Free GMAT Practice Problems with Solutions
Sample 1

Detailed answer to GMAT problems in sample 1.

Solution to Question 1

The length of the third side of a triangle is always less than the sum and greater than the difference of the lengths of other two sides.
The sum S and the difference D of the given sides is.
S = 3 + 5 = 8
D = 5 - 3 = 2
None of the values suggested in (I), (II) or (III) satisfy the above condition. Answer E

Solution to Question 2

Square both sides of the given equation
(√ x)2 = 32
Simplify to find x
x = 9
Elevate both sides to the power 4
x4 = 94
Simplify
x4 = 6561

Solution to Question 3

Since n is odd integer, it can be written as
n = 2 k + 1 , where k is an integer
Let us express n2 in terms of k as follows
n2 = (2 k + 1)2 = 4 k2 + 4 k + 1
Let rewrite n2 as follows
n2 = 2(2 k2 + 2 k) + 1
Hence n2 is odd.
We now express n2 + 1 in terms of k.
n2 + 1 = 2(2 k2 + 2 k) + 1 + 1 = 2(2 k2 + 2 k + 1)
Hence n2 + 1 is even.
We now express 3 n2 - 1 in terms of k.
3 n2 - 1 = 3 [2(2 k2 + 2 k) + 1 ] - 1
= 3 [2(2 k2 + 2 k) ] + 3 - 1
= 6(2 k2 + 2 k) + 2
Hence 3 n2 - 1 is even.
Answer E.

Solution to Question 4

Use the fact that 8 = 23 and 4 = 22 to rewrite the given expression as follows
8×2100 + 4×2101 = 23×2100 + 22 ×2101
Use rules of exponent to simplify
= 2103 + 2103
= 2×2103 = 2104

Solution to Question 5

Add the left and right hand sides of the given equations to obtain a new equation
(3x + 5y) + (x + 3y) = (5) + (20)
Simplify
4x + 8y = 25
Divide all terms of the above equation by 2
2x + 4y = 25 / 2

Solution to Question 6

n is positive and less than 1 is translated as follows
0 < n < 1
Multiply all terms of the above inequality by n to obtain
0 < n2 < n
which also gives
n2 - n < 0
Hence statement (I) is true
Multiply all terms of the above inequality by n to obtain
0 < n3 < n2
Since n2 < n, we have
0 < n3 < n
Hence statement (II) is true
For n = 0.75, statement (III) is not true
Hence statement (I) and (II) only are true.

Solution to Question 7

Use decimal numbers to rewrite the above expressions
A) 250% = 250/100 = 2.5
B) 2 + 1/2 = 2 + 0.5 = 2.5
C) 5 × 0.5 = 2.5
D) 1 / 0.1 = 10
E) 4 = 4
The expression 1 / 0.1 has the greatest value.

Solution to Question 8

Factor numerator as follows
(4x2 - 4) = 4(x2 - 1) = 4(x - 1)(x + 1)
Factor denominator as follows
(- 3x + 3) = - 3(x - 1)
Substitute in the given expression and simplify
(4x2 - 4) / (- 3x + 3) = [ 4(x - 1)(x + 1) ] / [ - 3(x - 1)]
= (- 4/3)(x + 1)

Solution to Question 9

Since x = 2 is a solution of x2 + bx = c, then
(2)2 + b(2) = c or 4 + 2b = c
Since x = -3 is a solution of x2 + bx = c, then
(-3)2 + b(-3) = c or 9 - 3b = c
We now have a system of simultaneous equations in b and c to solve. Combining the above equation, we obtain
4 + 2b = 9 - 3b
Solve for b
5b = 5 , b = 1
Substitute b by 1 in the equations 4 + 2b = c and solve for c
4 + 2(1) = c or c = 6
Values for b and c
b = 1 , c = 6

Solution to Question 10

Rewrite as
(√12 - √3)(-√12 + √3) = - (√12 - √3) (√12 - √3)
and simplify
= - (12 - 3) = -9

More References and Links to Maths Practice Tests

Free Practice for GAMT Math tests
Free Compass Math tests Practice
 Free Practice for SAT, ACT Math tests
Free GRE Quantitative for Practice
Free AP Calculus Questions (AB and BC) with Answers

{ezoic-ad-1}
{ez_footer_ads}