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You are here: Home / QA Grade 5 Math / Solving Math textbook exercises 5 Lessons: Practice

Solving Math textbook exercises 5 Lessons: Practice

30/07/2021 //  by admin//  Leave a Comment

Solving Math textbook exercises 5 Lessons: Practice

1. Solve problem 1 page 171 Math textbook 5

In the figure below, the area of ​​quadrilateral ABED is larger than the area of ​​triangle BEC by \(13.6cm^2\). Calculate the area of ​​quadrilateral ABCD, given that the ratio of the area of ​​triangle BEC and the area of ​​quadrilateral ABED is \(\dfrac{2}{3}\).

Solution method

– Find the area of ​​triangle BEC and the area of ​​quadrilateral ABED mathematically to find two numbers when the difference and ratio are known.

(draw a diagram showing the area of ​​triangle BEC with 2 parts and the area of ​​quadrilateral ABED with 3 such parts)

– Area of ​​quadrilateral ABCD \(=\) area of ​​triangle BEC \(+\) area of ​​quadrilateral ABED.

Solution guide

According to the topic, we have a diagram:

According to the diagram, the equal part difference is:

\(3 – 2 = 1\) (part)

The area of ​​triangle BEC is:

\(13.6 : 1 × 2 = 27.2\;(cm^2)\)

The area of ​​quadrilateral ABED is:

\(27.2 + 13.6 = 40.8\;(cm^2)\)

The area of ​​quadrilateral ABCD is:

\(40.8 + 27.2 = 68\;(cm^2)\)

Answer: \(68cm^2 \).

2. Solve problem 2 page 171 Math textbook 5

Class \(5A\) has \(35\) students. The number of male students is equal to \(\dfrac{3}{4}\) the number of female students. How many more female students than male students?

Solution method

Step 1: Draw a diagram to represent the problem

Step 2: Calculate the number of boys and girls in class 5A

Step 3: Calculate the number of female students more than male students

Solution guide

According to the topic, we have a diagram:

According to the diagram, the total number of equal parts is:

\(3 + 4 = 7\) (part)

The number of boys in class \(5A\) is:

\(35 : 7 × 3 = 15\) (student)

The number of girls in class \(5A\) is:

\(35 – 15 = 20\) (student)

The number of female students more than the number of male students is:

\(20 – 15 = 5\) (student)

Answer: \(5\) students.

3. Solve problem 3 page 171 Math textbook 5

A car traveling \(100km\) consumes \(12l\) of gasoline. How many liters of gasoline did that car have traveled \(75km\)?

Solution method

Step 1: Calculate the amount of gasoline consumed when the car travels the distance \(1km\)

Step 2: Calculate the amount of gasoline consumed when the car travels the distance \(75km\)

Solution guide

The car goes \(1km\) then it consumes all:

\(12 : 100 = 0.12\) (liter)

The car goes \(75 km\) then it consumes all:

\(0.12 \times 75 = 9\) (liters)

Answer: \(9\) liters of gasoline

4. Solve problem 4 page 171 Math textbook 5

The figure on the right is a chart showing the academic performance rating of students in the \(5\) grade of Victory Primary School. Calculate the number of students in each category, knowing that the number of students with good academic standing is \(120\) students.

Solution method

Step 1: Calculate the percentage of good students in the block \(5\)

Step 2: Calculate the number of students in grade \(5\) by dividing \(120\) by \(60\%\)

Step 3: Calculate the number of students who are good and the average number of students in the block \(5\)

Solution guide

The percentage of students with good grades \(5\) of Thang Loi school is:

\(100\% – (25\% + 15\%) = 60\%\)

The school’s total number of students \(5\) is:

\(120 : 60 \times 100 = 200\) (student)

The number of excellent students in block \(5\) of the school is:

\(200 \times 25 : 100 = 50\) (student)

The average number of students in class \(5\) of the school is:

\(200 \times 15 : 100 = 30\) (student)

Answer: Good: \(50\) students; Good: \(120\) students; Average: \(30\) students.

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Category: QA Grade 5 MathTag: 5 Maths Textbook Prize, 5th Grade Math Chapter 5, Math 5, Solve Maths 5

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