mathewhe/aqua_rat huggingface dataset:
Question: Of the 7 distinguishable wires that lead into an apartment, 4 are for cable television service, and 3 are for telephone service. Using these wires, how many distinct combinations of 3 wires are there such that at least 1 of the wires is for cable television?
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10
Rationale: Given:
Total number of wires = 5
Number of cable wires = 2
Number of telephone wires = 3
To find:
Number of combination which has at least one cable wires
Solution:
No of ways of selecting'at least'1 cable wire means, we can select more than one as well. The minimum we can select is one and the maximum we can select, given the constraints that 3 wires need to be selected in total and there are 2 cable wires, is 2
Since it is a combination of wires, the arrangement is not important
Approach 1:
Number of ways of selecting at least one cable wire in a selection of 3 wires from 5 wires =Selection 1(Number of ways of selecting one cable wire and two telephone wires )+Selection 2(Number of ways of selecting two cable wires and 1 telephone wire)
Selection 1
Number of ways of selecting one cable wire = 2C1 = 2
Number of ways of selecting 2 telephone wires = 3C2 = 3
Total = 2C1 * 3C2 = 6 ( m ways of doing something and n ways of doing something else together give m*n ways of doing - the holy grail rule in Combinatorics)
Selection 2
Number of ways of selecting one cable wire = 2C2 = 1
Number of ways of selecting 2 telephone wires = 3C1 = 3
Total = 2C2 * 3C1 = 3 ( m ways of doing something and n ways of doing something else together give m*n ways of doing - the holy grail rule in Combinatorics)
Selection 1 + Selection 2 = 9 ways of selecting 3 wires out of 5 such that at least one is a cable wire
Approach 2
Number of ways of selecting 3 wires out of 5 such that at least one is a cable wire =Selection X(Total number of ways of selecting 3 wires from the 5) -Selection Y(total ways of selecting 3 wires such that none is a cable i.e all the three are telephone wires)
Total number of ways of selecting 3 wires out of 5 = 5C2 = 10
Number ways of selecting 3 wires such that none is a cable i.e all the three are telephone wires = 3C3 ( 3 telephone wires and we are selecting all the three at once) = 1
Selection X - Selection Y =10
Answer is Option E
Answer: (E) 10
Question: 80 370 ? 1550 2440 3530
(A) 900
(B) 840
(C) 750
(D) 860
(E) 960
Rationale: 10^2 - 20 = 80
20^2 - 30 = 370
30^2 - 40 = 860
40^2 - 50 = 1550
50^2 - 60 = 2440
60^2 - 70 = 3530.
ANSWER:D
Answer: (D) 860
Question: If a clock takes seven seconds to strike seven, how long will it take to strike ten ?
(A) 7 seconds
(B) 9 seconds
(C) 10 seconds
(D) 10.5 seconds
(E) None of these
Rationale: for 7 strikes it have 6 interval so for 1 interval it takes 7/6sec
therefor for 10 strike it require 9 interval so time required=(7/6)*9=10.5sec
ANSWER:D
Answer: (D) 10.5 seconds
Question: The measurements obtained for the interior dimensions of a rectangular box are 120 cm by 120 cm by 180 cm. If each of the three measurements has an error of at most 1 centimeter, which of the following is the closes maximum possible difference, in cubic centimeters, between the actual capacity of the box and the capacity computed using these measurements?
(A) 55,600
(B) 53,600
(C) 54,600
(D) 58,600
(E) 57,600
Rationale: The options are well spread so we can approximate.
Changing the length by 1 cm results in change of the volume by 1*120*180 = 21600 cubic centimeters;
Changing the width by 1 cm results in change of the volume by 120*1*180 = 21600 cubic centimeters;
Changing the height by 1 cm results in change of the volume by 120*120*1 = 14400 cubic centimeters.
So, approximate maximum possible difference is 21600 + 21600 + 14400 = 57600 cubic centimeters.
Answer: E
Answer: (E) 57,600
Question: 3^14 + 3^14 + 3^14 =
(A) 9^12
(B) 3^16
(C) 9^14
(D) 3^15
(E) 3^17
Rationale: Since we have 3^14 added together 3 times, we can factor out a 3 and we get 3(3^14) which is equal to 3^15.
The answer is D.
Answer: (D) 3^15
Question: A certain manufacturer sells its products	to stores in 112 different	regions worldwide, with an average (arithmetic mean) of 181 stores per region. If last year these stores sold an	average of 51,752 units of the manufacturer’s product per store, which of the following is closest to the total number of units of manufacturer’s product	sold	worldwide last	year?
(A) 10^4
(B) 10^5
(C) 10^6
(D) 10^8
(E) 10^9
Rationale: 51752*181*112 or
51000*180*112 or
50000*200*100
=10^9
E
Answer: (E) 10^9
Question: A ship sails 4 degrees north, then 13 S. then 17 N. then 22 S. and has finally 11 degrees of south latitude. What was her latitude at starting ?
(A) 0
(B) 1
(C) 2
(D) 3
(E) None of these
Rationale: Let x = the latitude sought.
Then marking the northings +, and the southings -;
By the statement x + 4 - 13 + 17 - 22 = -11
And x = 3.
Answer D
Answer: (D) 3
Question: Rs.630/- distributed among A,B and C such that on decreasing their shares by RS.10,RS.5 and RS.15 respectively, The balance money would be divided among them in the ratio 3:4:5. Then, B’s share is:?
(A) Rs.310/-
(B) Rs.305/-
(C) Rs.205/-
(D) Rs.210/-
(E) Rs.215/-
Rationale: A:B:C = 3:4:5
Total parts = 12
B's share is = 4 parts
12 -----> Rs.600/-
4 -----> Rs.200/-
B's total = 200 + 5 = Rs.205/-
C)
Answer: (C) Rs.205/-
Question: A can do a work in 21 days and B alone can do it 28 days, they begin the work together, but A left after some days. B completed the remaining work in 21 days. After how many days did A leave ?
(A) 10
(B) 8
(C) 3
(D) 15
(E) 16
Rationale: Explanation:
(A+B) 1 day work = {1/21+1/28}=1/12 = 12days is required for A&B
B’S 1 day work = 1/28
B’s 21 days work = 21/28
1 - 21/28 = 7/28 - 1/4
1/4 *12=3 days
Answer: Option C
Answer: (C) 3
Question: A train travels from city C to city D. The average speed of the train is 60 miles/hr and it travels the first quarter of the trip at a speed of 90 mi/hr. What is the speed of the train in the remaining trip?
(A) 30
(B) 45
(C) 54
(D) 72
(E) 90
Rationale: Using the RTD chart, with A being the first quarter, B the rest of the trip and All the combinedtrip.
_______R_____T______D
C.........90........3..........270
D.........54.......15.........810
All........60.......18........1080
Let me explain how we fill in the chart:
1) Starting with what we know, we add 90 and 60
2) Picking an easy number for the total distance (6*9=54), so I chose 540. Multiply by 2 to get the whole trip, and we get 1080. Add 1080 for all-distance.
3) 1/4 of the whole distance happened at a Rate of 90. So, 1080/4=270, add 270 under A-D.
4) 2080 - 270 = 810, for the rest of the trip. Add 810 under B-D.
5) Claculate the individual Times for A and All, using T=D/R. Add the results, 3 and 18, under A-T and All-T.
6) 18-3=15, this is the remaining Time for B. Add 15 under B-T.
7) Finally, 15R=810 -->R=810/15 --> R= 54ANS C
Answer: (C) 54
Question: A man spends Rs. 1600 per month on an average for the first three months, Rs 1550 for next four months and Rs. 1800 per month for the last five months and saves Rs . 2800 a year. What is his average monthly income?
(A) 1900
(B) 2010
(C) 2100
(D) 2200
(E) 2300
Rationale: Total expenditure for the first 3 months = 3 × 1600 = 4800
Total expenditure for 4 months = 4 × 1550 = 6200
Total expenditure for 5 months = 5 × 1800 = 9000
Total expenditure and saving (which is income for one year)
= 4800 + 6200 + 9000 + 2800 = Rs. 22800
So, Average monthly income = 22800/12 = Rs. 1900
A
Answer: (A) 1900
Question: A bag contains 64 balls of eight different colours. There are eight of each colour (including red). What is the least number you would have to pick, without looking, to be sure of selecting 3 red balls?
(A) 49
(B) 37
(C) 59
(D) 72
(E) 60
Rationale: C
59
The first 56 balls could be of all colours except red. This would leave 8 balls, all of which are red. so any three chosen would be red.
Answer: (C) 59
Question: A student chose a number, multiplied it by 6, then subtracted 138 from the result and got 102. What was the number he chose?
(A) 40
(B) 120
(C) 130
(D) 140
(E) 150
Rationale: Solution:
Let xx be the number he chose, then
6⋅x−138=102
6x=240
x=40
Answer A
Answer: (A) 40
Question: In a lottery there are 20 prizes and 35 blanks. A lottery is drawn at random. What is the probability of getting a prize
(A) 2/13
(B) 4/11
(C) 5/11
(D) 6/11
(E) 3/13
Rationale: total draws = prizes+blanks= 20+35=55
probability of getting a prize = 20/55 = 4/11
correct option is B
Answer: (B) 4/11
Question: 4 dice are thrown simultaneously on the board. Find the probability show the same face.
(A) 2/113
(B) 2/141
(C) 1/216
(D) 2/115
(E) 3/140
Rationale: The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:
=6×6×6×6=64=6×6×6×6=64
n(S)=64n(S)=64
Let XX be the event that all dice show the same face.
X={(1,1,1,1,),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)}X={(1,1,1,1,),(2,2,2,2),(3,3,3,3),(4,4,4,4),(5,5,5,5),(6,6,6,6)}
n(X)=6n(X)=6
Hence required probability,
=n(X)n(S)=664=n(X)n(S)=664
=1/216
C
Answer: (C) 1/216
Question: Of the 200 students at University XYZ majoring in one or more of the engineering disciplines, 160 are majoring in electrical and 150 are majoring in mechanical. If at least 30 of the students are not majoring in either electrical or mechanical, then the number of students majoring in both electrical and mechanical could be any number from
(A) 50 to 70
(B) 70 to 100
(C) 110 to 130
(D) 130 to 150
(E) 150 to 170
Rationale: if there are 1630 students majoring in electrical then there must be 70 person not majoring electrical at all, since both of the student not majoring electrical and mechanical is at least 30, so the number of the student who are not majoring electrical but majoring mechanical will be at least 10.
if there are 150 students majoring in mechanical there must be 50 students who are not majoring mechanical at all, since the number of the student who are not majoring electrical but majoring mechanical will be at least 10, hence the number of students both majoring mechanical and electrical will be at least 50
so there must be at least 50 students who major in both
see the answers option the only possible answer is A !
Answer: (A) 50 to 70
Question: The average of 11 results is 60. If the average of first 6 results is 58 and that of the last six is 63, find the sixth result?
(A) 50
(B) 52
(C) 65
(D) 42
(E) 66
Rationale: Sixth result = 58*6 + 63*6 - 60*11 = 66
Answer is E
Answer: (E) 66
Question: n∑n=1 1/log2n(a) =?
(A) n(n + 1)/2 loga2
(B) n(n - 1)/2 log2a
(C) n(n + 1)2n2/4 loga2
(D) n(n + 1)2n2/4 loga3
(E) None of these
Rationale: n∑n=11/log2n(a) = n∑n=1loga(2n)
=n∑n=1nloga2
=n(n + 1)/2 loga2
ANSWER:A
Answer: (A) n(n + 1)/2 loga2
Question: A sum of money deposited at compound interest (compounded yearly) amounts to $3200 in 2 years and to $3296 in 3 years. What is the interest rate?
(A) 2%
(B) 3%
(C) 4%
(D) 5%
(E) 6%
Rationale: 3296 - 3200 = 96
96 / 3200 = 0.03 = 3%
The answer is B.
Answer: (B) 3%
Question: Two brothers took the GMAT exam, the higher score is X and the lower one is Y. If the difference between the two scores is equal to their average, what is the value of X/Y ?
(A) 3.
(B) 2.
(C) 1/2.
(D) 1/3.
(E) There isn't enough data to answer the question.
Rationale: Answer is A : 3
X - Y = (X + Y)/2
Solving for X/Y =3
Answer: (A) 3.
Question: A person goes to his office at 1/3rd of the speed at which he returns from his office. If the avg speed during the whole trip is 15m/h. what is the speedof the person while he was going to his office?
(A) 8km/h
(B) 10km/h
(C) 12km/h
(D) 13km/h
(E) 14km/h
Rationale: u = k , v= 3k
\inline \therefore \frac{2uv}{u+v}\: \: \Rightarrow \frac{2\times k\times 3k}{(k+3k)}=15
\inline \Rightarrow 1.5k = 15
\inline \Rightarrow k=10km/h
B
Answer: (B) 10km/h
Question: What is the maximum value of x, in the equation 2x+y=42,if y>x
(A) 10
(B) 11
(C) 12
(D) 13
(E) 14
Rationale: x will have max value when y will have least possible value which can be a little greater than x as y>x.
when x=y , then
2x+y=3x = 42
x=14.
ANSWER:E
Answer: (E) 14
Question: A man walks at a speed of 5 km/hr and runs at a speed of 10 km/hr. How much time will the man require to cover a distance of 10 km, if he completes half of the distance, i.e., (5 1/4) km on foot and the other half by running?
(A) 2.5
(B) 3.5
(C) 4
(D) 2
(E) 1.5
Rationale: Required time = (5)/5 + (5)/10 = 1.5 hours.
Answer:E
Answer: (E) 1.5
Question: The “connection” between any two positive integers a and b is the ratio of the smallest common multiple of a and b to the product of a and b. For instance, the smallest common multiple of 8 and 12 is 24, and the product of 8 and 12 is 96, so the connection between 8 and 12 is 24/96 = 1/4
The positive integer y is less than 20 and the connection between y and 6 is equal to 1/1. How many possible values of y are there?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Rationale: Since “connection” between y and 6 is 1/1 then LCM(6, y)=6y, which means that 6 and y are co-prime (they do not share any common factor but 1), because if the had any common factor but 1 then LCM(6, y) would be less than 6y.
So, we should check how many integers less than 20 are co-prime with 6, which can be rephrased as how many integers less than 20 are not divisible by 2 or 3 (6=2*3).
There are (18-2)/2+1=9 multiples of 2 in the range from 0 to 20, not inclusive;
There are (18-3)/3+1=6 multiples of 3 in the range from 0 to 20, not inclusive;
There are 3 multiples of 6 in the range from 0 to 20, not inclusive (6, 12, 18) - overlap of the above two sets;
Total multiples of 2 or 6 in the range from 0 to 20, not inclusive is 9+6-3=12;
Total integers in the range from 0 to 20, not inclusive is 19;
Hence, there are total of 19-12=7 numbers which have no common factor with 6 other than 1: 1, 5, 7, 11, 13, 17 and 19.
Answer: A.
Answer: (A) 7
Question: Mrs. Rodger got a weekly raise of $135. If she gets paid every other week, write an integer describing how the raise will affect her paycheck.
(A) $ 145.
(B) $ 146
(C) $ 135
(D) $ 148
(E) None
Rationale: Solution:
Let the 1st paycheck be x (integer).
Mrs. Rodger got a weekly raise of $ 135.
So after completing the 1st week she will get $ (x+135).
Similarly after completing the 2nd week she will get $ (x + 135) + $ 135.
= $ (x + 135 + 135)
= $ (x + 270)
So in this way end of every week her salary will increase by $ 135.
Answer C
Answer: (C) $ 135
Question: How many revolutions will it take for a circular wheel of 1.95 m radius make in traveling a distance of 1 km?
(A) 14557.1
(B) 24257.1
(C) 11257.1
(D) 12257.1
(E) 12567.1
Rationale: 2 * 22/7 * 1.95 * x = 12257.1
x = 1000
Answer: D
Answer: (D) 12257.1
Question: A customer pays 50 dollars for a coffee maker after a discount of 20 dollars
What is the original price of the coffe maker?
(A) $50
(B) $60
(C) $70
(D) $80
(E) $90
Rationale: Let x be the original price.
x - 20 = 50
x - 20 + 20 = 50 + 20
x + 0 = 70
x = 70
correct answer is C)$70
Answer: (C) $70
Question: The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence?
(A) 22
(B) 25
(C) 70
(D) 27
(E) 21
Rationale: A + B = 60, A = 2B
2B + B = 60 => B = 20 then A = 40.
5 years, their ages will be 45 and 25.
Sum of their ages = 45 + 25 = 70.Answer:C
Answer: (C) 70
Question: Ms. Harris invested in Fund X and Fund B. The total amount she invested, in both funds combined, was $100,000. In one year, Fund X paid 23% and Fund B paid 17%. The interest earned in Fund B was exactly $200 greater than the interest earned in Fund X. How much did Ms. Morris invest in Fund X?
(A)  $32,000
(B)  $36,000
(C)  $40,000
(D)  $42,000
(E)  $45,000
Rationale: 0.17B-0.23X = 200 | * 100
17B-23X=20,000
X+B=100,000
B=100,000 - X
17(100,000 - X) - 23X = 20,000
1,700,000 - 17X - 23X = 20,000
1,680,000 = 40X
divide by 40
first two digits: 42...so answer is D.
Answer: (D)  $42,000
Question: The speed of a boat in still water is 60kmph and the speed of the current is 20kmph. Find the speed downstream and upstream?
(A) 35, 25 kmph
(B) 80, 40 kmph
(C) 40, 60 kmph
(D) 50, 55 kmph
(E) 50, 65 kmph
Rationale: Speed downstream = 60 + 20 = 80 kmph
Speed upstream = 60 - 20 = 40 kmph
ANSWER:B
Answer: (B) 80, 40 kmph
Question: Working together, Tim and Tom can type 15 pages in one hour. If they would be able to type 18 pages in one hour if Tom increases his typing speed by 60%, what is the ratio of Tom's normal typing speed to that of Tim?
(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6
Rationale: Lets say Tim types x pages an hour and Tom types y pages an hour.
We know that x+y=15
Tom increase speed by 60% means he will type 1.6 y pages an hour.
So we get x+1.6 y=18
We need to know the ratio of Tom's speed to Tim's speed. This is going to be proportional to the number of pages each can type in an hour, hence (y/x).
Subtracting both : 0.6 y=3 so y=5 ... so x=10
(y/x)=5/10 =1/2
Answer is (A)
Answer: (A) 1/2
Question: Bob finishes the first half of an exam in two-sixth the time it takes him to finish the second half. If the whole exam takes him an hour, how many minutes does he spend on the first half of the exam?
(A) 15
(B) 24
(C) 27
(D) 36
(E) 40
Rationale: Many times, it is easiest to think of problems like this conceptually (as opposed to formulaically). Conceptually, if the first half of the exam takes 2/6's the time of the second half (6/6's), we can see that the hour the entire exam took to finish can be broken down into eight's. (Another way to look at this problem is via ratio's - first half is 2:8 and the second half is 6:8). With each sixth of an hour being 15/2 minutes (60/8), the first half of the exam would have taken 15 minutes. Correct answer is A.
Answer: (A) 15
Question: Find the prodect of the local value and absolute value of 4 in 23468
(A) 400
(B) 1600
(C) 1200
(D) 80
(E) 70
Rationale: Explanation:
Local value of 4 = 4 x 100 = 400
Place value of 4 = 4
There fore = 4 x 400 = 1600
Answer: Option B
Answer: (B) 1600
Question: How many kg of salt at 42 paise per kg must a man mix with 25 kg of salt at 24 paise per kg so that he may, on selling the mixture at 40 paise per kg gain 25% on the outlay?
(A) 15 kg
(B) 18 kg
(C) 20 kg
(D) 24 kg
(E) None of these
Rationale: Here, cost price of mixture = 100×100/100+25 = 32 paise
∴q1q2=32−24/42−32=810=45
and hence q1 = 4⁄5 × 25 = 20 kg
Answer C
Answer: (C) 20 kg
Question: At what rate percent per annum will the simple interest on a sum of money be 2/5 of the amount in 10 years?
(A) 5%
(B) 4%
(C) 2%
(D) 6%
(E) 7%
Rationale: Let sum = x. Then, S.I. = 2x/5, Time = 20 years.
Rate = (100 * 2x) / (x * 5 * 20) = 2%
ANSWER:C
Answer: (C) 2%
Question: In how many different ways can the letters of the word ‘ARMOUM’ be arranged?
(A) 200
(B) 210
(C) 220
(D) 240
(E) 360
Rationale: ARMOUR = 6 letter whereas R repeated twice 6! 6xSx4x3x2x1 => 360
E
Answer: (E) 360
Question: There are 200 female managers in a certain company. Find the total number R of female employees in the company, if 2/5 of all the employees are managers and 2/5 of all male employees are managers.
(A) R=300
(B) R=400
(C) R=500
(D) 600
(E) None of these
Rationale: {Managers} = {Female Managers} + {Male Managers};
We are told that the total number of managers in the company is 2/5 of all the employees , thus{Managers} = 2/5 (M + F), where M and F are number of female and male employees, respectively.
Also, we know that 2/5 of all male employees are managers:{Male Managers} = 2/5 * Mas well as there are total of 200 female managers:{Female Managers} = 200;
Thus:2/5 (M + F) = 200 + 2/5 * M-->F = 500.
Answer: C.
Answer: (C) R=500
Question: The wages of 24 men and 16 women amounts to Rs.11600 per day. Half the number of men and 37 women earn the same amount per day. What is the daily wage of a man?
(A) 2378
(B) 277
(C) 208
(D) 270
(E) 350
Rationale: Let the wage of a man is m and woman be w.
24m+16w=11600
12m+37w = 11600
Solving we get m = 350
Answer:E
Answer: (E) 350
Question: A committee of 6 people is to be selected from 8 people. How many different committees are possible?
(A) 28
(B) 40
(C) 56
(D) D.168
(E) 336
Rationale: Because the order does not matter in selecting 6 people for the committee, this is a combination question.
The number of ways to select 6 people from 8 = 8C6 = 8!/6!*2! = 28. Hence A.
Answer: (A) 28
Question: A student has 7 trousers and 10 shirts. In how many different ways can the student select a trouser and a shirt?
(A) 52
(B) 58
(C) 64
(D) 70
(E) 76
Rationale: The student can select one trouser in 7 ways.
The student can select one shirt in 10 ways.
The number of ways to select one trouser and one shirt is 7 * 10 = 70 ways
The answer is D.
Answer: (D) 70
Question: The average mark obtained by 22 candidates in an examination is 47. The average of the first ten is 55 while the last eleven is 40 .The marks obtained by the 11th candidate is ?
(A) 22
(B) 44
(C) 49
(D) 68
(E) 76
Rationale: It is clear that 22 x 47 = 10 x 55 + K + 11 x 40 ⇒⇒ K = 44
Answer:B
Answer: (B) 44
Question: Out of three numbers, the first is twice the second and is half of the third. If the average of the three numbers is 63, the three numbers in order are?
(A) 27, 54, and 108
(B) 54, 27, and 108
(C) 108, 27, and 54
(D) 108, 54, and 27
(E) 54, 108, and 27
Rationale: Let the numbers be 2x, x and 4x
(2x + x + 4x)/3 = 63
7x/3 = 63
x = 27
The numbers in order are 54, 27, and 108
The answer is B.
Answer: (B) 54, 27, and 108
Question: A’s speed is 51/44 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat?
(A) 1/51
(B) 4/51
(C) 7/51
(D) 3/44
(E) 7/44
Rationale: We have the ratio of A’s speed and B’s speed. This means, we know how much distance A covers compared with B in the same time.
This is what the beginning of the race will look like:
(Start) A_________B______________________________
If A covers 51 meters, B covers 44 meters in that time. So if the race is 51 meters long, when A reaches the finish line, B would be 7 meters behind him. If we want the race to end in a dead heat, we want B to be at the finish line too at the same time. This means B should get a head start of 7 meters so that he doesn’t need to cover that. In that case, the time required by A (to cover 51 meters) would be the same as the time required by B (to cover 44 meters) to reach the finish line.
So B should get a head start of 7/51th of the race.
Answer (C)
Answer: (C) 7/51
Question: Andrea and Zach are both waiting for an
appointment with a guidance counselor.
When they arrived, each received a card
from the secretary, telling the hour and
minute of his or her arrival. Two minutes
ago, Andrea had been waiting exactly as
many minutes as Zach. Three minutes from
now, Andrea will have been waiting exactly
2/3 as long as Zach. If the time is now 11:30,
at what time did Andrea arrive?
(A) 11:16
(B) 11:18
(C) 11:20
(D) 11:23
(E) 11:25
Rationale: Let a equal the number of minutes that Andrea had been waiting two minutes ago. At
that time, Zach had been waiting for twice as long as Andrea, so he had been waiting for
2a minutes. Three minutes from now, both Andrea and Zach will each have been waiting
for 5 minutes longer:
Two minutes ago Three minutes from now
Andrea a a + 5
Zach 2a 2a + 5
Three minutes from now, Andrea will have been waiting for 23
as long as Zach, so make an equtation as follows:
a+5=2/3(2a+5)
3a+15=2(2a+5)
3a+15=4a+10
3a+5=4a
5=a
Thus, two minutes ago, at 11:28, Andrea had been waiting for 5 minutes. So she arrived at 11:23.
correct answer D)11.23
Answer: (D) 11:23
Question: A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?
(A) 63
(B) 511
(C) 1023
(D) 15
(E) 31
Rationale: Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 45 - 1.
= 210 - 1 = 1024 - 1 = 1023
ANSWER:C
Answer: (C) 1023