**Question 1 :**

[(2/3) + (1/4)] ÷ 2 =

(A) 22/12 (B) 1/12 (C) 11/24 (D) 1/3 (E) 12

**Solution :**

** = [(2/3) + (1/4)] ÷ 2**

** = [(2/3) **⋅ (4/4)** + (1/4) **⋅ (3/3)**] ÷ 2**

** = [(8/12) ****+ (3/12)****] ÷ 2**

** = [(8 + 3)/12****] ÷ 2**

** = (11/12)**** ÷ (2/1)**

** = (11/12) **⋅ (1/2)

= 11/24

Hence the answer is 11/24.

**Question 2 :**

Michelle bought a dress that costs $103.00, a pair of shoes that costs $73.00 and a bag that costs $111.00. There is a 7% sales tax on all items priced at $90.00 and higher. There is no sales tax on items under $90.00. How much did Michelle spend on the following items, including tax ?

(A) $287.00 (B) 266.00 (C) 295.47

(D) 300.60 (E) $301.98

**Solution :**

Michelle bought three items.

A dress costs = $103.00

A pair of shoes costs = $73.00

A bag costs = $111.00

The cost of dress and bag are more than $90. So, we have to pay sales tax for those items.

Sales tax = 7%

107% of cost of dress = 1.07 (103.00) = 110.21

107% of cost of bag = 1.07(111.00) = 118.77

cost of shoes = 73

Total cost = 110.21 + 118.77 + 73

= 301.98

**Question 3 :**

How many terms are in the sequence

0, 3, 6.............57, 60 ?

(A) 20 (B) 21 (C) 23 (D) 30 (E) 60

**Solution :**

a = 0, d = 3 - 0 ==> 3 and l = 60

Total number of terms :

n = [(l - a)/d] + 1

n = [(60 - 0)/3] + 1

n = (60/3) + 1

n = 20 + 1

n = 21

**Question 4 :**

The larger of two consecutive even integers is two times the smaller. What is their sum ?

(A) 2 (B) 3 (C) 4 (D) 6 (E) 8

**Solution :**

Let "x" be a even number

Consecutive even number = x + 2

x + 2 = 2x

2x - x = 2

x = 2

So, the two even numbers are 2 and 4.

Their sum = 2 + 4 = 6

**Question 5 :**

3^{x} = 27 ^{a+b }and (a^{2} - b^{2}) /(a-b) = 5, what is x ?

(A) 6 (B) 9 (C) 12 (D) 15 (E) 27

**Solution :**

First, let us simplify (a^{2} - b^{2}) / (a-b)

5 = [(a + b) (a -b)] / (a - b)

5 = (a + b)

So, the value of a + b is 5.

3^{x} = 27^{5 }

3^{x} = (3^{3})^{5 }

3^{x} = 3^{15}

Hence the value of x is 15.

**Question 6 :**

In Mr.Farmer's class there are 30 kids. If there are twice as many boys as there are girls in the English club, then what percentage of the English club are boys ?

(A) 2 (B) 3 (C) 4 (D) 6 (E) 8

**Solution :**

Let "x" be the number of girls in English club

"2x" be the number of boys

x + 2x = 30

3x = 30

x = 10

2x = 20 (number of boys)

Percentage of boys = (20/30) ⋅ 100

= 66.6%

**Question 7 :**

a⊗b = a^{3}-3a^{2}b + 3ab^{2} - b^{3 }and a ⊕b = (a-b)(a-b), what is (a⊗b) / (a⊕b)

(A) a^{2} + b^{2} (B) a - b (C) a^{2} + 3b^{2} + 3ab

(D) a^{2} - b^{2} (E) b^{2} + 3a

**Solution :**

a⊗b = a^{3}-3a^{2}b + 3ab^{2} - b^{3 } = (a - b)^{3}

a ⊕ b = (a - b) (a - b) = (a-b)^{2}

a⊗b/a⊕b = (a - b)^{3}/ (a-b)^{2}

= (a - b)

**Question 8 :**

What is the prime factorization of 752 ?

(A) 2^{3 }⋅ 48 (B) 2^{3 }⋅ 49 (C) 3^{3 }⋅ 49

(D) 2^{4 }⋅ 47 (E) 3^{4 }⋅ 47

**Solution :**

752 = 2^{4} ⋅ 47

**Question 9 :**

If x can be any integer, what is least possible value of the expression 4x^{2} - 10 ?

(A) -10 (B) -4 (C) 4 (D) 10 (E) ∞

**Solution :**

By applying any negative value for x, we get positive value for x^{2}

The least value may appear, when x = 0

= 4(0)^{2} - 10

= -10

Hence the answer is -10.

**Question 10 :**

There is a total of 5 bicycles and tricycles in a park. There are 12 wheels. How many tricycles are there ?

(A) 2 (B) 3 (C) 6 (D) 7 (E) 8

**Solution :**

Let us use the concept system of equations to solve this problem.

Total number of vehicles = 5

Let x be the number of bicycles

Let y be the number of tricycles.

Each bicycle will have 2 wheels and tricycle will have 3 wheels.

x + y = 5 -----(1)

2x + 3y = 12 -----(2)

(1) ⋅ 2 ==> 2x + 2y = 10

(1) - (2)

2x + 2y - (2x + 3y) = 10 - 12

-y = -2

y = 2 (Number of tricycles)

So, there are two tricycles.

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