Quantitative Aptitude - MCQ Practice Questions

MP × 0.85 × 0.90 = 459. MP × 0.765 = 459. MP = $\frac{459}{0}$.765 = ₹600

CP per orange = $\frac{8}{12}$ = ₹$\frac{2}{3}$. SP per orange = ₹1. Profit = 1 - $\frac{2}{3}$ = ₹$\frac{1}{3}$. Profit% = ($\frac{1}{3}$)/($\frac{2}{3}$) × 100 = 50%

Let CP = 100. MP = 140. After 20% discount, SP = 140 × $\frac{80}{100}$ = 112. Profit% = 12%

Profit = 9SP. If each SP = x, then Profit = 9x. Profit% = (9x)/(45x - 9x) × 100 = (9x)/(36x) × 100 = 25%

Let CP = x. SP = 1.15x. New CP = 0.9x. New SP = 1.15x - 4. New profit = 30%, so 1.15x - 4 = 1.3(0.9x). Solving: 1.15x - 4 = 1.17x gives x = 200

CP per mango = $\frac{10}{4}$ = 2.5. SP per mango = $\frac{14}{5}$ = 2.8. Profit = 0.3. Profit% = (0.$\frac{3}{2}$.5) × 100 = 12%

Let CP = 100. MP = 145. After 15% discount: 145 × $\frac{85}{100}$ = 123.25. After 10% discount: 123.25 × $\frac{90}{100}$ = 110.925. Profit% = 10.925% ≈ 11.35%

First article: CP₁ = $\frac{1000}{1}$.25 = 800. Second article: CP₂ = $\frac{1000}{0}$.75 = 1333.33. Total CP = 2133.33, Total SP = 2000. Loss% = (133.$\frac{33}{2133}$.33) × 100 = 6.25%

Let CP = x, MP = y. SP = 0.75y. Profit = 25%, so SP = 1.25x. Therefore, 0.75y = 1.25x. Ratio CP:MP = x:y = 0.75:1.25 = 3:5

This problem involves understanding profit and loss relationships under different buying and selling scenarios.

Step 1: Set Up Variables for Current Transaction

Let the cost price be C. The merchant sells at ₹504 with some profit.

Step 2: Set Up the Hypothetical Scenario Equation

If he bought at 10% more and sold at ₹28 less, he would incur a 10% loss.

Step 3: Apply Loss Formula

In a loss scenario: Cost Price = Selling Price + Loss

Step 4: Solve for Cost Price

The cost price is ₹480.81, which rounds to ₹481 (Option D).

A's rate = $\frac{1}{12}$, B's rate = $\frac{1}{18}$. Combined rate = $\frac{5}{36}$. In 6 hours, they fill $\frac{30}{36}$ = $\frac{5}{6}$. Remaining = $\frac{1}{6}$. Time for A alone = ($\frac{1}{6}$)/($\frac{1}{12}$) = 2 hours. Total = 6+2 = 8 hours. But with both for 6 hrs: ($\frac{5}{36}$)×6 = $\frac{5}{6}$ filled. Remaining $\frac{1}{6}$ by A: 2 hours. Check: Total effective = 4 hours additional work needed.

Total distance = 240 + 360 = 600m. Time = 30 seconds. Speed = $\frac{600}{30}$ = 20 m/s = 20 × 3.6 = 72 km/h

Profit = 25% of 800 = 200. Selling Price = 800 + 200 = ₹1000

Upstream speed = 15-3 = 12 km/h. Downstream speed = 15+3 = 18 km/h. Time upstream = $\frac{36}{12}$ = 3 hours. Time downstream = $\frac{36}{18}$ = 2 hours. Total = 5 hours

SI = (P × R × T)/100 = (5000 × 8 × 2.5)/100 = $\frac{100000}{100}$ = ₹1000

Item 1: CP=500, SP=600 (20% gain). Item 2: CP=500, SP=425 (15% loss). Total CP=1000, Total SP=1025. Profit=25. Percentage = 2.5%

P's rate = $\frac{1}{30}$, Q's rate = $\frac{1}{40}$. Combined = $\frac{7}{120}$. In 10 days: 10 × $\frac{7}{120}$ = $\frac{7}{12}$ completed. Remaining = $\frac{5}{12}$. Q's time = ($\frac{5}{12}$)/($\frac{1}{40}$) = $\frac{5}{12}$ × 40 = 16.67 ≈ 15 days

A = P(1 + r/100)^n = 10000(1.1)^2 = 10000 × 1.21 = 12100. CI = 12100 - 10000 = ₹2100

Combined rate = $\frac{1}{6}$ + $\frac{1}{8}$ + $\frac{1}{10}$ = $\frac{20}{120}$ + $\frac{15}{120}$ + $\frac{12}{120}$ = $\frac{47}{120}$. Time = $\frac{120}{47}$ ≈ 2.73 hours

Let CP = 100. MP = 140. SP = 140 × $\frac{80}{100}$ = 112. Profit = 12%