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A ranch has horses and ponies. Five-sixths of the ponies have horseshoes. Half of the ponies with horseshoes are Icelandic. If there are three more horses than ponies, what is the minimum number of horses and ponies on the ranch?
9 15 27 48 54
Let the number of horses be h, and the number of ponies be p.
1) 5/6 of the ponies have horseshoes, so 5p/6 have horseshoes
2) 1/2 of the ponies with horseshoes are icelandic, so 1/2(5p/6) = 5p/12 are Icelandic
3) There are 3 more horses than ponies, so p+3 = h
The total number of horses and ponies on the ranch would be equal to (h+p). This is equivalent to (p+3+p) = (2p+3).
We can now quickly calculated a vaule for p, by equating 2p+3 to the values given in the choices.
If 2p+3 = 9, p = 3, 5p/6 = 2.5 ponies with horseshoes (nonsensical)
If 2p+3 = 15, p = 6, 5p/6 = 5 ponies with horseshoes, 5p/12 = 2.5 Icelandic ponies (nonsensical)
If 2p+3 = 27, p = 12, 5p/6 = 10 ponies with horseshoes, 5p/12 = 5 Icelandic ponies
If 2p+3 = 48, p = 22.5 ponies (nonsensical)
If 2p+3 = 54, p = 25.5 ponies (nonsensical)
So the answer is C.
Note: We can actually rule out 48 and 54 immediately, since we know 2p must be an even number, and if we add this even number to 3 (odd number), we would get an odd result.
The rectangular garden with dimensions x feet by y feet, is surrounded by a walkway 2 feet wide. Which of the following represents the area of the walkway, in square feet? A. 2x + 2y + 4 B. 2x + 2y +16 C. 4x - 4y + 8 D. 4x + 4y + 16 E. 4x + 4y + 32
Let's assume the lenght is x feet and the width is y feet. If there is a walkway 2 feet wide, the size of the rectangle inclusive of the walkway would be x+4 feet by y+4 feet. The area would then be (x+4)(y+4) = (xy+4x+4y+16) square feet.
The area of the walkway would be the area of the rectangle inclusive of the walkway, minus the area of the rectangle which is xy square feet.
A certain cake recipe states that the cake should be baked in a pan 8 inches in diameter. If Jules wants to use the recipe to make a cake of the same depth but 12 inches in diameter, by what factor should he multiply the recipe ingredients?
2 1/2 2 1/4 1 1/2 1 4/9 1 1/3
Since we're told that the pan has a diamter of 8 inches, I'll assume the pan is circular in shape. (Diamter is not used in any other shapes)
Let's assume the dept to be x inches. Thus, the original receipe would give a cake of volume (pi)(4*4)(x), and the new receipe would give the cake a volume (pi)(6*6)(x) [Note: Volume = pi*r^2^height]
Thus, the new volume is bigger by a factor of (pi)(6*6)(x)/(pi)(4*4)(x) = 36/16 = 2 1/4
If xy = z , where x ,y and z are positive integers and x and y are prime number, what is the value of y?
1. z is even 2. x is odd
This question is testing your knowledge of prime numbers, and also your knowledge of what happens when evens and odds are operated on.
We're told x and y are prime numbers.
If z is even, then either x or y must be even, so x or y must be 2, the only even prime number. However, we do not have any information to allow us to tell if x or y holds the value 2 and so statement (1) is insufficient.
From statement (2), we can't do anything since all we know is x is odd. y could be an even prime and thus z would be even. Odd y could be any prime number other than 2, and z would be odd. So statement (2) is insufficient.
Using both statements, we know z is even, so x or y is 2. Since x is odd, then y must be 2.
Thus, the answer is C.
Last edited by ywilfred on 22 Jun 2005, 19:50, edited 1 time in total.
Melting the three cheese balls and molding them together into one cheese ball will have one thing common. The volume of the big cheese ball will be the combined volume of the three cheese balls. To solve this, we need to use the formula for the volume of a sphere = 4/3*pi*(r^3)
For the cheese ball with dimeter of 1, V = 4/3 * pi * (1/2)^3 = pi/6
For the cheese ball with diameter of 2, V = 4/3 * pi * (1^3) = 4pi/3
For the cheese ball with diamater of 6, V = 4/3 * pi * (3^3) = 36pi
The total volume is therefore pi/6 + 4pi/3 + 36pi = 75pi/2
If the diamater of the big cheese cake = D, then the radius = D/2
So 4/3 * pi * (D/2)^3 = 75pi/2
D^3 = 225
D = 225^(1/3)
A sum of $385 was divided among Jack, Pollock and Gibbs. Who received the minimum amount?
(1) Jack received 2/9 of what Pollock and Gibbs together received.
(2) Pollock received 3/11 of what Jack and Gibbs together received
From statement (1), we do not have the proportion of how Pollock and Gibbs will split their combined sum, so we can't tell who got the minimum ammount. Statement (1) is therefore insufficient.
From statement (2), we again do not know how Jakc and Gibbs will split their sum, so we can't tell who got the minimum amount. Statement (2) alone is also insufficient.
Using both statements, we know from (1) that J = 2/9 * (P + G) (Note: Using initials to represent the amount each individual gets)
Since we know that the total sum is $385, we can write the equation:
J + P + G = 385
Substituting (P+G) with 9J/2, we have J + 9J/2 = 11J/2 = 385 => J = $70.
From (2), we know P = 3/11 * (J + G)
Using J + P + G = 385 and substituting J + G with 11P/3, we can find the sum Pollock gets. And using the result we got earlier for Jack, we can also calculate what Gibbs received for his share. Thus the question can be answered if we use both statements.
From statement (1), We're told q = 50 + r. However, this will not let us find the sum of q + r since we do not have any value for q or for r. Thus, statement (1) alone is insuffcient.
From statement (2), we're todl r = q/3 => 3r = q. Again, this does not let us answer the question since we have no way of knowing what the values for q or r are. So statement (2) alone is insufficient.
Using both statements, We can solve for q by substituting r with q/3 in the equation q = 50 + r. After solving for q, we can also solve for r. The sum of q + r can thus be calculated.
The price of postage stamps has increased 5 cents per year every year since 1990. If 10 stamps were purchased every year from 1998 to 2002, the total cost would be $35. How much did a stamp cost in 1995?
A. 45 cents B. 35 cents C. 40 cents D. 48 cents E. 52 cents
We're told the stamps increased by 5 cents every year since 1990. So, this is a Arithmetic Sequence, where the frist term is the cost of a stamp in 1990, which we will cal a. The arithmetic difference is the increase in price, which is 5 cents. Thus, the nth term (or the nth year) would be defined as a + (n-1)d
For 1998, a stamp would cost a + (9-1)5 = (a + 40) cents (Note: use 9 as n since the 2008 is actually the ninth term)
For 1999, a stamp would cost a + (10-1)5 = (a + 45) cents
For 2000, a stamp would cost a + (11-1)5 = (a + 50) cents
For 2001, a stamp would cost a + (12-1)5 = (a + 55) cents
for 2002, a stamp would cost a + (13-1)5 = (a + 60) cents
The cost of 10 stamps in 1998 would be (10a + 400) cents
The cost of 10 stamps in 1999 would be (10a + 450) cents
The cost of 10 stamps in 2000 would be (10a + 500) cents
The cost of 10 stamps in 2001 would be (10a + 550) cents
The cost of 10 stamps in 2002 would be (10a + 600) cents
Thus, the total cost of the stamps between 1998 to 2002 = (50a + 2500) cents = $35 = 3500 cents
Solving for a gives us a = 20 cents
The cost of a stamp in 1995 = a + (6-1)5 = 20 + 25 = 45 cents
A motorcycle covers a total distance of 588 km in 9 hours. It covers part of the distance at 60 km/hr speed and the balance at 72 km/hr speed. Find the time it travelled at 72 km/hr speed?
a. 3hrs b. 4 hrs c. 3 hrs 30 min d. 3 hrs 24 min e. 3 hrs 15 min
This question tests simple distance-speed-time relationships.
Let's assume the part covered by the 72km/h speed is x km, then the part covered by the 60km/hr speed would be (588 - x)km. [Hint: Use x to denote the 72 km/h stretch so that you do not need to do extra calculatiosn once you solve for x]
Since time = distance/speed, therefore (time spent travelling at 60km/h) + (time spent travelling at 72 km/h) = total time spent
(588 - x)/60 + x/72 = 9
[(3528-6x) + 5x]/360 = 9
3528 - x = 3240
x = 288 km
The time spend travelling at 72km/h is therefore = 288/72 = 4 hours.
By how many dollars has the cost of tuition increased since last year?
(1) This year, tuition is 10 percent higher than it was last year. (2) Two years ago, tuition was $11,000.
Statement (1) tells us that if tuition cost $100 last year, then it would cost $110 dollars this year. However, if tuition was $80 dollars last year, then it would cost $88 dollars last year. So unless we know the tuition cost for the previous year, we can't calculate how much the tuition cost increased. Statement (1) is therefore insufficient.
Statement (2) is useless to us. Knowing how much tuition costs two years ago will not help us to know how much the tuition has increased today. Thus statement (2) is insufficient.
Using both statements, we still can't find how much tuition costs today and so we can't answer the question.