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A is x% less than B, A is y% less than C. C is k% more than B. Express k in terms of x and y
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Answer A
Comparison of A, B and C in terms of x and k.
A = B(1 - x100x100) -------- Equation (1)
A = C(1 - y100y100) ------Equation (2)
C = B(1 + k100k100) -------- Equation (3)
Substituting equations (1) and (3) in (2), we get
B(1 + k100k100) (1 - y100y100) = B(1 - x100x100)
1 + k100k100 - y100y100 - ky1002ky1002 = 1 - x100x100
k100k100 (1 - y100y100) = y−x100y−x100
k = ((y−x)100100−y(y−x)100100−y)
The question is " Express k in terms of x and y. "
Hence, the answer is k = ((y−x)100100−y(y−x)100100−y)
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Here is the question for today, please solve it and mark your answer in the pole. Let us know your approach and answer in the comments section
A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is:
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Answer A
Comparison of A, B and C in terms of x and k.
A = B(1 - x100x100) -------- Equation (1)
A = C(1 - y100y100) ------Equation (2)
C = B(1 + k100k100) -------- Equation (3)
Substituting equations (1) and (3) in (2), we get
B(1 + k100k100) (1 - y100y100) = B(1 - x100x100)
1 + k100k100 - y100y100 - ky1002ky1002 = 1 - x100x100
k100k100 (1 - y100y100) = y−x100y−x100
k = ((y−x)100100−y(y−x)100100−y)
The question is " Express k in terms of x and y. "
Hence, the answer is k = ((y−x)100100−y(y−x)100100−y)
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Here is the question for today, please solve it and mark your answer in the pole. Let us know your approach and answer in the comments section
How many isosceles triangles with integer sides are possible such that sum of two of the side is 12?
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Answer C
Sum of two sides are given, how can we figure out what the third side could be?
Two possibilities: Two equal sides could add up to 12 or sum of 2 unequal sides = 12.
i.e. Sum of 2 equal sides = 12
Sum of 2 unequal sides = 12
=> If sum of two equal sides were 12, sides of the triangle should be 6, 6, x.
What are the values x can take?
x could range from 1 to 11.
11 integer values exist.
Now 2 unequal sides adding to 12. This could be 1 + 11, 2 + 10, 3 + 9, 4 + 8 or 5 + 7.
How many isosceles triangles are possible with the above combinations?
Isosceles triangles with the above combination:
1, 11, 11 
1, 1, 11 ☒
2, 10, 10 2, 2, 10 ☒
3, 9, 9 3, 3, 9 ☒
4, 8, 8 4, 4, 8 ☒
5, 7, 7 5, 5, 7
6 possibilities. Triplets such as (1, 1, 11), (2, 2, 10), etc are eliminated as sum of the two smaller values is less than the largest value. These cannot form a triangle.
The question is " How many isosceles triangles with integer sides are possible such that sum of two of the side is 12?"
11 + 6 = 17 possibilities totally.
Hence, the answer is 17.
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Here is the question for today, please solve it and mark your answer in the pole. Let us know your approach and answer in the comments section.
There is a set of parallel lines with x lines in it and another set of parallel lines with y lines in it. The lines intersect at 12 points. If x > y, find the maximum number of parallelograms that can be formed.
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Answer C
To get 12 points, either 6 and 2 lines can intersect or 4 and 3 lines can intersect!
Given: The lines intersect at 12 points
Therefore x * y = 12
Since x > y possibilities are x = 6 and y = 2 or x = 4 and y = 3
We need 4 points to form a paralleogram, 2 points forming 1 line and the other 2 points forming the other line.
When x = 6 and y = 2, the number of parallelograms formed = 6C2 * 2C2
= 15 * 1 = 15
When x = 4 and y = 3, the number of parallelograms formed = 4C2 * 3C2 = 6 x 3 = 18
Maximum = 18
The question is “find the maximum number of parallelograms that can be formed.”
Hence, the answer is 18.
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Here is the question for today, please solve it and mark your answer in the pole. Let us know your approach and answer in the comments section.
Four institutes, A, B, C, and D, had contracts with four vendors W, X, Y, and Z during the ten calendar years from 2010 to 2019. The contracts were either multi-year contracts running for several consecutive years or single-year contracts. No institute had more than one contract with the same vendor. However, in a calendar year, an institute may have had contracts with multiple vendors, and a vendor may have had contracts with multiple institutes. It is known that over the decade, the institutes each got into two contracts with two of these vendors, and each vendor got into two contracts with two of these institutes.
The following facts are also known about these contracts.
I. Vendor Z had at least one contract in every year.
II. Vendor X had one or more contracts in every year up to 2015, but no contract in any year after that.
III. Vendor Y had contracts in 2010 and 2019. Vendor W had contracts only in 2012.
IV. There were five contracts in 2012.
V. There were exactly four multi-year contracts. Institute B had a 7-year contract, D had a 4-year contract, and A and C had one 3-year contract each. The other four contracts were single-year contracts.
VI. Institute C had one or more contracts in 2012 but did not have any contract in 2011.
VII. Institutes B and D each had exactly one contract in 2012. Institute D did not have any contract in 2010.
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Answer for Question 1
Answer - D
From statement 1, 2, 3 and 4 it is clear that W and X doesn’t had any contract after the year 2015. Y had contract only in 2010 and 2019. Z had contract in all the years.
From the given option , only year 2015 will have 2 or more contracts ( at least 1 for Z and at least 1 for X)
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Answer for Question- 2
Answer - D
From point V) Institute B had a 7-year contract so it should be with Z as no other vendor worked for more than 5 years. From this statement it can also be concluded that Z had one more contract of either 4 years or 3 years and X will have one contract of 3 years and other can be of 4 years or 3 years . Each of W and Y will have 2 contracts each of 1 years.
From point IV)There were 5 contracts in 2012 and we also know that W had both of its contract in 2012, and Y do not have any contract in 2012 so either X or Z must have exactly 2 contract in 2012 which is not possible for Z . it mean X will have 2 contracts in 2012.
If X had contracts of 3 years and 4 years , Z will have contracts of 7 years and 3 years.
As X has 2 contract in 2012 so he should start his 2nd contract of 4 years in 2012 and first in 2010. Now D had a contract in 2012 so it should be with X and B must have a 7 year contract in 2012 which starts in 2010 with Z. Which also means the three year contract of Z will start in 2017.
Now from point VI) Institute C had one or more contracts in 2012 but did not have any contract in 2011 , It mean the three year contract of X starts with A. It means 3 year contract of Z will be with C.
Also C will have a one year contract with W in 2012.
Now all these given information’s can be tabulated as below .
Vendors
Institutes Vendor 1 Vendor 1
A X – 2010 to 2012 Y – 2010
B Z – 2010 to 2016 W –2012
C W – 2012 Z – 2017 to 2019
D X – 2012 to 2015 Y- 2019
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Answer for Question- 3
Answer - A
From point V) Institute B had a 7-year contract so it should be with Z as no other vendor worked for more than 5 years. From this statement it can also be concluded that Z had one more contract of either 4 years or 3 years and X will have one contract of 3 years and other can be of 4 years or 3 years . Each of W and Y will have 2 contracts each of 1 years.
From point IV)There were 5 contracts in 2012 and we also know that W had both of its contract in 2012, and Y do not have any contract in 2012 so either X or Z must have exactly 2 contract in 2012 which is not possible for Z . it mean X will have 2 contracts in 2012.
If X had contracts of 3 years and 4 years , Z will have contracts of 7 years and 3 years.
As X has 2 contract in 2012 so he should start his 2nd contract of 4 years in 2012 and first in 2010. Now D had a contract in 2012 so it should be with X and B must have a 7 year contract in 2012 which starts in 2010 with Z. Which also means the three year contract of Z will start in 2017.
Now from point VI) Institute C had one or more contracts in 2012 but did not have any contract in 2011 , It mean the three year contract of X starts with A. It means 3 year contract of Z will be with C.
Also C will have a one year contract with W in 2012.
Now all these given information’s can be tabulated as below .
Now all questions can be answered.
There was only one contract in the years years 2016, 2017 and 2018 .
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Answer for Question- 4
Answer - A
From point V) Institute B had a 7-year contract so it should be with Z as no other vendor worked for more than 5 years. From this statement it can also be concluded that Z had one more contract of either 4 years or 3 years and X will have one contract of 3 years and other can be of 4 years or 3 years . Each of W and Y will have 2 contracts each of 1 years.
From point IV)There were 5 contracts in 2012 and we also know that W had both of its contract in 2012, and Y do not have any contract in 2012 so either X or Z must have exactly 2 contract in 2012 which is not possible for Z . it mean X will have 2 contracts in 2012.
If X had contracts of 3 years and 4 years , Z will have contracts of 7 years and 3 years.
As X has 2 contract in 2012 so he should start his 2nd contract of 4 years in 2012 and first in 2010. Now D had a contract in 2012 so it should be with X and B must have a 7 year contract in 2012 which starts in 2010 with Z. Which also means the three year contract of Z will start in 2017.
Now from point VI) Institute C had one or more contracts in 2012 but did not have any contract in 2011 , It mean the three year contract of X starts with A. It means 3 year contract of Z will be with C.
Also C will have a one year contract with W in 2012.
Now all these given information’s can be tabulated as below.
Now all questions can be answered.
There were exactly 3 contract in 2010.
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Answer for Question- 5
Answer - D
From point V) Institute B had a 7-year contract so it should be with Z as no other vendor worked for more than 5 years. From this statement it can also be concluded that Z had one more contract of either 4 years or 3 years and X will have one contract of 3 years and other can be of 4 years or 3 years . Each of W and Y will have 2 contracts each of 1 years.
From point IV)There were 5 contracts in 2012 and we also know that W had both of its contract in 2012, and Y do not have any contract in 2012 so either X or Z must have exactly 2 contract in 2012 which is not possible for Z . it mean X will have 2 contracts in 2012.
If X had contracts of 3 years and 4 years , Z will have contracts of 7 years and 3 years.
As X has 2 contract in 2012 so he should start his 2nd contract of 4 years in 2012 and first in 2010. Now D had a contract in 2012 so it should be with X and B must have a 7 year contract in 2012 which starts in 2010 with Z. Which also means the three year contract of Z will start in 2017.
Now from point VI) Institute C had one or more contracts in 2012 but did not have any contract in 2011 , It mean the three year contract of X starts with A. It means 3 year contract of Z will be with C.
Also C will have a one year contract with W in 2012.
Now all these given information’s can be tabulated as below.
Now all questions can be answered.
So A and B are the only institutes had multiple contracts during the same year.
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Answer for Question- 6
Answer - C
From point V) Institute B had a 7-year contract so it should be with Z as no other vendor worked for more than 5 years. From this statement it can also be concluded that Z had one more contract of either 4 years or 3 years and X will have one contract of 3 years and other can be of 4 years or 3 years . Each of W and Y will have 2 contracts each of 1 years.
From point IV)There were 5 contracts in 2012 and we also know that W had both of its contract in 2012, and Y do not have any contract in 2012 so either X or Z must have exactly 2 contract in 2012 which is not possible for Z . it mean X will have 2 contracts in 2012.
If X had contracts of 3 years and 4 years , Z will have contracts of 7 years and 3 years.
As X has 2 contract in 2012 so he should start his 2nd contract of 4 years in 2012 and first in 2010. Now D had a contract in 2012 so it should be with X and B must have a 7 year contract in 2012 which starts in 2010 with Z. Which also means the three year contract of Z will start in 2017.
Now from point VI) Institute C had one or more contracts in 2012 but did not have any contract in 2011 , It mean the three year contract of X starts with A. It means 3 year contract of Z will be with C.
Also C will have a one year contract with W in 2012.
Now all these given information’s can be tabulated as below .
Now all questions can be answered.
So A,D,W and Z are institutes and vendors had more than one contracts in any year.
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Here is the question for today, please solve it and mark your answer in the pole. Let us know your approach and answer in the comments section.
Five vendors are being considered for a service. The evaluation committee evaluated each vendor on six aspects – Cost, Customer Service, Features, Quality, Reach, and Reliability. Each of these evaluations are on a scale of 0 (worst) to 100 (perfect). The evaluation scores on these aspects are shown in the radar chart. For example, Vendor 1 obtains a score of 52 on Reliability, Vendor 2 obtains a score of 45 on Features and Vendor 3 obtains a score of 90 on Cost.
1- On which aspect is the median score of the five vendors the least?
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2- A vendor’s final score is the average of their scores on all six aspects. Which vendor has the highest final score?
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3- List of all the vendors who are among the top two scorers on the maximum number of aspects is:
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4- List of all the vendors who are among the top three vendors on all six aspects is:
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Answer for Question- 1
Answer - D
The scores of venders given in graph can be tabulated as below:
Median of reliability = 52
Median of quality = 60
Median of customer s = 50
Median of cost = 80
Clearly the median of customer s is minimum.
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Dear Aspirants!
Answer for Question- 2
Answer - B
The scores of venders given in graph can be tabulated as below:
The vender which has highest total will have highest average. Vendor 3 has highest total thus highest average.
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