(a) A box contains 40 identical discs which are either red or white. If the probability of picking a red disc is \(\frac{1}{4}\); Calculate the number of (i) white discs ; (ii) red discs that should be added such that the probability of picking a red disc will be \(\frac{1}{3}\).
(b) A salesman bought some plates at N50.00 each. If he sold all of them for N600.00 and made a profit of 20% on the transaction, how many plates did he buy?
(b) A salesman bought some plates at N50.00 each. If he sold all of them for N600.00 and made a profit of 20% on the transaction, how many plates did he buy?
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Correct Answer: Option n
Explanation:
(a) (i) No of red discs = \(\frac{1}{4} \times 40 = 10\)
Total no of discs = 40
\(\therefore\) No of white balls = 40 - 10 = 30
(ii) Let the number of red balls to be added = z
\(\therefore \frac{10 + z}{40 + z} = \frac{1}{3}\)
\(3(10 + z) = 40 + z\)
\(30 + 3z = 40 + z \implies 3z - z = 40 - 30\)
\(2z = 10 \implies z = 5\)
Therefore, adding 5 red discs, the probability of picking a red disc = \(\frac{1}{3}\).
(b) CP of a plate = N50.00
Let the number of plates be x
\(\therefore \text{CP of x plates} = N50x\)
\(\text{SP of all plates} = N600\)
\(\frac{100 + 20}{100} \times N50x = N600 \implies N60x = N600\)
\(x = \frac{600}{60} = 10 plates\)
(a) (i) No of red discs = \(\frac{1}{4} \times 40 = 10\)
Total no of discs = 40
\(\therefore\) No of white balls = 40 - 10 = 30
(ii) Let the number of red balls to be added = z
\(\therefore \frac{10 + z}{40 + z} = \frac{1}{3}\)
\(3(10 + z) = 40 + z\)
\(30 + 3z = 40 + z \implies 3z - z = 40 - 30\)
\(2z = 10 \implies z = 5\)
Therefore, adding 5 red discs, the probability of picking a red disc = \(\frac{1}{3}\).
(b) CP of a plate = N50.00
Let the number of plates be x
\(\therefore \text{CP of x plates} = N50x\)
\(\text{SP of all plates} = N600\)
\(\frac{100 + 20}{100} \times N50x = N600 \implies N60x = N600\)
\(x = \frac{600}{60} = 10 plates\)