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A dealer mixes two grades of coffee, grade x costing sh. 140 per kg and grade y costing sh. 160 per kg. He sells the mixture at sh. 190 per kg and makes a profit of 25%. Find the ratio in which the tw

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2 parts
a

32−40x+20x2−5x3+58x4−132x532 - 40x + 20x^2 - 5x^3 + \frac{5}{8}x^4 - \frac{1}{32}x^5

b
31.602
2.4

Step-by-step solution

3 steps

:

Step 1: Define variables and set up the cost and selling price.
Let xx be the quantity (in kg) of grade x coffee and yy be the quantity (in kg) of grade y coffee in the mixture.
Cost of xx kg of grade x coffee = 140x140x sh.
Cost of yy kg of grade y coffee = 160y160y sh.
Total cost of the mixture = 140x+160y140x + 160y sh.
Total quantity of the mixture = x+yx+y kg.
Selling price of the mixture per kg = 190 sh.
Total selling price of the mixture = 190(x+y)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 = 1.25×1.25 \times Cost Price

190(x+y)=1.25(140x+160y)190(x+y) = 1.25(140x + 160y)

Step 3: Expand and simplify the equation.

190x+190y=1.25×140x+1.25×160y190x + 190y = 1.25 \times 140x + 1.25 \times 160y 190x+190y=175x+200y190x + 190y = 175x + 200y

Step 4: Group terms with xx and yy to find the ratio.

190x−175x=200y−190y190x - 175x = 200y - 190y 15x=10y15x = 10y

To find the ratio x:yx:y, divide both sides by yy and by 15:

xy=1015\frac{x}{y} = \frac{10}{15} xy=23\frac{x}{y} = \frac{2}{3}

The ratio of the mixture (grade x : grade y) is 2:32:3.
The final answer is 2:3\boxed{\text{2:3}}.

:

Part a

Expand and simplify (2−12x)5(2 - \frac{1}{2}x)^5.

Using the binomial theorem (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k, with a=2a=2, b=−12xb=-\frac{1}{2}x, and n=5n=5:

(2−12x)5=(50)(2)5(−12x)0+(51)(2)4(−12x)1+(52)(2)3(−12x)2+(53)(2)2(−12x)3+(54)(2)1(−12x)4+(55)(2)0(−12x)5(2 - \frac{1}{2}x)^5 = \binom{5}{0}(2)^5(-\frac{1}{2}x)^0 + \binom{5}{1}(2)^4(-\frac{1}{2}x)^1 + \binom{5}{2}(2)^3(-\frac{1}{2}x)^2 + \binom{5}{3}(2)^2(-\frac{1}{2}x)^3 + \binom{5}{4}(2)^1(-\frac{1}{2}x)^4 + \binom{5}{5}(2)^0(-\frac{1}{2}x)^5

Calculate each term:
Term 1: 1⋅32⋅1=321 \cdot 32 \cdot 1 = 32
Term 2: 5⋅16⋅(−12x)=−40x5 \cdot 16 \cdot (-\frac{1}{2}x) = -40x
Term 3: 10⋅8⋅(14x2)=20x210 \cdot 8 \cdot (\frac{1}{4}x^2) = 20x^2
Term 4: 10⋅4⋅(−18x3)=−5x310 \cdot 4 \cdot (-\frac{1}{8}x^3) = -5x^3
Term 5: 5⋅2⋅(116x4)=1016x4=58x45 \cdot 2 \cdot (\frac{1}{16}x^4) = \frac{10}{16}x^4 = \frac{5}{8}x^4
Term 6: 1⋅1⋅(−132x5)=−132x51 \cdot 1 \cdot (-\frac{1}{32}x^5) = -\frac{1}{32}x^5
So, the expansion is:

32−40x+20x2−5x3+58x4−132x5\boxed{32 - 40x + 20x^2 - 5x^3 + \frac{5}{8}x^4 - \frac{1}{32}x^5}

Part b31.602

Hence evaluate (1.995)5(1.995)^5 correct to 3 decimal places.

We need to find the value of xx such that 2−12x=1.9952 - \frac{1}{2}x = 1.995.

2−1.995=12x2 - 1.995 = \frac{1}{2}x 0.005=12x0.005 = \frac{1}{2}x x=2×0.005x = 2 \times 0.005 x=0.01x = 0.01

Substitute x=0.01x=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 \approx 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 \approx 32 - 0.4 + 20(0.0001) - 5(0.000001) (1.995)5≈32−0.4+0.002−0.000005(1.995)^5 \approx 32 - 0.4 + 0.002 - 0.000005 (1.995)5≈31.6+0.002−0.000005(1.995)^5 \approx 31.6 + 0.002 - 0.000005 (1.995)5≈31.602−0.000005(1.995)^5 \approx 31.602 - 0.000005 (1.995)5≈31.601995(1.995)^5 \approx 31.601995

Rounding to 3 decimal places:

(1.995)5≈31.602(1.995)^5 \approx \boxed{31.602}

:
Solve for xx in the equation log⁡512−log⁡5x=1\log_5 12 - \log_5 x = 1.

  1. Use the logarithm property log⁡bA−log⁡bB=log⁡b(AB)\log_b A - \log_b B = \log_b \left(\frac{A}{B}\right).

    log⁡5(12x)=1\log_5 \left(\frac{12}{x}\right) = 1

  2. Convert the logarithmic equation to an exponential equation.

    Recall that if log⁡bA=C\log_b A = C, then bC=Ab^C = A.

    51=12x5^1 = \frac{12}{x}

  3. Solve for xx.

    5=12x5 = \frac{12}{x}

    Multiply both sides by xx:

    5x=125x = 12

    Divide both sides by 5:

    x=125x = \frac{12}{5} x=2.4x = 2.4

    The final answer is 2.4\boxed{2.4}.

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