PS question a Day

Ajith asked me to post in this thread till he gets back.

Answers in 3 days time as usual.

PS005

In the formula w = p/(v^(1/t)), integers p and t are positive constants. If w =2 when v = 1 and if w = 1/2 when v = 64, then t =
(A) 1
(B) 2
(C) 3
(D) 4
(E) 16

PS006

Working alone, Maria can complete a task in 100 minutes. Shaniqua can complete the same task in two hours. They work together for 30 minutes when Liu, the new employee, joins and begins helping. They finish the task 20 minutes later. How long (in minutes) would it take Liu to complete the task alone?

A. 180
B. 240
C. 200
D. 160
E. 220

PS 007

What is the probability that a card selected from a deck will be either an ace or a spade?
A. 2/52
B. 2/13
C. 7/26
D. 4/13
E. 17/52

rajs.kumar wrote:PS006

Working alone, Maria can complete a task in 100 minutes. Shaniqua can complete the same task in two hours. They work together for 30 minutes when Liu, the new employee, joins and begins helping. They finish the task 20 minutes later. How long (in minutes) would it take Liu to complete the task alone?

A. 180
B. 240
C. 200
D. 160
E. 220

Answer is B

x is the total task

In 30 minutes Maria can complete 3x/10, and Shaniqua can complete x/4. Combined together 11x/20.

=> in 20 minutes all 3 can do 9x/20 of the job of which Maria and Shaniqua complete 11x/30 => x/12 is done by Liu in 20 minutes so the total time taken for x is 240 minutes.

PS 008

The number of passengers on a certain bus at any given time is given by the equation P = -2(S – 4)^2 + 32, where P is the number of passengers and S is the number of stops the bus has made since beginning its route. If the bus begins its route with no passengers, how many passengers will be on the bus two stops after the stop where it has its greatest number of passengers?

1. 32
2. 30
3. 24
4. 14
5. 0

rajs.kumar wrote: PS005

In the formula w = p/(v^(1/t)), integers p and t are positive constants. If w =2 when v = 1 and if w = 1/2 when v = 64, then t =
(A) 1
(B) 2
(C) 3
(D) 4
(E) 16

Answer: 3

w = 2, v = 1 => p = 2

w = 1/2, v = 64 => 64 ^ (1/t) = 4 or t = 3

rajs.kumar wrote:PS 008

The number of passengers on a certain bus at any given time is given by the equation P = -2(S – 4)^2 + 32, where P is the number of passengers and S is the number of stops the bus has made since beginning its route. If the bus begins its route with no passengers, how many passengers will be on the bus two stops after the stop where it has its greatest number of passengers?

1. 32
2. 30
3. 24
4. 14
5. 0

There are two ways to approach this problem. Answer C.

Method 1
======

(S - 4)^2 is always positive so -2(S - 4)^2 is always negative. For P to be maximum the when the first term is 0 since 32 is always constant. This means S = 4. So out desired value is 4 + 2 = 6. With this value we get # passengers as 24.

Method 2
======

For this problem this method may not be necessary. However, for other kind of maxima/minima problems where the maximum value is not direct this method can be used.

At maxima the rate of change is equal to 0

dP/dS = 0
-4(S - 4) . 1 = 0 => S = 4

=> P = 24.

For more complex functions this method is quicker. For example to find the maximum of y = x^2 - 100x. dy/dx = 2x - 100 = 0 => x = 50 or maximum y for a x value of 50.

ajith wrote:PS004


3 boxes of supplies have an average (arithmetic mean) weight of 7 kilograms and a median weight of 9 kilograms. What is the maximum possible weight, in kilograms, of the lightest box?

1
2
3
4
5

(X +Y + Z )/3 = 7 ==>X + Y + Z = 21

Median Y = 9

For Y to be the Median X,Y & Z should be in ascending order

==> X + 9 + Z = 21

Z >= 9

If Z = 9 then X = 3
If Z = 10 then X = 2

Therefore the max possible weight of the lightest a box can be is 3 kgs

x+y+z = 21
y = 9

So x + z must be = 12

If x = 5 then z = 7 ... which invalidates the statement that median = 9.

If x = 4 then z = 8 ... which invalidates the statement that median = 9.

So x = 3 ... in which case z = 9 and it still does not invalidate the
statement that median - 9.

I've just joined this forum... can I post questions too?

Also I notice no new question has been posted for a while - is this thread still active?

Hi Eric,

I would love to start a new thread - however I am not entirely sure how to go about this... point me in the right direction?

Pramila

rajs.kumar wrote:PS 007

What is the probability that a card selected from a deck will be either an ace or a spade?
A. 2/52
B. 2/13
C. 7/26
D. 4/13
E. 17/52

there are 52 cards in a deck, and 13 spades 4ace, but there is a card which is both ace and spade, therefore, 13+4-1/52=4/13 so choose D

rajs.kumar wrote:PS 007

What is the probability that a card selected from a deck will be either an ace or a spade?
A. 2/52
B. 2/13
C. 7/26
D. 4/13
E. 17/52

Answer is E. 17/52
4 aces and 13 spade cards... this is an OR condition so add the two probabilities:

4/52 + 13/52 = 17/52 !