RRB NTPC Graduate CBT 1 2024 PYQs

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

Concept: A rectangle is a parallelogram with all angles 90°: opposite sides equal, diagonals equal and bisecting each other.

1. All four angles are 90° – true.
2. Diagonals of a rectangle are equal (PR = SQ) – true.
3. Opposite sides are equal (PQ = RS, PS = QR) – true.
4. Diagonals of every parallelogram, hence of a rectangle, bisect each other – so “do not bisect” is FALSE.

Trick: Every parallelogram’s diagonals bisect each other; rectangle adds “equal diagonals”.

Answer: (A) The diagonals of rectangle PQRS do not bisect each other

Concept: If $a:x::x:b$, then $x$ is the mean proportional: $x^2=ab$.

1. $A^2=18\times 128=2304$
2. $A=\sqrt{2304}=48$

Trick: $18\times 128=(9\times 2)\times (2\times 64)=36\times 64$, so $A=6\times 8=48$.

Answer: (B) 48

Concept: Successive discounts multiply: $SP=MP\times (1-d_1)(1-d_2)(1-d_3)$.

1. $500\times 0.80=400$
2. $400\times 0.75=300$
3. $300\times 0.81=243$

Trick: $19\%$ of 300 $=57$; $300-57=243$.

Answer: (D) ₹243

Concept: Total = average × number of items; share the total in the given ratio.

1. Total $=15165\times 3=45495$
2. Ratio sum $=3+5+7=15$; one part $=\frac{45495}{15}=3033$
3. Costliest $=7\times 3033=21231$

Trick: Average of 3 items = 5 parts (middle term), so 1 part $=\frac{15165}{5}=3033$.

Answer: (C) ₹21,231

Concept: For 3 years: $CI-SI=P{\left(\frac{R}{100}\right)}^2(3+\frac{R}{100})$.

1. $R=10\%$: $CI-SI=P\times \frac{1}{100}\times 3.1=0.031P$
2. $0.031P=217$
3. $P=\frac{217}{0.031}=7000$

Trick: At 10% for 3 years the difference is always $3.1\%$ of P: $\frac{217}{3.1}\times 100=7000$.

Answer: (C) ₹7,000

Concept: Overtaking a person moving in the same direction: train length $=(v_{train}-v_{man})\times t$.

1. $12.6\text{km/h}=3.5\text{m/s}$, $30.6\text{km/h}=8.5\text{m/s}$
2. $(v-3.5)\times 19.8=(v-8.5)\times 25.8$
3. $25.8v-19.8v=219.3-69.3\Rightarrow 6v=150\Rightarrow v=25\text{m/s}$
4. $25\times \frac{18}{5}=90\text{km/h}$

Trick: Option check: 90 km/h gives $(25-3.5)\times 19.8=425.7$ and $(25-8.5)\times 25.8=425.7$ – same length ✓.

Answer: (C) 90 km/hr

Concept: Weighted average: surplus of one group above the overall average = deficit of the other group.

1. Skilled: 80 workers, each ₹1,000 above average → surplus = $80\times 1000=80000$.
2. Others are ₹2,000 below average → number $=\frac{80000}{2000}=40$.
3. Total workers $=80+40=120$.

Trick: Alligation: $15000$ and $12000$ around $14000$ give ratio $2:1$ → $80:40$.

Answer: (D) 120

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

1. Expression $=\frac{8{\cos }^2\theta }{7{\sin }^2\theta }=\frac{8}{7}{\cot }^2\theta$.
2. ${\cot }^2\theta =\frac{49}{64}$.
3. Value $=\frac{8}{7}\times \frac{49}{64}=\frac{7}{8}$.

Trick: Always reduce $(1\pm \sin )(1\mp \sin )$ type products to squares first – no need to find $\sin \theta$.

Answer: (A) $\frac{7}{8}$

Concept: Least number which leaves remainder $r$ on division by several numbers $=\text{LCM}+r$.

1. $20=2^2\times 5,15=3\times 5,14=2\times 7,22=2\times 11$.
2. $\text{LCM}=2^2\times 3\times 5\times 7\times 11=4620$.
3. Required number $=4620+5=4625$.

Trick: Check options: only $4625-5=4620$ ends in 0 and is divisible by 3, 7 and 11.

Answer: (C) 4625

Concept: Pipes × time = constant work (pipe-minutes).

1. Work $=4\times 48=192$ pipe-minutes.
2. With 26 pipes: time $=\frac{192}{26}=7.3846\dots$
3. Rounded to two decimals → $7.38$ minutes.

Trick: More pipes → less time (inverse proportion): $T_2=\frac{P_1{T}_1}{P_2}$.

Answer: (C) $7.38$