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## GRE Test Prep : Quantitative Comparison VI

 Formats Worksheet / Test Paper Quiz Review

 Compare the two quantities and choose the best answer from four choices given.

1. Amy, Beth and Charlie divided a pizza amongst themselves.
Amy took 30% of the pizza and ate (3/4) of what she took.
Beth took 20% of the pizza.
Charlie ate (2/5) of what he took.
Quantity A = The amount Amy ate
Quantity B = The amount Charlie ate
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given

If Amy took 30% and Beth took 20%, then Charlie took 50% of the pizza.
Let P denote the size of the total pizza.
Amount Amy ate = (3/4) (0.3 P) = 0.225 P.
Amount Charlie ate = (2/5) (0.5 P) = 0.2 P.

2. Quantity A = (55 + 59)(61 - 79)
Quantity B = (55 - 59)(61 - 79)
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given

Be careful! Note that (61 - 79) is a negative number.
Quantity A is negative because it is the product of a positive number and a negative number.
Quantity B is positive because it is the product of two negative numbers.

3. a, b, c and d are four consecutive integers.
Quantity A = The arithmetic mean of a and d
Quantity B = The arithmetic mean of b and c
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given
Answer: The two quantities are equal

If a, b, c and d are four consecutive integers,
then b = a + 1 ; c = a + 2 ; and d = a + 3.
Quantity A = (a + d)/2 = (a + a + 3)/2 = (2a + 3)/2.
Quantity B = (b + c)/2 = (a + 1 + a + 2)/2 = (2a + 3)/2.

4. 2 < z < 4
Quantity A = π2z3
Quantity B = π3z2
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given
Answer: The relationship cannot be determined from the information given

You could substitute values for z between 2 and 4 to compare.
An alternative simple strategy is to divide both Quantity A and B by π2z2.
Then Quantity A = z and Quantity B = π.
We note that Quantity A lies between 2 and 4, whereas Quantity B equals about 3.14

5. yz < 0
Quantity A = (yz)2
Quantity B = y2 + z2
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given

(yz)2 = (yz)(yz) = y2 + z2 − 2yz.
Since yz is negative, −2yz is positive. So, Quantity A > Quantity B.

6. 11y4/(3y2) = 11/3
Quantity A = y
Quantity B = 1
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given
Answer: The relationship cannot be determined from the information given

Dividing both sides by 11/3 gives y4/y2 = 1 or y4 = y2.
Now, dividing both sides by y2 gives y2 = 1.
So, y can equal 1 or -1.

7. y > 0
Quantity A = 100 y
Quantity B = 100 / y
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given
Answer: The relationship cannot be determined from the information given

When y = 1, Quantity A equals Quantity B.
When y > 1, Quantity A is greater. When y < 1, Quantity B is greater.

8. Quantity A = The perimeter of a rectangle whose area is 33.
Quantity B = 28
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given
Answer: The relationship cannot be determined from the information given

Area of rectangle = Length x Breadth
Perimeter of rectangle = 2 (Length + Breadth)
If the area of the rectangle is 33, then there are many possible values for the length and breadth as given below.
Length = 11, Width = 3 and Perimeter = 28.
Length = 33, Width = 1 and Perimeter = 68.
Length = 6, Width = 5.5 and Perimeter = 23.

9. n(n + 2)(n + 4) = 480
Quantity A = n + 2
Quantity B = 6
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given

Expanding the left-hand side will give a cubic polynomial and the resulting cubic equation will be difficult to solve. The following strategy may be
If n + 2 = 6, then n = 4, n + 4 = 8 and 4 x 6 x 8 = 192.
Since the product must be 480, n + 2 must be larger than 6.

10. Andrew stitches shirts thrice as fast as Bill.
Bill charges 40% more per shirt than Andrew.
Quantity A = The amount Andrew earns in 8 days
Quantity B = The amount Bill earns in 15 days
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given

Let Bill stitch b shirts per day. Then, Andrew stitches 3b shirts per day.
Let Andrew charge a dollars per shirt. Then, Bill charges 1.4a dollars per shirt.
Amount Andrew earns in 8 days = 8 (3b) a = 24 ab.
Amount Bill earns in 15 days = 15 b (1.4a) = 21 ab.

11. p = 3z/4, q = 4r/5, and r = 5z/6
Quantity A = 12p
Quantity B = 12q
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given
Answer: The relationship cannot be determined from the information given

12p = 12 (3z/4) = 9z.
q = 4r/5 = 4z/6.
12q = 12 (4z/6) = 8z.
If z = 0, then 9z = 8z = 0.
If z > 0, then 9z > 8z.
If z < 0, then 9z < 8z.

12. For any positive integer n,
n! is the product of all positive integers less than or equal to n.
Quantity A = 20! / 17!
Quantity B = 80! / 78!
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given

20! / 17! = (20 x 19 x 18 x 17 x 16 x ...)/(17 x 16 x 15 x ...) = 20 x 19 x 18.
80! / 78! = (80 x 79 x 78 x 77 x ...)/(78 x 77 x 76 x ...) = 80 x 79 = 20 x 4 x 79.
Now, 19 x 18 > 4 x 79.

13. 0 < a < 1
Quantity A = a3
Quantity B = a2
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given

When a > 1, then a3 > a2.
When a = 1, then a3 = a2 = 1.
When 0 < a < 1, then a3 < a2.
For example, (1/2)3 = 1/8, (1/2)2 = 1/4, and 1/8 < 1/4.

14. m is a negative number.
Quantity A = m3
Quantity B = -m2
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given
Answer: The relationship cannot be determined from the information given

On dividing by m2 (a positive quantity),
Quantity A = m and Quantity B = -1.
When m = -1, then Quantity A equals Quantity B.
When -1 < m < 0, then Quantity A is greater.
When m < -1, then Quantity B is greater.
For example, (-1/2)3 = -1/8, -(-1/2)2 = -1/4, and -1/8 > -1/4.

15. y is a positive integer less than 200.
Quantity A = The number of multiples of 3 between y and 200.
Quantity B = The number of multiples of 5 between y and 200.
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given
Answer: The relationship cannot be determined from the information given

Every third integer is a multiple of 3 and every fifth integer is a multiple of 5. The number of multiples of 3 will typically be greater than the number
of multiples of 5 in a large enough interval (if y is small). However, this will not be the case if y is large, e.g., if y = 199, then there is no multiple of 3 between 199 and 200, and there is only one multiple of 5.

16. In the correctly worked out addition problem given below, each letter represents a different digit.
BC + BC = CAA
Quantity A = A
Quantity B = 6
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given
Answer: The two quantities are equal

BC is a two-digit number, so it is less than 100.
BC + BC gives a three-digit number which is therefore between 100 and 198.
So, C in CAA must represent 1.
Adding the digits in the units place gives 1 + 1 = 2. So, A must represent 2.
Finally, C + C = 12 and therefore C must represent 6.

17. Quantity A = The average of the measures of the angles of a scalene triangle
Quantity B = The average of the measures of the angles of an isosceles triangle
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given
Answer: The two quantities are equal

The average of the three angles of a triangle is the sum divided by 3.
The sum of the angles of any triangle is always 180o.
So, the average is 60o irrespective of the kind of triangle.

18. Quantity A = Twice the area of an equilateral triangle whose sides are a
Quantity B = The area of a square whose sides are a
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given

Let h be the height of the equilateral triangle.
By Pythagoras' Theorem, h2 = a2 - (a/2)2 = 3a2/4.
So, the height h is less than a.
Area of triangle = (base x height)/2.
2 (Area of triangle) = base x height < a2 = Area of square.

19. A small circle is drawn inside a large circle, whose radius is 40% more than the radius of the small circle.
Quantity A = The area of the small circle
Quantity B = The area between the two circles
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given

If r is the radius of the small circle, then 1.4r is the radius of the large circle.
Area of small circle = π r2.
Area between circles = π (1.4r)2π r2 = (1.42 − 1) π r2 = 0.96 π r2.

20. m and n are prime numbers.
m < n and m + n = 16
Quantity A = n
Quantity B = 12
• Quantity A is greater
• Quantity B is greater
• The two quantities are equal
• The relationship cannot be determined from the information given
Answer: The relationship cannot be determined from the information given

3 + 13 = 16 and 5 + 11 = 16 are the two possibilities.
So, n can be either 11 or 13.

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