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In a plane there are two lines, a and b. a has 4 points on it while b has 5 points on it. Beside this there are six other points in this plane, no three of which are collinear. Also, apart from the points that lie on lines a and b, no three points are collinear. What is the maximum number of quadrilaterals that can be drawn using these points?
  • a)
    435
  • b)
    495
  • c)
    1215
  • d)
    915
  • e)
    None of these
Correct answer is option 'C'. Can you explain this answer?
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In a plane there are two lines, a and b. a has 4 points on it while b ...
In all there are 15 points in the plane. Number of ways in which four points can be selected from these 15 points is 15C4 = 1365.
But to form a quadrilateral, not more than 2 points can be on the same line.
So the cases when a selection of 4 out of 15 points, made in the above manner, will not result in a quadrilateral are:
Case (i):
When all the 4 points selected are on line a.
This selection can be made in 4C4 = 1 way.
Case (ii):
When all the 4 points selected lie on line b.
This selection can be made in 5C4 = 5 ways.
Case (iii):
When 3 points lie on line a, while the 4th point lies on either line b or is one of the 6 points in the plane.
This selection can be made in 4C
3
 x 5C3 + 4C3 x 6C1 = 44 ways.
Case (iv): When 3 points lie on line b, while the 4th point lies on either line a or is one of the 6 points in the plane.
This selection can be made in 5C3 x 4C3 + 5C3 x 6C1 = 100 ways.
So the total number o f quadrilaterals that can be formed is 1 3 6 5 - (1 + 5 + 44 + 100) =1215.
Hence, option 3.
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Most Upvoted Answer
In a plane there are two lines, a and b. a has 4 points on it while b ...
To find the maximum number of quadrilaterals that can be formed using the given points, we need to consider all possible combinations of the points. Let's break down the solution step by step.

Step 1: Count the number of ways to choose 4 points from the given 6 points (excluding the points on lines a and b).
This can be calculated using the combination formula: C(n, r) = n! / (r! * (n-r)!), where n is the total number of points and r is the number of points to be chosen.
In this case, n = 6 and r = 4.
C(6, 4) = 6! / (4! * (6-4)!) = 6! / (4! * 2!) = (6 * 5 * 4 * 3) / (4 * 3 * 2 * 1) = 15.

Step 2: Count the number of ways to choose 2 points from the 4 points on line a.
This can be calculated using the combination formula: C(n, r) = n! / (r! * (n-r)!), where n is the total number of points and r is the number of points to be chosen.
In this case, n = 4 and r = 2.
C(4, 2) = 4! / (2! * (4-2)!) = 4! / (2! * 2!) = (4 * 3) / (2 * 1) = 6.

Step 3: Count the number of ways to choose 2 points from the 5 points on line b.
This can be calculated using the combination formula: C(n, r) = n! / (r! * (n-r)!), where n is the total number of points and r is the number of points to be chosen.
In this case, n = 5 and r = 2.
C(5, 2) = 5! / (2! * (5-2)!) = 5! / (2! * 3!) = (5 * 4) / (2 * 1) = 10.

Step 4: Calculate the maximum number of quadrilaterals.
To form a quadrilateral, we need to choose 4 points. Two of these points must be from line a, and two must be from line b. So, the total number of quadrilaterals can be calculated by multiplying the results from step 1, step 2, and step 3.
Maximum number of quadrilaterals = (number of ways to choose 4 points from 6 points) * (number of ways to choose 2 points from 4 points on line a) * (number of ways to choose 2 points from 5 points on line b)
Maximum number of quadrilaterals = 15 * 6 * 10 = 900.

Therefore, the maximum number of quadrilaterals that can be drawn using these points is 900. None of the given options match this answer, so the correct answer is None of these.
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Worst Price, a leading retailer, was running a special sales promotion at its outlet in which, for every 1000 rupees that a customer spent on buying any items at the shop, the customer would get 1 reward point. The marketing department classified the total number of customers into seven different sections – G1, G2, G3,G4, G5, G6, G7 according to their respective number of reward point. The following pie chart gives the percentage wise break up of the number of customers classified into each of these seven sections. In the pie chart, the values given in the brackets alongside each section give the minimum and maximum number of reward point got by any customer classified into that section. To encourage customers to spend more at their outlets the management decided to give cash return to some of the customers from among whose who had at least 18 reward points. While deciding the cash return the management decided to classify the customers with different reward points into different groups-P to U- according to their number of reward points , as given in the table below: Group No. of reward point Cash Return (Rs.) P 18,19,20 1000 Q 21,22,23,24,25 1200 R 26,27,28,29,30 1400 S 31,32,33,34,35 1600 T 36,37,38,39,40 1800 U 41,42,43,44,45 2000 The number of customers selected for receiving a cash return were different for different groups However, within each group, the management selected an equal number of customers with each of the different number of reward points classified into that group. For example, in group P, of the total number of customers selected for receiving a cash return, the number of customers with 18 reward points is same as that with 19 reward points, which, in turn, is same as that with 20 reward points. The percentage wise distribution of the total value of the cash returns offered by the management to the customers belonging to the different groups is given in the following graph: Total amount paid as cash return Rs. 300000 For which section of customers is the number of customers selected for receiving a cash return the highest, when expressed as a percentage of the total number of customers in that section?

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In a plane there are two lines, a and b. a has 4 points on it while b has 5 points on it. Beside this there are six other points in this plane, no three of which are collinear. Also, apart from the points that lie on lines a and b, no three points are collinear. What is the maximum number of quadrilaterals that can be drawn using these points?a)435b)495c)1215d)915e)None of theseCorrect answer is option 'C'. Can you explain this answer?
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In a plane there are two lines, a and b. a has 4 points on it while b has 5 points on it. Beside this there are six other points in this plane, no three of which are collinear. Also, apart from the points that lie on lines a and b, no three points are collinear. What is the maximum number of quadrilaterals that can be drawn using these points?a)435b)495c)1215d)915e)None of theseCorrect answer is option 'C'. Can you explain this answer? for CAT 2024 is part of CAT preparation. The Question and answers have been prepared according to the CAT exam syllabus. Information about In a plane there are two lines, a and b. a has 4 points on it while b has 5 points on it. Beside this there are six other points in this plane, no three of which are collinear. Also, apart from the points that lie on lines a and b, no three points are collinear. What is the maximum number of quadrilaterals that can be drawn using these points?a)435b)495c)1215d)915e)None of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for CAT 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a plane there are two lines, a and b. a has 4 points on it while b has 5 points on it. Beside this there are six other points in this plane, no three of which are collinear. Also, apart from the points that lie on lines a and b, no three points are collinear. What is the maximum number of quadrilaterals that can be drawn using these points?a)435b)495c)1215d)915e)None of theseCorrect answer is option 'C'. Can you explain this answer?.
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