Question

Two cats, Tom and Bob, caught together 42 mice in three days. Each day, Tom caught twice as many mice as in the previous day, while Bob caught two more mice than in the previous day. Nonetheless, both cats caught the same number of mice for the three days. How many mice did Tom and Bob catch together during the first day?

Answer 1

Answer:

During the first day, they caught 8 mice together.

Step-by-step explanation:

This exercise can be expressed using simple equations:

First, it says that Tom and Bob caught 42 mice in three days, this can be expressed as:

$T + B =42$ (0)

Then we have how many mice caught Tom each day, expressed as:

$T= T1+T2+T3$ (1)

Being T1 the amount of mice Tom caught in the first day, T2 the amount of mice Tom caught in the second day and T3 the amount of mice Tom caught in the third day

"Tom caught twice as many mice as in the previous day" can be expressed by:

$T3= 2T2\\T2=2T1$

Replacing in equation (1)

$T= T1+2T1+2T2=T1+2T1+4T1$

Thus, $T=7T1$ (2)

Then we have how many mice caught Bob:

$B= B1+B2+B3$ (3)

Being B1 the amount of mice Bob caught in the first day, B2 the amount of mice Bob caught in the second day and B3 the amount of mice Bob caught in the third day

"while Bob caught two more mice than in the previous day", can be expressed by:

$B3=B2+2\\B2=B1+2$

Replacing in equation (3)

$B=B1+B1+2+B2+2=B1+B1+2+B1+2+2=3B1+6$

Finally we know that they both caught the same amount of mice during the three days:

$T=B$, Thus replacing in equation (0)

$T+B=T+T=B+B=2T=2B=42$, Thus T=21

Replacing now with (2)

$T=7T1=21\\Thus, \\T1=3$

Using the same

$B=3B1+6=21,\\Thus, \\B1=5$

So the amount of mie caught the first day is:

$T1+B1=3+5=8$

I hope you could follow it!