, 2013

Pulling out two of each type of bill gives Rad, $2\times(\$5+\$10+\$20+\$50)=2\times(\$85)=\$170$.

Since his total sum of money is $175, the only other bill that Rad pulled out must have a value of $\$175-\$170=\$5$.

That is, Rad pulled out three $5 bills, two $10 bills, two $20 bills, and two $50 bills, for a total of $3+2+2+2=9$ bills.
Sandy pulls out at least one of each type of bill and so she must have at least $\$5+\$10+\$20+\$50=\$85$.

Thus, we know four of the five bills that Sandy pulls out and that these four bills total $85. The fifth bill that Sandy pulls could be any one of the four different types of bills.

If this fifth bill is a $5 bill, then Sandy’s total sum of money is $\$85+\$5=\$90$.

If this fifth bill is a $10 bill, then Sandy’s total sum of money is $\$85+\$10=\$95$.

If this fifth bill is a $20 bill, then Sandy’s total sum of money is $\$85+\$20=\$105$.

Finally, if this fifth bill is a $50 bill, then Sandy’s total sum of money is $\$85+\$50=\$135$.

Therefore, the sums of money that Sandy could have are $90, $95, $105, and $135.
Lino could have at most three $50 bills since four $50 bills exceeds his total sum of money ($\text{4 50= 200 gt; 160}$).

If Lino had no $50 bills, then the bills (6 bills at most) each have value at most $20 and would total $\$160$.

However, this is not possible since $6\times\$20=\$120$ which is less than the required $160.

If Lino had one $50 bill, then the remaining bills (5 bills at most) would total $\$160-\$50=\$110$.

However, this is not possible since the largest denomination of the remaining bills is $20 and $5\times\$20=\$100$ which is less than the required $110.

Therefore, we proceed by considering the cases where Lino has two or three $50 bills.

These two cases are summarized in the table below.

In each case from the table above, Lino has a total sum of $160 and has pulled out 6 or fewer bills.

Since we are given that there are only four possibilities, then we have found them all.