Free DI and LR Practice Test - CAT

Question 1

In the financial category, the number of articles published in September 2002 as compared to March 2003

was larger

was smaller.

was equal.

Cannot be determined

SOLUTION

Solution : D

We know that the total number of articles published in Sept 02 $>$ Mar 03. However, we cannot compute if 25% of Sept 02 $>$ 37% of Mar 03. The exact values of total number of articles published in the two months is required to solve this question.

Question 2

In the health category, the number of articles published in December 2002 as compared to June 2003

was larger.

was smaller

was equal.

Cannot be determined

SOLUTION

Solution : A

Let the total number of articles published in December 2002 and june 2003 be 'x' and 'y' respectively Then we know that x $>$ y = $>$ .19x $>$ .18y

Question 3

In which category was the percentage of articles published increasing but at a decreasing rate?

Financial

Scams

Products

None of these

SOLUTION

Solution : C

In products category percentage is increasing with a decreasing rate i.e

$\frac{7 - 3}{3}, \frac{10 - 7}{7}, \frac{11 - 10}{10}$ which is decreasing

Question 4

What is the percentage share of total tax collected in total revenue?

A. 10.88

B. 12.54

C. 8.25

D. 9

SOLUTION

Solution : A

$\begin{matrix} B r a n d 1 & B r a n d 2 & B r a n d 3 & B r a n d 4 & B r a n d 5 U n i t s s o l d & 4500 & 3500 & 5500 & 6500 & 5000 T a x C o l l e c t e d b y v a l u e & 9135000 & 6160000 & 6600000 & 7605000 & 7425000 T o t a l c o l l e c t e d a s a % o f i t s r e v e n u e & 14 & 11 & 10 & 9 & 11 T o t a l r e v e n u e = T a x b y v a l u e / % v a l u e \times 100 & 65250000 & 56000000 & 66000000 & 84500000 & 67500000 V a l u e o f e a c h u n i t = R e v e n u e / U n i t s s o l d & 14500 & 16000 & 12000 & 13000 & 13500 \end{matrix}$
Total tax collected = 36925000
Total revenue = 339250000
Hence the required answer is $\frac{36925000}{339250000} \times 100 = 10.88 %$

Question 5

Which brand has the highest unit price value?

A. Brand 1

B. Brand 2

C. Brand 4

D. Brand 5

SOLUTION

Solution : B

Brand 2 has the highest unit price value( from table).

Question 6

What should be the new tax percentage on Brand 2 so that the tax collected by value for Brand 1 and 2 are the same?

A. 12%

B. 16.3%

C. 20.25%

D. 24.56%

SOLUTION

Solution : B

New value of tax collected by brand 2 =9135000.

Hence new percent value of tax should be $\frac{9135000}{56000000} \times 100 = 16.3125 %$ .

Question 7

If no of units of Brand 3 increases by 20% and there is decrease of 16.66% in its price per unit then tax collected by brand 3 will be more or less as compared to brand 1 by how much percentage approximately?

A. 22%

B. 25%

C. 28%

D. 32%

SOLUTION

Solution : D

Tax collected = No. of units $\times$ Price per unit $\times$ % of tax collected on value.

Now as No of units increases by ${\frac{1}{5}}^{t h}$ and price per unit decreases by ${\frac{1}{6}}^{t h}$ , the tax collected remains the same.

Hence required answer is = $= \frac{9135000 - 6160000}{9135000} \times 100 = 32.56 %$

Question 8

If a new brand , say brand 6 enters the market with total revenue same as brand 1 and tax percentage (on value) as 15%. Volume of brand 6 should be same as which brand to have unit price 50 more than any one of the given 5 brands?

A. Brand 5

B. Brand 4

C. Brand 3

D. Brand 2

SOLUTION

Solution : A

Now we have revenue is 65250000.

Price per unit $= \frac{65250000}{Volume of brand 6}$

Go from answer options

Price per unit = option a)13050

b)10038.5

c)11863.63

d)18462.85

Hence option a) is a correct answer.

Question 9

What is the minimum number of workers required to finish the job in one day?___

SOLUTION

Solution :
For the minimum number of workers to complete the job in one day, the job must have been completed in the minimum possible number of man-hours in 6 days.
This is possible if the number of workers working on Day 1,2,3,4,5, and 6 are 3,2,1,2,3 and 4 respectively. Thus, the minimum number of man-hours required for the job $= (3 + 2 + 1 + 2 + 3 + 4) \times 8 = 120$ .
The minimum number of workers required to complete the job in one day $= (\frac{120}{8}) = 15$ .

Question 10

If Mr. Shyam pays the workers Rs. 3040 in all for this job, how many workers work on day 3?

Cannot be determined

SOLUTION

Solution : A

The man-hours required for the job $= (\frac{3040}{20}) = 152$ .

The Total number of workers who worked on this job for 6 days $= (\frac{152}{8}) = 19$ .

We get the following possibilities for the number of workers:

$\begin{matrix} D a y 1 & D a y 2 & D a y 3 & D a y 4 & D a y 5 & D a y 6 3 & 2 & 3 & 4 & 3 & 4 3 & 4 & 3 & 2 & 3 & 4 \end{matrix}$

Question 11

If only 2 workers were employed on this job on Day 4, what could be the minimum amount that Mr. Shyam pays the workers for this job for 6 days?

Rs. 4000

Rs. 2400

Rs. 3680

Rs. 3660

SOLUTION

Solution : B

We get the following possibilities:

$\begin{matrix} D a y 1 & D a y 2 & D a y 3 & D a y 4 & D a y 5 & D a y 6 & M a n - h r s & R s . 3 & 2 & 1 & 2 & 3 & 4 & 120 & 2400 3 & 4 & 3 & 2 & 3 & 4 & 152 & 3040 3 & 2 & 3 & 2 & 3 & 4 & 136 & 2720 \end{matrix}$

Question 12

If not more than 4 workers were employed on this job on any day, what CANNOT be the total amount that Mr.Shyam pays the workers for this job for 6 days?

Rs. 2720

Rs. 3040

Rs. 2400

None of these

SOLUTION

Solution : D

If not more than 4 workers work on this job on any day, the following are the possibilities:

$\begin{matrix} D a y 1 & D a y 2 & D a y 3 & D a y 4 & D a y 5 & D a y 6 & R s . 3 & 4 & 3 & 4 & 3 & 4 & 3360 3 & 2 & 3 & 4 & 3 & 4 & 3040 3 & 2 & 1 & 2 & 3 & 4 & 2400 3 & 4 & 3 & 2 & 3 & 4 & 3040 3 & 2 & 3 & 2 & 3 & 4 & 2720 \end{matrix}$

Question 13

In order to complete this job in 6 days, Mr. Shyam employs 6 workers on some day. If he had employed exactly 3 workers on each day and paid them the same amount, the job would have been completed in:

6 days

7 days

8 days

9 days

SOLUTION

Solution : D

If, on some day, 6 workers employed on this job, then the only possibility is:

$\begin{matrix} D a y 1 & D a y 2 & D a y 3 & D a y 4 & D a y 5 & D a y 6 & M a n - h r s 3 & 4 & 5 & 6 & 5 & 4 & 216 \end{matrix}$

Since, Mr. soni paid the same amount as wages in both the cases, the man-hours in both the cases must be the same.

Three workers would have taken $= (\frac{120}{3} \times 8) = 9$ days to complete the same job.

Question 14

Which of the following combinations is NOT possible?

2 experts in population studies from the Americas and 2 health experts from Africa attended the conference.

2 experts in population studies from the Americas and 1 health expert from Africa Attended the conference.

3 experts in refugee relocation from the Americas and 1 health expert from Africa attended the conference.

Africa and America each had only 1 expert in population studies attending the conference.

SOLUTION

Solution : D

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 2 & 2 & 1 & 1 & 6 P S & 1 & 2 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 1 & 3 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 1 & 3 & 1 & 1 & 6 P S & 1 & 2 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 2 & 2 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 1 & 3 & 1 & 1 & 6 P S & 2 & 1 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 1 & 3 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

Question 15

If Ramos is the lone American expert in population studies, which of the following is NOT true about the numbers of experts in the conference from the four continents?

There is one expert in health from Africa.

There is one expert in refugee relocation from Africa.

There are two experts in health from the Europe.

There are three experts in refugee relocation from the Americas.

SOLUTION

Solution : C

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 2 & 2 & 1 & 1 & 6 P S & 1 & 2 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 1 & 3 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 1 & 3 & 1 & 1 & 6 P S & 1 & 2 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 2 & 2 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 1 & 3 & 1 & 1 & 6 P S & 2 & 1 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 1 & 3 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

Question 16

Rob, an American expert in refugee relocation, was the first keynote speaker in the conference. What can be inferred about the number of American experts in refugee relocation in the conference, excluding Rob?

i. At least one

ii. At most two

A. Only i and not ii

Only ii and not I

Both i and ii

Neither i nor ii

SOLUTION

Solution : C

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 2 & 2 & 1 & 1 & 6 P S & 1 & 2 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 1 & 3 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 1 & 3 & 1 & 1 & 6 P S & 1 & 2 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 2 & 2 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 1 & 3 & 1 & 1 & 6 P S & 2 & 1 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 1 & 3 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

Question 17

Which of the following numbers cannot be determined from the information given?

Number of labour experts from the Americas.

Number of health experts from Europe.

Number of health experts from Australasia.

Number of experts in refugee relocation from Africa.

SOLUTION

Solution : D

(i) As the labour expert is half of each of the other, so the only possible combination is

(ii) Statement (d): If the number of Australasia expert is 1 less, i.e. total experts are 20

American experts will be twice as each of other.

The only combined possible is Americas 8.

Australasia 4 + 1 = 5

Europe 4

Africa 4

Now, we need to workout the various options possible in the blank cells.

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 2 & 2 & 1 & 1 & 6 P S & 1 & 2 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 1 & 3 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 1 & 3 & 1 & 1 & 6 P S & 1 & 2 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 2 & 2 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

$\begin{matrix} A f r i c a & A m e r i c a & A u s t r a l i a & E u r o p e & T o t a l L a b o u r & 0 & 1 & 1 & 1 & 3 H e a l t h & 1 & 3 & 1 & 1 & 6 P S & 2 & 1 & 2 & 1 & 6 \frac{R e f u g e e}{A l l o c a t i o n} & 1 & 3 & 1 & 1 & 6 T o t a l & 4 & 8 & 5 & 4 & 21 \end{matrix}$

Question 18

All the required pairs cannot be determined if it is known that BLUE kept in a box together with

GREEN

YELLOW

PINK

PURPLE

SOLUTION

Solution : B

If blue is kept together with yellow, then two cases are possible.
1. (Blue, Yellow), (Green, Pink) and (Red, Purple).
2. (Blue, Yellow), (Green, Purple) and (Red, Pink).

Question 19

If one of the pairs is given, then what is the probability that all the other required pairs can be determined?

$\frac{4}{10}$

$\frac{5}{10}$

$\frac{7}{10}$

$\frac{3}{10}$

SOLUTION

Solution : B

The correct option is (b).

If the pairs (BLUE, GREEN), (BLUE, PINK), (BLUE, PURPLE), (RED, PURPLE) and (YELLOW, PURPLE) are given, then all other pairs can be determined, while if the pairs (BLUE, YELLOW), (RED, YELLOW), (RED, PINK), (GREEN, PINK) and (GREEN, PURPLE) are given, then all other pairs cannot be determined. Therefore, required probability.

$= \frac{T o t a l n u m b e r o f f a v o u r a b l e c a s e s}{t o t a l n u m b e r o f p o s s i b l e c a s e s} = \frac{5}{10}$

Question 20

Which of the following must be a sock pair?

A, E

F, A

C,H

G,I

SOLUTION

Solution : D

At least two of the pairs must be (B,D), (E,A) or (C,H). Therefore, the pairs can be

1) (B,D), (E,A), (C,H), (G,I)... F

2) (B,D), (E,A), (F,C), (G,I)... H

3) (B,D), (C,H), (F,A), (G,I)... E

Question 21

Which of the following cannot be the single sock?

SOLUTION

Solution : A

At least two of the pairs must be (B,D), (E,A) or (C,H). Therefore, the pairs can be

1) (B,D), (E,A), (C,H), (G,I)... F

2) (B,D), (E,A), (F,C), (G,I)... H

3) (B,D), (C,H), (F,A), (G,I)... E

Question 22

Pebbles can buy eggs from two stores: a new Relance supermarket that sells eggs only in crates of 4, and a road side shoppe that sells single eggs without a crate. If Pebbles wants to ensure that the total number of eggs she buys is a multiple of 5, what is the minimum number of eggs she must buy from the road side shoppe?

none of these

SOLUTION

Solution : D

Pebbles can buy 0 eggs from the road side shoppe and 20 eggs from the supermarket

Question 23

Aditi bought some bananas, mangoes, guavas, pears and sapotas. She bought atleast 5 of each. All the numbers bought were distinct. Given that she has bought the least number of bananas, the number of bananas bought can be exactly determined if the total number of items bought is at most?

SOLUTION

Solution : C

5 + 6 + 7 + 8 + 9 = 35 and 6 + 7 + 8 + 9 + 10 = 40 so if he buys 39 items, he has to buy 5 bananas.

Question 24

A league tournament was played between the teams Australia, Bangladesh, Canada and Zimbabwe in which each team played every other team exactly once. There were no ties or abandoned matches. All the results followed the rule that in a match between two teams, the team which had won a larger number of matches out of all their previous encounters won this match. Which team won the largest number of matches?

A. Australia

B. Bangladesh

C. Canada

D. Zimbabwe

SOLUTION

Solution : A

$\begin{matrix} TEAM 1 & TEAM 2 & TEAM 3 A & B & A A & C & A A & Z & A B & C & B B & Z & B C & Z & Z \end{matrix}$

Question 25

In the league tournament with the conditions as described in the first question, a bookie Charlie follows the following system. In a match between two teams "a” and "b”, the team "a” wins more of their previous matches. If team "a” wins this match, he will pay Rs. 1.5 for every Re. 1 bet on team "a”. If team "b” wins this match, he will pay Rs. 2 for every Re. 1 bet on team "b”. In every match, equal money was bet on both the teams playing in that match. What was Charlie's gain, as a fraction, on the total money that he bet?

$\frac{- 1}{4}$

$\frac{1}{4}$

$\frac{1}{2}$

SOLUTION

Solution : C

Based on the information in the previous question, Charlie gets, say Rs 1 each from the betters of the two teams. In 6 matches, Charlie gets Rs 12. We know that the better team is winning in each case; hence his net gain is 50 paise in 6 matches $\Rightarrow$ his total gain $= \frac{(0.5 \times 6)}{12} = \frac{1}{4}$

Question 26

If the matches played by Australia are represented in a pie chart, what will be the angle subtended by the matches played between Australia and Canada?

10^{\circ}

20^{\circ}

30^{\circ}

40^{\circ}

SOLUTION

Solution : D

Total number of matches played by A =

$\begin{matrix} 51414255643215739039502 \end{matrix}$
Number of matches played between A & C = 41 + 15 = 56.
$502 \to 360^{\circ}$
$56 = 56 \times \frac{360}{502} = 40^{\circ}$ .

Question 27

The teams are going to be seeded based on the following rule. The country that has won more matches in all the matches played between the two countries is given 2 points; the other country does not get any points. All pairs of countries are taken into consideration and the teams are ranked according to the total number of points obtained. If two countries get the same number of points, the country that has won a larger number of matches, in the matches played between the two countries, is given a higher ranking. Based on this, which country would be ranked second?

A. Australia

B. Bangladesh

C. Denmark

D. Cannot be decided based on the criteria given.

SOLUTION

Solution : D

This is because after the points are allotted to the countries, there are 3 countries coming $2^{n d}$ (Australia, Denmark and England). There is no rule mentioned as to the winner in this situation. Hence, the country which comes in second position cannot be determined

Question 28

What is the ratio of the matches played by either Denmark or Zimbabwe (not both) to the total number of matches played between all countries?

A. 0.512

B. 0.353

C. 0.995

D. 0.616

SOLUTION

Solution : D

Total number of matches played between all countries

$\begin{matrix} Australia & Bangladesh & Canada & Denmark & England & Zimbabwe Australia & 0 & 51 & 41 & 42 & 55 & 64 & 253 Bangladesh & 32 & 0 & 38 & 55 & 41 & 53 & 219 Canada & 15 & 7 & 0 & 11 & 9 & 21 & 63 Denmark & 73 & 30 & 31 & 0 & 64 & 38 & 236 England & 90 & 54 & 49 & 41 & 0 & 50 & 284 Zimbabwe & 39 & 20 & 54 & 57 & 32 & 0 & 202 249 & 162 & 213 & 206 & 201 & 226 \end{matrix}$
Total = 1257
Matches played by Denmark or Zimbabwe =
$\begin{matrix} 236206202226 - 57 - 38 \end{matrix}$
=775
Ratio = $\frac{775}{1257} = 775 \times 8 \approx 0.62 (s i n c e \frac{1}{8} = 12.33)$

Question 29

What is the least cost of sending one unit from any refinery to any district?___

SOLUTION

Solution :
Observe table A & B

Min. Cost of sending one unit from refinery to depot : BC to AC = 0

Min. Cost of sending one unit from depot to district : AC to AAC = 0

BC to AC to AAC = 0

Question 30

What is the least cost of sending one unit from any refinery to the district AAB?___

SOLUTION

Solution :
BD to AE to AAB = 95.2

The least cost to reach to AAB is for AE. And that is BD to AE is zero.

Question 31

What is the least cost of sending petrol from refinery BB to district AAA?
___

SOLUTION

Solution :
First you check the minimum cost for receiving at AAA.

This is 0 for AE. But BB to AE is very high.

Next is AC [314.5] BB to AC is 451.1.

After AC the others are high.

Hence, 314.5 + 451.1 = 765.6

Question 32

How many possible ways are there for sending petrol from any refinery to any district?___

SOLUTION

Solution :
Number of refineries = 6

Number of depots = 7

Number of districts = 9

Therefore, number of possible ways to send petrol from any refinery to any district is 6 × 7 × 9 = 378.

Free DI and LR Practice Test - CAT

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Question 16

SOLUTION

Question 17

SOLUTION

Question 18

SOLUTION

Question 19

SOLUTION

Question 20

SOLUTION

Question 21

SOLUTION

Question 22

SOLUTION

Question 23

SOLUTION

Question 24

SOLUTION

Question 25

SOLUTION

Question 26

SOLUTION

Question 27

SOLUTION

Question 28

SOLUTION

Question 29

SOLUTION

Question 30

SOLUTION

Question 31

SOLUTION

Question 32

SOLUTION

Related ExamsDI and LRDI and LRDI and LR

Related Exams
DI and LR
DI and LR
DI and LR