hardest word problems solutions

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1. (18) Our goal is to find how old Ann is now. We are

given that Mary is 24 years old. Therefore we have 24nowMary =

Now we are given that Mary is twice as old as Ann was when Mary was as old as Ann is now. There are four ages for both Mary and Ann that we need to consider.

before before

now now

Mary Ann

Mary Ann

= =

= =

We already have one age, and we are looking for another.

24 ?before before

now now

Mary Ann

Mary Ann

= =

= =

Now we are told the Mary ( )Mary is twice as old as

2Mary⎛ ⎞=⎜ ⎟

⎝ ⎠ Ann was

2 beforeMary Ann⎛ ⎞=⎜ ⎟

⎝ ⎠, when

Mary ( )Mary was as old as ( )Mary = Ann is now

( )nowMary Ann= . We have

2

2

before

nowbefore

Mary Ann

Ann Ann

⎛ ⎞=⎜ ⎟⎝ ⎠⎛ ⎞=⎜ ⎟⎝ ⎠

Alternate explanation: The problem is stating that Mary is twice as old as Ann was when Mary was Ann’s age. In other words, when Mary was Ann’s age, Ann was 12. Since the difference in age between Mary and Ann will always be the same, Ann must be 18. So now Mary is 24 and Ann is 18. When Mary was Ann’s age (18), Ann was 12. 2. ( 19 ) Our goal is to find how many members are in set Y.

Now we are given that sets X, Y and Z have 4 members in common. We can use a Venn-diagram to display this.

Now we are told that sets Y and Z have a total of 8 members in common. Note that since 4 are common in

all three sets and 8 total have sets Y and Z in common, there are 4 left that ONLY have sets Y and Z in common.

Now we are also given that each member of set Y is contained in at least one of the other two sets. So that means there is nobody who is ONLY in set Y. So we have

Now we are also given that sets X and Y are a total of 15 members in common. Since 4 are common in all three sets, then there are 11 people (15-4=11) who are ONLY in sets X and Y. So we have

Since we are looking for the the total number of members in set Y, we can add all four components in the Y circle above. So we have

11 4 4 019

YY= + + +=

The answer is 19.

4

X

Z

Y

4

0 11

4

X

Z

Y

4

0

4

X

Z

Y

4

4

X

Z

Y

COMPLETE SOLUTIONS TO 8 Hardest SAT Word Problems Ever

8 HARDEST SAT WORD PROBLEMS Version 1.1


3. (B) Our goal is to find how far the fly has traveled. Now we have two trucks 20 miles apart that head towards each other at a rate of 10mph.

Also we also have a fly that is sitting on the hood of the first truck and flies at 15mph towards the second truck. Once it reaches the second truck, it turns and flies toward the first truck. This continues until the two truck meet. Note that the fly will travel at a rate of 15mph throughout the described event, regardless of the direction it is traveling. Now find the time it will take for the two trucks to meet. In one hour, the first truck will travel 10 miles, and the second truck will travel 10 miles also.

1

2

10 *1 10

10 *1 10

milestruck hour mileshour

milestruck hour mileshour

= =

= =

Now on our diagram, in one hour, both trucks will travel 10 miles. Since the total distance between them is 20 miles, the two trucks will meet in one hour.

During this one hour the fly will be flying at 15mph constantly. Note the direction will change when he reaches the trucks, but the speed does not change. Therefore, since the two trucks will meet in one hour, the fly will be traveling also for one hour only. So the distance that the fly will cover is as follows

15 *1

15

milesDistance hourhour

Distance miles

=

=

So the distance the fly travels is 15 miles. Note that the direction changes but the distance traveled is the same regardless. The answer is (B).

4. ( 207 ) Our goal is to find the greatest possible value of one of the integers. We are given that the average of three different positive integers is 70. Since the integers are positive, the integers must be greater than zero. Note that if we want one value to be large, the other two values must be as small as possible. Now the two smallest

positive integers at 1 and 2. Note that we cannot use any number twice because all three numbers must be different. Suppose that the largest value of one of the numbers is x. Since the average of 1, 2 and x is 70, we have

1 2 7033 210

207

x

xx

+ +=

+ ==

The greatest possible value of x is 207. 5. (B) Our goal is to find how many hours it will take to

weed the garden. We are given that John and Paul each

weed 14

of their garden in an hour, and their father can

weed 13

of the garden in an hour. Note that they are all

working together on the garden. Since the rates of each of the three people are given in one hour, we can according

add the rates together. So in one hour, John will do 14

,

Paul will do 14

, and the father will do 13

of garden. So

total, in one hour we will have done 1 1 1 3 3 4 10 54 4 3 12 12 12 12 6+ + = + + = =

of the garden. Now since 5/6 of the garden will be done in one hour, and we are looking for the time, t, it would take to do one full garden, we can setup the following equation

( ) ( )( )

( ) ( )( )

( )

516

15 1 16

1 156

156

65

garden gardenhour t hours

garden t hours garden hour

garden hourt hours

garden

hourt hours

t hours hours

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

=⎛ ⎞⎜ ⎟⎝ ⎠

=

=

So the time it would take all three t work on the garden is 6/5 (or 1.2) hours. The answer is (B).

10mph 10mph

20 miles

10 miles 10 miles

10mph 10mph

20 miles

8 HARDEST SAT WORD PROBLEMS Version 1.1


6. () Our goal is to find what percent his final weight was of his initial weight. We are given that Joe’s weight first increased by 10 percent and then his new weight decreased by 25 percent. Suppose that Joe’s initial weight is x. Now his weight first increased by 10 percent. Note that 10 percent is simply the same as “.1”. So we have

( )

( )( )

10% .1*1 .1

1 .1

1.11.1

x x x xx x

x

xx

+ = +

= +

= +

=

=

Next we are told that his weight decreased by 25 percent. So we have

( )

( )( )

1.1 25% 1.1 1.1 .25*(1.1 )1 .275

1 .275

1.2751.275

x x x xx x

x

xx

− = +

= +

= +

=

=

So this final weight is 1.275x. Remember that we

hardest word problems solutions

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