RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Write every base as a power of the same prime, then add/subtract exponents: $a^m\times a^n=a^{m+n}$, $a^m÷a^n=a^{m-n}$.

1. ${125}^{1.3}=5^{3\times 1.3}=5^{3.9}$, ${25}^{1.8}=5^{2\times 1.8}=5^{3.6}$.
2. $5^{3.9+3.6-2.5}=5^5$ ⇒ $x=5$.
3. $18x-5=90-5=85$.

Trick: $125=5^3$, $25=5^2$ — just multiply the decimal exponents by 3 and 2 mentally.

Answer: (A) 85

Concept: To join ratios, make the common term equal by cross-multiplying step by step.

1. $a:b=1:3$, $b:c=5:7$ ⇒ $a:b:c=5:15:21$.
2. $c:d=9:5$ with $c=21$: multiply first by 9 and second by 21 ⇒ $a:b:c:d=45:135:189:105$.
3. Divide by 3: $15:45:63:35$.

Trick: $a:b:c:d=(1·5·9):(3·5·9):(3·7·9):(3·7·5)=45:135:189:105$ — one line.

Answer: (C) 15 : 45 : 63 : 35

Concept: $V=\frac{4}{3}\pi r^3$ with $r=\frac{\text{diameter}}{2}$.

1. $r=\frac{37.8}{2}=18.9$ cm; ${18.9}^3=6751.269$.
2. $V=\frac{4}{3}\times \frac{22}{7}\times 6751.269=\frac{88\times 6751.269}{21}=\frac{594111.672}{21}$.
3. $V\approx 28291.03{\text{cm}}^3$.

Trick: $18.9=\frac{189}{10}=\frac{7\times 27}{10}$, so the 7 cancels with $\pi =\frac{22}{7}$: $V=\frac{88\times 49\times {27}^3}{3\times 1000}=\frac{88\times 49\times 6561}{1000}\approx 28291.03$.

Answer: (B) 28291.03

Concept: Discount is on MP, profit is on CP: $\text{SP}=\text{MP}(1-d)=\text{CP}(1+p)$.

1. Let CP $=100$. MP $=100+150=250$.
2. Required SP for 60% gain $=160$.
3. Discount $=250-160=90$ ⇒ $\frac{90}{250}\times 100=36\%$.

Trick: $\frac{\text{SP}}{\text{MP}}=\frac{160}{250}=0.64$ ⇒ discount $=36\%$.

Answer: (A) $36\%$

Concept: Complete the square: $p^2-2p+4=(p-1{)}^2+3\ge 3$ and $-p^2+2p-4=-[(p-1{)}^2+3]\le -3$.

1. The first expression is always $\ge 3$ and the second always $\le -3$, so $max\{·\}$ is always the first one.
2. Minimum of $(p-1{)}^2+3$ on $0\le p\le 1$ occurs at $p=1$: value $=3$.

Trick: The two expressions are negatives of each other, so the max is simply the positive one, $(p-1{)}^2+3$.

Answer: (D) 3

Concept: If a sum becomes $n$ times in $T$ years at SI, interest $=(n-1)P$ and $R=\frac{(n-1)\times 100}{T}$.

1. 5 times ⇒ SI $=4P$ in 12 years.
2. $R=\frac{4\times 100}{12}=33.33\%$ per annum.

Trick: "$n$ times in $T$ years" ⇒ $R=\frac{100(n-1)}{T}$ — plug and go.

Answer: (C) 33.33

Concept: Monthly saving = salary − expenses; yearly saving = saving × number of months actually saved.

1. Monthly saving $=18000-5000-2000=11000$.
2. In the birthday month the whole saving is spent, so only 11 months count.
3. Yearly saving $=11000\times 11=121000$.

Trick: Read the twist — "12 months" gives 132000 (option A), the trap answer.

Answer: (B) 121000

Concept: LCM of monomials = each variable raised to its highest power among the terms.

1. Rewrite: $x^{3.5}{y}^{1.5}$; $\frac{y^{2.5}}{x^{-4.5}}=x^{4.5}{y}^{2.5}$; $\frac{x}{y^{-3}}=x{y}^3$.
2. Highest powers: $x^{4.5}$ and $y^3$ ⇒ LCM $=x^{4.5}{y}^3$.
3. $x^{4.5}{y}^3=x^3{y}^3·x^{1.5}=(xy{)}^3\sqrt{x^3}$.

Trick: A negative power in the denominator jumps to the numerator with a positive sign: $\frac{1}{x^{-a}}=x^a$.

Answer: (A) $(xy{)}^3\sqrt{x^3}$

Concept: $a^3-3a^{2}b+3a{b}^2-b^3=(a-b{)}^3$.

1. Take $a=4.23$, $b=0.23$: $0.69=3b$ and $12.69=3a$.
2. Numerator $=a^3-3a^{2}b+3a{b}^2-b^3=(a-b{)}^3=4^3=64$.
3. Denominator $=0.4\times 0.4\times 4=0.64$ ⇒ $\frac{64}{0.64}=100$.

Trick: Spot $0.69=3\times 0.23$ and $12.69=3\times 4.23$ — the sign pattern $+,-,+,-$ screams $(a-b{)}^3$.

Answer: (C) 100

Concept: When D is added, A, B, C keep their shares; the new total is 140% of the old total.

1. Old total $=3+5+7=15$ parts.
2. New total $=15\times 1.4=21$ parts.
3. $x=21-15=6$.

Trick: $40\%$ of 15 $=6$ — the increase itself is D's share.

Answer: (D) 6

Concept: Proportion of boys $=\frac{\text{boys}}{\text{boys}+\text{girls}}$; compare these fractions.

1. First $\frac{4}{9}\approx 0.444$; Second $\frac{3}{5}=0.6$; Third $\frac{6}{11}\approx 0.545$; Fourth $\frac{5}{11}\approx 0.455$.
2. Highest is $0.6$ — the second class.

Trick: Only $3:2$ and $6:5$ have more boys than girls; $3:2=1.5$ boys per girl beats $6:5=1.2$.

Answer: (A) Second class (3 : 2)

Concept: $\sqrt[3]{2197}=13$ (since ${13}^3=2197$) and $\sqrt[4]{10000}=10$ (since ${10}^4=10000$).

1. Expression $=\frac{13}{10}\times \frac{5}{39}\times 48$.
2. $\frac{13}{39}=\frac{1}{3}$, $\frac{5}{10}=\frac{1}{2}$ ⇒ $\frac{1}{6}\times 48=8$.

Trick: Cancel before multiplying: 13 with 39, 5 with 10 — the answer drops out as $\frac{48}{6}$.

Answer: (A) 8

Concept: Overall % $=\frac{\text{total SP}-\text{total CP}}{\text{total CP}}\times 100$.

1. CP each $=\frac{54000}{45}=1200$.
2. 30 at 25% profit: $30\times 1500=45000$; 15 at 20% loss: $15\times 960=14400$.
3. Total SP $=59400$; profit $=5400$ ⇒ $\frac{5400}{54000}\times 100=10\%$ profit.

Trick: Weighted: $\frac{30\times 25-15\times 20}{45}=\frac{750-300}{45}=10\%$ — CP per item is not even needed.

Answer: (B) 10% profit

Concept: In a right triangle, $\tan A=\cot C$ when $A+C={90}^{\circ }$ (complementary angles).

1. $BC=\sqrt{{13}^2-{12}^2}=\sqrt{25}=5$ cm.
2. $\tan A=\frac{BC}{AB}=\frac{5}{12}$; $\cot C=\frac{BC}{AB}=\frac{5}{12}$ (adjacent of C is BC, opposite is AB).
3. $\tan A-\cot C=0$.

Trick: $\angle A+\angle C={90}^{\circ }$ ⇒ $\cot C=\cot ({90}^{\circ }-A)=\tan A$ — the difference is 0 without finding BC.

Answer: (C) 0

Concept: If he uses $w$ g instead of 1000 g and sells at $80\%$ of the price of 1000 g, then $\frac{\text{SP}}{\text{CP}}=\frac{0.8\times 1000}{w}$.

1. Gain 40% ⇒ $\frac{800}{w}=1.4$.
2. $w=\frac{800}{1.4}=\frac{4000}{7}\approx 571.4$ g.

Trick: $\frac{\text{true weight}}{\text{false weight}}=\frac{100+\text{gain}}{100-\text{loss}}=\frac{140}{80}$ ⇒ false weight $=1000\times \frac{80}{140}=571.4$ g.

Answer: (C) 571.4 gm

Concept: A number is divisible by 3 if its digit sum is divisible by 3.

1. Digit sum of 445A0208 $=4+4+5+A+0+2+0+8=23+A$.
2. $23+A$ divisible by 3 ⇒ $A=1,4,7$.
3. Maximum $A=7$ ($23+7=30$).

Trick: $23\equiv 2\left(\mathrm{mod}3\right)$, so $A\equiv 1\left(\mathrm{mod}3\right)$: 1, 4, 7 — pick the largest digit.

Answer: (D) 7

Concept: Add the per-hour rates (filling positive, emptying negative).

1. Net rate $=\frac{1}{4}+\frac{1}{32}-\frac{1}{32}=\frac{1}{4}$ cistern per hour.
2. Full cistern in 4 hours ⇒ one-fourth in $\frac{1}{4}\times 4=1$ hour.

Trick: The 32-hour filler and 32-hour emptier cancel exactly — only the 4-hour tap matters.

Answer: (B) 1

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

1. $P\times {\left(\frac{4}{100}\right)}^2=53$ ⇒ $P\times \frac{16}{10000}=53$.
2. $P=\frac{53\times 10000}{16}=33125$.

Trick: Difference for 2 years = SI on the first year's interest; $P=\frac{\text{diff}\times {100}^2}{R^2}$.

Answer: (B) 33125

Concept: If a sum becomes $n$ times itself on simple interest, the interest earned is $(n-1)$ times the principal, so $R=\frac{(n-1)\times 100}{T}$.

1. Amount $=5P$, so $\text{SI}=5P-P=4P$.
2. $R=\frac{\text{SI}\times 100}{P\times T}=\frac{4P\times 100}{P\times 18}=\frac{400}{18}$.
3. $R=22.22\%$ (to two decimal places).

Trick: "$n$ times in $T$ years" on SI $\Rightarrow R\times T=(n-1)\times 100$. Here $R\times 18=400$.

Answer: (C) 22.22

Concept: Profit is shared in the ratio of (capital $\times$ time). When capital changes mid-year, add the products for each period.

1. Ratio $6:3:5$ and Z $=$ ₹12,000 $\Rightarrow$ 1 unit $=$ ₹2,400. So X $=$ ₹14,400, Y $=$ ₹7,200 (we only need the ratio units).
2. X: $6\times 6+3\times 6=54$; Y: $3\times 12=36$; Z: $5\times 12=60$. Total $=150$ units.
3. 1 unit of profit $=\frac{69000}{150}=$ ₹460.
4. X $-$ Y $=(54-36)\times 460=18\times 460=$ ₹8,280.

Trick: Z's actual amount is a distraction – only the ratio units matter. Difference $=18$ units straight away.

Answer: (D) ₹8280