### NTS Quantitative Reasoning Practice Test # 5

**INSTRUCTIONS:**

- This Quiz is related to
**NTS Quantitative Reasoning Practice Test # 5**in this series - The Quiz based on the Topic, “
**Quantitative Reasoning**“ - If you have good understanding about the subject, then you can attempt it
- There will be 10 Multiple Choice Questions (MCQs) in this test
- You have 60 Seconds to answer a single Multiple Choice Question
- Questions will be randomly changed every time you start this test
- You should practice more and more to get high marks
- Passing Criteria is 60 percent for this Quiz
- You can retake this test as many time as you like
- The progress bar at the top of screen will show your progress as well as the time remaining where timed quiz
- After the quiz, You will find Your Test Score and Grade
- If you feel that any Incorrect Answer to a Question, simply Comment us about the Question.

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#### If x = - 2 is a solution to the equation (1/2) a x2 + 3x - 4 = 0, where a is a constatnt, then what is the second solution of this equation?

x = - 2 is a solution and therefore satisfies the given equation. Hence (1/2) a (-2) 2 + 3x - 4 = 0 Simplify and solve for a (1/2) a (-2) 2 + 3(-2) - 4 = 0 2 a - 6 - 4 = 0 a = 10 / 2 = 5 We now substitute a by 5 in the given equation and then factor knowing that x + 2 is a factor since x = - 2 is a solution. (5/2) x2 + 3x - 4 = 0 (x + 2)( (5/2)x - 2 ) = 0 The second solution is found by setting the second factor equal to 0 and solve (5/2)x - 2 = 0 x =x = - 2 is a solution and therefore satisfies the given equation. Hence (1/2) a (-2) 2 + 3x - 4 = 0 Simplify and solve for a (1/2) a (-2) 2 + 3(-2) - 4 = 0 2 a - 6 - 4 = 0 a = 10 / 2 = 5 We now substitute a by 5 in the given equation and then factor knowing that x + 2 is a factor since x = - 2 is a solution. (5/2) x2 + 3x - 4 = 0 (x + 2)( (5/2)x - 2 ) = 0 The second solution is found by setting the second factor equal to 0 and solve (5/2)x - 2 = 0 x = 4 / 5

#### X and Y are two real numbers such that X + Y = 11 / 5, X · Y = 2 / 5 and X is greater than 1. Find X and Y.

The second equation X · Y = 2 / 5 may written as follows Y = 2 / (5 X) Substitute Y by 2 / (5 X) in the equation X + Y = 11 / 5 X + 2 / (5 X) = 11 / 5 Multiply all terms of the above equation and simplify 5 X2 + 2 = 11 x 5 X2 - 11 X + 2 = 0 Factor left hand side and solve (5X - 1)(X - 2) = 5X - 1 = 0 , X = 1 / 5 X - 2 = 0 , X = 2 We now calculate the value of Y using the equation X + Y = 11 / 5 For X = 1 / 5, Y = 11 / 5 - X = 11 / 5 - 1 / 5 = 10 / 5 = 2 For X = 2, Y = 11 / 5 - X = 11 / 5 - 2 = 11 / 5 - 10 / 5 = 1 / 5 The two numbers are X = 2 and Y = 1 / 5

#### If n is an odd integer, then which of the following is even? (I) n2 (II) n2 + 1 (III) 3n2 - 1

Since n is odd integer, it can be written as n = 2 k + 1 , where k is an integer Let us express n2 in terms of k as follows n2 = (2 k + 1)2 = 4 k2 + 4 k + 1 Let rewrite n2 as follows n2 = 2(2 k2 + 2 k) + 1 Hence n2 is odd. We now express n2 + 1 in terms of k. n2 + 1 = 2(2 k2 + 2 k) + 1 + 1 = 2(2 k2 + 2 k + 1) Hence n2 + 1 is even. We now express 3 n2 - 1 in terms of k. 3 n2 - 1 = 3 [2(2 k2 + 2 k) + 1 ] - 1 = 3 [2(2 k2 + 2 k) ] + 3 - 1 = 6(2 k2 + 2 k) + 2 Hence 3 n2 - 1 is even.

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