PSEG Nuclear Salem County Math Showcase
2014 6th Grade
Written Challenge
Test Number «testnum»
Date:
29 May, 2013
Time:
30 minutes
Answer Key
Page 1 of 11
05/16/2014
Question 1 ( 10 points ):
If half the people in a room leave at the end of every five-minute interval and at
the end of twenty minutes the next to last person leaves, how many people were
in the room to start with? (Assume that no one enters the room once the process
begins.)
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Solution:
At the end of each five-minute interval, half the number of people in the room leave and
the other half remain. At the end of twenty minutes, this step of half leaving and half
remaining involved only 2 people, 1 leaving and 1remaining. This means that during the
preceding five-minute interval (the interval from minute 15 to minute 20), there were 2
people in the room.
At the end of minute 15, the step of half leaving and half remaining must have involved 4
people, 2 who left and 2 who remained.
By the same reasoning, at the end of minute 10 the step involved 8 people, and at the end
of minute 5 it involved 16 people. So we started with 16.
Place Answer Here
16 people
Page 2 of 11
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Question 2 ( 10 points ):
A geologist has 10 rocks of exactly equal weight. If 6 rocks and a 10-ounce weight
balance on one side of a scale with 4 rocks and a 22-ounce weight on the other,
what is the weight, in ounces, of one of these rocks?
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Solution:
Let R equal the weight of each rock, in ounces. The weight on one side of the
scale is 6R + 10 ounces and the weight on the other side is 4R + 22 ounces.
Since the two weights balance, you can write the equation
6R + 10 = 4R + 22.
Solving:
2R = 12
R = 6
Place Answer Here
1 rock weighs 6 ounces
Page 3 of 11
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Question 3 ( 10 points ):
Every student who studies art in a certain school receives exactly one of the
grades A, B, C, or D.
If one-fifth of the students receive A’s, one-quarter receive B’s, one-half receive
C’s, and 10 students receive D’s, how many students in the school study art?
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Solution:
The students who receive A’s, B’s, and C’s account for 1/5 + 1/4 + 1/2 of the
students:
1/5 + 1/4 + 1/2 = 4/20 + 5/20 + 10/20 = 19/20
This leaves 1/20 of the students receiving D’s.
Since 10 students receive D’s, 10 is 1/20 of the total.
Then the total number of students is 10 *20 = 200
Place Answer Here
200 total students study art
Page 4 of 11
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Question 4 ( 10 points ):
Insert mathematical notations in between the following numbers to make the equation
true.
You may use the following symbols: + - * / ( )
7 7 7 7=1
7 7 7 7=2
7 7 7 7 = 48
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Solution:
here is one possible solution, there may be others.
(7 + 7) / (7 + 7) = 1
(7 / 7) + (7 / 7) = 2
(7 X 7) - (7 / 7) = 48
Page 5 of 11
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Question 5 ( 10 points ):
In an 8-gram solution of water and alcohol, the ratio by mass of water to alcohol
is 3 to 1.
If 12 grams of a solution consisting of 2 grams water for each gram of alcohol is
added to the 8-gram solution, what fraction by mass of the new solution is
alcohol?
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Solution:
To find what fraction by mass of the new solution is alcohol, you can find how
many grams of alcohol are in the final solution of 20 grams.
Since the 8-gram solution is 1 part alcohol to 3 parts water, it is 1/4 alcohol. So it
contains 2 grams of alcohol.
Similarly, the 12-gram solution is 1/3 alcohol by mass, so it contains 4 grams of
alcohol.
Thus, the final 20-gram solution contains 6 grams of alcohol.
So 6/20 = 3/10 of it is alcohol.
Place Answer Here
3/10 of it is alcohol
Page 6 of 11
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Question 6 ( 10 points ):
At Central High School, the math club has 15 members and the chess club has 12
members.
If a total of 13 students belong to only one of the two clubs, how many students
belong to both clubs?
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Solution:
Let n stand for the number of students who belong to both clubs.
The 15 members of the math club can be broken down into two groups: those who
are in both clubs (there are n students in this category) and those who are in the
math club only (there are 15-n students in this category).
The 12 members of the chess club can also be broken down into two groups: n
students who are in both clubs and 12-n students who are in the chess club only.
Since a total of 13 students belong to only one of the two clubs,
you know that (15-n) + (12-n) = 13.
Solving this equation gives n = 7.
Place Answer Here
7students belong to both clubs
Page 7 of 11
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Question 7 ( 10 points ):
In a stationary store, pencils have one price and pens have another price. Two
pencils and three pens cost 78 cents. But three pencils and two pens cost 72 cents.
How much does one pencil cost?
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Solution #1:
If we add both lines of information we find that 5 pencils and 5 pens cost 150
cents
Then, one pencil and one pen must cost 30 cents.
And two pencils and two pens must cost 60 cents.
Since we know that three pencils and two pens cost 72 cents, we can conclude
that one pencil costs 12 cents
Solution #2:
let C = pencils and N = pens, then:
2C + 3N = 78 => multiply by 2 = > 4C + 6N = 156
3C + 2N = 72 => multiply by 3 = > 9C + 6N = 216
subtract the two equations:
5C
= 60
C
= 12
Place Answer Here
12 cents
Page 8 of 11
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Question 8 ( 10 points ):
There are many numbers that divide 109 with a remainder of 4.
List all two-digit numbers that have that property.
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Solution:
If 4 is subtracted from 109 the result is 105.
Then each of the two-digit numbers that will divide 109 with a remainder of 4 will
also divide 105 evenly.
Thus we need to find all two-digit factors of 105.
Factors of 105 are: 3, 5, 7, 15, 21, 35
Place Answer Here
15, 21, 35
Page 9 of 11
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Question 9 ( 10 points ):
The owner of a bicycle store has both bicycles (two wheels) and tricycles (three
wheels) on sale. Bicycles and tricycles each have only two pedals.
When he counted the total number of pedals, he got 50.
When he counted the total number of wheels, he got 64.
How many tricycles were on sale? How many bicycles were on sale?
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Solution #1:
Since the total number of pedals = 50 and there are 2 pedals per cycle, then there
are 25 bicycles and tricycles total.
If there were only bicycles, then there would only be 50 wheels instead of 64, so
there are 14 extra wheels for tricycles. Therefore there are 14 tricycles.
Since there are only 25 total then 25 – 14 = 11 bicycles on sale.
Solution #2:
Let B = number of bicycles and T = number of tricycles.
Then,
2B + 2T = 50 since there are 2 pedals per bike and 50 pedals total
Also,
2B + 3T = 64 since there are 2 wheels on bicycles, 3 on tricycles and 64 total
Subtracting,
2B + 3T = 64
2B + 2T = 50
T = 14
Solving for B,
2B + 2(14) = 50
2B = 22
B = 11
Place Answer Here
14 tricycles, 11 bicycles
Page 10 of 11
05/16/2014
Question 10 ( 10 points ):
A 4 inch by 4 inch wooden cube is painted completely red on all 6 sides. The
painted cube is then cut into 1 inch by 1 inch wooden cubes.
How many of the 1 inch by 1 inch wooden cubes do not have any red paint on
them?
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Solution:
Each and every of the outer 1 inch by 1 inch wooden cubes have red paint on
them.
It is only the inner 1 inch by 1 inch wooden cubes that do not have red paint.
If we remove the outer layer of 1 inch by 1 inch wooden cubes we will have a 2
inch by 2 inch wooden cube.
A 2 inch by 2 inch wooden cube will contain eight 1 inch by 1 inch wooden cubes.
Place Answer Here
8 1 inch by 1 inch wooden cubes
Page 11 of 11
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