RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Find each pipe’s full-tank time; filling rates add, emptying rate subtracts.

1. P fills $\frac{2}{9}$ in $28$ h ⇒ full in $28\times \frac{9}{2}=126$ h; Q: $12\times \frac{7}{2}=42$ h
2. In $6$ h: $6(\frac{1}{126}+\frac{1}{42})=6\times \frac{4}{126}=\frac{4}{21}$ filled
3. R empties $\frac{4}{21}$ in $8$ h ⇒ full tank in $8\times \frac{21}{4}=42$ h
4. Together: $\frac{1}{126}+\frac{1}{42}-\frac{1}{42}=\frac{1}{126}$ ⇒ $126$ h

Trick: R empties exactly as fast as Q fills ($42$ h each), so they cancel and only P works.

Answer: (C) $126$ hours

Concept: Interior $+$ exterior $={180}^{\circ }$, and exterior angle of a regular $n$-gon $=\frac{{360}^{\circ }}{n}$.

1. Exterior $=\frac{2}{5}\times {180}^{\circ }={72}^{\circ }$
2. $n=\frac{{360}^{\circ }}{{72}^{\circ }}=5$

Trick: Interior : exterior $=3:2$ ⇒ $n=\frac{2(3+2)}{2}=5$ (formula: $n=\frac{2(a+b)}{b}$).

Answer: (C) $5$

Concept: For two numbers: $\text{HCF}\times \text{LCM}=\text{product of the numbers}$.

1. Other number $=\frac{29\times 174}{58}$
2. $=\frac{5046}{58}=87$

Trick: Check: $58=29\times 2$, $87=29\times 3$ ⇒ HCF $29$, LCM $29\times 6=174$ ✓

Answer: (D) $87$

Concept: If $2A=7B=4C$, then $A:B:C=\frac{1}{2}:\frac{1}{7}:\frac{1}{4}$.

1. Multiply by LCM $28$: $A:B:C=14:4:7$, total $25$ parts
2. $1$ part $=\frac{32100}{25}=1284$
3. $A=14\times 1284=17976$

Trick: Ratio is the reciprocal of the multipliers: $\frac{1}{2},\frac{1}{7},\frac{1}{4}$.

Answer: (A) ₹$17976$

Concept: Sum of first $n$ odd natural numbers $=n^2$, so their average $=n$.

1. Sum $={151}^2$
2. Average $=\frac{{151}^2}{151}=151$

Yaad rakhein: Average of first $n$ odd numbers $=n$; of first $n$ even numbers $=n+1$.

Answer: (A) $151$

Concept: With the same time, equal SI means $P_1{R}_1=P_2{R}_2$, so sums are in inverse ratio of rates.

1. $P_1:P_2=4:6=2:3$ (at $6\%$ : at $4\%$)
2. Sum at $6\%=\frac{2}{5}\times 22800=9120$

Trick: Time ($8$ years) cancels out – it is not needed.

Answer: (B) ₹$9120$

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

1. Total parts $=2+22+70=94$; $1$ part $=\frac{4700}{94}=50$
2. Difference R − Q $=70-22=48$ parts $=48\times 50=2400$

Trick: Find value of one part first, then multiply by the difference of parts.

Answer: (D) ₹$2400$

Concept: Convert each ratio to a decimal (or compare with $\frac{1}{2}$) to find the greatest.

1. $\frac{30}{53}\approx 0.57$, $\frac{48}{55}\approx 0.87$
2. $\frac{22}{51}\approx 0.43$, $\frac{40}{90}\approx 0.44$
3. Greatest: $\frac{48}{55}$

Trick: Only $48:55$ has the antecedent very close to the consequent.

Answer: (B) $48:55$

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

1. $lb+bh+hl=3\times 5+5\times 8+8\times 3=15+40+24=79$
2. TSA $=2\times 79=158{\text{cm}}^2$

Trick: $79$ (option A) is only half – don’t forget the factor $2$.

Answer: (B) $158$ square cm

Concept: Take numbers $n$ and $n+1$: $n(n+1)=90$.

1. $n^2+n-90=0\Rightarrow (n+10)(n-9)=0$
2. $n=9$; numbers $9$ and $10$; greater $=10$

Trick: $\sqrt{90}\approx 9.5$ – the two numbers lie on either side: $9\times 10$.

Answer: (B) $10$

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) $1369$