Numeric entry


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Numeric entry

  1. 1. REVISED GRE - Numeric Entry (No Choice Given) 1. The item was listed at $65 and sold at $56.16. Two successive discounts were given on the item and the first discount was 10%, Find the second discount percent. Solution; let second discount be x%. then, 65*0.9*(1-x%)=56.16 1-x%=0.96 x%=0.04 x=4% Hence the second discount was 4%. 2. Two numbers are in the ratio 6:13. The least common multiple of the two numbers is 312. Find the sum of the two numbers. Solution; Let the numbers be 6x and 13x. Then, LCM of 6x and 13x=78x=312 So, x=312/78=4 Sum=6*4+13*4=19*4=76 3. Find the value of x3 /y3 +y3 /x3 if x/y+y/x=6. Solution; x/y+y/x=6
  2. 2. Taking cube on both sides, (x/y+y/x)3 =63 (x/y)3 +3*(x/y)2 (y/x)+3*(x/y)(y/x)2 +(y/x)3 =216 since (a+b)3 =a3 +3a2 b+3ab2 +b3 x3 /y3 +y3 /x3 +3x/y+3y/x=216 x3 /y3 +y3 /x3 +3*6=216 x3 /y3 +y3 /x3 =216-18=198 4. How many sides does a polygon with 44 diagonals have? Solution; Number of diagonals in a polygon of sides n is given by n(n-3)/2 where n>3 For ex, for a rectangle, there are 4(4-3)/2=2 diagonals. For a hexagon, there are 6(6-3)/2=9 diagonals and so on. So by given, 44=n(n-3)/2 88=n2 -3n n2 -3n-88=0 n2 -11n+8n-88=0 n(n-11)+8(n-11)=0 (n-11)(n+8)=0 n=11 or -8 Sides cannot be negative so n=11.
  3. 3. 5. What is the greatest value of a positive integer n such that 3n is a factor of 1812 ? Solution; 1812 =(2*3*3)12 =324 *212 Here, the power of 3 is 24. So for 3n to be an integer of 324 *212 , the greatest possible value of n is also 24. 6. A merchant made a profit of $5 on the sale of a sweater that cost the merchant $15. What is the profit expressed as a percent of the merchant's cost? Give your answer to the nearest whole percent. Solution; Profit%=Profit/CP*100%=5/15*100=33.333….=$33 (to the nearest whole percent) 7. RESULTS OF A USED-CAR AUCTION Small Cars Large Cars Number of cars offered 32 23 Number of cars sold 16 20 Projected sales total for cars offered(in '000) $70 $150 Actual sales total(in '000) $41 $120 For the large cars sold at an auction that is summarized in the table above, what was the average sale price per car? $ Solution; $120*1000/20=$6,000 8. A list of numbers has a mean of 8 and a standard deviation of 2.5. If x is a number in the list that is two standard deviations above the mean, what is the value of x?
  4. 4. Solution; x=mean+(2*standard deviation)=8+(2*2.5)=13 9. Line k lies in the x-y plane. The x-intercept of line k is -4 and line k passes through the mid- point of the line segment whose end-points are (2,9) and (2,0). What is the slope of line k? Give your answer as a fraction. Solution; (2,9) (2,4.5) (-4,0) (2,0) Slope of line k=4.5-0/2-(-4)=4.5/6=9/12=3/4 (slope of a line passing through two points (x1,y1) and (x2,y2) is given by y2-y1/x2-x1) 10. Y Frequency 1/2 2 3/4 7 5/4 8 3/2 8 7/4 9 The table shows the frequency distribution of the values of a variable Y. What is the mean of the distribution? Give your answer to the nearest 0.01. Solution;
  5. 5. mean=(1/2*2+3/4*7+5/4*8+3/2*8+7/4*9)/(2+7+8+8+9)=1.2941…=1.29(to the nearest 0.01) 11. Of the 20 light-bulbs in a box, two are defective. An inspector will select two lightbulbs simultaneously and at random from the box. What is the probability that neither of the light- bulbs will be defective? Give your answer as a fraction. Solution; P(neither defective)=P(1st not defective)*P(2nd not defective)=(18/20)*(17/19)=153/190 Another method P(neither defective)=18C2/20C2=(18!/16!2!)/(20!/18!2!)=(9*17)/(10*19)=153/190 12. Twenty percent of the sweaters in a store are white. Of the remaining sweaters, forty percent are brown and the rest are blue. If there are 200 sweaters in the store, then how many more blue sweaters than white sweaters are in the store? Solution; white=20% of 200=40 remaining=200-40=160 brown=40% of 160=64 blue=160-64=96 blue-white=96-40=56 13. In a graduating class of236 students, 142 took algebra and 121 took chemistry. What is the greatest possible number of students that could have taken both algebra and chemistry? Solution; 21 121 A 142-121 0 C 236-21-121=94
  6. 6. For the greatest number of students that could have taken both algebra and chemistry, the number of students taking chemistry only must be zero. Then all 121 students who took chemistry took both classes. Hence maximum number of students who took both classes is 121. 14. In the given sequence, each term after the first term is equal to the preceding term plus the constant c. If a1+a2+a3=27, what is the value of a2+a4? a2+a4= Solution; a2=a1+c a3=a2+c=a1+2c a4=a3+c=a1+3c a5=a1+4c a1+a3+a5=27 a1+a1+2c+a1+4c=3a1+6c=3(a1+2c)=27 a1+2c=9 a2+a4=2a1+4c=2(a1+2c)=2*9=18 15. In the xy-plane, the point with coordinates (-6,-7) is the center of circle C. The point with coordinates (-6,5) lies inside C and the point with coordinates (8,-7) lies outside C. If m is the radius of C and m is an integer, what is the value of m? m= Solution;
  7. 7. (-6,5) O (-6,-7)C m (8,-7) From the figure, 5+7<m<8+6 12<m<14 m=13 16. Jordan has taken 5 math tests so far this semester. If he gets a 70 on his next test, it will lower the average (arithmetic mean) of his test scores by 4 points. What is his average now? Solution; Let current average be x. then new average=new total sum/new number of tests or, (5x+70)/6=x-4 or, 5x+70=6x-24 or, x=94
  8. 8. 17. A B D C In the figure above, if ABCD is a rectangle what is the sum of marked angles? degrees 180+180+90-180=270 18. Soltech employs 20 programmers for every 3 managers, and 5 managers for every director. If the total number of employees at company is between 300 and 400, what must the number of managers who work at the company equal to? Solution; P:M=20:3=100:15 M:D=5:1=15:3 P:M:D=100:15:3=200:30:6=300:45:9=400:60:12 So number of managers must be 45. 19. How many 2 digit integers can be chosen such that none of the digits appear more than once, and none of the digits equals zero? integers Solution; 9 choices*8 choices=72 choices The tens digit can be any integer from 1 to 9 i.e. 9 choices and the units digit has 8 choices left since repetition is not allowed. 20. 90 students represent x percent of the boys at the Ardmore Elementary School. If the boys at Ardmore Elementary make up 40 percent of the total population of x students, what is x?
  9. 9. Solution; total=x Boys=40% of x=4x/10 x% of boys=90 x/100*4x/10=90 4x2 =90*103 =9*104 2x=3*100 x=150 21. Given that the sum of the odd integers from 1 to 99 inclusive is 2500, what is the sum of the even integers from 2 to 100 inclusive? Solution; 1+3+5+……..+99=2500 (2-1)+(4-1)+(6-1)+……+(100-1)=2500 (2+4+6+…..+100)-(1+1+1+….50 terms)=2500 2+4+6+…..+100=2500+50=2550 22. A confectioner has 500 mint, 500 orange and 500 strawberry flavored sweets. He wishes to make packets containing 10 mint, 5 orange and 5 strawberry sweets. What is the maximum number of packets of this type can he make? Solution; Here each packet must have 10 mint and there are equal number of each flavored sweets. That means mint will be used up first and it is the limiting factor. So maximum number of packets that he can make=500/10=50
  10. 10. 23. What is the sum of all positive factors of 12? Solution; Positive factors of 12 are 1,2,3,4,6 and 12, the sum of which is 1+2+3+4+6+12=28 24. Of 60 students in a class 2/3 are girls and 2/5 of the class are taking music lessons. What is the maximum number of girls that are not taking music lessons? Solution; 2/3*60=40 girls and 60-40=20 boys 2/5*60=24 take music lessons. 60-24=36 don't take music lessons. Since there are 40 girls, it's possible that all 36 students that don't take music lessons are girls. So the maximum number of girls not taking music lessons is 36. 25. If |x+1|<=5 and |y-1|<=5, what is the least possible value of the product xy? |x+1|<=5 -5<=x+1<=5 -6<=x<=4 |y-1|<=5 -5<=y-1<=5 -4<=y<=6 for the product xy to be the least, x=-6 and y=6 so that xy=-36 26. What is the length of the largest stick that can be filled in a rectangular box of length 12 cm, breadth 5 cm and height 84 cm.
  11. 11. Solution; 84 longest stick length=/122 +52 +842 =85 cm 5 12 27. 35 percent of senior students in a certain college are absent. Two-fifth of those absentees went to watch a movie. What fraction of total senior students are neither present nor went for watching movie? Give your answer as fraction. Solution; Let senior students=100 then absentees=35 2/5*35=14 went to watch a movie. 35-14=21 seniors were neither present nor went for watching movie. So required fraction=21/100 28. The total amount of Judy's water bill for the last quarter of the year was $40.50. The bill consisted of a fixed charge of $13.50 plus a charge of $0.0075 per gallon for the water used in the quarter. For how many gallons of water was Judy charged for the quarter gallons Solution; Let he was charged for n gallons. total charge=fixed charge+variable charge=13.50+(n*0.0075)=40.50 27=n*0.0075 n=27/0.0075=3600
  12. 12. 29. The average (arithmetic mean) of 11 numbers in a list is 14. If the average of 9 of the numbers in the list is 9, what is the average of the other 2 numbers? Solution; Sum of 11 numbers=11*14=154 Sum of 9 numbers=9*9=81 Sum of remaining 2 numbers=154-81=73 and average=73/2=36.5 30. The circles shown are tangent at point B. Point A is the center of the larger circle, and line segment AB (not shown) is a diameter of the smaller circle. The area of the smaller circle is what fraction of the area of the larger circle? Solution; Area of smaller circle As=3.14*d2 /4 Area of larger circle Al=3.14*(2d)2 /4 As/Al=1/4
  13. 13. 31. 10,10,10,10,8,8,8,8,12,12,11,y The twelve numbers shown represent the ages, in years, of the twelve houses on a certain city block. What is the median age, in years, of the twelve houses on the block? years Solution; The possible positions of y are as follows y,8,8,8,8,10,10,10,10,11,12,12 median=(12+1)th/2=6.5th item=average of 10 and 10=10 8,8,8,8,y,10,10,10,10,11,12,12 median=10 8,8,8,8,10,10,10,10,y,11,12,12 median=10 8,8,8,8,10,10,10,10,11,y,12,12 median=10 8,8,8,8,10,10,10,10,11,12,12,y median=10 Hence median age of 12 houses is 10 years (no matter what y be) 32. Working alone at its constant rate, machine A produces k car parts in 10 minutes. Working alone at its constant rate, machine B produces k car parts in 15 minutes. How many minutes does it take machines A and B, working simultaneously at their respective constant rates, to produce k car parts? minutes Solution; In 1 min, A produces k/10 car parts. In 1 min, B produces k/15 car parts. In 1 min, A+B produce k/10+k/15=k/6 parts. In 6 min, A+B produce k/6*6=k car parts.
  14. 14. ANNUAL PERCENTAGE CHANGE IN DOLLAR AMOUNT OF SALES AT FIVE RETAIL STORES FROM 2006 TO 2008 Store Percent change from 2006 to 2007 Percent change from 2007 to 2008 P Q R S T 10 -20 5 -7 17 -10 9 12 -15 -8 33. At store T, the dollar amount of sales for 2007 was what percent of the dollar amount of sales for 2008? Give your answer to the nearest 0.1 percent. % Solution; Let the dollar amount of sales at store T for 2006 be 100. Then, dollar amount for 2007=100+17=117 dollar amount for 2008=92% of 117=107.64 117/107.64*100=108.69565…=108.7(to the nearest 0.1%) 34. List L: 2,x,y List M: 1,2,3,x,y If the average (arithmetic mean) of the 3 numbers in list L is 10/3, what is the average of the 5 numbers in list M? Give your answer as a fraction. Solution; 2+x+y=10/3*3=10
  15. 15. x+y+2=10 1+2+3+x+y=10+4=14 Therefore, average=sum/total number=14/5 35. Thirty five percent of senior students in a certain college were absent. Two-fifths of those absentees went to watch a movie. What fraction of total senior students were absent and did not go for movie? Give your answer as fraction. Solution; total seniors=100 absent=35 present=100-35=65 movie=2/5*35=14 no movie=35-14=21 Hence our answer is 21/100. 36. In the figure shown, what is the value of x? 30 40 x Solution; A B D E C Solution; In triangle CAE, angle AEC=180-90-30=60
  16. 16. In triangle DBC, angle BDC=180-90-40=50 x=360-60-50-90=160 37. What is the length of a diagonal of a rectangle that has width 5 and perimeter 34? Solution; P=2(l+b) 34=2(l+5) 17=l+5 l=12 Diagonal= /122 +52 =13 38. Among the people attending a convention in the Europe, 32 percent traveled from Asia and 45 percent of those who traveled from Asia are women. What percent of the people at the convention are women who traveled from Asia? Solution; Total=100 Asia=32 Asian Women=45/100*32=14.4 Hence women from Asia=14.4% 39. A candidate is required to answer 6 out of 10 questions which are divided into 2 groups each containing 5 questions and he is not permitted to attempt more than 4 from any group. In how many different ways can he make up his choice? ways Solution;
  17. 17. Group A GroupB 1 6 2 7 3 8 4 9 5 10 He can make up the choices as follows 4 from group A and 2 from group B=5C4*5C2=5*10=50 or 3 from group A and 3 from group B=5C3*5C3=10*10=100 or 2 from group A and 4 from group B=5C2*5C4=10*5=50 So total number of ways of choices=50+100+50=200 40. A 120 a O a P B In the figure above, PA is tangent to circle O at point A, PB is tangent to circle O at point B. Angle AOB measures 120 and OP=24/pi. What is the length of minor arc AB? Solution; angle PAO=angle PBO=90 ( the line joining the centre and point of contact of the tangent and the circle are at right angles) In triangle AOP,
  18. 18. 41. Average of five positive integers is 10. What is the greatest possible integer among them? Solution; a+b+c+d+e=10*5=50 1+1+1+1+46=50 So the greatest possible integer among them is 46. 42. 10,5,x,1,18 Median of the above 5 numbers is 5. What is the greatest possible value of x? Solution; x,1,5,10,18 median=5 and x<1 1,x,5,10,18 median=5 and 1<x<=5 1,5,x,10,18 For median to be 5, the value of x must be 5. 1,5,10,x,18 median=10 1,5,10,18,x median=10 From above results, the maximum possible value of x is 5. 43. If mean and median of 5 different positive integers are 10 and 12 respectively, what is the greatest possible integer among them? Solution; sum of 5 integers=10*5=50 1+2+12+13+x=50 x=50-28=22
  19. 19. 44. Susan received scores of 75,80 and 85 out of 100 on three math exams. The pass grade is 50 and the score is given only in integer values. If she appeared for 2 more math exams and passed all five exams with an average of 80, what was the highest possible score she received on her five math exams? Solution; For scoring an average of 80 on 5 exams, total=5*80=400 total on her initial 3 tests=75+80+85=240 total on her next 2 tests=400-240=160 If the scores on the next two tests be x and y, x+y=160 Since pass grade is 50, let x=50(minimum) for y to be maximum. then y=160-50=110 which is not possible since the maximum scores in an exam is 100. Next if x=59, then y=160-59=101 which is still not possible. if x=60, then y=160-60=100 which is the highest possible score obtained. 45. During a party attended by 3 females and 3 males, 3 people at random enter a previously empty room. What is the probability that there are exactly 2 males in the room? Solution; 3 males=A,B,C 3 females=D,E,F Possibilities: ABD+ABE+ABF+BCD+BCE+BCF+CAD+CAE+CAF 2 males+1 female ADE+AEF+AFD+BDE+BEF+BFD+CDE+CEF+CFD 1 male+3 females ABC all males
  20. 20. DEF all females P(exactly 2 males)=9/20 P(exactly 2 males)=P(2 males and 1 female)=3C2*3C1/6C3=3*3/20=9/20 46.