fill the 7, pour it into the 3, then throw it away.
Repeat. The 7 now holds 1 cup - pour it into the 10.
Fill the 7, pour it into the 3, leaves 4 in the 7.
Add the contents of the 7 to the 10.
Without further information we can't assume it's cylidrical though- it could be a truncated cone shape. We also can't assume we have a ruler. The bhg solution doesn't make these assumptions. Having said that, the solution does assume an adequate supply of liquid is available
Fill the 3, pour it into the 7
repeat
repeat until 7 is full, leaves 2 in the 3.
Pour this into the ten.
fill 3 again and pour into 10.
10 now contains 2 + 3 = 5.
I think you threw people when you stated the beakers can't be emptied out - do you mean you cannot pour from one to the other as offered in the solutions above?
Fill 10 cup
fill 3 cup from 10 cup
pour 3 cup into 7 cup
fill 3 cup from remainder of 10 cup
again pour 3 cup into 7 cup
fill 3 cup again (this leaves 1 cup in 10 cup)
pour from 3 cup to fill 7 cup( this leaves 2 cups in 3 cup. 1 cup in 10 cup and 7 cup full)
pour 7 cup into 10 cup beaker( leaving 7 cup empty 10 cup holding 8 cups and 3 cup holding 2 cups)
pour 2 cups from 3 cup into 7 cup
fill 3 cup from 10 cup leaving 5 cups in 10 cup, 3 cup full, and 7 cup holding 2 cups.