## Advanced Problems in Percentages

Example 1:

1.What percent is 4 kms of 5 miles?

1mile = 8/5 kms. 5 miles = 5 × 8/5 = 8 kms.
So, 4 kms of 5 miles = 4 kms of 8kms = (4/8) × 100% = 50%

2. What percent of 2.5 liters is 300 ml?

1 liter = 1000ml, 2.5 liters = 2.5 × 1000 = 2500 ml.
So, 300 ml of 2.5 liters = (300/2500) × 100% = 12%

Example 2:

If 50% of a number is added to 50, the result is the original number. Find the number.

Let N be the number.
So, 50% of N + 50 = N
First of all, 50% of N = (50/100) × N = (½)× N
So, N/2 + 50 = N, so N – N/2 = 50, so N/2 = 50 and N= 100

Example 3:

If x is 25% more than y, then express y in terms of x

x = y + 25% of y,
So x = y + (25/100) × y, So x = (125/100) × y,
So, x = (5/4) × y, So y = (4/5) × x
Now, multiply 100 to (4/5) x and insert the percent symbol % to express y in terms of percentage of x:
(4/5) × 100 % of x = 80 % of x.

Example 4:

A pen costs 10 times as much as a pencil. Find by what percent is the pencil’s cost less than the pen.

let pencil cost \$10. So, pen costs \$100.
From lesson no. 1:
% less = (difference/bigger number) × 100% = (90/100) × 100% = 90%

Example 5:

The price of a book reduces by 20%. By what percent can a book keeper buy more books for a same outlay on the books?

outlay is N × P, in which N is number of books bought and P is the price of each book.
Let Initial outlay be 100 × 100.
Final outlay will be n × 80; in which n is the increased number of books and 80 is the reduced price.
Since expenditures are same, so
100 × 100 = n × 80, so n = (100 × 100)/80 = 125
So, 25% more books can be bought for a same outlay.

Example 6:

Of all the employees in a company, 40% are graduate officers and 20% are junior graduate officers. What percent of the graduate officers are employees that are not junior graduate officers?

Solution:
Take number of employees as 100.
Officers are 40 and junior officers are 20
(Note that percent of junior officers is expressed out of employees in the company and not out of officers. Avoid error of proximity)
Officers that are not junior graduates = 40 – 20 = 20
Now, this 20 is what percent of 40 = (20/40) × 100% = 50%

Example 7:

In an alcohol solution of 8 liters the concentration of alcohol is 25%. To make alcohol 20% concentrate, how many liters of water must be added?

initially, alcohol = 25% of 8 = (25/100) × 8 = 2
New strength of alcohol is 20% of (N + 8), where N is liters of water to be added.
Volume of alcohol remains same, only concentration changes.
So, 20% of (N + 8) = 2, i.e. (20/100) × (N +8) = 2, So (1/5) ( N + 8) = 2, So N + 8 = 10
So, N = 10 – 8 = 2
So, 2 liters of water added will change alcohol concentration to 20%
Short-cut: C1 V1 = C2V2

• C1 and C2 are the initial and final concentrations of alcohol and
• V1 and V2 are the initial and final volumes of the solution.

25 × 8 = 20 × (n + 8),
So, n + 8 = 200/20 = 10, so n + 8 = 10, n = 2
So, add two liters of water.

Example 8:

In a class of 40 students, boys’ strength is increased by 30% and that of girls by 20%. The class size grows to 60. What percentage of boys are girls?

Solution:

let number of boys be x and number of girls be y.
So, x + y = 40, so y = 40 – x
Increased number of boys = x + 30% of x = x + 0.3x = 1.3x
Increased number of girls = y + 20% of y = y + 0.2y = 1.2y
1.3x + 1.2y = 60……………… (2)
Plugging y = 40 – x in 2:
1.3x + 1.2 (40 – x) = 60, So 1.3x + 48 – 1.2x = 60,
So 1.3x – 1.2x = 60 – 48, So 0.1x = 12, so x = 120

Example 9:

The price of a book decreases by 20%. By what percent now, can a book-keeper purchase more books for the same outlay as before?

Solution:
Short-cut: Apply successive percent change formula:
Let x = -20% and y be the percent increase. Net percent change is 0 as the expenditure should be same before and after price changes.
(x + y + xy/100)% = net percent change.
(- 20 + y – 20y/100)% = 0%
y – y/5 = 20, 4y/5 = 20, y = 25%
the book keeper can buy 25% more books for the same outlay.

Example 10:

The length of a rectangle increases by 25% and the breadth decreases by 20%. Find by what percent the area changes

Solution:
Apply net percent change:
(x + y + xy/100)% = net percent change.
% Increase in length is x = +25%,
And % decrease in breadth is y = -20%
Net percent change = 25 – 20 – (25 × 20/100) = 5 – 5 = 0%
Therefore, the area of the rectangle does not change.