sankarbalaji

I get the second one.. but how is the first one's answer C? Please explain.. cos I think the answer is E. Here is my reasoning for that: If ACB is bisected by CD, then AC = CB, so the triangle can be a isosceles one.. but it can also be a equilateral one. If CD is perpendicular to AB, combined with the (2) we can find that these are similar triangles with the same hypotenuse and the other side.. so AD=DB.. but still there is no proof to state that AB is not equal to AC or CB.

Hi Bobogee If the permiter is 16 + sq.root(16), the hypotenuse cannot be 16 as specified in the answer.. In fact if the perimeter is 16+sq.root(16), which is 20, and if the hypotenuse is 16, then the remaining 2 sides can have a combined length of 4 ( 20 - 16 ). Since this is a isosceles right triangle, their lengths are to be equal, which means each should of 2 units length. Now lets validate if this correct using the Pythagorus theorem. is (2)^2 + (2)^2 = (16)^2 ? No.. The perimeter must have been 16 + 16*sq.root(2). In this case, you can figure out that the hypotenuse is 16 and the sides are 8*sq.root(2) using the idea that the ratio of the sides of a isosceles right triangle is 1:1:sq.root(2) ( because it is a 45-45-90 triangle )

Method: i.) Now we know that 3 altitudes are part of the same triangle which means the area formed out of these altitudes and their respective bases ( sides ) will be the same.. ii.) For simplicity sake, lets consider for a factor 1, the altitudes are exactly 2,3 and 4 units.. iii.) And lets have their corresponding bases named as b1, b2 and b3. iv.) Now lets deal with the first 2 altitudes and bases. We know the area formed from them is the same.. So 1/2 * 2 * b1 = 1/2 * 3 * b2 b1/b2 = 3/2 ( which is same as b1:b2 is 3:2 ) iv.) Now lets deal with the second 2 altitudes and bases. We know the area formed from them is also the same.. So 1/2 * 3 * b2 = 1/2 * 4 * b3 b2/b3 = 4/3 ( which is same as b2:b3 is 4:3 ) v.) When dealing with merging 2 sets of ratio-of-two into a single set of ratio-of-three, easier method is to choose a pivot variable and covert.. In this case, lets choose b2 because in first ratio it is 2 and in second ratio it is 4.. so lets multiply both sides of the ratio of b1:b2 by 2, which makes it 6:4. Now b1:b2 = 6:4 b2:b3 = 4:3 So b1:b2:b3 = 6:4:3

This can be solved by the Venn Diagram too! Just that you need to factor that the 6 common advisors between the pair of schools includes the 5 common advisors who are already shared by all the 3 schools.. So you will get that there is only 1 distinct advisor being shared by a pair of school.. From this, you can find the distinct advisors exclusive for that school is 5+1+1+x=10.. x = 3 So finally the answer is (3*3) + (3 *1) + 5 = 17.

ok.. from the answer where it says the hypotenuse is 16, we can find that the perimeter was supposed to be 16+16*21/2

1.) you can simplify the equation to z = xz.. from where you can decipher that it is x = 1 or z = 0 2.) I am not sure if the perimeter is correctly expressed.. it looks like 16 + sq. root(16) which is 20.. is it right?

The answer should be E It can be easily solved like this.. i.) Square root of 4 is 2.. We all know this.. ii.) 4th root of 4 = Square root of 2 = 1.414.. By now we know the answer is above 3.414.. So knock out option B. iii.) 3rd root of 4 > 1.414. cos it will always be greater than 4th root of 4 and less than square root of 4.. (if you want to prove this, you can find out like this.. 3rd root of 4 = 3rd root of (1.414 * 1.414 *1.414 *1.414) = 1.414 * cube root(1.414).. Now we know that cube root(1.414) will be atleast 1 bcos cube root(1) is 1. So 1.414 * cube root(1.414) will be atleast 1.414) So from above you can decipher that M = 2 + 1.414 + ( a value > 1.414 ) ... which means M > 4.828.. which is > 4

Think it must be 6:4:3.. Am I correct?

The answer should be A. Reason: i.) verb tense of the first phrase "solar system that is populated" is present and so out of construct, the second phrase also should be in present. so remove the B(past) and C(past perfect). ii.) Solar system is a singular noun. so remove E cos "feature" is wrong.. iii.) Try putting the second phrase in place of the first place "a solar system that features many of the ...." against "a solar system that featuring many of the ....", where the latter is awkward. You can remove D also.. Thanks Sankar