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# Tabular DI(Data Interpretation) Shortcut Tricks & Tips

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# Tabular DI Tricks & Tips

Data Interpretation questions typically have large amounts of data given in the form of tables, pie-charts, line graphs or some non-conventional data representation format. The questions are calculation heavy and typically test your approximation abilities. A very large number of these questions check your ability to compare or calculate fractions and percentages. If you sit down to actually calculate the answer, you would end up spending more time than required. Here are few ideas that you can use for approximation.

Funda 1 Calculating (Approximating) Fractions

When trying to calculate (approximate) a fraction p/q, add a value to the denominator and a corresponding value to the numerator before calculating (approximating).

#### Get Best Maths Short Tricks Book With 2500+ Questions (50% Off)> Click Here

Example,

What is the value of 1789/762 ?

First the denominator. We can either take it close to 750 or to 800. Lets see how it works in both cases. We know that the answer is between 2 and 3, so for adding values / subtracting values from the denominator or the numerator, I will consider a factor of 2.5.

Case 1: 762 is 12 above 750, so I will subtract 12 from the denominator. Keeping the factor of 2.5 in mind, I will subtract 25 from the numerator.

My new fraction is,

(1789 – 25) / (762 – 12) = 1763 / 750 = 1763 ? (4 / 3000 ) = 7.052 / 3 = 2.350666

As you can see, with very little effort involved in approximation, we arrived really close to the actual answer.

Case 2: 762 is 38 below 800, so I will add 38 to the denominator. Keeping the factor of 2.5 in mind, I will add 95 to the numerator.

My new fraction is,

(1789 + 95) / (762 + 3 = 1884 / 800 = 2.355

As you can see, even this is close to the actual answer. The previous one was closer because the magnitude of approximation done in the previous case was lesser.

Funda 2 Comparing Fractions

If you add the same number to the numerator and denominator of a proper fraction, the value of the proper fraction increases.

If you add the same number to the numerator and denominator of an improper fraction, the value of the improper fraction decreases.

Note: You can remember this by keeping in mind that,

1/2

and

3/2 > 4/3 > 5/4 > 6/5 …

Example,

Arrange the following in increasing order: 117/229, 128/239, 223/449.

Lets first compare 117/229 & 128/239.

If we added 11 to the numerator and the denominator of the first proper fraction, the resulting proper fraction would be 128/240, which will be bigger in value than the original (as per Funda 2).

We know that 128/240 is smaller than 128/239, as the latter has a lower base.

So, 117/229

? 117/229

Now lets compare 117/229 and 223/449.

If we added 11 to the numerator and the denominator of the second proper fraction, the resulting proper fraction would be 234/460, which will be bigger in value than the original.

If we doubled the numerator and denominator of the first proper fraction, the resulting proper fraction would be 234/458.

We know that 234/460 is smaller than 234/458, as the latter has a lower base.

So, 223/449

? 223/449

Using the above two results, we can say that 223/449

Note: This question can be solved much simply by just looking at the numbers and approximately comparing them with 12. I used this long explanation to illustrate the funda given above.

Following are a few other shortcuts that might come in handy during DI-related calculations.

Funda 3 Percentage Growth

If the percentage growth rate is r for a period of t years, the overall growth rate is approximately: rt + t * (t-1) * r2 / 2

Note: Derived from the Binomial theorem, this approximation technique works best when the value of ‘r’ is small. If the rate is above 10%, then this approximation technique yields bad results. Also, if the rate is 5% then r = 0.05; if the rate is 7.2% then r = 0.072.

Funda 4 Comparing Powers

Given two natural numbers a and b such that a > b > 1,

ab will always be less than ba

Note: There are only two exceptions to this funda. I hope someone in the comments will point them out (anyone?).

## Tabular Form

Tabular form or Tables is an easy area to score marks in the aptitude section of IBPS, SBI PO and SSC exams. One or two problems in the exam are asked on Tabular form of Data Interpretation.

A tabular form is a representation of data in a table format. It is easy and convenient to represent data in a tabular form. The data is present in rows and columns and one can draw conclusions from it easily. It is the most organized method to represent data.

Let us now look at the different types of Data Interpretation questions that are asked from the tabular form.

Problem I: Total 24500 people who are in the given profession and (of these) percentage of female and males.

Question 1: What is the ratio of the total number of males in the medical and teaching profession together to the ratio of females in the same professions together?

*Tip: Ratio is the comparison of like terms in its simplest form.

Solution:

Step 1:

As we need to find the ratios, we use the formula

Males ( Medical + Teaching): Female ( Medical + Teaching)

Step 2:

By substituting the values in the above formula

(40% x 11%x 24500)+(20% x 21% x 24500) : (60% x 11% x 24500)+(80% x 21% x 24500)

Step 3:

By eliminating 24500 and percentage as it is common, according to the concept of ratio we get,

(40 x 11) + (20 x 21) : (60 x 11) + (80 x 21)

Step 4:

By Simplification,

22 + 21 : 33 + 84

= 43 : 117

Therefore, The ratio of total number of Males: Females in the medical and teaching profession together is 43:117

*Note: Don’t write the steps during examination as it consumes a lot of time. Directly jump to the step 3 as 24,500 is common and step 2 can directly be eliminated.

Question 2: Total number of people in the teaching profession is what percent of total number of people in law?

*Tip-In this question we are suppose to find the percentage change

Solution:

Step 1:

As we need to find the percentage change, we need to use percentage formula

x is what of % of y  = (x/y)100 [Formula]

X-Total number of people in teaching program

Y- Total number of people in law

Step 2:

By substitution,

(Total number of people in teaching program/ total number of people in law) x100

Step 3:

By substituting values in the above formula,

(21% x 24500/24% of 24500) x 100

Step 4:

By simplification,

21/24 x 100

Step 5:

By simplification,

7/8 x 100

= 87.5%

Therefore, the total percentage of people in the teaching profession is 87.5% of the total percentage of total people in law.

Question 3: What is the total number of males from all the profession together?

1) 11472  2)12784  3)12348  4)12453  5) None of these

Solution:

Step 1:

As we need to find the total number of males in all the profession, we are suppose to take (the percentage of males in one profession x percentage of the total number of people in that profession x total number of people) and thus we need to add all the values together.

(40% of 11% of 24500)+(70% of 18%of 24500)+(55% of 24% of 24500)+(20% of 21% of 24500)+ (65% of 16% of 24500)+(56% of 10% of 24500)

Step 2:

By Simplification,

(40 x 11) + (70 x 18) + (55 x 24) + (20 x 21) + (65 x 16) + (56 x 10)

= 5040

Therefore the answer is Option 5 – None of these.

In examination on average about 30 to 36 seconds are allotted to solve a question. Though this question appears to be simple the process of calculation is really lengthy as we cannot eliminate any number. The quickest way to solve this question is to skip the question or else leave the question for the end.

Question 4: The female in the Engineering professions are what percent of males in the management profession.

1) 71.71  2) 96.43 3) 83.16 4)68.54 5) None of these

Solution:

Step 1:

As we need to find the percentage change, we use the percentage formula,

x is what percent of y = x/y x 100

x- Females of Engineering

y-Males of Management

Step 2

By using the percentage formula in finding out the female engineering profession percentage to male percentage in management profession

Females of Engineering/Males of Management x 100

Step 3:

By substituting the values in the formulas,

30% of 18% of 24500/56 x 10 x 24500

Step 4:

By simplification,

30 x 18 /56 x 10 x 100

Step 5

By simplification

54/56 X 100

= 96.43%

Therefore, the answer is option 2 – 96.3%.

Remember don’t waste your time in writing down steps. Directly jump to step 4 and eliminate all unnecessary steps.Solve this sum on the bases of assumption, it will help you to save some time.

Question 5: What is the ratio of number of males in the banking profession to the number of males in the engineering profession.

Solution:

Step 1:

We use the formula as we need to find the ratio of males in banking to ratio of males in Engineering

Males (Banking) : Males ( Engineering)

Step 2:

By substituting the values in the formulas,

65% x 16% of 24500 : 70% x 18% of 24500

Step 3:

By simplification,

65 x 16 : 70 x 18

Step 4:

By Simplification,

52 : 63

Therefore, the ratio of number of males in the banking profession to the number of males in engineering profession is 52:63.

Tables refer to the arrangement of data in the form of rows and columns.

Positives Negatives
1. Data is available in a compiled form; hence there is no ambiguity in interpretation. 1. Trends cannot be easily established in the table.
2. Data Values are directly given and hence one need not spend time finding the accurate Values. 2. One can get confused over the sheer volume of the data.

#### Shortcuts to crack DI sets containing Tables

1. Do not get carried away by the sheer amount of data, the set may be easy for all you know!!

Check out this table from CAT 2002.

At the first glance, it seems that this table has too data intensive and hence should not be attempted. But on second thoughts if you look at the questions, you will find that this is a simple set pertaining to counting some values. So rather than getting carried away by the volume of data, you need to have a look at the questions as well.

2. Modify the question such that the answers can be easily calculated.

Check out this table pertaining to CAT 2005.

In the first question, we need to find the ratio of ‘production to population’ i.e. divide the second-last column with the last one and find out how many of these values are greater than that of Gujarat. However one may find that this division results in values in fractions and hence difficult to compute. Instead, if we were to modify the question and calculate the ratio of ‘population to production’ and find out how many of these values are less than that of Gujarat, the entire calculation becomes oral. For example, for Gujarat this value is between 2 and 3. We can find that the only states for which this value is less than that of Gujarat are Haryana (1…), Punjab (1), Maharashtra (2…) and Andhra Pradesh (0…).

Similarly in the second question, we need to simply figure out that the values of the second last column need to be multiplied by 10 and this needs to be divided by the values of the last column. The states where this value is more than 4 is Haryana, Gujarat, Punjab, Madhya Pradesh, Tamil Nadu, Maharashtra, Uttar Pradesh and Andhra Pradesh.

Now Let us come to the most Important Thing.

” Demystifying Data Interpretation ”

Look anyone can solve DI question, if there is no time limit given. But in Exam we are required to solve the question in limited Time.

Few Necessary Skills which are required for Mastering DI are Listed as below:

1.Read the data very carefully. Even the minutest word must not be overlooked since many a times even single word/group of words could become critical.
2.If there are more than one graphs/charts/tables, understand the relationship between them clearly before you proceed to solve the questioned asked.
4.Be careful to use proper units and beware of charts and tables with non-uniform units.
5.Avoid simple calculation mistake and revise your answer at least once before moving to next question.

Tips to reduce calculation time
In DI most of the questions are usually based on percentage increase and decrease, ratios and averages. A simple trick for solving data interpretation problems quickly is:

• Learn tables till 20.
• learn fractions till 1/20 to improve your speed
• Waste little time in finding averages: Here if you have to find averages for sales in branches B1 to B6 for the year 2014, use this shortcut:
What you usually do: [80+75+95+85+75+70]/6
Instead: 70 + [10+5+25+15+5]/6=70+ [60/6]= 80
With this technique, you will reduce calculation time and are also bound to make fewer mistakes.

Types of Data Analysis & Interpretation

1. Tabular representation
2. Bar charts
3. Line Graphs
4. Pie  charts
5. Caselets
6. Spider charts
7. Missing DI
8. Miscellaneous charts.

We will now go on and analyse each of the types of charts.

## Tabular Representation

Some important Tricks and Concepts generally used in tabular DI:
1. Average :
Average =total of data/No.of data

2. Percentage :
If we have to find y% of x, then
y% of x=(x*y)/100

3. Ratio & Proportion :

• The ratio of a to b is written as a : b = a / b
• The idea of proportions is that two ratio are equal.      If a : b = c : d, we write a : b : : c : d
The following exercise will help you to clearly understand Tables and the kind of questions that might be created on tables.

Ex-1: Directions (Q. 1-6): These questions are based on the following information regarding the price changes that a certain pharmaceutical company is considering for its products.

1. A man is prescribed a combination of Antacid and Anti-Hypertensive in the ratio 2 : 3 for the first week and of Anti-Hypertensive and Anti-Flatulent in the ratio 3 : 4 for the second week. The purchased all the medicines under the existing price. His expenditure in the second week is what % more than in the first week?
1) 24% more
2) 18% less
3) 26% more
4) 25% less

2. If a family has a hypertensive and an asthmatic patient, where the person with hypertension has to consume three tablets of Anti-Hypertensive per day and the asthmatic patient has to take two tablets of Anti-Asthmatic every alternate day, what will be the increase in expenditure on the two patients for 30 days?
1) Rs 37.50
2) Rs 42.75
3) Rs 46.50
4) Rs 38.50
5) Rs 39.25

3. What is the percentage increase in the expenditure of a person for one year if he consumes 32 tablets of Antacid in one week?

4. A person is prescribed to take two spoonfuls of Expectorant thrice everyday for a period of 20 weeks. Assuming that each bottle of Expectorant contains 90 spoonfuls, find the expenditure according to the existing prices.
1) Rs 210
2) Rs 200
3) Rs 168
4) Rs 240
5) Rs 220

5. In the question no. (1), average cost per tablet for the first week is what % less than the average cost per tablet for the second week?
1) 17.9%
2) 17.02%
3) 24.5%
4) 25.6%
5) Can’t say

Ex:2. Directions (Q. 6-10): Read the table and answer the questions that follows

Distribution of Students at Harvard University according to Professional Courses

6. If 60% of the boys and 70% of the girls are successful in the courses taken by them, then what is the combined pass percentage? (Approximately)
1) 65.9
2) 64.2
3) 62
4) 67
5) 66.8

7. In which course is the percentage of girls (among the total number of students) higher than the percentage of girls in any other course?
2) Computers
3) finance
4) Others
5) Cannot be determined

8. By what percentage is the number of the students doing Computers more than the number of students doing Business Management?
1)67.2
2)63.1
3)62
4)68.5
5)65.8

9. The percentage of girls engineers doing Business Management are:
1) 11.2
2) 2.2
3) 15
4) 14
5) None of these

10. Talking all the courses together, by what percentage do the number of boys exceed the number of girls?
1) 521.4%
2) 421.4%
3) 321.4%
4) 221.4%
5) None of these

1. 4;  The constant present in the first week combination is either similar or different from the constant present in the second week combination. Hence we can’t find the expenditures of first and second weeks.

2. 1; Three tablets/day would mean the hypertensive has to be taken 90 times in 30 days.
The increase is of Rs 2.50 for 10 tablets
Hence increase will be of 2.50 × 9 = 22.50 for 90 tablets
Asthmatic has to consume 30 tablets.
∴ increase = 5.00 × 3 = 15.00
∴ total increase = 22.50 + 15 = 37.50.

6.  c;  Take the Weighted average.
Total number of boys=611
Total number of girls=145
So, Ratio of boys to girls= 611:145= 4.21:1≈4:1

## Tips on Solving Table Chart Problems:

A: Read the data very carefully, as the smallest detail may change the meaning of the question completely. Similarly, the instructions have to
be understood carefully to prevent wasting time in calculating data that is not required, and also to find out exactly what is the answer that
is sought.
B: Try to understand the data provided carefully, before jumping to answer the questions. The questions are designed to be deceptive, and proper understanding of the requirements is a must. If the Data provided is of the combined variety or if there are more than one data table/charts/graphs, try to understand the relation between the given tables.

For Example, one table may talk about absolute sales figures, while the other table may talk of sales as a percentage of production. Hence, any question on excess production or Goods in stock, will require data from both tables.

C: Be very careful of the units used in the tables, and the units in which the answers (options) are provided. A mistake in the units may yield an
entirely different answer. Also be careful of whether the answer is required in decimal or percentage. Such errors are common and easily
avoidable.

Here is an example consisting tabular data:

### Example 1:

 Category of Assistance Average number receiving per month Total cost per help year(in crores of Rs.) Cost paid by Centre for the year (in crores of Rs.) 1995 1996 1995 1996 1995 1996 A 36097 38263 38.4 34.8 18.4 17.4 B 6632 5972 5.0 3.2 2.6 1.6 C 32545 31804 76.4 59.4 13.0 10.0 D 13992 11782 26.4 42.6 6.6 10.6 E 21275 228795 216.6 242.8 55.0 62.6

Example 1.1:
The category receiving the least percentage help from the centre (in the entire data) is:
(A) Category B in 1995                         (B) Category C in 1996 (C) Category B in 1996     (D)Category D in 1995
Solution:
In this type of question, it is better to examine the alternatives given rather than trying to ﬁnd the least percentage from the table. Let us now calculate the required percentage of the given alternatives:
(A) Category B in I995 =(2.65.0)×100=52% (Even without calculation, you can eliminate this choice.)
(B) Category C in 1996 was =(10.059.4)×100=16.8%
(C) Category B in 1996 was =(1.63.2)×100=50% (Even without calculation, you can eliminate this choice.)
(D) Category D in 1995 =(55.0216.6)×100=25.4%
From this we arrive at the answer (B) since this is the least percentage.

Example 1.2:
The difference between the average costs paid by the Centre during 1995 and 1996 is
(A) Rs. 66 lakh                        (B) Rs. 13.2 crore (C) Rs. 132 lalth                      (D) Rs. 13.2 lakh
Solution:
Adding all the cost ﬁgures in the 1995 column, i.e. 18.4+2.6+13.0+6.6+55.0, you get 95.6.
The average in 1995: =95.6+ Number of categories =95.6+5 = Rs. 19.12 Crore
Similarly, the average in 1996: =(17.4+1.6+10.0+10.6+62.6)5 = Rs. 20.44 Crore
The difference = Rs. (20.44−19.12) Cr = Rs. 1.32 Cr = Rs. 132 lakh
(Note how the answer needed conversion from crores to lakhs).

Example 1.3:
Monthly cost to the city receiving E category assistance in 1996 is most nearly:
(A) Rs. 1.8 crore less than that in 1995      (B) Rs. 2.1 crore more than that in 1995 (C) Rs. 2.1 cnore less than that in 1995     (D) Rs. 1.8 crore more than that in 1995
Solution:
Here, straight calculation is only needed. We need to look at the total assistance ﬁgures.
In 1995: 216.612=18.05
In 1996: 242.812=20.23
Difference = 2.183 crore ≈ Rs. 2.1 crore

Example 1.4:
Assuming
that 50% of the persons receiving category B help in 1995 were adults
caring for minor children, but the city’s contribution towards
maintaining these adults was 40% of the total contribution to B program
in 1995, average amount paid by the city for each adult per year in 1995
is most nearly:
(A) Rs. 5900                   (B) Rs. 6000 (C) Rs. 7500                   (D) Rs. 3000.
Solution:
50% of persons receiving B category help during 1995 =3316
City’s contribution to maintenance: =5.0×0.4
= Rs. 2 crore =2,00,00,0003316
= Rs. 6031.36
Rs. 6000 nearly
The correct choice is (B).

Example 1.5:
Monthly costs to the city of category D during 1995 and 1996 bear a ratio (most nearly)
(A) 2 : 3               (B) 5 : 3              (C) 3 : 2              (D) 3 : 5
Solution:
Again, we can straightaway determine the answer through simple calculation.
Since a ratio is required to be calculated, we can avoid the division by 12.
Directly from the table we have, total assistance in 1995 and 1996 for Category D as 26.4 and 42.6.
Hence the ratio is 26.4:42.6=3:5 nearly.

Sample Question

Direction: Refer to the following table and answer the given questions.

Number of cars sold by 6 Stores in 5 different months

The above Table shows:

• The number of cars sold by store P (In Jan = 133, Feb = 183, March = 278, April = 178, May = 264)

Like this we can see the others. Lets do solve some questions.

1. Number of cars sold by store T in march is what percent less then number of cars sold by Store P in may? (Rounded off to nearest integer)

(a) 29%

(b) 31%

(c) 37%

(d) 33%

Solution:

Number of  cars sold by Store T in March = 178

Number of cars sold by store P in May = 264

Required percentage = (264 – 178 /264) * 100 (in question asked less then number that’s why we deducted) = (86/264) * 100 = 32.57%

So rounded figure it will be 33%, Answer D

2. What is the average number of cars sold by all the given stores in Feb?

(a) 207

(b) 211

(c) 219

(d) 223

Solution:

To find average we have to add all the figures of Feb month and then divided by 6

= 183 + 123 + 277 + 176 + 239 + 268 / 6 = 1266/6 = 211, Answer B

3. Total number of cars sold by store Q during  all the given months together is what percent of the total number of cars sold by store S during all the given month together?

(a) 82%

(b) 88%

(c)92%

(d) 86%

Solution:
Total number of cars sold by store Q during  all the given months together = 161 + 123 + 154 + 272 + 107 = 817

Total number of cars sold by store S during  all the given months together = 225 + 176 + 98 + 284 + 167 = 950

Required percentage = (817/950) * 100= 86%, Answer D

4. What is the difference between total number of cars sold by all the given stores together in Jan and total number of cars sold by all the given stores together in April?

(a) 353

(b) 379

(c) 363

(d) 347

Solution:

Total number of cars sold by all the given stores together in Jan = 133 + 161 + 213 + 225 + 282 + 196 = 1210

Total number of cars sold by all the given stores together in April = 178 + 272 + 269 + 284 + 293 + 277 = 1573

Required difference = 1573 – 1210 = 363, Answer C

5. What is the respective ratio between total number of cars sold by stores P and R together in March and total number of cars sold by stores T and U together in May?

(a) 9:11

(b) 11:13

(c) 5:7

(d) 13:17

Solution:

Total number of cars sold by stores P and R together in March = 278 + 226 = 504

Total number of cars sold by stores T and U together in May = 379 + 237 = 616

Ratio= 504 : 616 = 9 : 11, Answer A

Directions: Study the table and answer the questions that follows:

Data Related to Human Resource of a Multinational Company (X) which has 146 Offices across 8 Countries.

Question 1: If the number of male post-graduate employees in country H is 1800, what percent of female employees in that particular country is post-graduate?

[1] 76
[2] 74
[3] 72
[4] 64
[5] 68

In country H, 80% are post-graduate. That is = [80/100]*3360 = 2688
Male is given 1800. Hence, female post-graduate employees = 2688 – 1800 = 888
Total female employees is = [5/14]*3360 = 1200
Hence, required percentage = [888/1200]*100 = 74 Percent. Answer [2] is correct.

Question 2: In which of the given countries is the percentage of women employees to the number of employees (both male and female) in that country the second lowest?

[1] G
[2] B
[3] E
[4] H
[5] D

These types of question are too much calculative. But you can apply the reasoning process to solve these questions a little more quickly. The question asks the percentage of female to total employees. This can be arrived at from the ratio that’s given in the table under the third column. So, just focus upon that ratio and focus upon the countries given in the options. (I.e. Countries G, B, E, H and D only).

Country B = 11:5 à [5/16]*100
Now, let’s say this is approximately 30% (16*3 = 48 which is close to 50).
Country D = [2/5]*100 = 40%
Country E = [6/13]*100 = approx. 45%
Country G = [7/15]*100 = approx. 45%
Country H = [5/14]*100 = approx. 35%
Now, second highest is Country H. Question solved. Answer [4] is correct.

Question 3: What is the respective ratio between total number of male employees in countries B and H together and total number of female employees in countries C and D together?

[1] 63:52
[2] 51:38
[3] 77:64
[4] 69:44
[5] 57:40

These sort of questions requires faster calculation. No other alternative is there!

Total male employees from countries B and H = [11/16]*2880 + [9/14]*3360 = 1980 + 2160 = 4140
Total female employees from Countries C and D = [11/21]*2310 + [2/5]*3575 = 1210 + 1430 = 2640
Required ratio = 4140:2640 = 69:44. Answer [4] is correct.

Question 4: What is the difference between average number of post-graduate employees in countries A, B and D together and average number of post-graduate employees in countries F, G and H together?

[1] 282
[2] 276
[3] 316
[4] 342
[5] 294

Again. Mastery at calculation is required. But here’s a reasoning approach to simplify the calculations.

75% of 2568 is required. Divide 2788 in four parts and add three parts. Thus 2568/4 is 642 and 642*3 = 1926
65% of 2880 is required. Divide 2880 in ten parts and add six parts and half of 7th part. Thus 288*6 + 288/2 = 1728 + 144 = 1872
60% of 3575 is required. Divide 3575 in 5 parts and add three parts. Thus 3575/5 = 715 and 715*3 = 2145
Average of these three is = [1/3]*[1926+1872+2145] = 1981
Using similar procedures, average of other three is calculated as = 2275
Difference = 2275 – 1981 = 294. Answer [5] is correct.

Question 5: Which of the given countries has the highest number of average employees per office?

[1] F
[2] H
[3] B
[4] C
[5] D

This question is similar to question no. 2. Calculating the averages of the given options only gives

B = 2880/18 = 1440/9 = 160
C = 2310/14 = 330/2 = 165
D = 3575/22 = 325/2 = 162.5
F = 2788/17 = 164
H = 3360/21 = 480/3 = 160
Highest is in country C.

Study the following table chart and answer the questions based on it.Expenditures of a Company(in Lakh Rupees) per Annum Over the given Years.

 Year Salary FuelandTransport Bonus InterestonLoans Taxes 1998 288 98 3.00 23.4 83 1999 342 112 2.52 32.5 108 2000 324 101 3.84 41.6 74 2001 336 133 3.68 36.4 88 2002 420 142 3.96 49.4 98
• #### 1. What is the average amount of interest per year which the company had to pay during this period ?

1. Rs. 36.66 lakhs
2. Rs. 36.36 lakhs
3. Rs. 36.26 lakhs
4. Rs. 36.06 lakhs

Option A

Explanation:

Average amount of interest paid by the Company during the given period will be

23.4+32.5+41.6+36.4+49.45lakh=36.66lakhs23.4+32.5+41.6+36.4+49.45lakh=36.66lakhs

• #### 2. The total amount of bonus paid by the company during the given period is approximately what percent of the total amount of salary paid during this period ?

1. .5%
2. 1%
3. 1.5%
4. 2%

Option B

Explanation:

Here we simply need to calculate that bonus is what percent of salary.
We will just sum all bonus and salary to get the percentage as below,

(3.00+2.52+3.84+3.68+3.96288+342+324+336+420100)%=(171710100)%=1%approx.(3.00+2.52+3.84+3.68+3.96288+342+324+336+420∗100)%=(171710∗100)%=1%approx.

• #### 3. Total expenditure on all these items in 1998 was approximately what percent of the total expenditure in 2002 ?

1. 61%
2. 47%
3. 59%
4. 69%

Option D

Explanation:

Required percentage we can easily calculate from the above table chart.
Required percentage will be,

(288+98+3.00+23.4+83420+142+3.96+49.4+98100)%=(495.4713.36100)%=69.45%(288+98+3.00+23.4+83420+142+3.96+49.4+98∗100)%=(495.4713.36∗100)%=69.45%

• #### 4. Calculate the total expenditure of the company over these items during the year 2000 from the table chart given.

1. Rs. 543.44 lakhs
2. Rs. 544.44 lakhs
3. Rs. 545.44 lakhs
4. Rs. 546.44 lakhs

Option B

Explanation:

Total expenditure of the Company during 2000

= Rs. (324 + 101 + 3.84 + 41.6 + 74) lakhs

= Rs. 544.44 lakhs.

• #### 5. The ratio between the total expenditure on Taxes for all the years and the total expenditure on Fuel and Transport for all the years respectively is approximately?

1. 4:13
2. 7:13
3. 10:13
4. 11:13

Option C

Explanation:

Required Ratio will be

(83+108+74+88+9898+112+101+133+142)=(451586)=11.3=1013

Study the following table carefully and answer the questions given below it:Number of Different categories of vehicles sold in the country over the years (in thousands)

 Year Heavy Light Commercial Vehicles Cars Jeeps Two Wheelers 1990 26 64 232 153 340 1991 45 60 242 172 336 1992 72 79 248 210 404 1993 81 93 280 241 411 1994 107 112 266 235 442 Total 331 408 1268 1011 1933
• #### 1. In which of the following years was the number of light commercial vehicles sold approximately 25% of the number of 2-wheelers sold?

1. 1991
2. 1992
3. 1993
4. 1994

Option D

Explanation:

In the year 1994,
Number of light commercial vehicles sold were 112
Number of 2-wheelers sold were 442

So,

(112442100)=25.33%(112442∗100)=25.33%

• #### 2. If the same percentage increase in the number of Heavy Vehicle as in 1994 over 1993 is expected in 1995, approximately how many heavy vehicles will be sold in 1995?

1. 141
2. 156
3. 176
4. 181

Option A

Explanation:

In this question, by referring the table chart, first we need to calculate the increase(in percentage) in sale of Heavy Vehicle in 1994 over 1993

Heavy Vehicle sold in 1993 = 81
Heavy Vehicle sold in 1993 = 107

Increase = 107 – 81 = 26

Now 26 is what percent of 81 ?

(2681100)%=32% aprox.(2681∗100)%=32% aprox.

So increase in sale were 32% in 1994, so get the answer we need to calculate the 132% of 107.

132100107=141.24132100∗107=141.24

• #### 3. The number of Heavy Vehicles sold in 1993 was approximately what percent of the total number of Vehicles sold in 1992 ?

1. 6%
2. 7%
3. 8%
4. 9%

Option C

Explanation:

Required percentage =(8172+79+248+210+404100)%=(811013100)%=7.99%=8%approxRequired percentage =(8172+79+248+210+404∗100)%=(811013∗100)%=7.99%=8%approx

1. 1990
2. 1991
3. 1992
4. 1993

Option A

• #### 5. The percentage increase in the sales in 1993 over the previous year was maximum for which of the following categories of vehicles ?

1. Two Wheeler
2. Jeeps
3. Light Commercial Vehicles
4. Cars

Option C

The following table gives the sales of batteries manufactured by a company over the years. Study the table chart and answer the questions based on it.

 Type Of Batteries(In thousand’s) Year 4AH 7AH 32AH 35AH 55AH Total 1992 75 144 114 102 108 543 1993 90 126 102 84 126 528 1994 96 114 75 105 135 525 1995 105 90 150 90 75 510 1996 90 75 135 75 90 465 1997 105 60 165 45 120 495 1998 115 85 160 100 145 605
• #### 1. What was the approximate percentage increase in the sales of 55AH batteries in 1998 compared to that in 1992 ?

1. 31%
2. 33%
3. 34%
4. 36%

Option C

Explanation:

Increase is = 145-108 = 37

Now we need to calculate that 37 is what percent of 108.

(37108100)%=34.25%(37108∗100)%=34.25%

• #### 2. In the case of which battery, there was a continuous decrease in sales from 1992 to 1997 ?

1. 35 AH
2. 4 AH
3. 32 AH
4. 7 AH

Option D

Explanation:

After analysing table chart, it is clear that the sales of 7AH batteries have been decreasing continuously from 1992 to 1997.

• #### 3. What is the difference in the number of 35AH batteries sold in 1993 and 1997 ?

1. 39000
2. 40000
3. 43000
4. 49000

Option A

Explanation:

Required difference = [(84 – 45) * 1000] = 39000.

• #### 4. The total sales of all the seven years is the maximum for which battery ?

1. 35 AH
2. 4 AH
3. 7 AH
4. 32 AH

Option D

Explanation:

The total sales (in thousands) of all the seven years for various batteries
are : a t.
For 4AH = 75 + 90 + 96 + 105 + 90 + 105 + 115 = 676
For 7AH = 144 + 126 + 114 + 90 + 75 + 60 + 85 = 694
For 32AH = 114 + 102 + 75 + 150 + 135 + 165 + 160 = 901
For 35 AH = 102 + 84 + 105 + 90 + 75 + 45 + 100 = 601
For 55 AH = 108 + 126 + 135 + 75 + 90 + 120 + 145 = 799.
Clearly, sales are maximum in case of 32AH batteries.

# Practice Problems on Table DI

1. Directions(6-10): Study the following table carefully and answer the question given below it.
2. If farmer  D and farmer E, both sell 240 kgs. of Bajra each what would be the respective ratio of their earnings?
A) 9 : 13
B) 14 : 15
C) 17 : 19
D) 11 : 13
E) 10 : 19
Option B
Solution:
Required ratio = 240 * 28 : 240 * 30 = 14 : 15
3. What is the average price per kg of Bajra sold by all the farmers together?
A) 22
B) 25.10
C) 23.3
D) 33.5
E) 41.15
Option
Solution:
Average price of Bajra = (22 + 24.5 + 21 + 28 + 30)/5 = Rs. 25.10 per kg
4. If farmer C sells 180 kgs. each of Corn , Paddy and Jowar grains how much would be earn?
A) Rs. 15200
B) Rs.12690
C)Rs.11050
D) Rs.19500
E) Rs.14500
Option B
Solution:
Farmer C’s earnings = (180*24  + 180*26 + 180*20.5) = Rs. 12690
5. If farmer A sells 350 kg of Rice , 150kg of Corn and 250 kg of Jowar , how much would he earn?
A) Rs.24510
B) Rs.11452
C) Rs. 15420
D) Rs. 18375
E) Rs.11450
Option D
Solution:
Farmer’s A earnings = (350 * 30  + 150 * 22.5 + 250 * 18) = Rs.18375
6. Earnings on 150 kg of Paddy sold by farmer B are approximately what per cent of the earnings on the same amount of Rice sold by the same farmer?
A) 48%
B) 52%
C) 69%
D) 70%
E) 65%
Option C
Solution:
Required % = (25/36)*100 = 69%(approx.)

Directions(1-5): Study the following table to answer the given questions.

1. What is the difference between total number posts and clerks?
A) 42563
B) 45278
C) 32690
D) 25478
E) 52480
Option C
Solution:
Number of posts = 72760

Number of clerks = 105450
Difference = 105450 – 72760 = 32690
2. In Kolkata , number of Specialist Officers is approximately what per cent of that of Officers?
A) 12%
B) 10%
C) 6%
D) 8%
E) 11%
Option D
Solution:
Required % = (1200/14900)*100 = 8%
3. In Chennai, the number of Clerks is approximately how much per cent more than that of Officers?
A)  10%
B) 22%
C) 18%
D) 21%
E) 30%
Option B
Solution:
Required % = (2000/9000)*100 = 22%
4. Which centre has the highest number of candidates?
A) Delhi
B) Bangalore
C) Mumbai
D) Kolkata
E) Chennai
Option C
Solution:
Number of candidates:

Bangalore -> 2000 + 5000 + 50 + 2050 + 750 = 9850
Delhi -> 15000 + 17000 + 160 + 11000 + 750 = 43910
Mumbai -> 170000 + 19500 + 70 + 7000 + 900 = 44470
Kolkata -> 14900 + 17650 + 70 + 1300 + 1200 = 35120
5. Which centre has 300% more number of Clerks as compared to Bangalore?
A) Delhi
B) Mumbai
C) Bangalore
D) Chennai
Option E
Solution:
Number of Clerks in Hyderabad  = 20000 which is 300% more than 5000 at Bangalore.

Directions(6-10): Study the following table and answer the given questions.

Note:
I). Data relate to the number of students studying in colleges A,B, C and D in the year 2009. The mentioned colleges offer courses in three streams only – Arts, Commerce and
Science.
II). Total students strength = students studying (Arts + Commerce + Science)

1. Number of students studying Arts in college C is what percent of number of students studying Arts in college D?
A) 18(2/3)
B) 21(2/3)
C) 17(1/2)
D) 20(3/2)
E) 11(1/3)
Option B
Solution:
In college C , Percentage of students who study science = 100 – 25 – 25 = 50%

Therefore , 50% == 1040
Then, Students who study Arts = (1040/50)*25 = 520
In college D , Percentage of students who study Science  = 100 – 50 – 30 = 20
Therefore, 20% == 960
Then, Students who study Arts  = (960/20)*50 = 2400
Required % = (520/2400)*100 = 65/3 = 21(2/3)
2. What is the total number of students studying Commerce in colleges C and D together?
A) 1078
B) 2000
C)1850
D) 1960
E) 2144
Option D
Solution:
Students who study Commerce :

College C = (1040/50)*25 = 520
College D = (960/20)*30 = 1440
Sum = 520 + 1440 = 1960
3. What is the average number of students studying Science in all the mentioned colleges?
A) 780
B) 875
C) 750
D) 800
E) 700
Option B
Solution:
Average number of students who study Science = (750 + 750 + 1040 + 960)/4
= 3500/4 = 875
4. What is the difference between the total student strength of college A and B together and that of colleges C and D together ?
A) 1850
B) 1999
C) 1480
D) 2250
E) 2380
Option E
Solution:
Total number of students:

College A ,
Therefore , 25% == 750
Then, 100% == (750/25)*100 = 3000
College B,
Therefore, 50% == 750
Then, 100% == (750/50)*100 = 1500
College C,
50%== 1040
Then , 100% == 1040*2 = 2080
College D,
20% == 960
Then, 100% == (960/20)*100 = 4800
Required difference  = (4800+2080) – (3000+1500)
= 6880 – 4500 = 2380
5. Total number of students studying Arts and Commerce together in college A is what percent more than that of college B ?
A) 156
B) 110
C) 200
D) 150
E) 198
Option C
Solution:
In college A,

Percentage of students who study Science = 100 – 20 – 55 = 25%
Therefore, 25% == 750
Students who study Arts = (750/25)*20 = 600
Students who study Commerce  = (750/25)*55 = 1650
Sum = 600+1650 = 2250
In college B,
Percentage of students who study in Science = 100 – 30 – 20 = 50%
Therefore , 50% == 750
Students who study Arts = (750/50)*30 = 450
Students who study Commerce  = (750/50)*20 = 300
Sum = 450+300 = 750
Required % = [(2250-750)/750]*100 = 200

Directions(1-5): Study the following table carefully to answer the questions given below it.

1. What is the difference between total number of male officers in Advertising and Public Relations Departments and the total number of female managers in these two departments?
A) 110
B) 150
C) 200
D) 180
E) 205
Option C
Solution:

Male Officers = (600*7)/12 = 350
Female Managers = (900*5)/9 = 500
Public Relations Department:
Male Officers = (1500*7)/15 = 700
Female Managers  = (800*7)/16 = 350
Required Difference = (350+700) – (500+350) = 1050 – 850 = 200
2. What is the respective ratio between total number of female managers from operations and finance departments and the total number of male officers from these two departments?
A) 2:3
B) 3:4
C) 2:5
D) 2:7
E) 1:2
Option B
Solution:

Female Managers :
Operations Department = (11*1200)/24 = 550
Finance Department = (13*1500)/30 = 650
Male Officers :
Operational Department = (1800*7)/18 = 700
Finance Department = (2200*9)/22 = 900
Required Ratio = (650+550) : (700+900) = 3 : 4
3. Total number of female employees (Managers and Officers) in procurement department is by what percent more than their male counter part?
A) 8%
B) 10%
C)20%
D) 15%
E) 11%
Option E
Solution:
Females in procurement department :

Managers = [17/(17+18)]*700 = 340
Officers = (11/20)*1200 = 660
Total = 340 + 660 = 1000
Males in Procurement department :
Managers = 700 – 340 = 360
Officers = 1200 – 660 = 540
Total = 360 + 540 = 900
Required Percent = [(1000 – 900)/900]*100 = 11%
4. What is the ratio between total number of managers in public relation, finance and sales and operation departments and the total number of officers in finance, advertising sales and procurement departments respectively?
A) 7:8
B) 10:13
C) 9:11
D) 8:7
E) 10:17
Option A
Solution:
Total number of managers in public relation, finance and sales and operation departments = 800 + 1500 + 1400 + 1200 = 4900

Officers in finance,  advertising sales and procurement departments
=2200 + 600 + 1600 + 1200 = 5600
Required Ratio = 4900 : 5600 = 7:8
5. Total number of female managers in finance department is what percent of the total number of male employees in sales department?
A) 42.5
B) 39.4
C) 33.2
D) 40.15
E) 24.8
Option B
Solution:
Female managers in Finance department = 650

Total male employees  in sales department = (1400*4)/7 + 850 = 1650
Required Percent  = (650/1650)*100 = 39.4

Directions(6-10): Data regarding number of books sold in either hard bound or paperback editions and also the categories of books sold in fiction and non- fiction categories by four different shops, in a particular mouth (Feb. 2015).

6. What is the average number of fiction books sold by shop A and B together?
A) 900
B) 990
C)1000
D) 1500
E) 1100
Option B
Solution:
Fiction books sold:

Shop A = 60% of [(5/3)*1200]= 1200
Shop B = (1200*65)/100= 780
Required % = (1200+780)/2 = 990

7. What is the respective ratio between the number of non-fiction books sold by shop C and number of non- fiction books sold by shop D?
A) 14(1/7)%
B) 11(1/9)%
C) 9(1/11)%
D) 8(1/17)%
E) 15(1/8)%
Option B
Solution:
Total books sold :

Shop B = 1200
Shop D = (8/5)*675 = 1080
Required % = [(1200-1080)/1080]*100
= 100/9 = 11(1/9)%
8. In March 2015, the number of paperback editions sold by shop D was 20% more than the same sold by the same shop in the previous month. The number of paperback editions sold in March 2015 by shop D constituted 50% of the total number of books sold by shop D in March 2015. What was the total number of books sold in March 2015 by shop D ?
A) 1620
B) 1600
C) 1550
D) 1490
E) 1560
Option A
Solution:
Paperback books sold by shop D in March 2015

= (675*120)/100 = 810
Total number of books sold in March 2015 = 810*2 = 1620

9. The number of hard bound editions sold by shop C  is what percent less than that sold by shop A ?
A) 77%
B) 65%
C) 75%
D) 80%
E) 60%
Option C
Solution:
Hard bound editions sold by shop C = (1/3)*600 = 200

Hard bound editions sold by shop A = (2/3)*1200 = 800
Required % = [(800-200)/800]*100 = 75%
10. The number of non-fiction sold by shop B is what % of the number of non-fiction books sold by shop A?
A) 33.15%
B) 39.99%
C)48.5%
D) 50%
E) 52.5%
Option E
Solution:
Total books sold by shop B = (6/5)*1000 = 1200

Non-fiction  books sold  = 1200 *(35/100) = 420
Total books sold by shop A = (5/3)*1200 = 2000
Non-fiction books sold = 2000*(40/100) = 800
Required % = (420/800)*100 = 52.5%

Study the following table carefully and answer the questions that follow:
Monthly Expenditure (In thousand ) by 5 persons on Rent, Food, Children’s Education, Clothes and Travelling.

1. What is the respective ratio between the expenditure of person-A on food and the expenditure of person-E on clothes?
A) 15:18
B) 10:19
C) 11:14
D) 16:25
E) 10:15
Option D
Solution:
Required ratio = 4.8 : 7.5 = 16:25
2. Total expenditure on rent by all the persons together is what per cent of expenditure of D on children’s education?
A) 550
B) 320
C) 416
D) 500
E) 450
Option C
Solution:
Total expenditure on rent = (12.4 + 6 + 5.6 + 13.6 + 14.4)thousands

= Rs.52 thousands
Required thousands = (52/12.5)*100 = 416
3. What is difference between the expenditure of person-B on Travelling  and the expenditure of person-A on food?
A) Rs.3000
B) Rs.1500
C) Rs.1000
D) Rs.1100
E) Rs.2000
Option D
Solution:
Required difference = 5800 – 4800 = Rs.1000
4. What is the average expenditure of person-C on all the five commodities together?
A) Rs.7580
B) Rs.6500
C) Rs.8050
D) Rs.9000
E) Rs.7660
Option E
Solution:
Required average expenditure by person-C = [(5.6 + 6.4 + 14.6 + 6.4 + 5.3)*1000]/5

= Rs.7660
5. Expenditure of which person on all the five commodities together is second highest?
A) E
B) D
C) B
D) A
E) C
Option A
Solution:
Total expenditure of five commodities:

A => (12.4+4.8+7.5+5.4+4.5)*1000 = Rs. 34600
B => (6+7.8+12.4+12.6+5.8)*1000 = Rs.44600
C => Rs.38300
D => (13.6 + 7.8+12.5+16.4+9.5)*1000 =Rs.59800
E => (14.4+8.4+13.2+7.5+7.4)*1000 = Rs.50900

Study the table carefully to answer the questions that follow:
Distance (in kms) travelled by six trucks on six different days of the week

1. What is the average distance travelled by truck S in all the days together?
A) 299(1/6)
B) 250(1/3)
C) 310(1/5)
D) 111(1/3)
E) 90(1/5)
Option A
Solution:
Required average distance covered by truck S = (325+314+312+278+292+274)/6

= 1795/6 = 299(1/6)km
2. If to travel the given distance , the time taken by truck Q on Friday was 8 hours , what was its speed on that day?
A) 40.5kmph
B) 37.75kmph
C) 30.45kmph
D) 42kmph
E) 50kmph
Option B
Solution:
Speed of truck Q on Friday =302/8 = 37.75kmph
3. If the speed of truck P on Monday was 19.2 kmph, what was the time taken by it to cover the given distance?
A) 13hrs.
B) 12.5 hrs.
C) 15.6hrs.
D) 9.5hrs.
E) 10.14hrs.
Option B
Solution:
Required time = 240/19.2 = 12.5 hours
4. If on Tuesday truck R and truck T travelled  at the same speed, what was the respective ratio of time taken by truck R and time taken by truck T to cover their respective distances?
A) 147:164
B) 144:171
C) 150:161
D) 154:159
E) 110:113
Option D
Solution:
Required ratio = 308:318 = 154:159
5. What is the total distance travelled by all the trucks together on Saturday ?
A) 1245km
B) 1450km
C) 1632km
D) 1200km
E) 1550km
Option C
Solution:
Total distance travelled by the trucks on Saturday

= (292+284+260+274+280+242)km = 1632km

Directions(1-5): Study the following table carefully to answer the questions. The table given the percentage of marks obtained by six students in six different subjects. Here, P, Q, R, S, T, U are the six different subjects.

1. What is the average percentage of marks obtained by all the students in subject ‘P’?
A) 72(1/3)
B) 72(1/2)
C)78(1/5)
D)77(1/4)
E) 77(1/5)
Option A
Solution:
Required average % = (80+55+74+68+75+82)/6 = 72(1/3)
2. What is the overall percentage of marks obtained by B in all the subjects?
A) 70%
B) 78%
C) 72%
D) 88%
E) 82%
Option C
Solution:
Required % = (55 + 70 + 67 + 74 + 88 + 78)/6 = 72%
3. What are the total marks obtained by all the students together in subject T ?
A)  527
B) 625
C)600
D)534
E) 550
Option D
Solution:
Total marks obtained = 90 + 105.6 + 74.4 + 86.4 + 96 + 81.6  = 534
4. What are the average marks obtained by all the students in subject ‘R’ (approx.)?
A)  60
B) 56
C) 50
D) 64
E) 78
Option B
Solution:
Required average marks = (46.4 + 53.6 + 57.6 + 65.6 + 51.2 + 60)/6 = 55.7 = 56(approx.)
5. What are the total marks obtained by student C in subjects Q, S and T ?
A)  130
B) 140
C) 138
D) 155
E) 145
Option C
Solution:
Marks obtained by C in subject Q = 54*(40/100) = 21.6

subject S = 84*(50/100) = 42
subject T = 62*(120/100) = 74.4
Therefore ,required total marks = 21.6 + 42+ 74.4 = 138

Directions(6-10): Study the table carefully and answer the questions that follow:

Description of Literate and Illiterate population of six villages.

1. The number of literate women in village B is 39760 and that of illiterate women in village C is 25600. By what percent is the population of village B is less that that of village C?
A)  13.07
B) 13.25
C)10.10
D) 11.25
E) 21.5
Option D
Solution:
Literate men in village B = (8/7)*39760 = 45440

Literate males and females = 45440 + 39760 = 85200
If 60% ==== 85200
then 100% ==== (85200/60)*100 = 142000
Illiterate males in village C = (3/4)*25600 = 19200
Illiterate population = 25600 + 19200 = 44800
If 28% ====44800
then 100%====(44800/28)*100 = 160000
Therefore ,required % = (160000 – 142000)*100/160000 = 11.25
2. If the number of literate women in village D is 32200, what is the number of illiterate population in the same village ?
A)  48575
B) 57042
C)45000
D) 54050
E) 48300
Option E
Solution:
Literate males in village D = (5/4)*32200 = 40250

Total literates = 32200 + 40250 = 72450
60%====72450
40%====(72450*40)/60 = 48300
3. If the population of village F is 168000, what is the number of literate males ?
A)76800
B) 67200
C)45072
D) 57480
E) 65770
Option B
Solution:
Literate population in village F = (168000*64)/100 = 107520

Number of literate males = (5/8)*107520 = 67200
4. If the illiterate female population of village E be 77000, what is the total population of that village ?
A)  257850
B) 331250
C) 350000
D) 425600
E) 324560
Option C
Solution:
Illiterate males in village E = (14/11)*77000 = 98000

Total illiterate = 77000 + 98000 = 175000
Therefore ,total population = 2 *175000 = 350000
5. If the number of literate males in village A be 35840, the number of illiterate males in the same village ?
A)  45022
B) 43264
C) 41500
D) 40000
E) 42781
Option B
Solution:
Literate females in village A =(5/7)*35840 = 25600

Total literate population = 25600 + 35840 = 61440
48% ====61440
52%====(61440/48)*52 = 66560
Illiterate males = (13/20)*66560 = 43264

Directions(1-5): The table shows the number of people staying in five different localities and the percentage-wise breakup of Men,Women and Children. Study the table carefully and answer the related questions.

1. If 20% of the men staying in locality D are working in Infosys and the 40% of the women staying in locality B are working in HCL, then what is the difference between the number of men staying in locality D who are working in Infosys and the number of women staying in locality B who are working in HCL?
A) 900
B) 1300
C)1000
D)1400
E) 800
Option D
Solution:
Number of men staying in locality D who are working in Infosys = 12000*(35/100)*(20/100) = 840

Number of women staying in locality B who are working in HCL = 16000*(35/100)*(40/100) = 2240
Required difference = 2240 – 840 = 1400
2. Number of children staying in localities A and B together is approximately what percent of the number of children staying in localities D & E together?
A)  48%
B) 68.42%
C)50%
D) 42.54%
E) 60.15%
Option B
Solution:
Number of children staying in A and B together = 18000*(20/100) + 16000*(10/100) = 5200

Number of children staying in D & E together = 12000*(25/100) + 2000*(23/100) = 7600
Required % = (5200/7600)*100= 68.42%
3. If  40% of the women staying in locality C are graduate and 60% of the men staying in locality E are post graduate then what is the ratio between the number of graduate women staying in locality C and the number of post graduate men staying  in locality E?
A) 1:2
B) 3:2
C)1:4
D) 5:2
E) 4:5
Option A
Solution:
Number of graduate women  staying in C = 15000*(45/100)*(40/100)= 2700

Number of post graduate men staying in E = 20000*(45/100)*(60/100)= 5400
Required ratio = 1:2
4. What is the difference between the total number of men staying in all the localities together and that of women staying in all the localities together ?
A) 6500
B) 6300
C) 6000
D) 6500
E) 6250
Option E
Solution:
Number of men staying in A, B, C, D and E =

18000*(45/100) + 16000*(55/100) + 15000*(40/100) + 12000*(35/100) + 20000*(45/100) = 8100+8800+6000+4200+9000= 36100
Number of women staying in A, B, C, D and E =18000*(35/100) + 16000*(35/100)  + 15000*(45/100) + 12000*(40/100) + 20000*(32/100) = 6300+5600+6750+4800+6400 = 29850
Required Difference = 36100 – 29850 = 6250
5. If 40% of the men staying in locality A are self employed, then the self employed men staying in locality A is approximately what percent of the number of women staying in locality B ?
A) 60%
B)58%
C)45%
D) 42%
E) 55%
Option B
Solution:
Number of self-employed men staying in locality A = 18000*(45/100)*(40/100) = 3240

Number of women staying in locality B = 16000*(35/100) = 5600
Required % = (3240/5600)*100 = 57.86%= 58%(approx.)

Directions(6-10): The table shows the distribution of total number of phones manufactured and percentage of defective phones in five different companies in six different years . Study the table carefully and answer the related questions.

1. Number of non-defected phones manufactured by company D in 2017 is approximately what % of the number of non-defected phones manufactured by company C in 2014?
A) 60%
B) 58%
C) 74%
D) 55%
E) 42%
Option B
Solution:
Number of non-defected phones manufactured by company D in 2017 = 17800*(98.5/100) = 17533

Number of non-defected phones manufactured by company C in 2014
= 31000*(98/100) = 30380
Required % = (17533/30380)*100 = 57.71% = 58%(approx.)
2. What is the ratio between the number of non-defected phones manufactured by company B in 2013 and that of company D in 2016?
A) 8347:15826
B) 8475:10457
C) 4757:7458
D) 5784:8453
E) 6456:11450
Option A
Solution:

Number of non-defected phones manufactured by company B in 2013 = 17000*(98.2/100) = 16694
Number of non-defected phones manufactured by company D in 2016 = 32800*(96.5/100) = 31652
Required ratio = 8347:15826
3. What is the total number of non-defective phones manufactured by company B in all the years together?
A) 147855
B) 247850
C) 221748
D) 117500
E) 117147
Option E
Solution:
Number of non-defective phones manufactured by B in all the  years = 15000*(98.4/100)+ 17000*(98.2/100)+ 14000*(99/100)+ 22000(98.5/100)+ 26000*(98.8/100)+ 25000*(97.9/100)

= 14760+ 16694+ 13860+ 21670+ 25688+ 24475 = 117147
4. What is the difference between the total number of phones manufactured by company C and that of company E in all the years together?
A) 17650
B) 15000
C) 16600
D) 15850
E) 18570
Option C
Solution:
Required difference = 186000 – 169400 = 16600
5. Total number of defective phones manufactured by company A is approximately what % of the total number of defective phones manufactured by company D in all the  years together?
A) 75%
B) 80%
C) 50%
D) 100%
E) 60%
Option B
Solution:
Total number of defective phones by A in all the years = 12000*(2/100)+ 64000*(1.4/100)+ 72000*(2.1/100)+ 8000*(2.4/100)+ 16000*(1.8/100)+ 7800*(1/100)

= 240+896+1512+192+288+78 = 3206
Total number of defective phones by D in all the years = 41000*(2.3/100)+ 28500*(3/100)+ 27500*(2/100)+ 24500*(1/100)+ 32800*(3.5/100)+ 17800*(1.5/100)
=943+855+550+245+1148+267 = 4008
Required % = (3206/4008)*100 =79.99%= 80%