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valentinak56 [21]

1 year ago

7

6. Jones covered a distance of 50 miles on his first trip. On a later trip he traveled 300 miles while going three times as fast

. His new time compared with the old time was: (a) three times as much, (b) twice as much, (c) the same, (d) half as much, (e) a third as much

1 answer:

nata0808 [166]1 year ago

5 0

<h3><em><u>given</u></em><em><u>:</u></em></h3>

<em><u>first</u></em><em><u> </u></em><em><u>trip</u></em><em><u>=</u></em><em><u> </u></em><em><u>5</u></em><em><u>0</u></em><em><u> </u></em><em><u>miles</u></em><em><u>.</u></em>

<em><u>second</u></em><em><u> </u></em><em><u>trip</u></em><em><u>=</u></em><em><u> </u></em><em><u>3</u></em><em><u>0</u></em><em><u>0</u></em><em><u> </u></em><em><u>miles</u></em><em><u>.</u></em>

<h3><em><u>to</u></em><em><u> </u></em><em><u>find</u></em><em><u>:</u></em></h3>

<em><u>his</u></em><em><u> </u></em><em><u>new</u></em><em><u> </u></em><em><u>time</u></em><em><u> </u></em><em><u>compared</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>old</u></em><em><u> </u></em><em><u>time</u></em><em><u>.</u></em>

<h3><em><u>solution</u></em><em><u>:</u></em></h3>

<em><u>Let speed of the 1st trip x miles / hr. and speed of the 2nd trip 3x / hr.</u></em>

<em><u>Speed = Distance/Time</u></em>

<em><u>So, times taken to covered a distance of 50 miles on his first trip = 50/x hr.</u></em>

<em><u>= 100/x hr</u></em>

<em><u>Therefore</u></em><em><u>,</u></em><em><u> </u></em><em><u>his new time compared with the old time was twice as much.</u></em>

<em><u>answer</u></em><em><u>=</u></em><em><u> </u></em><em><u>b</u></em><em><u>)</u></em><em><u> </u></em><em><u>twice</u></em><em><u> </u></em><em><u>as</u></em><em><u> </u></em><em><u>much</u></em>

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