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100 100#= Simplify by multiplying numerator and denominator by 4: 250 100 100 , , 40 x 4 4 1 000 40 000 # # #= = = example 4 John is now m years old and Sally is 4 years older than John. Which represents Sally\u2019s age 6 years ago? (A) m 1 10 (B) m 2 10 (C) m 2 2 (D) m 2 4 (E) 4m 2 6 Choice C is correct. Translate: John is now m years old \u2193 \u2193 \u2193 J 5 m Sally is 4 years older than John \u2193 \u2193 \u2193 \u2193 \u2193 S 5 4 1 J Sally\u2019s age 6 years ago \u2193 \u2193 S 2 6 So we get: J 5 m S 5 4 1 J and find: S 2 6 5 4 1 J 2 6 S 2 6 5 J 2 2 S 2 6 5 m 2 2 (substituting m for J) See Math Strategy 7, Example 2 (page 92) for an alternate approach to solving this problem, using a different strategy: Use Specific Numerical Examples to Prove or Disprove Your Guess. example 5 Phil has three times as many DVDs as Sam has. Even after Phil gives Sam 6 DVDs, he still has 16 more DVDs than Sam has. What was the original number of DVDs that Phil had? (A) 20 (B) 24 (C) 28 (D) 33 (E) 42 Choice E is correct. Translate: Phil has three times as many DVDs as Sam has \u2193 \u2193 \u2193 \u2193 \u2193 P 5 3 3 S Even after Phil gives Sam 6 DVDs, he still has 16 { \u2193 \u2193 \u2193 \u2193 \u2193 P 2 6 5 16 more DVDs than Sam has { \u2193 \u2193 1 S 1 6 Sam now has (S 1 6) DVDs because Phil gave Sam 6 DVDs. So we end up with the equations: P 5 3S P 2 6 5 16 1 S 1 6 Find P; get rid of S: P 5 3S; P S3 = 6 16 6 62 P P P P 3 3 48 18 2 = + + = + + 3P 2 18 5 48 1 P 1 18 2P 5 84 P 5 42 example 6 If q is 10% greater than p and r is 10% greater than y, qr is what percent greater than py? (A) 1% (B) 20% (C) 21% (D) 30% (E) 100% Choice C is correct. GRE2015_P04.indd 78 4/18/14 11:28 AM STRATEGY SECTION \u2022 79 Translate: If q is 10% greater than p { \u2193 \u2193 \u2193 q 5 100 10 p 1 p and r is 10% greater than y { \u2193 \u2193 \u2193 r 5 100 10 y 1 y qr is what percent greater than py? \u2193 \u2193 \u2193 qr 5 x 100 py 1 py So we have three equations: q p p p r y y y qr x py py x py 100 10 100 10 1 100 10 100 10 1 100 100 1 = + = + = + = + = + = + c c ` m m j 1 2 3 Multiply 1 and 2 : qr py100 10 1 2 = +c m 4 Now equate 4 with 3 : qr x py py100 1 100 10 1 2 = + = +` cj m You can see that 1x100 10 1 100 2 + = +c m , canceling py. , , , x x x 100 10 1 10 000 100 2 100 10 100 1 10 000 100 100 20 100 21 100 21 So 2 + = + = + + = = = c m 1+c m The answer is x 5 21. Alternate approach: Choose numbers for p and for y: Let p 5 10 and y 5 20 Then, since q is 10% greater than p: q 5 10% greater than 10 10 10 11q 100 10 = + =c m Next, r is 10% greater than y: r 5 10% greater than 20 Or, r 5 10 10 11q 100 10 = + =c m20 1 20 5 22 Then: qr 5 11 3 22 and py 5 20 3 10 So, to find what percent qr is greater than py, you would need to find: 11 22 20 102 py qr py 100 20 10 100 or2 # # # # # This is: 100 21200 42 # = example 7 Sales of Item X Jan2Jun, 2004 Month Sales ($) Jan 800 Feb 1,000 Mar 1,200 Apr 1,300 May 1,600 Jun 1,800 According to the above table, the percent increase in sales was greatest for which of the following periods? (A) Jan2Feb (B) Feb2Mar (C) Mar2Apr (D) Apr2May (E) May2Jun Choice A is correct. The percent increase from Month A to Month B 5 Sales (Month A) Sales (Month B) Sales (Month A) 100# - Month Sales ($) Periods % Increase in Sales Jan 800 Jan2 Feb 1,000 800 100 100800 800 2002 # #= Feb 1,000 Feb2 Mar , 1,200 1,000 100 , 100 2 1 000 1 000 200 # #= Mar 1,200 Mar2 Apr , 1,300 1,200 100 , 100 2 1 200 1 200 100 # #= Apr 1,300 Apr2 May , 1,600 1,300 100 , 100 2 1 300 1 300 300 # #= May 1,600 May2 Jun , 1,800 1,600 100 , 100 2 1 600 1 600 200 # #= Jun 1,800 You can see that 800 200 100# (Jan2Feb) is the greatest. GRE2015_P04.indd 79 4/18/14 11:28 AM 80 \u2022 GRUBER\u2019S COMPLETE GRE GUIDE 2015 Know how to Find Unknown Quantities (areas, Lengths, arc and angle Measurements) from Known Quantities (the Whole Equals the Sum of Its Parts) When asked to find a particular area or length, instead of trying to calculate it directly, find it by subtracting two other areas or lengths\u2014a method based on the fact that the whole minus a part equals the remaining part. This strategy is very helpful in many types of geometry problems. A very important equation to remember is The whole 5 the sum of its parts 1 Equation 1 is often disguised in many forms, as seen in the following examples: example 1 X Y Z In the diagram above, \u394XYZ has been inscribed in a circle. If the circle encloses an area of 64, and the area of \u394XYZ is 15, then what is the area of the shaded region? (A) 25 (B) 36 (C) 49 (D) 79 (E) It cannot be determined from the information given. Choice C is correct. Use equation 1 . Here, the whole refers to the area within the circle, and the parts refer to the areas of the shaded region and the triangle. Thus, Area within circle 5 Area of shaded region 1 Area of \u394XYZ 64 5 Area of shaded region 1 15 or Area of shaded region 5 64 2 15 5 49 (Answer) Math StratEg y 3 example 2 In the diagram below, \u2014 AE \u2014 is a straight line, and F is a point on \u2014 AE \u2014 . Find an expression for m+DFE. A B C D E x° y° ? 60° F (A) x 1 y 2 60 (B) x 1 y 1 60 (C) 90 2 x 2 y (D) 120 2 x 2 y (E) 180 2 x 2 y Choice D is correct. Use equation 1 . Here, the whole refers to the straight angle, +AFE, and its parts refer to +AFB, +BFC, +CFD, and +DFE. Thus, m+AFE 5 m +AFB 1 m+BFC 1 m+CFD 1 m+DFE 180 5 x 1 60 1 y 1 m+DFE or m+DFE 5 180 2 x 2 60 2 y m+DFE 5 120 2 x 2 y (Answer) example 3 In the diagram below, AB 5 m, BC 5 n, and AD 5 10. Find an expression for CD. (Note: Diagram represents a straight line.) B C DA (A) 10 2 mn (B) 10 2 m 2 n (C) m 2 n 1 10 (D) m 1 n 2 10 (E) m 1 n 1 10 GRE2015_P04.indd 80 4/18/14 11:28 AM STRATEGY SECTION \u2022 81 Choice B is correct. Use equation 1 . Here, the whole refers to AD, and its parts refer to AB, BC, and CD. Thus, AD 5 AB 1 BC 1 CD 10 5 m 1 n 1 CD or CD 5 10 2 m 2 n (Answer) example 4 The area of triangle ACE 5 64. The sum of the areas of the shaded triangles ABF and FDE is 39. What is the side of square BFDC? A B C D E F (A) 5 (B) 4 (C) 5 (D) 44 (E) Cannot be determined. explanatory answer Choice A is correct. Since we are dealing with areas, let\u2019s establish the area of the square BFDC, which will then enable us to get its side. Now, the area of square BFDC 5 area of triangle ACE 2 (area of triangles ABF 1 FDE) Area of square BFDC 5 64 2 39 5 25 Therefore, the side of square BFDC 5 5. example 5 B A 3 O In the figure above, O is the center of the circle. Triangle AOB has side 3 and angle AOB 5 90°. What is the area of the shaded region? (A) 9 4 2 12rc m (B) 9 122 r` j (C) 9(p 2 1) (D) 9 24 4 1rc m (E) Cannot be determined. explanatory answer Choice A is correct. Subtract knowns from knowns: Area of shaded region 5 area of quarter circle AOB 2 area of triangle AOB Area of quarter circle AOB 5 ( )4 3 2r (since OA 5 3 and area of a quarter of a circle 5 4 1 3 p 3 radius2). Area of triangle AOB 5 2 3 3# (since OB 5 3 and area of a triangle 5 2 1 base 3 height). Thus, area of shaded region 5 9 24 9 2 9 4 2 12 =r rc m. example 6 The sides in the square above are each divided into five equal segments. What is the value of Area of shaded region Area