# Free DI and LR Practice Test - CAT

### Question 1

What is the ratio of number of matches played by India in Sydney to number of matches won by India in

#### SOLUTION

Solution :B

Let number of matches played by India in Australia be 10,000. In this caselet, you can solve all questions based on percentages and ratios, where the exact values are not required

VenueNo.of matchesindia′s win The WACA 1000100Melboume2000400Adelaide3000900Sydney2500625Bellerive1500225Total10,0002250From the above table, the ratio of number of matches played by India in Sydney to number of matches won by India in Melbourne is 25:4;

### Question 2

What is the difference between number of matched won by India in The WACA and in Bellerive?

5

125

25

#### SOLUTION

Solution :B

We cannot determine exact number of matches played by India in Australia. option (d).

### Question 3

What is the percentage of India's winning in Sydney against India's winning in Adelaide?

#### SOLUTION

Solution :A

From the above table, we can conclude (625/900)*100=69.4%of India's winning in Sydney against India's winning in Adelaide. Option (a).

### Question 4

**In 2004, sixty percent of population in Delhi and Mumbai were male. Given that the population of females increases at the rate of 15 percent per annum and that of males at the rate of 20 percent per annum, what will be the approximate percentage growth of population between 2004 and 2011 in Delhi and Mumbai? **

150%

155%

221%

171%

#### SOLUTION

Solution :C

Let total population of Delhi and Mumbai in 2004 = 100

Number of men in Delhi and Mumbai = 60

Therefore number of males in 2011 = 60 × (1.2)7 = 60 × 3.583 ≈215

Therefore number of females in 2011= 40 × (1.15)7 = 40 × 2.66 ≈106

Therefore total population of Delhi and Mumbai in 2011 = 321% growth of population = (321−100)100 × 100 ≈ 221%.

### Question 5

**While the population in Kolkata and Bangalore has been growing steadily towards that of the Delhi and Mumbai, the growth rate in Kolkata and Bangalore seems to be declining. Which of the following is the closest to the percent change in growth rate of 2006 (over 2005) relative to the growth rate of 2008 (over 2007) for Kolkata and Bangalore together?**

8.57 %

20%

22%

60%

#### SOLUTION

Solution :A

Growth rate of 2006 =10900−1050010500 × 100 = 3.8%

Growth rate of 2008 =11800−1140011400 × 100 = 33%

Therefore % change in growth rate of 2006 relative to growth rate of 2008 is = 3.8−3.53.5 × 100 = 8.57%.

### Question 6

**Which of the following cities have seen the maximum population growth rate in the period (2005 - 2010)?**

Delhi

Kolkata

Mumbai

Bangalore

#### SOLUTION

Solution :A

From the given graph, we can easily conclude "Maximum population growth is for Delhi.”

### Question 7

**Which is the year in which the absolute value of population change in Bangalore and Mumbai together is the highest over the previous year?**

2004

2008

2006

2007

#### SOLUTION

Solution :D

From the given graph, we can easily conclude the absolute value of population change in Bangalore and Mumbai is the highest for 2007 over 2006.

### Question 8

**If the population in the country in the year 1980 was 800 mn, how many million votes did the party secure in this election?**

162

360

266.4

None of these

#### SOLUTION

Solution :A

In 1989,out of 800 million people, 60 percent were eligible to vote, out of them, around 75% turned up for voting, out of those 75%,only 45% voted for the party. Hence the number of votes secured by the party=(35)×(34)×(45100)×800=162mn.

### Question 9

**If the population growth from 1998 to 2000 was 10 %, what is the ratio of total population in 1998 to the number of people who voted for the party in 2000(approx)?**

125 : 11

2500 : 231

1250 : 231

Cannot be determined

#### SOLUTION

Solution :C

Let the population in 1998 be 100.

Then population in 2000 is 110.

So number of people who voted for the party in 2000 = 110 × (40100) × (60100) × (70100)

So the required ratio is 100110×25×35×70100 = 1250231

### Question 10

**What is the ratio of the number of people who did not vote to the number of people who voted for the party in 2000?**

5 : 3

145 : 42

3 : 5

Cannot be determined

#### SOLUTION

Solution :B

Let the population be 100.

The number of people who did not vote = 30 + ((25)×70) = 58

Number of people who voted for the party = 100 × (70100) × (35) × (25)

So the required ratio = 145 : 42

### Question 11

Number of shirts bought by B

53

23

47

none of these

#### SOLUTION

Solution :B

From the given condition we can conclude :-

NameNo.of shirtsNo.of trousersSumAR90BRPc84C78DQ84cEPQ68

The possible values of P and Q are 37 and 31 or 61 and 7 respectively.

1st condition:-

If B had 84 garments. Then:-

NameNo.of shirtsNo.of trousersSumA434790B473784C78D31CE373168Now, we can use only 5 prime numbers. So, we put 43 (in the column of No. of trousers). We will not get a fifth prime number. So, this case is not possible If D had 84 garments. Then: -

NameNo.of shirtsNo.of TrousersSumAR90BR37CC78D315384E373168

Now, we can use only 5 prime numbers. So, we put 53 (in the column of No.of shirts). We will not get fifth prime number. So, this is not possible

2nd Condition:-

NameNo.of shirtsNo.of trousersSumAR90BR61CC78D77784E61768

Here, we are getting 77, which is not prime number. So, this case is not possible

The possible values are shown below

NameNo.of shirtsNo.of trousersSumA672390B236184C116778D71118E61768

### Question 12

Number of trousers bought by C

67

43

53

none of these

#### SOLUTION

Solution :A

From the given condition we can conclude :-

NameNo.of shirtsNo.of trousersSumAR90BRPc84C78DQ84cEPQ68

The possible values of P and Q are 37 and 31 or 61 and 7 respectively.

1st condition:-

If B had 84 garments. Then:-

NameNo.of shirtsNo.of trousersSumA434790B473784C78D31CE373168Now, we can use only 5 prime numbers. So, we put 43 (in the column of No. of trousers). We will not get a fifth prime number. So, this case is not possible If D had 84 garments. Then: -

NameNo.of shirtsNo.of TrousersSumAR90BR37CC78D315384E373168

Now, we can use only 5 prime numbers. So, we put 53 (in the column of No.of shirts). We will not get fifth prime number. So, this is not possible

2nd Condition:-

NameNo.of shirtsNo.of trousersSumAR90BR61CC78D77784E61768

Here, we are getting 77, which is not prime number. So, this case is not possible

The possible values are shown below

NameNo.of shirtsNo.of trousersSumA672390B236184C116778D71118E61768

### Question 13

Who had bought minimum numbers of garments?

B

C

D

E

#### SOLUTION

Solution :C

From the given condition we can conclude :-

NameNo.of shirtsNo.of trousersSumAR90BRPc84C78DQ84cEPQ68

The possible values of P and Q are 37 and 31 or 61 and 7 respectively.

1st condition:-

If B had 84 garments. Then:-

NameNo.of shirtsNo.of trousersSumA434790B473784C78D31CE373168Now, we can use only 5 prime numbers. So, we put 43 (in the column of No. of trousers). We will not get a fifth prime number. So, this case is not possible If D had 84 garments. Then: -

NameNo.of shirtsNo.of TrousersSumAR90BR37CC78D315384E373168

Now, we can use only 5 prime numbers. So, we put 53 (in the column of No.of shirts). We will not get fifth prime number. So, this is not possible

2nd Condition:-

NameNo.of shirtsNo.of trousersSumAR90BR61CC78D77784E61768

Here, we are getting 77, which is not prime number. So, this case is not possible

The possible values are shown below

NameNo.of shirtsNo.of trousersSumA672390B236184C116778D71118E61768

### Question 14

Which one of the following is an acceptable listing of films to show this week?

Shout, Mist, Trek, QuePasa, Fly, andJealousy

Shout, Mist, Trek, Fly, Jealousy, and AbraCadabra

QuePasa, Lessons, Mist, Shout, AbraCadabra, and Trek

Shout, Lessons, Mist, Trek, Fly, and Jealousy

#### SOLUTION

Solution :B

In option A, there are two foreign films and this violating the third statement.

Option C and D, violating first statement.

Option B is the correct answer.

### Question 15

If Shout starts at 8:30, Mist at 8:15, Trek at 8:00, Fly at 7:45, Jealousy at 7:30, and AbraCadabra at 7:15, and each movie is exactly two hours long, at what time will the next showing of Trek start?

10:20

10:15

10:30

10:45

#### SOLUTION

Solution :C

The first showing of Trek will be over at 10:00. Then, the employees will need 20 minutes to clean the theater, which is 10:20. Since the movies always start on the quarter hour, the second showing of Trek will be 10:30. Option (c)

### Question 16

The movies this week are showing in the following theaters:

Theatre 1ShoutTheatre 2AbraCadabraTheatre 4JealousyTheatre 5FlyTheatre 6Mist

Shout is doing the most business, followed by Trek and, to the management's surprise, Mist. The management wants to move Mist to a larger theater. Which theater is the most logical?

theater 1

theater 2

theater 3

theater 5

#### SOLUTION

Solution :D

Since Shout is doing the most business and Trek the second most, they should remain in the two largest theaters. Also, the theater never shows a foreign film in the largest theater. Theaters 3 and 4 must show the movies that are rated G and PG, so the movies that are there must stay there. The most logical choice is to put Mist in theater 5 and Fly in theater 6. Option (d)

### Question 17

In a certain perfume mixture consisting of only two flavors, for every 3 milliliters of Flavour A, there are 7 milliliters of Flavour B. After 10 milliliters of Flavour C are added to this perfume mixture, what is the ratio of the quantities of Flavour A to Flavour C?

1. Before flavour C was added, there were 50 milliliters of perfume mixture.

2. After flavour C was added, there were 60 milliliters of perfume mixture.

#### SOLUTION

Solution :B

The question stem already tells us that there are 10 milliliters of Flavour C in the final perfume mixture. We also know that the original perfume mixture consists of only Flavours A and B in the ratio of 3 to 7. Thus, we need the original volume of the perfume mixture to determine the amount of FlavourA contained in it.

SUFFICIENT: This tells us that original perfume mixture was 50 milliliters. Thus, there must have been 15 milliliters of Flavour A (to 35 milliliters of Flavour B). Theratio of A to C is 15 to 10 (or 3 to 2).

SUFFICIENT: This tells us that the final perfume mixture was 60 milliliters. We know that this includes 10 milliliters of Flavour C. This means the original perfume mixture contained 50 milliliters. Thus, there must have been 15 milliliters of Flavour A (to 35 milliliter of Flavour B). The ratio of A to C is 15 to 10 (or 3 to 2).The correct answer is B.

### Question 18

What is the perimeter of isosceles triangle ABC?

1. The length of side AB is 9

2. The length of side BC is 4

#### SOLUTION

Solution :C

The perimeter of a triangle is equal to the sum of the three sides.

(1) INSUFFICIENT: Knowing the length of one side of the triangle is not enough to find the sum of all three sides.

(2) INSUFFICIENT: Knowing the length of one side of the triangle is not enough to find the sum of all three sides.

Together, the two statements are SUFFICIENT. Triangle ABC is an isosceles triangle which means that there are theoretically 2 possible scenarios for the lengths of the three sides of the triangle:(1)AB = 9, BC = 4 and the thirdside, AC = 9 or (1)AB = 9,BC = 4 and the third side AC = 4.

These two scenarios lead to two different perimeters for triangle ABC, HOWEVER, upon careful observation we see that the second scenario is an IMPOSSIBILITY. A triangle with three sides of 4, 4, and 9 is not a triangle. Recall that any two sides of a triangle must sum up to be greater than the third side. 4 + 4 < 9 so these are not valid lengths for the side of a triangle.

Therefore the actual sides of the triangle must be AB = 9, BC = 4, and AC = 9. The perimeter is 22.The correct answer is C.

### Question 19

**Which of the following is true?**

C shot 8 balloons and 4 coins but no needle

The person who shot 5 balloons and one coin did not shoot any needle.

The person who shot an equal number of balloons and coins also shot needles.

The person who shot 4 balloons and 2 coins also shot needles.

#### SOLUTION

Solution :B

From condition (i), we can conclude that A shot 6 coins( as only 3 *2 is 6).Proceeding in similar lines we can arrive at the following distribution which satisfies all the conditions.

FriendsBalloonsCoinsNeedlesA66NA(but not 0)B100C42NA(but not 0)D510E840

### Question 20

**Who shot an equal number of coins and balloons?**

A

B

C

D

#### SOLUTION

Solution :A

From condition (i), we can conclude that Ashot 6 coins( as only 3 *2 is 6).

Proceeding in similar lines we can arrive at the following distribution which satisfies all the conditions.

FriendsBalloonsCoinsNeedlesA66NA(but not 0)B100C42NA(but not 0)D510E840

### Question 21

**Which of the following is true?**

D shot 5 balloons

A shot 8 balloons

E shot 1 balloon

E shot 6 balloons

#### SOLUTION

Solution :A

From condition (i), we can conclude that A shot 6 coins( as only 3 *2 is 6).Proceeding in similar lines we can arrive at the following distribution

which satisfies all the conditions.

### Question 22

The number of students studying in the school during 1999 was?

#### SOLUTION

Solution :3000 + (350-250)+ (300-450) + (450-400) +(500-350)= 3150

### Question 23

**The school had its highest strength in the year :**

2000

2001

1996

1999

#### SOLUTION

Solution :B

For 1996, number of students = 3000+350-250 = 3100

Proceeding likewise, we can find out the number of students of various years as

For 1997 = 2950

For 1998 = 3000

For 1999 = 3150

For 2000 = 3100

For 2001 = 3200

### Question 24

**What is the average savings of the two persons whose income data is not given in the graph(in 1000s)?**

35

55

Insufficient data

#### SOLUTION

Solution :B

From the graphs given we can arrive at the following:

PersonIncomeExpensesSavings as % of expenseSavingsA1501202530B1609077.7770C2001404360D2402002040E2001801120F16014014.2820

So we do not have any information regarding incomes of B and F. So their average savings is 70+202= 45.

### Question 25

**If E's income saw an increase of 10% in 2012 and if his expenses remained same, what is the percentage change in his savings as a percentage of his incomes for the two years?**

18.18%

27.27%

9.1%

11.11%

#### SOLUTION

Solution :C

E's income in 2011 = 200

Savings as a percentage of income in 2011= 10%

E's income in 2012 = 220

Savings as a percentage of income in 2012 = (40220) × 100=18.18%So percentage change = 81.8%

### Question 26

If expense on rent as a percentage of income of F is 12.5%, then the other expenses as a percentage of his savings is

#### SOLUTION

Solution :Income of C = 160

So expense on rent= 20

Remaining expenses = 120

So the required answer = (12020) × 100=600%

### Question 27

**The above pie chart shows the contribution of various factors towards the price in 2008**

**If the manufacturing cost and the profit remained constant for 2008 and 2009, and if the logistics cost for 2008 and 2009 were Rs 200 and Rs 270, what was the percentage increase in raw material cost?**

40

50

60

cannot be determined

#### SOLUTION

Solution :C

Logistics cost for 2008 = Rs 200

So, total cost = Rs 1000

Hence, Raw material cost in 2008 = Rs 300

Total cost in 2009 =Rs 1250

So increase in raw material cost = Rs 180Therefore, percentage increase = 60%

### Question 28

How many female employees joined in 2004?

#### SOLUTION

Solution :Given that six employees left in 2002.

246 + 6 - 217 = 35 employees joined in 2002.

Also 6×5=30 employees joined in 2000.

Similarly, 306+355−293

i.e., 20 employees joined in 2004.

Given that four female employees left in 2000.

102 + 4 - 82 = 24 female employees joined in2000.

Also 4×4=16 female employees joined in 1998.

Similarly, 115 + 24 - 105 =16 female employees

Let the number of employees left in 2003 and 2005 be x and y respectively.

293 + x - 246 = 47 + x is the number of employees who joined in 2003.

Similarly, 152 + x - 115 = 37 + x is the number of female employees who joined in 2003.

(47+x)5=(37+x)4

[As number of employees and female employees left in 2005 are equal]

x = 3

50 employees and 40 female employees joined in 2003.

As 3 left in 2003, 3×5=15 employees and 3×4=12 female employees joined in 2001.

Also, 219 + 15 - 217 = 17 employees and 102 + 12 - 105 = 9 female employees left in 2001.

85 employees of which 36 were females joined in 1999.

The following table gives the number of employees who joined each year:

Year199920002001200220032004Employees853015355020Female Employees362412164016

16 female employees joined in 2004

### Question 29

In which of the years from 1999 to 2004, did the maximum number of employees join the company?

#### SOLUTION

Solution :Given that six employees left in 2002.

246 + 6 - 217 = 35 employees joined in 2002.

Also 6×5=30 employees joined in 2000.

Similarly, 306+355−293

i.e., 20 employees joined in 2004.

Given that four female employees left in 2000.

102 + 4 - 82 = 24 female employees joined in2000.

Also 4×4=16 female employees joined in 1998.

Similarly, 115 + 24 - 105 =16 female employees

Let the number of employees left in 2003 and 2005 be x and y respectively.

293 + x - 246 = 47 + x is the number of employees who joined in 2003.

Similarly, 152 + x - 115 = 37 + x is the number of female employees who joined in 2003.

(47+x)5=(37+x)4

[As number of employees and female employees left in 2005 are equal]

x = 3

50 employees and 40 female employees joined in 2003.

As 3 left in 2003, 3×5=15 employees and 3×4=12 female employees joined in 2001.

Also, 219 + 15 - 217 = 17 employees and 102 + 12 - 105 = 9 female employees left in 2001.

85 employees of which 36 were females joined in 1999.

The following table gives the number of employees who joined each year:

Year199920002001200220032004Employees853015355020Female Employees362412164016

16 female employees joined in 2004

The maximum number of employees joined the company in 1999.

### Question 30

**How many male employees left in 2004? ___**

#### SOLUTION

Solution :Given that six employees left in 2002.

246 + 6 - 217 = 35 employees joined in 2002.

Also 6×5=30 employees joined in 2000.

Similarly, 306+355−293

i.e., 20 employees joined in 2004.

Given that four female employees left in 2000.

102 + 4 - 82 = 24 female employees joined in2000.

Also 4×4=16 female employees joined in 1998.

Similarly, 115 + 24 - 105 =16 female employees

Let the number of employees left in 2003 and 2005 be x and y respectively.

293 + x - 246 = 47 + x is the number of employees who joined in 2003.

Similarly, 152 + x - 115 = 37 + x is the number of female employees who joined in 2003.

(47+x)5=(37+x)4

[As number of employees and female employees left in 2005 are equal]

x = 3

50 employees and 40 female employees joined in 2003.

As 3 left in 2003, 3×5=15 employees and 3×4=12 female employees joined in 2001.

Also, 219 + 15 - 217 = 17 employees and 102 + 12 - 105 = 9 female employees left in 2001.

85 employees of which 36 were females joined in 1999.

The following table gives the number of employees who joined each year:

Year199920002001200220032004Employees853015355020Female Employees362412164016

16 female employees joined in 2004

355=7 employees left in 2004,

4 female employees left in 20047 - 4 = 3 male employees left in 2004

### Question 31

How many female employees left the company from 2000 to 2006?

#### SOLUTION

Solution :Given that six employees left in 2002.

246 + 6 - 217 = 35 employees joined in 2002.

Also 6×5=30 employees joined in 2000.

Similarly, 306+355−293

i.e., 20 employees joined in 2004.

Given that four female employees left in 2000.

102 + 4 - 82 = 24 female employees joined in2000.

Also 4×4=16 female employees joined in 1998.

Similarly, 115 + 24 - 105 =16 female employees

Let the number of employees left in 2003 and 2005 be x and y respectively.

293 + x - 246 = 47 + x is the number of employees who joined in 2003.

Similarly, 152 + x - 115 = 37 + x is the number of female employees who joined in 2003.

(47+x)5=(37+x)4

[As number of employees and female employees left in 2005 are equal]

x = 3

50 employees and 40 female employees joined in 2003.

As 3 left in 2003, 3×5=15 employees and 3×4=12 female employees joined in 2001.

Also, 219 + 15 - 217 = 17 employees and 102 + 12 - 105 = 9 female employees left in 2001.

85 employees of which 36 were females joined in 1999.

The following table gives the number of employees who joined each year:

Year199920002001200220032004Employees853015355020Female Employees362412164016

16 female employees joined in 2004

Female employees who left from 2000 to 2006= 4 + 9 + 6 + 3 + 4 + 10 + 4 = 40

### Question 32

There are three boxes, one contains only apples, one contains only oranges, and one contains both apples and oranges. The boxes have been incorrectly labeled such that no label identifies the actual contents of the box it labels. Opening just one box, and without looking in the box, you take out one piece of fruit. By looking at the fruit, you can immediately label all of the boxes correctly. Which box did you open?

Containing apple

Containing oranges

Containing both apples and oranges

Cannot be determined

#### SOLUTION

Solution :C

The box that must be opened is the one labeled "apples and oranges." By definition, whichever fruit is inside; is the only fruit type that that box contains. Let's say that you found an apple in that box that was labeled with both apples and oranges; because you know it must therefore only contain apples, then you conclude that the box that is labeled "oranges" cannot contain only oranges, as all boxes have been said to be mislabeled. Thus, the box labeled "oranges" must contain both apples and oranges, leaving the box labeled "apples" to contain only oranges.