SOLUTION: Kindly, help me solve this question.
There are 38 twenty cent coins, x ten cent coins and (2x-15)fifty cent coins in a box. A coin is selected at random and the probability that i
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Question 206307This question is from textbook New Syllabus Mathematics D
: Kindly, help me solve this question.
There are 38 twenty cent coins, x ten cent coins and (2x-15)fifty cent coins in a box. A coin is selected at random and the probability that it is fifty cent coins is 1/8. Find the value of x. Hence, find the probability that the selected coin is a
(a) Twenty cent coin,
(b) Ten cent coin.
This question is from textbook New Syllabus Mathematics D
Found 2 solutions by mickclns, stanbon:
Answer by mickclns(59) (Show Source): You can put this solution on YOUR website!
There are 38 + x + 2x-15 = 3x + 23 total coins and 1/8 of them are 50c coins
so (1/8)(3x+23) = 2x-15, which we are told is the number of 50c coins
Multiplying both sides of the above eqn by 8, we get:
3x +23 = 16x - 120 Subtracting 3x and adding 120 to both sides, we get:
143 = 13x Dividing both sides by 13, we have x = 11. So,
There are 38 twenties, 11 tens, 7 fifties and 56 total coins
a) p(twenty) = 38 / 56 = 19 / 28 and
b) p(ten) = 11/56
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
There are 38 twenty cent coins, x ten cent coins and (2x-15)fifty cent coins in a box. A coin is selected at random and the probability that it is fifty cent coins is 1/8. Find the value of x. Hence, find the probability that the selected coin is a
(a) Twenty cent coin,
P(50 cent coin) = (2x-15)/[38+x+2x-15] = 1/8
Solve for "x":
8(2x-15) = 23+3x
16x-120 = 23+3x
13x = 143
x = 11
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So, P(20 cent coin) = 38/[38+x+2x-15] = 38/[38+11+22-15] = 38/56
---------------------
(b) Ten cent coin.
And, P(10 cent coin) = 11/56
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Cheers,
Stan H.
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