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Some problems I cannot solve, please help!!!

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Some problems I cannot solve, please help!!!

by Veronica » Wed Sep 22, 2010 11:18 pm
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8?
a. 1/4; b.3/8; c.1/2; d.5/8; e. 3/4

The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4 and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?
I.3; II.7; III. 12
A. I only; B. II only; C. III only; D. I and III only; E. I, II and III

At a garage sale, all of the prices of the items sold were different. If the price of a radio sold at the garage sale was both the 15th highest price and the 20th lowest price among the prices of the item sold, how many items were sold at the garage sale?
a. 33; b.34; c.35; d.36; e. 37

The numbers x and y are three-digit positive integers, and x+y is a four-digit integer. The tens digit of x equals 7 and the tens digit of y equals 5. If x<y, which of the following must be true?
I.The units digit of x+y is greater than the units digit of either x or y.
II. The tens digit of x+y equals 2
III. The hundreds digit of y is at least 5
A. II only; B.III only; C.I and II; D. I and III; E. II and III

A teacher gave the same test to three history classes: A, B and C. The average (arithmetic mean) scores for the three classes were 65, 80 and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?
A.74; B.75; C.76; D.77; E.78

At a certain company, each employees has a salary grade s that is at least 1 and at most 5. Each employee receives an hourly wage p, in dollars, determined by the formula p = 9.50 + 0.25(s-1). An employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 1?
A. $0.5; B.$1.00; C.$1.25; D.$1.5; E.$1.75

For this problem, I found that p1 = 9.5, p5 = 10.5, therefore, I chose B. But the answer key is C. Please explain this and show me how to solve the others. thank you!!!
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by sgmuthukumar » Thu Sep 23, 2010 12:25 am
Veronica wrote: The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4 and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?
I.3; II.7; III. 12
A. I only; B. II only; C. III only; D. I and III only; E. I, II and III
3,4,6,7,9,10=39/6... Mean =6.5 and Median=6.5

4,6,7,7,9,10=43/6....Mean= 7.1 and Median=7

4,6,7,9,10,12=48/6... Mean =8 and Median =8

Answer is D. I and III only
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by limestone » Thu Sep 23, 2010 12:33 am
Well, you posted so many at once. I'll solve the first
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8?
a. 1/4; b.3/8; c.1/2; d.5/8; e. 3/4
From 1 to 96 there are 12 no. that are divisible to 8 ( such as 8,16,24,...,88,96), 12 no. if divided by 8 will give out 1 as their remainder. (such as 9,7,25,33, ...,81,89), and so on ... 12 no. if divided by 8 will give out 7 as their remainder (such as 7,15,23,...,87,95)
List all the cases:
n if divided by 8 will give out 0 as its remainder, then n(n+1)(n+2) is divisible to 8

n if divided by 8 will give out 1 as its remainder, then n(n+1)(n+2) is (8*a +1)(8*a+2)(8*a+3) (where a is an integer) which is not divisible to 8.

n if divided by 8 will give out 2 as its remainder, then n(n+1)(n+2) is (8*a +2)(8*a+3)(8*a+4)which is divisible to 8, why? as 8*a+2 is the multiple of 2, 8*a+4 is the multiple of 4, then (8*a +2)(8*a+4) is the multiple of 8.

n if divided by 8 will give out 3 as its remainder, then n(n+1)(n+2) is (8*a +3)(8*a+4)(8*a+5) (where a is an integer) which is not divisible to 8.

n if divided by 8 will give out 4 as its remainder, then n(n+1)(n+2) is (8*a +4)(8*a+5)(8*a+6) (where a is an integer) which is divisible to 8. As 8*a+6 is the multiple of 2, 8*a+4 is the multiple of 4, then (8*a +6)(8*a+4) is the multiple of 8

n if divided by 8 will give out 5 as its remainder, then n(n+1)(n+2) is (8*a +5)(8*a+6)(8*a+7) (where a is an integer) which is not divisible to 8.

n if divided by 8 will give out 6 as its remainder, then n(n+1)(n+2) is (8*a +6)(8*a+7)(8*a+8) (where a is an integer) which is divisible to 8. As (8*a+8) is a multiple of 8.

n if divided by 8 will give out 7 as its remainder, then n(n+1)(n+2) is (8*a +7)(8*a+8)(8*a+9) (where a is an integer) which is divisible to 8. As (8*a+8) is a multiple of 8.

Because the possibility that n has each remainder is the same. We get 5 over 8 cases accepted. So the possibility is 5/8.
Pick D.
Sorry for my long approach, I just want to make it clear.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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by limestone » Thu Sep 23, 2010 12:42 am
At a garage sale, all of the prices of the items sold were different. If the price of a radio sold at the garage sale was both the 15th highest price and the 20th lowest price among the prices of the item sold, how many items were sold at the garage sale?
a. 33; b.34; c.35; d.36; e. 37
Take a simple example. There're 5 items A,B,C,D,E priced at $1,2,3,4,5 respectively.
If D is the second highest priced, then D will be the forth lowest price. Note that 2+4 = 6, however, we get totally 5 items.
So the rule is the total number of items = rank of a specific item in ascendent + rank of that item in descendent - 1
In this case: 15+20 - 1 = 34.
Pick B.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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by limestone » Thu Sep 23, 2010 1:19 am
The numbers x and y are three-digit positive integers, and x+y is a four-digit integer. The tens digit of x equals 7 and the tens digit of y equals 5. If x<y, which of the following must be true?
I.The units digit of x+y is greater than the units digit of either x or y.
II. The tens digit of x+y equals 2
III. The hundreds digit of y is at least 5
A. II only; B.III only; C.I and II; D. I and III; E. II and III

A teacher gave the same test to three history classes: A, B and C. The average (arithmetic mean) scores for the three classes were 65, 80 and 77, respectively. The ratio of the numbers of students in each class who took the test was 4 to 6 to 5, respectively. What was the average score for the three classes combined?
A.74; B.75; C.76; D.77; E.78

At a certain company, each employees has a salary grade s that is at least 1 and at most 5. Each employee receives an hourly wage p, in dollars, determined by the formula p = 9.50 + 0.25(s-1). An employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 1?
A. $0.5; B.$1.00; C.$1.25; D.$1.5; E.$1.75
For the first quoted question:
Call X a7b, Y c5d , where a,b,c,d are the value of digits
a7b + c6d are four digit no.
hence a+c+1 >=10 ( 6+7 is 13, so 1 is added to the hundreds digit)
or a+c>=9,

I. if the units digit of X is 9 of Y is 8, that of x+y is 7 -which is smaller than both 8 & 9 --> not a must be
II. At first, it looks like 5+7 in tens digit will give out a "2", however, if the units digit of X is 9 of Y is 8, then the tens digit of x+y must be 3 ( 5+7+1 =13) --> not a must be
III. from the condition above a+c>=9,however, x<y then a<c
then c+c >a+c>=9, or 2c>=9, c>=4.5
c is an integer, then c>=5 --> a must be

The second
Plug in method:
Let class A,B,C have 4,6,5 students respectively.
Average grade: (4*65+6*80+5*77)/(4+5+6) = 75. Pick B

The third:
I come up with the same answer and pick B. Please check OA again, and if the answer is C, please copy the explanation to this topic for me.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
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