RRB NTPC Undergraduate CBT 1 2024 PYQs

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) $69$

Concept: If $a:b=b:x$, then $x$ is the third proportional of $a$ and $b$, so $x=\frac{b^2}{a}$.

1. $9:57=57:x$
2. $x=\frac{{57}^2}{9}=\frac{3249}{9}$
3. $x=361$

Trick: $\frac{{57}^2}{9}={\left(\frac{57}{3}\right)}^2={19}^2=361$ – divide first, then square.

Answer: (B) $361$

Concept: Two successive discounts $a\%$ and $b\%$ equal a single discount of $(a+b-\frac{ab}{100})\%$. The marked price does not matter.

1. $a=10,b=15$
2. Single discount $=10+15-\frac{10\times 15}{100}=25-1.5$
3. $=23.5\%$

Trick: Multiplier method: $0.90\times 0.85=0.765$, so discount $=1-0.765=23.5\%$.

Answer: (A) $23.5\%$

Concept: Curved surface area of a cylinder $=2\pi rh$, so $r=\frac{\text{CSA}}{2\pi h}$.

1. $2\times \frac{22}{7}\times r\times 68=1500$
2. $r=\frac{1500\times 7}{2\times 22\times 68}=\frac{10500}{2992}$
3. $r\approx 3.509\approx 3.51\text{cm}$

Trick: $2\pi h\approx \frac{44}{7}\times 68\approx 427.4$; $\frac{1500}{427.4}$ is a little above 3.5 → only 3.51 fits.

Answer: (A) $3.51$

Concept: Compound amount $A=P{(1+\frac{r}{100})}^n$. Take the ratio $\frac{A}{P}$ and find its $n$-th root.

1. ${(1+\frac{r}{100})}^2=\frac{841}{625}$
2. $1+\frac{r}{100}=\sqrt{\frac{841}{625}}=\frac{29}{25}$
3. $\frac{r}{100}=\frac{4}{25}\Rightarrow r=16\%$

Trick: $625={25}^2$ and $841={29}^2$; rate $=\frac{29-25}{25}\times 100=16\%$.

Answer: (C) $16\%$

Concept: Distance $=$ Speed $\times$ Time (units must match). Ignoring the car's length, the distance covered while crossing is the bridge length.

1. Time $=1.6\text{min}=\frac{1.6}{60}\text{h}$
2. Distance $=84\times \frac{1.6}{60}$
3. $=1.4\times 1.6=2.24\text{km}$

Trick: $84\text{km/h}=\frac{84}{60}=1.4\text{km}$ per minute; $1.4\times 1.6=2.24$.

Answer: (A) $2.24\text{km}$

Concept: A number is divisible by 5 only if its last digit is 0 or 5.

1. 61,75,753 ends in 3 – no
2. 65,67,359 ends in 9 – no; 71,32,954 ends in 4 – no
3. 70,48,030 ends in 0 – yes

Yaad rakhein: Just look at the units digit: 0 or 5 means divisible by 5.

Answer: (D) 70,48,030

Concept: Write present ages as $F$ and $S$, form equations from the given conditions.

1. 2 years ago: $F-2=2(S-2)+22\Rightarrow F=2S+20$
2. After $x$ years: $F+x=2(S+x)\Rightarrow x=F-2S$
3. $x=(2S+20)-2S=20$

Trick: Today the father is $2S+20$, i.e. 20 years more than twice the son; that excess vanishes after exactly 20 years.

Answer: (D) 20 years

Concept: Let the consecutive even numbers be $n$ and $n+2$; form a quadratic from the sum of squares.

1. $n^2+(n+2{)}^2=2180\Rightarrow 2n^2+4n-2176=0$
2. $n^2+2n-1088=0\Rightarrow (n-32)(n+34)=0$
3. $n=32$; numbers 32 and 34; sum $=66$

Trick: Each square ≈ $\frac{2180}{2}=1090\approx {33}^2$, so the numbers are 32 and 34: $1024+1156=2180$.

Answer: (C) 66

Concept: Profit is shared in the ratio of (capital $\times$ time).

1. M + N $=1860$, M − N $=372$ ⇒ M $=1116$, N $=744$, ratio $3:2$
2. $\frac{7(N+35000)}{5N}=\frac{3}{2}\Rightarrow 14N+490000=15N$
3. $N=490000$, so M's capital $=490000+35000=525000$

Trick: Check option: M = 5,25,000 ⇒ N = 4,90,000; $\frac{525000\times 7}{490000\times 5}=\frac{3}{2}$ ✓.

Answer: (D) ₹5,25,000

Concept: Area of a triangle $=\frac{1}{2}\times \text{base}\times \text{height}$.

1. Base $=28\text{cm}$, height $=15\text{cm}$
2. Area $=\frac{1}{2}\times 28\times 15=14\times 15$
3. $=210{\text{cm}}^2$

Trick: Halve the even number first: $14\times 15=210$.

Answer: (C) $210{\text{cm}}^2$