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

Hi Satwik,

The amount of practice that is needed for DI or any other section varies from person to person. A good way of figuring out how much percentage of time you need to allot for a particular section is by analysing how well you are performing in that section in the mock tests. In DI, there as some sets that are straight-forward like sets based on line charts, bar charts etc. that are generally calculation intensive. If you are good with calculations in general, you need not spend a lot of time on these type of sets. Then there are other types of sets like games and tournaments, balancing quantities etc. that are more logic based and less calculation based. If you find that these are troubling you in mocks, you need to spend more time on these type of sets. On the whole, it is a good idea to solve at least 2-3 DI sets everyday in the last two months.

Check out a few of our DI tests here:

https://cracku.in/cat/quotient/dailytest-day141

https://cracku.in/cat/quotient/dailytest-day124

https://cracku.in/cat/quotient/dailytest-day110

Average of Ajmer = A; number of students in Ajmer = a
Average of Delhi = D; number of students in Delhi = d
Average of Chennai = C; number of students in Chennai = c

(aA+dD)/(a+d) = 71;(cC+dD)/(c+d) = 76;(cC+aA)/(c+a) = 79;

Cross-multiplying and adding the equations,
2Aa + 2Dd + 2Cc = 150a + 147d + 155c

Because a, d and c are independent, equate the like terms
So, 2A = 150 => A = 75; 2C = 155 => C = 77.5; 2D = 147 => D = 73.5
So, the maximum and minimum values of (aA+dD+cC)/(a+d+c) are 77.5 and 73.5 respectively
73.5 < N < 77.5

The first statement can be written as "No yellow is pen"
The second statement says "Some pencils are yellow"
So some pencils(which are yellow) are not pens.

Hi Anjum,

Improvement in accuracy/percentile takes time and consistent hard work. We advise students to spend at least a couple of hours every day working on their areas of weakness. The most important section in Verbal is RC - proficiency in all other areas of Verbal can be improved if one improves in RC. To improve in RC, initially make a small summary of the RC that contains the gist of the article. After writing the summary, go back and read the article to see if you missed any points. Follow this method consistently for a couple of weeks, and you will find yourself improving in RC.

Apart from this, concentrate on Grammar questions because this is an area where the accuracy can be close to 100%. Most importantly, every time you get a question wrong, do a thorough analysis of where you went wrong in the thought process. Try to incorporate these learnings into your next test/question.

Take a look at these links for practise questions in Verbal Ability:

Para Jumbles

Reading Comprehension

Critical Reasoning