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%