Solving Math textbook exercises 5 Lessons: General practice
1. Solve problem 1 page 176 Math textbook 5
Calculate:
a) \(683 \times 35\); \(1954 \times 425\); \(2438 \times 306\).
b) \(\dfrac{7}{9} \times \dfrac{3}{35}\); \(\dfrac{9}{22} \times 55\); \(\dfrac{11}{17} : \dfrac{33}{34}\).
c) \(36,66 : 7,8\); \(15.7 : 6.28\); \(27.63 : 0.45\).
d) \(16 \, \text{hour} \, 15 \, \text{minute} : 5\); \(14 \, \text{minute} \, 36 \, \text{second} : 12\)
Solution method
– Set the calculation and then calculate according to the rules learned about multiplication or division of decimal numbers.
To multiply two fractions, multiply the numerator by the numerator and the denominator by the denominator.
To divide two fractions, multiply the first fraction by the reverse of the second fraction.
Solution guide
a)
b) \( \dfrac {7}{9} \times \dfrac {3}{35} = \dfrac {7 \times 3}{9 \times 35}\) \(=\dfrac {7 \times 3} {3 \times 3\times 7 \times 5}=\dfrac {1}{15};\)
\( \dfrac{9}{22} \times 55 =\dfrac {9\times 55}{22}\) \(=\dfrac {9 \times 5 \times 11}{2 \times 11}=\dfrac {45}{2} ;\)
\( \dfrac {11}{17} : \dfrac {33}{34} = \dfrac {11}{17} \times \dfrac {34}{33}\) \(=\dfrac {11 \times 34 }{17 \times 33}=\dfrac {11 \times 2 \times 17}{17 \times 3\times 11 }=\dfrac {2}{3}; \)
c)
d)
2. Solve problem 2 page 176 Math textbook 5
Find \(x\):
\(a)\; 0.12 \times x = 6\)
\(b) \;x : 2.5 = 4\)
\(c) \;5,6 : x = 4\)
\(d) \;x \times 0.1 = \dfrac{2}{5}\)
Solution method
Apply the rules:
To find the unknown factor, divide the product by the known factor.
– To find the divided number, we take the merchant with the divisor.
To find the divisor, divide the divisor by the quotient.
Solution guide
a) \(0.12 \times x = 6 \)
\(x = 6 : 0.12\)
\(x = 50\)
b) \(x : 2.5 = 4\)
\(x = 4 \times 2.5\)
\(x = 10\).
c) \(5,6 : x = 4 \)
\(x = 5.6 : 4\)
\(x = 1.4\)
d) \( x \times 0.1 = \dfrac{2}{5}\)
\(x = \dfrac{2}{5}: 0.1 \)
\(x = \dfrac{2}{5} : \dfrac{1}{10}\)
\(x = \dfrac{2}{5} \times \dfrac{10}{1}= \dfrac{20}{5}= 4. \)
3. Solve problem 3 page 176 Math textbook 5
In three days a shop sells \( 2400kg\) sugar. The first day sold \(35\%\) that number, the second day sold \(40\%\) that number. How many kilograms of sugar did the store sell on the third day?
Solution method
Step 1: Calculate the number of lines of goods sold on the first day
As a percentage multiplied by the total number of sugars sold in three days and then divided by \(100\)
Step 2: Calculate the number of selling lines in the second day (same step 1)
Step 3: Calculate the number of lines the store sells on Tuesday
Solution guide
The number of streets the store sold on the first day was:
\(2400 \times 35 : 100 = 840 \, (kg)\)
The number of streets the store sold on the second day was:
\(2400 \times 40 : 100 = 960 \, (kg)\)
The number of streets the store sold on the third day was:
\(2400 – (840 + 960) = 600 \, (kg)\)
Answer: \(600kg\) sugar.
4. Solve problem 4 page 176 Math textbook 5
A shop selling fruit (fruit) earns \(1 \, 800\, 000\) VND. Calculate the amount of profit equal to \(20\%\) the purchase amount. How much is the capital to buy that amount of fruit?
Solution method
Step 1: If you consider the purchase amount as \(100\%\), find the percentage of the sale amount to the purchase amount
Step 2: Calculate the amount of capital spent to buy fruit
By the amount of sales divided by \(120\%\) (or divided by \(120\) and then multiplied by \(100\))
Solution guide
Consider the purchase amount as \(100\%\).
The ratio of the sales amount to the purchase amount is:
\(100 \% + 20\% = 120\% \) (purchase amount)
The capital amount to buy that amount of fruit is:
\(1 \,800 \,000 : 120 \times 100 = 1 \, 500\, 000\) (VND)
Answer:\( 1 \, 500\, 000\) (VND)
.
=============