Makumajon

far from doing x, does y... opposite connection. It means that "[some may think that Hampshire's assertion would do x. But let alone doing x, the assertion in fact did the opposite: y." B is the answer. x= dismiss about objectivity; y=reveal that they are relavant, x & y show opposite connection. "far from dismissing about objectivity, actually shows more relevance."

The main problem here is not about a/an, but about especially/specially. And two standard idioms are able/ability + to capable/capability + of

If N = (p^a)(q^b)(r^c)..., in which p, q, r, ... are different primes, then the total no. of factors of N, including N itself and 1, is given by No. of total factors = (a+1)(b+1)(c+1)... So, to find the total no. of factors of an integer, first factorize it into the lowest form, and then get (a+1)(b+1)(c+1)..., in which a, b, c, ... are the powers of prime factors.

It is WW, so nations must not state anything directly that might backfire. So they are resorting to jargon and verbal delicacy. Euphemism is "the act or an example of substituting a mild, indirect, or vague term for one considered harsh, blunt, or offensive." So, euphemism requires verbal delicacy. For example, "the man has been killed" may sound blunt; the euphemism here might be "the man has gone to the heaven."

First WW began in...jargon & verbal delicacy...that were skillfully made by language & literature. When language & literature create jargon & verbal delicacy: it is not literal. Moreover, something literal does not go with circumlocution. it cannot be lucid. it may be impenetrable [because of jargon & verbal delicacy.] D & E have word pairs that do not clearly match with each other.

All the choices miss a required pronoun, its, in the introductory phrase: Because of its less availability and greater demand in scientific research, ....

Exactly 4 times winning means exactly 1 time not-winning. So [according to binomial theorem] you put the desired outcome not only for 4 times but also for the remaining times.

A-->99^9/9^99 = (9*11)^9/9^99 [looking at B, you do not need to break 9 further] = 9^9*11^9/9^99 = 11^9/9^90 [recall some rules of power] Base Same Rule1: p^a*p^b=p^(a+b) [if bases are same, powers add in multiplication] Rule2: p^a/p^b=p^(a-b) or 1/p^(b-a) [if bases are same, one power is subtracted from another in division, carefuly see the side in which the subtraction to keep: numerator side or denominator side] Power Same Rule 3: (p*q)^a = p^a*q^a [if power is same, then power applies equally to bases in multiplication] Rule 4: (p/q)^a = p^a/q^a [if power is same, then power applies equally to bases in division]

Sorry, rajatmeh. Yes, you are right. Perhaps I need to have my eyes checked. :)

Use binomial distribution: (a) p(exactly 4 times) = 5C4*(0.6)^4(1-0.6)^1 (b) p(at least 4 times) = p(4 times) + p(5 times)

For all x, x^2+ax+t=(x+c)^2. Therefore, this equation is an identity, i.e., x^2+ax+t≡(x+c)^2. So, genius is right in equating equal parts, but I wonder why genius missed statement 1. By equating equal parts, we get a=2c & t=c^2 (1) t=0, so c=0, and finally a=0. Sufficient. (2) c=-3, so a=-6. Sufficient. D it is.

Dear pras, in an equation, never cancel a variable [i.e., divide by a variable] if it is in the numerator part. Doing so would reduce the no. of true solutions. For example, in stem 2, x=9 is one solution. Another one is x=0, which you could get if you factorize by side changing rather than divide both sides.

Genius, you are right. However, this is also applicable for any triangle whose two sides are fixed, regardless of the third side. Imagine three sides are as in the pic A______________B__________________C. Side AC & AB are fixed, while BC can vary. In the hypothetical case of the picture, the area is 0. Now keep AC horizontal but raise AB inclined with AC. You get a new BC, and a new height, and some area for the triangle. Continue raising AB. At 90 degree inclination of AB on AC, you get the highest height [AB itself] and the highest area 1/2*AC*AB. As you further move AB, it will now go to opposite side of AC. The third side is still becoming increasing larger still but height is now falling. Therefore, from maximum area, it is now deceasing. You can also think of the formula 1/2*a*b*SinC. When C is 90 degree, area is the maximum, regardless of the third side as long as a & b are fixed.

Well, x=7a+1 & x=3b+2 Combining, you get [say] x=21c+r Now this last expression (21c+r) must also leave remainder 1 when divided by 7 and leave remainder 2 when divided by 3. Since 21c part is divisible by both 7 and 3 [in fact this is why you were searching for combined 21c], you now must play with r. So, considering the higher divisor 7, first value of r=8. Now test with 3, hmm gives 2 as remainder. So, r=8 fits. You can try with higher values not exceeding 21, although that may not be required.

Hi, Carthik, height of isosceles triangle is 5 and base is 2√2. We need to find one of the equal sides. Drop a perpendicular from vertex (i.e., the point at which the equal sides meet) to the base. The triangle will be split into two congruent right triangles. Consider one of the right triangle, in which Hypotenuse=side that we want to know Perpendicular = height, i.e., 5 in this case Third side = √2 [half of the base of complete triangle]