Set P comprises all multiples of 4 less than 500 Set Q comprises all odd multiples of 7 less than 50...........

Quantitative Aptitude ( CCS) Set theory

Set P comprises all multiples of 4 less than 500. Set Q comprises all odd multiples of 7 less than 500, Set R comprises all multiples of 6 less than 500. How many elements are present in P ∪ Q ∪ R?

202

243

228

186

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Explanation:

This question involves both Number Systems and Set Theory.
Set P = {4, 8, 12, ….496} ? 124 elements {all elements from 1 * 4 to 124 * 4}
Set Q = {7, 21, 35, 49,……497} ? {7 * 1, 7 * 3, 7 * 5, ….. 7 * 71} ? 36 elements.
Set R = {6, 12, 18, 24, …..498} ? {6 * 1, 6 * 2, 6 * 3, ….. 6 * 83} ? 83 elements.
Sets P and R have only even numbers; set Q has only odd numbers. So,
P ∩ Q = Null set
Q ∩ R = Null set
P ∩ Q ∩ R = Null set
So, If we find P ∩ R , we can plug into the formula and get P ∪ Q ∪ R
P ∩ R = Set of all multiples of 12 less than 500 = {12, 24, 36,…..492}
= {12 * 1, 12 * 2 , 12 * 3, …12 * 41} ? This has 41 elements
P ∪ Q ∪ R = P + Q + R – (P ∩ Q) – ( Q ∩ R) – (R ∩ P) + (P ∩ Q ∩ R)
P ∪ Q ∪ R = 124 + 36 + 83 – 0 – 0 – 41 + 0 = 202