Upcoming SlideShare
×

# Revised gre quantitative complete!!!

6,516 views

Published on

Revised GRE quantitative questions by Rejan Chitrakar. This ebook is sufficient to be able to tackle all types of revised gre questions. All the best for your GRE!!!

Published in: Education, Technology
1 Comment
9 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• do you have any for the verbal section?

Are you sure you want to  Yes  No
Views
Total views
6,516
On SlideShare
0
From Embeds
0
Number of Embeds
5
Actions
Shares
0
428
1
Likes
9
Embeds 0
No embeds

No notes for slide

### Revised gre quantitative complete!!!

1. 1. 2014 Worldlink Communications Rejan Chitrakar Rejan Chitrakar [REVISED GRE QUANTITATIVE] Complete revised GRE quantitative problems and solutions. The only book you will need to score at or above 90th percentile .
2. 2. Quantitative Comparison Questions - Revised GRE 1. Lionel is younger than Maria. QtyA QtyB Twice Lionel's age Maria's age Solution; here, L's age<M's age, let L's age=5 and M's age=10. then QtyA=2*5=10=Maria's age=QtyB which implies our answer is C. (since QtyA=QtyB) But we must check with other numbers also to be sure. So now let L's age=5 and M's age=6. then QtyA=2*5=10 and QtyB=6 which implies our answer is A. (since QtyA>QtyB) Since different assumptions are leading to different answers, our answer is D. That means the relationship cannot be determined from the given information. 2. y=2x2+7x-3 QtyA QtyB x y Solution; Put x=0, then y=2*0+7*0-3=-3. That means x=0 and y=-3 in which case our answer would be A(since 0>-3) Next, put x=1, then y=2*1+7*1-3=6. That means x=1 and y=6 in which case our answer would be B(since 6>1). Hence our answer is D. 3. y>4 Qty A QtyB (3y+2)/5 y 2
3. 3. Solution; (3y+2)/5 ? y or, 3y+2 ? 5y or, 2 ? 5y-3y or, 2 ? 2y or, 1 ? y Since we are given, y>4, we also know, y>1. So, 1<y If we reverse our process, we get 1<y or, 2<2y (multiplying both sides by 2) or, 2<5y-3y or, 3y+2<5y (shifting 3y to the left) or, (3y+2)/5<y (Dividing both sides by 5) That means QtyB is greater and that's our answer. 4. QtyA QtyB (230-229)/2 228 Solution; QtyA=(230-229)/2=229(2-1)/2=229/2=229-1=228=QtyB So our answer is C. 5. QtyA QtyB x2+1 2x-1 Solution; x2+1 ? 2x-1 3
4. 4. or, x2-2x+1 ? -1 or, (x-1)2 ? -1 Since square of a number cannot be negative, (x-1)2 cannot be negative. So it's greater than -1. Now reversing the process, we get x2+1>2x-1. That means our answer is A. 6. w>1 QtyA QtyB 7w-4 2w+5 Solution; 7w-4 ? 2w+5 or, 7w-2w ? 5+4 or, 5w ? 9 or, w ? 9/5 Now we are given w>1. So let w=2. Since 2>9/5, we get 7w-4>2w+5. That means our answer would be A. But we can also assume w=3/2 since w has to be greater than 1. Now for this case, 3/2<9/5. That means our answer would be B. Hence our answer is D since different assumptions are leading to different answers. 7. O is the center of the Circle above. 3 x O QtyA QtyB 3 x 5 4 Solution; radius of the circle, r=/32+42=5 (since in a right angled triangle, p2+b2 =h2) Now x=/52-32=4 That means QtyB is greater. Hence our answer is B. 4
5. 5. 8. x<y<z QtyA QtyB (x+y+z)/3 y Solution; Let x=1,y=2 and z=3. i.e., QtyB=2 then QtyA=(1+2+3)/3=6/3=2 which implies our answer is C.(since QtyA=QtyB=2) To be sure, let x=0,y=1 and z=7 i.e., Qty B=1 then QtyA=(0+1+7)/3=8/3 which implies our answer is A.(since QtyA>QtyB) Hence our answer is D. 9. QtyA x/y x QtyB 1 50 y Solution; remaining angle=180-90-50=40 Therefore, x>y (since side opposite to greater angle is always greater in a triangle) That means, x/y>1 Hence our answer is A. 10. 0<x<y<1 QtyA QtyB 1-y y-x Solution; Let x=0.2 and y=0.3 then, QtyA=1-0.3=0.7 and QtyB=0.3-0.2=0.1 in which case our answer would be A. 5
6. 6. Now let x=0.2 and y=0.9 then, QtyA=1-0.9=0.1 and QtyB=0.9-0.2=0.7 in which case our answer would be B. Hence the relationship cannot be determined from the given information. That is our answer is D. 11. 'p' is the probability that event E will occur and 's' is the probability that event E will not occur. QtyA QtyB p+s ps Solution; Here p+s=1 or, s=1-p or, ps=p-p2 Now p+s ? p-p2 or, s ? -p2 Here right side is always negative and left side is a number between 0 and 1 since it is a probability of non occurrence of an event. So s>-p2. Hence, p+s is greater i.e., A is the answer. 12. X is the set of all integers n that satisfy the inequality 2<=|n|<=5 QtyA QtyB The absolute value of the greatest The absolute value of the least integer integer in X in X Solution; X={2,3,4,5,-2,-3,-4,-5} A=|5|=5 B=|-5|=5 6
7. 7. Hence our answer is C. 13. x and m are positive numbers and m is a multiple of 3. QtyA QtyB xm/x3 xm/3 Solution; xm-3 ? xm/3 or, m-3 ? m/3 or, m-m/3 ? 3 or, 2m/3 ? 3 or, 2m ? 9 or, m ? 9/2 Since m is a multiple of 3, let m=3. then 3<9/2 in which case our answer would be B. Again if we assume m=6, then 6>9/2 in which case our answer would be A. Hence our answer is D. 14. k is a digit in the decimal 1.3k5 and 1.3k5 is less than 1.33. QtyA QtyB k 1 Solution; 1.3k5<1.33 that means, 0<=k<3. If k=0, B would be greater. If k=2, A would be greater. 7
8. 8. Hence our answer is D. 15. st=/10 QtyA QtyB s2 10/t2 Solution; s2 ? 10/t2 or, s2t2 ? 10 or, st ? /10 st=/10 Hence our answer is C. 16. Three consecutive integers have a sum of -84. QtyA QtyB The least of the three integers -28 Solution; x+(x+1)+(x+2)=-84 or, 3x+3=-84 or, 3x=-87 or, x=-87/3=-29 Therefore the three consecutive integers are -29,-29+1 and -29+2 i.e.,-29, -28 and -27 The least of three integers is -29 which is less than -28. Hence our answer is B. 17. In the xy-plane, the equation of line k is 3x-2y=0. 8
9. 9. QtyA QtyB x-intercept of line k y-intercept of line k Solution; for x-intercept, put y=0. So 3x-2*0=0 i.e., x=0. for y-intercept, put x=0. So 3*0-2y=0 i.e., y=0. hence A=B and our answer is C. OR we can directly know the answer if we see the equation clearly. 3x-4y=0 or, y=3/4*x which is in the form y=mx and which passes through origin. That means it does not have x and y intercepts or their values are zero. 18. n is a positive integer that is divisible by 6. QtyA QtyB The remainder when n is The remainder when n is divided by 18 divided by 12 Solution; Since n is divisible by 6, let n=6. then when 6 is divided by 12, the remainder will be 6 and when it is divided by 18, the remainder will be still 6. That indicates answer C. But if we suppose n=12, since 12 is also divisible by 6. then, when 12 is divided by 12, the remainder is 0 and when it is divided by 18, the remainder is 12. That indicates answer B. Therefore our answer is D. 19. 1-x/1+x=1/x Qty A QtyB x -1/2 9
10. 10. Solution; 1-x/x-1=1/x or, x-x2=x-1 (cross multiplication) or, -x2=-1 or, x2=1 or, x=+-1 Here you might think the answer is D since, if x=+1, then A would be greater and if x=-1, then B would be greater. But that is not the case. Try to plug in the obtained values of x back into the given equation, then you will see that x=+1 will give 1-x/x-1=1-1/1-1=0/0, which is an indeterminate form. So x=+1 is eliminated. and the only value of x remained is x=-1. Since -1<-1/2, our answer is B. That means QtyB is greater. 20. In a set of 24 positive integers, 12 of the integers are less than 50. The rest are greater than 50. QtyA QtyB The median of the 24 integers 50 Solution; Here, median=value of (N+1)th/2 item=(24+1)th/2=12.5th item=average of 12th and 13th items. We just know that the 12th item is less than 50 and the 13th item is greater than 50 since half of the numbers are less than 50 and half of the numbers are greater than 50. But we don't know their exact values. So let 12th number=40 and 13th number=60 then median=(40+60)/2=50 That means QtyA=QtyB i.e., our answer could be C. But if we assume 12th number=45 and 13th number=65, then median=(45+65)/2= 55 in which case our answer would be A. Hence D is the answer. 10
11. 11. 21. QtyA QtyB (40% of 50)+60 (60% of 50)+70 Solution; Don't try to waste your time by computing on this type of questions. Just by seeing, you can decide which side is greater. It's B since 60% of 50>40% of 50. In addition, 70>60. So our answer is B. 22. 0<r<t QtyA QtyB r/t t/r Solution; Since r<t, r/t<1 and t/r>1. Hence QtyB is greater. 23. m,p and x are positive integers and mp=x. QtyA QtyB m x Solution; If p=1, m=x. That means our answer would be C. But if p=2, then 2m=x. That means QtyB would be greater or our answer would be B. Hence the answer is D. 24. Triangular regions T1 and T2 have equal areas and have heights h1 and h2 respectively. QtyA QtyB (The area of T1)/h1 (The area of T2)/h2 11
12. 12. Solution; area of T1=b1h1/2 area of T2=b2h2/2 So QtyA=b1h1/2h1=b1/2 and QtyB=b2h2/2h2=b2/2 Now we don't know what relation exists between b1 and b2. So we can't determine which is greater. Hence our answer is D. 25. x2y>0, xy2<0 QtyA QtyB x y Solution; x2y>0. This means y must be positive since x2 cannot be negative. xy2<0. This means x must be negative since y2 cannot be negative. This means, x is negative and y is positive. Hence y is greater. Our answer is B. 26. The diameter of the circle is 10. QtyA QtyB The area of the region 40 enclosed by the quadrilateral Solution; Area of the circle=3.14*52=78.5 So area enclosed by quadrilateral<78.5 12
13. 13. Now we could draw a small quadrilateral whose area is less than 40 as follows. Or we could draw a big quadrilateral whose area is greater than 40 with the same diameter of 10. We can even draw a quadrilateral with area exactly 40 inside a circle of diameter 10. That means we cannot determine which is greater. Hence our answer is D. 27. B The circle shown has a radius of 5. AB=8. A C QtyA QtyB The perimeter of triangle ABC 24 Solution; BC=/AC2-AB2=/102-82=/100-64=/36=6 Therefore perimeter of triangle ABC=AB+BC+CA=8+6+10=24 Hence QtyA=QtyB=24. So our answer is C. 13
14. 14. 28. x, y and z are negative integers. QtyA QtyB The product of x, y and z The sum of x, y and z Solution; Let x=y=z=-1 then QtyA=x*y*z=(-1)*(-1)*(-1)=-1 and QtyB=x+y+z=(-1)+(-1)+(-1)=-3 That means QtyA is greater. To be sure, let's check with x=y=z=-2 then we have, QtyA=(-2)*(-2)*(-2)=-8 and QtyB=(-2)+(-2)+(-2)=-6 That means QtyB is greater. Hence our answer is D. 29. Team X scored 10 points in the first half of a certain game. In the second half of the game, team Y scored 15 points more than team X. QtyA QtyB The number of points scored The number of points scored by team X in the first half of by team Y in the first half of the game the game Solution; We don't know what team Y scored in the first half of the game. That means Qty B is unknown. Hence our answer is D. 30. x>y>w>0 QtyA QtyB xy/w yw/x 14
15. 15. Solution; xy/w ? yw/x or, x/w ? w/x Since x>y>w, x>w. That means, x/w>1 and w/x<1. So QtyA is greater. Hence our answer is A. 31. n>1 QtyA QtyB (n/n+1)+1 1-(1/n+1) Solution; (n/n+1)+1 ? 1-(1/n+1) or, n/n+1 ? -1/n+1 or, n ? -1 Since n>1, n>-1. Therefore, QtyA is greater. 32. y x p q QtyA QtyB x+y p+q Solution; Let the remaining two angles be z and r. Then x+y+z=180 and p+q+r=180. (sum of angles in a triangle is 180) Again, z=r (vertically opposite angles made by two straight lines are equal) So, x+y=p+q That is our answer is C. 15
16. 16. 33. N Q 40 M R T MN//PQ and PR//ST 55 x y P S QtyA QtyB y-x 15 Solution; angle QPM=40 (corresponding angles made by two parallel lines are equal) y=angle RPM=55+40=95 x+y=180 (sum of co-interior angles made by parallel lines is 180) or, x=180-95=85 So y-x=95-85=10. That means QtyB is greater. 34. x is not zero. Qty A QtyB |x|+|-2| |x-2| Solution; if x=2, then, QtyA=|2|+|-2|=2+2=4 QtyB=|2-2|=0. In this case, QtyA is greater. But if x=-2, then, QtyA=|-2|+|-2|=2+2=4 QtyB=|-2-2|=|-4|=4. In this case, QtyA=Qty B. Hence our answer is D. 16
17. 17. 35. QtyA QtyB The standard deviation of a The standard deviation of 5 set of 5 different integers, each of integers, each of which is between 10 and 20 which is between 0 and 10. Solution; Standard deviation is a measure of dispersion and is not affected by the size of data members. So we cannot determine the answer from the given information. 36. A power station is located on the boundary of a square region that measures 10 miles on each side. Three substations are located inside the square region. QtyA QtyB The sum of the distances from 30 miles the power station to each of the substations. Solution; We could make the sum less than 30 miles as shown in the diagram below. 10 miles In the figure alongside, A B PS(X) AX+BX+CX<30, in this case our answer would be B. C We could make the sum greater than 30 miles by putting the substations far away from the power house as below. In the figure alongside, A PS(X) AX+BX+CX>30, in this case our answer would be A. B C Likewise, we can make the sum equal to 30 miles. That means our answer is D. 17
18. 18. 37. 6<x<7 , y=8 QtyA QtyB x/y 0.85 Solution; 6<x<7 or, 6/y<x/y<7/y or, 6/8<x/y<7/8 or, 0.75<x/y<0.875 therefore, x/y=0.84 in which case our answer would be B. Also x/y=0.87 in which case our answer would be A. Hence our answer is D. 38. 'n' is the total number of numbers under 1000 that can be formed by using the digits 3,7,0,4. QtyA QtyB n The lowest prime number greater than 48 Solution; n=(number of choices 1000s digit has)*(number of choices 100s digit has)*(number of choices 10s digit has)*(number of choices unit digit has) or, n=1*4*4*4=64 The lowest prime number greater than 48=53 Hence our answer is A. 39. The average of 21 consecutive integers is M. A new set of numbers is created by adding 23 to the smallest number, 22 to the next, 21 to the third etc. The average of the new set is N. QtyA QtyB N-M 13 18
19. 19. Solution; N-M=average of 21 new numbers =(23+22+21+……21 terms)/21 =21/2{2*23+(21-1)(-1)}/21=(46-20)/2=13 (sum of n terms in an arithmetic progression Sn=n/2{2a+(n-1)d} that means our answer is C. 40. 4v4-1024=0 QtyA QtyB v 0 Solution; 4v4=1024 or, v4=1024/4=256 or, v=+-4 If v=+4, our answer would be A. If v=-4, our answer would be B. Hence our answer is D. 41. ABC is an isosceles triangle. AB=3 inches and BC=4 inches. QtyA QtyB the length of CA 3 inches Solution; A CA could be either equal to AB or BC. 3 inches B That means, if CA=3 inches, our answer would be C. C 4 inches and if CA=4 inches, our answer would be A Hence our answer is D. 19
20. 20. 42. p>0 QtyA QtyB p2+1 1-p Solution; p2+1 ? 1-p or, p2 ? -p or, p ? -1 Since p>0, p>-1. Hence reversing the process, we get, p2+1>1-p. Hence our answer is A. OR we can see, in the expression p2 ? -p, p2 is always positive while -p is always negative. That means p2>-p. 43. A ball is dropped from a certain height, bounces one half the original height. If this ball is dropped from 80 ft, it will travel a distance of k ft when it hits the ground for the third time. QtyA QtyB k 160 ft Solution; 80 ft 40 ft 20ft k=80+(40+40)+(20+20)=200 ft 44. x is a positive integer and y is a negative integer. Qty A QtyB x-y y-x Solution; 20
21. 21. Since x=+ve and y=-ve, x-y becomes addition of x and y. and y-x becomes addition of -y and -x. So QtyA is greater since it's +ve while QtyB will be -ve. So our answer is A. 45. The probability that both events E and F will occur is 0.42. QtyA QtyB The probability that event E 0.58 will occur. Solution; P(E and F)=P(E)*P(F) , which is the probability of occurrence of both events E and F or, P(E)*P(F)=0.42 or, P(E)=0.42/P(F) When P(F)=max i.e. 1, P(E)=min i.e. 0.42 and when P(F)=min i.e. 0.42 (since if P(F)<0.42, P(E)>1, which is not possible), P(F)=max i.e. 1 So 0.42<=P(E)<=1. Hence our answer is D. 46. a and b are positive integers. QtyA QtyB a/b a+3/b+3 Solution; Put a=1 and b=1, then QtyA=1/1=1 and QtyB=1+3/1+3=4/4=1 That implies answer C. Let's check with other numbers also to be sure about the answer. Put a=1 and b=2, then QtyA=1/2 and QtyB=1+3/2+3=4/5 That implies answer B. Hence our answer is D. 21
22. 22. 47. The arithmetic mean of 100 measurements is 23 and the average of 50 additional measurements is 27. QtyA QtyB Average of 150 measurements 25 Solution; Average of 150 measurements=(100*23+50*27)/150=24.33 That means our answer is B. 48. QtyA QtyB Slope of line k 1 k .(3,4) Solution; The line passing from the origin and the point (3,4) has slope=4/3, which is greater than 1. Also, we can see from the figure that line k has greater slope than the line passing through (3,4). This implies, slope of line k is greater than 1. Hence our answer is A. 49. The original price of a suit was 30% less than the suit's \$250 suggested retail price. The price at which the suit was sold was 20% less than the original price. QtyA QtyB Price at which the suit was sold 50% of the suit's suggested price Solution; Original price=70% of \$250=\$175 Price at which suit was sold=80% of \$175=140 22
23. 23. And 50% of suit's suggested price=0.5*\$250=\$125 Hence our answer is A. 50. B 8 F 3 A C 4 D 5 E G 7 QtyA QtyB Area of rectangular region Area of trapezoidal region ABCD H EFGH Solution; QtyA=length*breadth=8*3=24 QtyB=1/2*height(sum of the bases)=0.5*4*(5+7)=24 Hence our answer is C. 51. QtyA QtyB The sum of all the integers The sum of all the integers from 19 to 59 inclusive from 22 to 60 inclusive Solution; QtyA=19+20+……………………………59 QtyB=22+23+……………………………60 Subtracting 22+23+…………………59 on both quantities, QtyA=19+20+21=60 QtyB=60 Hence our answer is C. (Note: while comparing, we can add or subtract equal quantities on both) 23
24. 24. 52. R QRS is an isosceles triangle. QtyA 7 Q x 4 QtyB Perimeter of triangle QRS 17 S Solution; Since QRS is an isosceles triangle, the value of x could be either 7 or 4. If x=7, then perimeter of the triangle=7+7+4=18 in which case answer would be A and if x=4, perimeter=7+4+4=15 in which case answer would be B. Hence the answer is D. 53. The average of a set of mean daily temperatures for x days is 70 degrees. When a mean daily temperature of 75 degrees is added to the set, the average increases to 71 degrees. QtyA QtyB x 5 Solution; New average= New sum of temperatures/new sum of days or, 71=70x+75/x+1 or, 71x+71=70x+75 or, x=4 Hence QtyB is greater. 54. The scores for the 500 students who took Mr Johnson's final exam had a normal distribution. There were 80 students who scored at least 92 points out of a possible 100 total points and 10 students who scored at or below 56. QtyA QtyB The mean score on the final 87 exam 24
25. 25. Solution; 34% 34% 14% 14% 2% 56 2% mean 92 Fig: Standard Bell Curve 80/500*100=16% scored at least 92 points. 10/500*100=2% scored at or below 56. That means 92 is within 1 standard deviation from the mean and 56 is within 2 standard deviations from the mean. Hence we can conclude that 56 and 92 are within 3 standard deviations. i.e.,92-56=3*sd or, sd=12 therefore mean=92-12=80 So QtyB is greater. 55. In 2009, the property tax on each home in Town X was p percent of the assessed value of the home, where p is a constant. The property tax in 2009 on a home in Town X that had an assessed value of \$125,000 was \$2,500. QtyA QtyB The property tax in 2009 on \$3,000 a home in Town X that had an assessed value of \$160,000 Solution; p% of 125,000=2,500 or, p/100*125,000=2,500 25
26. 26. or, p=2% Therefore, QtyA=2% of 160,000=\$3,200 Hence our answer is A. 56. r, s and t are three consecutive odd integers such that r<s<t. QtyA QtyB r+s+1 s+t-1 Solution; r+s+1 ? s+t-1 or, r+1 ? t-1 or, 2 ? t-r Since r , s and t are three consecutive odd integers, t-r=4 (for ex 1,3,5 in which 5-1=4) therefore, 2<t-r i.e., QtyB is greater. 57. Machine R, working alone at a constant rate, produces x units of a product in 30 minutes and machine S, working alone at a constant rate, produces x units of the product in 48 minutes, where x is a positive integer. QtyA QtyB The number of units of The number of units of the product that machine the product that machine R, working alone at its S, working alone at its constant rate, produces in constant rate, produces in 3 hours. 4 hours. Solution; R x units 30 minutes(0.5 hours) R 6x units 0.5*6=3 hours 26
27. 27. S S x units 48 minutes(48/60 hours) x*60/48*4=5x units 4 hours Since number of units i.e. x can't be negative, 6x>5x. That means QtyA is greater. 58. Frequency Distribution for List X Number 1 2 3 5 Frequency 10 20 18 12 Frequency Distribution for List Y Number 6 7 8 9 Frequency 24 17 10 9 List X and list Y each contain 60 numbers. Frequency distributions for each list are given above. The average of the numbers in list X is 2.7 and the average of the numbers in list Y is 7.1. List Z contains 120 numbers: the 60 numbers in list X and the 60 numbers in list Y. Qty A QtyB The average of the 120 The median of the 120 numbers in list Z numbers in list Z Solution; average=total sum of list X + total sum of list Y/total number of items in X and Y total sum of list X=1*10+2*20+3*18+5*12=10+40+54+60=164 total sum of list Y=6*24+7*17+8*10+9*9=144+119+80+81=424 total number of items in X and Y=60+60=120 So, avg=164+424/120=4.9 Now, median=value of (N+1)th/2 item=(120+1)/2=60.5th item=average of 60th and 61st items=(5+6)/2=5.5 27
28. 28. We can see from the given table that 60th item belongs to list X and its value is 5 and 61st item belongs to list Y and its value is 6. That means our answer is B. 59. Among the 9,000 people attending a football game at college C, there were x students from college C and y students who were NOT from college C. QtyA QtyB The number of people 9000-x-y attending the game who were not students Solution; There are three categories in which 9000 people have been divided. 9000=number of students from college C + number of students not from college C + number of people who were not students That means, number of people who were not students=9000-x-y. Hence our answer is C. 60. x is an integer greater than 1. QtyA QtyB 3x+1 4x Solution; Put x=2, then QtyA=32+1=33=27 and QtyB=42=16 this implies answer is A. Again put x=4, then QtyA=34+1=35=243 and QtyB=44=256 this implies answer is B. Hence our answer is D. 28
29. 29. 61. A, B and C are three rectangles. The length and width of rectangle A are 10 percent greater and 10 percent less, respectively, than the length and width of rectangle C. The length and width of rectangle B are 20 percent greater and 20 percent less respectively, than the length and width of rectangle C. QtyA QtyB The area of rectangle A The area of rectangle B Solution; length of C=100, breadth of C=50 length of A=110% of 100=110 and breadth of A=90% of 50=45 area of A=110*45=4950 length of B=120% of 100=120 and breadth of B=80% of 50=40 area of B=120*40=4800 area of A>area of B Hence our answer is A. 62. Set S consists of all positive integers less than 81 that are NOT equal to the square of an integer. QtyA QtyB The number of integers 72 in set S Solution; Set of all positive integers less than 81 that are square of an integer={1,4,9,16,25,36,49,64} Therefore, S={1,2,3……..80} - {1,4,9,16,25,36,49,64} Hence, number of integers in set S=80-8=72 That means our answer is C. 29
30. 30. 63. A certain punch is created by mixing two parts soda and three parts ice-cream. The soda is 4 parts sugar, 5 parts citric acid and 11 parts other ingredients. The ice-cream is 3 parts sugar, 2 parts citric acid and 15 parts other ingredients. Solution; QtyA QtyB Parts sugar in the punch Parts citric acid in the punch Solution; Punch = 2Soda+3Ice-cream eqn(i) Soda = 4Sugar+5Citric acid+11Other eqn(ii) Ice-cream = 3Sugar+2Citric acid+15Other eqn(iii) From eqns(i and ii), we can see Parts sugar in the punch=parts sugar in the soda + parts sugar in the Ice-cream =2*4+3*3=8+9=17 Parts citric acid in the punch=parts citric acid in the soda + parts citric acid in the Ice-cream =2*5+3*2=10+6=16 Hence our answer is A. 64. A random variable Y is normally distributed with a mean of 200 and a standard deviation of 10. QtyA QtyB The probability of the event 1/6 that the value of Y is greater than 220. Solution; We draw a bell curve for this problem. 30
31. 31. 2% 14% 34% 34% 200 14% 210 2% 220 Here the curve has been divided into six regions and the value above 220 covers one region. So you might think that the probability of the event that Y>220 i.e., P(Y>220)=1/6 and the answer might be C. But that's not so. The percentages you are seeing on the bell curve refer to the percentages of areas covered by different regions of the curve. Therefore 2% area covered just after the value 220 means there are 2% values in the distribution that are greater than 220. Therefore, P(Y>220)=2%=2/100=1/50 which is less than 1/6. Hence our answer is B. 65. In a decimal number, a bar over one or more consecutive digits means that the pattern of digits under the bar repeats without end. For example, 0.387=0.387387387…. QtyA QtyB 0.717 0.71 Solution; Although both look alike when we expand them, QtyA has repetition of 717 after the decimal whereas QtyB has repetition of 71 after the decimal and since 717/1000>71/100, QtyA is greater. 66. Of 30 theater tickets sold, 20 tickets were sold at prices between \$10 and \$30 each and 10 tickets were sold at prices between \$40 and \$60 each. QtyA QtyB The average of the prices \$50 of the 30 tickets Solution; The highest possible average=(20*30+10*60)/30=\$40 31
32. 32. That means even if all 30 tickets were sold at maximum prices, the average would be less than \$50. Hence our answer is B. Note: you might be tempted to choose answer D since we don't know how much each ticket was sold for. So you must consider the maximum and minimum possibilities for this type of questions. 67. Car X can come with any of these 5 additional features: sunroof, stereo, tinted windows, leather seats and cruise control. QtyA QtyB Number of different 25 combinations possible Solution; The car has 2 choices for each 5 additional features. That means it will either have a sunroof or not. It will either have a stereo or not. It will either have a tinted windows or not. It will either have leather seats or not and finally it will either have a cruise control or not. Hence total number of possible combinations=2*2*2*2*2=32 That means our answer is A. We can solve this problem by another way also. The car can have no additional features at all or 1 or 2 or 3 or 4 or all 5 additional features. So, total number of possible combinations=5C0+5C1+5C2+5C3+5C4+5C5=32 if the car has no additional feature at all if it has 1 additional feature if it has 2 additional features if it has 3 additional features if it has 4 additional features if it has all 5 additional features Note: Combination= Number of ways of selection of 'r' objects out of 'n' objects at a time and is given by the formula: 32
33. 33. nCr=n!/{r!(n-r)!} For example, There are three objects A, B and C and we have to select 2 objects at a time. Then we could select the objects as follows: AB BC CA That means, we have 3 choices for selecting 2 objects out of 3 objects at a time. If we use our formula, then we get nCr=3C2=3!/{2!(3-2)!}=3!/2!1!=3*2*1/2*1*1=3 68. QtyA QtyB Average of integers from Average of integers from -50 to -1 inclusive -50 to 0 inclusive Solution; QtyA={(-50)+(-49)+……………………………………….+(-1)}/50=-(1+2+3+……….+50)/50 -1, there will be 50 terms. QtyB={(-50)+(-49)+………………………………………….+(-1)+0}/51=-(1+2+3+…………+50)/51 to 0, there will be 51 terms. From -50 to From -50 Comparing the two quantities, we see that QtyB is less negative than QtyA as its denominator is higher. That means QtyB must be greater. Hence our answer is B. 69. QtyA QtyB The average of 34 32 ,34 and 36 Solution; 33
34. 34. Average=(32+34+36)/3=32(1+32+34)/3=3+33+35 We can clearly see QtyA is greater. 70. a, b, c and d are different positive numbers. The average of a and b is 30. The average of a, b, c and d is 40. QtyA QtyB The greatest possible 99 value of d Solution; (a+b)/2=30 a+b=60 (a+b+c+d)/4=40 a+b+c+d=160 Solving the two equations, c+d=100 Since c and d are different positive numbers, the greatest possible value of d is 99.99…………. which is greater than 99. So answer is A. Note: If a, b, c and d were different positive INTEGERS, then QtyA=99 and our answer would be C. 71. D 3030 A B C QtyA QtyB AB BC Solution; We can see, if DB were perpendicular to AC, AB=BC in which case our answer would be C. 34
35. 35. Since only given quantities are two equal angles, we could draw the figure without affecting the given angles as follows: D 30 30 C B In the figure alongside, we can see the two angles are still 30 A degrees but now AB>BC. Hence our answer is D. 72. O QtyA QtyB Area of semi-circular region A B Area of triangular region ABC C O is the center of the circle and AO=OB Solution; Area of the semi-circle=1/2*3.14*r2=1.57r2, where r is the radius of the semicircle Area of the triangular region ABC=1/2*2r*r=r2. That means Qty A is greater. 73. 0<a<b<c<1 QtyA QtyB ab/c 1 Solution; ab/c ? 1 or, ab ? c or, ab<c Illustration for this: Let a=0.5, b=0.6 and c=0.7 then ab=0.5*0.6=0.3 and since 0.3<0.7, ab<c. That means our answer is B. 35
36. 36. 74. 0<abc<1 QtyA QtyB ab/c 1 Solution; Let a=1, b=1 and c=1/2 so that abc=1*1*1/2=1/2<1 then, QtyA=1*1/0.5=2>1 The answer is A Again let a=1/2, b=1/2and c=1 so that abc=0.5*0.5*1=0.25 < 1 then QtyA=0.5*0.5/1=0.25 < 1 The answer is B Hence our answer is D. 75.QtyA QtyB The number of primes The number of primes that are divisible by 9 that are divisible by 19 Solution; QtyA=0 (prime numbers are divisible by 1 and by themselves only) QtyB=1 (19 is a prime number and it is divisible by 19) Hence our answer is B. 76.The first term of a sequence is 2 and each succeeding term is one more than three times the preceding terms. Qty A QtyB The term just greater than 200 150 in the sequence Solution; t1=2 t2=3*2+1=7 36
37. 37. t3=3*7+1=22 t4=3*22+1=67 t5=3*67+1=202 That means 5th term is just greater than 150 and its value is 202 which is greater than 200. Hence our answer is A. 77. x<90 y QtyB y 20 QtyA 70 x Solution; By the property of triangles, we have 20+y=x or, y=x-20 If x were exactly 90, y=90-20=70 But since x<90, y<70 Hence QtyB is greater. 78. x2=16 and y3=64 QtyA QtyB x y Solution; x2=16 x=+-4 y3=64 y=4 If x=+4 then C would be the answer. If x=-4 then B would be the answer. 37
38. 38. Hence our answer is D. 79. Rectangular region R has an area 30. QtyA QtyB The perimeter of R 25 Solution; A=30 l*b=30 Let l=5 and b=6 (l*b=30), then P=2(l+b)=2(5+6)=22 which would lead us to answer B Now, Let l=3 and b=10 (l*b=30), then P=2(l+b)=2(3+10)=26 which would lead us to answer A. Hence our answer is D. 80. n is an odd positive integer. 700<n<800 QtyA QtyB The number of prime The number of prime factors of n factors of 2n Solution; Here, since 2 is already a prime number, the number of prime factors of 2n is always 1 greater than the number of prime factors of n Ex let n=750. then 750=5*5*5*2*3 and 2n=2*750=2*5*5*5*2*3 5. number of prime factors of 750=5 number of prime factors of 2*750=6, which is 1 more than 81. ab<0, bc>0 QtyA QtyB ac 0 38
39. 39. Solution; ab<0 When a=+ve, b=-ve and when a=-ve, b=+ve bc>0 When b=+ve, c=+ve and when b=-ve, c=-ve Therefore when a=+ve, b=-ve and when b=-ve, c=-ve and when a=-ve, b=+ve and when b=+ve, c=+ve ac=+ve*-ve=-ve ac=-ve*+ve=-ve In any case, the product ac will be negative which is less than 0. So QtyB is greater. m 82. O is the center of the above circle, with radius r (not shown). 0 The circle is tangent to both x and y axes. QtyA QtyB r m/2/2 Solution; from the figure, r2+r2=m2 or, 2r2=m2 or, r=m//2=QtyB Hence our answer is C. 83. xy>0 ; x>y QtyA QtyB x/y y/x Solution; xy>0 when x=+ve, y=+ve and when x=-ve, y=-ve 39
40. 40. x>y Let x=2, y=1, then x/y=2/1=2 and y/x=1/2 A let x=-1 and y=-2, then x/y=-1/-2=1/2 and y/x=-2/-1=2 B Hence our answer is D. 84. Jim is 3 years older than Jonathan. Myra is 5 years older than Melisa. Jonathan is 2 years older than Melisa. QtyA QtyB Jim's age Myra's age Solution; Jim=Jon+3 Myra=Melisa+5 Melisa=Myra-5 Jon=Melisa+2 therefore, Jim=Jon+3=Melisa+2+3=Melisa+5=Myra-5+5=Myra Hence our answer is C 85. 750<n<1500 QtyA QtyB 1500-n n-750 Solution; 1500-n ? n-750 1500+750 ? n+n 2250 ? 2n 1125 ? n Let n=850, then 1125>850 A let n=1450, then 1125<1450 B 40
41. 41. Hence our answer is D. 86. t is an integer QtyA QtyB 1/(1+2t) 1/(1+3t) Solution; 1/(1+2t ) ? 1/(1+3t ) 1+3t ? 1+2t 3t ? 2t Let t=0, then 30=20 C Let t=1, then 31 > 21 A Hence our answer is D. 87. x>1 QtyA QtyB x/x+1 -x/1-x Solution; x/x+1 ? -x/1-x 1/x+1 ? -1/1-x 1/x+1 ? 1/x-1 x-1 ? x+1 -1 ? +1 -1<+1 Hence QtyB is greater. Thinking logically, QtyB=-x/1-x=x/x-1 41
42. 42. Now when x/x+1 is compared with x/x-1, we can see numerator is same in both and denominator is less in QtyB. Hence QtyB must be greater. 88. For each positive integer n, the nth term of the sequence S is 1+(-1)n. QtyA QtyB The sum of the first 39 terms 39 of S Solution; S1=1+(-1)1=1-1=0 S2=1+(-1)2=1+1=2 S3=1+(-1)3=1-1=0 S4=1+(-1)4=1+1=2 and so on. That means for all odd powers of -1, the term is 0 and for all even powers of -1, the term is 2. From 1 to 39, there are 20 odd terms and 19 even terms. The sum of 20 odd terms will be zero and the sum of remaining 19 terms will be 19*2=38 Therefore, QtyB is greater. 89. n is a positive integer. QtyA QtyB The remainder when n is divided by 5 The remainder when n+10 is divided by 5 Solution; Since 10 is exactly divisible by 5, the remainder when n is divided by 5 will be equal to the remainder when n+10 is divided by 5. For ex, let n=2. Then remainder when 2 is divided by 5 is 2. 42
43. 43. And 10+2=12. The remainder when 12 is divided by 5 is also 2. Hence our answer is C. 90. k is an integer for which 1/21-k<1/8. QtyA QtyB k -2 Solution; 1/21-k<1/8 8<21-k 23<21-k 3<1-k k<1-3 k<-2 Hence QtyB is greater. 91. In city X, the range of the daily low temperatures during June 2012 was 20 Fahrenheit and the range of the daily low temperatures during July 2012 was 25 Fahrenheit. QtyA QtyB The range of the daily 30 Fahrenheit low temperatures in city X for the two month period from June 1,2012 through July 31,2012. Solution; for JUNE, highest daily low temperature(A)-lowest daily low temperature(B)=20 F Let A=60 F and B=40 F so that A-B=20 F for JULY, highest daily low temperature(C)-lowest daily low temperature(D)=25 F 43
44. 44. Let C=50 F and D=25 F so that C-D=25 F Here, in this case, Range of daily low temperatures for the two months=40-25=15 That means our answer would be B. But, if A=100F, B=80 F and C=60 F , D=35F then Range of daily low temperatures for the two months=80-35=45 That means our answer would be A. So our answer is D. 92. x>0 and x4=625 QtyA QtyB The greatest prime factor of 36x x Solution; x4=625=(+-5)4 x=+-5 Since x>0, x=+5 Now QtyA= The greatest prime factor of 36x=36*5 is 5. (since 36*5=2*2*3*3*5) QtyB=x=5 Hence our answer is C. 93. n is an integer and n2<39 QtyA QtyB The greatest possible value of 12 n minus the least possible value of n. Solution; n=0,+-1,+-2+-3,+-4,+-5,+-6 The greatest possible value of n-the least possible value of n=+6-(-6)=6+6=12 44
45. 45. Hence our answer is C. 94. A rectangle is inscribed in a circle of radius r. QtyA QtyB Half of the perimeter of 2r the rectangle Solution; l+b > 2r (In any triangle, the sum of two sides is greater than the third side) b r . Perimeter of rectangle=2(l+b) l And 1/2*2(l+b)=l+b=half perimeter of rectangle Hence our answer is A. 95. The figure below shows a regular octagon. A diagonal of an octagon is any line segment connecting two non-adjacent vertices. QtyA QtyB The number of diagonals of The number of diagonals of the octagon that are parallel the octagon that are not parallel to at least one side of the octagon to any side of the octagon Solution; 45
46. 46. We can see, For each of 4 sides of the octagon, there will be two lines that are parallel to sides of the octagon. So, QtyA=2*4=8 For each of 4 corners of the octagon, there will be three lines that are not parallel to any side of the octagon. So, QtyB=3*4=12 Hence QtyB is greater. 96. QtyA QtyB 678987*12345 678986*12346 Solution; Don't try to calculate the product on the calculator because the result will be too large for the calculator to display. So, 678987*12345 ? 678986*12346 or, 12345/12346 ? 678986/678987 or, 12345/12346 < 678986/678987 Illustration: If we compare 7/8 with 8/9 then 8/9 will be greater. Although the differences between numerator and denominator of the fractions are same, the fraction with larger values will be larger. Likewise, if we compare 100/101 and 10000/10001, the later fraction will be greater. Hence, QtyB is greater. 97. 13613-9216=x QtyA QtyB The unit digit of x 0 Solution; 1361 will have unit digit 6 46
47. 47. 1362 will also have unit digit 6 1363 will also have unit digit 6 That means no matter what power of 136 be, the result will have the unit digit 6. Now, 921=92 has unit digit 2 922=5704 has unit digit 4 923=524768 has unit digit 8 924=………….6 has unit digit 6 925=………….2 has unit digit 2 926=………….4 has unit digit 4 That means the pattern of unit digit of powers of 92 is 2,4,8,6,2,4,8,6……….. If we observe this pattern carefully, we'll find that 9216 will have unit digit 6 since it's in 4th place in the pattern and the 16th digit in the pattern will also be 6. Hence, QtyA=unit digit of x=unit digit of 13613-unit digit of 9216=6-6=0. Thus, QtyA=QtyB and our answer is C. 98. x is a positive integer. When x is divided by 2,3,5 and 6, the remainder is 1. QtyA QtyB x 30 Solution; Here, clearly x=31,61,91 and so on because when these are divided by 2,3,5 and 6, the remainder will be 1. Does that mean we select choice A immediately? Now there lies a TRAP within this question. You might not have thought of any other numbers which when divided by 2,3,5 and 6 could yield a remainder of 1. But there is indeed a number. and that's 1. When you divide this number by 2,3,5 and 6, the quotient will be 0 and the remainder will be 1. 47
48. 48. That means, we cannot determine the answer. Hence, our answer is D. 99. A coin is flipped 5 times. QtyA QtyB Probability of getting 2 heads Probability of getting 3 heads Solution; P(HH)=P(TTT)=P(HHH) That means probability of getting 2 heads is equal to probability of getting 3 heads. Hence our answer is C. OR we can solve this problem by using combination rule also. number of ways of occurrence of 2 heads out of 5 flips=5C2 =5!/{2!(5-2)!}=5*4*3*2*1/2!3! =10 total number of ways of outcomes=25=32 Therefore probability of getting 2 heads, P(HH)=10/32 number of ways of occurrence of 3 heads out of 5 flips=5C3=5!/2!3!=10 Therefore probability of getting 3 heads, P(HHH)=10/32 That means QtyA=QtyB REMEMBER: Try to minimize calculations as much as possible during GRE exam since you'll have lots of questions to answer and only little time. So try to use logic as much as possible rather than wasting time on calculations. So the first method I showed is better as it consumes no or little time. 100. In a certain college class, each female student has 3 pencils and each male student has 1 pencil and the average number of pencils per student in the class is 1.8. In the same class, each female student has 1 pen and each male student has 2 pens. 48
49. 49. QtyA QtyB The average number of pens 1.5 per student in the class Solution; average number of pencils per student=(1*M+3*F)/(M+F)=(M+3F)/(M+F)=1.8 or,M+3F=1.8M+1.8F or,1.2F=0.8M or,F/M=8/12=2/3 That means, there are 3 male students for every 2 female students. Now, the average number of pens per student=(2*M+1*F)/(M+F) =(2M+F)/(M+F)=(2*3+2)/(3+2)=8/5=1.6 Hence our answer is A. 101. QtyA QtyB 220 185 Solution; 185=(2*3*3)5=(2*2*1.5*2*1.5)5=215*(2.25)5 Now, 220 ? 215(2.25)5 or, 220/215 ? (2.25)5 or, 25 ? (2.25)5 We know, 25 < (2.25)5 Hence our answer is B. 102. Triangle ABC is scalene. QtyA QtyB Length of altitude to side AC Length of side AB 49
50. 50. Solution; Let's draw figure for the problem. B A M C Here, BM has to be compared with AB. Clearly, AB>BM (In right angled triangle AMB, hypotenuse i.e. side AB is the largest side) Hence our answer is B. 103. Three circles touch each other externally. QtyA QtyB The perimeter of the The circumference of the triangle connecting the centers largest circle Solution; The largest possible value of the triangle connecting the centers of circles=6R(when all triangles have same radius R, where, R is the radius of the largest circle) And at that time, the circumference of the largest circle would be 2*3.14*R>6R. Hence QtyB is always greatest. 50
51. 51. 104. Alex has a six-sided die with faces numbered 1 through 6. He rolls the die twice. QtyA QtyB The probability that both The probability that neither rolls are even roll is a multiple of 3 Solution; P(EE)=P(E)*P(E)=3/6*3/6=1/4 P(neither roll multiple of 3)=4/6*4/6=4/9 Hence QtyB is greater. 105. A polygon known as a "polyhedron" has 10,11 or 12 equal sides and angles. QtyA QtyB The degree measure of any 155 interior angle of a polyhedron Solution; Each angle of a regular polygon=180(n-2)/n for n=10, =180(10-2)/10=144<155 for n=11, =180(11-2)/11=147.2<155 for n=12, =180(12-2)/12=150<155 Hence for all figures, QtyB is greater. Hence our answer is B. 106. rt<0<-r QtyA QtyB t 0 Solution; rt<0<-r Since 0<-r, r should be -ve. 51
52. 52. Since rt<0, r=-ve, t=+ve which is greater than 0. Hence our answer is A. 107. One person is to be selected at random from a group of 25 people. The probability that the selected person will be a male is 0.44, and the probability that the selected person will be a male who was born before 1960 is 0.28. QtyA QtyB The number of males in the 4 group who were born in 1960 or later Solution; P(Male)=0.44 Number of males=0.44*25=11 P(Male and born before 1960)=0.28 Number of males born before 1960=0.28*25=7 Hence, Number of males born in 1960 or later=11-7=4. That means our answer is C. 108. T is a list of 100 different numbers that are greater than 0 and less than 50. The number x is greater than 60 percent of the numbers in T and the number y is greater than 40 percent of the numbers in T. QtyA QtyB x-y 20 If we arrange the numbers in ascending order, then x will be in 61st position while y will be in 41st position. therefore, x-y=61st number-41st number Since we don’t know the values of 61st and 41st number, we don't know the value of x-y. For ex, if x=40 and y=30, then x-y=10, in which case our answer would be B 52
53. 53. and if x=45 and y=25, then x-y=20, in which case our answer would be C That means our answer is D. 109. n5-n3<0 QtyA QtyB n n2 Solution; n5-n3<0 n5< n3 n2<1 Let n=0.5(since 0.52<1) then A=0.5 and B=0.52=0.25 answer is A To be sure, let's check with other number also. Let n=-0.5((since (-0.5)2<1) then A=-0.5 and B=(-0.5)2=0.25 answer is B Hence our answer is D. 110. Joe is twice as old as Ann. Eight years ago, Joe was four times older than Ann. QtyA QtyB Joe's age 6 years ago Ann's age 6 years from now Solution; J=2A (now) 8 years ago, Joe=J-8 Ann=A-8 By given, J-8=4(A-8) or, 2A-8=4A-32 53
54. 54. or, 24=2A or, A=12 Hence J=2*12=24 QtyA=J-6=24-6=18 QtyB=A+6=12+6=18 Hence our answer is C. 111. Julius has a piece of string that is 24 inches long. He is stretching the string around pegs on a pegboard to create different shapes. QtyA QtyB The area of a regular hexagon The area of a square created by the string created by the string Solution; 2 /(42-22)=/12 4 Area of the regular hexagon=2*(area of one of trapezoid)=2*6/12=12/12=41.56 (area of trapezoid=0.5*h(b1+b2)=0.5*/12(4+8)=6/12) Area of the square=l2=62=36 Hence QtyA is greater. 112. Shaundra drove the same route to work each morning, Monday through Friday, in a particular week. On Monday and Tuesday, she averaged 20 miles per hour and on her three remaining work days, she averaged 30 miles per hour. 54
55. 55. QtyA QtyB Shaundra's average speed for 26 miles per hour all five morning commutes Solution; Let Shaundra's distance of the route be x. Then Shaundra drove a total distance of 2x for two days at 20 miles/hr and a total distance of 3x for three days at 30 miles/hr. Now, time to drive 2x distance=2x/20 hrs (speed=distance/time)=x/10hrs time to drive 3x distance=3x/30=x/10 hrs Average speed for 5 days=(2x+3x)/(x/10+x/10)=5x/(2x/10)=5x*10/2x=25 miles/hr Hence QtyB is greater. 113. In the figure below, the area of the smaller square region is half the area of the larger square region. 1 inch QtyA QtyB The positive difference in /2/2 length between the diagonal of the larger square and the diagonal of the smaller square Solution; Area of small square=1/2*Area of large square=1/2*12=1/2 or, (length of small square)2=1/2 i.e., length of smaller square=1//2 Diagonal(larger)=/12+12=/2 55
56. 56. Diagonal(smaller)=/(1//2)2+(1//2)2=/1/2+1/2=/1=1 /2-1 ? /2/2 /2-/2/2 ? 1 /2/2 ? 1 1//2 < 1 That means our answer is B. 114. H is the mid-point of IG. J is the midpoint of IK. IG=GK=10. I H J G K QtyA QtyB IJ 5 Solution; We could draw the figure as below. I In the figure alongside, 5 IJ > 5 (In a right angled triangle, hypotenuse is the largest side H J A 5 G 10 K OR we could draw the figure as below. 5 H 60 I In the figure alongside, J IJ=5 (Triangle IJH being an equilateral triangle) 5 G 60 C K That means our answer is D. 56
57. 57. 115. In the figure below, ABCD is a parallelogram. B 125 C 6 A 4 D QtyA QtyB The area of ABCD 24 Solution; Area of the parallelogram ABCD=base*height=4*height B 125 6 C height A D base=4 From the figure, height<6 (since in any right angled triangle, hypotenuse is the largest side) That means area of ABCD=4*height<24 (since height<6) Hence our answer is B. 116. x>y QtyA QtyB |x+y| |x-y| solution; Her, Let x=2 and y=1. Then, A=|2+1|=|3|=3 B=|2-1|=|1|=1 which points to answer A. 57
58. 58. And now let x=1 and y=0. Then A=|1+0|=|1|=1 and B=|1-0|=|1|=1 which points to answer C. Hence our answer is D. 117. Three circles with their centers on line segment PQ are tangent at points P,R and Q, where point R lies on line segment PQ. QtyA QtyB The circumference of the largest The sum of the circumferences of the two circle. smaller circles. Solution; From the figure, we can see PR+RQ=PQ or, Diameter of smaller circle1+Diameter of smaller circle2=Diameter of largest circle or, d1+d2=D Now Circumference of the largest circle=3.14D and Circumference of circle C1=3.14d1 and Circumference of circle C2=3.14d2 So, C1+C2=3.14(d1+d2)=3.14D, which means our answer is C. 58
59. 59. 118. Line k is parallel to line m. QtyA QtyB x+y w+z Solution; Since alternate angles made by parallel lines are equal, x=z and y=w Hence adding the two, we get x+y=w+z Hence our answer is C. 119. The length of each side of equilateral triangle T is 6 times the length of each side of equilateral triangle X. QtyA QtyB The ratio of the length of one The ratio of the length of one side of T to the length of another side of X to the length of another side of T. side of X Solution; A=1 and B=1 (since the lengths of an equilateral triangle are equal) So our answer is C. 120. A right circular cylinder with radius 2 inches has volume 15 cubic inches. QtyA QtyB the height of the cylinder 2 inches 59
60. 60. Solution; Volume of a cylinder=3.14*r2*h 15=3.14*22*h h=15/(4*3.14)=1.19 inches < 2 inches Hence our answer is B. 121. The frequency distributions shown above represent two groups of data. Each of the data values is a multiple of 10. QtyA QtyB The standard deviation of The standard deviation of distribution A distribution B Solution; Average of A=(10*3+20*3+30*6+40*3+50*3)/18=(30+60+180+120+150)/18=30 Average of B=(10*5+20*3+30*2+40*3+50*5)/18=(50+60+60+120+250)/18=30 That means average of both are same. now square of deviation from mean for A={3(10-30)2+3(20-30)2+6(30-30)2+3(40-30)2+3(50-30)2}/18=1200+300+300+1200=3000 for B={5(10-30)2+3(20-30)2+2(30-30)2+3(40-30)2+5(50-30)2}/18=2000+300+300+2000=5000 That means standard deviation of B is greater. 60
61. 61. 122. D A C O B O is the center of the circle. QtyA QtyB The average of BC and BD AB Solution; AB is greater than BC or BD (diameter is the longest chord of a circle) i.e. AB>BC and AB>BD So, AB+AB>BC+BD or, AB>(BC+BD)/2 Hence Qty B is greater. 123. Q x P y x R x>y QtyA QtyB Length of arc PRQ Length of arc QPR Solution; Since angle PQR=angle PRQ, arc QP=arc PR (i) 61
62. 62. (arcs opposite to equal circumscribed angles of a circle are equal) Now arc PRQ= arc PR + arc RQ arc QPR=arc QP + arc PR (ii) (iii) From (i),(ii) and (iii), arc PRQ ? arc QPR arc PR + arc RQ ? arc QP + arc PR arc RQ ? arc QP arc RQ < arc QP (arc opposite to greater angle is greater than the arc opposite to smaller angle in the case of circumscribed angles of a circle and we are given x>y) That means, our answer is B. 124. xy<0 and yz>0 QtyA QtyB xz 0 Solution; xy<0 When x=-ve,y=+ve and when x=+ve,y=-ve yz>0 When y=+ve,z=+ve and when y=-ve,z=-ve That means, When x=-ve, z=+ve the product xz is -ve when x=+ve, z=-ve the product xz is -ve Hence QtyB is greater. 62
63. 63. 125. One of the roots of the equation x2+kx-6=0 is 3, and k is a constant. QtyA QtyB The value of k -1 Solution; Since one of the roots is 3, it must satisfy the given equation. So, 32+k*3-6=0 or, 9+3k=6 or, 3k=-3 or, k=-1 Hence our answer is C. 126. x<y QtyA The average of x and y QtyB The average of x, y and y Solution; A=(x+y)/2 B=(x+y+y)/3=(x+2y)/3 Now, (x+y)/2 ? (x+2y)/3 or, 3x+3y ? 2x+4y or, x ? y x<y Hence QtyB is greater. 63
64. 64. 127. ABCD and EFGH are both squares whose sides are 12. All the circles are tangent to one another and the sides of the square. QtyA QtyB The area of the shaded region The area of the shaded region in figure 1 in figure 2 Solution; radius of each circle in figure 1=12/4=3 So area of each circle in figure 1=3.14*32=3.14*9 area of 4 circles in figure 1,A1=4*3.14*9=36*3.14 area of square ABCD,A2=l2=122=144 Therefore, area of the shaded region in figure 1=A2-A1=144-(36*3.14) radius of each circle in figure 2=12/6=2 So area of each circle in figure 2=3.14*22=3.14*4 area of 9 circles in figure 2,A3=9*3.14*4=36*3.14 area of square ABCD,A4=l2=122=144 Therefore, area of the shaded region in figure 1=A4-A3=144-(36*3.14) Hence our answer is C. 64
65. 65. 128. The price of a large pizza is 30% more than the price of a small pizza. QtyA QtyB The price of a large pizza The price of a small pizza when it is on sale for 30% off Solution; Let price of a small pizza be 100. Then, price of a large pizza=130 A=(100-30)% of 130=0.7*130=91 and B=100 Hence, QtyB is greater. 129. QtyA QtyB The area of an equilateral The area of an isosceles right triangle whose sides are 6 triangle whose legs are 6 Solution; Area=/3/4*(side)2 Area=1/2(base)*(height)=1/2*(base)2 6 6 6 A=(/3/4)*62=/3*9 B=1/2*62=18=3*9 We can see, B>A. So our answer is B. 65
66. 66. 130. A bag contains four slips of paper, two of which have the number 1 written on them and two of which have the number -1 on them. Two of the slips are chosen at random. QtyA QtyB The probability that the product The probability that the product of the two numbers chosen is -1 of the two numbers chosen is 1 Solution; The product will be -1 if the two numbers chosen are (-1,1) or (1,-1). So P(product=-1)=P(-1,1)+P(1,-1)=(2/4*2/3)+(2/4*2/3)=1/3+1/3=2/3 the product will be 1 if the two numbers chosen are (1,1) or (-1,-1). So P(product=1)=P(1,1)+P(-1,-1)=(2/4*1/3+2/4*1/3)=1/6+1/6=2/6=1/3 Hence QtyA is greater. 131. The test scores for a class have a normal distribution, a mean of 50, and a standard deviation of 4. QtyA QtyB Percentage of scores at or above 58 Percentage of scores at or below 42 Solution; 34% 34% 14% 14% 42 46 50 54 58 mean=50, sd=4 So, mean + 1sd=50+4=54 and mean - 2sd=50-(2*4)=42 Hence from the normal distribution curve, we can see Both 58 and 42 are at two standard deviation from mean. So percentage of scores at or above 58=2% 66
67. 67. and percentage of scores at or below 42=2% Hence our answer is C. 132. In the xy-plane, the point (1,2) is on line j, and the point (2,1) is on line k. Each of the lines has a positive slope. QtyA QtyB The slope of line j The slope of line k Solution; y line j (1,2) line k (2,1) The lines could be parallel. That means their slopes could be equal. x line j line k (1,2) (2,1) Here angle made by line j with x-axis is greater than the angle made by line k with x-axis. That means slope of line j could be made greater than slope of line k. Similarly, slope of line k could be made greater by making its angle with x-axis greater than that made by line j. Hence our answer is D. 133. For each positive integer n, the nth term of the sequence S is 1+(-1)n. QtyA QtyB The sum of the first 39 terms of S 39 Solution; 67
68. 68. Here, t1=1+(-1)1=0 t2=1+(-1)2=1+1=2 t3=1+(-1)3=0 That means terms with even powers of -1 have value 2 and those with odd powers of -1 have value 0. We know, from 1 to 40, there are 20 even numbers and 20 odd numbers. So from 1 to 39, there are 19 even numbers and 20 odd numbers. So our sum=2*19+0*20=38 Hence our answer is B. 134. For all positive numbers p, the operation ♥ is defined by p♥=p+1/p. QtyA QtyB ((2/7)♥)♥ 3.5 Solution; ((2/7)♥)♥=(2/7+7/2) ♥={(2*2+7*7)/14}♥=(53/14)♥=53/14+14/15=3.78+0.93 Hence our answer is A. 135. x>0 and x4=625 QtyA QtyB The greatest prime factor of 36x x Solution; x4=625 x=+-5 Since x>0, x=+5 i.e.,B=5 Now, A=the greatest prime factor of 36*5(=3*3*2*2*5) is also 5. Hence our answer is C. 68
69. 69. 136. n is an integer, and n2<39. QtyA QtyB The greatest possible value 12 of n minus the least possible value of n Solution; Since n2<39, possible values of n are 0,+-1,+-2,+-3,+-4,+-5 and +-6. Now A=+6-(-6)=6+6=12 B=12 Hence our answer is C. 137. The function f is defined by f(x)=5x+1 for all numbers x. QtyA QtyB f(t+54)-f(t+50) 20 Solution; A={5(t+54)+1}-{5(t+50)+1}=5t+271-(5t+251)=271-251=20 Hence our answer is C. 138. The radius of the circle is 0.1. 69
70. 70. QtyA QtyB AB+BC+CD+DE+EA 1 Solution; Diameter=0.2 Since diameter is the longest chord, each side of pentagon is less than 0.2. that means, sum of lengths of 5 sides of pentagon<5*0.2 or 1. Hence our answer is B. 139. QtyA QtyB The perimeter of the pentagon The circumference of the circle Solution; Since each chord is shorter than the arc it forms, so the circumference of the circle will be greater than the perimeter of the pentagon. Hence our answer is B. 140. The median income of a group of college C graduates six months after graduation was \$3,000 higher than the median income of a group of College D graduates six months after graduation. 70
71. 71. QtyA QtyB The 75th percentile of the The 75th percentile of the incomes of the group of incomes of the group of college C graduates six college D graduates six months after graduation. months after graduation. Solution; Let there be 4 graduates each from college C and college D. Let incomes of graduates from college C={1000,2000,8000,9000} Here, median=(4+1)th/2=2.5th item=(2000+8000)/2=5000 75th percentile(C)=8000 Incomes of graduates from college D={1000,2000,2000,9000} Here, median=(4+1)th/2=2.5th item=(2000+2000)/2=2000 Difference between the incomes=5000-2000=3000 which is as per given. 75th percentile(D)=2000 In this case, our answer would be choice A since 75th percentile of incomes of college C graduates is greater. Next, Let there be 5 graduates from college D. Let incomes of graduates from D={1000,1500,2000,9000,10000} so that median income=2000 and incomes of graduates from C is same as above so that median income=5000 75th percentile(D)=3(5+1)th/4 item=4.5th item=(9000+10000)/2=9500 In this case, our answer would be choice B since 75th percentile of incomes of college D graduates is greater. Hence our answer is D. 71
72. 72. 141. In a quality-control test, 50 boxes-each containing 30 machine parts-were examined for defective parts. The number of defective parts was recorded for each box, and the average (arithmetic mean) of the 50 recorded numbers of defective parts per box was 1.12. Only one error was made in recording the 50 numbers: "1" defective part in a certain box was incorrectly recorded as "10". QtyA QtyB The actual average number 0.94 of defective parts per box Solution; Defective/box(incorrect)=1.12 Total defective(incorrect)=1.12*50=56 That means 56 parts were defective while examining 50 boxes Since "1" part was incorrectly recorded as "10", the actual total of defective parts=56-10+1=47 and actual average number of defective parts per box=47/50=0.94 Hence our answer is C. 142. QtyA QtyB The sum of the odd integers The sum of the even integers from 1 to 199 from 2 to 198 Solution; 1+3+5+……….197+199 (100 terms) ? 2+4+6+……..+198 (99 terms) 199 ? (2-1)+(4-3)+(6-5)+…..+(198-197) (99 terms) 199 ? 1+1+1+….99 terms 199 ? 99 199 > 99 Hence choice A is greater. 72
73. 73. 143. Before Maria changed jobs, her salary was 24% more than Julio's salary. After Maria changed jobs, her new salary was 24% less than her old salary. QtyA QtyB Julio's salary Maria's new salary Solution; Before Julio's salary=100 Maria's salary=124 After Maria's salary=0.86*124=106.64 Hence our answer is B. 144. p and q are different prime numbers. r is the least prime number greater than p, and s is the least prime number greater than q. QtyA QtyB r-p s-q Solution; p=2, q=7 r=3, s=11 r-p=3-2=1 s-q=11-7=4 in which case our answer would be B. p=7, q=2 r=11, s=3 r-p=11-7=4 s-q=3-2=1 in which case our answer would be A. Hence our answer is D. 73
74. 74. 145. Team X scored p points more than team Y, and the two teams together scored a total of 10 points. QtyA QtyB Twice the number of 10-p points team Y scored Solution; Let team X scored X points and team Y scored Y points. Then, X=Y+p and X+Y=10 So, (Y+p)+Y=10 2Y=10-p Hence our answer is C. 146. QtyA QtyB (x-1)(x)(x+1) (x)(x)(x) Solution; If x=1, then QtyA=0*1*2=0 and QtyB=1*1*1=1 In this case, our answer would be B. If x=0, then QtyA=-1*0*1=0 and QtyB=0*0*0=0 In this case, our answer would be C. Hence our answer is D. 147. QtyA QtyB The number of distinct prime The number of distinct prime factors of x factors of 4x Solution; 74
75. 75. Here, the number of prime factors of 4x cannot be less than the number of prime factors of x since 4x contains all the factors of x. So, the answer must be that either they are equal or there is not enough information. let x=2 then prime factors of 2=2 in which number of distinct prime factors is 1. prime factors of 4*2=2*2*2 in which the number of distinct prime factors is 1. Here number of distinct prime numbers is same in both cases. i.e., our answer would be C. Next let x=21 then prime factors of 21=3*7 in which the number of distinct prime factors is 2. prime factors of 4*21=2*2*3*7 in which the number of distinct prime factors is 3. Here number of distinct prime numbers is greater in B i.e., our answer would be B. Hence D is our answer. 148. x is an even integer. QtyA QtyB The number of distinct prime The number of distinct prime factors of 4x factors of x Solution; Let x=2, then factors of 4*2=2*2*2 and factors of 2=2 That means our answer would be C. Next let x=6 then factors of 4*6=2*2*2*3 factors of 6=2*3 Since 4 is an even number which has factors of 2 and 2 and which are not distinct, so the number of distinct factors of 4x will be equal to the number of distinct prime factors of x. 75
76. 76. Hence our answer is C. 149. x>y QtyA QtyB z 60 Solution; A x k x z B y C z>k 2z+k=180 let k=30, then 2z=180-30=150 z=75 in which case our answer would be A. let k =70, then 2z=180-70=110 z=55 which cannot be, since z>k and here, z<k That means z cannot be less than 60. If z=60, then z=k which cannot be. So our answer is A. 76
77. 77. 150. Three tennis balls of identical size are stacked one on the top of the other so that they fit exactly inside a closed right cylindrical can, as shown. QtyA QtyB The height of the stack of 3 balls The circumference of one of the balls Solution; A=3d B=3.14*d Hence QtyB is greater. 151. S is the set of all fractions of the form n/n+1, where n is a positive integer less than 20. QtyA QtyB The product of all the fractions 1/20 that are in S S={1/2, 2/3, 3/4, 4/5,………….19/20} A=1/2*2/3*3/4*4/5*……………*19/20=1/20=B Hence our answer is C. 77
78. 78. 152. QtyA QtyB The length of minor arc of The length of minor arc WX of the circle. YZ of the circle. Solution; Angles circumscribed on equal arcs (or same arcs) of a circle are equal. So arc WX=arc YZ Hence our answer is C. 153. A retail business has determined that its net income, in terms of x, the number of items sold, is given by the expression x2+x-380. QtyA QtyB The number of items that must 10 be sold for the net income to be zero. Solution; Net Income= x2+x-380=0 x2+20x-19x-380=0 x(x+20)-19(x+20)=0 (x+20)(x-19)=0 x=-20 or 19 78
79. 79. Since number of items sold cannot be negative, x=19. Hence our answer is A. 154. PQRS is a parallelogram QtyA QtyB x y x-5=y+10 (opposite angles of a parallelogram are equal) x-y=10+5=15 Hence A is greater. 155. S is the midpoint of segment PR. QtyA QtyB The length of segment QT The length of segment QR Solution; QP=/QS2+SP2=/QS2+SR2=QR QP>QT (common sense) So QR>QT Hence B is greater. 79
80. 80. 156. Segments PA, PB and PC are the angle bisectors of triangle ABC. QtyA QtyB x+y 57 Solution; A+B+C=180 A/2+B/2+C/2=90 x+33+y=90 x+y=90-33=57 Hence our answer is C. 157. Q and T are the midpoints of opposite sides of square PRSU. QtyA QtyB The area of region PQST 3/2 Solution; Since PRSU is a square, 80
81. 81. angle R=90 So in right angled triangle QRS, QR=1 and RS=2 since 12+22=(/5)2 Area of triangle QRS=area of triangle TUP=1/2*1*2=1 Area of square=22=4 Area of region PQST=4-(1+1)=2 Hence A is greater. 158. In circles C1 and C2, the length of segment PR equals the length of segment QR. QtyA QtyB The circumference of circle C1 The circumference of circle C2 Solution; A=2*3.14*PR B=2*3.14*QR PR=QR So A=B hence our answer is C. 159. QtyA x(x+1)(x+2) QtyB x*x*x Solution; 81
82. 82. For positive numbers, A will be greater. For x=0, A=0 and B=0 So our answer is D. 160. The probability that both events E and F will occur is 0.25. QtyA QtyB The probability that event E 0.24 will occur P(E)*P(F)=0.25 P(E)=0.25/P(F) When P(F) is maximum i.e. 1, P(E) will be minimum. If P(F)=1, then, P(E)=0.25 When P(F) is minimum i.e. 0.25, P(E) will be maximum If P(F)=0.25, then P(E)=1 P(E) lies from 0.25 to 1. So A is greater. 161. The mean of five distinct positive integers is 10. QtyA QtyB The largest possible value of 46 one of the integers Solution; a+b+c+d+e+f=10*5=50 1+2+3+4+f=50 f=50-10=40 A=40 So B is greater. Note: If the positive integers given were not distinct, then 1+1+1+1+f=50 f=50-4=46 i.e. A and B would be equal. 82
83. 83. 162. M={4,6,3,7,x,x} N={1,5,8,y} x>y>0 QtyA QtyB The mean of set M The mean of set N Solution; A=(4+6+3+7+x+x)/6=(20+2x)/6 B=(1+5+8+y)/4=(14+y)/4 (20+2x)/6 ? (14+y)/4 80+8x ? 84+6y 8x-6y ? 4 4x-3y ? 2 If x=2 and y=1, then 4*2-3*1=8-3=5>2 which points to answer A. if x=0.5 and y=0.4, then 4*0.5-3*0.4=2-1.2=0.8<2 which points to answer B. Hence our answer is D. 83
84. 84. MULTIPLE CHOICE QESTIONS - Select one answer choice 1. If 15!/3m is an integer, what is the greatest possible value of m? A. 4 B. 5 C. 6 D. 7 E. 8 Solution; 15!/3m=15*14*13*…..*1/3m=36*5*4*2*……../3m Since, 15=3*5, 12=3*4, 9=3*3, 6=3*2, 3=3*1 We can see, for 15!/3m to be an integer, the maximum possible value of m could be 6. Hence answer is C. 2. Which of the following numbers is farthest from the number 1 on the number line? A. -10 B. -5 C. 0 D. 5 E. 10 Solution; Clearly, the answer is either A or E. distance between numbers 1 and -10=1-(-10)=1+10=11 distance between numbers 1 and 10=10-1=9 84
85. 85. Hence answer is A. 3. A certain jar contains 60 jelly beans - 22 white, 18 green, 11 yellow, 5 red and 4 purple. If a jelly bean is to be chosen at random, what is the probability that the jelly bean will neither be red nor purple? A. 0.09 B. 0.15 C. 0.54 D. 0.85 E. 0.91 Solution; P(neither red nor purple)=P(White or Green or Yellow) =22/60+18/60+11/60=51/60=0.85 Hence answer is D. 4. A certain store sells two types of pens: one type for \$2 per pen and the other type for \$3 per pen. If a customer can spend up to \$25 to buy pens at the store and there is no sales tax, what is the greatest number of pens the customer can buy? A. 9 B. 10 C. 11 D. 12 E. 20 Solution; For buying greatest number of pens, the customer must buy maximum number of low cost pens. Hence she can buy 12 low cost pens for \$24 and take \$1 in return or she can buy 11 \$2 pens for \$22 and a \$3 pen for \$3. In both cases she can buy a maximum of 12 pens. 85
86. 86. Hence D is answer. 5. If y=3x and z=2y, what is x+y+z in terms of x? A. 10x B. 9x C. 8x D. 6x E. 5x Solution; x+y+z=x+3x+2(3x)=10x Hence A is the answer. 6. A certain shipping service charges an insurance fee of \$0.75 when shipping any package with contents worth \$25.00 or less, and an insurance fee of \$1.00 when shipping any package with contents worth over \$25.00. If Dan uses the shipping company to ship three packages with contents worth \$18.25, \$25.00 and \$127.50, what is the total insurance fee that the company charges Dan to ship three packages? A. \$1.75 B. \$2.25 C. \$2.50 D. \$2.75 E. \$3.00 Solution; 0-\$25 \$0.75 \$25.00+ \$1.00 Total charge=(\$0.75*2)+\$1.00=\$2.50 Hence our answer is C. 86
87. 87. 7. If 55 percent of the people who purchase a certain product are female, what is the ratio of females who purchase the product to the number of males who purchase the product? A. 11 to 9 B. 10 to 9 C. 9 to 10 D. 9 to 11 E. 5 to 9 Solution; Total=100 Number of females purchasing the product=55 Number of males purchasing the product=100-55=45 So, Females/Males=55/45=11/9 Hence our answer is A. 8. The figure above shows the graph of the function f in the x-y plane. What is the value of f(f(-1))? A. -2 B. -1 C. 0 D. 1 E. 2 87
88. 88. Solution; From the graph, f(-1)=2 therefore, f(f(-1))=f(2)=1 Hence D is our answer. 9. By weight, liquid A makes up 8 percent of solution R and 18 percent of solution S. If 3 grams of solution R are mixed with 7 grams of solution S, the liquid A accounts for what percent of the weight of the resulting solution? A. 10% B. 13% C. 15% D. 19% E. 26% Solution; In 100 grams solution of R, there are 8 grams of liquid A. In 1 grams solution of R, there are 8/100 grams of liquid A. In 3 grams solution of R, there are 8*3/100 grams of liquid A. Similarly, In 100 grams solution of S, there are 18 grams of liquid A. In 1 gram solution of S, there are 18/100 grams of liquid A. In 7 grams solution of S, there are 18*7/100 grams of liquid A. That means, when the two solutions are mixed, the resulting 3+7=10 grams of solution will have (8*3/100)+(18*7/100)=0.24+1.26=1.50 grams of liquid A. Hence percentage of liquid A=1.50/10*100=15% Hence our answer is C. 88
89. 89. 10. Of the 700 members of a certain organization, 120 are lawyers. Two members of the organization will be selected at random. Which of the following is closest to the probability that NEITHER of the members selected will be a lawyer? A. 0.5 B. 0.6 C. 0.7 D. 0.8 E. 0.9 Solution; P(neither lawyer)=P(first not lawyer)*P(second not lawyer) =580/700*579/699=0.686(=0.7) Hence our answer is C. Another method!!! P(neither lawyer)=580C2/700C2=0.68 (=ways of selection of both non-lawyers/total selection) 11. A manager is forming a 6-person team to work on a certain project. From the 11 candidates available for the team, the manager has already chosen 3 to be on the team. In selecting the other 3 team members, how many different combinations of 3 of the remaining candidates does the manager have to choose from? A. 6 B. 24 C. 56 D. 120 E. 462 Solution; From the 11 candidates, 3 have already been chosen. So for forming a 6 person team, the manager has to select (6-3)=3 candidates out of (11-3)=8 candidates. So, 8C3=8!/3!5!=56. That means our answer is C. 89
90. 90. A. 0 B. 0 C. 0 D. 0 E. 0 12. Which of the following could be the graph of all values of x that satisfy the inequality 2-5x<=-(6x-5)/3? Solution; 2-5x<=-(6x-5)/3 6-15x<=-6x+5 6-5<=15x-6x 1<=9x 1/9<=x Hence our answer is C. 13. If 1+x+x2+x3=60, then the average of x,x2,x3 and x4 is equal to which of the following? A. 12x B. 15x C. 20x D. 30x E. 60x Solution; (x+x2+x3 +x4)/4=x(1+x+x2+x3)/4=x*60/4=15x Hence our answer is B. 90
91. 91. 14. The sequence of numbers a1, a2, a3,…..an,…. is defined by an=(1/n)-(1/n+2) for each integer n>=1. What is the sum of the first 20 terms of this sequence? A. (1+1/2)-1/20 B. (1+1/2)-(1/21+1/22) C. 1-(1/20+1/23) D. 1-1/22 E. 1/20-1/22 Solution; a1=1/1-1/3 a2=1/2-1/4 a3=1/3-1/5 ……………….. a1+a2+a3+……….a20=(1+1/2+1/3+…….+1/20)-(1/3+1/4+…….1/22)=(1+1/2)-(1/21+1/22) Hence our answer is B. 15. What is the least positive integer that is NOT a factor of 25! and is NOT a prime number? A. 26 B. 28 C. 36 D. 56 E. 58 Solution; 25!=25*24*23*22*…………………1 26=13*2 28=14*2 36=12*3 91
92. 92. 56=7*8 58=29*2 That means 58 is not a factor of 25! since none of the factors of 25! is divisible by 29. Hence our answer is E. 16. If 0<a<1<b, which of the following is true about the reciprocals of a and b? A. 1<1/a<1/b B. 1/a<1<1/b C. 1/a<1/b<1 D. 1/b<1<1/a E. 1/b<1/a<1 Solution; a<1<b Here a and b are both positive. Taking reciprocals, 1/a>1>1/b Ex,0.1<1<2 taking reciprocals, 1/0.1>1/1>1/2 i.e., 10>1>0.5 Hence our answer is D. 17. Of the 750 participants in a professional meeting, 450 are female and 1/2 of the female and 1/4 of the male participants are less than 30 years old. If one of the participants will be randomly selected to receive a prize, what is the probability that the person selected will be less than 30 years old? A. 1/8 B. 1/3 C. 3/8 D. 2/5 92
93. 93. E. 3/4 Solution; Female=450, Male=300 1/2 of female=1/2*450=225 (<30 years old) 1/4 of male=1/4*300=75 (<30 years old) P(less than 30)=(225+75)/750=300/750=2/5 Hence our answer is D. 18. From the even numbers between 1 and 9, two different even numbers are to be chosen at random. What is the probability that their sum will be 8? A. 1/6 B. 3/16 C. 1/4 D. 1/3 E. 1/2 Solution; Even numbers between 1 and 9 are 2,4,6 and 8. There is just one combination for the sum to be 8 i.e. combination of 2 and 6. Total number of combinations=4C2=6 Hence P(sum=8)=1/6 Hence A is our answer. 19. If x and y are the tens and the unit digit respectively of a product 725,278*67,066, what is the value of x+y? A. 12 B. 10 C. 8 93
94. 94. D. 6 E. 4 Solution; 725278 *67066 .……….68 ...……68 + 48 Now 4+8=12 Hence our answer is A. 20. What is the least possible value of x+y/xy if 2<=x<y<=11 and x and y are integers? A. 22/121 B. 5/6 C. 21/110 D.13/22 E. 1 Solution; x+y/xy=1/x+1/y value of 1/x+1/y is least when both x and y are greatest. From the given inequality, the greatest possible value of x is 10 and that of y is 11. So least possible value of x+y/xy is (10+11)/10*11=21/110. Hence our answer is C. 94
95. 95. 21. If 998*1002>106-x, x could be A. 1 B. 2 C. 3 D. 4 E. 5 Solution; 998*1002=(1000-2)(1000+2)=10002-4=(103)2-4=106-4 So, 106-4>106-x i.e., -4>-x i.e., x>4 Hence our answer is E. 22. To reproduce an old photograph, a photographer charges x dollars to make a negative, 3x/5 dollars for each of the first 10 prints and x/5 dollars for each print in access of 10 prints. If \$45 is the total charge to make a negative and 20 prints from an old photograph, what is the value of x? Solution; A. 3 B. 3.5 C. 4 D. 4.5 E. 5 Solution; x+3x/5*10+x/5*10=45 x+6x+2x=45 x=5. So our answer is E. 95
96. 96. 23. A certain cake recipe states that the cake should be baked in a pan 8 inches in diameter. If Jules wants to use the recipe to make a cake of the same depth but 12 inches in diameter, by what factor should he multiply the recipe ingredients? A. 5/2 B. 9/4 C. 3/2 D. 13/9 E. 4/3 Solution; Volume of cake1=3.14*d12*h1/4 Volume of cake2=3.14*d22*h2/4 Now required factor of multiplication=V2/V1=(d2/d1)2=(12/8)2=(3/2)2=9/4 Hence our answer is B. 24. A reading list for humanities course consists of 10 books, of which 4 are biographies and the rest are novels. Each student is required to read a selection of 4 books from the list, including 2 or more biographies. How many selections of the 4 books satisfy the requirements? A. 90 B. 115 C. 130 D. 144 E. 195 Solution; Biographies Novels Combinations 2 2 4C2*6C2=90 3 1 4C3*6C1=24 4 0 4C4=1 96
97. 97. Total ways=90+24+1=115 Hence our answer is B. 25. If the probability of choosing 2 red marbles without replacement from a bag of only red and blue marbles is 3/55 and there are 3 red marbles in the bag, what is the total number of marbles in the bag? A. 8 B. 11 C. 55 D. 110 E. 165 Solution; Let n be the number of blue marbles. Then P(RR)=P(R)*P(R)=(3/3+n)*(2/2+n)=6/(3+n)(2+n) or, 3/55=6/(3+n)(2+n) or, 1/55=2/(3+n)(2+n) or, 6+3n+2n+n2=110 or, n2+5n-104=0 or, n(n+13)-8(n+13)=0 or, (n+13)(n-8)=0 Since n cannot be negative, n=8. Hence total number of marbles in the bag=8+3=11. So answer is B. 26. If 1/(211)(517) is expressed as a terminating decimal, how many non-zero digits will the decimal have? A. 1 B. 2 97
98. 98. C. 4 D. 6 E. 11 Solution; 1/211517=1/211*511*56=1/1011*56=(1/5)6*10-11=0.26*10-11=(2*10-1)6*10-11=64*10-17 Hence there will be 2 non-zero digits in the decimal value. Our answer is B. 27. Distance from Centerville(miles) Freight train -10t+115 Passenger train -20t+150 The expressions in the table above give the distance from Centerville to each of two trains t hours after 12:00 noon. At what time after 12:00 noon will the trains be equivalent from Centerville? A. 1:30 B. 3:30 C: 5:10 D. 8:50 E. 11:30 Solution; -10t+115=-20t+150 10t=35 t=3.5 hours. Hence, they will be equivalent from Centerville at 3:30. Hence our answer is B. 98
99. 99. 28. In state X, all vehicle license plates have 2 letters from the 26 letters of the alphabet followed by 3 one-digit numbers. How many different license plates can State X have if repetition of letters and numbers is allowed? A. 23,400 B. 60,840 C. 67,600 D. 608,400 E. 676,000 Solution; Total combinations=26*26*10*10*10=676000 Hence our answer is E. 29. A developer has land that has x feet of lake frontage. The land is to be subdivided into lots, each of which is to have either 80 feet or 100 feet of lake frontage. If 1/9 of the lots are to have 80 feet of frontage each and the remaining 40 lots are to have 100 feet of frontage each, what is the value of x? A. 400 B. 3,200 C. 3,700 D. 4,400 E. 4,760 Solution; 1/9 of lots=40/8=5lots 8/9 of lots=40lots 80 feet of frontage each 100 feet of frontage each x=5*80+40*100=400+4000=4,400 Hence our answer is D. 99
100. 100. 30. Which of the following is not a factor of 1030? A. 250 B. 125 C. 32 D. 16 E. 6 Solution; 1030=(2*5)30=230*530 250=53*2 125=53 32=25 16=24 6=2*3 Our answer is E since 230*530 does not have a factor of 3. 31. R y Q T x S In the figure, QRS is an equilateral triangle and QTS is an isosceles triangle. If x=47, what is the value of y? A. 13 B. 23 C. 30 D. 47 E. 53 100
101. 101. Solution; =60-47=13 y 47 60 Hence our answer is A. 32. If k is an integer and 0.0010101*10k>1000, what is the least possible value of k? A. 2 B. 3 C. 4 D. 5 E. 6 Solution; 0.0010101*10k>1000 If k=4, 0.0010101*10k=0.0010101*104=10.101<1000 Now k=5 too won't be working because it would become 101.01 So if k=6, 0.0010101*10k=0.0010101*106=1010.1 Any value of k greater than 6 would also satisfy the given equation. But 6 is the least integer. hence our answer is E. 33.If n is a positive integer and k+2=3n, which of the following could NOT be a value of k? A. 1 B. 4 C. 7 D. 25 101
102. 102. E. 79 Solution; k=3n-2 for n=1, k=31-2=1 for n=2, k=32-2=7 That means, 4 could not be the value of k since there are no integers in between 1 and 2. Hence our answer is B. 34. -10 A B -2 C -1 0 D 2 E 10 Five points A,B,C,D and E are shown on a number line. What is the probability of all numbers being negative if three numbers are selected at random? A. 1/10 B. 3/5 C. 2/5 D. 4/5 E. 3/10 Solution; A,B and C are 3 negative numbers and D,E are 2 positive numbers. P(all negative)=3C3/5C3=number of ways of selection of 3 negative numbers/total ways =1/10 Hence our answer is A. 35. Square T is formed by joining the mid-points of sides of square S. The perimeter of square S is 40. What is the area of square T? A. 45 B. 48 C. 49 102
103. 103. D. 50 E. 52 Solution; 5 5 Square S Square T /52+52=/50 10 Perimeter of S=40 4l=40 l=10 Area=(/50)2=50 Hence our answer is D. 36. How many different positive integers are there in which the tens digit is greater than 6 and the units digit is less than 4? A. 7 B. 9 C. 10 D. 12 E. 24 Solution; tens digit needs to be greater than 6 i.e. it can be 7,8 or 9 so it has 3 choices. units digit needs to be less than 4 i.e. it can be 0,1,2 or 3 so it has 4 choices. Hence total number of ways=3*4=12. Our answer is D. 103
104. 104. 37. If in 1998 there were 10,000 bias-motivated offenses based on ethnicity, how many more offenses were based on religion than on sexual orientation? A. 4 B. 40 C. 400 D.4000 E. 40,000 Solution; 10.0% 10000 1% 1000 16% 16000(religion) 15.6% 15600(sexual orientation) 16000-15600=400 Answer C. 104
105. 105. 38. A rectangular game board is composed of identical squares arranged in a rectangular array of r rows and r+1 columns. The r rows are numbered from 1 through r, and the r+1 columns are numbered from 1 through r+1. If r>10, which of the following represents the number of squares on the board that are neither in the 4th row nor in the 7th column? A. r2-r B. r2-1 C. r2 D. r2+1 E. r2+r Solution; r+1 columns r rows Solution; Here, r(r+1)=r2+r is the area of the board. Our answer must be less than this. So we can eliminate choice E. Sum of squares of 4th row and 7th column=(r+1)+r-1=2r So required number of squares=r2+r-2r=r2-r Hence our answer is A. 105
106. 106. 39. S is a set containing 9 different numbers. T is a set containing 8 different numbers, all of which are members of S. Which of the following statements CANNOT be true? A. The mean of S is equal to the mean of T. B. The median of S is equal to the median of T. C. The range of S is equal to the range of T. D. The mean of S is greater than the mean of T. E. The range of S is less than the range of T. Solution; S={-4,-3,-2,-1,-0.5,0,1,2,3} and T={-4,-3,-2,-1,0,1,2,3} mean of S=-4.5/9=-1/2 mean of T=-4/8=-1/2 Choice A can be true. Next, let S={-4,-3,-2,-1,0,1,2,3,4} and T={-4,-3,-2,-1,0,1,2,3} mean of S=0/9=0 mean of T=-4/9 Choice D can also be true. If S={-4,-3,-2,-1,0,1,2,3,4} and T={-4,-3,-2,-1,1,2,3,4} then range of S=4-(-4)=8 and range of T=4-(-4)=8 Choice C can be true as well. Next, let S={-4,-3,-2,-1,1,2,3,4} and T={-4,-3,-2,-1,1,2,3,4} then range of S=5-(-4)=8 and range of T=5-(-4)=9 Choice E can also be true. 106
107. 107. 40. What is the greatest positive integer n such that 2n is a factor of 1210? A. 10 B. 12 C. 16 D. 20 E. 60 Solution; 1210=(2*2*3)10=210*210*310=220*310 If n=20, then 1210/220=220*310/220=310 But if n=21, then 1210/221=220*310/221=310/2 Hence the greatest possible value of n is 20. Hence our answer is D. 41. If x is an integer and y=9x+13, what is the greatest value of x for which y is less than 100? A. 12 B. 11 C. 10 D. 9 E. 8 Solution; Try with x=10 then y=9*10+13=103 That means we can eliminate choices A,B and C because, if x=10 gives y>100, then x=11 or 12 will also give y>100. Now try x=9, then y=9*9+13=94(<100) x=8 will also give y<100. But since we need the greatest value of x for which y<100, our answer is D. 107
108. 108. 42. Each month, a certain manufacturing company's total expenses are equal to a fixed monthly expense plus a variable expense that is directly proportional to the number of units produced by the company during that month. If the company's total expenses for a month in which it produces 20,000 units are \$570,000 and the total expenses for a month in which it produces 25,000 units are \$705,000, what is the company's fixed monthly expense? A. \$27,000 B. \$30,000 C. \$67,500 D. \$109,800 E. \$135,000 Solution; total expenses=fixed expense+variable expense \$570,000=fixed expense+k*20,000 \$705,000=fixed expense+k*25,000 (variable expense=k*number of units produced) Solving above equations, we get 135,000=5000k k=27 So 570,000=fixed expense+(27*20,000) fixed expense=30,000 Hence our answer is B. 43. A team has a record of 12 wins and 13 losses for the season. Three games remain. If the probability of winning each remaining game is 1/2 and there are no draws, what is the probability that the team will finish the season with a winning record? A. 1/5 B. 1/4 C. 3/8 108
109. 109. D. 1/2 E. 5/8 Solution; The possible results of remaining 3 games are WWW (12+3=15 wins and 13 losses win) WWL (12+2=14 wins and 13+1=14 losses, no draws) WLW LWW LWL (12+1=13 wins and 13+2=15 losses loss) LLW (loss) WLL (loss) LLL (12 wins and 13+3=16 losses loss) 1 winning possibility out of 5 possibilities so probability of winning the season=1/5, which is answer A. 44. How many three digit integers are odd and do not contain the digit 5? A. 360 B. 320 C. 288 D. 256 E. 252 Solution; 8*9*4=288 can be from 1 to 9 except 5 can be from 0 to 9 except 5 can be 1,3,5 or 7 8 choices 9 choices 4 choices 109
110. 110. Hence our answer is C. 45. When the fraction 1/37 is converted to a decimal, what is the 24th digit to the right of the decimal place? A. 0 B. 2 C. 3 D. 5 E. 7 Solution; 1/37=0.027027027….. The pattern is repeating after the 3rd digit of the decimal i.e. 7. So the 24th digit after the decimal is also 7 since 24 is divisible by 3. Hence our answer is E. 46. What is the ratio of surface area of a cube to the surface area of a rectangular solid identical to the cube in all ways except that its length has been doubled? A. 1:4 B. 3:8 C. 1:2 D. 3:5 E. 2:1 Solution; Let length of each edge of the cube be 1. Then, Surface area of cube, S1=6l2=6 for the rectangular solid, length=2, breadth=1 and height=1 Surface area of rectangular solid, S2=2(lb+bh+lh)=2(2*1+1*1+2*1)=10 110
111. 111. S1/S2=6/10=3/5 which is choice D. 47. 12,2732+12,2742= A. 299,235,509 B. 300,568,327 C. 301,277,605 D. 302,435,782 E. 303,053,291 Solution; 12,2732+12,2742=…………..9+…………6=………………5 The result must have unit digit 5. Hence our answer is C. 48. The quantity 22335566 will end in how many zeroes? A. 0 B. 2 C. 3 D. 5 E. 6 Solution; 22335566=223355(2*3)6=2233552636=283955=23253955=2310539 Hence there will be 5 zeroes. That means our answer is D. 49. Employee at a company is paid fixed salary of \$90,000 annually plus 100 shares of that company. After 9 months, he leaves the job and receives \$65,000 and 80 shares. What is the price of the share? A. \$450 B. \$500 111
112. 112. C. \$700 D. \$750 E. Cannot be determined from the given information. Solution; In 12 months, he gets \$90,000 plus 100 shares. In 1 month, he gets \$90,000/12=\$7,500 plus 100/12 shares. In 9 months, he gets \$7,500*9=\$67,500 plus 100/12*9=75 shares. Let each share worth \$x. Then 67,500+75x=65,000+80x x=\$500 Hence our answer is B. 50. Paint needs to be thinned to a ratio of 2 parts paint to 1.5 parts water. The painter has by mistake added water so that he has 6 liters of paint which is half water and half paint. What must he add to make the proportions of the mixture correct? A. 1 liter paint B. 1 liter water C. 1/2 liter water and 1 liter paint D. 1/2 liter paint and 1 liter water E. 1/2 liter paint Solution; Paint:water=2/1.5=4/3(required) In 6 liters of paint mixture, there are 1/2*6=3 liters of paint and 1/2*6=3 liters of water. Now to make the ratio 4:3, he must add 1 liter paint to the mixture. i.e., paint:water=(3+1)/3=4/3 Hence our answer is A. 112
113. 113. 51. Kelly took three days to travel from City A to City B by automobile. On the first day, Kelly traveled 2/5 of the distance from City A to City B and on the second day, she traveled 2/3 of the remaining distance. Which of the following is equivalent to the fraction of the distance from City A to City B that Kelly traveled on the third day? A. 1-2/5-2/3 B. 1-2/5-2/3(2/5) C. 1-2/5-2/5(1-2/3) D. 1-2/5-2/3(1-2/5) E. 1-2/5-2/3(1-2/5-2/3) Solution; 1st day=2/5 remain=1-2/5 2nd day=2/3*(1-2/5) 3rd day=1-2/5-2/3(1-2/5) Hence our answer is choice D. 52. If x and y are integers and x=50y+69, which of the following must be odd? A. xy B. x+y C. x+2y D. 3x-1 E. 3x+1 Solution; 50y is always even. 69 is odd. So x must be odd. (even+odd=odd) and y can be either odd or even. xy=odd*(odd or even)=odd or even 113
114. 114. x+y=odd+(odd or even)=even or odd x+2y=odd+even=odd 3x-1=3*odd-1=odd-1=even 3x+1=3*odd+1=odd+1=even Hence our answer is choice C. 53. In the first half of the last year, a team won 60 percent of the games it played. In the second half of the last year, the team played 20 games, winning 3 of them. If the team won 50 percent of the games it played last year, what was the total number of games the team played last year? A. 60 B. 70 C. 80 D. 90 E. 100 Solution; Let number of games played in 1st half be x. 1st half=60% of x won=0.6x 2nd half=3 won out of 20 games Number of games won in 1st half and 2nd half=50% of total number of games played 0.6x+3=0.5(x+20) 0.1x=7 x=70 Total number of games played=70+20=90 Hence our answer is D. 114
115. 115. 54. In the sequence a1,a2,a3,……a100, the kth term is defined by ak=1/k-1/k+1 for all integers k from 1 through 100. What is the sum of the 100 terms of this sequence? A. 1/10,100 B. 1/101 C. 1/100 D. 100/101 E. 1 Solution; a1=1-1/2 a2=1/2-1/3 a3=1/3-1/4 a4=1/4-1/5 ………………… ………………… a100=1/100-1/101 a1+a2+a3+a4+……..+a100=1-1/101=100/101 Hence our answer is choice D. 55. Eight hundred insects were weighed, and the resulting measurements in milligrams, are summarized in the box plot below. 100 105 110 114 120 126 130 140 146 If the 80th percentile of the measurements is 130 mgs, about how many measurements are between 126 mgs and 130 mgs? A. 30 B. 32 115
116. 116. C. 35 D. 40 E. 42 Solution; 75th percentile=126 80th percentile=130 So between 126 mgs and 130 mgs, there are 5% of total data. 5% of 800=5/100*800=40 Hence our answer is choice D. 56. There is a leak in the bottom of tank. This leak can empty a full tank in 8 hours. When the tank is full, a tap is opened into the tank which intakes water at rate of 6 gallons per hour and the tank is now emptied in 12 hours. What is the capacity of tank? A. 28 gallons B. 36 gallons C. 144 gallons D. 150 gallons E. cannot be determined from the information given. Solution; Let capacity of tank be x. In 1 hour, the leak empties x/8 gallons water. In 1 hour, the tank intakes 6 gallons water. In 1 hour, x/8-6 gallons water is emptied. In 12 hours, 12(x/8-6) gallons water is emptied. We are given, x gallon tank(full tank) is emptied in 12 hours. So, 12(x/8-6)=x 1.5x-72=x 116
117. 117. 0.5x=72 x=144 Hence C is our answer. 57. Set S includes elements {8,2,11,x,3,y} and has an average of 7 and a median of 5.5. If x<y, then which of the following is the maximum possible value of x? A. 0 B. 1 C. 2 D. 3 E. 4 Solution; (8+2+11+x+3+y)/6=7 24+x+y=42 x+y=18 Median=5.5 if x=1, then y=17 1,2,3,8,11,17 median=(3+8)/2=5.5 if x=2, then y=16 2,2,3,8,11,16 median=(3+8)/2=5.5 if x=3, then y=15 2,3,3,8,11,15 median=(3+8)/2=5.5 if x=4, then y=14 117
118. 118. 2,3,4,8,11,14 median=(4+8)/2=6 Hence maximum possible value of x is 3. So our answer is D. 58. If x+y=10, and xy=20, what is the value of 1/x+1/y? A. 1/20 B. 1/15 C. 1/10 D. 1/2 E. 2 Solution; 1/x+1/y=(y+x)/xy=10/20=1/2 Hence our answer is D. 59. In a normal distribution, 68 percent of scores lie within one standard deviation of the mean. If the SAT scores of all the high school juniors in the Center City followed a normal distribution with a mean of 500 and a standard deviation of 100, and if 10,200 students scored between 400 and 500, approximately how many students scored above 600? A. 2,400 B. 4,800 C. 5,100 D. 7,200 E. 9,600 118
119. 119. Solution; 34% 34% 2%14% 400 500 14% 2% 600 400-500 10,200 34% 10,200 1% 300 16% 4,800 Hence our answer is B. 60. John bought a \$100 DVD player on sale at 8% off. How much did he pay including 8% sales tax? A. \$84.64 B. \$92.00 C. \$96.48 D. \$99.36 E. \$100.00 Solution; SP=92% of \$100=\$92 (excluding tax) SP=108% of \$92=\$99.36 Hence our answer is D. 119
120. 120. 61. For how many positive integers m<=100 is (m-5)(m-45) positive? A. 45 B. 50 C. 58 D. 59 E. 60 Solution; (m-5)(m-45)>0 Two possibilities m-5>0 and m-45>0 So m>5 and m>45 m>45 and m<=100 So number of possible values of m in this case=100-45=55 m-5<0 and m-45<0 So m<5 and m<45 m<5 So number of possible values of m in this case=4 Hence our answer is 55+4=59, that means D. 62. If the average (arithmetic mean) of 3a and 4b is less than 50, and a is twice b, what is the largest possible integer value of a? A. 9 B. 10 C. 11 D. 19 E. 20 Solution; (3a+4b)/2<50 120
121. 121. 3a+4b<100 a=2b 2a=4b So 3a+2a<100 5a<100 a<20 Since a needs to be an integer, a=19 So our answer is D. 63. x,y,a and b are positive integers. When x is divided by y, the remainder is 6. When a is divided by b, the remainder is 9. Which of the following is NOT a possible value of y+b? A. 24 B. 21 C. 20 D. 17 E. 15 Solution; x divided by y gives remainder 6 means y>6 Ex, 12)18(1 -12 6 a divided by b gives remainder 9 means b>9 Ex, 10)19(1 -10 9 If y>6 and b>9, then y+b>15 121
122. 122. Thus, 15 cannot be the sum of y and b. So our answer is E. 64. A pair of dice is tossed twice. What is the probability that the first toss gives a total of either 7 or 11 and the second toss gives a total of 7 ? A. 1/27 B. 1/18 C. 1/9 D. 1/6 E. 7/18 Solution; 1 3 4 5 6 (1,2) (1,3) (1,4) (1,5) (1,6) 2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) 6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) 1 2 (1,1) There are a total of 62=36 possible outcomes in the first toss of a pair of coins as shown above. P(total=7 or 11 in the first toss)=8/36 Again, there are same number of 36 possible outcomes in the second toss. P(total=7 in the second toss)=6/36 P(total=7 or 11 in the first toss) and P(total=7 in the second toss)=8/36*6/36=2/9*1/6=1/27 Hence our answer is A. 122
123. 123. 65. Three dice are rolled simultaneously. What is the probability that exactly two of the dice will come up as the same number? A. 5/12 B. 11/24 C. 25/54 D. 13/27 E. 1/2 Solution; If we roll three dice simultaneously, then there will be a total of 63=216 possible outcomes. Let A,B and C be 3 dice. If A and B come up same number, then there are 5 choices left with C. Ex If A and B are both 1 or treating A and B as one unit, then C will have 5 choices left for exactly two of the dice to come up as same number. (that means C cannot be 1) So there are 6*5=30 ways that A and B come up as same number. Likewise there are 30 ways that (B and C) and (C and A) come up as same number. So there are 30*3=90 favorable outcomes. So probability=90/216=45/108=15/36=5/12 Hence our answer is A. 66. Which of the following could be the median for a set of integers {97, 98, 56, x, 86}, given that 20 < x < 80? A. 71 B. 86 C. 91.5 D. 97 E. 397.5 Solution; 123
124. 124. x,56,86,97,98 56,x,86,97,98 Hence our answer is B. 67. In the coordinate plane, rectangle WXYZ has vertices at (–2, –1), (–2, y), (4, y), and (4, –1). If the area of WXYZ is 18, what is the length of its diagonal? A. 3/2 B. 3/3 C. 3/5 D. 3/6 E. 3/7 Solution; (-2,y) (-2,-1) (4,y) (4,-1) Area=18=6*(y+1) y+1=3 y=2 Diagonal length=/(62+32)=/(36+9)=/45=/(5*9)=3/5 Hence answer is C. 68. In the repeating decimal 0.0653906539..., the 34th digit to the right of the decimal point is A. 9 B. 6 124
125. 125. C. 5 D. 3 E. 0 Solution; The pattern is repeating after 5th digit. The 5th digit after the decimal is 9. So multiples of 5 will have the digit 9. Hence 35th digit is also 9 and a digit before it will be 3. Hence our answer is D. 69. The numbers in data set S have a standard deviation of 5. If a new data set is formed by adding 3 to each number in S, what is the standard deviation of the numbers in the new data set? A. 2 B. 3 C. 5 D. 8 E. 15 Solution; Let S={0,5,10} mean=15/3=5 sd= (0-5)2+(5-5)2+(10-5)2 = 50/3 3 S1={0+5,5+5,10+5}={5,10,15} mean=30/3=10 sd= (5-10)2+(15-10)2+(15-15)2 = 50/3 3 That means, standard deviation of a set of numbers is not affected by adding same number to each number. Hence our answer is C. 70. Aisha's income in 2004 was 20 percent greater than her income in 2003. What is the ratio of Aisha's income in 2004 to her income in 2003? A. 1 to 5 B. 5 to 6 125
126. 126. C. 6 to 5 D. 5 to 1 E. 20 to 1 Solution; In 2003, income=100 In 2004, income=120 income(2004)/income(2003)=120/100=6/5 Hence our answer is C. 71. Jacob's weekly take-home pay is n dollars. Each week he uses 4n/5 dollars for expenses and saves the rest. At those rates, how many weeks will it take Jacob to save \$500, in terms of n? A. 500/n B. 2,500/n C. n/625 D. n/2,500 E. 625n Solution; Expenses=4n/5 \$ Save=(n-4n/5) \$=n/5 \$ In 1 week, save=n/5 \$ in 5/n weeks, save=1 \$ in 5/n*500=2,500/n weeks, save=500\$ Hence our answer is B. 72. The operation ♥ is defined for all integers x and y as x♥y=xy-y. If x and y are positive integers, which of the following CANNOT be zero? A. x♥y B. y♥x C. (x-1)♥y D. (x+1)♥y E. x♥(y-1) Solution; 126
127. 127. x♥y=y(x-1) If x=1, then its value can be zero. y♥x=yx-x=x(y-1) If y=1, then its value can be zero. (x-1)♥y=(x-1)y-y=y(x-1-1)=y(x-2) If x=2, its value also can be zero. (x+1)♥y=(x+1)y-y=xy+y-y=xy Its value cannot be zero since x and y are positive integers and the product of two positive integers cannot be zero. If we see choice E, x♥(y-1)=x(y-1)-(y-1)=(y-1)(x-1) If x=y=1, its value can be zero. Hence our answer is D. 73. P, Q and R are three points in a plane, and R does not lie on line PQ. Which of the following is true about the set of all points in the plane that are the same distance from all three points? A. It contains no points. B. It contains one point. C. It contains two points. D. It is a line. E. It is a circle. Solution; P Q S R If any three non co-linear points are given, then we can find a point such that it is at same distance from all given points. A circle can be drawn through points P,Q and R so that S is the centre and SP=SQ=SR=radii of the circle. But note that the circle is not at same distance from points P,Q and R. It's point S or the center of the circle which is at same distance from the given points. That means, the set of points that are the same distance from all three points contain just a point. Hence our answer is B. 127
128. 128. 74. X and Y are two points on a plane. Which of the following is NOT true about the set of all points in the plane that are the same distance from the given two points? A. It is a straight line. B. It contains infinite points. C. It bisects the line joining X and Y. D. It is perpendicular to the line joining X and Y. E. It contains two points. Solution; X Y The set of points that are same distance from X and Y will be perpendicular bisector of the line joining X and Y. Hence choice E is not true. 75. If x<y<0, which of the following inequalities must be true? A. y+1<x B. y-1<x C. xy2<x D. xy<y2 E. xy<x2 Solution; x<y<0 Let x=-2 and y=-1 then, -1+1>-2 or y+1>x So eliminate A. and -1-1=-2 or y-1=x So eliminate B. and -2*(-1)2=-2 or xy2=x So eliminate C. and -2*(-1)>(-1)2 or xy>y2 So eliminate D. and -2*(-1)<(-2)2 or xy>x2 So our answer is E. 128
129. 129. 76. For all integers x, the function f is defined as follows. f(x)=x-1 if x is even =x+1 if x is odd If a and b are integers and f(a)+f(b)=a+b, which of the following statements must be true? A. a=b B. a=-b C. a+b is odd. D. Both a and b are even. E. Both a and b are odd. Solution; a=b Let a=b=1 then f(a)+f(b)=f(1)+f(1)=(1+1)+(1+1)=4 and a+b=1+1=2 So choice A cannot be true. Next, a=-b Let a=1 and b=-1 then f(a)+f(b)=f(1)+f(-1)=(1+1)+(-1+1)=2 and a+b=1-1=0 So choice B cannot be true. Next, a+b is odd Let a=1 and b=2 so that a+b=3 which is odd. then f(a)+f(b)=f(1)+f(2)=(1+1)+(2-1)=3 and a+b=1+2=3 So choice A can be true. Next, both a and b are even Let a=2 and b=4 then f(2)+f(4)=2-1+4-1=4 and a+b=2+4=6 So choice D is not true. Next, both a and b are odd Let x=1 and y=3 then f(1)+f(3)=1+1+3+1=6 and a+b=1+3=4 So choice E is not true. Hence, our answer is C. 129
130. 130. 77. If y-2+2y-1-15=0, which of the following could be the value of y? A. 3 B. 1/5 C. -1/5 D. -1/3 E. -5 Solution; 1/y2+2/y-15=0 1+2y-15y2=0 15y2-2y-1=0 15y2-5y+3y-1=0 5y(3y-1)+1(3y-1)=0 (3y-1)(5y+1)=0 y=1/3 or -1/5 Hence our answer is C. 78. The figure shows the standard normal distribution, with mean 0 and standard deviation 1, including approximate percents of the distribution corresponding to the six regions shown. Ian rode the bus to work last year. His travel times to work were approximately normally distributed, with a mean of 35 minutes and a standard deviation of 5 minutes. According to the figure shown, approximately what percent of Ian's travel to work last year were less than 40 minutes? A. 14% B. 34% C. 60% D. 68% E. 84% 130