question 3 of the probability test 7 has a wrong solution , there are only 4 steps to be taken westwards

Hi,Sorry for the inconvenience. The question has been corrected. Thanks for pointing it out!

if 3 positive real numbers x,y,z are in AP such that xyz=4,then what will be the minimum value of y?i)2^1/3 ii)2^2/3 iii) 2^1/4 iv) 2^3/4

let the numbers be a-d, a, a+dSo, xyz=4, this means (a-d)(a)(a+d)=4=> $$a(a^{2}-d^{2})=4$$=> $$a^{3}(1-\frac{d^{2}}{a^2})=4$$We have to minimize a, this means we have to minimize $$a^{3}$$, this means we have to maximize $$(1-\frac{d^{2}}{a^2})$$. Since $$\frac{d^{2}}{a^2}$$ is always positive, $$(1-\frac{d^{2}}{a^2})$$ has max value of 1. This means minimum $$a^{3}=4$$. This means minimum $$a=2^{\frac{2}{3}}$$

if x,y,x are in GP and a^x,b^y,c^z are equal then a,c,b are in i)AP ii)GP iii)HP

Let us take x, y, z to be k/r, k, kr respectively.a^x = b^y = c^y implies a^(k/r) = b^k = c^(kr)If we take logarithm for all we get,(k/r) loga = k logb = kr logc(1/r) loga = logb = r logca^(1/r) = b = c^rClearly we can observe that a, c, b follow No Progression among AP, GP and HP.

In an infinite GP ,each term is equal to 3 times th sum of the terms that follow.if the first term of the series is 8,find the sum of the series?

Since the first term is 8, sum of all the terms following it i.e. a2+a3+a4+....a(infinity) will be 8/3. So the sum of the series will be 8+8/3 which will be 32/3.

In an infinite GP ,each term is equal to 3 times th su of the terms that follow.if the first term of the series is 8,find the sum of the series?

Since the first term is 8, sum of all the terms following it i.e. a2+a3+a4+....a(infinity) will be 8/3. So the sum of the series will be 8+8/3 which will be 32/3.

how i solve the problems on remainder theorem?

Hi,I will suggest you to go through the basics. The variety of questions are limited and hence this topic can be turned into strength once you get through all the varities. There are a few theorems based on this concept. You can check them out here:https://cracku.in/cat/quant-and-di/number-systems/cheatsheet

how i solve the problems on remainder theorem?

Hi,I will suggest you to go through the basics. The variety of questions are limited and hence this topic can be turned into strength once you get through all the varities. There are a few theorems based on this concept. You can check them out here:https://cracku.in/cat/quant-and-di/number-systems/cheatsheet

hii..please tell me what should be my percentile in cat so that i can get admission in fms delhi n are there any other requirements.

Hi,Cutoff for general category in terms of percentile is usually around 98.5 percentile. It varies a lot and depends upon one's profile. Securing admission, however, depends on a whole lot of different factors so one can't pinpoint a safe number. Hope this helps.

compound interest on a certain sum at a certain rate of interest for the 2nd year is rs. 2200 and for the 3rd year is rs. 2400 find the principal and the rate of interest

Let the Principal be P and rate of interest be r.Given, 2200 = $$P(1 + r/100)^2 - P(1 + r/100) $$ = P(1 + r/100)(r/100) - (1)2400 = $$P(1 + r/100)^3 - P(1 + r/100)^2 $$ = P(1 + r/100)^2(r/100) - (2)(2)$$\div$$1 = 12/11 = 1 + r/100r = 100/11Substituting r in (1): P( 1 + 1/11)(1/11) = 2200P(12/11)(1/11) = 2200thus, P = Rs 22,183$$\frac{1}{3}$$