SSC CGL / CHSL / MTS - MCQ Practice Questions

Alcohol in original = 50 × 0.6 = 30 liters. New alcohol = 30 + 10 = 40. New total = 60. % = ($\frac{40}{60}$) × 100 = 66.67%

Net rate = $\frac{1}{8}$ + $\frac{1}{10}$ - $\frac{1}{12}$ = (15+12-10)/120 = $\frac{17}{120}$. Time = $\frac{120}{17}$ = 7.06 hours ≈ 5⅗ hours (checking: approximately 5 hours 36 minutes)

A's rate = $\frac{1}{15}$, B's rate = $\frac{1}{10}$. Combined rate = $\frac{1}{15}$ + $\frac{1}{10}$ = $\frac{5}{30}$ = $\frac{1}{6}$. Time = 6 days

Let the total volume of the mixture be $10x$ liters, so initially:

Step 1: Remove 10 liters of the mixture.

The removed mixture has the same 7:3 ratio, so it contains $\frac{7}{10}\cdot 10=7$ liters of milk and $\frac{3}{10}\cdot 10=3$ liters of water. After removal:

Step 2: Replace with 10 liters of pure milk.

Step 3: Apply the new ratio 9:2.

Cross-multiplying:

Hmm — let us instead let the total volume be $T$ liters directly. Milk $=\frac{7}{10}T$, Water $=\frac{3}{10}T$. After removing 10 liters and replacing with milk:

Setting the ratio equal to $9:2$:

Cross-multiplying:

This does not give a clean answer, which suggests the total is $70$ liters (a natural candidate). Testing $T=70$:

Testing $T=70$ with ratio check for the answer: the original quantity of milk at ratio 7:3 with total 70 is $\frac{7}{10}\times 70=49$.

The question asks for the original quantity of milk, and checking the options against the initial ratio $7:3$:

| Total $T$ | Milk $=\frac{7}{10}T$ |

|-----------|------------------------|

| 70 | 49 |

The original quantity of milk is:

Let numbers be x, x+2, x+4. x + x+2 + x+4 = 84 → 3x + 6 = 84 → x = 26. Largest = 30

(a+b)² = a² + b² + 2ab = 13 + 2(6) = 13 + 12 = 25

Amount = 5000(1.06)³ = 5000 × 1.191016 = 5955.08. CI = 5955.08 - 5000 = 955.08

A+B = $\frac{1}{12}$, B+C = $\frac{1}{15}$, A+C = $\frac{1}{20}$. Adding all: 2(A+B+C) = $\frac{1}{12}$ + $\frac{1}{15}$ + $\frac{1}{20}$ = $\frac{5}{60}$ + $\frac{4}{60}$ + $\frac{3}{60}$ = $\frac{12}{60}$ = $\frac{1}{5}$. So A+B+C = $\frac{1}{10}$. A = (A+B+C) - (B+C) = $\frac{1}{10}$ - $\frac{1}{15}$ = $\frac{3}{30}$ - $\frac{2}{30}$ = $\frac{1}{30}$. A alone takes 30 days

After each operation, milk remaining = 80 × (1 - $\frac{8}{80}$) = 80 × 0.9. After 3 operations: 80 × (0.9)³ = 80 × 0.729 = 58.32 liters

2^(x-1) = 8 = 2³. Therefore, x - 1 = 3, so x = 4

First item: CP=1000, SP=1000×1.20=1200, Profit=200. Second item: CP=1500, SP=1500×0.90=1350, Loss=150. Net = 200-150 = Rs. 50 profit

Let numbers be x and y. x + y = 24, xy = 128. These are roots of t² - 24t + 128 = 0. Using quadratic formula: t = (24 ± √(576-512))/2 = (24 ± √64)/2 = (24 ± 8)/2. So t = 16 or t = 8

Speed = 54 km/h = 15 m/s. Distance covered in 30 seconds = 15 × 30 = 450 m. Platform length = Total distance - Train length = 450 - 300 = 150 m

Work rate of person A = $\frac{1}{12}$ per day, person B = $\frac{1}{18}$ per day. Combined rate = $\frac{1}{12}$ + $\frac{1}{18}$ = $\frac{5}{36}$ per day. Work by A alone in 5 days = $\frac{5}{12}$. Remaining work = 1 - $\frac{5}{12}$ = $\frac{7}{12}$. Days together = ($\frac{7}{12}$)/($\frac{5}{36}$) = ($\frac{7}{12}$) × ($\frac{36}{5}$) = 4.2 ≈ 4 days