Hai Tejashree,

I believe the doubt is in reference to the fourth parajumble question asked in CAT QUOTIENT - parajumble test 8.
Sentence D introduces the subject of sense and must clearly precede all the other four sentences given. As you mentioned, D is clearly preceded by some sentences, but as they are not present in the given set of five sentences, D would be the first sentence among the five given sentences.
Thus, the first sentence of a parajumble need not be the first sentence of a passage, article, book etc. ,it should just be the first sentence among the given set of questions.

The radius of the cylinder and its height is same. Thus, its curved surface area is $$2\pi rh=2\pi r^{2}$$. Since the cone is placed on the cylinder, its radius must be equal to that of the cylinder. Its curved surface area becomes $$2\pi rl$$. Now the ratio of area is 8:5, this means :
$$\frac{2\pi r^{2}}{\pi rl}=\frac{8}{5}=>\frac{r}{l}=\frac{4}{5}$$. We need to find the ratio of height of cylinder to height of cone. Lets suppose height of the cone is h. We want the ratio r/h since height of cylinder is equal to r.
$$\frac{r}{l}=\frac{r}{\sqrt{{r^{2}+h^{2}}}}=\sqrt{\frac{r^{2}}{r^{2}+h^{2}}}$$. Let r/h be x. Dividing numerator and denominator by $$h^{2}$$, we get:
$$\sqrt{\frac{x^{2}}{x^{2}+1}}=\frac{4}{5}=>\frac{x^{2}}{x^{2}+1}=\frac{16}{25}=>x^{2}=\frac{16}{9}=>x=\frac{4}{3}$$.

are

Hi Arun Kanade,

For RC preparation, it is important to read as much as possible and from as many sources as possible. You have to read widely and deeply. Do not limit your reading to only one particular subject. Read from different subjects among Politics, Philosophy, Psychology, History etc. Once you read an article, make it a point of summarizing the article in a few sentences. This will improve your comprehension skills.

Check out a few RC practice tests here:
https://cracku.in/cat/quotient

Either the data is inadequate or the answer is DECBA

iii) and is 0

iii)

Although in the question it hasn't been specified how many cells to start with. I am assuming it to be 1 considering the options.
It is essentially a GP with a=1, r=2 and n=9. So, the number of cells by the end of 9 hours will be-
[a(r^n-1)]/(r-1) which will give us (i) as the answer.

Sum of first n terms is half of sum of next n terms. This means, sum of first 2n terms is 3 times the sum of first n terms. Let the first term be a and common difference be d.
This means that $$\frac{\frac{2n}{2}(2a+(2n-1)d)}{\frac{n}{2}(2a+(n-1)d}=3$$
=> 4a+4nd-2d=6a+3nd-3d
=>2a-nd-d=0
Now the ratio of sum of 3n terms to sum of n terms is:
$$\frac{\frac{3n}{2}(2a+(3n-1)d)}{\frac{n}{2}(2a+(n-1)d}$$
$$\frac{6a+9nd-3d}{2a+nd-d}$$
$$\frac{6a+3nd-3d+6nd}{2a+nd-d}$$
$$3+\frac{6nd}{2a+nd-d}$$
Now, 2a=nd+d
this means 2a+nd-d=2nd
Thus, answer is 3+3=6

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

let the numbers be a-d, a, a+d
So, 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 $$a=2^{\frac{2}{3}}$$