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
x
4 = 94
Simplify
x
4 = 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
n
2 = (2 k + 1)2 = 4 k2 + 4 k + 1
Let rewrite n2 as follows
n
2 = 2(2 k2 + 2 k) + 1
Hence n2 is odd.
We now express n2 + 1 in terms of k.
n
2 + 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 n
2 - 1 = 3 [2(2 k2 + 2 k) + 1 ] - 1
= 3 [2(2 k
2 + 2 k) ] + 3 - 1
= 6(2 k
2 + 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
82
100 + 42101 = 232100 + 22 2101
Use rules of exponent to simplify
= 2
103 + 2103
= 22
103 = 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 < n
2 < n
which also gives
n
2 - n < 0
Hence statement (I) is true
Multiply all terms of the above inequality by n to obtain
0 < n
3 < n2
Since n2 < n, we have
0 < n
3 < 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
(4x
2 - 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
(4x
2 - 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

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