# Free GMAT Practice Problems with Solutions

Sample 1

Detailed answer to GMAT problems in sample 1.
## Solution to Question 1The 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 2Square both sides of the given equation(√ x) ^{2} = 3^{2}Simplify to find x x = 9 Elevate both sides to the power 4 x ^{4} = 9^{4}Simplify x ^{4} = 6561
## Solution to Question 3Since n is odd integer, it can be written asn = 2 k + 1 , where k is an integer Let us express n ^{2} in terms of k as followsn ^{2} = (2 k + 1)^{2} = 4 k^{2} + 4 k + 1
Let rewrite n ^{2} as followsn ^{2} = 2(2 k^{2} + 2 k) + 1
Hence n ^{2} is odd.We now express n ^{2} + 1 in terms of k.n ^{2} + 1 = 2(2 k^{2} + 2 k) + 1 + 1 = 2(2 k^{2} + 2 k + 1)
Hence n ^{2} + 1 is even.We now express 3 n ^{2} - 1 in terms of k.3 n ^{2} - 1 = 3 [2(2 k^{2} + 2 k) + 1
] - 1
= 3 [2(2 k ^{2} + 2 k) ] + 3 - 1
= 6(2 k ^{2} + 2 k) + 2
Hence 3 n ^{2} - 1 is even.Answer E. ## Solution to Question 4Use the fact that 8 = 2^{3} and 4 = 2^{2}
to rewrite the given expression as follows8×2 ^{100} + 4×2^{101} = 2^{3}×2^{100} + 2^{2}
×2^{101}Use rules of exponent to simplify = 2 ^{103} + 2^{103}= 2×2 ^{103} = 2^{104}## Solution to Question 5Add 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 6n is positive and less than 1 is translated as follows0 < 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} < n^{2}Since n ^{2} < n, we have0 < 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 7Use decimal numbers to rewrite the above expressionsA) 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 8Factor numerator as follows(4x ^{2} - 4) = 4(x^{2} - 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 9Since x = 2 is a solution of x^{2} + bx = c, then(2) ^{2} + b(2) = c or 4 + 2b = c
Since x = -3 is a solution of x ^{2} + 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 10Rewrite as(√12 - √3)(-√12 + √3) = - (√12 - √3) (√12 - √3) and simplify = - (12 - 3) = -9 ## More References and Links to Maths Practice TestsFree Practice for GAMT Math testsFree 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 |