Word problems on percentage will help us to solve various types of problems related to percentage. Follow the procedure to solve similar type of percent problems.
Word problems on percentage:
1. In an exam Ashley secured 332 marks. If she secured 83 % makes, find the maximum marks.
Solution:
Let the maximum marks be m.
Ashley’s marks = 83% of m
Ashley secured 332 marks
Therefore, 83% of m = 332
⇒ 83/100 × m = 332
⇒ m = (332 × 100)/83
⇒ m =33200/83
⇒ m = 400
Therefore, Ashley got 332 marks out of 400 marks.
2. An alloy contains 26 % of copper. What quantity of alloy is required to get 260 g of copper?
Solution:
Let the quantity of alloy required = m g
Then 26 % of m =260 g
⇒ 26/100 × m = 260 g
⇒ m = (260 × 100)/26 g
⇒ m = 26000/26 g
⇒ m = 1000 g
3. There are 50 students in a class. If 14% are absent on a particular day, find the number of students present in the class.
Solution:
Number of students absent on a particular day = 14 % of 50
i.e., 14/100 × 50 = 7
Therefore, the number of students present = 50  7 = 43 students.
4. In a basket of apples, 12% of them are rotten and 66 are in good condition. Find the total number of apples in the basket.
Solution:
Let the total number of apples in the basket be m
12 % of the apples are rotten, and apples in good condition are 66
Therefore, according to the question,
88% of m = 66
88/100 × m = 66
m = (66 × 100)/88
m = 3 × 25
m = 75
Therefore, total number of apples in the basket is 75.
5. In an examination, 300 students appeared. Out of these students; 28 % got first division, 54 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed.
Solution:
The number of students with first division = 28 % of 300 = 28/100 × 300 = 8400/100 = 84
And, the number of students with second division = 54 % of 300
= 54/100 × 300
=16200/100
= 162
Therefore, the number of students who just passed = 300 – (84 + 162) = 54
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