Moth 00 Okanagan Collees ne slutions in the form (e.b), 12] 11. Solve for e and

Question

Moth 00 Okanagan Collees ne slutions in the form (e.b), 12] 11. Solve for e and 6Le w. 2a-36 24 2-4-28 2] 12. The markup os an $87i (2) 12. The markup on an 887 itern is 30% of out. Ir the lten is marked down xn. thes is the reduced selling price of the iten 12] 13. Hom many days will it take for an investment of 840000 to grow to S420 00 atinterest rate of 800% per year? Assume simple interest. Round your answer to the bearest one day, annually, how many years will it take for the enrollment to double? Round your answer to the nearest year. 2 14. If the student enrollment at Olanagan College grows at a rate of 7.50% eomp unded 2) 15. What amount is required to fund a perpetuity that pays $10,000 at the beginning of each quarter? The funds can be invested to earn 5% compounded quarterly Page 5 of 17

Explanation / Answer

11) 2a-3b=24-----------------(1)

      2a-4b=28------------------(2)

substracting (2) from (1)

4b-3b= 24-28

b=-4

substituting b in (1)

2a-3(-4)=24

2a= 24-12

2a =12

a=6

solution of the equations is (6,-4)

12) The reduced selling price is

(1-0.20)*87

= 0.8*87

= 69.6

The reduced selling price of this item is $69.6

13) simple interest= (P*R*T)/100

2000= 40000(0.08)t

2/40 = 0.08t

0.05/0.08 =t

0.625 =t

now converting the time from year to days i.e 0.625*365

It will take 228.125 i.e 229 days for an investment of $40,000 to become 42,000.

14) Let the enrollment be x, so

2x= x (1+0.075)n

2= 1.075n

taking log at both sides

log2= n log(1.075)

n= 0.301/0.0314

= 9.58

i.e 10 years

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