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Select all that apply

  1. 1. MULTIPLE CHOICE QESTIONS - Select all that apply 1. y Q(1,3) P(1,1) x Figure shows line segment PQ and a circle with radius 1 and center (5,2) in the x-y plane. Which of the following values could be the distance between a point on line segment PQ and a point on the circle? A. 2.5 B. 3.0 C. 3.5 D. 4.0 E. 4.5 F. 5.0 G. 5.5 H. 6.0 Solution; (4,2) O(5,2) Q(1,3) radius=1 (1,2)M N R(6,2) P(1,1) From the figure, the maximum distance=QR=/(6-1)2 +(2-3)2 =/(25+1)=/26<5.5
  2. 2. That means we eliminate choices G and H which are greater than /26. And the minimum distance=MN=4-1=3 That means we eliminate choice A which is less than minimum possible distance 3. Hence our answers are B,C,D,E and F. 2. Which of the following could be the unit digit of 57n , where n is a positive integer? A. 0 B. 1 C. 2 D. 3 E. 4 F. 5 G. 6 H. 7 I. 8 J. 9 Solution; 571 =57 572 =…….9 573 =…….3 574 =…….1 575 =…….7 The sequence of the unit digits of 57n is 7,9,1,3,7,9,1,3….. Hence, our answers are B,D,H and J.
  3. 3. 3. If the lengths of two sides of a triangle are 5 and 9 respectively, which of the following could be the length of the third side of the triangle? A. 3 B. 5 C. 8 D. 15 Solution; A 5 9 B C BC<5+9 or 14 (sum of two sides of a triangle is always greater than the third side) Eliminate D. Likewise, BC+5>9 or, BC>4 Eliminate A. Hence our answers are B and C. 4. Sandy has a husband and two children. She brings at least 1 member of her family to a dinner for lunch every day. The dinner offers 10 lunch specials. If no one orders the same thing, how many orders can Sandy's family make for lunch. (Indicate all possible answers) A. 45 B. 90 C. 120 D. 210 E. 720 F. 5040 Solution;
  4. 4. If she brings 1 family member, 10C2=10!/8!2!=45 If she brings 2 family members, 10C3=10!/7!3!=120 If she brings all 3 family members, 10C4=10!/6!4!=210 Hence our answers are A,C and D. 5. In triangle ABC, angle A=25, angle B>90. Which of the following could be the measure of angle C? Indicate all such measures. A. 12 B. 15 C. 45 D. 50 E. 70 Solution; A+B+C=180 25+B+C=180 C=155-B If B=90, C=155-90=65 But, since B>90, C<65 Hence our answers are A,B,C and D. 6. The total amount that Mary paid for a book was equal to the price of the book plus a sales tax that was 4 percent of the price of the book. Mary paid for the book with a $10 bill and
  5. 5. received the correct change, which was less than $3.00. Which of the following statements must be true? Indicate all such statements. A. The price of the book was less than $9.50 B. The price of the book was greater than $6.90 C. The sales tax was less than $0.45 Solution; Let price of the book be $x. Total price=104% of x=1.04x By given, 7<1.04x<10 or, 6.73<x<9.61 Price of the book was less than $9.61 (let $9.55>$9.50 Choice A is incorrect) Price of the book was greater than $6.73 (let $6.75<$6.90 Choice B is incorrect) Now, for sales tax, 6.73*0.04<x*0.04<9.61*0.04 (since sales tax=0.04x) or, 0.2692<sales tax<0.3844 Sales tax was less than 0.3844 which means it was less than $0.45 also. That means the only correct choice is C. 7. The length of AB is 10/3. B A D C Which of the following statements individually provides sufficient additional information to determine the area of triangle ABC above? Indicate all such statements.
  6. 6. A. DBC is an equilateral triangle. B. ABD is an isosceles triangle C. The length of BC is equal to the length of AD. D. The length of BC is 10. E. The length of AD is 10. Solution; Area of triangle ABC=1/2* BC* AB=1/2*10/3*BC=5/3*BC Now if we know the length of BC, then we know the area of the triangle. So choice D is one of the correct choices. Now in case of choice A, if triangle DBC is an equilateral triangle, then angle BCD=60. That means triangle ABC would be a 30-60-90 triangle in which side AB=10/3. That means we could determine the other sides BC and CA also. That means we could determine the area of the triangle. So choice A is also one of the correct choices. Now in case of choice B, knowing that triangle ABD is an isosceles triangle would not provide any value of side BC. Hence it does not provide enough information to determine the area of triangle ABC. Similar is the case with choice C. BC=AD. Since the length of AD is not known, the length of BC will also be unknown. Hence it is not the correct choice either. Now our last choice E says AD=10. Just knowing the length of side AD would not provide any information about length of side BC. Hence it is not the correct choice. So our answers are A and D. Note something: here, choices C and E combined would provide enough information to determine the area of triangle ABC. Since, AD=10 and BC=AD=10. That means we could determine the area of triangle ABC if choices C and E were combined. So you can see the question has emphasized the word 'individually'. 8. In triangle ABC, the measure of angle B is 90. The length of side AB is 4 and the length of side BC is x. If the length of hypotenuse AC is between 4 and 8, which of the following could be the value of x? Indicate all such values.
  7. 7. A. 1 B. 2 C. 3 D. 4 E. 5 F. 6 Solution; A 4 B x C AC=/(x2 +42 =/(x2 +16) We are given, 4<AC<8 or, 4</(x2 +16)<8 or, 16< x2 +16<64 or, 0<x2 <48 or, 0<x</48 or, 0<x<6.92 Hence all given choices are correct. 9. 3.7, 4.1, a, 8.5, 9.2, 2a The six numbers shown are listed in increasing order. Which of the following values could be the range of the six numbers? Indicate all such values.
  8. 8. A. 4.0 B. 5.2 C. 7.3 D. 11.6 E. 12.9 F. 14.1 Solution; Range=2a-3.7 We are given 4.1<a<8.5 We are also given, 9.2<2a or, 4.6<a Hence we get 4.6<a<8.5 or, 2*4.6<2a<2*8.5 or, 9.2<2a<17 or, 9.2-3.7<2a-3.7<17-3.7 or, 5.5<Range<13.3 Hence our answers are C, D and E. 10. Every year between 70% and 85% of the students at Central High School attend the homecoming rally. If one year, 1435 students attended the rally, which of the following could have been the number of students at Central High School that year? Indicate all possible numbers of students. A. 1675 B. 1775 C. 1875 D. 1975 E. 2075 Solution;
  9. 9. 70% of total=1435 or, 0.7*total=1435 or, total=1435/0.7=2050 And 85% of total=1435 or, 0.85*total=1435 or, total=1435/0.85=1688 That means the range of students=1688-2050 Hence our answers are B,C and D. 11. If a-b>a+b, where a and b are integers, which of the following must be true? Indicate all that apply. A. a<0 B. b<0 C. ab<0 Solution; a-b>a+b -b>b b is negative. choice B is true. a could be positive or negative. Ex 3-(-4)>3+(-4) or -3-(-4)>-3+(-4) That means choice A is not necessarily true. Similar is the case with choice C. If a=-ve and b=-ve, then ab=+ve if a=+ve and b=-ve then ab=-ve That means choice C is not necessarily true either.
  10. 10. Hence our correct answer is choice B. 12. x, y and z are positive integers. The average (arithmetic mean) of x, y and z is 11. If z two greater than x, which of the following must be true? indicate ALL that apply. A. x is even B. y is odd C. z is odd Solution; x+y+z=3*11=33 x+y+(x+2)=33 2x+y=31 In this equation, the quantity 2x is always even no matter what x be. Since 2x(even)+y=31(odd), so y is odd. (even+odd=odd….ex 4+3=7) x can be either even or odd and z=x+2=(even or odd)+even=even or odd Hence only choice B is true. 13. A quality control analyst collected 200 measurements of length. The range of 200 measurements is 17 centimeters, and one of the measurements is 49.5 centimeters. Which of the following could be another one of the 200 measurements, in centimeters? Indicate ALL such measurements. A. 31.0 B. 32.0 C. 33.0 D. 34.0 Solution; Range=L-S=17 centimeters
  11. 11. {……………………………………….49.5 cms…………………………………………………………} 200 data Here the possibility is that 49.5 cms could be either the largest among 200 data or the smallest among 200 data. If the given one data 49.5 cms was the largest, Range=49.5-S 17=49.5-S S=49.5-17=32.5 If it was the smallest, Range=L-49.5 17=L-49.5 L=17+49.5=66.5 That means the largest possible data is 66.5 cms and the smallest possible data is 32.5 cms. Hence our answers are C and D. 14. USED CARS SOLD LAST MONTH AT CAR DEALERSHIP X Price Range number of cars sold under $5,000 7 $5,000-$7,409 10 $7500-$10,000 8 over $10,000 6 For the 31 used cars sold last month at Car Dealership X, which of the following could be the median price? Indicate ALL such prices. A. $5,500 B. $6,500 C. $7,000 D. $8,500
  12. 12. E. $9,000 F. $10,500 Solution; Median=value of (N+1)th/2 item=value of (31+1)th/2=value of 16th item 16th item lies in the range $5,000-$7,409 Hence our answers are A,B and C. 15. A survey asked 1,150 people to choose their favorite laundry detergent from brands A,B and C. Of the people surveyed, x percent chose A as their favorite brand. If x is rounded to the nearest integer, the result is 3. Which of the following could be the number of people who chose A as their favorite brand? Indicate ALL such numbers. A. 20 B. 25 C. 30 D. 35 E. 40 F. 45 G. 50 Solution; 2.5<=x<3.5 Any number greater than or equal to 2.5 or less than 3.5 will be rounded to 3. 2.5% of 1150=28.75 3.5% of 1150=40.25 28.75<=x<40.25 Hence our answers are C,D and E.
  13. 13. 16. If x,y and z are positive integers such that 3x<2y<4z, which of the following statements could be true? Indicate All such statements. A. x=y B. y=z C. y>z D. x>z Solution; 3x<2y If x=y, 3x>2y Ex x=1 and y=1, then 3*1>2*1 That means choice A is incorrect. 2y<4z if y=z, 2y<4z Ex y=z=2, then 2*2<4*2 if y>z, 2y<4z Ex y=5, z=3, then 2*5<4*3 That means choice B and C could be correct. 3x<4z if x>z, 3x<4z Ex x=6, z=5, then 3*6<4*5 That means choice D could also be correct. Hence correct answers are B,C and D. 17. Roney's policy is such that his spending does not exceed 35% of his income. If one month, he spent $1000, what could be his income per year? Indicate ALL such values. A. $30,000 B. $34,000 C. $35,000 D. $36,000 E. $36,500
  14. 14. Solution; spending<=0.35*income 1000<=0.35*income income>=1000*12/0.35 per year income>=$34285.7 Hence our answers are C,D and E. 18. If a positive integer n, divided by 5 has a remainder 2, which of the following must be true? Select ALL such statements. A. n is odd B. n+1 cannot be a prime number C. n+2 divided by 7 has remainder 2 D. n+3 is divisible by5 Solution; n=5k+2 (k=0,1,2…..) Let k=0. then n=5*0+2=2 which is even. Choice A is incorrect. Let k=0. then n+1=(5*0+2)+1=3 which is a prime number. So choice B is incorrect. Let k=0. then n+2=(5*0+2)+2=4 and 4 divided by 7 gives a remainder 4 not 2. So choice C is incorrect. Let k=0. then n+3=(5*0+2)+3=5 which is divisible by 5. So choice D is correct. 19. One side of a triangle has length 8 and a second side has length 5. Which of the following could be the area of the triangle? Select ALL that apply.
  15. 15. A. 24 B. 20 C. 5 Solution; C 5 N A M 8 B Area of triangle ABC=1/2*AB*CM=1/2*8*CM=4*CM<20 CM<5 (in right angled triangle AMC, AC=5 which is hypotenuse is the longest side) or, area of triangle ABC=1/2*AC*BN=1/2*5*BN=2.5*BN<20 BN<8 (in right angled triangle ANB, AB=5 which is hypotenuse is the longest side) So eliminate choice A. We could draw a triangle with sides 5 and 8 and height 5 so that area will be 20. C A B and we could draw a triangle with sides 5 and 8 and height 1.25 so that area will be 5. In the above triangle, AB=8,AC=5 and AB is perpendicular to CA so that area=1/2*8*5=20 C A M B In the above triangle, AB=8,AC=5 and CM=1.25 so that area=1/2*8*1.25=5 Hence our answers are B and C. Another simple way to solve this problem is to have a little knowledge about trigonometry.
  16. 16. C A M B The area of the triangle ABC in trigonometric form=1/2*AB*CM=1/2*AB*CA*sinA So area=1/2*8*5*sinA=20*sinA The value of sine of a function ranges from from 0 to 1. So minimum possible area=20*sin0=0(when the angle between two given lines is 0) and maximum possible area=20*sin90=20*1=20(when the angle between two given lines is 90) So choices B and C are true. 20. A triangle has a perimeter 13. The two shorter sides have integer lengths equal to x and x+1. Which of the following could be the length of the other side? Select as many as are correct. A. 4 C B. 6 x x+1 C. 8 A y B Solution; Let y be the longest side. x+x+1+y=13 2x+1+y=13 2x+y=12 x=(12-y)/2 If y=4, then x=4 and x+1=5 sides will be y=4,x=4 and x+1=5 which don't satisfy the question because we need y to be the longest. Choice A is not true.
  17. 17. If y=6, then x=3 and x+1=4 Here, y is the longest and the sum of any two sides is also greater than the remaining side. So choice B is true. If y=8, then x=2 and x+1=3 here, 2+3<8 i.e. the sum of two sides of is not greater than the third side. So the sides don't form a triangle. Hence our answer is B only. 21. In the xy-plane, line k is a line that does not pass through the origin. Which of the following statements individually provide(s) sufficient additional information to determine whether the slope of line k is negative? Indicate ALL such statements. A. The x-intercept of line k is twice the y-intercept of line k. B. The product of the x-intercept and the y-intercept of line k is positive. C. Line k passes through the points (a,b) and (r,s), where (a-r)(b-s)<0. Solution; x/a+y/b=1 x/2b+y/b=1(x-intercept=2*y-intercept) x+2y=2b y=-(1/2)x+b in which slope=-1/2 i.e. negative. So choice A is true. a*b=positive When a=+ve,b=+ve and when a=-ve,b=-ve x/a+y/b=1 bx+ay=ab y=(-bx+ab)/a=-(b/a)x+b in which the slope is -(b/a) i.e. -ve. x/(-a)+y/(-b)=1
  18. 18. -bx-ay=ab y=(-bx-ab)/a=-(b/a)x-b in which the slope is -(b/a) i.e. -ve. So choice B is true. Slope of the line k=b-s/a-r=-ve (a-r)(b-s)<0 When a-r=+ve, then b-s=-ve when a-r=-ve, then b-s=+ve So choice C is also true. Hence all are correct choices. 22. Each employee of a certain company is in either Department X or Department Y, and there are more than twice as many employees in Department X as in Department Y. The average salary is $25,000 for the employees in Department X and is $35,000 for the employees in Department Y. Which of the following amounts could be the average salary for all of the employees in the company? Indicate ALL such amounts. A. $26,000 B. $28,000 C. $29,000 D. $30,000 E. $31,000 F. $32,000 G. $34,000 Solution; The combined average salary would be greater than $25,000 since greater average salary has been combined with less average salary and the combined average will be greater than the less average salary.
  19. 19. If there were exactly twice as many employees in Department X as in Department Y, then combined average salary=(2*25,000+1*35,000)/(2+1)=$28,333 But, since there are more employees with less average salary, the combined salary of employees of two Departments must be less than $28,333. Eliminate choices C,D,E,F and G. Hence our answers are A and B. 23. The integer v is greater than 1. If v is the square of an integer, which of the following numbers must also be the square of an integer? Indicate ALL such statements. A. 81v B. 25v+10/v+1 C. 4v2 +4/v+1 Solution; v=22 ,32 ,42 ,52 ,…….. so let v=k2 (where k=2,3,4…..) Now, 81v=92 *k2 =(9*k)2 Choice A is square of an integer. 25v+10/v+1=52 *k2 +10*k+1=(5k)2 +2*5k*1+12 =(5k+1)2 Choice B is also square of an integer. Let v=22 then 4v2 +4/v+1=4*(22 )2 +4*2+1=64+8+1=73 Choice C is not the square of an integer. Hence A and B are the correct answer choices. 24. A and B are two independent events and the probability that both events occur is 1/2. Which of the following could be the probability of occurrence of event A? Select ALL that apply.
  20. 20. A. 0 B. 1/4 C. 1/2 D. 3/4 Solution; P(A and B)=P(A)*P(B)=1/2 We know, 0<P(A)<=1 and 0<P(B)<=1 If P(A)=0, then P(A)*P(B)=0 so it cannot be the correct answer. If P(A)=1/4, then, 1/4*P(B)=1/2 P(B)=1/2*4=2, which is not possible. So it is not correct choice either. If P(A)=1/2, then, 1/2*P(B)=1/2 P(B)=1, which is possible. So it can be the probability of occurrence of event A. If P(A)=3/4, then, 3/4*P(B)=1/2 P(B)=1/2*4/3=2/3, which is also possible. So it also can be the probability of occurrence of event A. Hence our answers are C and D. 25. If the area of a rectangle is 40, which of the following could be the perimeter of the rectangle? Indicate all such perimeters. A. 20 B. 40 C. 200 D. 400 E. 2,000
  21. 21. F. 4,000 Solution; Perimeter will be the smallest when the rectangle is a square. i.e., when l=b So A=l2 =40 or, l=/40=6.324 So, smallest perimeter possible=4*6.324=25.29 Eliminate A. Now let's see the largest perimeter possible. If l=0.001 and b=40,000 so that l*b=40 as given. Then Perimeter=2(0.001+40,000)=80,000.002 That means, we can make the perimeter as large as we desire, by making the difference between length and breadth of the rectangle large. So all choices except A are correct. 26. Which of the following is an equation of a line that is perpendicular to the line whose equation is 2x+3y=4? Indicate ALL such equations. A. 3x+2y=4 B. 3x-2y=4 C. 2x-3y=4 D. 4-3x=-2y E. 4+2x=3y Solution; Slope of given line=-2/3 So the line perpendicular to given line must have slope=3/2 (m1*m2=-1 for two perpendicular lines)
  22. 22. Slope of line A=-3/2 Slope of line B=-3/-2=3/2 Slope of line C=-2/-3=2/3 Slope of line D=-(-3)/2=3/2 Slope of line E=-2/-3=2/3 Note: Slope of a line in general form Ax+By+C=0 is -A/B. Hence our answers are B and D. 27. Which of the following inequalities have at least one positive solution and at least one negative solution? Indicate all such inequalities. A. 5x/3<x B. x3 <x C. x-6<x-7 Solution; 5x/3<x 5x<3x For this inequality, x cannot be positive otherwise 5x>3x For ex, let x=1, then 5*1>3*1 which means it has negative solution only. A is not the answer. x3 <x Here, x can be positive as well as negative. Ex, let x=0.5, then 0.53 =0.125<0.5 which means x can be positive. now, let x=-2, then (-2)3 =-8<-2 which means c can be negative also. So B is the answer. x-6<x-7
  23. 23. -6<-7 which is not true. This inequality doesn't have any solution. Hence our answer is B only.

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