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ETF2700/ETF5970 Mathematics for Business, S1, 2020 Week 2 Tutorial Solutions

1 In-class Exercises 1.1

Ex 1.3

(a) To simplify this fraction we express the two component fractions in terms of a common denominator. This is achieved by multiplying the numerator and denominator of the first fraction by 8 and the second fraction by 5 so that the common denominator is 40: 2×8 3×5 16 + 15 31 2 3 . + = + = = 40 5 8 5×8 8×5 40 (b) To simplify this fraction we express the two component fractions in terms of a common denominator. This is achieved by multiplying the numerator and denominator of the first fraction by 2 so that the common denominator is 32: 3 5×2 3 10 − 3 7 5 − = − = = 16 32 16 × 2 32 32 32 (c) To simplify this fraction we note that 5 divides both 15 and 35, and 18 divides both 18 and 54. Therefore 15 18 3 18 3 1 1 × = × = × = 7 54 35 54 7 3 7 (d) We write

32 21 32 56 ÷ = × 49 21 49 56

and then note that 7 divides 21 and 49, and 8 divides 32 and 56. Therefore, 32 21 32 3 4 3 12 32 56 ÷ = × = × = × = . 49 21 49 56 7 56 7 7 49

1

1.2

Ex 1.4

(a) To compare the relative sizes of two fractions, we express them in terms of a common denominator. Now 32 = 4 × 8 and 24 = 3 × 8, so the common denominator is 3 × 4 × 8 = 96. Therefore,

11 × 3 33 7 7×4 28 11 = = and = = . 32 32 × 3 96 24 24 × 4 96

So 11/32 > 7/24, since 33/96 > 28/96.

(b) The decimal representations of the two fractions, rounded to 2 decimal places, 11 7 ≈ 0.29. ≈ 0.34 and 24 32 So 11/32 > 7/24, since 0.34 − 0.29 = 0.05 is larger than the maximal rounding error 0.02.

1.3

Ex 1.5

(a) 51.24 (b) 7.90 (c) 326.28

1.4

Question 2

Convert the average price of a Big Mac in US into AUD by solving the equation 5.28 × 1.25 = 6.6AUD. So it is cheaper to buy a Big Mac in Australia, since 6.6 > 5.90.

1.5

Question 3

Suppose the USD/AUD exchange rate is x (x > 0), that is 1 USD = x AUD. A Big Mac is equally expensive if and only if 5.28x = 5.90 Dividing both side by 5.28, we solve x = 1.12 rounded to two decimal places.

2

1.6

Question 4

1021000 − 785750 × 100% ≈ 29.94% 785750 The percentage change (rounded to 2 decimal places) in median price of houses in Melbourne %Change =

city from 2013 to 2015 was 29.94%

1.7

Question 5

Adding 5 to both sides of the inequality yields 3x − 5 + 5 > x − 3 + 5, or 3x > x + 2. Subtracting x to both sides yields 3x − x > x − x + 2, so 2x > 2, and after dividing by the positive number 2, we get x > 1. The solution to this inequality is (1, ∞).

1.8

Question 6

Step 1: a = 1, b = −1, c = −1 Step 2: Evaluate the discriminant ∆ = b2 − 4ac = (−1)2 − 4 × 1 × (−1) = 1 + 4 = 5 > 0 Step 3: We have two solutions: √ √ −b + ∆ 1+ 5 x1 = , = 2 2a

x2 =

√ √ −b − ∆ 1 − 5 = . 2a 2

2 Post-class Exercises 2.1

Ex 2.1

(a) Add 4 to both sides of the equation 3x − 4 = 2, so that all the constant terms are on the right-hand side: 3x − 4 + 4 = 2 + 4, or 3x = 6. Finally divide both sides by 3: 3x 6 = , so x = 2. 3 3 Therefore, the solution to this equation is x = 2. 3

(b) To simplify this equation we multiply both sides by the lowest common multiple of 3 and 4, i.e., 12. Therefore, we have 12



2x − 1 3



= 12



 3x − 1 + 12. 4

This simplifies to give 4(2x − 1) = 3(3x − 1) + 12, or, on removal of the brackets, 8x − 4 = 9x − 3 + 12. This equation is now rearranged so that all terms involving x appear on the left-hand side and all the constant terms appear on the right-hand side. Subtracting 9x from both sides, we obtain −x − 4 = 9. Adding 4 to both sides yields −x = 13. Finally, dividing by −1 on both sides yields x = −13. Therefore, the solution to this equation is x = −13.

2.2

Ex 3.4

(a) Comparing the coefficients of this equation with those of the general quadratic equation we have a = 1, b = −4 and c = 3. The discriminant is therefore ∆ = (−4)2 − 4 × 1 × 3 = 16 − 12 = 4 > 0. We have two solutions: √ √ 4+ 4 −b + ∆ 4+2 = 3, = x1 = = 2 2a 2

√ √ −b − ∆ 4− 4 4−2 x2 = = = 1. = 2a 2 2

Optional: you may plot the quadratic function on the left-hand-side of the equation in WolframAlpha via this link. (b) Comparing the coefficients of this equation with those of the general quadratic equation we have a = 3, b = 5 and c = −8. The discriminant is therefore ∆ = 52 − 4 × 3 × (−8) = 25 + 96 = 121 > 0. 4

We have two solutions: √ √ −5 + 121 −b + ∆ −5 + 11 = x1 = = = 1, 2a 6 6

√ √ −5 − 121 −5 − 11 8 −b − ∆ = = =− . x2 = 2a 3 6 6

Optional: you may plot the quadratic function on the left-hand-side of the equation in WolframAlpha via this link. (c) Comparing the coefficients of this equation with those of the general quadratic equation we have a = 2, b = −19 and c = −10. The discriminant is therefore ∆ = (−19)2 − 4 × 2 × (−10) = 361 + 80 = 441 > 0. We have two solutions: √ √ 19 + 441 −b + ∆ 19 + 21 = 10, = x1 = = 4 2a 4

√ √ −b − ∆ 19 − 441 19 − 21 1 x2 = = = =− . 2a 2 4 4

Optional: you may plot the quadratic function on the left-hand-side of the equation in WolframAlpha via this link.

2.3

Question 7

If you cannot obtain the answer(s): change your calculator into Algebraic mode and try again. For help on switching between Chain and Algebraic modes please read the ‘HP 10bII+ Manual’ in Moodle.

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