PSAT PRACTICE QUESTIONS IN MATH

PSAT Practice Questions in Math :

Here we are going to see some sample questions for PSAT exams. For each and every questions, you will have solutions with step by step explanation.

PSAT Practice Questions in Math - Practice questions

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) x² + y²

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) y²/x²(B) (x + y)²/y²(C) x²/(y²-x²)

(D) (y²-x²)/y²(E) π

Solution :

Sample space = n(S) = πy²

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

Area of shaded region = πy²- πx²

n(A) = π(y²- x²)

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

= π(y²- x²)/πy²

= (y²- x²)/y²

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 (a² + 2ab + b²)/(a + b)³ ?

(A) a + b (B) a² + b² (C) 1/(a + b)

(D) 1/(a+ b)²(E) (a + b)²

Solution :

(a² + 2ab + b²)/(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 = πr²h 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 :

V₁ = π(2r)² (h/2)

V₁= π(4r²) (h/2)

V₁ = 2πr² h

Ratio of the original volume to the new volume :

= V : V₁

= πr² h : 2πr² 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