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Question 1 :
Five apples and 6 bananas together cost as much as 8 bananas and 9 pears. One apple costs as much as 2 pears. For the same price as 1 pear, how many bananas could be bought?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 6
Solution :
Let "x", "y" and "z" be the cost pf one apple, one banana and one pears.
5x + 6y = 8y + 9z
5x + 6y - 8y = 9z
5x - 2y = 9z ---(1)
Given that :
One apple costs as much as 2 pears.
x = 2z
Applying the value of x in (1), we get
5(2z) - 2y = 9z
10z - 9z = 2y
z = 2y
Cost of 1 pear is equal too 2 bananas.
Question 2 :
Let R = 8 · 9 · 7 · 13 and S = 4 · 6 · 11 · 49. The greatest common factor of R and S is
(A) 12 (B) 42 (C) 56 (D) 168 (E) 504
Solution :
R = 8 · 9 · 7 · 13 and S = 4 · 6 · 11 · 49
R = 23 · 32 · 7 · 13, S = 23 · 3 · 11 · 72
Greatest common factor of R and S = 23 · 3 · 7
= 168
Hence the required GCF is 168.
Question 3 :
What is the value of 4|x – y| - 3|xy| if x = -5 and y = 2?
(A) -42 (B) -18 (C) -2 (D) 42 (E) 58
Solution :
= 4|x – y| - 3|xy|
= 4|-5 – 2| - 3|(-5)(2)|
= 4|-7| - 3|(-10)|
= 4(7) - 3(10)
= 28 - 30
= -2
Question 4 :
In the following sequence of eight numbers, each number has one more “1” than the number before it:
2, 12, 112, 1112,. . . . , 11111112
What is the hundreds digit in the sum of all eight of these numbers?
(A) 0 (B) 6 (C) 7 (D) 8 (E) 9
Solution :

In 8th column, we have 8 two's. By adding these two's we get 16.
We write 6 in the unit place and carry 1.
In ten's place, we have 7 one's. So we have to write 7 + 1 = 8.
In hundreds place, we have six 1's.So the required answer is 6.
Question 5 :
In Ms. Hsiao’s class there are twice as many boys as there are girls. Half of the boys are in the math club and all of the girls are in the math club. What percent of Ms. Hsiao’s class is in the math club?
(A) 25% (B) 33 (1/3)% (C) 50%
(D) 66 (2/3)% (E) 75%
Solution :
Let "x" be the number of girls
"2x" be the number of boys.
Total number of students = x + 2x = 3x
Number of boys in math club = (1/2) 2x = x
Number of students in math club = x + x = 2x
Percentage of students in math club
= (2x/3x)⋅100%
= 200/3
= 66 2/3 %
Question 6 :
The total cost to rent an inflatable slide for a birthday party is $23 per hour with an initial $15 setup fee. Which of the following equation represents the relationship between the total cost c in dollars, and the number of hours h, to rent the inflatable slide ?
(A) c = 15 h (B) c = 23h (C) c = 15h + 23
(D) c = 23h + 15
Solution :
Initial setup fee = $15
Charge per hour = $23
c = 23h + 15
So, option D is correct.
Question 7 :
-3x = y
(5x + y)2 = 64
If (x, y) is a solution to the system of equations above, what is the value of xy ?
(A) -3 (B) -24 (C) -48 (D) -96
Solution :
-3x = y -----(1)
(5x + y)2 = 64 -----(2)
Applying y = -3x in (2), we get
(5x - 3x)2 = 64
(2x)2 = 64
4x2 = 64
x2 = 16
x = -4 and x = 4
When x = -4, y = -3(-4) ==> 12
When x = 4, y = -3(4) ==> -12
xy = -4(12) ==> -48
So, option C is correct.
Question 8 :
The function f is defined by f(x) = 5x2 + 2x + 5. What is the value of f(2) ?
Solution :
f(x) = 5x2 + 2x + 5
To find f(2), we apply x = 2
f(2) = 5(2)2 + 2(2) + 5
= 5(4) + 4 + 5
= 20 + 4 + 5
= 29
So, the answer is 29.
Question 9 :
10 - 3 (x + 2) = 3x + 4
How many different values of x satisfy the equation above ?
(A) None (B) One (C) Two (D) More than two
Solution :
10 - 3 (x + 2) = 3x + 4
10 - 3x - 6 = 3x + 4
-3x + 4 = 3x + 4
-3x - 3x = 4 - 4
-6x = 0
Dividing by -6, we get
x = 0
So, there is only one solution for x. Option B is correct.
Question 10 :
In the xy-plane, line l contains the point (1, 3) and has slope 2. Which of the following is an equation of line l ?
(A) y = 2x + 1 (B) y = 3x + 2 (C) y = 2x + 3
(D) y = x + 2
Solution :
Equation of the line :
(y - y1) = m(x - x1)
y - 3 = 2(x - 1)
y = 2x - 2 + 3
y = 2x + 1
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