# Free GMAT Practice Problems with Solutions Sample 1

 Detailed answer to GMAT problems in sample 1. ABC is a triangle whose sides AB and BC have length of 3 and 5 respectively. Which of the following could be the length of side AC? (I) 8 (II) 12 (III) 0.5 A) (I) only B) (II) only C) (III) only D) (I) and (II) only E) None Solution 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 Given the following If √ x = 3, then x4 = Solution 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 If n is an odd integer, then which of the following is even? (I) n2 (II) n2 + 1 (III) 3n2 - 1 A) (I) only B) (II) only C) (III) only D) (I) and (II) only E) (II) and (III) only Solution 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. 8×2100 + 4×2101 = Solution 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 If 3x + 5y = 5 and x + 3y = 20, then 2x + 4y = Solution 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 taerms of the above equation by 2 2x + 4y = 25 / 2 If n is positive and less than 1, then which of the following is true? (I) n2 - n < 0 (II) n3 < n (III) n + 1 < 1 A) (I) only B) (II) only C) (III) only D) (I) and (II) only E) (II) and (III) only Solution 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. Which of the following has the greatest value? A) 250% B) 2 + 1/2 C) 5 × 0.5 D) 1 / 0.1 E) 4 Solution 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. (4x2 - 4) / (- 3x + 3) = Solution 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) For what values of b and c will the equation x2 + bx = c have the solutions 2 and -3? Solution 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 (√12 - √3)(-√12 + √3) Solution Rewrite as (√12 - √3)(-√12 + √3) = - (√12 - √3) (√12 - √3) and simplify = - (12 - 3) = -9 More Math 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