Afzal sat for aptitude certification exam. The exam had no negative marking. He correctly attempted 19 questions and got 113 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 19 questions in which number of correctly attempted Type A and Type B questions by Afzal are interchanged
Answers
Answer:
134
Step-by-step explanation:
for 19 questions-113 marks
11 questions * 3 marks-33 total
8 questions * 10 marks-80 total
after interchange
11questions *10 Mark's =110
8 questions* 3 marks = 24
therefore total 110+24 =134
Given : correctly attempted 19 questions and got 113 marks.
. Questions were of two types. Type A awarded 10 marks and Type B awarded 3 marks to him.
To Find : total marks of another lot of 19 questions in which number of correctly attempted Type A and Type B are interchanged
Solution:
Type A = x
Type B = 19 - x
Type A Marks = 10x
Type B Marks = 3(19 - x) = 57 - 3x
10x + 57- 3x = 113
=> 7x = 56
=> x = 8
Type A = 8
Type B = 11
Interchanged
Type A =11
Type B = 8
Marks = 10 (11) + 3(8)
= 110 + 24
= 134
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