RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: $1+{\cot }^2\theta ={\csc }^2\theta$ and $1+{\tan }^2\theta ={\sec }^2\theta$.

1. $\frac{1+{\cot }^2\theta }{1+{\tan }^2\theta }=\frac{{\csc }^2\theta }{{\sec }^2\theta }$.
2. $=\frac{{\cos }^2\theta }{{\sin }^2\theta }={\cot }^2\theta$.
3. ${\cot }^2\theta ={\csc }^2\theta -1$.

Trick: Put $\theta ={45}^{\circ }$: expression $=1$; only ${\csc }^2\theta -1=2-1=1$ fits.

Answer: (A) ${\csc }^2\theta -1$

Concept: Successive change: net $\%=a-b-\frac{ab}{100}$ for $+a\%$ then $-b\%$.

1. $24-13-\frac{24\times 13}{100}=11-3.12=7.88\%$.
2. Check: $100\to 124\to 124\times 0.87=107.88$.

Trick: Take CP = 100; MP = 124; SP = $124-16.12=107.88$.

Answer: (C) 7.88%

Concept: For equal distances at speeds $a$ and $b$: average speed $=\frac{2ab}{a+b}$.

1. $\frac{2\times 9\times 18}{9+18}=\frac{324}{27}=12$.

Trick: Take distance 18 km: time $2+1=3$ h, total 36 km → $\frac{36}{3}=12$.

Answer: (B) 12

Concept: Form two linear equations in mother's age $M$ and daughter's age $D$.

1. $2M=5D+25$ and $3D=M-9\Rightarrow M=3D+9$.
2. $2(3D+9)=5D+25\Rightarrow D=7$.
3. $M=30$; difference $=30-7=23$.

Trick: Check options quickly using $M=3D+9$: difference $=2D+9$ must be odd.

Answer: (D) 23

Concept: A number is divisible by 9 if its digit sum is divisible by 9.

1. Digit sum $=4+8+8+3+0+9+1+3+9=45$, divisible by 9 ✓.
2. Not by 10 (last digit 9).
3. By 11: alternating sum $=(9+1+0+8+4)-(3+9+3+8)=22-23=-1$, not divisible.

Trick: Digit sum 45 → divisible by both 3 and 9.

Answer: (A) 9

Concept: Make decimal places equal, find LCM of the integers, then put back the decimal places.

1. $2.600$ and $0.084$ → 2600 and 84 (3 decimal places).
2. $2600=2^3\times 5^2\times 13$, $84=2^2\times 3\times 7$.
3. LCM $=2^3\times 3\times 5^2\times 7\times 13=54600$ → $54.600=54.6$.

Trick: LCM of decimals must be a multiple of the bigger number 2.6: $54.6=2.6\times 21$ ✓.

Answer: (D) 54.6

Concept: For the same time period, profit is shared in the ratio of investments.

1. Total investment $=1220+1370+1110=3700$.
2. Kiran's share $=\frac{1110}{3700}\times 1160=0.3\times 1160=348$.

Trick: $\frac{1110}{3700}=\frac{3}{10}$ exactly — 30% of 1160.

Answer: (D) ₹348

Concept: Overall average = total runs ÷ total matches.

1. Runs in 23 matches $=23\times 43=989$.
2. Runs in 10 matches $=10\times 13=130$.
3. Average $=\frac{989+130}{33}=\frac{1119}{33}\approx 33.91$.

Trick: Deviation method: $43-\frac{10\times 30}{33}=43-9.09=33.91$.

Answer: (D) 33.91

Concept: Overhead (decoration) is added to cost price. $\text{Loss}\%=\frac{\text{Loss}}{\text{CP}}\times 100$.

1. CP $=670+10=680$.
2. SP $=290$ → loss $=680-290=390$.
3. $\text{Loss}\%=\frac{390}{680}\times 100=\frac{975}{17}\%$ (≈ 57.35%).

Trick: SP is less than CP → only the 'loss' options; then divide 39000 by 680.

Answer: (D) $\frac{975}{17}\%$ loss

Concept: Together, rates add: $\frac{1}{T}=\frac{1}{16}+\frac{1}{21}+\frac{1}{11}$.

1. LCM $=16\times 21\times 11=3696$.
2. $\frac{231+176+336}{3696}=\frac{743}{3696}$ tank per hour.
3. $T=\frac{3696}{743}=4\frac{724}{743}$ hours.

Trick: All options share the fraction $\frac{724}{743}$; only the whole part $3696÷743=4$ is needed.

Answer: (C) $4\frac{724}{743}$ hours

Concept: Follow BODMAS: division and multiplication before addition and subtraction.

1. $(-60)÷(-10)=6$ and $(-2)\times 7=-14$
2. Expression $=-9-6+(-14)$
3. $=-9-6-14=-29$

Trick: Signs: $(-)÷(-)=+$, $(-)\times (+)=-$.

Answer: (D) $-29$

Concept: Simplify the complex fraction first, then take the LCM.

1. $1+\frac{3}{8}=\frac{11}{8}\Rightarrow \frac{1}{11/8}=\frac{8}{11}$
2. $\frac{5}{2}+\frac{8}{11}-\frac{5}{3}$, LCM $=66$
3. $=\frac{165+48-110}{66}=\frac{103}{66}$

Trick: Denominator must be a factor of 66 (2, 11, 3) – only $\frac{103}{66}$ fits.

Answer: (B) $\frac{103}{66}$

Concept: 54,10,96,943 in the Indian system = 541096943. Test each option by division.

1. $541096943÷97=5578319$ exactly (remainder 0).
2. $541096943÷103$ leaves remainder 39; $÷87$ leaves 8; $÷107$ leaves 83.
3. Also 87 = $3\times 29$: digit sum $=4+1+0+9+6+9+4+3+5=41$, not divisible by 3, so 87 is out at once.

Trick: Eliminate fast: digit-sum rule kills 87; then divide by the remaining primes.

Answer: (B) 97

Concept: Use ${\tan }^2\theta =\frac{1}{{\cot }^2\theta }$, ${\sec }^2\theta =1+{\tan }^2\theta$, ${\text{cosec}}^2\theta =1+{\cot }^2\theta$.

1. ${\tan }^2\theta =1-a^2$, so $\sec \theta =\sqrt{2-a^2}$
2. ${\text{cosec}}^2\theta =1+\frac{1}{1-a^2}=\frac{2-a^2}{1-a^2}$, so $\text{cosec}\theta =\frac{\sqrt{2-a^2}}{\sqrt{1-a^2}}$
3. $\text{cosec}\theta {\tan }^3\theta =\frac{\sqrt{2-a^2}}{\sqrt{1-a^2}}\times (1-a^2{)}^{3/2}=\sqrt{2-a^2}(1-a^2)$
4. Sum $=\sqrt{2-a^2}[1+1-a^2]=(2-a^2{)}^{\frac{3}{2}}$

Trick: Put $a=0$: $\theta ={45}^{\circ }$, value $=\sqrt{2}+\sqrt{2}=2\sqrt{2}=2^{3/2}$ – only option A gives this.

Answer: (A) $(2-a^2{)}^{\frac{3}{2}}$

Concept: For successive discounts, multiply by $(1-\frac{d_1}{100})(1-\frac{d_2}{100})$.

1. SP $=7500\times \frac{92}{100}\times \frac{95}{100}$
2. $7500\times 0.92=6900$
3. $6900\times 0.95=6555$

Trick: Net discount $=8+5-\frac{8\times 5}{100}=12.6\%$; $7500\times 0.874=6555$.

Answer: (B) ₹6,555

Concept: Open the brackets carefully and combine like terms.

1. $3(\frac{4}{3}{x}^2-24x+18)=4x^2-72x+54$
2. $-4(x^2+7x-18)=-4x^2-28x+72$
3. Add: $(4x^2-4x^2)+(-72x-28x)+(54+72)=-100x+126$

Trick: The $x^2$ terms cancel, so check only the $x$ and constant terms: $-72-28=-100$, $54+72=126$.

Answer: (B) $-100x+126$

Concept: If SP of $a$ articles = CP of $b$ articles, then gain/loss $\%=\frac{b-a}{a}\times 100$.

1. SP of 50 = CP of 16, so SP is less than CP → loss.
2. Loss on 50 books $=$ CP of $(50-16)=34$ books
3. $\text{Loss}\%=\frac{34}{50}\times 100=68\%$

Trick: Smaller number on the CP side → loss; $\frac{\text{difference}}{\text{SP-side count}}\times 100$.

Answer: (C) 68% loss

Concept: Angle sum of a triangle is ${180}^{\circ }$; an exterior angle equals the sum of the two interior opposite angles.

1. In right $△BDC$: $\angle C={180}^{\circ }-{90}^{\circ }-{29}^{\circ }={61}^{\circ }$
2. In $△AEC$, $\angle AEB$ is the exterior angle at E.
3. $\angle AEB=\angle CAE+\angle ACE={25}^{\circ }+{61}^{\circ }={86}^{\circ }$

Trick: Exterior angle = sum of the two far interior angles – no need to find $\angle AEC$.

Answer: (B) ${86}^{\circ }$

Concept: Metal volume = outer volume − inner (hollow) volume. Open box: thickness is reduced from 2 sides in length and width, but only from the bottom in height.

1. Outer volume $=60\times 40\times 23=55200{\text{cm}}^3$
2. Inner: $54\times 34\times 20=36720{\text{cm}}^3$
3. Metal $=55200-36720=18480{\text{cm}}^3$
4. Weight $=18480\times 0.5=9240\text{g}=9.24\text{kg}$

Trick: Open box → height reduces by thickness only once ($23-3=20$).

Answer: (B) 9.24

Concept: Profit share ∝ capital × time.

1. $M+N=4064$, $M-N=508\Rightarrow M=2286,N=1778$; ratio $=9:7$
2. Let N's capital $=x$: $\frac{4(x+39000)}{7x}=\frac{9}{7}$
3. $4x+156000=9x\Rightarrow x=31200$
4. M's capital $=31200+39000=70200$

Trick: $2286:1778$ – divide both by 254 to get $9:7$ quickly.

Answer: (D) ₹70,200