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.
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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.