Question

Afzal sat for aptitude certification exam. The exam had no negative marking. He correctly attempted 20 questions and got 116 marks. Questions were of two types. Type A awarded 10 marks and Type B awarded 3 marks to him. Can you tell us the total marks of another lot of 20 questions in which number of correctly attempted Type A and Type B questions by Afzal are interchanged?

Answer 1

Answer:

Type A Questions answered by Afzal 8

Type B Questions answered by Afzal 12

8×10+12×3=116

If numbers of Type A & Type B questions answered by Afzal are interchanged then Afzal's total marks will be=

12×10+8×3

=144

Answer 2

Given: Afzal sat for aptitude certification exam. The exam had no negative marking. He correctly attempted 20 questions and got 116 marks. Questions were of two types. Type A awarded 10 marks and Type B awarded 3 marks to him.

To find: The total marks of another lot of 20 questions in which number of correctly attempted Type A and Type B questions by Afzal are interchanged.

Solution:

Let the number of type A questions attempted by Afzal be x and the number of type B questions attempted by Afzal be y. According to the question, the marks obtained by solving correctly a particular number of type A and type B questions is 116. Also, the total number of questions solved correctly is 20. These two statements can be represented in the form of two equations.

$10x + 3y = 116$

$x + y = 20$

The two linear equations are solved simultaneously and the value of x and y are found to be

$x = 8$

$y = 12$

Now, the number of type A and type B questions are interchanged. So the total marks of those questions are

$3x + 10y = 3(8) + 10(12)$

              $= 144$

Therefore, the total marks of another lot of 20 questions in which the number of correctly attempted Type A and Type B questions by Afzal are interchanged are 144.