cos }^2{22}^{\circ }$ ⇒ numerator $={\cos }^2{22}^{\circ }+{\sin }^2{22}^{\circ }=1$
2. ${\cos }^2{73}^{\circ }={\sin }^2{17}^{\circ }$ ⇒ bracket $={\cos }^2{17}^{\circ }+{\sin }^2{17}^{\circ }=1$
3. Value $=\frac{1}{2\times 1}=\frac{1}{2}$

Trick: Angles adding to ${90}^{\circ }$ in ${\sin }^2+{\sin }^2$ or ${\cos }^2+{\cos }^2$ always give 1.

Answer: (D) $\frac{1}{2}$

Concept: Form two equations in bag price $b$ and pen price $p$.

1. $3b+6p=744\Rightarrow b+2p=248$
2. $9b+9p=1854\Rightarrow b+p=206$
3. Subtract: $p=42$, $b=164$
4. $6b-10p=984-420=564$

Trick: Divide each equation by its common factor first — the numbers become small.

Answer: (C) ₹564

Concept: Number of diagonals of an $n$-sided polygon $=\frac{n(n-3)}{2}$.

1. Heptagon: $n=7$
2. Each vertex joins $7-3=4$ non-adjacent vertices
3. Diagonals $=\frac{7\times 4}{2}=14$

Trick: Divide by 2 because each diagonal is counted from both ends.

Answer: (C) 14

Concept: Net rate $=$ filling rate $-$ emptying rate; time $=\frac{\text{work}}{\text{rate}}$.

1. Net rate $=\frac{1}{15}-\frac{1}{17}=\frac{17-15}{255}=\frac{2}{255}$ per hour
2. Time for $\frac{2}{3}$ tank $=\frac{2}{3}÷\frac{2}{255}=\frac{255}{3}=85$ hours

Trick: Full tank takes $\frac{15\times 17}{17-15}=127.5$ h; two-third of it $=85$ h.

Answer: (B) 85

Concept: $SI=\frac{P\times R\times T}{100}$ and Amount $=P+SI$.

1. $SI=\frac{50500\times 6\times 2}{100}=505\times 12=6060$
2. Amount $=50500+6060=56560$

Trick: 2 years at 6% means 12% in total: $12\%$ of 50500 $=6060$. Option ₹6,060 is only the interest, not the amount.

Answer: (C) ₹56,560

Concept: Diagonal of a cuboid $d=\sqrt{l^2+b^2+h^2}$, Volume $=l\times b\times h$.

1. $d^2=(17\sqrt{29}{)}^2=289\times 29=8381$
2. $l^2+b^2=2500+5625=8125$
3. $h^2=8381-8125=256\Rightarrow h=16\text{cm}$
4. $V=50\times 75\times 16=60000{\text{cm}}^3$

Trick: Square the diagonal first; ${17}^2=289$ and $289\times 29=8381$.

Answer: (A) $60000$

Concept: For $a{x}^2+bx+c=0$, discriminant $D=b^2-4ac$: $D>0$ two distinct real roots, $D=0$ equal roots, $D<0$ complex roots.

1. $a=3,b=-5,c=2$
2. $D=25-24=1>0$
3. So the roots are real and distinct ($x=1,\frac{2}{3}$).

Trick: $a+b+c=0$ means $x=1$ is a root; other root $=\frac{c}{a}=\frac{2}{3}$ – two different real roots.

Answer: (C) two distinct real roots

Concept: A number is divisible by 6 if it is divisible by both 2 and 3.

1. Last digit 2 is even → divisible by 2 for any p.
2. Digit sum $=5+9+0+4+4+p+2+2=26+p$.
3. $26+p$ must be a multiple of 3 → $p=1,4,7$.
4. Maximum p = 7.

Trick: Next multiple of 3 after 26 is 27, then 30, 33 → $p=1,4,7$; pick the largest single digit.

Answer: (C) 7

Concept: Net work per hour = sum of filling rates − emptying rate.

1. Rates: $+\frac{1}{2},+\frac{1}{81},-\frac{1}{81}$
2. Net $=\frac{1}{2}+\frac{1}{81}-\frac{1}{81}=\frac{1}{2}$
3. Time $=2\text{hours}$

Trick: The second filling tap and the emptying tap have the same time (81 h), so they cancel – only the 2-hour tap matters.

Answer: (D) 2

Concept: Solve innermost brackets first; inside a bracket do multiplication before subtraction.

1. $30\times 4=120$, so $(30-120)=-90$
2. $\{30-(-90)\}=120$
3. $[30-120]=-90$
4. $30-(-90)=120$

Trick: The value keeps alternating between −90 and 120 – count the brackets carefully.

Answer: (C) 120

Concept: Evaluate a nested root from the innermost root outward.

1. $\sqrt{144}=12$
2. $\sqrt{213+12}=\sqrt{225}=15$
3. $\sqrt{385+15}=\sqrt{400}=20$

Trick: Each step should give a perfect square – if not, recheck the previous step.

Answer: (B) 20

Concept: Sum of exterior angles of any polygon $={360}^{\circ }$, so for a regular polygon $n=\frac{{360}^{\circ }}{\text{exterior angle}}$.

1. $n=\frac{{360}^{\circ }}{{30}^{\circ }}=12$
2. Check: interior angle $={180}^{\circ }-{30}^{\circ }={150}^{\circ }$ (regular dodecagon).

Trick: Exterior angle × sides $={360}^{\circ }$ always.

Answer: (C) 12

Concept: Savings % $=100\%-$ expenditure %.

1. Savings $=100\%-55\%=45\%$ of income
2. $45\%\text{of income}=36000$
3. Income $=\frac{36000\times 100}{45}=80000$

Trick: $\frac{36000}{45}=800$ is 1%, so 100% = 80,000.

Answer: (A) ₹80,000

Concept: Area of rectangle $=l\times b$; here $l=b+30$.

1. $b(b+30)=18271$
2. Since $\sqrt{18271}\approx 135$, $b\approx 135-15=120$; try $b=121$.
3. $121\times 151=18271$ ✓
4. So breadth = 121 m.

Trick: Check options by unit digit: $b\times (b+30)$ must end in 1 → $b$ ends in 1 → only 131 or 121; $131\times 161=21091$ ✗.

Answer: (B) 121