Solving Math textbook exercises 5 Lessons: General practice
1. Solve problem 1 page 175 Math textbook 5
Calculate:
a) \(85793 – 36841 + 3826 \)
b) \(\dfrac{84}{100} – \dfrac{29}{100} + \dfrac{30}{100}\)
c) \(325.97 + 86.54 + 103.46\)
Solution method
Expressions that have only addition and subtraction are calculated from left to right, respectively.
Solution guide
a) \(85793 – 36841 + 3826 \)
\(= 48952 + 3826 = 52778\)
b) \(\dfrac{84}{100} – \dfrac{29}{100}+ \dfrac{30}{100} \)
\(=\dfrac{55}{100} + \dfrac{30}{100} =\dfrac{85}{100} = \dfrac{17}{20}\)
c) \(325.97 + 86.54 + 103.46\)
\(= 412.51 + 103.46 = 515.97\).
2. Solve problem 2 page 175 Math textbook 5
Find \(x\):
\(a) \;x + 3.5 = 4.72 + 2.28\)
\( b) \;x – 7.2 = 3.9 + 2, 5\)
Solution method
– Calculate the value of the right hand side.
– Find \(x\) based on learned rules:
To find the unknown term, subtract the known term from the sum.
+ To find the number to be subtracted, we take the difference plus the minus number.
Solution guide
a) \(x + 3.5 = 4.72 + 2.28 \)
\(x + 3.5 = 7\)
\(x = 7 – 3.5 \)
\(x = 3.5 \)
b) \(x – 7.2 = 3.9 + 2, 5\)
\(x – 7.2= 6.4\)
\( x = 6.4 + 7.2\)
\(x = 13.6\)
3. Solve problems 3 pages 175 Math textbook 5
A trapezoidal plot has a small base of \(150m\), a large base equal to \(\dfrac{5}{3}\) a small base, and a height equal to \(\dfrac{2}{5}\) a large base . Ask the area of the land is how many square meters, how many hectares?
Solution method
– Calculate big bottom = small bottom \(\times \,\dfrac{5}{3}\).
– Calculate height = big bottom \(\times \,\dfrac{2}{5}\).
– Calculate area = (big bottom \(+\) small bottom) \(\times \) height \(:2\).
– Convert the area measurement to hectare, noting that \(1ha =10000m^2\).
Solution guide
The large base of a trapezoidal plot is:
\(150 × \dfrac{5}{3} = 250\;(m)\)
The height of the trapezoidal plot is:
\(250 × \dfrac{2}{5} = 100\;(m)\)
The area of a trapezoidal plot is:
\(\dfrac{(250 + 150) \times 100}{2} = 20000\;(m^2)\)
\(20000m^2 = 2ha\)
Answer: \(20000m^2\) ; \(2ha\).
4. Solve problems 4 pages 175 Math textbook 5
At 6 o’clock, a freight car leaves from A at a speed of 45 km/hr. At 8 o’clock, a tourist car also goes from A at a speed of 60km/h and goes in the same direction as the cargo car. What time does the tourist car catch up with the cargo car?
Solution method
Step 1: Calculate the time the cargo box goes ahead of the tourist car
Step 2: Calculate the distance traveled by the freight car in front of the tourist car
Step 3: Calculate the distance after each hour travel car near the truck (speed difference)
Step 4: Calculate the time it takes the passenger car to catch up with the freight car (take the distance of the freight car ahead and divide it by the speed difference)
Step 5: Calculate the time when the tourist car catches up with the cargo car
Solution guide
The time taken by the freight car to go ahead of the tourist car is:
\(8 – 6 = 2\) (hour)
The distance traveled by the car in \(2\) is now:
\(45 \times 2 = 90\) (km)
After every hour the tourist car approaches the freight car is:
\(60 – 45 = 15\) (km)
The time it takes the traveling car to catch up with the freight car is:
\(90 : 15 = 6\) (hour)
Tourist cars catch up with cargo cars at:
\(8 + 6 = 14\) (hour)
Answer: \(14\) hours.
5. Solve problem 5 page 175 Math textbook 5
Find the appropriate natural number of \(x\) such that:
\(\dfrac{4}{x} = \dfrac{1}{5}\)
Solution method
Apply the basic property of fractions: If you multiply both the numerator and the denominator of a fraction by the same other natural number \(0\), you get a fraction equal to the given fraction.
Solution guide
Method 1:
We have: \(\dfrac{1}{5}= \dfrac{1 \times 4}{5 \times 4} = \dfrac{4}{20}\)
Hence: \(\dfrac{4}{x}= \dfrac{4}{20}\).
Infer: \(x = 20\) (If two equal fractions have the same numerator, the denominator is also the same).
Method 2:
\(\dfrac{4}{x}= \dfrac{1}{5}\)
or \(4 : x= \dfrac{1}{5}\)
\(x = 4 : \dfrac{1}{5}\)
\(x = 20\)
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