Title

GRE Math Practice Test II
P. 2 (Problems 10 - 19)

WTAMU > Virtual Math Lab > GRE Math

 Because of the enormity of this part, I split the answer/discussion into three pages.  P. 1 has the answer/explanations to all levels of the first 9 problems. This page, P. 2, has the answer/explanations to all levels of problems 10 - 19.  P. 3 has the answer/explanations to all levels of problems 20 - 28. This page also has links to the mathematical area(s) that each question comes from in case you need to go back and review. Printing warning: Note that because of all of the levels per problem, there are a lot of questions on this page.  So if you wish to print this out, note that there will be a lot of pages to print.

 10 - 1.

 Column A Column B 6

 Answer: B Since the square root of 36 is 6, then 6 is greater than . Need more help on these topics?  Square Root

 10 - 2.

 Column A Column B

 Answer: A Since you will not be able to use a calculator on the test, let me show you how you can solve this algebraically.  I'm going to rewrite Column A so that it is under one radical:

 *Add like radicals *2 is the same as square root of 4

 is greater than . Need more help on these topics?   Square Root

 10 - 3.

 Column A Column B

 Answer: B Since you will not be able to use a calculator on the test, let me show you how you can solve this algebraically.  I'm going to rewrite Column B so that it is under one radical:

 *2 is the same as square root of 4

 is greater than . Need more help on these topics?   Square Root

 10 - 4.

 Column A Column B

 Answer: C Rationalizing the denominator in Column B we get:

 *Multiply num. and den. by sq. root of 7

 is equal to . Need more help on these topics?   Square Root

 10 - 5.

 Column A Column B

 Answer: B Rewriting Column B we get:

 Approx. .775 *Simplify the fraction

 Since is approximately .775 and = .6,  has a greater value than . Need more help on these topics?   Square Root

 11 - 1.  If the radius of the equal semicircles is r and the length of the rectangle is 3r, which of these expressions represents the perimeter of this object?

 Answer: B Since we need to find the perimeter, we need to find the measurement around the figure.  The two straight lines each have a measurement of 3r, for a total of 6r. The other part of the figure is two semicircles.  These two parts make a one whole circle.  The circumference of a circle is  Plugging in r for radius we have:  Adding all of the parts together we get: The perimeter of this figure is. Need more help on these topics?   Formulas for Two-Dimensional Figures

 11 - 2.  Using the figure shown, find the area of the middle region inside the square.

 Answer: D Since part of the problem involves the area of a square, we can use the formula : Plugging in 10 for the side we get the following area for the square: Since part of the problem involves the area of a circle, we can use also use the formula: Plugging in 5 for the radius we get the following area for the four quarter circles (whole circle): In this problem,  if we take the area of the square and subtract out the area of the four quarter circles (whole circle), we will have the area of the middle region of the given figure above: The area of the middle region is . Need more help on these topics?   Formulas for Two-Dimensional Figures

 11 - 3.  A ramp 5 feet long is leaning against a raised platform which is 3 feet above the ground. What is the distance from the ramp’s contact point with the ground and the base of the platform?

 Answer: B Anytime you are missing one side of a right triangle, you can use the Pythagorean theorem to find that long lost side: The variables in this formula represent the following: a and b = legs of the right triangle c = hypotenuse of the right triangle In this problem,  a = ? = this is the variable we are looking for b = 3 c = 5  Plugging these values into the formula and solving for a we get:

 *Plug in 3 for b and 5 for c   *Inv. of add 9 is sub. 9 *Square root method

 Since a represents the length of the side of a triangle, then we can only use the positive value. The distance from the ramp’s contact point with the ground and the base of the platform is 4 feet. Need more help on these topics?   Formulas for Two-Dimensional Figures

 11 - 4. A ramp 10 feet long is leaning against a raised platform which is 5 feet above the ground. What is the distance from the ramp’s contact point with the ground and the base of the platform?

 Answer: C Anytime you are missing one side of a right triangle, you can use the Pythagorean theorem to find that long lost side: The variables in this formula represent the following: a and b = legs of the right triangle c = hypotenuse of the right triangle   In this problem,  a = ? = this is the variable we are looking for b = 5 c = 10  Plugging these values into the formula and solving for a we get:

 *Plug in 5 for b and 10 for c   *Inv. of add 25 is sub. 25 *Square root method

 Since a represents the length of the side of a triangle, then we can only use the positive value. The distance from the ramp’s contact point with the ground and the base of the platform is  feet. Need more help on these topics?   Formulas for Two-Dimensional Figures

 11 - 5.  What is the area of the rectangle above?

 Answer: A   The area of a rectangle is length times width.  So, we need the length and the width to get our answer.  It looks like we have the length which is 5, but we still need the width.  Note how we have the diagonal of the rectangle, which is 7.  The diagonal, length and width of the rectangle make up an right triangle.  Anytime you are missing one side of a right triangle, you can use the Pythagorean theorem to find that long lost side: The variables in this formula represent the following: a and b = legs of the right triangle c = hypotenuse of the right triangle   In this problem,  a = ? = this is the variable we are looking for b = 5 c = 7  Plugging these values into the formula and solving for a we get:

 *Plug in 5 for b and 7 for c   *Inv. of add 25 is sub. 25   *Square root method

 2 square root of 6 is the width of the rectangle.  Putting the length and the width values of the rectangle into the area formula we get:

 *Plug in 5 for length and 2 square root of 6 for width

 The area of the rectangle is . Need more help on these topics?   Formulas for Two-Dimensional Figures

 12 - 1.  =

 Answer: E Using the order of operations we get:

 *(  )  *Multiplication  *Add and subtract

 The answer is 25. Need more help on these topics?   Order of Operations

 12 - 2.  =

 Answer: B Using the order of operations we get:

 The answer is 29. Need more help on these topics?   Order of Operations

 12 - 3.  =

 *Writing equivalent fractions with LCD of 32     *Add

 The answer is . Need more help on these topics?    Fractions and Order of Operations

 12 - 4.    =

 Answer: E Using the order of operations we get:

 *exponents     *Writing equivalent fractions with LCD of 16   *Add

 The answer is . Need more help on these topics?   Fractions, Exponents and Order of Operations

 12 - 5.    =

 Answer: B Using the order of operations we get:

 *Exponents *Rewrite division as mult. of the reciprocal                   *Writing equivalent fractions with LCD of 20 *Add

 The answer is . Need more help on these topics?   Fractions, Exponents and Order of Operations

 13 - 1.

 Answer: C Using the FOIL method we get:

 *First Outside Inside Last *Add like terms

 The answer is . Need more help on these topics?   Multiplying Polynomials

 13 - 2.

 Answer: E Using the FOIL method and distributive property we get:

 *First Outside Inside Last *Distribute the 7 *Add like terms

 The answer is . Need more help on these topics?   Multiplying Polynomials

 13 - 3.

 Answer: C Squaring the binomial we get:

 *Square the binomial *First Outside Inside Last *Add like terms

 The answer is . Need more help on these topics?   Multiplying Polynomials

 13 - 4.

 Answer: A Squaring the binomial and then using distributive property we get:

 *Square the binomial   *First Outside Inside Last *Distribute the 5 *Add like terms

 The answer is . Need more help on these topics?   Multiplying Polynomials

 13 - 5.

 Answer: A Cubing the binomial we get:

 *Cube the binomial *Multiply last two binomials     *Add like terms

 The answer is . Need more help on these topics?   Multiplying Polynomials

 14 - 1.   A student got 80% of 40 questions correct.  How many questions did he get correct?

 Answer: E When you take a percent of a number you multiply the decimal form of that percent times the number. Finding 80% of 40:

 *Move decimal two to the left to write 80% as .8 *Multiply

 He got 32 questions correct. Need more help on these topics?   Decimals and Percents

 14 - 2.   Quiz scores on a history exam are 9, 9, 7, 10, and 5.  If there is one more quiz, what would need to be made on that quiz to have an overall average of 7?

 Answer: A In this problem we are given the mean and we need to find one of our values.  Keep in mind that this is still a mean problem.  We will still use the idea that we need to sum up the exams and then divide it by 6 to get the mean.  We can let our unknown exam be x. Using the mean formula we get:

 *(sum of tests)/(# of tests) = mean *Solve for x (missing quiz) *Mult. both sides by LCD of 6       *Inverse of add 40 is sub. 40

 A score of 2 would be needed to have an overall average of 7. Need more help on these topics?   Averages (Mean)

 14 - 3.   A student has quiz scores of 70, 75 and 55.  If tests count as 2 quiz grades, what would he have to make on a test, so that the overall average is 70?

 Answer: D In this problem we are given the mean and we need to find one of our values.  Keep in mind that this is still a mean problem.  We will still use the idea that we need to sum up the exams and then divide it by 5 to get the mean.  We can let our unknown exam be x.  Keep in mind that a test score counts twice. Using the mean formula we get:

 *(sum of tests)/(# of tests) = mean *Need 2 x's since tests count twice *Solve for x (missing test) *Mult. both sides by LCD of 5     *Inverse of add 200 is sub. 200 *Inverse of mult. by 2 is divide by 2

 The test score would have to be at least 75. Need more help on these topics?   Averages (Mean)

 14 - 4.   If 30 students in one class had an average of 80% and 10 students from another class had another average of 90%, what is the average in % of all 40 students?

 Answer: C In general, when finding the average of tests, you add the tests together and then divide by the number of tests.  On this problem we have two separate averages to start with. So we need to find the total  number of points we have using that information and then divide by the number of students, which, in this case, is 40. If 30 students averaged 80, then the total number of points would be (30)(80) = 2400. If 10 students averaged 90, then the total number of points would be (10)(90) = 900. So the total number of points is 2400 + 900 = 3300. To find the average, take the number of points and divide by the number of students: The overall average percent is 82.5%. Need more help on these topics?   Averages (Mean)

 14 - 5.   If 5 students in one class had an average of 55%, 15 students in one class had an average of 80% and 7 students from another class had another average of 92%, approximately, what is the average in % of all 27 students?

 Answer: A In general, when finding the average of tests, you add the tests together and then divide by the number of tests.  On this problem we have three separate averages to start with. So we need to find the total  number of points we have using that information and then divide by the number of students, which in this case, is 27. If 5 students averaged 55, then the total number of points would be (5)(55) = 275.  If 15 students averaged 80, then the total number of points would be (15)(80) = 1200. If 7 students averaged 92, then the total number of points would be (7)(92) = 644.   So the total number of points is 275 + 1200 + 644 = 2119. To find the average, take the number of points and divide by the number of students: The overall average percent is 78.5%. Need more help on these topics?   Averages (Mean)

 15 - 1.   and

 Column A Column B

 Answer: B Since x is less than 0, then it is a negative number. Since y is greater than 0, then it is a positive number. With that in mind, when you multiply x times y, you will have a negative times a positive.  That means the product xy will be negative. When you square x, you will end up with a positive value. Since ALL positive numbers are greater than ALL negative numbers,  will be greater than .   Need more help on these topics?   Exponents and Linear Inequality

 15 - 2. and

 Column A x Column B y

 Answer: D Solving the first inequality for x we get:

 *Inv. of add 1 is sub. 1

 Solving the second inequality for y we get:

 *Inv. of sub. 1 is add 1

 Even though 6 is greater than 4, we cannot determine the relationship because x can be ANY number less than 4 and y can be ANY number less than 6.  So y could be 2 and x could be 3 OR y could be 2 and x could be -5, etc...   Need more help on these topics?  Linear Inequality

 15 - 3. and

 Column A x Column B y

 Answer: D Solving the first inequality for x we get:

 *Inv. of mult. by 2 is div. by 2

 Solving the second inequality for y we get:

 *Inv. of mult. by 3 is div. by 3

 We cannot determine the relationship because x can be ANY number greater than 2 and y can be ANY number less than 10.  So y could be 2 and x could be 3 OR y could be 5 and x could be 3, etc.... Need more help on these topics?  Linear Inequality

 15 - 4.

 Column A Column B 9

 Answer: A Note how 15x + 6y is three times 5x + 2y. Taking 3 times the given inequality we get:

 *Multiply BOTH sides by 3

 15x + 6y is greater than 9. Need more help on these topics?  Linear Inequality

 15 - 5.

 Column A Column B 8

 Answer: B Note how -2x + 3y is a negative four times 1/2x - 3/4y.  Taking -4 times the given inequality we get:

 *Multiply BOTH sides by -4 *Since we are mult. both sides by a negative number we need to change the inequality sign

 8 is greater than -2x + 3y. Need more help on these topics?  Linear Inequality

 16 - 1.  How much was spent on rent in 2002?

 Answer: B Since 45% of the basic expenditures was spent on rent, then we will need to take 45% of \$15000 to find the answer to this question.  Whenever you take a percent of a number you take the decimal form of that percent times the number:

 *Move decimal on 45% two places to the left

 \$6750 was spent on rent in 2002. Need more help on these topics?   Percent and Problem Solving

 16 - 2.  What was the total amount spent on food and clothing in 2002?

 Answer: B Together food and clothing is 35% + 15% = 50% of the basic expenditures. Since 50% of the basic expenditures was spent on food and clothing, then we will need to take 50% of \$15000 to find the answer to this question.  Whenever you take a percent of a number you take the decimal form of that percent times the number:

 *Move decimal on 50% two places to the left

 \$7500 was spent on food and clothing in 2002. Need more help on these topics?  Percent and Problem Solving

 16 - 3.  How much more money was spent on clothing than insurance in 2002?

 Answer: A We need to find the difference between what was spent on clothing versus insurance.  There was 15% - 5% = 10% more spent on clothing than insurance. Since 10% more of the basic expenditures was spent on clothing than on insurance, then we will need to take 10% of \$15000 to find the answer to this question.  Whenever you take a percent of a number you take the decimal form of that percent times the number:

 *Move decimal on 10% two places to the left

 \$1500 more was spent on clothing than insurance in 2002. Need more help on these topics?  Percent and Problem Solving

 16 - 4.  How much more was spent on rent and insurance than on food in 2002?

 Answer: E Together, rent and insurance is 45% + 5% = 50% of the basic expenditures We need to find the difference between what was spent on rent and insurance versus food.  There was 50% - 35% = 15% more spent on rent and insurance than food. Since 15% more of the basic expenditures was spent on rent and insurance than on food, then we will need to take 15% of \$15000 to find the answer to this question.  Whenever you take a percent of a number you take the decimal form of that percent times the number:

 *Move decimal on 15% two places to the left

 \$2250 more was spent on clothing than insurance in 2002. Need more help on these topics?  Percent and Problem Solving

 16 - 5.  If the amount that was spent on food is 5% higher in 2002 than 2001, approximately what amount was spent on food in 2001?

 Answer: B We need to be careful here, the temptation is to just take 40% - 5% = 35% of 15000.   However, keep in mind that we are talking about 5% higher in terms of amount, not  compared to the other expenditures and the total amount  may or may not have been the same overall for 2001. The first thing we need to find is the actual amount spent on food in 2002.  According to the chart, food made up 35% of the 15,000 spent.  Taking 35% of 15,000 we get:

 *Move decimal on 35% two places to the left

 5250 is 5% higher than what was spent on food  in 2001.  Letting x represent the 2001 amount spent on food and putting this into an equation we get:

 *5% higher means we ADD 5% (or .05) of the amount   *Divide BOTH sides by 1.05

 \$5000 was spent on food  in 2001. Need more help on these topics?   Percent and Problem Solving

 17 - 1.

 Column A The ratio of food to clothing. Column B The ratio of clothing to food.

 Answer: A You can do this problem two different ways.  You can either take the ratio of the percents for each expenditure or you can find the amount that go with each one and take that ratio.  Either way you would end up with the same answer. I’m going to use the percents that are given to set up each ratio. For the ratio of food to clothing, we need to make sure that the percent attached to food goes on top and the percent for clothing goes on bottom: For the ratio of clothing to food, we need to make sure that the percent attached to clothing goes on top and the percent for food goes on bottom: It looks like the ratio of food to clothing is greater than the ratio of clothing to food. Need more help on these topics?   Percent and Problem Solving

 17 - 2.

 Column A The ratio of food to rent. Column B The ratio of rent to food.

 Answer: B You can do this problem two different ways.  You can either take the ratio of the percents for each expenditure or you can find the amount that go with each one and take that ratio.  Either way you would end up with the same answer. I’m going to use the percents that are given to set up each ratio. For the ratio of food to rent, we need to make sure that the percent attached to food goes on top and the percent for rent goes on bottom: For the ratio of rent to food, we need to make sure that the percent attached to rent goes on top and the percent for food goes on bottom: It looks like the ratio of rent to food is greater than the ratio of food to rent.   Need more help on these topics?  Percent and Problem Solving

 17 - 3.

 Column A The ratio of food to insurance. Column B The ratio of insurance to food.

 Answer: A You can do this problem two different ways.  You can either take the ratio of the percents for each expenditure or you can find the amount that go with each one and take that ratio.  Either way you would end up with the same answer. I’m going to use the percents that are given to set up each ratio. For the ratio of food to insurance, we need to make sure that the percent attached to food goes on top and the percent for insurance goes on bottom: For the ratio of insurance to food, we need to make sure that the percent attached to insurance goes on top and the percent for food goes on bottom: It looks like the ratio of food to insurance is greater than the ratio of insurance to food.   Need more help on these topics?  Percent and Problem Solving

 17 - 4.

 Column A The ratio of insurance and rent to food and clothing. Column B The ratio of food and clothing to insurance and rent.

 Answer: C The total percent for insurance and rent is 5% + 45% = 50%. The total percent for food and clothing is 35% + 15% = 50%. You can do this problem two different ways.  You can either take the ratio of the percents for each expenditure or you can find the amount that go with each one and take that ratio.  Either way you would end up with the same answer. I’m going to use the percents that are given to set up each ratio. For the ratio of insurance and rent to food and clothing, we need to make sure that the percent attached to insurance and rent goes on top and the percent for food and clothing goes on bottom: For the  ratio of food and clothing to insurance and rent, we need to make sure that the percent attached to food and clothing goes on top and the percent for insurance and rent goes on bottom: It looks like the two ratios are the same.   Need more help on these topics?  Percent and Problem Solving

 17 - 5.

 Column A The ratio of food and insurance  to clothing and rent. Column B The ratio to clothing and rent to food an insurance.

 Answer: B The total percent for food and insurance is 35% + 5% = 40%. The total percent for clothing and rent is 15% + 45% = 60%. You can do this problem two different ways.  You can either take the ratio of the percents for each expenditure or you can find the amount that go with each one and take that ratio.  Either way you would end up with the same answer. I’m going to use the percents that are given to set up each ratio. For the ratio of food and insurance to clothing and rent, we need to make sure that the percent attached to food and insurance goes on top and the percent for clothing and rent goes on bottom: For the ratio of clothing and rent to food and insurance, we need to make sure that the percent attached to clothing and rent goes on top and the percent for food and insurance goes on bottom: The ratio of clothing and rent to food and insurance is greater than the ratio of food and insurance to clothing and rent.   Need more help on these topics?  Percent and Problem Solving

 18 - 1.  Find the area of rectangle ABCD shown above.

 Answer: E The area of a rectangle is length times width. Side BC can be found by taking the absolute value of the difference of the x values of their ordered pairs, which would be is |6 - 2| = 4.  Side BA can be found by taking the absolute value of the difference of the y values of their ordered pairs, which would be |1 - (-2)| =  3.  Putting these values into the area formula for a rectangle we get: The area of the rectangle is 12.   Need more help on these topics?  Formulas for Two-Dimensional Figures and Coordinate Geometry

 18 - 2.  Find the area of the circle shown above.

 Answer: B The area of a circle is . Note how the circle’s widest part goes from (3, 4) to (-3, 4).  This means that the diameter spans 6 units.  If the radius is half of the diameter, that means the radius is 3. Putting 3 in for the radius in the area of a circle formula we get: The area of the circle is .   Need more help on these topics?  Formulas for Two-Dimensional Figures and Coordinate Geometry

 18 - 3.  Find the area of triangle ABC shown above.

 Answer: B The area of a triangle is .     The base, side AC, can be found by taking the absolute value of the difference of the x values of their ordered pairs, which would be is |4 - (-4)| = 8.  The height can be found by taking the absolute value of the difference of the y values of B and the origin, which would be |7 - 0| =  7.  Putting in 8 for the base and 7 for the height into the area of a triangle formula we get:   The area of the triangle is 28.   Need more help on these topics?  Formulas for Two-Dimensional Figures and Coordinate Geometry

 18 - 4.  Find the area enclosed by BCDE shown above.

 Answer: D From the picture, we can see that figure BCDE is a quadrilateral but it doesn’t fit into a form like rectangle, parallelogram, etc...  We will have to take the area of rectangle ABCD and subtract from it the area of triangle ABE, this will leave us with the area of figure BCDE.  Lets look at rectangle ABCD first.  The area of a rectangle is length times width.  Side AB can be found by taking the absolute value of the difference of the y values of their ordered pairs, which would be |5 - 0| = 5. Side BC can be found by taking the absolute value of the difference of the x values of their ordered pairs, which would be |4 - (- 4)| = 8. Plugging 8 in for the length and 5 in for the width into the formula for the area of a rectangle we get: The area of rectangle ABCD is 40.   Now lets look at triangle ABE.  The area of a triangle is .  So we need to know the base and the height of this triangle.  The base of this triangle is side AE.  The height of this triangle is side AB. Side AE can be found by taking the absolute value of the difference of the x values of their ordered pairs, which would be |-2 - (-4)| = 2. Side AB can be found by taking the absolute value of the difference of the y values of their ordered pairs, which would be |5 - 0| = 5. Plugging 2 in for the base and 5 in for the height into the formula for the area of a triangle we get: The area of triangle ABE is 5. Last, we need to take the difference of the area of the rectangle and  triangle: 40 - 5 = 35 The area of the figure BCDE is 35.   Need more help on these topics?  Formulas for Two-Dimensional Figures and Coordinate Geometry

 18 - 5.  Find the area enclosed by BCDE shown above.

 Answer: C From the picture, we can see that figure BCDE is a quadrilateral but it doesn’t fit into a form like rectangle, parallelogram, etc...  We will have to take the area of triangle ABC and subtract from it the area of triangle AED, this will leave us with the area of figure BCDE.  Let's look at triangle ABC first.  The area of a triangle is .  So we need to know the base and the height of this triangle.  Since it is a right triangle, the base and height are going to be the lengths of the two legs, which are sides AB and BC.  Side BC can be found by taking the absolute value of the difference of the y values of their ordered pairs, which would be |-7 - 0| = 7. Side AB can be found by taking the absolute value of the difference of the x values of their ordered pairs, which would be |-7 - 0| = 7. Plugging 7 in for the base and 7 in for the height into the formula for the area of a triangle we get: The area of triangle ABC is 24.5.   Now let's look at triangle AED.   So we need to know the base and the height of this triangle.  The base of this triangle is side AE.  The height of this triangle is side DE. Side AE can be found by taking the absolute value of the difference of the x values of their ordered pairs, which would be |-7 - (-3)| = 4. Side DE can be found by taking the absolute value of the difference of the y values of their ordered pairs, which would be |0 - (-4)| = 4. Plugging 4 in for the base and 4 in for the height into the formula for the area of a triangle we get: The area of triangle AED is 8. Last, we need to take the difference of the areas of the two triangles: 24.5 - 8 = 16.5 The area of the figure BCDE is 16.5.   Need more help on these topics?  Formulas for Two-Dimensional Figures and Coordinate Geometry

 19 - 1.   What is the solution(s) to  ?

 *Factor the difference of squares   *Set 1st factor = 0         *Set 2nd factor = 0

 This problem has two solutions, x = -3 and x = 3. Need more help on these topics?  Solving Quadratic Equations

 19 - 2.   What is the solution(s) to  ?

 *Factor out the GCF of 2 *Factor the difference of squares *Set 1st factor = 0         *Set 2nd factor = 0

 This problem has two solutions, x = -7 and x = 7. Need more help on these topics?  Solving Quadratic Equations

 19 - 3.   What is the solution(s) to  ?

 *Solve for x         *Square root method *Square root of -1 is i

 There is no real number solution to this problem. Need more help on these topics?  Solving Quadratic Equations

 19 - 4.   What is the solution(s) to  ?

 *Factor the trinomial *Set the 1st factor = 0 *Inv. of add 2 is sub. 2 *Inv. of mult. by 5 is div. by 5             *Set the 2nd factor = 0 *Inv. of sub. 1 is add 1 *Inv. of mult. by 3 is div. by 3

 This problem has two solutions, x = -2/5 and x = 1/3. Need more help on these topics?  Solving Quadratic Equations

 19 - 5.   What is the solution(s) to  ?

 *Inv. of mult. by 2/5 is mult. by 5/2 *Square root method

 This problem has two solutions, x =  and x = . Need more help on these topics?  Solving Quadratic Equations

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