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 General Mathematics Paper 2,Nov/Dec. 2013
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
Weakness/Remedies
Strength

Question 6
1. If y2 – x2  = 5(y – x)2, find  x:y
2. In a partnership, Ajayi contributed N500,000 .00 more than Kunle.  The total profit made was 15% of their total contribution.  If Kunle received  of the total profit which amounted to N84,000.00, how much was:

(i)      Ajayi’s share of the profit?

(ii)      Kunle’s contribution to the partnership

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Observation

This question was also reported to be attempted by majority of the candidates and they performed well in it.

In part (a) candidates were reported to have expanded the right hand side to have 52-10+52  i.e 2 – 2 = 52 – 10 + 52.  Bringing like terms together gave 4y2 – 10xy + 6x2 = 0. Factorizing gave 4y2 – 4 – 6 + 62 = 0 i.e (4y-6)(y-) = 0 which implied that either y = x or 4y = 6.  Hence, the ratio x:y = 1:1 or 2:3.

In part(b), candidates were reported to have shown that if  of the total profit was N84,000.00, then the total profit =  = N210,000.00.  Ajayi’s share would then be gotten as  of 210,000 = N126,000.00.  Let Kunle’s contribution to the partnership be x, then Ajayi’s contribution =  + 500,000.  From the second statement,  x  = 210,000.  Solving this equation gave  = N450, 000.00.  This meant that Kunle contributed N450, 000.00 to the partnership.