## Lesson No. 2: Problems in Percents:

In this lesson, we will solve numerous problems on Percents. For basic concepts and formulas in Percentages, this web site reader is suggested to read the Lesson No. 1 on Percents.

Example 1:

John was 150 cms tall 2 years ago. Today he is 160 cms tall. What is the percent increase in his height?

Solution:

From lesson no. 1,

% increase = (increase/initial value) × 100%

Increase = 160 – 150 = 10, initial value = 150

% increase = (10/150) × 100% = 6.66%

Example 2:

Two years from today, the price of an object rises by 25% to \$250. What is the price today?

Solution:

From speed calculations in lesson no. 1:

25% increase yields fraction 125/100.

Multiply this fraction 125/100 to initial price, say P and equate it to final price 250.

(125/100) × P = 250, P = 200

So, price today is 200

Short-cut: take original price as 100.

100 becomes 125 (after a 25% increase)

? will become 250

On cross-multiplying: 100 × 250 =? × 125,

? = (100 × 250)/125 = 200

Example 3:

The length and breadth of a rectangular paper are 20 and 15 cms respectively. If the paper is cut by 5cms each along length and breadth, what is the percent decrease in the area of the rectangle?

Solution: Area of a rectangle = L × B

Before cutting, area = 20 × 15 = 300, and

After cutting, area = (20 – 5) × (15 – 5) = 15 × 10 = 150

% decrease in area is

(Amount of decrease/initial area) × 100%

= [(300 – 150)/300] × 100% = (150/300) × 100% = 50%

Example 4:

Ryan increased 10% in height from 1980 to 1985 and 15% from 1985 to 1990 What is the percent increase in his height from 1980 to 1995?

Solution: Apply net percent change formula during successive percent changes:

(x + y + xy/100)%

So, Net percent increase in Ryan’s height from 1980 to 1990 is

[10 + 15 + (10 × 15)/100]% = (25 + 150/100) % = (25 + 1.5) % = 26.5%

Example 5:

A is 30% taller than B. B is what percent shorter than A?

Solution:

Take B’s height as 100. Then, A’s height is (130/100) × 100 = 130

Now, find 130 is what % less than 100?

From lesson no. 1:

% Less = (Difference/Bigger Number) × 100%

= (30/130) × 100 % = 23.1% or 23% approximately

Example 6:

Mike scored 20% less than Jack in a Maths test. By what percentage is Jack’s score more than Mike’s?

Solution:

Let Jack’s score be 100. So, mike’s score is 20% less than 100, i.e. 80

Now, find 100 is what percent more than 80?

From “% more than” formula from lesson no. 1:

% More = (Difference/Smaller Number) × 100%

= (20/80) × 100% = 25%

Example 7:

What percent of 5 yards is 5 feet?

Solution: know that 1 yard is 3 feet.

5 yards is 5 × 3 = 15 feet.

Now, to find 5 feet is what % of 15 feet, write it as below:

(5/15) × 100% = (1/3) × 100% = 33.33%

Example 8:

In a physics experiment, a student measured the length of a line segment as 10 cm instead of 8. What is the percent error in the calculation?

Solution:

On what actual value is the 2cms excess written.

On 8cm.

Therefore, percent error needs to be expressed out of 8cms, the actual value.

% error = (2/8) × 100% = 25%

Example 9:

In a class test in English, Jim scored 20% more than Kim and 25% more than Tim. By what percent did Tim score less than Kim?

Solution:

Let Kim’s score be 100.

Note: Among the three, it is best to take Kim’s score as 100, because the question is “percent less than Kim

Then, Jim’s score is 20% more than 100 i.e. 120.

Jim’s score 120 is 25 more than Tim’s.

i.e. 120 = (125/100) × Tim’s score

So, Tim’s score = (120 × 100)/125 = 96

So, Tim’s score 96 is 4% less than Kim’s 100.

Example 10:

The price of a shirt was discounted by p% twice. The final price is 36% of the original price. Find x

Solution: Let us apply net percent change formula for successive percent changes:

(x + y + xy/100)%

Write a “―” in both x and y as each is a discount, which is a decrease.

(x + y + xy/100)% = [– p –p + (p2/100)] % = [–2p + (p2/100)] %

Final price is 36% of original price.

So net percent change in price is = 100 – 36 = 64%

So, [–2p + (p2/100)] % = -64%

{Note: Write a minus sign before 64 in R.H.S to denote decrease}

–2p + (p2/100) = –64

On simplifying, you will have to solve a quadratic equation.

–200p + p2 = –6400, after rearranging, the equation is:

p2 -200p +6400 = 0,

p2 -160p -40p +6400 = 0,

p (p – 160) – 40(p – 160) = 0

(p – 160) (p – 40) = 0,

So, either p – 160 = 0 or p – 40 = 0

p = 160 or p = 40

Since p denotes discount and the price of any article cannot be discounted by more than 100%, therefore 160 for p is rejected.

So, p is 40

Example 11:

The radius of a circle is 10 cms. It is decreased by 20%. Find Percent decrease in area   By what percent reduced area must be increased to original area.

Solution:

Area of a circle is ∏r2.

After 20% decrease, reduced radius is 80% of original radius, i.e. 80% of 10

(80/100) × 10 = 8

Reduced area of the circle is = ∏r2 = ∏ (82) = 64∏

So, percent decrease in area = (36∏/100∏) × 100% = 36%

Reduced area 64∏ should be increased to 100∏, the original area of the circle.

So, % increase = (36∏/64∏) × 100% = (36/64) × 100% = 56.25%