# Question #0d824

Jan 5, 2017

1.Michelle left for school at 6:30 A.M driving her car traveling at a constant speed of 30 kph. At 6:40 A.M, her brother Jake noticed that she left her research paper on the table. Driving his car at a speed of 75 kph, he left their house at exactly 6:45 A.M. How far has Jake traveled when he reaches Michelle?

2.A 130-meter length of fence is used to enclose a rectangular parking lot. The length of the garden is 5 meters more than its width. Find the length and the width of the parking lot to be enclosed.

3.The supplement of an angle measures 25 degrees more than twice its complement. Find the measure of the angle.

Ans of Q1
Jake started @75 kmph to reach his sister after $15 \min \mathmr{and} \frac{1}{4} t h \text{ } h r$ . His sister drove @30 kmph and was $\frac{1}{4} \times 30 = 7.5 k m$ ahead.When her brother Jake started. If Jake reached his sister after t hr of his start, then we can write

$75 \times t - 30 \times t = 7.5$

$\implies t = \frac{7.5}{45} h r = \frac{1}{6} h r$

So in this period Jake traveled $75 \times \frac{1}{6} k m = 12.5 k m$

Ans to Q2

Let the width of the garden be $x$ m and its length be $x + 5$ m.

Perimeter of the parking lot

$2 \left(\text{length"+"width}\right) = 130 m$

$2 \left(x + 5 + x\right) = 130$

$\implies x = \frac{65 - 5}{2} = 30 m$

So width =30m

Length=30+5=35m

Ans to Q3

Let the measure of the angle be ${x}^{\circ}$

Its supplement $= {\left(180 - x\right)}^{\circ}$

And its complement $= {\left(90 - x\right)}^{\circ}$

By the given condition

$180 - x = \left(90 - x\right) \times 2 + 25$

$\implies 180 - x = 180 - 2 x + 25$

$\implies x = {25}^{\circ}$