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# Tutor profile: Imran D.

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Imran D.
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## Questions

### Subject:Pre-Algebra

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Question:

Evaluate 8x + 7 given that x - 3 = 10

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Imran D.

x - 3 = 10 (given equation) x = 10 + 3 = 13 (On solving the given equation) On substituting x by 13 in the given expression, we get 8(13) + 7 = 111

### Subject:Basic Math

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Question:

What is 0.4% of 36?

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Imran D.

0.4% of 36 = X 0.4/100 of 36 = X 0.4/100 = X/36 0.4 x 36 = 100 X 14.4 = 100 X so, divide 14.4 by 100 which moves the decimal place 2 places to the left i.e. 0.144 so, X = 0.144

### Subject:Set Theory

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Question:

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?

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Imran D.

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

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