Challenge equations (word problems) (1 Viewer)

kpad5991

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a) As a reward for a week of good behaviour, Tommy was given $7 to spend at the school canteen. By the time Tommy got to the canteen, there were only chocolate bars, meat pies and pizza pieces left. The prices of a chocolate bar, a meat pie and a pizza piece were 75 c, $1.05 and $1.65 respectively. What is the largest amount Tommy could spend?

b) In a mathematical competition consisting of 12 problems, 8 marks are given for each correct response, 0 marks for each incorrect response and each no response is awarded 3 marks. Vicki scored 35 marks in this competition. What is the largest amount of incorrect responses she could have made?

c) Steve has a broken calculator. When just turned on, it displays 0. If the + key is pressed it adds 51. If the – key is pressed it subtracts 51. If the × key is pressed it adds 85 and if the ÷ key is pressed it subtracts 85. The other keys do not function. Steve turns the calculator on. What is the number closest to 2003 that he can get using this calculator?

d) If x and y are non-negative integers and 3x+4y = 96, how many pairs (x, y) are there?
 

Eagle Mum

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a) The highest common factor of 75, 105 and 165 is 15. Therefore, the sum of Tommy‘s purchases must be a multiple of 15. The largest multiple of 15 that is less than 700 is 690. When 690 is divided by 15, the quotient is 46. The respective quotients when 75, 105 & 165 are divided by 15 are 5, 7 & 11. Since it is possible to find a combination of multiples of 5, 7 & 11 which add to 46 (namely 2x(5+7+11)=46), then it is possible to purchase items which total $6.90 (namely (2x $0.75)+(2x $1.05)+(2x $1.65)).

In equation/inequality form:
75a + 105b + 165c = 15m < 700
5a + 7b + 11c = m < 46.7 Therefore m = 46
By observation, 2x(5+7+11) = 46
Therefore, (2x $0.75)+(2x $1.05)+(2x $1.65) = $6.90

b) To maximise the number of incorrect responses which scored zero, the total of 35 should comprise the maximum number of correct answers worth 8 points. As it happens, when 35 is divided by 8 the quotient & remainder are 4 r 3, so Vicki could score 35 with four correct answers, one no response and seven incorrect responses. Therefore, the maximum number of incorrect responses with a score of 35 is seven.

c) The highest common factor of 85 and 51 is 17. Steve can produce any multiple of 17 by repeatedly pressing +, +, ÷ (+51+51-85 = +17). Since he can only ever add or subtract a multiple of 17, all of the numbers he is able to produce are multiples of 17. When 2003 is divided by 17, the quotient and remainder are 117 r 14. Since 14 is closer to 17 than to zero, the multiple of 17 closest to 2003 is 118 x 17 = 2006.

d) Since 4y and 96 are both divisible by 4, 3x must also be divisible by 4. Since 3 and 4 have no common factors (they are co-prime), ‘x’ must be divisible by 4. For x as each multiple of 4, including zero, the corresponding value of y is as follows:
(0,24), (4,21), (8,18), (12,15), (16,12), (20,9), (24,6), (28,3), (32,0). There are nine possible pairs of (x,y).

Perhaps someone who is currently more familiar with the appropriate notation could post the corresponding equations/inequalities for b), c) & d).
 

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