# GMAT Math Bible by Jeff J. Kolby, Derrick Vaughn PDF

By Jeff J. Kolby, Derrick Vaughn

Complete Prep for GMAT Math each year, scholars pay \$1,000 and extra to check prep businesses to organize for the mathematics component to the GMAT. you can now get an identical training in a publication. even though the GMAT math part is tough, it's very learnable. GMAT Math Prep direction provides a radical research of GMAT math and introduces a number of analytic strategies that can assist you immensely, not just at the GMAT yet in enterprise university besides. good points: finished overview: Twenty-three chapters supply entire evaluate of GMAT math. perform: contains 164 examples and greater than six hundred routines! prepared from effortless to medium to not easy to very challenging Diagnostic try out: The diagnostic attempt measures your strengths and weaknesses and directs you to parts you want to research extra. functionality: in case your objective is seven hundred+ ranking, this is often the booklet. if you happen to seeking to organize for GMAT math with an emphasis on information sufficiency difficulties, Nova Press is happy to announce GMAT facts Sufficiency Prep direction. on hand right here at Amazon.

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Additional info for GMAT Math Bible

Example text

The sum of the positive integers from 1 through n can be calculated by the formula n(n + 1)/2. Which one of the following equals the sum of all the even numbers from 0 through 20, inclusive? (A) (B) (C) (D) (E) 50 70 90 110 140 26. a, b, c, d, and e are five consecutive numbers in increasing order of size. Deleting one of the five numbers from the set decreased the sum of the remaining numbers in the set by 20%. Which one of the following numbers was deleted? (A) (B) (C) (D) (E) a b c d e 27. A set has exactly five consecutive positive integers starting with 1.

The answer is (D). 1 bh, where b is the base and h is the height. Sometimes the base must be extended 2 in order to draw the altitude, as in the third drawing directly below: The area of a triangle is h h A= h b b 1 bh 2 b In a triangle, the longer side is opposite the larger angle, and vice versa: a 100˚ b 50˚ 50˚ is larger than 30˚, so side b is longer than side a. 30˚ c Pythagorean Theorem (right triangles only): The square of the hypotenuse is equal to the sum of the squares of the legs. c a c 2 = a2 + b 2 b Pythagorean triples: The numbers 3, 4, and 5 can always represent the sides of a right triangle and they appear very often: 52 = 32 + 4 2 .

Then by the Difference of Squares formula a2 – b 2 = (a – b)(a + b) = 4(a + b) [Given that difference between the digits is 4]. This is maximum when (a + b) is maximum. If a and b are single digit numbers, then a + b would be a maximum when a is equal to 9 and b = 9 – 4 = 5 (given the numbers differ by 4). Hence, the maximum value of a 2 – b2 is 9 2 – 52 = 56. The answer is (D). 43 44 GMAT Math Bible  Very Hard 31. We are given that the numbers m and n, when divided by 6, leave remainders of 2 and 3, respectively.