Download Spring Level 4 | 08/21/2010

Spring Level 4 | 08/21/2010
(1)
Tomas enters a classroom at exactly 9:00 a.m. and notices that the 12-hour
analog clock on the wall is behaving very strangely. The clock reads 4:20 and the second
hand is racing very fast. In fact, the second hand makes one complete circle every 4
seconds. The minute hand and hour hand continue to behave as if every full rotation of the
second hand indicates that a minute has passed. When Tomas leaves the class at 9:50
a.m., what time will the clock on the wall read?
(2)
Three congruent rectangular banners are sewn so that each banner consists
of three adjacent, congruent, vertical stripes-a middle white stripe bordered on each side by
a gray stripe. The three banners are displayed so that each overlaps the other by half their
width and half their length. Find the ratio of the visible white areas to the visible gray
areas. Express your answer as a common fraction.
(3)
The greatest number named by the ancient Greek mathematicians was a
myriad which was the word for ten-thousand. Write a myriad of myriads as a power of 10.
(4)
Express the slope of the line determined by the equation 4y
common fraction.
(5)
In a magic square, numbers are arranged in an n n array so that the sum
of the numbers listed in each row and column is the same. This sum is called the "magic
sum". Use the nine numbers -10, -8, -6, -4, -2, 0, 2, 4 and 6 to form a 3 3 magic square.
What is the magic sum for this magic square?
3x = 10 as a
(6)
Steve's bathtub, when full, will drain in 12 minutes. The hot water tap takes
6 minutes to ll the tub, and the cold water tap takes 4 minutes to ll the tub. If Steve
opens both taps but forgets to put in the plug, in how many minutes will the tub be lled?
(7)
If a phonograph record makes 33 revolutions per minute, how many
revolutions does it make in playing a record 3 minutes and 36 seconds long?
(8)
(9)
1
3
decimal.
Find the value of
1
100
+
2
100
+
3
100
+ +
100
100
. Express your answer as a
The areas of three faces of a rectangular prism are 55, 77, and 35 square
units. Find the number of cubic units in the volume of the prism.
(10)
The volume of a given sphere is 36 cubic inches. How many square inches
are in its surface area? Express your answer in terms of .
(11)
A particular geometric sequence has strictly decreasing terms. After the rst
term, each successive term is calculated by multiplying the previous term by m . If the rst
term of the sequence is positive, how many possible integer values are there for m?
7
(12)
What is the 101st term of the sequence 1; 2; 2; 3; 3; 3; :::; in which each
positive integer n occurs in blocks of n terms?
(13)
When a number x is divided by 8, the remainder is 5. What is the remainder
when x is divided by 4?
(14)
An auditorium with 30 rows of seats has 10 seats in the rst row. Each
successive row has one more seat than the previous row. If students taking an exam are
permitted to sit in any row, but not next to another student in that row, what is the
maximum number of students that can be seated for an exam?
(15)
A golf resort oers the following two special packages. The Par 3 Package
includes four nights of lodging and three rounds of golf for $443.50. The Par 5 Package
includes seven nights of lodging and ve rounds of golf for $765.50. The rate for each
night of lodging and for each round of golf are the same in both packages. How many
dollars does one night's lodging cost?
(16)
A non-zero fraction's denominator is increased by 1 and its numerator is
decreased by 50%. If the resulting fraction is of the original fraction, what is the original
fraction's denominator?
1
3
(17)
Tamyra is making four cookies and has exactly four chocolate chips. If she
distributes the chips randomly into the four cookies, what is the probability that there are
no more than two chips in any one cookie? Express your answer as a common fraction.
(18)
Given 2a = 32 and ab = 125 nd ba .
(19)
How many even divisors does 7! have?
(20)
How many ordered triples of three prime numbers exist for which the sum of
the members of the triple is 24?
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Answer Sheet
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Answer
4:50
1/2
10
8
3
4
-6
3
120 revolutions
50.5
385
36
6
14
1
375 students
79
2
51/64
243
48 even divisors
15
Problem ID
0052
513C
3B1B
304A1
10AD
3BDD
C15D1
54041
12DD
52DD
1023
5B311
B4211
32A4
31BA
AA3
A5D1
4B4D1
5B322
12C31
Copyright MATHCOUNTS Inc. All rights reserved
Solutions
(1)
4:50
ID: [0052]
Since the second hand would normally take 60 seconds to make a full revolution, the clock
is running = 15 times faster than normal. Therefore, in 50 minutes the clock advances
50 15 = 750 minutes. Divide 750 by 60 to nd that 750 minutes is 12 hours and 30
minutes. Twelve hours and 30 minutes after 4:20 is 4:50 .
60
4
(2)
1/2
ID: [513C]
No solution is available at this time.
(3) 10
ID: [3B1B]
No solution is available at this time.
8
(4)
3
4
ID: [304A1]
No solution is available at this time.
(5)
-6
(6)
3
(7)
120 revolutions
(8)
50.5
ID: [10AD]
No solution is available at this time.
ID: [3BDD]
No solution is available at this time.
ID: [C15D1]
1 revolutions
as
Working with mixed numbers is never pleasant, so we rewrite 33
3 minute
100 revolutions
36 3
. Since there are 60 seconds in a minute, 36 seconds is
= minutes
3 minutes
60 5
3 18
and we can rewrite 3 minutes 36 seconds as 3 =
minutes. Therefore our answer is
5 5
18
100 revolutions
going to be given by
minutes = 120 revolutions.
5
3 minutes
ID: [54041]
No solution is available at this time.
(9)
385
ID: [12DD]
No solution is available at this time.
(10) 36
ID: [52DD]
No solution is available at this time.
(11)
6
ID: [1023]
Since the geometric sequence is strictly decreasing, the common ratio m=7 must be a
positive number between 0 or 1. For, if it were greater than 1, the sequence would keep
increasing, since the rst term is positive. If the ratio was 0, then the sequence would
consist of 0's after the rst term and would not be strictly decreasing. Finally, if the ratio
was negative, the sequence would alternate between positive and negative terms, and thus
would not be decreasing. So we have 0 < m < 1, or 0 < m < 7. There are 6 possible
integer values of m: 1 , 2, 3, 4, 5, 6.
7
(12)
14
(13)
1
ID: [5B311]
No solution is available at this time.
ID: [B4211]
No solution is available at this time.
(14)
375 students
ID: [32A4]
If n is even, then at most n=2 students may sit in a row with n seats. If n is odd, then at
most (n + 1)=2 students may sit in a row with n seats. So 10=2 = 5 students can sit in the
rst row, 6 students can sit in the second row, 6 students can sit in the third row, 7
students in the fourth row, 7 students in the fth row, and so on. The last row has
10 + 29 = 39 seats, so the total number of students who can sit in the room is
5 + 6 + 6 + 7 + 7 + + 19 + 19 + 20
= 5 + 20 + 2 (6 + 7 + + 19)
(6 + 19)(14)
= 25 + 2 2
= 25 + 25 14
= 25 15
= 375 ;
where we have used the formula
(rst term + last term)(number of terms)=2
for the sum of an arithmetic series.
(15)
79
(16)
2
(17)
51/64
(18)
243
ID: [31BA]
No solution is available at this time.
ID: [AA3]
No solution is available at this time.
ID: [A5D1]
No solution is available at this time.
ID: [4B4D1]
We note that 32 = 2 2 2 2 2 = 2 , so a = 5. This leaves us with
5b = 125 = 5 5 5 = 5 , which means that b = 3. Therefore our answer is ba = 3 = 243 .
5
3
5
(19)
48 even divisors
ID: [5B322]
Count the number of even divisors of 7! by counting the number of ways to form the prime
factorization of an even divisor of 7!. Suppose that 7! is divisible by an even positive
integer r . Since the prime factorization of 7! is 7 (2 3) 5 (2 2) 3 2 = 2 3 5 7, the
prime factorization of r does not include any primes other than 2, 3, 5, and 7. Express r in
terms of its prime factorization as 2a 3b 5c 7d . Then 7!=r = 2 a 3 b 5 c 7 d . Since 7!=r is
an integer, d must equal 0 or 1, c must equal 0 or 1, and b must equal 0, 1 or 2. Finally, a
may be no larger than 4, but it must be at least 1 since r is even. Altogether there are
2 2 3 4 = 48 total possibilities for the four exponents a, b, c , and d , and hence 48 even
divisors.
4
4
(20)
15
2
1
2
1
ID: [12C31]
No solution is available at this time.
Copyright MATHCOUNTS Inc. All rights reserved