2700 = $$2^2 * 3^3 * 5^2$$ = 1*2*2*3*3*3*5*5
These are 8 numbers (not distinct). If we combine any two of them, the number becomes less than 8.
So, the number of ways of writing 2700 as a product of 8 distinct integers is 0.

And would be 6 as we can also take into account the negative numbers.

(x+y)(x-y) = 225

225 = 3^2 * 5^2

Number of factors = 3*3 = 9 = 225*1, 45*5, 25*9, 75*3, 15*15

x+y = 225 , x-y = 1 -> 1 solution pair. Similarly x+y = 1, x-y = 225 -> 1 solution pair

15*15 gives only 1 solution pair

Total number of pairs = 4*2 + 1 = 9

Hi Laxmi,

It is almost impossible for anybody to remember every detail given in the passage. Rather than trying to do that, a good way to approach the RC would be to identify the main point of each paragraph as you read through the passage. If you do this, you will have a good idea about which paragraph deals with which idea. Now, If you encounter a question specific to a paragraph, you can easily go back to that specific paragraph and answer the question. Generic questions are mostly on the main points of the passage and if such a question is asked, you will be able to answer it without having to go back to the passage since you have already identified the main points.

Let a% and b% be the concentrations of M and N respectively => Initial volumes of acid in the solutions M and N are 130a and 180b respectively.
Final volumes of acid in the solution after transferring x ml from each of the flasks to the other flask are 130a-ax+bx and 180b-bx+ax.
Final concentrations are equal => (130-ax+bx)/130 = (180b-bx+ax)/180
On solving this equation you get the value of x as 75.48 ml.

Thanks for the solution. But how to solve the equation (130-ax+bx)/130 = (180b-bx+ax)/180?

Hi Vishal,

Implementing the following strategy can help you increase your accuracy in Reading Comprehension:

1. Identify the theme of every paragraph while reading the RC for the first time.

2. Identify the specific questions. These are questions that pertain to a specific paragraph in the essay. These questions can be answered with greater accuracy.

3. For the 'generic questions', try to eliminate the least likely options. As a general strategy, choose the RC whose topic is familiar to you.

Try to answer at least three of the four questions in the RC. You can practise RC questions at the following links:

https://cracku.in/cat/quotient/dailytest-day86

https://cracku.in/cat/quotient/dailytest-day72

If there is no restriction on the variables (such as non-negative, positive etc.), the number of solutions is infinite.

a+b+c+d = 12
a, b, c and d must be at least 1 => So give 1 to each of the variables => a + b + c + d = 8
Number of non negative solutions = (8+4-1)C(4-1) = 11c3 = 165
But none of the variables can be more than 6. So, we have to remove certain cases.
Give 6 more to a => a + b + c + d = 2 =>a is 7 which is wrong according to the given information. So, all these cases are to be removed.
Number of non negative solutions = (2+4-1)C(4-1) = 5c3 => But 6 can be given to a, b, c or d
So, the total number of cases to be removed = 4 * 5c3 = 4*10 = 40
Answer = 165 - 40 = 125

In decimal system, numbers of the form 1/n are terminating if n is of the form 2^a * 5^b. So, in base 57, n should be of the form 3^a * 19^b. In base 323, n should be of the form 17^k * 19^p. Since the number should terminate in both the bases, n should be of the form 19^p (Common prime factor). Since n is less than 10000, the only possible values of p are 0, 1, 2 and 3 - a total of 4 values.

Answer = 4

Number of multiples of 3 = 121, number of multiples of 4 = 91, number of multiples of 7 = 52, number of multiples of 15 = 24.

Number of multiples of LCM of (15,3) = 24,

Number of multiples of LCM of (15,7) = 3,

Number of multiples of LCM of (15,4) = 6,

Number of multiples of LCM of (7,3) = 17,

Number of multiples of LCM of (7,4) = 13,

Number of multiples of LCM of (4,3) = 30,

Number of multiples of LCM of (15,4,3) = 6,

Number of multiples of LCM of (15,3,7) = 3,

Number of multiples of LCM of (15,4,7) = 0,

Number of multiples of LCM of (3,4,7) = 4,

Number of multiples of LCM of (15,3,4,7) = 0

Exactly 2 newspapers = 3+6+24+13+17+30 - (3+4+6) = 93 - 13 = 80

No newspaper = 365-[(121+91+52+24) - 93 + 13] = 365 - 208 = 157

66
All prime.nos +2

1*3*5*...*99 = 25k
25k mod 100 = 25 if k mod 4 = 1
25k mod 100 = 75 if k mod 4 = 3
In this case, 1*3*...*23*27*...*99 mod 4 = 1*-1*1*-1*...*1 (25 pairs) = -1 = 3
So, remainder = 75