? about answer key for Math Mammoth

[1GirlTwinBoys]

Does it have the problems worked out with the solution or is it just the answers.

[jlovebaker]

the word problems - there's a little more on those. However, I've found it completely adequate and never have needed more - especially with the examples (we're using 5th grade).

[lisabees]

Hopefully, Maria won't mind I copied this little bit of a 5th grade answer key. Sometimes, it is just the answer, sometimes she includes the problems and describes how she thought it out and sometimes she adds a personal note. Definitely more than adequate.

 

 

 

1. a. 46,795 b. 93,252 c. 33,702 d. 589,245 e. 166,221 f. 82,750

2. a. 14,500 b. 71,400 c. 196,000 d. 89,100 e. 249,000 f. 171,500

3. a. 100,000 b. 6,300 c. 10,000 d. 800.000 e. 960,000 f. 60,000 g. 3,200,000 h. 1,200,00

 

4. 49,307 days.

There are 135 full years. Out of those, every fourth year is a leap year. Since 135 / 4 = 33 R3, this means there would be

33 leap years (the years 1872, 1876, 1880, and so on, till 2004) in this time interval. However, since the year 1900 was not

leap year, actually there have been just 32 leap years. The number of days is therefore 135 × 365 + 32 = 49,307.

5. The exact amount saved in one year is $6240.

The exact amount saved in five years is $31,200.

The error of estimation for one year is $240.

The error of estimation for five years is $1,200.