RRB NTPC Undergraduate CBT 1 2024 PYQs

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) $210{\text{cm}}^2$

Concept: Simplify the inner (continued) fraction first, then combine like fractions.

1. $1+\frac{4}{7}=\frac{11}{7}\Rightarrow \frac{1}{11/7}=\frac{7}{11}$
2. $\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{2}$
3. $\frac{1}{2}+\frac{7}{11}=\frac{11+14}{22}=\frac{25}{22}$

Trick: Group $\frac{5}{4}$ and $-\frac{3}{4}$ first – same denominator, instant $\frac{1}{2}$.

Answer: (C) $\frac{25}{22}$

Concept: Increasing a number by $100\%$ doubles it.

1. $x+100\%\text{ of }x=2x$
2. $2x=2790$
3. $x=1395$

Trick: Increase by $100\%$ ⇒ ×2, so just halve the result: $\frac{2790}{2}=1395$.

Answer: (D) 1395

Concept: For equal time, profit is shared in the ratio of capitals.

1. Ratio $1110:1760:1230=111:176:123$
2. Sum of ratio $=410$
3. Naveen's share $=\frac{123}{410}\times 1620=0.3\times 1620=486$

Trick: $\frac{1230}{4100}=0.3$ exactly – Naveen gets $30\%$ of the profit.

Answer: (B) ₹486

Concept: Selling price $=\text{MP}\times (1-\frac{d_1}{100})(1-\frac{d_2}{100})$.

1. After 12%: $2860\times 0.88=2516.80$
2. $\frac{2390.96}{2516.80}=0.95$
3. Second discount $=1-0.95=5\%$

Trick: Check option: $2516.80\times 5\%=125.84$ and $2516.80-125.84=2390.96$ ✓.

Answer: (C) $5\%$

Concept: In a two-candidate contest, margin $=$ (winner's % − loser's %) of valid votes.

1. Let registered $=x$; polled $=0.8x$; valid $=85\%$ of $0.8x=0.68x$
2. Margin $=(60-40)\%$ of $0.68x=0.136x$
3. $0.136x=1394\Rightarrow x=10250$

Trick: Chain multipliers: $0.8\times 0.85\times 0.2=0.136$; $\frac{1394}{0.136}=10250$.

Answer: (B) 10250

Concept: One day's work $=\frac{1}{\text{days}}$; subtract combined rates to get an individual rate.

1. Anjani $=\frac{1}{24}$; Anjani + Khushbu $=\frac{1}{20}$; all three $=\frac{1}{8}$
2. Sushmita $=\frac{1}{8}-\frac{1}{20}=\frac{5-2}{40}=\frac{3}{40}$
3. Anjani + Sushmita $=\frac{1}{24}+\frac{3}{40}=\frac{5+9}{120}=\frac{7}{60}$ ⇒ $\frac{60}{7}$ days

Trick: LCM method: total work 120 units; A = 5, A+K = 6, A+K+S = 15 ⇒ S = 9, A+S = 14 ⇒ $\frac{120}{14}=\frac{60}{7}$.

Answer: (C) $\frac{60}{7}$ days

Concept: For a decimal, the square root has half as many decimal places: $\sqrt{0.0324}=0.18$ since ${18}^2=324$.

1. $\sqrt{196}=14$
2. $\sqrt{0.0324}=0.18$, $\sqrt{6.76}=2.6$
3. $14+0.18-2.6=11.58$

Trick: 4 decimal places → root has 2; 2 decimal places → root has 1.

Answer: (B) 11.58

Concept: For complementary angles: $\sin ({90}^{\circ }-\theta )=\cos \theta$, $\sec ({90}^{\circ }-\theta )=\csc \theta$; and ${\sin }^2\theta +{\cos }^2\theta =1$, ${\csc }^2\theta -{\cot }^2\theta =1$.

1. ${\sin }^2{50}^{\circ }={\cos }^2{40}^{\circ }$ ⇒ numerator $={\sin }^2{40}^{\circ }+{\cos }^2{40}^{\circ }=1$; denominator $={\cos }^2{40}^{\circ }+{\sin }^2{40}^{\circ }=1$
2. ${\sec }^2{50}^{\circ }={\csc }^2{40}^{\circ }$ ⇒ ${\csc }^2{40}^{\circ }-{\cot }^2{40}^{\circ }=1$
3. ${\cos }^2{70}^{\circ }={\sin }^2{20}^{\circ }$ ⇒ denominator $=1$; value $=\frac{1}{1}-\frac{1}{1}=0$

Trick: Angles adding to ${90}^{\circ }$ in each pair → every part collapses to 1, so $1-1=0$.

Answer: (C) 0

Concept: Expand each bracket (watch the sign of the multiplier) and combine like terms.

1. $2(\frac{3}{2}{x}^2-25x+14)=3x^2-50x+28$
2. $-3(x^2+4x-13)=-3x^2-12x+39$
3. Sum: $(3x^2-3x^2)+(-50x-12x)+(28+39)=-62x+67$

Trick: The $x^2$ terms cancel; only check signs: $x$-term $-50-12=-62$, constant $28+39=67$.

Answer: (C) $-62x+67$

Concept: Combined average $=\frac{\text{total runs}}{\text{total matches}}$.

1. Total in 23 matches $=23\times 41=943$
2. Total in 10 matches $=10\times 15=150$
3. Average $=\frac{943+150}{33}=\frac{1093}{33}\approx 33.12$

Trick: Deviation method: from 41, the 10 matches are 26 short each → $41-\frac{260}{33}=41-7.88=33.12$.

Answer: (D) 33.12