**Question 1 :**

If x and y are integers, which is always even ?

(A) (x + y)/2 (B) 2(x + y)/x (C) x - y

(D) 2(x + y) (E) x^{2} + y^{2}

**Solution :**

Since x and y are integers, the sum of those integers will be odd or even. The multiple of 2 will not be odd. So the answer is 2 (x + y).

**Question 2 :**

What is the probability that a point chosen will be in the shaded region ?

(A) y^{2}/x^{2 }(B) (x + y)^{2}/y^{2 }(C) x^{2}/(y^{2}-x^{2})

(D) (y^{2}-x^{2})/y^{2 }(E) π

**Solution :**

Sample space = n(S) = πy^{2}

Let "a" be the event that a point chosen in a shaded region.

Area of shaded region = πy^{2 }- πx^{2}

n(A) = π(y^{2 }- x^{2})

P(A) = n (A) / n (S)

= π(y^{2 }- x^{2})/πy^{2}

= (y^{2 }- x^{2})/y^{2}

Let us look into the next example on "PSAT Practice Questions in Math".

**Question 3 :**

If q = x + y and x = y + z, what is z in terms of y and q?

(A) q - 2x (B) q - 2y (C) 2x - q (D) 2y + q (E) 2q + x

**Solution :**

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

x = y + z ------(2)

z = x - y

From the first equation, let us find the value of x in terms of y and q

x = q - y

z = (q - y) - y

z = q - y - y

z = q - 2y

**Question 4 :**

What is (5 !)!/ 5! ?

(A) 1 (B) 5 (C) 120 (D) 720 (E) None of these

**Solution :**

Let x be 5 !

(5 !)!/ 5! = x!/x

= x (x- 1)!/x

= (x - 1)!

= (5! - 1)!

= (120 - 1)!

= 119!

Hence none of the above is the answer.

**Question 5 :**

What is (a^{2} + 2ab + b^{2})/(a + b)^{3} ?

(A) a + b (B) a^{2} + b^{2} (C) 1/(a + b)

(D) 1/(a+ b)^{2 }(E) (a + b)^{2}

**Solution :**

(a^{2} + 2ab + b^{2})/(a + b)^{3}

= (a + b)^{2}/(a + b)^{3}

= 1/(a + b)

**Question 6 :**

It takes 3 cats 3 minutes to catch 3 mice. How many cats are needed to catch 99 mice in 99 minutes ?

(A) 3 (B) 6 (C) 11 (D) 33 (E) 99

**Solution :**

**Given :** It takes 3 cats 3 minutes to catch 3 mice.

That is,

3 cats -----> 3 minutes -----> 3 mice

Let us assume,

3 cats = 1 man

Then, we have

1 man -----> 3 minutes -----> 3 mice

Based on the above statement, we can consider the following situations also.

1 man -----> 1 minute -----> 1 mouse

1 man -----> 5 minutes -----> 5 mice

1 man -----> 10 minutes -----> 10 mice

In this way, we can have

1 man -----> 99 minutes -----> 99 mice

But, we know that,

1 man = 3 cats

Then, we have

3 cats -----> 99 minutes -----> 99 mice

Therefore, 3 cats are needed to catch 99 mice in 99 minutes.

**Question 7 :**

The sum of seven consecutive odd integers is 749. What is the largest of the seven integers ?

(A) 99 (B) 103 (C) 11 (D) 113 (E) 115

**Solution :**

Let x be the odd integer.

6 consecutive odd integers will be (x+2), (x+4), (x+6), (x+8), (x+10) and (x+12).

Sum of seven consecutive integers = 749

x + x+2 + x+4 + x+6 + x+8 + x+10 + x+12 = 749

7x + 42 = 749

7x = 749 - 42

7x = 707

x = 707/7 ==> 101

Largest number of seven consecutive odd numbers

= x + 12

= 101 + 12

= 113

**Question 8 :**

V = πr^{2}h Using the formula, if r is doubled and h is divided by 2, what is the ratio of the original volume to the new volume ?

(A) 1 : 4 (B) 1 : 2 (C) 1 : 1 (D) 2 : 1 (E) 4 : 1

**Solution :**

r is doubled means r = 2r

h is divided by 2 means h = h/2

New volume :

V_{1} = π(2r)^{2} (h/2)

V_{1 }= π(4r^{2}) (h/2)

V_{1} = 2πr^{2} h

Ratio of the original volume to the new volume :

= V : V_{1}

= πr^{2} h : 2πr^{2} h

= 1 : 2

Hence the required ratio is 1 : 2.

**Question 9 :**

A yellow cab has the base fare of $3.50 per ride plus $0.20 for each 1/4 of mile ridden. If a yellow cab costs $ 22.50, how many miles long was the ride ?

(A) 23.75 miles (B) 42.5 miles (C) 47.5 miles

(D) 95 miles (E) 112.5 miles

**Solution :**

Let "x" be the number of miles ridden

To find the (1/4)^{th} miles of x miles, we have to multiply 4 by x.

So, there are 4x (1/4)^{th} miles in x miles.

To understand this, let us consider the following example in the picture given below.

Cost of yellow cab = $22.50

3.50 + 0.20 (4x) = 22.50

0.80x = 22.50 - 3.50

0.80x = 19

(80x/100) = 19

x = 19(100)/80

x = 23.75 miles

Hence he traveled 23.75 miles.

**Question 10 :**

John works 40 hours a week, and his monthly salary in June was $4000. In the month July, John got 4% raise on his monthly salary. In the month of July, what was John's hourly rate ?

(A) $25 (B) $26 (C) $40 (D) $100 (E) $104

**Solution :**

June month salary of John = $4000

Percentage of salary raised = 4%

New salary in the month July = 4000 + 4% of 4000

= 4000 + 160

= 4160

John's 1 week salary = 4160/4 = 1040

He works 40 hours per week.

Hourly rate for the month July = 1040/40 = $26

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