RRB NTPC Undergraduate CBT 1 2024 PYQs

Concept: Take distance $MN=D$; speed $=\frac{D}{\text{journey time}}$; after both start, gap closes at the sum of speeds.

1. A takes $9$ h (speed $\frac{D}{9}$), B takes $6$ h (speed $\frac{D}{6}$).
2. By 8:35 AM, A has run $2$ h: covered $\frac{2D}{9}$, gap $=\frac{7D}{9}$.
3. Relative speed $=\frac{D}{9}+\frac{D}{6}=\frac{5D}{18}$; time $=\frac{7}{9}÷\frac{5}{18}=2.8$ h $=2$ h $48$ min.
4. Meeting time $=8:35+2:48=11:23$ AM.

Trick: Distance cancels – always use $D$ or LCM of times ($18$ units).

Answer: (D) 11:23 AM

Concept: Write both numbers in terms of one variable.

1. Let numbers be $5x$ and $x$: $5x^2=1805$ ⇒ $x^2=361$ ⇒ $x=19$.
2. Numbers $95$ and $19$; sum $=114$.

Trick: Sum $=6x$, so it must be a multiple of $6$: only $114$ and $126$ qualify; $x=19$ gives $114$.

Answer: (C) $114$

Concept: Form two equations and solve by substitution.

1. $F+5S=42$ and $2.8F+S=76$.
2. $F=42-5S$ ⇒ $2.8(42-5S)+S=76$ ⇒ $117.6-13S=76$ ⇒ $S=3.2$.
3. $F=42-16=26$.

Trick: Check options: $F=26$ ⇒ $S=3.2$, and $2.8\times 26+3.2=76$ ✓.

Answer: (A) $26$

Concept: Total work in man-days stays constant; convert women into men.

1. Work $=9\times 8=72$ man-days $=4\times 9=36$ woman-days.
2. So $1$ woman $=2$ men.
3. $2$ women $+4$ men $=8$ men ⇒ days $=\frac{72}{8}=9$.

Trick: Equate total work: $72M=36W$ ⇒ $W=2M$.

Answer: (C) $9$ days

Concept: Savings $=$ Income $-$ Expenditure; increase income and expenditure separately.

1. Savings $=18.5\%$ of $44000=8140$; expenditure $=35860$.
2. New income $=44000\times 1.17=51480$; new expenditure $=35860\times 1.4=50204$.
3. New savings $=51480-50204=1276$.
4. Change $=8140-1276=6864$ decrease.

Trick: Expenditure grows faster ($40\%$ vs $17\%$) on a big base – savings must fall, ruling out A and D.

Answer: (B) ₹6,864 decrease

Concept: Median of an even number of observations $=$ average of the two middle values (after sorting).

1. Sorted: $3,6,18,32,37,48$.
2. Middle values: $18$ and $32$.
3. Median $=\frac{18+32}{2}=25$.

Trick: Always sort first – never take the middle of the given order.

Answer: (A) $25$

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

1. $4\times 2600=10400$.
2. Other number $=\frac{10400}{104}=100$.

Trick: Check: $\text{HCF}(104,100)=4$ ✓.

Answer: (D) $100$

Concept: For equal distances, average speed $=\frac{2xy}{x+y}$.

1. $\frac{2\times 15\times 30}{15+30}=\frac{900}{45}=20\text{km/h}$.

Trick: Average speed is always below the simple average $22.5$.

Answer: (C) $20$

Concept: Third proportional of $a$ and $b$ is $x$ where $a:b=b:x$, i.e. $x=\frac{b^2}{a}$.

1. $x=\frac{{70}^2}{28}=\frac{4900}{28}=175$.

Trick: $\frac{70}{28}=2.5$, so $x=70\times 2.5=175$.

Answer: (B) $175$

Concept: $a^m÷a^n=a^{m-n}$ and $a^m\times a^n=a^{m+n}$ (left to right).

1. ${55}^{-6}÷{55}^5={55}^{-11}$.
2. ${55}^{-11}\times {55}^{-8}={55}^{-19}$.

Trick: Just add/subtract powers: $-6-5-8=-19$.

Answer: (B) ${55}^{-19}$

Concept: Write the decimals with the same number of decimal places, find the LCM of the whole numbers, then put back the decimal point.

1. $6.3=\frac{630}{100}$ and $0.07=\frac{7}{100}$.
2. $\text{LCM}(630,7)=630$ (since $630=7\times 90$).
3. $\text{LCM}=\frac{630}{100}=6.3$

Trick: If one number is a multiple of the other, the larger one is the LCM: $6.3=0.07\times 90$.

Answer: (D) $6.3$

Concept: Equal simple interest for the same time means $P_1{R}_1=P_2{R}_2$, so principals are in the inverse ratio of rates.

1. Let ₹$x$ at 10% and ₹$y$ at 9%. $\frac{x\times 10\times 8}{100}=\frac{y\times 9\times 8}{100}\Rightarrow 10x=9y$.
2. $x:y=9:10$, total parts $=19$.
3. $x=\frac{9}{19}\times 20900=9\times 1100=9900$

Trick: Sum at 10% gets the 'other' rate as its share: $\frac{9}{9+10}$ of total.

Answer: (D) ₹$9,900$

Concept: Find each quantity separately and subtract.

1. $65\%\text{ of }65=\frac{65\times 65}{100}=\frac{4225}{100}=42.25$
2. $\frac{2}{5}\text{ of }55=22$
3. $\text{Difference}=42.25-22=20.25$

Trick: ${65}^2=4225$ (numbers ending in 5: $6\times 7=42$, append 25).

Answer: (D) $20.25$

Concept: Apply the divisibility tests for 10, 11, 3 and 8.

1. Last digit is 9, so not divisible by 10.
2. Last three digits 059 are not divisible by 8.
3. For 11: (sum of odd-place digits from right) $-$ (even-place) $=(9+0+7+8+6)-(5+9+6+4)=30-24=6$, not a multiple of 11.
4. Digit sum $=6+4+8+6+7+9+0+5+9=54$, divisible by 3 (and by 9) ✔

Trick: Digit sum divisible by 3 ⇒ number divisible by 3.

Answer: (C) 3

Concept: $a^m÷a^n=a^{m-n}$ and $a^m\times a^n=a^{m+n}$; work left to right.

1. ${57}^{-9}÷{57}^4={57}^{-9-4}={57}^{-13}$
2. ${57}^{-13}\times {57}^{-10}={57}^{-13-10}={57}^{-23}$

Trick: Add the exponents with signs: $-9-4+(-10)=-23$.

Answer: (C) ${57}^{-23}$

Concept: Surface area of a sphere $=4\pi r^2$, so area $\propto r^2$.

1. Before: $S_1=4\pi r^2$
2. After: $S_2=4\pi (5r{)}^2=25\times 4\pi r^2$
3. $S_2:S_1=25:1$

Trick: Radius $\times k$ ⇒ area $\times k^2$, volume $\times k^3$.

Answer: (C) $25:1$

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

1. $60:30=30:x$
2. $x=\frac{{30}^2}{60}=\frac{900}{60}=15$

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

Answer: (C) $15$

Concept: Sum of the first $n$ even numbers $=n(n+1)$.

1. $\text{Sum}=156\times 157$
2. $\text{Average}=\frac{156\times 157}{156}=157$

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

Answer: (D) $157$

Concept: Take CP = 100 units and follow the mark-up and discount.

1. List price $=125$ units; $\text{SP}=125\times \frac{84}{100}=105$ units.
2. Profit $=105-100=5$ units $=$ ₹64 ⇒ 1 unit $=$ ₹$12.8$.
3. List price $=125\times 12.8=$ ₹$1600$

Trick: Net effect $=25-16-\frac{25\times 16}{100}=5\%$ profit; $\text{CP}=\frac{64}{5\%}=$ ₹$1280$, list $=1280\times 1.25$.

Answer: (B) ₹$1,600$

Concept: Let the numbers be $x$ and $x+2$.

1. $x^2+(x+2{)}^2=2452\Rightarrow 2x^2+4x+4=2452\Rightarrow x^2+2x-1224=0$
2. $(x+36)(x-34)=0\Rightarrow x=34$
3. Numbers 34 and 36: ${34}^2+{36}^2=1156+1296=2452$ ✔; sum $=70$

Trick: Half of 2452 is 1226 ≈ ${35}^2=1225$ ⇒ numbers around 35, i.e. 34 and 36.

Answer: (A) $70$