CAT Online 2009 - Data Sufficiency - CAT 2009 Online Preparation

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CAT 2009 Online Preparation

CAT Online 2009 - Data Sufficiency

Author: Sureshbala

"The ultimate goal of mathematics is to eliminate any need for intelligent thought."-A. N. Whitehead

Well, for starters try to think what's the meaning of the above quote, it has a nice and beautiful meaning. To make things so simple that an average mind is able to see it. So lets put this into practice, with another concept lesson.

Today, we will discuss DATA SUFFICIENCY, one of the very scoring problems in cat and other MBA entrance exams. The good thing about DS is we get DS both in quant and DI, and it makes for 6-8 problems in almost every paper. As there are no theorems in DS, we will take things to note

Things To Note:

1) DS problems, we need to answer if it is sufficient information to answer, means, there should be one conclusive answer. We do not need to find the answer, just if it can be found or not?

2) Some questions ask , is this true? So if we can find that the information available is enough to prove that it is not, we are still able to answer the question, that it is not true. Hence we are able to answer the question.

3) Check for all possibilties, that is using one statement, using second, then only combine the two.

Lets take up an example.

Instructions For DS Questions

Each question is followed by two statements, X and Y. Answer each question using the following instructions:
Mark (A) If the question can be answered by using the statement X alone but not by using the statement Y alone.
Mark (B) If the question can be answered by using the statement Y alone but not by using the statement X alone.
Mark (C) If the question can be answered by using either of the statements alone.
Mark (D) If the question can be answered by using both the statements together but not by either of the statements alone.
Mark (E) If the question cannot be answered on the basis of the two statements.

Example 1

In a triangle ABC, D is a point on the side BC and M, N are length of perpendicular dropped on line AD from the vertices B and C respectively. Is M > N?

(X)AB>AC
(Y)BD<DC

The problem is simple, but again see this, this question asks is M>N? The answer may be yes or no, but what we are concerned with is our ability to answer it and not the actual answer.

AB>AC gives us no idea, just imagine a few figures and you will know this.

Look at the other statement it says BD<DC, if BP and CQ are the perpendiculars then, BDP and CDQ are similar

so $\frac {BD}{DC}=\frac {BP}{CQ}$ so we know the ratio and thus are able to answer and see this the answer came NO.

so, we don't really want the answer, but the ability to give a unique answer.

So, answer is B)

Let D' be the midpoint of BC and let X' and Y' be the feet of the perpendicular from B and C to AD' respectively => X' = Y'. As D' shifts to right or left, we can know which of M or N is bigger. Thus, (Y) answers

(X) doesn't tell us anything

=> choice (B) is the right answer

Practice Problem 1

Is x=y?

X:(x+y)(1/x+1/y)=4?

Y:(x-50)^2=(y-50)^2

Practice Problem 2

What is the value of m and n?

X: n is an even number, m is an odd number, m>n

Y: mn=30

Note: Both the practice problems are old cat problems, enjoy!

Example 2

The distance of point P=(x,y,z) from the origin is $\sqrt{(62)}units$ , then find the coordinates of point P.

X: x+y+z=12

Y: x,y, and z are positive integers.

From original questions x^2+y^2+z^2=62

From statement X

(x+y+z)^2=144

2(xy+yz+zx)=62

can't say for sure

From statement Y:

positive integers

but we can easily find two pairs (1,5,6), (2,3,7). can't find a unique solution

Option E

Practice Problem 3

Rahim plans to draw a square JKLM with a point O on the side JK, but is not successful. Why is Rahim unable to draw the square?

X: The length of OM is twice that of OL

Y: The length of OM is 4 cm

This problem is taken from CAT 07 paper.

Example 3 Let p(x)=x^2+40 . Then for any two positive integers i and j where i > j, is p(i) + p(j) a composite number?

(X) p(i) - p(j) is not a composite number
(Y) p(2i) + p(2j) is a composite number

One of my favorite problems!

p(i)-p(j) is not a composite number

= i^2-j^2 is a prime as i, j are positive integers and i>j,(i^2-j^2) can't be 1

=(i+j)(i-j)=prime

so i-j=1

let p be the prime so $i=\frac {(p+1)}{2}$

$j=\frac {(p-1)}{2}$

clearly p is not 2 hence all p is odd

$p(i)+p(j)=80+\frac {(p^2+1)}{2}$

now p^2=6k+1 (Refer my lesson on prime numbers for this)

therefore

$p(i)+p(j)=80+\frac {(p^2+1)}{2}$

becomes

$80+\frac {(6k+2)}{2}=81+3k=3(27+k)$

so not a prime => can be answered by using X

now, p(2i) + p(2j) is a composite number

4(i^2+j^2+20) is composite

now i and j can be anything, can't make any conclusions

=> choice (A) is the right answer