RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: For decimals, make the decimal places equal, find the LCM of the numbers without the decimal point, then put back the same number of decimal places.

1. Write both with 2 decimal places: $10.80$ and $0.06$ → $1080$ and $6$.
2. $\text{LCM}(1080,6)=1080$ (since $1080÷6=180$).
3. Put back 2 decimal places: $10.80=10.8$.

Trick: If the larger number is a multiple of the smaller ($10.8÷0.06=180$), the LCM is the larger number itself.

Answer: (A) $10.8$

Concept: Simplify the bracket first, open brackets by multiplying, then combine like terms.

1. $\frac{6}{3}=2$, so the first part is $3(2x^2-21x+18)=6x^2-63x+54$.
2. Second part: $-6(x^2+5x-18)=-6x^2-30x+108$.
3. Add: $x^2$ terms $6x^2-6x^2=0$; $x$ terms $-63x-30x=-93x$; constants $54+108=162$.
4. Result: $-93x+162$.

Trick: Watch the sign: $-6\times (-18)=+108$, so the constant is positive and $x$-term negative → only (D) fits.

Answer: (D) $-93x+162$

Concept: In 'buy $x$ get $y$ free', the customer takes $x+y$ items but pays for $x$; the free items are the discount: $\text{Discount}\%=\frac{y}{x+y}\times 100$.

1. Buy 5, get 3 free → total items $=5+3=8$.
2. Free items $=3$ out of $8$.
3. $\text{Discount}\%=\frac{3}{8}\times 100=37.5\%$.

Trick: $\frac{\text{free}}{\text{total taken}}$ – never divide by items paid ($\frac{3}{5}=60\%$ is the trap).

Answer: (D) $37.5\%$

Concept: $\text{Average}=\frac{\text{Sum of observations}}{\text{Number of observations}}$.

1. Sum $=40+49+56+75+36=256\text{kg}$.
2. Number of members $=5$.
3. Average $=\frac{256}{5}=51.2\text{kg}$.

Trick: Assume 50: deviations $-10,-1,+6,+25,-14$ sum to $+6$; $50+\frac{6}{5}=51.2$.

Answer: (D) $51.2\text{kg}$

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

1. Brijesh's share $=\frac{4400}{16000}=\frac{11}{40}$; Kushal's share $=\frac{40-11}{40}=\frac{29}{40}$.
2. So investments are in ratio $29:11$, total $40$ parts $=₹75100$.
3. Difference $=29-11=18$ parts $=\frac{18}{40}\times 75100=₹33795$.

Trick: Difference of investments = total investment × $\frac{\text{difference of profit shares}}{\text{total profit}}=75100\times \frac{7200}{16000}$.

Answer: (D) ₹$33,795$

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

1. In right $△BDC$ ($\angle BDC={90}^{\circ }$): $\angle C={180}^{\circ }-{90}^{\circ }-{72}^{\circ }={18}^{\circ }$.
2. In $△AEC$: $\angle EAC={56}^{\circ }$ and $\angle ECA={18}^{\circ }$.
3. $\angle AEB$ is the exterior angle of $△AEC$ at $E$ (since $B$, $E$, $C$ are collinear).
4. $\angle AEB={56}^{\circ }+{18}^{\circ }={74}^{\circ }$.

Trick: $\angle AEB=\angle CAE+({90}^{\circ }-\angle DBC)={56}^{\circ }+{18}^{\circ }$.

Answer: (D) ${74}^{\circ }$

Concept: If two consecutive numbers multiply to $N$, they lie just around $\sqrt{N}$.

1. Let the numbers be $n$ and $n+1$: $n(n+1)=240$.
2. $\sqrt{240}\approx 15.5$, so try $15\times 16=240$. ✓
3. (Or $n^2+n-240=0\Rightarrow (n+16)(n-15)=0\Rightarrow n=15$.)
4. Greater number $=16$.

Trick: Check options: only $16\times 15=240$; $19\times 18$, $23\times 22$ are far bigger.

Answer: (D) $16$

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

1. $7\frac{2}{3}=\frac{23}{3}$ years: $4347=\frac{P\times 5.4\times 23}{3\times 100}=P\times 0.414$.
2. $P=\frac{4347}{0.414}=₹10500$.
3. New SI at $7.5\%$ for $4\frac{4}{5}=4.8$ years: $\frac{10500\times 7.5\times 4.8}{100}=₹3780$.
4. Amount $=10500+3780=₹14280$.

Trick: $R\times T$ first: $5.4\times \frac{23}{3}=41.4\%$ of $P$ is ₹4347; then $7.5\times 4.8=36\%$, so amount $=136\%$ of $10500$.

Answer: (B) ₹$14,280$

Concept: The mean proportional $x$ between $a$ and $b$ satisfies $a:x=x:b$, so $x=\sqrt{ab}$.

1. $x=\sqrt{26\times 104}$.
2. $104=4\times 26$, so $26\times 104=4\times {26}^2$.
3. $x=\sqrt{4\times {26}^2}=2\times 26=52$.

Trick: When $b=4a$, mean proportional $=2a$.

Answer: (B) $52$

Concept: Area of four walls of a room $=2h(l+b)$; for a cube with side $a$ it becomes $4a^2$ (floor and ceiling are not painted).

1. Side $a=18.5\text{ft}$.
2. $a^2={18.5}^2=342.25{\text{ft}}^2$.
3. Four walls $=4\times 342.25=1369{\text{ft}}^2$.

Trick: $4a^2=(2a{)}^2={37}^2=1369$.

Answer: (D) $1369$

Concept: Savings = Income − Expenditure; apply each % change separately.

1. Old savings $=31\%\times 44200=13702$, old expenditure $=44200-13702=30498$.
2. New income $=44200\times 1.22=53924$.
3. New expenditure $=30498\times 1.5=45747$.
4. New savings $=53924-45747=8177$; change $=13702-8177=5525$ (a fall).

Trick: Change in savings $=22\%\text{ of }44200-50\%\text{ of }30498=9724-15249=-5525$ – one line!

Answer: (B) decrease by ₹5,525

Concept: A number is divisible by 11 if (sum of odd-place digits) − (sum of even-place digits) is 0 or a multiple of 11.

1. Digits of N46M66: odd places N, 6, 6 → $N+12$; even places 4, M, 6 → $M+10$.
2. Difference $=(N+12)-(M+10)=N-M+2$.
3. For the smallest case, $N-M+2=0\Rightarrow M-N=2$.

Trick: Test option (D) quickly: $M=N+2$, e.g. $N=1,M=3$: 146366 $=11\times 13306$ ✓.

Answer: (D) $M-N=2$

Concept: Take distance MN as 1 unit; speeds = $\frac{1}{\text{time}}$. Time to meet = gap ÷ relative speed.

1. A takes 9 h ($\frac{1}{9}$ per h), B takes 5 h ($\frac{1}{5}$ per h).
2. By 8:30 AM, A has gone 2 h: $\frac{2}{9}$; gap left $=\frac{7}{9}$.
3. Relative speed $=\frac{1}{9}+\frac{1}{5}=\frac{14}{45}$ per h.
4. Time $=\frac{7}{9}÷\frac{14}{45}=\frac{7}{9}\times \frac{45}{14}=2.5$ h → 8:30 + 2:30 = 11:00 AM.

Trick: Take MN = LCM(9, 5) = 45 km: speeds 5 and 9 km/h; gap at 8:30 = 35 km; $35÷14=2.5$ h.

Answer: (B) 11:00 AM

Concept: SP $=$ CP × (1 + markup) × (1 − discount).

1. Let CP $=x$; list price $=1.5x$.
2. SP $=1.5x\times 0.7=1.05x$, so profit $=0.05x$.
3. $0.05x=86\Rightarrow x=1720$.
4. List price $=1.5\times 1720=2580$.

Trick: Net effect of +50% and −30%: $50-30-\frac{50\times 30}{100}=5\%$ profit → CP $=86\times 20$.

Answer: (B) 2,580

Concept: Sum = average × number. Common member = (sum of two pairs) − (total).

1. K + T + S $=46\times 3=138$.
2. K + T $=72$, T + S $=88$.
3. Adding pairs: K + 2T + S $=160$; T $=160-138=22$.

Trick: Common person = sum of both pairs − total sum.

Answer: (B) 22

Concept: $\sqrt{0.04}=0.2$ because ${0.2}^2=0.04$; decimal places in the root = half of those in the number.

1. $\sqrt{289}=17$.
2. $\sqrt{0.04}=0.2$.
3. $\sqrt{5.76}=2.4$ (since ${24}^2=576$).
4. $17+0.2-2.4=14.8$.

Trick: Remove decimals: $\sqrt{576}=24$, then put back one decimal place → 2.4.

Answer: (C) 14.8

Concept: For a fixed distance, speed ∝ $\frac{1}{\text{time}}$.

1. Distance $=99\times 20=1980$ km.
2. Required speed $=\frac{1980}{12}=165$ km/hr.

Trick: $99\times \frac{20}{12}=99\times \frac{5}{3}=165$.

Answer: (A) 165 km/hr

Concept: Form an equation from the past condition, then use it for the future.

1. 8 years ago: $F-8=2(S-8)+21\Rightarrow F=2S+13$.
2. After $x$ years: $F+x=2(S+x)\Rightarrow x=F-2S$.
3. $x=13$ years.

Trick: "Twice" condition: years needed $=F-2S$; here $F-2S=13$ directly.

Answer: (A) 13

Concept: Convert both to numbers and subtract.

1. $65\%\text{ of }55=\frac{65\times 55}{100}=35.75$.
2. $\frac{2}{5}\text{ of }25=10$.
3. Difference $=35.75-10=25.75$.

Trick: $65\%\text{ of }55=55\%\text{ of }65$: $0.65\times 55=35.75$.

Answer: (B) 25.75

Concept: Product of two numbers = HCF × LCM.

1. Other number $=\frac{3\times 276}{69}=\frac{828}{69}=12$.
2. Check: HCF(69, 12) = 3 and LCM = 276 ✓.

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

Answer: (A) 12