GRE GEOMETRY AND ARITHMETIC PROBLEMS
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This document contains multiple geometry problems and their solutions: 1) Calculating the area of a shaded portion using Pythagoras theorem. 2) Determining that the relationship between the area of a quadrilateral and a given quantity cannot be determined without more information. 3) Finding that quantity B is greater based on the relationships between variables x and y. 4) Calculating the areas of two rectangles where one length and breadth vary by 10% and 20% respectively from a third rectangle, and determining that the area of rectangle A is greater.
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GRE GEOMETRY AND ARITHMETIC PROBLEMS
- 2. 4cm 3 cm 2cm x Calculate area of the shaded portion By Pythagoras theorem x = 5 cm A(rectangle)= 5(2)=10 cm mathews.suman@gmail.com
- 3. A B C D Diameter of the circle is 10 cm Quantity A Quantity B A(quadrilateral ABCD) 40 a) Quantity A is greater b) Quantity B is greater c) The 2 quantities are equal d) The relationship cannot be determined from the information given. Answer is d Even if AC were the diameter of the circle, to determine the area , we need more than 1 side of the figure mathews.suman@gmail.com
- 4. 𝑥2 𝑦 > 0, 𝑥𝑦2 < 0 Quantity A x Quantity B y if 𝑥2 𝑦 > 0, 𝑦 > 0 If x𝑦2 < 0, 𝑥 < 0 Hence y is greater. Answer = B mathews.suman@gmail.com
- 5. Given 3 rectangles A, B, C, The length of rectangle A IS 10 % more than the length of C and the breadth of A is 10 % lesser than the breadth of C. The length of rectangle B is 20 % more than the length of rectangle C The breadth of rectangle B is 20 % lesser than the breadth of rectangle C. Quantity A A(rectangle A) Quantity B A(rectangle B) C x y A 1.1x 0.9y B 1.2x 0.8y mathews.suman@gmail.com
- 6. A(rectangle A ) = 1.1x(0.9y)=0.99xy A(rectangle B)= 1.2x (0.8y) = .96xy A (rectangle A ) is greater Ans : = Quantity A mathews.suman@gmail.com