(a) Solve the simultaneous equations 3y - 2x = 21 ; 4y + 5x = 5.
(b) Six identical cards numbered 1 - 6 are placed face down. A card is to be picked at random. A person wins $60.00 if he picks the card numbered 6. If he picks any of the other cards, he loses $10.00 times the number on the card. Calculate the probability of (i) losing ; (ii) losing $20.00 after two picks.
(b) Six identical cards numbered 1 - 6 are placed face down. A card is to be picked at random. A person wins $60.00 if he picks the card numbered 6. If he picks any of the other cards, he loses $10.00 times the number on the card. Calculate the probability of (i) losing ; (ii) losing $20.00 after two picks.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option n
Explanation:
(a) \(3y - 2x = 21 ... (1)\)
\(4y + 5x = 5 .... (2)\)
Multiply (1) by 4 and (2) by 3 so we have,
\(12y - 8x = 84 ... (3)\)
\(12y + 15x = 15 ... (4)\)
(3) - (4) : \(-8x - 15x = 69 \implies -23x = 69\)
\(x = \frac{69}{-23} = -3\)
Put x = -3 in (1),
\(3y - 2(-3) = 21\)
\(3y + 6 = 21 \implies 3y = 21 - 6 = 15\)
\(y = \frac{15}{3} = 5\)
\(x, y = -3, 5\).
(b)(i) Probability of losing
Probability of winning = \(\frac{1}{6}\)
\(\therefore\) Probability of losing = \(1 - \frac{1}{6} = \frac{5}{6}\).
(ii) Losing $20.00 after two picks = picking the card numbered 1 twice.
= \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)
(a) \(3y - 2x = 21 ... (1)\)
\(4y + 5x = 5 .... (2)\)
Multiply (1) by 4 and (2) by 3 so we have,
\(12y - 8x = 84 ... (3)\)
\(12y + 15x = 15 ... (4)\)
(3) - (4) : \(-8x - 15x = 69 \implies -23x = 69\)
\(x = \frac{69}{-23} = -3\)
Put x = -3 in (1),
\(3y - 2(-3) = 21\)
\(3y + 6 = 21 \implies 3y = 21 - 6 = 15\)
\(y = \frac{15}{3} = 5\)
\(x, y = -3, 5\).
(b)(i) Probability of losing
Probability of winning = \(\frac{1}{6}\)
\(\therefore\) Probability of losing = \(1 - \frac{1}{6} = \frac{5}{6}\).
(ii) Losing $20.00 after two picks = picking the card numbered 1 twice.
= \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)