19832659 Data Interpretation

8/3/2019 19832659 Data Interpretation

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DATA INTERPRETATION 1

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PresentsPresents

DI & LRDI & LR

ForForCAT 2009CAT 2009


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DATA INTERPRETATION


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INTRODUCTION


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TYPE 1 : AVERAGES TYPE

Questions are of the following type :

A rural child specialist has to determine the weight of different ages. He knows from past experience that each child

would weight less than 30 kg and each of them would havedifferent weights. Unfortunately, the scale in the villagecould measure only weights above 30 kgs. So, he measuredthe children in pairs and the weights are:35,36,37,39,40,41,42,45,46 & 47 kgs. He however forgot tonote down their names. What is the weight of the lightestchild? (XAT 2009)

y A) 15 kg

y B) 16 kg

y C) 17 kg

y D) 18 kg

y E) 20 kg


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SOLUTION

Let us assume there are 5 children such that

A


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SOLUTION

So, we have the weights as

A


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SOLUTION

Now, using (b) and (c)

(b) 16 < B < 20 < D < E

So, B should be 35 16 = 19

Now, we get 16 < 19 < 20

Examine all their combinations :

16+19 = 35

16+20 = 36

19+20 = 39


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SOLUTION

Lets check (c)

(c) 17


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SLIGHT MODIFICATION

There are 6 students in a class, where marks are arranged as

A>B>C>D>E>F (all integers)

Their teachers gave sum of marks of 5 of them as follows :

But, one of them is lying. Using the above info, answer the

following questions.

Teacher Marks

English 232

Math 238

Bio 244

Geo 245

Marathi 250

Hindi 256

History 235


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SLIGHT MODIFICATION

1. What is the difference between the highest and

lowest total score?

a) 21 b) 24 c) 18 d) 27 e) none

2. What is the second highest score?

a) 55 b) 56 c) 57 d) 55 or 56 e) none

3. What is the least possible score?

a) 34 b) 35 c) 36 d) 37 e) none

4. If sum of at least 2 scores is 80, which can be

sum of 2 or more scores?

a) 93 b) 104 c) 110 d) 102 e) none


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SOLUTION

No. of combinations = 6C5 = 6 combinations

But, we are given 7 values. So 1 value is redundant.

Now, let us consider the redundant value as X

5(A+B+C+D+E+F) + X = 232+238+244+245+250+256+235= 1700

So, A+B+C+D+E = (1700 X)/5

As all the marks are integers, 1700 X must be divisibleby 5.

This will happen only if X is ending with 5 or 0


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SOLUTION

So, only geography, marathi or hindi teachers

can be lying.

Consider Geo teacher is lying.

Then,(1700 245) / 5 = 291

Now, we obtain the various marks as,

(A+B+C+D+E) (A+B+C+D) = E

Ie.,

291 232 = 59 291 250 = 41

291 238 = 53 291 256 = 35

291 244 = 47 291 235 = 56


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SOLUTION

Arrange such that A>B>C>D>E

The marks can be tabulated for each lying

teacher as follows.

Teacher Total A B C D E F

Geo (1700-245)/5 = 291 59 56 53 47 41 35

Marathi (1700-250)/5 = 290 58 55 52 46 40 34

Hindi (1700-235)/5 = 293 61 58 55 49 43 37


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SOLUTION

Answering the questions :

1. What is the difference between the highest and

lowest total score?

a) 21 b) 24 c) 18 d) 27 e) none Solution: In each case

y 59-35 = 24 | 58-34 = 24 | 61-37 = 24

2. What is the second highest score?

a) 55 b) 56 c) 57 d) 55 or 56 e) none Solution:

y From each case, we get second highest scores as

56,55 and 58


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SOLUTION

3. What is the least possible score?

a) 34 b) 35 c) 36 d) 37 e) none

Solution:

y Least possible score is present in case with marathiteacher lying -> 34

4. If sum of at least 2 scores is 80, which can besum of 2 or more scores?

a) 93 b) 104 c) 110 d) 102 e) none Solution:

y Sum of least scores = 80, in case with Hindi Teacherlying. Here we get, 55 + 49 = 104. and no othercombination.


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EXERCISE

Visit the following link for exercises on this type of

Data Interpretation Problems.

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19832659 Data Interpretation