future_dentist

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These r the few hard problems.....for QR....

What is the median for a series of five consecutive odd integers where the sum of the smallest and largest numbers is 34?



A. 11
B. 13
C. 15
D. 17
E. 19

Ans: D

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Which of the following items is the best buy?



A. 12-diaper-pack for $4.00
B. 24-diaper-pack for $7.00
C. 28-diaper-pack for $8.00
D. 32-diaper-pack for $9.00
E. 40-diaper pack for $12.00

Ans: D

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What is the largest number from a series of four consecutive negative integers where the sum of the two smallest integers is -17?



A. -14
B. -12
C. -10
D. -8
E. -6

Ans: E
 

allstardentist

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for the first problem, it is possible to do it systematically but it is prolly faster to do by trial and error. But if you were to do it systematically, it would be:
set x as the median, (x-5) + (x+5) = 34 -> 2x=34 -> x=17

for the second problem, just divde the cost by the # of diapers and the answer is the one wit the highest value. there an easy way to compare among them.. but kinda hard to explain..

thrid problem, trial and error would be the fastest...

not too hard, just practice..
 

wigglytooth

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What is the median for a series of five consecutive odd integers where the sum of the smallest and largest numbers is 34?



A. 11
B. 13
C. 15
D. 17
E. 19

Ans: D

Translate to algebra terms:
5 consecutive odd integers would be x, x+2, x+4, x+6, x+8.
Sum of the largest and smallest = 34 ---> x + (x+8) = 34, x = 13. Check by plugging if you want.
And the median is the middle term when each term is ordered from smallest to largest: 13, 15, 17, 19, 21.

D = 17.

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Which of the following items is the best buy?



A. 12-diaper-pack for $4.00
B. 24-diaper-pack for $7.00
C. 28-diaper-pack for $8.00
D. 32-diaper-pack for $9.00
E. 40-diaper pack for $12.00

Ans: D

Someone else answered this before. I don't know if there's an easier way, but the other poster and I solved it by just seeing what the price per one diaper is by dividing the price for a pack by the number in a pack. The lowest unit cost is the best buy.

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What is the largest number from a series of four consecutive negative integers where the sum of the two smallest integers is -17?



A. -14
B. -12
C. -10
D. -8
E. -6

Ans: E[/QUOTE]

Kind of the same as the other one but different algebraic terms.
Four consecutive negative integers would be x, x + 1, x + 2, x + 3. The two smallest would be x and x + 1, where x would equal -9.
If -9 is the smallest, then the series of four numbers would look like -9, -8, -7, -6... since we're looking at negative numbers, -6 is the largest in the series.

E = -6
 
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Banana Berry

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Ok, median is the number in the middle of a series of numbers. If all the numbers are odd integers, the numbers should be X, X+2, X+4, X+6, X+8. The sum of the smallest and largest is X+ (X+8)= 34. Solve for it, you get X=13. Therefore, the numbers are 13,15,17,19,21 and the median is the one in the middle, which is 17.

#2 is asking for the cheapest one (best buy). Just use the money divide the total # in the pack and compare values. Pick the smallest value, which is 28 cents each for choice D.

#3 is similiar to #1. The numbers are X, X+1, X+2, X+3, and the sum of the 2 smallest ones is X+ (X+1)= -17. Solve for X, you will get -9. Therefore, the largest one would be X+3=(-9)+3= -6.

I hope this helps. Good luck study!
 

wigglytooth

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allstardentist said:
for the first problem, it is possible to do it systematically but it is prolly faster to do by trial and error. But if you were to do it systematically, it would be:
set x as the median, (x-5) + (x+5) = 34 -> 2x=34 -> x=17
Good trick :thumbup:
 

wigglytooth

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lol I like how 3 people answered this question at once :cool:
 
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future_dentist

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Thank u so much....u guys r great :) ....
 
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