This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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You're on a roll — Problem 4: Step 1: Define variables and set up the cost and selling price. Let x be the quantity (in kg) of grade x coffee and y be the quantity (in kg) of grade y coffee in the mixture. Cost of x kg of grade x coffee = 140x sh. Cost of y kg of grade y coffee = 160y sh. Total cost of the mixture = 140x + 160y sh. Total quantity of the mixture = x+y kg. Selling price of the mixture per kg = 190 sh. Total selling price of the mixture = 190(x+y) sh. Step 2: Use the profit percentage to relate cost price and selling price. The dealer makes a profit of 25%. Selling Price = Cost Price + 25% of Cost Price Selling Price = 1.25 × Cost Price 190(x+y) = 1.25(140x + 160y) Step 3: Expand and simplify the equation. 190x + 190y = 1.25 × 140x + 1.25 × 160y 190x + 190y = 175x + 200y Step 4: Group terms with x and y to find the ratio. 190x - 175x = 200y - 190y 15x = 10y To find the ratio x:y, divide both sides by y and by 15: (x)/(y) = (10)/(15) (x)/(y) = (2)/(3) The ratio of the mixture (grade x : grade y) is 2:3. The final answer is 2:3. Problem 5: a) Expand and simplify (2 - (1)/(2)x)^5. Using the binomial theorem (a+b)^n = _k=0^n nk a^n-k b^k, with a=2, b=-(1)/(2)x, and n=5: (2 - (1)/(2)x)^5 = 50(2)^5(-(1)/(2)x)^0 + 51(2)^4(-(1)/(2)x)^1 + 52(2)^3(-(1)/(2)x)^2 + 53(2)^2(-(1)/(2)x)^3 + 54(2)^1(-(1)/(2)x)^4 + 55(2)^0(-(1)/(2)x)^5 Calculate each term: Term 1: 1 · 32 · 1 = 32 Term 2: 5 · 16 · (-(1)/(2)x) = -40x Term 3: 10 · 8 · ((1)/(4)x^2) = 20x^2 Term 4: 10 · 4 · (-(1)/(8)x^3) = -5x^3 Term 5: 5 · 2 · ((1)/(16)x^4) = (10)/(16)x^4 = (5)/(8)x^4 Term 6: 1 · 1 · (-(1)/(32)x^5) = -(1)/(32)x^5 So, the expansion is: 32 - 40x + 20x^2 - 5x^3 + (5)/(8)x^4 - (1)/(32)x^5 b) Hence evaluate (1.995)^5 correct to 3 decimal places. We need to find the value of x such that 2 - (1)/(2)x = 1.995. 2 - 1.995 = (1)/(2)x 0.005 = (1)/(2)x x = 2 × 0.005 x = 0.01 Substitute x=0.01 into the expansion. We only need the first few terms for accuracy to 3 decimal places. (1.995)^5 ≈ 32 - 40(0.01) + 20(0.01)^2 - 5(0.01)^3 (1.995)^5 ≈ 32 - 0.4 + 20(0.0001) - 5(0.000001) (1.995)^5 ≈ 32 - 0.4 + 0.002 - 0.000005 (1.995)^5 ≈ 31.6 + 0.002 - 0.000005 (1.995)^5 ≈ 31.602 - 0.000005 (1.995)^5 ≈ 31.601995 Rounding to 3 decimal places: (1.995)^5 ≈ 31.602 Problem 6: Solve for x in the equation _5 12 - _5 x = 1. Step 1: Use the logarithm property _b A - _b B = _b ((A)/(B)). _5 ((12)/(x)) = 1 Step 2: Convert the logarithmic equation to an exponential equation. Recall that if _b A = C, then b^C = A. 5^1 = (12)/(x) Step 3: Solve for x. 5 = (12)/(x) Multiply both sides by x: 5x = 12 Divide both sides by 5: x = (12)/(5) x = 2.4 The final answer is 2.4. Drop the next question.