<p>Solve this and explain your answer,</p>
<p>Graham walked to school at an average speed of 3 miles an hour and returned back at 5 miles an hour. If the toatl time he took for a roundtrip was 1 hour, how much distance did he cover in all?</p>
<p>(2<em>rate1</em>rate2) / ( rate1+rate 2) = Avg Speed</p>
<p>2<em>3</em>5=30
3+5 = 8</p>
<p>30/8 = 2.25 mph</p>
<p>Avg. speed = total distance / total time
2.25 = x/1
2.25=x
2.25 miles</p>
<p>Might be wrong, its 5am, who knows :o</p>
<p>Sorry i tried it. the answer is 3.75</p>
<p>anyone else on the prev ques. Also, another one, a 50 foot wire runs from the roof of a house to a 14 foot pole on the other side of the street 14 feet across. Howmuch taller would the pole have to be if the street was 16 feet wider and the wire lenght remained the same.</p>
<p>ok the ans to first ques is 3.75 and to the second ques is 8 ft. also </p>
<p>try this </p>
<p>3) If the average of x y and 80 is 6 more than avg of y z and 80, find the value of x-z?</p>
<p>ans is 18</p>
<p>Rememeber! I need explanations to all these. Please someone help!!!</p>
<p>3)
(x+y+80)/3=a+6
(y+z+80)/3=a</p>
<p>*3 to both sides of both equations</p>
<p>x+y+80=3a+18
y+z+80=3a</p>
<p>-(y+80) from both equations</p>
<p>x=3a+18
z=3a</p>
<p>Difference is 18.</p>
<p>Huh, that was a failure of math on my part. 30/8 is 3.75 lol</p>
<p>And uh</p>
<p>Avg = Object/# of objects
Avg1 = ((x + y + 80)/3) + 6
Avg2 = (y + z + 80)/3 </p>
<p>((x + y + 80)/3) + 6 = (y + z + 80)/3 [Multiple both sides by 3]
x + y + 80 + 18 = y + z + 80 [Cancel 80]
x + y + 18 = y + z [Cancel y]
x + 18 = z [Subtract 18]
x = z - 18 [Subtract y]
x - z = -18
Difference has to be positive, so 18</p>
<p>The first one, you just have to realize that it takes longer to get to school at 3 mph than 5 mph. So he spent the majority of his time at 3 mph and the rest at 5 mph. thus the average will be slightly under 4 mph.</p>
<p>Thanks so much for all your replies</p>
<p>1) how come the formula of avg speed = 2(rate1*rate2) / (rate1 + rate2)</p>
<p>2) plz ans my second ques.</p>
<p>Again thanks so much all of you.</p>
<p>1) This is the harmonic mean, which is used to average rates (check out [Harmonic</a> mean - Wikipedia, the free encyclopedia](<a href=“http://en.wikipedia.org/wiki/Harmonic_mean]Harmonic”>Harmonic mean - Wikipedia)). Basically, it’s the number of items being averaged divided by the sum of the reciprocals of said items. For two speeds, it would be HM = 2/((1/n1) + (1/n2)) = 2/((n1 + n2)/(n1<em>n2)) = (2</em>n1*n2)/(n1 + n2). </p>
<p>2) The house is 14 + (50^2 - 14^2)^(1/2) = 62 feet tall. If the street was sixteen feet wider, then the pole would have to be (62-x)^2 + 30^2 = 50^2 -> 3844 - 124x + x^2 + 900 = 2500 -> x^2 - 124x + 2244 = 0 -> (x - 102)(x - 22) = 0. The pole cannot be 102 ft tall (it would be taller than the house); thus, the pole is 22 ft tall, which is 22 - 14 = 8 ft taller than the previous pole.</p>
<p>1) how come the formula of avg speed = 2(rate1*rate2) / (rate1 + rate2)</p>
<p>In other way let us say that R1=d/T1 and R2=d/T2
average speed= Total distance/total time taken = 2d/(T1+T2)=2d/ [ (d/R1)+(d/R2)]
=2/ [(1/R1) + (1/R2)] = (2<em>R1</em>R2)/(R1+R2)</p>
<p>I think the easiest way to do 1 is a unit conversion. To do this, you need to convert 3 miles/hour to 1/3 hours/mile and the other speed ot hours/mile. So, say the distance one way is x. x miles<em>(1 hour/ 3 miles) + x miles</em>(1 hour/5 miles)= 1 hour. The miles cancel out, so It becomes x/3 hours + x/5 hours = 1 hour or x/3+x/5=1. So, 5x/15+3x/15=1 and 8x/15=1 and 8x=15 and x=15/8. Since x is the one way distance, you multiply it by two to get the total distance. 2x=15*2/8=15/4=3.75 miles. </p>
<p>Generalizing, the formula is x/mph first way (mph 1) + x/mph second way (mph 2)= time total (t). Combining fractions: (x<em>mph1+x</em>mph2)/(mph1<em>mph2)=t. x</em>(mph1+mph2)/(mph1<em>mph2)=t. x=t</em>(mph1<em>mph2)/(mph1+mph2). Since the distance (d) is 2x, d=(2</em>t<em>mph1</em>mph2)/(mph1+mph2). Since distance/time is average speed, average speed=(2<em>mph1</em>mph2)/(mph1+mph2). Notice that the t (or time) has canceled out due to the division.</p>
<p>For the rope/pole question, it helps to draw it out. But, I’ll try to explain it to you. First, it helps to find the height of the building. Set up a right triangle, where the hypotenuse is the 50 foot rope and the base is the 14 foot wide street extended upward to the level of the top of the pole. 50^2-14^2=2304, which is the square of 48 (if you don’t have a calculator, recognize that this is similar to the 7,24,25 triangle and multiply 24 by 2 to get 48). This is the height of the building measured from the top of the pole, thus the height of the building is 48+14=62 feet. Now, the street is now 14+16 feet wide, or 30 feet. Extend the street width upward to the base of the unknown pole height. This is now the base of another right triangle. The hypotenuse of said triangle is again 50 feet. You need to determine the height (or length of the vertical triangle side) which is measured from the top of the pole to the top of the building. This is 50^2-30^2=40^2 (3, 4, 5 triangle). So, the height of the building measured FROM THE TOP OF THE NEW POLE is 40 feet. You know the height of the building is actually 62 feet, thus the difference is 22 feet, which is the height of the pole. Since the original pole was 14 feet tall, the difference in height is 8 feet.</p>
<p>^my bad, I didn’t realize people already had explanations for stuff. But I am leaving this up because the OP might as well read this anyway, as I did the problems with a thought process that differed somewhat (1 I back derived, 2 I did it without real algebra).</p>