<p>1)there are two different cylindrical tanks whose sides are perpendicular to a flat base. If some amount of water is poured into the first cylindrical tank, the water level rises 0.5ft. If the same amount of water is poured into the second cylindrical tank, the water level rises 2ft. What is the ration of the radius of the base of the first tank to the radius of the base of the second tank?</p>
<p>answer= 2:1</p>
<p>2)a boat can reach its destination in 3hrs. if it goes twice as fast it can reach 10 miles more than its destinations in 2hrs. how fast does it originally go?</p>
<p>answer= 10miles/hr</p>
<p>3)a man can drive a certain distance in 5hrs. if he increased his speed by 10miles/hr he could travel the same distance in 4+(1/6)hrs. what is the distance he travels?</p>
<p>answer= 250miles</p>
<p>4)Sam can do a job in x hours. Denise can do the same job in x+2 hrs. how long will it take Sam and Denise to do the job together?</p>
<p>answer= (x^2+2x)/(2x+2)</p>
<p>5)a man can run uphill a distance of 500 yards in 8 min. he can run downhill a distance of 500 yards in 4 min. what is his average speed for both the uphill and downhill trips?</p>
<p>answer=83+(1/3)yards/min</p>
<p>I am really sorry and thankful to post this thread!!! Please understand that I'm really bad at math and I need help T_T! Thank you so much!! </p>
<p>PS: could someone give me tips for solving Distance/rate/time problem? I really have hard time solving those kind of probs!! </p>
<p>"1)there are two different cylindrical tanks whose sides are perpendicular to a flat base. If some amount of water is poured into the first cylindrical tank, the water level rises 0.5ft. If the same amount of water is poured into the second cylindrical tank, the water level rises 2ft. What is the ration of the radius of the base of the first tank to the radius of the base of the second tank?</p>
<p>answer= 2:1"</p>
<p>The rate of increase in a cylinder’s water level is dependent only on the area of any circle taken as a cross-section perpendicular to the base. They are related inversely (i.e., as circle area increases, rate of increase in water level decreases). </p>
<p>So, because cylinder B’s water level rises four times as much as that of cylinder A, we know the ratios of the areas of the circles I referred is A:B::4:1 (because the relationship is inverse). </p>
<p>Because the area of a circle involves the product of the square of the radius and a constant (~3.14), we know that the square root of this ratio yields the ratio of the radii: 2:1.</p>
<p>"2)a boat can reach its destination in 3hrs. if it goes twice as fast it can reach 10 miles more than its destinations in 2hrs. how fast does it originally go?</p>
<p>answer= 10miles/hr:</p>
<p>Let’s use a system of equations (all derived from distance= rate x time (d=rt))</p>
<p>For the first scenario, time=3: d=3r > 3=d/r</p>
<p>For the second scenario, rate = 2r (it’s twice as fast), distance changes to d+10 (it went 10 miles further), and time = 2 (it took two hours): d+10=(2)(2r) > d=4r-10</p>
<p>Substitute 4r-10 for d in the first equation to get: 3=(4r-10)/r. Solve for r and get 10.</p>
<p>"3)a man can drive a certain distance in 5hrs. if he increased his speed by 10miles/hr he could travel the same distance in 4+(1/6)hrs. what is the distance he travels?</p>
<p>answer= 250miles"</p>
<p>Again, let’s use a system of equations (all derived from distance= rate x time (d=rt))</p>
<p>For the first scenario, time = 5: d=5r</p>
<p>For the second scenario, rate = r+10 (he sped up by 10), distance is unchanged, and time = 25/6: d=(25/6)(10+r) </p>
<p>Substitute 5r for d in the second equation and get 5r=(25/6)(10+r). Solve for r and get 250.</p>
<p>"4)Sam can do a job in x hours. Denise can do the same job in x+2 hrs. how long will it take Sam and Denise to do the job together?</p>
<p>answer= (x^2+2x)/(2x+2)"</p>
<p>Sam can do 1/x of the job per hour. Denise can do 1/(x+2) of the job per hour. By adding these expressions, we get, in terms of x, how much they can do per hour: </p>
<p>1/x + 1/(x+2) = (2x+2)/(x^2+2x). (Make sure you find a common denominator when solving this.)</p>
<p>To get how long it will take them, we take the reciprocal of the above, which yields: (x^2+2x)/(2x+2).</p>
<p>"5)a man can run uphill a distance of 500 yards in 8 min. he can run downhill a distance of 500 yards in 4 min. what is his average speed for both the uphill and downhill trips?</p>
<p>answer=83+(1/3)yards/min"</p>
<p>First, let’s find the speeds for each leg of the trip:</p>
<p>Going up: distance = 500 and time = 8, so 500=8r. r=62.5</p>
<p>Going down: distance = 500 and time = 4, so 500=4r. r=125</p>
<p>Because the speeds are over equal distances, we can now apply a very convenient formula:</p>
<p>2ab/(a+b) = average speed, where a and b are speeds over the same distance.</p>
<p>^ That is very true. I thought it would be a good opportunity, though, to use the average speed formula so that the OP could solve problems that aren’t as straightforward.</p>
<p>Thanks Silverturtle…I had used about two equation, then equated the two to the volume of the cylider in the First problem. Thanks for the easy way out.</p>