### MC Review - Kilmer Math Counts Club, 2011

```Math Counts Test Review
2010 Chapter
7. A copy of the Sunday Times costs \$0.95 more
than a copy of the Sunday News. Ms. Jones buys
both papers for \$7.25. How much does the Sunday
News cost?
Let X be the cost of Sunday News
X + (X + 0.95) = 7.25
2 X = 7.25 – 0.95 = 6.3
X = 3.15
11. Joshua has taken four tests. His median score is
80 points and the difference between his greatest
score and his least score is 12 points. What is the
maximum possible value of the mean of his
scores?
Let A,B, C, D be the test scores. A  B  C  D
(B + C)/2 = 80, D – A = 12, C  D & A  B
Largest possible D = 80 + 12 = 92 with A = 80
Largest mean = (80*2+92+80)/4
= 40 + 23 + 20 = 83
17. A grandfather clock chimes at the start of each
hour. Beginning precisely at six o’clock the
grandfather clock chimes 6 times with the final
chime beginning exactly 10 seconds after six
o’clock. If the chimes are uniformly spaced, how
long after 12 o’clock does the twelfth chime begin?
Time in between chimes: 10/5 = 2 seconds
At 12 o’clock, there are 12 chimes
The time before the 12th chime: 11*2 = 22
18. A bag contains exactly six balls; two are red
and four are green. Sam randomly selects one of
the six balls and puts it on the table. Then he
randomly selects one of the five remaining balls.
What is the probability that the two selected balls
common fraction.
Total # of possible selection: 6 * 5
# of desired outcome: 2 (r) x 4 (g) + 4(g) x 2(r)
Probability = (8 + 8) / 30 = 16/30 = 8/15
Alternatively: P(r)(g) + P(g)(r)
= 2/6 * 4/5 + 4/6 * 2/5 = 8/15
23. In rectangle ABCD, side AB measures 6 units and
side BC measures 3 units, as shown. Points F and G
are on side CD with segment DF measuring 1 unit
and segment GC measuring 2 units, and lines AF and
BG intersect at E. What is the area of triangle AEB?
Note that EFG  EAB
FG = 6 - 1 - 2 = 3, AB = 6
Let H be the height of EFG
H / (H + 3) = 3/6 = ½
2H = H + 3  H = 3
Area of EAB = 6 *( H + 3)/2 = 18
27. If f(x) = 5x – 3, g(x) = 3x2 + 1 and h(x) = f(x) + g(x),
what is the sum of the x-values for which h(x) = 0?
h(x) = 3x2 + 1 + 5x – 3 = 3x2 + 5x – 2 = 0
Let X1, X2 be the roots, by sum of the root,
We have: X1 + X2 = -5/3
28. Two real numbers have an average of 7. The
average of their squares is 54. What is the product of
the two numbers?
(X + Y) / 2 = 7  X + Y = 14 ----- (1)
(X2 + Y2) /2 = 54  X2 + Y2 = 108 --- (2)
From (1), (X + Y)2 = X2 + 2XY + Y2 = 196 --- (3)
(3) – (2): 2 XY = 196 – 108 = 88
Hence XY = 88/2 = 44
29. In triangle ABC, AB = AC and D is a point on AC so
that BD bisects angle ABC. If BD = BC, what is the
A
measure, in degrees, of angle A?
D
By AB = AC, ABC = ACB
B
By BD = BC, BDC = ACB
Hence ABC  BCD
We get: CBD = A
Since BD bisects ABC, A = ½ ABC
We have: 5 * A = 180  A = 180/5 = 36
C
X + 43 = m * m
----- (1)
X + 16 = n * n
------ (2)
(1) – (2)  m*m – n*n = 27
(m + n) (m – n) = 27
Since m, n (m > n) are integers, m+n & m – n
are also positive integers
Possible answers are: m+n = 27, m-n =1 -- (3)
or
m+n = 9, m-n =3 -- (4)
From (3), m = 14, n = 13  mn = 14*13=182
From (4), m = 6, n = 3  mn = 18, but wrong X
G3. The distribution of the 37 test scores in a math
class is given in the stem and leaf plot where 5|6
represents 56 points. What percent of the scores are
at most 5 points from the value of the median?
There are total of 37 scores
Median is the 19th -- 78
There are 8 scores that are
between 73 and 83:
Answer: 8/37 = 0.216 = 22
G4. If a + b + c + d = 11, 2a + 3c = 19, b + 4d = 22, 4a
+ d = 14 and 5b + 3c = 5, what is the value of d?
Listing out the equations:
a + b + c + d = 11
------- (1)
2a + 3c = 19
------- (2)
b + 4d = 22
------- (3)
4a + d = 14
------- (4)
5b + 3c = 5
------- (5)
(2) + (3) + (4) + (5), we have:
6a + 6b + 6c + 5d = 19+22+14+5 = 60 ------- (6)
6 * (1) – (5): d = 66 – 60 = 6
G6. A shopper notes that a box of brand A cornflakes
costs 50% more than a box of brand B and weighs
25% more than a box of brand B. According to this
information, for equal weights, the cost of brand A is
what percent more than the cost of brand B?
Let Ca – cost of A, Cb – cost of B
Let Wa – weight of A, Wb – weight of B
Unit-Cost-A = Ca /Wa = (150% Cb)/(125% Wb )
= 6/5 Cb/Wb
Difference in unit-cost:
Ca /Wa - Cb/Wb = 6/5 Cb/Wb - Cb/Wb = 1/5 Cb/Wb
Percentage of the difference = (1/5 Cb/Wb )/ (Cb/Wb)
= 1/5 = 20%
G8. How many ordered pairs of positive integers
satisfy the equation, 1/X + 1/Y = 1/6 with x < y?
From 1/x + 1/y = 1/6, we get: (x+y)/xy = 1/6
6 x + 6 y = xy  xy – 6x – 6y = 0
xy - 6x – 6y + 36 – 36 = 0
(x – 6) (y – 6) = 36
Since X, Y are positive integers, x-6, y-6 are integers
(x – 6) and (y – 6) are integer factors of 36
Since X < Y, we have (x-6) < (y-6)
Possible values for (x-6): 1, 2, 3, 4,
Answer: 4 (7, 42), (8, 24), (9, 18), (10, 15)
M5. A set of five positive integers has a mean,
median and range of 7. How many distinct sets
whose members are listed from least to greatest
could have these properties? Note: {4, 6, 7, 7, 11} is
one such set to include.
By 7 be the mean, median, 7 must be in the set
a
7
b
d
c
Let a, b, c, d, be differences of the other 4 numbers to 7, as shown.
We have a + b = c + d, and a + d = 7
We have a + b = c + d, and a + d = 7
If a >= 5  d <=2  c + d <= 4 < a + d  no fit answers
If a = 4,  d = 3  c = 1, 2, 3  b = 0, 1, 2  3 answers
If a = 3,  d = 4  b = 1, 2, 3  c = 0, 1, 2  3 answers
If a = 2,  d = 5  a + b <= 2+2 = 4 < 5 no fit answers
M7. What is the greatest common factor of 20! and
200,000?
200000 = 26 * 55
20! = 5*10*15*20 * 2*4*8*12*14*16*…
Common factors: 26 * 54 = 200000/5 = 40000
M8. Harry has 49 coins totaling \$1.00. Harry has only
pennies, nickels, dimes and/or quarters. What is the
smallest number of dimes he could have?
Need to remove 51 coins
Every replace of 1 dime removes 9 coins
Every replace of 1 nickel removes 4 coins
Every replace of 1 quarter removes 24 coins
We can’t remove 51 coins with zero or even number of
dimes as nickels and quarters remove even # of coins
If use 1 dime, we need to remove 51 – 9 = 42
coins, which can’t be done with multiple 4 & 24
If use 3 dimes, we need to remove 51 – 27 = 24 coins – can
be achieved with 1 quarter OR 6 nickels
M9. A bag contains 3 red balls, 4 green balls, and 5
yellow balls. If balls are drawn one at a time without
replacement, what is the probability that the first
yellow ball is drawn on the eighth draw? Express
There are total of 7 balls that are
not of color yellow.
Prob. = 7/12 * 6/11 * 5/10 * 4/9 * 3/8 * 2/7 * 1/6 * 1
= 7/12 * 6/11 * 5/10 * 4/9 * 3/8 * 2/7 * 1/6 * 1
= 7/12 * 6/11 * 5/10 * 4/9 * 3/8 * 2/7 * 1/6 * 1
= 1/3 * 1/11 * 1/3 * 1/8
= 1/792
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