Math (sat) problem!

Test Preparation
1

<p>Number # Number
of Throws # of people
------------------------------
1 # 7<br>
2 # 6<br>
3 # 6<br>
4 # 4<br>
5 # 2 </p>

<p>in a certain game , each person threw beanbag at a target until the person missed the target .The table shows the results for 25 people who played the game. for example , 4 people hit the target on their first 3 throws and missed on their 4th throw. Based on the information in the table, which of the following must be true?</p>

<pre><code> I.More than half the people hit the target on their first throw.
II. For all of the throws attempted, more hit the target than missed the target.
III. o one hit the target 5 times.
</code></pre>

<p>A)I only.
B)II only.
C)I and III only.
D)II and III only.
E)I , II, and III .</p>

2

<p>Casework.</p>

<p>I. The number of times the target is hit is n-1 where n is the number of throws (because a person stops throwing when he/she misses the target) so a new chart of the number of times the target was actually hit vs. the number of people (derived from this chart would be)</p>

<p>0 # 7
1 # 6
2 # 6
3 # 4
4 # 2</p>

<p>Clearly, 7 is less than half of 25, so I is true.</p>

<p>That eliminates choices B and D and choices C and E both included III. so let’s test III. </p>

<p>If “o one hit the target 5 times” means “only one hit the target 5 times” than III is false because the most times anyone hit the target is 5-1 = 4 times = A,</p>

<p>If “o one hit the target 5 times” means “0 hit the target 5 times” than III is true.</p>

<p>If III is true, than that leaves choices C and E. Testing II, the number of throws attempted is 7(1)+6(2)+6(3)+4(4)+2(5)=7+12+18+16+10= 63. By the new chart above, the sum of the number of times the target was hit is 7(0)+6(1)+6(2)+4(3)+2(4)= 0+6+12+12+8= 38. 38 is more than half of 63, so II is true. Since there is no “I and II” choice, E must be the correct answer.</p>

3

<p>thanx cortana431 that is the correct answer :)</p>

4

<p>Note that in this game </p>

<h1>of misses equals # of people eliminated</h1>

<p>and </p>

<h1>of hits equals # of people left.</h1>

<p>Let’s expand the table.</p>

<p>Throw #…throws…misses…hits
1…25…7…18
2…18…6…12
3…12…6…6
4…6…4…2
5…2…2…0</p>

<p>I.
18 people - more than half - hit the target on their first throws.
I is true.</p>

<p>II.
Total misses 7+6+6+4+2=25
Total hits 18+12+6+2+0=38
More people hit the target than missed it.
II is true.</p>

<p>Out of 5 answers only E includes both I and II.
E is the answer.</p>

5

<p>Let “.” be a person who hit a target, and “!” one who missed it.
The results of the game may be represented by the following diagram:</p>

<p>1 …!!!
2 …!!!
3 …!!!
4 …!!!
5 !!</p>

<p>It’s quite easy to see that
I. more than half of people hit the target on their first throw -> I is true,
and
II. there are exactly 25 misses and many more hits -> II is true.</p>

<p>E is the correct answer.</p>

<p>Just as a bonus: no one hit the target on their 5-th throw -> III is true too.</p>