Mark has fewer than \(9\) coins. Each coin is either a quarter (worth \(\$0.25\)) or a dime (worth \(\$0.10\)). How many dimes does Mark have?
(1) The total value of Mark’s coins is \(\$1.70.\)
(2) Mark has three times as many quarters as he has dimes.
Answer: A
Source: Magoosh
Mark has fewer than \(9\) coins. Each coin is either a quarter (worth \(\$0.25\)) or a dime (worth \(\$0.10\)). How many
This topic has expert replies

 Legendary Member
 Posts: 823
 Joined: 01 Mar 2018
 Followed by:2 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats

 Legendary Member
 Posts: 1974
 Joined: 02 Mar 2018
 Followed by:4 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Total coins < 9
Let total quarter = Q and total dimes = D
Q + D < 9
Target question => How many dimes does Mark have?
Find D
Statement 1 => The total value of Mark's coins is $1.70
$$0.25Q+0.1D=1.70$$
$$\frac{0.1D}{0.1}=\frac{\left(1.700.25Q\right)}{0.1}$$
$$D=\left(\frac{1.700.25Q}{0.1}\right)$$
From the question stem Q+D < 9. Substituting the value of D
$$Q+\left(\frac{1.700.25Q}{0.1}\right)<9$$
$$\frac{0.1Q+1.700.25Q}{0.1}<9$$
$$1.700.15Q<0.9$$
$$0.15Q<0.91.70$$
$$\frac{0.15Q}{0.15}<\frac{0.8}{0.8}$$
$$Q>5.3$$
$$\sin ce\ Q\ as\ a\ coin\ cannot\ be\ a\ fraction\ Q\ge6$$
$$if\ Q=6\ then\ D\ <\ 96=>D<3$$
$$if\ Q=6\ and\ D=2\ then\ 0.25Q+0.1D=1.70$$
$$if\ Q=7\ then\ \ D<97=>D<2$$
$$if\ Q=7\ and\ D=1\ then\ 0.25Q\ +0.1D\ne1.70$$
$$statement\ 1\ is\ SUFFICIENT$$
$$so\ Q=6\ and\ D=2$$
$$statement\ 2\ =>\ Mark\ has\ three\ times\ as\ many\ quarters\ as\ he\ has\ \dim es$$
$$Q=3D$$
$$From\ question\ stem\ Q+D<9\ where\ Q=3D$$
$$3D+D<9$$
$$4D<9$$
$$D<\frac{9}{4}$$
$$D<2.25$$
Since D is a coin, it cannot be a fraction so D can be either 0,1 or 2. The information provided was not enough to arrive at a definite answer hence statement 2 is NOT SUFFICIENT
Since only statement 1 is SUFFICIENT,
Answer = A
Let total quarter = Q and total dimes = D
Q + D < 9
Target question => How many dimes does Mark have?
Find D
Statement 1 => The total value of Mark's coins is $1.70
$$0.25Q+0.1D=1.70$$
$$\frac{0.1D}{0.1}=\frac{\left(1.700.25Q\right)}{0.1}$$
$$D=\left(\frac{1.700.25Q}{0.1}\right)$$
From the question stem Q+D < 9. Substituting the value of D
$$Q+\left(\frac{1.700.25Q}{0.1}\right)<9$$
$$\frac{0.1Q+1.700.25Q}{0.1}<9$$
$$1.700.15Q<0.9$$
$$0.15Q<0.91.70$$
$$\frac{0.15Q}{0.15}<\frac{0.8}{0.8}$$
$$Q>5.3$$
$$\sin ce\ Q\ as\ a\ coin\ cannot\ be\ a\ fraction\ Q\ge6$$
$$if\ Q=6\ then\ D\ <\ 96=>D<3$$
$$if\ Q=6\ and\ D=2\ then\ 0.25Q+0.1D=1.70$$
$$if\ Q=7\ then\ \ D<97=>D<2$$
$$if\ Q=7\ and\ D=1\ then\ 0.25Q\ +0.1D\ne1.70$$
$$statement\ 1\ is\ SUFFICIENT$$
$$so\ Q=6\ and\ D=2$$
$$statement\ 2\ =>\ Mark\ has\ three\ times\ as\ many\ quarters\ as\ he\ has\ \dim es$$
$$Q=3D$$
$$From\ question\ stem\ Q+D<9\ where\ Q=3D$$
$$3D+D<9$$
$$4D<9$$
$$D<\frac{9}{4}$$
$$D<2.25$$
Since D is a coin, it cannot be a fraction so D can be either 0,1 or 2. The information provided was not enough to arrive at a definite answer hence statement 2 is NOT SUFFICIENT
Since only statement 1 is SUFFICIENT,
Answer = A