mother buys some shares of company A on day 0. on day 7, the share price of company a is $4.60. if she sells all her shares pf company A and buys 2000 shares of company B on day 7, she would receive $7400. on day 12, the share price of company A is $4.80 and the share price of company B is $0.50 less than that on day 7  If she sells all her shares of company A and buys 5000 shares of company B on day 12, she would have to pay $5800. find a. the number of shares of company A huixian mother has. b. the share price of company B on day 12

Please use silmutaneous linear equations either elimination method or substitution
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2 Answers

DAY 0: a0A is the cost of A shares at $a0 per share.

DAY 7: sale of A shares at $a7=$4.60 per share=a7A=4.60A; cost of 2000 shares of company B=2000b7; 4.60A-2000b7=7400.

DAY 12: a12=4.80; b7=4.80-0.50=4.30;

5000b12-a12A=5800=5000b12-4.80A: 5000b12-4.80A=5800.


So A=16000/4.6=3478.26 shares.

5000b12=5800+4.80A=5800+16695.65=22495.65, so b12=$4.50

The number of shares=3478.26; company B's share price on Day 12 is $4.50.


If company A shares sold, result of sale of A shares on Day 7 would be 3478.26*4.60=$16000.

Cost of B shares on Day 7 is 2000*4.30=$8600, so 16000-8600=$7400 in pocket.

If company A shares are sold on Day 12, sale=3478.26*4.80=$16700; 5000 B shares cost 5000*4.50=$22500. To pay: 22500-16700=$5800.

(Some of the figures are rounded up.)



by Top Rated User (608k points)
Let x be the no of company A's share. Let y be the cost of a company B's share. On day 7, 4.6x = 2000y+$7400 On day 12, 4.8x=5000(y-$0.5)-$5800 Upon rearranging and simplification, Day 7 ==> 4.6x-2000y=$7400 ---(1) Day 12==> 4.8x-5000y= -($8300)---(2) (1)*5==> 23x-10000y=$37000 ---(3) (2)*2==>9.6x-10000y= -($16600) ---(4) (3)-(4) ==> 13.4x=$53600. x=53600/13.4, x=4000 shares. Sub, x =4000 shares in (2), 4.8(4000)-5000y=-$8300 -5000y=-$8300-$19200 y=$27500/5000 y=$5.50 each share on day 7. Therefore, on day 12, y-$0.5= $5.5-$0.5 =$5 I hope you got the answer by now.

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