DATA INTERPRETATION 1
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Presents
DI & LR
For
CAT 2009

DATA INTERPRETATION

INTRODUCTION

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 have different
weights. Unfortunately, the scale in the village could measure
only weights above 30 kgs. So, he measured the children in
pairs and the weights are: 35,36,37,39,40,41,42,45,46 & 47 kgs.
He however forgot to note down their names. What is the weight
of the lightest child? (XAT 2009)
 A) 15 kg
 B) 16 kg
 C) 17 kg
 D) 18 kg
 E) 20 kg

SOLUTION
 Let us assume there are 5 children such that
A<B<C<D<E
 Now, A+B/A+C/A+D/A+E would be the possible
combinations. So, each child is weghed 4 times.
 So,
4(A+B+C+D+E)= 35+36+37+39+40+41+42+45+46 +47
A+B+C+D+E = 408/4 = 102
 We know, A+B = min = 35, D+E = max = 47
(35)+C+(47) = 102 => C = 20 kgs.

SOLUTION
 So, we have the weights as
A<B< 20< D< E
 Examining the options, we can rule out (e) as, if 20
was least weight, A and B cannot be lesser.
 Also, 18 + 17 = 35, so 18 cannot be least weight! So,
(d) is out.
 Also, 15 + 20 = 35, but 20 is supposed to be 3rd
largest. So, (a) is ruled out.

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

SOLUTION
 Let’s check (c)
 (c) 17<B<20<D<E
 17 + 18 = 35
 So, 17<18<20<D<E
Examine all their combinations :
17+18 = 35
17+20 = 37
18+20 = 38 (wrong)
So, (c) is wrong, correct answer is (b)

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 :
Teacher Marks
English 232
Math 238
Bio 244
Geo 245
Marathi 250
Hindi 256
History 235
 But, one of them is lying. Using the above info, answer the following
questions.

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

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

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

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

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
 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:
 From each case, we get second highest scores as
56,55 and 58

SOLUTION
3. What is the least possible score?
a) 34 b) 35 c) 36 d) 37 e) none
 Solution:
 Least possible score is present in case with marathi
teacher lying -> 34
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
 Solution:
 Sum of least scores = 80, in case with Hindi Teacher
lying. Here we get, 55 + 49 = 104. and no other
combination.

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