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Test: Pie Chart - 1 (January 3) - CAT MCQ


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Test: Pie Chart - 1 (January 3) - Question 1

100 students appeared for two examinations. 60 passed the first, 50 passed the second and 30 passed both. Find the probability that a student selected at random has failed in both the examinations?

Detailed Solution for Test: Pie Chart - 1 (January 3) - Question 1


From the figure it is evident that 80 students passed at least 1 exam. Thus, 20 failed both and the required probability is 20/100 = 1/5.

Test: Pie Chart - 1 (January 3) - Question 2

In rolling two dices, find the probability that there is at least one ‘6’

Detailed Solution for Test: Pie Chart - 1 (January 3) - Question 2

With a six on the first dice, there are 6 possibilities of outcomes that can appear on the other dice (viz. 6 & 1, 6 & 2, 6 & 3, 6 & 4, 6 &5 and 6&6).At the same time with 6 on the second dice there are 5 more possibilities for outcomes on the first dice: (1 & 6, 2 & 6, 3 & 6, 4 & 6, 5 & 6)
Also, the total outcomes are 36. Hence, the required probability is 11/36.
 

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Test: Pie Chart - 1 (January 3) - Question 3

From a bag containing 4 white and 5 black balls a man draws 3 at random. What are the odds against these being all black?

Detailed Solution for Test: Pie Chart - 1 (January 3) - Question 3

Odds against an event = p(E’)/p(E)
In this case, the event is: All black, i.e., First is black and second is black and third is black.
P(E) = 5/9 X 4/8 X 3/7 = 60/504 = 5/42.
Odds against the event = 37/5.

Test: Pie Chart - 1 (January 3) - Question 4

Phoebe throws three dice in a special game of Ludo. If it is known that he needs 15 or higher in this throw to win then find the chance of his winning the game

Detailed Solution for Test: Pie Chart - 1 (January 3) - Question 4

Event definition is: 15 or 16 or 17 or 18.
15 can be got as: 
5 and 5 and 5 (one way)            
Or
6 and 5 and 4 (Six ways)        
Or 6 and 6 and 3 (3 ways)        
Total of 10 ways. 16can be got as: 6 and 6 and 4 (3ways)
Or 6 and 5 and 5 (3ways) = Total 6 ways.
17 has 3 ways and 18 has 1 way of appearing. Thus, the required probability is: (10 + 6 + 3 + 1)/216 = 20/216 = 5/54. 

Test: Pie Chart - 1 (January 3) - Question 5

Find out the probability of forming 187 or 215 with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 when only numbers of three digits are formed and when repetitions are not allowed.

Detailed Solution for Test: Pie Chart - 1 (January 3) - Question 5

Positive outcomes = 2 (187 or 215) Total outcomes = 9 X 8 X 7
Required probability = 2/504

Test: Pie Chart - 1 (January 3) - Question 6

Instructions: On the basis of the information given below, answer the questions that follow.

Maruti Suzuki has tie-ups with five dealers in Allahabad to sell their cars. The first pie chart given below gives the dealer-wise break-up of the number of cars sold by Maruti Suzuki in Allahabad in 2015. The second pie chart shows the model-wise breakup of the total cars sold by Maruti Suzuki in 2015 in Allahabad.

Q. If for any dealer, the percentage of Wagon R’s sold by him did not exceed 25% of his total sales, then at least how many dealers sold Wagon R?

Detailed Solution for Test: Pie Chart - 1 (January 3) - Question 6

We have to minimize the number of dealers who sold Wagon R.

So to minimize the number of dealers, the dealers who have largest share in total sales should have maximum share of Wagon R.

So if we assume that Q and R sold Wagon R the maximum number of Wagon R’s they can sell is 25% of 51 which is equal to 12.75%.

But the overall share of Wagon R is 16% so there should be at least 1 more dealer who sells Wagon R.

Hence the correct answer is 3.

Test: Pie Chart - 1 (January 3) - Question 7

Instructions: On the basis of the information given below, answer the questions that follow.

Maruti Suzuki has tie-ups with five dealers in Allahabad to sell their cars. The first pie chart given below gives the dealer-wise break-up of the number of cars sold by Maruti Suzuki in Allahabad in 2015. The second pie chart shows the model-wise breakup of the total cars sold by Maruti Suzuki in 2015 in Allahabad.

Q. If the dealer T sold an equal number of all five models of Maruti Suzuki then what is the maximum share(in %) that T can have in total sales of any model?(Enter only the integer value rounded off to nearest integer. Don’t put % sign)

Detailed Solution for Test: Pie Chart - 1 (January 3) - Question 7

If dealer T sold an equal number of each model, then for each model his share would be 4.2%.

Hence his share would be maximum for the model which had least share among the total sales.

The least share is for Zen which is 9%

Hence, the maximum share that T can have for any model = 4.2/9 × 100

= 46.67% which on rounding off becomes 47%.

Hence the required integral value to be entered is 47.

Test: Pie Chart - 1 (January 3) - Question 8

Instructions: On the basis of the information given below, answer the questions that follow.

Maruti Suzuki has tie-ups with five dealers in Allahabad to sell their cars. The first pie chart given below gives the dealer-wise break-up of the number of cars sold by Maruti Suzuki in Allahabad in 2015. The second pie chart shows the model-wise breakup of the total cars sold by Maruti Suzuki in 2015 in Allahabad.

Q. If the dealers R and S sold only Zen, Baleno and swift then at least what percentage of the Swifts sold, were sold by these two dealers alone? (Enter only the integer value rounded off to nearest integer. Don’t put % sign)

Detailed Solution for Test: Pie Chart - 1 (January 3) - Question 8

If the number of swifts sold by R and S is to be minimum, R and S should sell a maximum number of Baleno and Zen.

The share of R and S in total sales is 40%.

So even if they sell all Baleno’s and Zen that will account to only 24%, so they together will sell 16% Swifts.

The total share of swifts is 27% so they sold 16%.

Hence percentage share would be 16/27 × 100

= 59.25% which on rounding off becomes 59%. So correct answer is 59.

Test: Pie Chart - 1 (January 3) - Question 9

Instructions: The pie chart shown below gives the percentage distribution of car sales (in
numbers) between five companies – Maruti, Honda, Hyundai, Ford and Tata during the year 2014 – 2015. The total number of cars sold by the five companies together in 2014-2015 is four hundred thousand.

The line graph shown below gives the percentage distribution of the number of cars sold by each of the five companies in the four different Quarters of the year.

Q. Which company sold the highest number of cars in the second quarter?

Detailed Solution for Test: Pie Chart - 1 (January 3) - Question 9

The data given in the pie chart and the line chart can be tabulated into the following table (Data is in 000’s):

From the table, we can see that Ford sold the highest number of cars in the second quarter.

Test: Pie Chart - 1 (January 3) - Question 10

Instructions: The pie chart shown below gives the percentage distribution of car sales (in
numbers) between five companies – Maruti, Honda, Hyundai, Ford and Tata during the year 2014 – 2015. The total number of cars sold by the five companies together in 2014-2015 is four hundred thousand.

The line graph shown below gives the percentage distribution of the number of cars sold by each of the five companies in the four different Quarters of the year.

Q. In how many quarters were the number of cars sold by any two companies in that quarter the same?

Detailed Solution for Test: Pie Chart - 1 (January 3) - Question 10

The data given in the pie chart and the line chart can be tabulated into the following table (Data is in 000’s):

We can see that in Q1 and Q4 there are companies which sold equal number of cars. Thus, D is the right choice.

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