# Data Interpretation Capsules - Edwin Jose, CAT DILR 100 Percentiler - Set 1

• DIRECTIONS : Answer the questions on the basis of the information given below.

The two graph above gives me crime statistics for the USA. Fig 1 give the variation of the number of property crimes per 1000 households with time. Property crimes consist of motor vehicle theft, theft and burglary. Fig 2 gives the number of violent crimes per 1000 population with time. Violent following can be divisible into three categories – aggravated assault, simple assault and robbery. The following facts are also given and may be used in answering the questions that follow.

• The population of the USA between 1975 and 2050 is by the equation P= 2.3 (T-1950) + 157, where P is the population in millions in the year T.
• The number of persons per household can assumed to remain constant for the period 1975 to 2050.

(1) Let x1 and x2 be the number of property crimes in 1975 and in 2000 expressed as a percentage of the population, respectively. What is the ratio of x1 to x2?
(a) 3 : 1
(b) 27 : 10
(c) 1 : 3
(d) 18 : 5

(2) Assume that the total number of property crimes per year follow the following trend after 2000. The total number of property crimes per year at the end of every 25 year is 0.71 times the number at the beginning. What is the number of property crimes per 1000 households in the year 2050?
(a) 63
(b) 90
(c) 129
(d) 180

(3) In 2000, the number of aggravated assaults/1000 population was greater the number of robberies/1000 population by 1.8 and the number of simple assaults was three time the number of aggravated assaults. What was the total number of robberies in 2000?
(a) 1.1×10^6
(b) 1.1×10^3
(c) 3.9
(d) 3.9×10^3

(4) Let y1 and y2 be the total number of violent crimes per year in 1975 and 2000, respectively. What is y2 – y1
(a) 22
(b) -22
(c) 4.2
(d) none of these

(5) The total number of property rimes in the year 200 was 1.45 times the total number of violent crimes in the year 1975. what is the average number of person per household?
(a) 2.1
(b) 3.2
(c) 4.2
(d) 4.9

Solutions:

(1) a
Since the number of persons per household can be assumed to be a constant, the ratio of x1 to x2 is the same as the ratio of the number of property crimes per 1000 households in 1975 to that in 2000. This value is 560/180 = 3:1.
(2) a
Let the number of households in thousands be T1 in 2000 and T2 in 2050. The total number of property crimes in 2000 is, therefore, T1 × 180. This will reduce to T1 × 180 × 0.71 × 0.71. The number of property crimes per thousand households in 2050 is, therefore, T1 × 180 × 0.71 × 0.71/T2
Since the number of persons per households remains constant, the ratio of T1/ T2 is the same as the ratio of the populations. The ratio of the populations is (157+ 2.3 × 50)/ (157 +2.3 × 100). The required answer comes out to be approximately 63.
(3) a
Let x be the number of aggravated assaults per thousand population. The number of robberies is x – 1.8 and the number of simple assaults is 3x. From the graph, we get the sum of all the three as 27.
Hence x – 1.8 + x+ { 3x = 27, giving x = 5.76 and x – 1.8 = 3.96.
The total number of robberies is, therefore, (3.96/1000) × (1.57 + 2.3 × 50) ×10^6 = 1.1 × 10^6.
(4) c
(5) b
The number of property crimes in 2000 was 180 × H where H is the number of households in the thousands in 2000.

DIRECTIONS : Answer the questions on the basis of the information given below.

ABC is a firm which deals with furniture. Manufacturing of table requires three levels of assembly. The finished table is at first level. The leg assembly and table top are second level. The pieces that go into the leg assembly are at the third level which consist of short rails, long rails and legs. One unit of table requires one unit of tabletop and one unit of leg assembly. One unit of leg assembly requires 2 units of short rails, 2 units of long rails and 4 units. Orders are placed just in time to minimise storage.

The lead time for activities are (Lead time is waiting time required to complete one activity)

The availability of part at present time

Demand of finished Tables

(1) For meeting the demand of 200 units of finished table of week 4, when would the first order of tabletops be placed?
(a) Week 1
(b) Week 3
(c) Week 4
(d) Week 5

(2) What is the net requirement of legs for meeting the demand of week 4 finished table?
(a) 200
(b) 50
(c) 400
(d) 800

(3) When and how many units of shortrail would be placed for meeting the demand of finished table of week 6?
(a) 100 units in week 1
(b) 200 units in week 3
(c) 300 units in week 6
(d) Data insufficient

(4) If in-hand units of legs are increased from 150 to 300, then what would be the net requirements of legs for meeting the demand of finished table of week 5?
(a) 1800, 900
(b) 2200, 1100
(c) 1600, 800
(d) 800, 400

(5) The supplier of long rails has shifted his manufacturing unit to its new location. Because of this the delivery time of long rail has been increased by 1 week. When would the order of long rails be placed to meet the week 5 demand of finished table?
(a) Week 1
(b) Week 4
(c) Week 5
(d) None of above

Solutions:

(1) a
Week 1, for detailed solution refer the table below.
(2) b 50, for detailed solution refer the table below.
(3) b 200 units’ week 3, for detailed solution refer the table below.
The gross requirement of leg is 200 and 600 in week 3 out of 300 in – hand units of legs, 200 units would be used for week 2 requirement and the rest units would be used for meeting the requirements of week 3. Therefore net requirement of week would be 600 – 100 = 500 units of legs for meeting the demand of finished table of week 5.
(4) b
For meeting the additional demand of 200 tables Short rails = 4 ×200 = 800
Legs = 8 × 200 = 1600
Total short rails = 300 + 800 = 1100
Total legs = 600 + 1600= 2200.
(5) a
Details of components available on a particular week.

Figure in bracket represent the week to which components belong.

Details of component ordered on a particular week.
Week 1
As lead – time increased by 1 week so planned order release would be by one week ahead.

DIRECTIONS:

In a group of 200 people, number of people having at least primary education: number of people having at least middle school education : number of people having at least high school education :: 7 : 3 : 1. 90 of these play football and 60 play hockey. 5 people in category III (defined as people having high school education) and one fourth each in category I and II (defined as people having primary school education only and people having middle school education but not high school education, respectively) do not play any game. In each of the above category the number of people playing only hockey equal the number of people playing only football. 2 people each in categories I and II and 1 persons in category III play both the games. 2 people playing both games are uneducated (category IV). 5 people in category III play only hockey.

Assume middle school education can be had only after completing primary school and high school education can be had only after completing middle school. Also all people in the group fall under the four categories described above.

(1) How many people have middle school education?
(a) 16
(b) 32
(c) 48
(d) 64

(2). How many high school educated people do not play football?
(a) 6
(b) 8
(c) 10
(d) 12

(3) How many people having middle school, but not high school, education play only football?
(a) 12
(b) 7
(c) 11
(d) 15

(4) How many people who completed primary school could not finish middle school?
(a) 48
(b) 64
(c) 80
(d) 96

(5) How many uneducated people play neither hockey nor football?
(a) 15
(b) 20
(c) 23
(d) 28

Solutions:

DIRECTIONS: Refer to the following data for the following solutions.

Let ‘x’ people have high school education.
∴ 3x have middle school education and 7x have primary school education.
Also as all and middle school educational people have primary school education and all high school educated people have middle school education, number of people in
Category I = 4x,
Category II = 2x, Category III = x.
∴ Number of people Category I and II, who do not play any game = x and x/2 respectively.
∴ Number of people playing only hockey = Number of people playing only football.
= (3x-2)/2 and ( 3x/2-2)/2=(3x-4)/4 respectively
Also number of people in category III playing only hockey = (x-6)/2=5 = > x=16

1. (c) 3x = 48 people have middle school education.
2. (c) Number of high school education who do not play football = (x-6)/2 + 5 = 10
3. (c) Number of people having middle school education but not high school education who play only football = (x-6)/2 = 11
4. (b) Number of such people = Number of people having primary school education - No. of people having middle school education = 7x – 3x = 4x = 64.
5. (d) Number of educated people playing football only = 39.
Number of educated people playing hockey only = 39.
∴Number of an uneducated people playing football = 90 – (39 + 5) = 46 and number of uneducated people playing hockey
= 60 – (39 + 5) = 16. Out of these 2 play both games.
∴ No. of uneducated people playing at least one game = 46 + 16 - 2 = 60.
No. of uneducated people = 200 – 7x = 200 – 112 = 88.
∴ 88 – 60 = 28 uneducated people do not play any game.

• Sir I did not get solution for third set. please will you pls explain this condition
Number of people playing only hockey = Number of people playing only football.
Also as all and middle school educational people have primary school education and all high school educated people have middle school education, number of people in

1. Category I = 4x,
Category II = 2x, Category III = x.

2. (3x-2)/2 and ( 3x/2-2)/2=(3x-4)/4 respectively

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