RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: If $8A=4B=3C=k$, then $A:B:C=\frac{1}{8}:\frac{1}{4}:\frac{1}{3}$.

1. Multiply by LCM 24: $A:B:C=3:6:8$, total 17 parts.
2. 1 part $=\frac{30600}{17}=1800$.
3. A $=3\times 1800=5400$.

Trick: Take $k=24$: A = 3, B = 6, C = 8 directly.

Answer: (A) ₹5,400

Concept: $V=\pi r^{2}h$, so new volume factor $=(\text{radius factor}{)}^2\times \text{height factor}$.

1. Radius factor $=0.82$, height factor $=2.41$.
2. ${0.82}^2\times 2.41=0.6724\times 2.41=1.6205$.
3. Increase $\approx 62.05\%\approx 62\%$.

Trick: Percent change formula for $r^2$: $-18-18+\frac{324}{100}=-32.76\%$; then with $+141\%$: $-32.76+141-46.19\approx 62\%$.

Answer: (D) $62\%$

Concept: Walk one way + drive one way = 4 h 45 min; drive both ways is 2 h less.

1. Drive both ways $=4.75-2=2.75$ h → drive one way $=1.375$ h.
2. Walk one way $=4.75-1.375=3.375$ h.
3. Walk both ways $=2\times 3.375=6.75$ h $=6$ h 45 min.

Trick: Walk both $=2\times (\text{mixed})-\text{drive both}=9.5-2.75=6.75$ h.

Answer: (D) 6 hours 45 minutes

Concept: Overall average = total runs ÷ total matches.

1. Runs in 23 matches $=23\times 43=989$; next 10 matches $=10\times 14=140$.
2. Total $=1129$ runs in 33 matches.
3. Average $=\frac{1129}{33}\approx 34.21$.

Trick: $43-\frac{10\times (43-14)}{33}=43-\frac{290}{33}=43-8.79=34.21$.

Answer: (B) $34.21$

Concept: Solve innermost bracket first, then multiplication, addition and division (BODMAS).

1. $(7-3)=4$.
2. $\{25+5\times 4\}=25+20=45$.
3. $135÷45=3$.

Trick: Brackets order: ( ) → { } → [ ].

Answer: (A) 3

Concept: Angle of depression from the top = angle of elevation at the boat; $\tan \theta =\frac{\text{height}}{\text{distance}}$. Boats are on opposite sides, so add the distances.

1. $AO=\frac{150}{\tan {30}^{\circ }}=150\sqrt{3}$ m.
2. $OB=\frac{150}{\tan {60}^{\circ }}=\frac{150}{\sqrt{3}}=50\sqrt{3}$ m.
3. $AB=150\sqrt{3}+50\sqrt{3}=200\sqrt{3}$ m.

Trick: For 30° and 60° on opposite sides: distance $=h(\sqrt{3}+\frac{1}{\sqrt{3}})=\frac{4h}{\sqrt{3}}$.

Answer: (D) $200\sqrt{3}\text{m}$

Concept: Successive changes multiply: net factor $=(1-0.25)(1+0.12)$.

1. $0.75\times 1.12=0.84$.
2. Final salary is 84% of initial → lower by $16\%$.

Trick: Net % $=a+b+\frac{ab}{100}=-25+12-\frac{300}{100}=-16\%$.

Answer: (B) $16\%$

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

1. $4.70$ and $0.04$ → 470 and 4 (two decimal places).
2. $\text{LCM}(470,4)=940$.
3. Put back 2 decimal places → $9.40=9.4$.

Trick: LCM of decimals = LCM of numerators ÷ common power of 10.

Answer: (C) $9.4$

Concept: Third proportional of $a$ and $b$: $a:b=b:x$ → $x=\frac{b^2}{a}$.

1. $x=\frac{{40}^2}{16}=\frac{1600}{16}$.
2. $x=100$.

Yaad rakhein: Third proportional $=\frac{(\text{second}{)}^2}{\text{first}}$.

Answer: (D) 100

Concept: A decimal with 4 decimal places has a square root with 2 decimal places.

1. $1.0201=\frac{10201}{10000}$.
2. $\sqrt{10201}=101$ since ${101}^2=10201$.
3. $\sqrt{1.0201}=\frac{101}{100}=1.01$.

Trick: $(1.01{)}^2=1.0201$ — check the option by squaring.

Answer: (D) $1.01$

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

1. In right $△BDC$: $\angle BDC={90}^{\circ }$, $\angle DBC={39}^{\circ }$, so $\angle C={180}^{\circ }-{90}^{\circ }-{39}^{\circ }={51}^{\circ }$.
2. In $△AEC$: $\angle EAC={16}^{\circ }$ and $\angle ECA={51}^{\circ }$.
3. $\angle AEB$ is the exterior angle of $△AEC$ at $E$: $\angle AEB={16}^{\circ }+{51}^{\circ }={67}^{\circ }$.

Trick: $\angle AEB=\angle CAE+({90}^{\circ }-\angle DBC)={16}^{\circ }+{90}^{\circ }-{39}^{\circ }={67}^{\circ }$.

Answer: (C) ${67}^{\circ }$

Concept: For three equal distances at speeds $a,b,c$: $\text{Average speed}=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ (harmonic mean).

1. $\frac{1}{2}+\frac{1}{87}+\frac{1}{53}=0.5+0.01149+0.01887=0.53036$
2. $\text{Average}=\frac{3}{0.53036}\approx 5.656\text{km/h}$
3. Rounded to 1 decimal place: $5.7\text{km/h}$.

Trick: The slow third alone takes $\frac{d}{2}$ hours, so the average is a little below $3\times 2=6\text{km/h}$ (the fast parts add very little time) – hence $5.7$.

Answer: (B) $5.7$

Concept: Same power: $a^m\times b^m\times c^m=(abc{)}^m$, and $x^{-m}=\frac{1}{x^m}$.

1. $5^{-23}\times {16}^{-23}\times {21}^{-23}=(5\times 16\times 21{)}^{-23}$
2. $5\times 16\times 21=80\times 21=1680$
3. ${1680}^{-23}=\frac{1}{{1680}^{23}}$

Trick: Only the base product matters: $5\times 16=80$, $80\times 21=1680$ – match it with the denominators.

Answer: (B) $\frac{1}{{1680}^{23}}$

Concept: Test each option with its divisibility rule.

1. By 8: last three digits $541$ is odd, so not divisible by $8$ (nor by $2$, so not by $12$).
2. By 15: last digit is $1$, not $0$ or $5$, so not divisible by $5$ or $15$.
3. By 7: $814242541÷7=116320363$ exactly ($7\times 116320363=814242541$), so it is divisible by $7$.

Trick: An odd number ending in $1$ cannot be divisible by $8$, $12$ or $15$ – so eliminate them and $7$ is left.

Answer: (C) $7$

Concept: No overall profit/loss means the loss on one equals the gain on the other in rupees.

1. Let CP of first $=x$; second $=1200-x$.
2. $0.08x=0.22(1200-x)\Rightarrow 0.30x=264\Rightarrow x=880$
3. Check: loss $=0.08\times 880=70.4$; gain $=0.22\times 320=70.4$ – equal.

Trick: Alligation: $-8\%$ and $+22\%$ around $0\%$ give ratio $22:8=11:4$; first $=\frac{11}{15}\times 1200=880$.

Answer: (C) ₹$880$

Concept: Pipes working together: add their one-hour work, $\frac{1}{T}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

1. $\frac{1}{17}+\frac{1}{21}+\frac{1}{19}=\frac{399+323+357}{6783}=\frac{1079}{6783}$ (LCM $=17\times 21\times 19=6783$)
2. $T=\frac{6783}{1079}$ hours
3. $6783=6\times 1079+309$, so $T=6\frac{309}{1079}$ hours.

Trick: All three pipes take about 19 h each, so together roughly $\frac{19}{3}\approx 6.3$ h – only $6\frac{309}{1079}$ fits.

Answer: (C) $6\frac{309}{1079}$ hours

Concept: Express everything in terms of CP: a loss of $30\%$ means $\text{SP}=0.7\text{CP}$.

1. Let actual SP $=S$, CP $=C$. Then $\frac{4}{14}S=\frac{2}{7}S=0.7C$.
2. $S=\frac{0.7\times 7}{2}C=2.45C$
3. New SP $=54\%$ of $S=0.54\times 2.45C=1.323C$
4. $\text{Profit}\%=(1.323-1)\times 100=32.3\%$

Trick: Take $C=100$: $\frac{2}{7}S=70\Rightarrow S=245$; $54\%$ of $245=132.3$ → profit $32.3\%$.

Answer: (D) $32.3\%$

Concept: Least number of square tiles: tile side $=\text{HCF}$ of length and breadth; number $=\frac{\text{Area of floor}}{\text{Area of tile}}$.

1. $12.15\text{m}=1215\text{cm}$, $11.34\text{m}=1134\text{cm}$.
2. $\text{HCF}(1215,1134)$: $1215-1134=81$ and $1134=81\times 14$, so HCF $=81\text{cm}$.
3. Tiles $=\frac{1215}{81}\times \frac{1134}{81}=15\times 14=210$

Trick: Difference $1215-1134=81$ – check if it divides both: $81\times 15$ and $81\times 14$ ✔.

Answer: (C) $210$

Concept: Same time, equal interest: $P_1{R}_1=P_2{R}_2$, so principals are in the inverse ratio of rates.

1. $P_1\times 13\times 7=P_2\times 3\times 7\Rightarrow P_1:P_2=3:13$
2. Sum at $13\%$ $=\frac{3}{16}\times 22400=4200$
3. Check: $\frac{4200\times 13\times 7}{100}=3822$ and $\frac{18200\times 3\times 7}{100}=3822$ – equal.

Trick: Equal SI → swap the rates: amounts in ratio $3:13$ (the higher rate gets the smaller sum).

Answer: (B) ₹$4,200$

Concept: Median: arrange data in order; for odd $n$, median $={\left(\frac{n+1}{2}\right)}^{\text{th}}$ value.

1. Ascending order: 16, 50, 55, 57, 60, 69, 72, 73, 78, 91, 96.
2. $n=11$, so median $={\left(\frac{11+1}{2}\right)}^{\text{th}}=6^{\text{th}}$ value.
3. The 6th value is $69$.

Trick: Never take the middle of the unsorted list (that would be $50$) – always sort first.

Answer: (C) $69$