RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Overall literacy % is a weighted average of male and female literacy %.

1. Females $=368000-220000=148000$
2. Total literate $=62\%$ of 368000; male literate $=62\%$ of 220000
3. Female literate $=62\%\times (368000-220000)=62\%$ of 148000
4. So female literacy $=62\%$

Trick: If the overall average and one group's value are equal (62%), the other group must also be 62%.

Answer: (D) 62%

Concept: Write the condition as an equation and divide.

1. $\frac{20}{100}A=\frac{4}{5}B$
2. $\frac{A}{5}=\frac{4B}{5}\Rightarrow A=4B$
3. $A:B=4:1$

Trick: $\frac{A}{B}=\frac{4/5}{1/5}=4$.

Answer: (B) 4 : 1

Concept: Curved surface area $=2\pi rh$; volume $=\pi r^{2}h$.

1. $2\times \frac{22}{7}\times 21\times h=5280\Rightarrow 132h=5280\Rightarrow h=40\text{cm}$
2. $V=\frac{22}{7}\times {21}^2\times 40=1386\times 40$
3. $V=55440{\text{cm}}^3$

Trick: $V=\frac{\text{CSA}\times r}{2}=\frac{5280\times 21}{2}=55440$.

Answer: (B) 55,440

Concept: Use ${\sec }^2\theta -1={\tan }^2\theta$ and $\tan \theta =\frac{\sin \theta }{\cos \theta }$.

1. $1-{\sec }^2\theta =-{\tan }^2\theta$
2. $\frac{-{\tan }^2\theta }{{\sin }^2\theta }\times {\cos }^2\theta =-\frac{{\sin }^2\theta }{{\cos }^2\theta }\times \frac{{\cos }^2\theta }{{\sin }^2\theta }$
3. $=-1$

Trick: Put $\theta ={45}^{\circ }$: $\frac{1-2}{1/2}\times \frac{1}{2}=-1$.

Answer: (C) $-1$

Concept: Add the hourly filling rates of all pipes.

1. Rates: $\frac{1}{16}+\frac{1}{28}+\frac{1}{8}$
2. LCM 112: $\frac{7+4+14}{112}=\frac{25}{112}$ tank per hour
3. Time $=\frac{112}{25}=4\frac{12}{25}$ hours

Trick: Take tank $=112$ units: A $=7$, B $=4$, C $=14$ units/h, total 25.

Answer: (A) $4\frac{12}{25}$ hours

Concept: Average of the first $n$ even numbers $=n+1$.

1. Sum of first $n$ even numbers $=n(n+1)$
2. Average $=\frac{n(n+1)}{n}=n+1$
3. For $n=155$: $156$

Yaad rakhein: First $n$ even → $n+1$; first $n$ odd → $n$; first $n$ natural → $\frac{n+1}{2}$.

Answer: (A) 156

Concept: Expand, the $x^2$ terms cancel, leaving a linear equation.

1. $6x^2-12-6x^2-24x+15=14$
2. $-24x+3=14$
3. $-24x=11\Rightarrow x=-\frac{11}{24}$

Trick: Watch the sign: $-3\times (-5)=+15$.

Answer: (D) $-\frac{11}{24}$

Concept: Find one man's and one woman's one-day work.

1. 1 man's 1-day work $=\frac{1}{6\times 9}=\frac{1}{54}$
2. 1 woman's 1-day work $=\frac{1}{6\times 6}=\frac{1}{36}$
3. $2W+3M=\frac{2}{36}+\frac{3}{54}=\frac{1}{18}+\frac{1}{18}=\frac{1}{9}$
4. Days $=9$

Trick: $1M=\frac{2}{3}W$, so $3M=2W$; $2W+2W=4W$ → $\frac{6\times 6}{4}=9$ days.

Answer: (D) 9 days

Concept: Test only odd numbers not ending in 5 for divisibility by primes up to $\sqrt{438}\approx 20.9$.

1. Candidates: 431, 433, 437 (others are even or end in 5)
2. 431 and 433 are not divisible by 2, 3, 5, 7, 11, 13, 17, 19 → prime
3. $437=19\times 23$ → not prime
4. Count $=2$

Trick: Check primes only up to $\sqrt{N}$.

Answer: (B) 2

Concept: Mode is the value that occurs most often.

1. Frequencies: 13 → 3 times, 12 → 2 times
2. All others (10, 11, 15, 19, 20) occur once
3. Mode $=13$

Yaad rakhein: Mode = most frequent; Median = middle value; Mean = sum ÷ count.

Answer: (A) 13

Concept: When two trains move towards each other, $\text{time to meet}=\frac{\text{gap}}{\text{sum of speeds}}$.

1. Let $MN=D$. A takes 7:50 a.m. to 2:50 p.m. $=7$ h, so speed $\frac{D}{7}$; B takes 9:50 a.m. to 2:50 p.m. $=5$ h, speed $\frac{D}{5}$.
2. By 9:50 a.m. A has run 2 h: $\frac{2D}{7}$; gap $=\frac{5D}{7}$.
3. Relative speed $=\frac{D}{7}+\frac{D}{5}=\frac{12D}{35}$.
4. Time $=\frac{5D}{7}÷\frac{12D}{35}=\frac{25}{12}\text{ h}=2\text{ h }5\text{ min}$.
5. Meeting time $=$ 9:50 a.m. $+2\text{ h }5\text{ min}=$ 11:55 a.m.

Trick: Take $D=35$ (LCM of 7, 5): speeds $5$ and $7$; gap $25$; time $\frac{25}{12}$ h after 9:50.

Answer: (B) 11:55 a.m.

Concept: Number to add $=\text{divisor}-\text{remainder}$.

1. $4519÷30$: $30\times 150=4500$, remainder $=19$.
2. Number to add $=30-19=11$.
3. Check: $4519+11=4530=30\times 151$.

Trick: Last two digits decide: $19\to 30$ needs $11$ (multiples of 30 end in a multiple of 3 tens: $4530$).

Answer: (D) $11$

Concept: Write both numbers in one variable and use the product.

1. Let second number $=x$, first $=5x$.
2. $5x\times x=2880\Rightarrow x^2=576\Rightarrow x=24$.
3. First $=120$; sum $=120+24=144$.

Trick: Sum $=6x$, so the answer must be a multiple of 6 – only $144$ fits ($6\times 24$).

Answer: (D) $144$

Concept: If the whole and one part have the same percentage, the remaining part has that percentage too.

1. Literate total $=14\%\times 351000=49140$.
2. Literate males $=14\%\times 164000=22960$.
3. Females $=351000-164000=187000$; literate females $=49140-22960=26180$.
4. Female literacy $=\frac{26180}{187000}\times 100=14\%$.

Trick: Weighted average of $14\%$ and $x\%$ equals $14\%$ ⇒ $x=14$ without any calculation.

Answer: (B) $14\%$

Concept: If HCF $=h$, numbers are $ha$ and $hb$ with $a,b$ co-prime, and $\text{LCM}=hab$.

1. $6ab=36\Rightarrow ab=6$; co-prime pairs: $(1,6)$ or $(2,3)$.
2. Numbers: $6,36$ (difference $30$) or $12,18$ (difference $6$).
3. Difference $30$ ⇒ numbers are $6$ and $36$.

Trick: Check options: product of numbers $=\text{HCF}\times \text{LCM}=216$; only $6\times 36=216$.

Answer: (C) $6$ and $36$

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

1. $P_1\times 10=P_2\times 9\Rightarrow P_1:P_2=9:10$.
2. Total parts $=19$; 1 part $=\frac{20900}{19}=1100$.
3. Sum at $10\%$ $=9\times 1100=$ ₹9,900.

Trick: Rates $10:9$ ⇒ sums $9:10$; $20900\times \frac{9}{19}=9900$. (Time 5 years cancels.)

Answer: (C) ₹9,900

Concept: Expand each bracket and then combine like terms.

1. $3(\frac{4}{3}{x}^2-30x+18)=4x^2-90x+54$.
2. $-4(x^2+8x-14)=-4x^2-32x+56$.
3. Add: $(4x^2-4x^2)+(-90x-32x)+(54+56)=-122x+110$.

Trick: $x^2$ terms cancel ($4-4$); then only signs matter: $x$-term negative, constant positive ⇒ option (B).

Answer: (B) $-122x+110$

Concept: Compare ratios by turning them into fractions or decimals.

1. $\frac{40}{51}\approx 0.78$, $\frac{44}{74}\approx 0.59$, $\frac{42}{70}=0.6$, $\frac{22}{76}\approx 0.29$.
2. Largest is $\frac{40}{51}$.

Trick: Only $40:51$ has the antecedent close to the consequent (above $\frac{3}{4}$); all others are $\le 0.6$.

Answer: (A) $40:51$

Concept: Overhead (decoration) cost is added to CP: $\text{Profit}\%=\frac{\text{SP}-\text{CP}}{\text{CP}}\times 100$.

1. Effective $\text{CP}=456+20=476$.
2. $\text{SP}=612>476$, so profit $=612-476=136$.
3. $\text{Profit}\%=\frac{136}{476}\times 100=\frac{2}{7}\times 100=\frac{200}{7}\%$.

Trick: $476=68\times 7$, $136=68\times 2$ ⇒ profit $=\frac{2}{7}$ of CP.

Answer: (B) $\frac{200}{7}\%$ profit

Concept: Interest of the 2nd year $=P\times \frac{R}{100}\times (1+\frac{R}{100})$.

1. Here $=P\times 0.2\times 1.2=0.24P$.
2. $0.24P=6786$.
3. $P=\frac{6786}{0.24}=28275$.

Trick: 2nd-year CI $=$ 1st-year SI $+$ interest on it: $20\%+4\%=24\%$ of $P$.

Answer: (A) ₹28,275