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The company bought 9 computers and 6 printers
<em><u>Solution:</u></em>
Let "x" be the number of computers bought
Let "y" be the number of printers bought
cost of 1 computer = $ 585
cost of 1 printer = $ 385
<em><u>The company buys a total of 15 machines</u></em>
Therefore, we can frame a equation as:
number of computers bought + number of printers bought = 15
x + y = 15 ------ eqn 1
<em><u>The company spends $7,575</u></em>
Therefore, we can frame a equation as:
number of computers bought x cost of 1 computer + number of printers bought x cost of 1 printer = 7575
585x + 385y = 7575 ------ eqn 1
<em><u>Let us solve eqn 1 and eqn 2</u></em>
From eqn 1,
x = 15 - y --------- eqn 3
<em><u>Substitute eqn 3 in eqn 2</u></em>
585(15 - y) + 385y = 7575
8775 - 585y + 385y = 7575
-200y = 7575 - 8775
-200y = -1200
y = 6
<em><u>Substitute y = 6 in eqn 3</u></em>
x = 15 - 6
x = 9
Thus the company bought 9 computers and 6 printers
Tricky question. This isn't a ratio question even though it seems like one.
First divide 7 by 4.
With more men, the hours taken to repair a road will be even shorter.
So next divide 6 by 1.75.
Answer : It'll take 3.43 hours to do the job if they work at the same rate.
Hope this helps. - M
First divide 7 by 4.
With more men, the hours taken to repair a road will be even shorter.
So next divide 6 by 1.75.
Answer : It'll take 3.43 hours to do the job if they work at the same rate.
Hope this helps. - M