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 x^{4} =
Solution
Square 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

If n is an odd integer, then which of the following is even?
(I) n^{2}
(II) n^{2} + 1
(III) 3n^{2}  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 n^{2} in terms of k as follows
n^{2} = (2 k + 1)^{2} = 4 k^{2} + 4 k + 1
Let rewrite n^{2} as follows
n^{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.

8×2^{100} + 4×2^{101} =
Solution
Use the fact that 8 = 2^{3} and 4 = 2^{2}
to rewrite the given expression as follows
8×2^{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}

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) n^{2}  n < 0
(II) n^{3} < 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 < 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 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.

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.

(4x^{2}  4) / ( 3x + 3) =
Solution
Factor 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)

For what values of b and c will the equation x^{2} + bx = c have the solutions 2 and 3?
Solution
Since 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

(√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

