Several math probs, pls help to explain me, 600+

[tracyyahoo]

(1) The average ( arithematic mean) cost per book for the 12 books on a certain table is k dollars. If a book that costs 18 dollars is removed from the table and replaced by a book that costs 42 dollars, then in terms of k, what will be the average cost per book, in dollars, for the books on the table???

a) k+2 b) k-2 c) 12+24/k d)12-24/k e)12k-6

I chose B, why OA A???




(2) 10000^100 is equivalent to which of the following??

I (100^2)(100^100)
II 100^200
III 10^400

A) None B)I only C)III only D)II and III only E)I,II,III

I chose B why the hell OA is D



(3)Acoording to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)+500 for 0=t=10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum???

a) 5:30 b) 7:00 c) 7:30 d) 8:00 e)9:00

[Anurag@Gurome]

(2) 10000^100 is equivalent to which of the following??

I (100^2)(100^100)
II 100^200
III 10^400

A) None B)I only C)III only D)II and III only E)I,II,III

I chose B why the hell OA is D

10000^100 = (10^4)^100 = 10^400

I (100^2)(100^100) = (10^4)(10^200) = 10^204
II 100^200 = (10^2)^200 = 10^400
III 10^400

II and III are true

The correct answer is D.

[shankar.ashwin]

Cost/book - k
No of books - 12

Total cost - 12k

Now a $18 book is removed - (12k-18)

And, $42 book is added - (12k-18+42) = 12k+24

Avg = total/ No of books

= 12k+24/2 = k+2

[sl750]

In the first problem, you made an error in the sign (-18+42) = 24. Therefore average cost is (12k+24)/12 = k+2

In second problem ii 100^200 = (10^2)^200 = 10^400 . (10^a)^b = (10)^ab

For problem 3

The tank reaches maximum when t=5
N(5) = 500. Since the estimate is made at t hours past 2:00, the time the tank reaches maximum capacity is 7 AM

[shankar.ashwin]

For problem 3

The tank reaches maximum when t=5
N(5) = 500. Since the estimate is made at t hours past 2:00, the time the tank reaches maximum capacity is 7 AM

Wont the tank reach max when t=1?

You then get; -20(1-5) + 500 = 80+500 = 580.

From answer choices earliest time would be max level, wont it be A?

[GMAT_1986_subha]

For problem 3

The tank reaches maximum when t=5
N(5) = 500. Since the estimate is made at t hours past 2:00, the time the tank reaches maximum capacity is 7 AM

Wont the tank reach max when t=1?

You then get; -20(1-5) + 500 = 80+500 = 580.

From answer choices earliest time would be max level, wont it be A?

Yes the answer is A as it's the only time in the choice when (t-5) is lowest and -20(t-5) is lowest value in -ve.

So the value at 5:30 will be N(t)= -20(5.5-5)+500 = -10+500 = 490

[tracyyahoo]

For no. 3
(3)Acoording to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)+500 for 0=t=10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum???

a) 5:30 b) 7:00 c) 7:30 d) 8:00 e)9:00

Could someone explain in details , tahnk you.

[sl750]

For problem 3

The tank reaches maximum when t=5
N(5) = 500. Since the estimate is made at t hours past 2:00, the time the tank reaches maximum capacity is 7 AM

Wont the tank reach max when t=1?

You then get; -20(1-5) + 500 = 80+500 = 580.

From answer choices earliest time would be max level, wont it be A?

Ok, you make a point. But how does that make t=5:30?, if t=1, then the earliest time would be 3:00 AM and not 5:30 AM

[sl750]

For no. 3
(3)Acoording to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)+500 for 0=t=10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum???

a) 5:30 b) 7:00 c) 7:30 d) 8:00 e)9:00

Could someone explain in details , tahnk you.

Can you confirm if the equation is correct? I suspect it should read N(t) = -20(t-5)^2 + 500

[shankar.ashwin]

No, I just gave an example as t=1. for 5.30 am, it woud obviously be 3.5 and still the depth would be higher than at 7am according to the equation.

For no. 3
(3)Acoording to a certain estimate, the depth N(t), in centimeters, of the water in a certain tank at t hours past 2:00 in the morning is given by N(t)= -20(t-5)+500 for 0=t=10. According to this estimate, at what time in the morning does the depth of the water in the tank reach its maximum???

a) 5:30 b) 7:00 c) 7:30 d) 8:00 e)9:00

Could someone explain in details , tahnk you.

Can you confirm if the equation is correct? I suspect it should read N(t) = -20(t-5)^2 + 500