Questions on Mensuration with Solutions

Problem 1:A basketball ring has a radius of 21 cm. Find its circumference and area. (Use
π = 22/7)

Solution: We know,

Circumference of circle = 2πr = 2 x (22/7) x 21 = 2 x 22 x 3 = 132 cm

Area of circle = πr2 = (22/7) x 212 = 22/7 x 21 x 21 = 22 x 3 x 21

Area of circle with radius, 21cm = 1386 cm2

Problem 2: If one side of a square box is 4 cm, then what will be its area and perimeter?

Solution: Given,

Length of side of square = 4 cm

Area = side2 = 42 = 4 x 4 = 16 cm2

Perimeter of square = sum of all its sides

Since, all the sides of the square are equal, therefore;

Perimeter = 4+4+4+4 = 16 cm

Problem 3: A kite having diagonals of length 10 cm and 16 cm, respectively. Find its
area.

Solution: d1 = 10 cm

D2 = 16 cm

Area of rhombus = ½ d1 d2

A = ½ x 10 x 16

A= 80 cm2

Problem 4: The area of a trapezium shaped field in our barangay is 480 m2, the
distance between two parallel sides is 15 m and one of the parallel sides is 20 m. Find
the other parallel side.

Solution: One of the parallel sides of the trapezium is a = 20 m, let another parallel side
be b, height h = 15 m.

The given area of trapezium = 480 m2

We know, by formula;

Area of a trapezium = ½ h (a+b)

480 = ½ (15) (20+b)

20 + b = (480×2)/15

B = 64 – 20 = 44 m

Problem 5: Height, length and width of a cuboidal storage box are 20 cm, 15 cm and 10
cm, respectively. Find its area.

Solution: Total surface area = 2 (20 × 15 + 20 × 10 + 10 × 15)

TSA = 2 ( 300 + 200 + 150) = 1300 cm2

AGE PROBLEM

Problem 1: The sum of the ages of John and Mary is 32. Four years ago, John was twice
as old as Mary.
Problem 2: The sum of the ages of a father and son is 56. Four years ago, the father
was 3 times as old as the son.

Problem 3: The sum of the ages of me and my brother is 20 years. Four years ago, my
brother was one-half of my age

Problem 4: My Tito is 36 years old and his daughter mina is 3. In how many years will
the man be 4 times as old as his daughter?

Problem 5: Bob’s age is twice that of Barry’s. Five years ago, Bob was three times older
than Barry. Find the age of both.

INVESTMENT PROBLEM

Problem 1: You have 6,000 invested at 9%. How much more money should you invest at
12% to have a yearly interest income of 1,980?

Solution:

We calculate interest on accounts using the formula 𝐼 = 𝑝𝑟𝑡

Write the following equation:

(9%)(6000)(1) + (12%)(𝑥)(1) = 1980

540 + (0.12)(𝑥) = 1980

0.12𝑥 = 1440

𝑥 = 12000

Answer: $12000

Problem 2: Tom invested 8,000 in two investments, part of it at 6.5% and the rest at
4.5%. If the total interest earned after 2 years was 880, how much was put in each
investment?

Solution:

We calculate interest on accounts using the formula 𝐼 = 𝑝𝑟𝑡,

Following equation:

(6.5%)(𝑥)(2) + (4.5%)(8000 – 𝑥)(2) = 880

(0.13)𝑥 + (0.09)(8000 – 𝑥) = 880

0.13𝑥 + 720 – 0.09𝑥 = 880

0.04𝑥 + 720 = 880

0.04𝑥 = 160

𝑥 = 4000

If 𝑥 = 4000, then 8000 – 4000 = 4000

Answer: $4000 at 6.5%, $4000 at 4.5%

Problem 3: I have 9,000 pesos invested at 15%. How much more money should you
invest at 24% to have a yearly interest income of 5, pesos190?

Solution:

Write the following equation:

(15%)(9000)(1) + (24%)(𝑥)(1) = 5190

1350 + (0.24)(𝑥) = 5190

0.24𝑥 = 3840
 𝑥 = 16000

Answer: 16000

Problem 4: Jane deposited 1,000 into a savings account with an annual interest rate of
5%. How much will she have in her account after one year?

Solution: To calculate the amount in the account after one year, we use the formula for
simple interest:

Interest = Principal × Rate × Time

Interest = $1000 × 0.05 × 1 year = $50

So, after one year, Jane will have $1000 + $50 = $1050 in her account.

Problem 5: Sarah invested $2000 in a government bond with an annual interest rate of
3%. How much interest will she earn after three years?

Solution: To find the interest earned, we use the formula for simple interest:

Interest = Principal × Rate × Time

Interest = $2000 × 0.03 × 3 years = $180

So, Sarah will earn $180 in interest after three years.

MIXTURE PROBLEM

Problem 1: Candy have two buckets of paint, one red and one blue. The red paint has
20% red pigment, and the blue paint has 30% blue pigment. How much of each paint do
you need to mix to get 10 gallons of paint with 25% red pigment?

Let \( x \) be the gallons of red paint and \( y \) be the gallons of blue paint.

We have the following equations:

\[ x + y = 10 \] (total gallons of paint)

\[ 0.2x + 0.3y = 0.25(10) \] (total red pigment in the mixture)

Solving these equations gives:

\( x = 4 \) gallons of red paint and \( y = 6 \) gallons of blue paint.

Problem 2: Louie have 50 ounces of a 20% alcohol solution. How many ounces of pure
alcohol must be added to make a 25% alcohol solution?

Let \( x \) be the ounces of pure alcohol added.

We have the following equations:

\[ 0.2(50) + x = 0.25(50 + x) \] (total alcohol in the mixture)

Solving this equation gives:

\( x = 5 \) ounces of pure alcohol added.

Problem 3: Lidia have 3 liters of juice that is 20% fruit juice and 5 liters of juice that is
30% fruit juice. How many liters of water should you add to get a mixture that is 25%
fruit juice?

Let \( x \) be the liters of water added.

We have the following equations:

\[ 0.2(3) + 0.3(5) = 0.25(3 + 5 + x) \] (total fruit juice in the mixture)

Solving this equation gives:

\( x = 2 \) liters of water added.

 Problem 4:Iya have 10 gallons of gasoline that is 85% pure gasoline and 8 gallons of
gasoline that is 90% pure gasoline. How many gallons of each type should you mix to get
a mixture that is 87% pure gasoline?

Let \( x \) be the gallons of 85% gasoline and \( y \) be the gallons of 90% gasoline.

We have the following equations:

\[ 0.85x + 0.9y = 0.87(10 + 8) \] (total pure gasoline in the mixture)

\[ x + y = 18 \] (total gallons of gasoline)

Solving these equations gives:

\( x = 6 \) gallons of 85% gasoline and \( y = 12 \) gallons of 90% gasoline.

Problem 5:A bartender wants to make 30 liters of a 25% alcohol solution by mixing a
20% alcohol solution with a 30% alcohol solution. How many liters of each solution
should be used?

Solution:

Let x be the liters of 20% solution and y be the liters of 30% solution.

Equation 1: x + y = 30 (total liters)

Equation 2: 0.20x + 0.30y = 0.25(30) (alcohol content)

Solving these equations, you find x = 15 liters and y = 15 liters.

WORK PROBLEM

Problem 1: One pipe can fill a swimming pool in 10 hours, while another pipe

can fill the same pool in 15 hours. How many hours will it take if both pipes are

used to fill the tank?

Problem 2: If a printer can print 80 pages in 4 minutes, how many pages can it print in
20 minutes?

Solution: If the printer prints 80 pages in 4 minutes, it prints 80/4 = 20 pages per minute.
In 20 minutes, it will print 20 pages/minute × 20 minutes = 400 pages.

Problem 3: If it takes 10 workers 8 hours to build a house, how long will it take 5
workers to build the same house?

Solution: More workers mean less time needed, so it’s inversely proportional. Using the
formula Work = Rate × Time, we can find the time. Since the amount of work is the
same, we can write: 10 workers × 8 hours = 5 workers × Time. Solving for Time, we get
Time = (10 × 8) / 5 = 16 hours.

Problem 4: A baker can bake 40 loaves of bread in 2 hours. How many loaves can the
baker bake in 6 hours?

Solution: The baker’s rate of baking is

40÷2=20

=20 loaves per hour. Therefore, in 6 hours, the baker can bake

20×6=120 loaves of bread.

Problem 5: A construction project can be completed by a team of 10 workers in 20

days. If 5 workers leave after 10 days, how many more days will it take to complete the
project?

MOTION PROBLEM

Problem 1: A car travels at a constant speed of 60 miles per hour. How far will it travel
in 3 hours?
Solution**: Distance = Speed × Time = 60 mph × 3 hours = 180 miles.

Problem 2: A train travels at a speed of 50 kilometers per hour. How long will it take to
travel 250 kilometers?

**Solution**: Time = Distance ÷ Speed = 250 km ÷ 50 km/h = 5 hours.

Problem 3: A cyclist travels at a speed of 15 miles per hour. How long will it take to
travel 45 miles?

**Solution**: Time = Distance ÷ Speed = 45 miles ÷ 15 mph = 3 hours.

Problem 4: A boat travels downstream at a speed of 10 kilometers per hour and

upstream at a speed of 6 kilometers per hour. If the river flows at a speed of 2 kilometers
per hour, what is the speed of the boat in still water?

**Solution**: Speed of the boat in still water = (Downstream speed + Upstream speed)
÷2

= (10 km/h + 6 km/h) ÷ 2

= 16 km/h ÷ 2

= 8 km/h.

Problem 5:A plane flies at a speed of 500 miles per hour. How long will it take to travel
2000 miles?

**Solution**: Time = Distance ÷ Speed = 2000 miles ÷ 500 mph = 4 hours.