RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: If divisor − remainder is the same $k$ for every divisor, number $=\text{LCM}\times n-k$.

1. $3-1=5-3=7-5=9-7=2$, so $k=2$.
2. LCM(3, 5, 7, 9) = 315.
3. Largest 6-digit multiple of 315: $315\times 3174=999810$ (since $999999=315\times 3174+189$).
4. Number $=999810-2=999808$.

Trick: Constant gap (divisor − remainder) ⇒ subtract that gap from the LCM multiple.

Answer: (A) 999808

Concept: Distance = Speed × Time (car length is negligible).

1. Time $=4.9\text{min}=\frac{4.9}{60}\text{h}$.
2. Distance $=72\times \frac{4.9}{60}=1.2\times 4.9=5.88\text{km}$.

Trick: 72 km/h = 1.2 km per minute; $1.2\times 4.9=5.88$.

Answer: (C) 5.88 km

Concept: Eliminate by quick rules, then test the remaining divisor.

1. 966181 is odd, so not divisible by 36, 38 or 30 (all even).
2. Test 37: $37\times 26113=966181$ exactly.

Trick: An odd number can never be divisible by an even number – three options go at once.

Answer: (D) 37

Concept: Write everything in $\sin \theta$ and $\cos \theta$ ($\theta ={23}^{\circ }$).

1. $\text{cosec}\theta -\sin \theta =\frac{1-{\sin }^2\theta }{\sin \theta }=\frac{{\cos }^2\theta }{\sin \theta }$.
2. $\tan \theta +\cot \theta =\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{1}{\sin \theta \cos \theta }$.
3. $\sec \theta -\cos \theta =\frac{{\sin }^2\theta }{\cos \theta }$.
4. Product $=\frac{{\cos }^2\theta ·{\sin }^2\theta }{\sin \theta ·\sin \theta \cos \theta ·\cos \theta }=1$.

Trick: The value does not depend on the angle – put $\theta ={45}^{\circ }$: $(\sqrt{2}-\frac{1}{\sqrt{2}})(2)(\sqrt{2}-\frac{1}{\sqrt{2}})=\frac{1}{2}\times 2=1$.

Answer: (A) 1

Concept: Product of two numbers = HCF × LCM.

1. Other number $=\frac{3\times 672}{96}=\frac{2016}{96}=21$.
2. Check: HCF(96, 21) = 3, LCM = 672.

Trick: Other number = $\frac{\text{HCF}\times \text{LCM}}{\text{given number}}$.

Answer: (D) 21

Concept: One-day work of A and B together $=\frac{1}{11}+\frac{1}{19}$.

1. $\frac{1}{11}+\frac{1}{19}=\frac{30}{209}$.
2. In 6 days: $\frac{180}{209}$ of the work.
3. Left $=1-\frac{180}{209}=\frac{29}{209}$.

Trick: Total work = LCM 209 units; A = 19, B = 11 units/day; 6 days = 180 units; left 29.

Answer: (D) $\frac{29}{209}$

Concept: Form two linear equations in present ages P and Q.

1. $4P-Q=19$ … (i)
2. After 7 years: $2(Q+7)=7(P+7)-47\Rightarrow 2Q+14=7P+2\Rightarrow 7P-2Q=12$ … (ii)
3. From (i) $Q=4P-19$: $7P-8P+38=12\Rightarrow P=26$.
4. $Q=4\times 26-19=85$.

Trick: Check: after 7 years P = 33, Q = 92; $7\times 33-47=184=2\times 92$ ✓.

Answer: (C) 85

Concept: Volume of shell $=\frac{4}{3}\pi (R^3-r^3)$; mass = density × volume.

1. $R^3-r^3=8-1=7$.
2. Volume $=\frac{4}{3}\times \frac{22}{7}\times 7=\frac{88}{3}{\text{cm}}^3$.
3. Weight $=7\times \frac{88}{3}=\frac{616}{3}\approx 205.33$, i.e. 205 g.

Trick: Here density 7 cancels the 7 of $\pi$: weight $=\frac{4}{3}\times 22\times 7=\frac{616}{3}$.

Answer: (B) 205

Concept: Find each person's one-day work from the given combinations.

1. Isha $=\frac{1}{22}$; Isha + Smriti $=\frac{1}{11}\Rightarrow$ Smriti $=\frac{1}{22}$.
2. Ashlesha $=\frac{1}{10}-\frac{1}{11}=\frac{1}{110}$.
3. Isha + Ashlesha $=\frac{1}{22}+\frac{1}{110}=\frac{6}{110}=\frac{3}{55}$.
4. Days $=\frac{55}{3}$.

Trick: Total work = 110 units: Isha 5, Isha+Smriti 10, all three 11 ⇒ Ashlesha 1; Isha+Ashlesha 6 ⇒ $\frac{110}{6}=\frac{55}{3}$.

Answer: (C) $\frac{55}{3}$

Concept: Simplify the inner (continued) fraction first; $+\frac{3}{4}$ and $-\frac{3}{4}$ cancel.

1. $1+\frac{3}{7}=\frac{10}{7}$, so $\frac{1}{10/7}=\frac{7}{10}$.
2. Expression $=\frac{3}{4}+\frac{7}{10}-\frac{3}{4}=\frac{7}{10}$.

Trick: Spot $+\frac{3}{4}$ and $-\frac{3}{4}$ – they cancel; only the bracket remains.

Answer: (D) $\frac{7}{10}$

Concept: Mean proportional of $a$ and $b$ is $x$ where $a:x=x:b$, so $x=\sqrt{ab}$.

1. $x=\sqrt{16\times 256}$
2. $=\sqrt{4096}=64$
3. Check: $16:64=1:4$ and $64:256=1:4$ ✓

Trick: $\sqrt{16}\times \sqrt{256}=4\times 16=64$ – no need to multiply first.

Answer: (A) $64$

Concept: Mode is the observation that occurs the maximum number of times.

1. Arrange: $10,11,12,12,12,14,14,15,16,19$
2. Frequencies: $12\to 3$ times, $14\to 2$ times, all others once.
3. Highest frequency is for $12$.

Trick: Just tally – no formula needed for ungrouped data.

Answer: (D) $12$

Concept: Total surface area of a cuboid $=2(lb+bh+hl)$.

1. $l=18,b=12,h=9$
2. $lb=216,bh=108,hl=162$
3. Sum $=486$; TSA $=2\times 486=972{\text{m}}^2$

Trick: Unit digit: $2(6+8+2)=2\times 16\to$ ends in $2$; only $972$ fits.

Answer: (D) $972{\text{m}}^2$

Concept: Find each value and subtract: $x\%$ of $N=\frac{x}{100}\times N$.

1. $65\%$ of $65=\frac{65\times 65}{100}=\frac{4225}{100}=42.25$
2. $\frac{3}{5}$ of $25=15$
3. Difference $=42.25-15=27.25$

Trick: ${65}^2=4225$ (number ending in 5: $6\times 7=42$, then $25$).

Answer: (C) $27.25$

Concept: Diagonals of a rectangle are equal and bisect each other, so $OP=OQ=OR=OS$.

1. $\angle POS=\angle QOR={36}^{\circ }$ (vertically opposite angles).
2. In $△OPS$, $OP=OS$, so $\angle OPS=\angle OSP$.
3. $\angle OPS=\frac{{180}^{\circ }-{36}^{\circ }}{2}={72}^{\circ }$

Trick: Base angle of the isosceles triangle $={90}^{\circ }-\frac{{36}^{\circ }}{2}={72}^{\circ }$.

Answer: (B) ${72}^{\circ }$

Concept: Work back from the margin: margin $=$ (winner $\%$ − loser $\%$) of valid votes.

1. Let registered voters $=N$. Votes polled $=0.9N$; valid $=82\%$ of $0.9N=0.738N$.
2. Winner $51\%$, loser $49\%$ → margin $=2\%$ of valid $=0.01476N$.
3. $0.01476N=1845\Rightarrow N=\frac{1845}{0.01476}=125000$

Trick: One line: $N\times 0.9\times 0.82\times 0.02=1845$.

Answer: (D) $125000$

Concept: BODMAS: do division and multiplication before addition and subtraction; take care of signs.

1. $(-60)÷(-6)=10$
2. $(-4)\times 6=-24$
3. $(-9)-10+(-24)=-9-10-24=-43$

Trick: Negative ÷ negative = positive; then the minus before it flips it back.

Answer: (A) $-43$

Concept: Divisibility by 13 (also 7, 11): split the number into 3-digit groups from the right; the alternating sum must be divisible by 13.

1. Not by $16$/$20$: number is odd. Groups: $716|710|501$
2. $501-710+716=507=13\times 39$ ✓ → divisible by $13$.
3. Indeed $716710501=13\times 55131577$; for $19$ the remainder is non-zero.

Trick: Odd number → kill 16 and 20 at once; test 13 by the 3-digit alternating rule.

Answer: (D) $13$

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

1. $1+\frac{3}{8}=\frac{11}{8}$, so $\frac{1}{\frac{11}{8}}=\frac{8}{11}$
2. Expression $=\frac{5}{2}+\frac{8}{11}-\frac{3}{5}$; LCM $=110$
3. $=\frac{275+80-66}{110}=\frac{289}{110}$

Trick: LCM of $2,11,5$ is $110$ – only option C has denominator $110$.

Answer: (C) $\frac{289}{110}$

Concept: The broken part, the standing part and the ground form a right triangle; standing part $h$ is opposite the ${30}^{\circ }$ angle and the broken part $18-h$ is the hypotenuse.

1. $\sin {30}^{\circ }=\frac{h}{18-h}=\frac{1}{2}$
2. $2h=18-h\Rightarrow 3h=18$
3. $h=6\text{ft}$ (broken part $=12\text{ft}$)

Trick: At ${30}^{\circ }$, standing part : broken part $=1:2$, so standing $=\frac{18}{3}=6$.

Answer: (B) $6\text{ft}$