divelab/Sys2Bench huggingface dataset:
Question: Liam is pulled over for speeding just as he is arriving at work.He explains to the police officer that he could not afford to be late today, and has arrived at work only four minutes before he is to start. The officer explains that if Liam had driven 5mph slower for his whole commute, he would have arrived at work exactly on time. If Liam's commute is 40 miles long,how fast was he actually driving?(Assume that Liam drove at a constant speed for the duration of his commute.)
Options:
A)50 mph
B)45 mph
C)48 mph
D)52 mph
E)60 mph
Rationale: Let t be the number of hours he would need to reach office on time.
when he is driving with over speed, he reached office 4 min earlier! so the equation for this is s(t - 4/60) = 40
where s is the speed and 30 is the distance.
if he decreases his speed by 5mph then he would have reached his office on time: (s-5)t = 40
if you solve above equations, you will arrive at t = 2/3 hr and s = 60mph
therefore answer is E
Answer: E
Question: Salesperson A's compensation for any week is $240 plus 6 percent of the portion of A's total sales above $1,000 for that week. Salesperson B's compensation for any week is 8 percent of B's total sales for that week. For what amount of total weekly sales would both salespeople earn the same compensation?
Options:
A)$7000
B)$9000
C)$11,000
D)$13,000
E)$15,000
Rationale: 240+0.06(x-1000) = 0.08x
0.02x = 180
x = $9,000
The answer is B.
Answer: B
Question: A train passes a station platform in 21 seconds and a man standing on the platform in 15 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
Options:
A)299
B)277
C)90
D)88
E)112
Rationale: Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 15) m = 225 m.
Let the length of the platform be x meters.
Then, x + 225 / 21 = 15
x + 225 = 315
x = 90 m.
Answer:C
Answer: C
Question: Evaluate combination
11C3
Options:
A)10000
B)1000
C)165
D)1
E)0
Rationale: Explanation:
nCn=1
nCr=n!/r!(n-r)!
11C3=11!/3!(11-3)!=165
Option C
Answer: C
Question: The present ratio of students to teachers at a certain school is 40 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of students to teachers would then be 25 to 1. What is the present number of teachers?
Options:
A)5
B)8
C)10
D)12
E)15
Rationale: We are given that the ratio of students to teacher is 40 to 1. We can rewrite this using variable multipliers.
students : teachers = 40x : x
We are next given that student enrollment increases by 50 and the number of teachers increases by 5. With this change the new ratio becomes 25 to 1. We can put all this into an equation:
Students/Teachers  25/1 = (40x + 50)/(x + 5)
If we cross multiply we have:
25(x + 5) = 40x + 50
25x + 125 = 40x + 50
5 = x
Since x is the present number of teachers, currently there are 5 teachers.
Answer A.
Answer: A
Question: There are 15 points in a given plane, no three of which are on the same line. If one of the points is represented as 'L', then how many triangles can be determined with the 15 points that contain the point L?
Options:
A)91
B)105
C)182
D)210
E)455
Rationale: Fist point is L which is fixed so can be selected in 1 way
Second point can be selected in 14 ways
Third point can be selected in 13 ways
so total ways = 1x14x13 = 182
but answer is 91 which is 182/2=C
Answer: C
Question: Choose the number pair/group which is different from others?
Options:
A)8 : 39
B)8 : 38
C)8 : 33
D)8 : 32
E)8 : 31
Rationale: Explanation:
In all other pairs, 2nd number = (1st number × 3) + 1.
Answer: C) 8 : 33
Answer: C
Question: A and B together can do a piece of work in 8 days. If A alone can do the same work in 12 days, then B alone can do the same work in?
Options:
A)22 days
B)77 days
C)24 days
D)17 days
E)18 days
Rationale: B = 1/8 – 1/2 = 1/24 => 24 days
Answer: C
Answer: C
Question: Which of the following CANNOT be weighed using a balance scale and these unit weights: 1, 4, 7, and 10
Options:
A)13
B)15
C)17
D)20
E)21
Rationale: A. 18 (can be weighed by keeping weights of 10,1 and 7 )
B. 15 (can be weighed by keeping weights of 10,4 and 1 kg)
C. 17 (can be weighed by keeping weights of 10 and 7 kg)
D. 20(cannot be weighed)
E. 21 (can be weighed by keeping weights of 10, 7 and 4 kg)
D
Answer: D
Question: If pqr = 1 , rqt = 0 , and tpr = 0, which of the following must be zero?
Options:
A)P
B)Q
C)R
D)S
E)T
Rationale: pqr = 1 which means neither p / q/ r is 0
rq x t = 0 , we know r and q are not 0 from above hence t must be 0.
similarly p and r are not 0
t x p x r = 0, p and r are not 0 hence t =0.
option E
Answer: E
Question: If q is an integer, what is the greatest possible value for q that would still make the following statement true: 11*10^q < 1/10 ?
Options:
A) –4
B) –3
C) –2
D) –1
E) 0
Rationale: Just putting values, the LHS becomes
q=0 --> 11
q=-1 --> 11/10
q=-2 --> 11/100
q=-3 --> 11/1000
Anything lower will be smaller than 11/1000. q=-2, equality does not hold but it does for q=-3.
Answer is B.
Answer: B
Question: Three 6 faced dice are thrown together. The probability that no two dice show the same number on them is
Options:
A)5/7
B)5/9
C)5/4
D)5/1
E)5/3
Rationale: No two dice show same number would mean all the three faces should show different numbers. The first can fall in any one of the six ways. The second die can show a different number in five ways. The third should show a number that is different from the first and second. This can happen in four ways.
Thus 6 * 5 * 4 = 120 favourable cases.
The total cases are 6 * 6 * 6 = 216.
The probability = 120/216 = 5/9.
Answer:B
Answer: B
Question: In a certain year, the population of a certain town was 9000. If in the next year the population of males increases by 3% and that of the females by 9% and the total population increases to 9600, then what was the ratio of population of males and females in that given year?
Options:
A)7 : 11
B)5 : 4
C)2 : 3
D)Data inadequate
E)None of these
Rationale: Let the population of males = x; then the population of females = 9000 – x
Now, 3% of x + 9% of (9000 – x)
= (9600 – 9000 ) = 600
or 0.03x + 720 – 0.09x = 600
or 720 – 600 = 0.09x – 0.03x
or, 120 = 0.06x
x = 3500
Reqd ratio of population of males and females
3500/9000−3500=3500/5500=7:11
Answer A
Answer: A
Question: Find the greatest number that will divide 54, 87 and 172 so as to leave the same remainder in each case
Options:
A)13
B)3
C)15
D)1
E)2
Rationale: Explanation:
Required number = (87-54),(172-87),(172-54)
= H.C.F of 33,85 and 118 is 1
Answer: Option D
Answer: D
Question: A dishonest milkman wants to make a profit on the selling of milk. He would like to mix water (costing nothing) with milk costing Rs.42 per litre so as to make a profit of 20% on cost when he sells the resulting milk and water mixture for Rs.48 In what ratio should he mix the water and milk?
Options:
A)1:20
B)1:10
C)1:8
D)1:4
E)6:11
Rationale: water = w (liter)
Milk = m (liter)
==> Cost = Price x Quantity = 0.42m
==> Revenue = Price x Quantity = 0.48(m + w)
==> Profit =0.48(m+w) - 0.42m = 0.2*(0.42m)[20% of cost]
==> 0.48m + 0.48w - 0.42m = 0.084m
==> 0.024m = 0.48w
==> m/w = 0.48/0.024 = 20
--OR--
w/m = 1/20
A is correct.
Answer: A
Question: If the average marks of three batches of 45, 68 and 55 students respectively is 50, 55, 60, then the average marks of all the students is
Options:
A)55.29
B)54.68
C)54.6
D)54.58
E)None of these
Rationale: Explanation:
(45Ã—50)+(68Ã—55)+(55Ã—60) / 45+68+55
9290 / 168=55.29
Option A
Answer: A
Question: The average of first five multiples of 3 is :
Options:
A)3
B)9
C)12
D)15
E)17
Rationale: Solution
Average = 3(1+2+3+4+5)/5 =45/5 =9 Answer B
Answer: B
Question: A bus takes 6 hours to cover a distance of 540 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2 th of the previous time?
Options:
A)60 Kmph
B)72 Kmph
C)80 Kmph
D)84 Kmph
E)86 Kmph
Rationale: Time = 6
Distance = 540
3/2 of 6 hours = 6 * 3/2 = 9 Hours
Required speed = 540/9 = 60 Kmph
A
Answer: A
Question: Evaluate 50!/47!
Options:
A)102500
B)112584
C)117600
D)118450
E)128450
Rationale: Evaluate 50!/47! = 50*49*48* (47!)/47!
= 50*49*48
= 117600
ANSWER:C
Answer: C
Question: In the next Worldcup of cricket there will be 12 teams, divided equally in 2groups. Teams of eachgroup will play a match against each other. From each group 3 top teams will qualify for the next round. In this round each team will play against each others once. 4top teams of this round will qualify for the semifinal round, where they play the best of 3matches. The Minimum number of matches in the next World cup will be
Options:
A)34
B)36
C)53
D)59
E)61
Rationale: The number of matches in first round,
= 6C2 +6C2;
Number of matches in next round,
= 6C2;
Number of matches in semifinals,
= 4C2;
Total number of matches,
= 6C2 +6C2+6C2+4C2+2.
= 53.
C
Answer: C
Question: In how many ways can 5 men & 4 women can be made from a gang of 7 men & 8 women ?
Options:
A)91
B)98
C)97
D)86
E)90
Rationale: Explanation:
7C5 * 8C4 = 91
A
Answer: A
Question: A group of 55 adults and 70 children go for trekking. If there is meal for either 70 adults or 90 children and if 35 adults have their meal, find the total number of children that can be catered with the remaining food.
Options:
A)33
B)54
C)18
D)45
E)01
Rationale: Explanation:
As there is meal for 70 adults and 35 have their meal, the meal left can be catered to 35 adults.
Now, 70 adults = 90 children
7 adults = 9 children
Therefore, 35 adults = 45 children
Hence, the meal can be catered to 45 children.
ANSWER: D
Answer: D
Question: On his first 5 tests, David received an average score of N points. If on his sixth test, he exceeds his previous average score by 36 points, what is his average score for his first 6 tests?
Options:
A)N +12
B)N +6
C)N +3
D)N +8
E)N +18
Rationale: Total of 1st 5 tests= 5N
6th test score = N + 36
Total of all 6 tests = 6N + 36
Average = N +6
Answer : B
Answer: B
Question: What number has a 5:1 ratio to the number 10?
Options:
A)11
B)50
C)99
D)77
E)21
Rationale: 5:1 = x: 10
x = 50
Answer: B
Answer: B
Question: Each side of a rectangular field diminished by 40%. By how much per cent is the area of the field diminished?
Options:
A)32
B)64
C)25
D)16
E)None
Rationale: Solution: Let the Original length of the rectangle be 20 unit and breadth be 10 unit. Then
Original Area = length *breadth = 20*10 = 200 Square unit.
40% decrease in each side, then
Length = (20 - 40% of 20) = 12 unit.
Breadth = (10 - 40% of 10) = 6 unit.
Now, Area = 12 *6 = 72 Square unit.
Decrease in area = 200 - 72 = 128 square unit.
% Decrease in Area = (128/200)*100 = 64%
Mind Calculation Method:
Let the original area be 100 square unit.
100 ===40%↓(decrease in length)===> 60 ===40%↓===(decrease in breadth)===> 36.
Diminished in area = 100-36 = 64%
Answer: Option B
Answer: B
Question: A certain club has 10 members, including Jarry. One of the 10 members is chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. What is the probability that Jarry will be either the member chose to be secretary or the member chose to be treasurer?
Options:
A)1/720
B)1/80
C)1/10
D)1/9
E)1/5
Rationale: In order to become a secretary the person cannot be chosen as a president. This can be done in 9/10 ways. Then to become a secretary this person's probability is 1/9
Probability to become a secretary = (9/10 * 1/9) = 1/10
Similar concept for treasurer. Probability of not president = 9/10, Probability of not secretary = 8/9, probability of treasurer 1/8
probability to become a treasurer = (9/10 * 8/9 * 1/8) = 1/10
Since, the problem is saying Jarry secretary OR Treasurer it is addition: 1/10 + 1/10 = 1/5=E
Answer: E
Question: In how many ways can be 8 boys and 8 girls sit around circular table so that no two boys sit next to each other?
Options:
A)(5!)^2
B)(6!)^2
C)7!8!
D)11!
E)(5!)^2*6!
Rationale: first fix one boy and place other 7 in alt seats so total ways is 7!
now place each girl between a pair of boys... total ways of seating arrangement of girls 8!
total is 7!*8!
ANS C
Answer: C
Question: Out of first 35 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is ?
Options:
A)17/35
B)7/35
C)37/70
D)27/35
E)27/70
Rationale: n(S) = 35
n(Even no) = 17 = n(E)
n(Prime no) = 11 = n(P)
P(EᴜP)
= 17/35 + 11/35 - 1/35
= 27/35
Answer:D
Answer: D
Question: A small pool filled only with water will require an additional 600 gallons of water in order to be filled to 80% of its capacity. If pumping in these additional 600 gallons of water will increase the amount of water in the pool by 30%, what is the total capacity of the pool in gallons?
Options:
A)1000
B)1250
C)1300
D)1600
E)2000
Rationale: Since pumping in additional 600 gallons of water will increase the amount of water in the pool by 30%, then initially the pool is filled with 1,000 gallons of water.
So, we have that 1,000 + 600 = 0.8*{total} --> {total} = 2,000.
Answer: E.
Answer: E
Question: Chris mixed 3 pounds of raisins with 4 pounds of nuts. If a pound of nuts costs 2 times as much as a pound of raisins, then the total cost of the raisins was what fraction of the total cost of the mixture?
Options:
A) 1/7
B) 1/5
C) 1/4
D) 1/3
E) 3/11
Rationale: 1 lbs of raisin = $1
3 lbs of raisin = $3
1 lbs of nuts = $2
4 lbs of nuts = $8
Total value of mixture = 8+3 = 11
Fraction of the value of raisin = 3/11
Ans:E
Answer: E
Question: A shopkeeper buys two articles for Rs.1000 each and then sells them, making 20% profit on the first article and 20% loss on second article. Find the net profit or loss percent?
Options:
A)200
B)768
C)276
D)280
E)279
Rationale: Profit on first article = 20% of 1000 = 200.
This is equal to the loss he makes on the second article. That, is he makes neither profit nor loss.
Answer: A
Answer: A
Question: The cost of 10 kg of mangos is equal to the cost of 24 kg of rice. The cost of 6 kg of flour equals the cost of 2 kg of rice. The cost of each kg of flour is $24. Find the total cost of 4 kg of mangos, 3 kg of rice and 5 kg of flour?
Options:
A)347.4
B)987.4
C)877.2
D)1027.2
E)1667.2
Rationale: Let the costs of each kg of mangos and each kg of rice be $a and $r respectively.
10a = 24r and 6 * 24 = 2r
a = 12/5 r and r = 72
a = 144
Required total cost = 4 * 144 + 3 * 72 + 5 * 24
= 691.2 + 216 + 120 = $1027.20
D
Answer: D
Question: Three partners A,B and C invest Rs.18000, Rs.22500 and Rs.27000 respectively in a business. If total profit is Rs.12000 then A's share is (in Rs.)
Options:
A)10000
B)6500
C)5000
D)3200
E)1500
Rationale: 18000:22500:27000
4:5:6
A's share=12000*4/15=3200
ANSWER:D
Answer: D
Question: A shipment of 1500 heads of cabbage, each of which was approximately the same size was purchased for $600.The day the shipment arrived 2/3 of the heads were sold, each at 25% above the cost per head.The following day the rest were sold at a price per head equal to 14% less than the price each head sold for the day before.what was the gross profit on this shipment?
Options:
A)$100
B)$115
C)$125
D)$130
E)$135
Rationale: Sol:
1500 heads -> $600
1 head -> $600/1500
1 head -> $(2/5)
25% more of (2/5) -> 125/100 * 2/5 = $(1/2)
He sold 2/3*1500 = 1000 heads for $(1/2) per head
Total revenue by selling 1000 heads = 1000 * 1/2 = $500
Heads left: 500
Cost per head: 86% of the previous price: 86/100 * 1/2 = $(3/7)
Total revenue by selling 500 heads = 3/7 * 500 = 215
Total revenue after selling 1500 cabbage heads - 215+500 = $715
Money spent on the purchase: $600
Profit = 715-600 = $115
Ans: B
Answer: B
Question: Each of three investments has a 20% of becoming worthless within a year of purchase, independently of what happens to the other two investments. If Simone invests an equal sum M in each of these three investments on January 1, the approximate chance that by the end of the year, she loses no more than 1/3 of her original investment is
Options:
A)90%
B)80%
C)70%
D)60%
E)40%
Rationale: The problem asks for the approximate chance that no more than 1/3 of the original investment is lost. We can apply the “1 – x” technique: what’s the chance that more than 1/3 of the original investment is lost? There are two outcomes we have to separately measure:
(a) All 3 investments become worthless.
(b) 2 of the 3 investments become worthless, while 1 doesn’t.
Outcome (a): The probability is (0.2)(0.2)(0.2) = 0.008, or a little less than 1%.
Outcome (b): Call the investments X, Y, and Z. The probability that X retains value, while Y and Z become worthless, is (0.8)(0.2)(0.2) = 0.032. Now, we have to do the same thing for the specific scenarios in which Y retains value (while X and Z don’t) and in which Z retains value (while X and Y don’t). Each of those scenarios results in the same math: 0.032. Thus, we can simply multiply 0.032 by 3 to get 0.096, or a little less than 10%.
The sum of these two probabilities is 0.008 + 0.096 = 0.104, or a little more than 10%. Finally, subtracting from 100% and rounding, we find that the probability we were looking for is approximately 90%.
The correct answer is A.
This problem illustrates the power of diversification in financial investments. All else being equal, it’s less risky to hold a third of your money in three uncorrelated (independent) but otherwise equivalent investments than to put all your eggs in one of the baskets. That said, be wary of historical correlations! Housing price changes in different US cities were not so correlated—and then they became highly correlated during the recent housing crisis (they all fell together), fatally undermining spreadsheet models that assumed that these price changes were independent.
Answer: A
Question: 0.007 x 0.45=?
Options:
A)3.15e-05
B)0.000315
C)0.315
D)0.0315
E)0.00315
Rationale: 7 x 45 = 315
Sum of decimal places =5
Therefore, 0.007 x 0.45 = 0.00315
Answer is E.
Answer: E
Question: Josiah had $42,895.39 in his savings account. He wanted to make a $10,000 deposit on a new car. The car was priced at $85,216.78. The interest rate that Josiah will be paying back, for having the vehicle financed, is 8.99%. What is the total amount that Josiah would have paid on the total cost (including financed amount (interest)) on the vehicle minus $10.000 deposit?
Options:
A)92,877.77
B)100,562.31
C)1.29
D)120,261.53
E)60,487.15
Rationale: A
Answer: A
Question: A, B and C can do a work in 6 days, 8 days and 12 days respectively. In how many days can all three of them working together, complete the work?
Options:
A)2 2/3
B)2 2/9
C)2 2/8
D)2 2/5
E)2 2/1
Rationale: Work done by all three of them in one day = 1/6 + 1/8 + 1/12 = 3/8.
The number of days required = 8/3 = 2 2/3 days.
Answer:A
Answer: A
Question: A line that passes through (–2, –4) and (3, k) has a slope = k. What is the value of k?
Options:
A) 3/4
B) 1
C) 4/3
D) 2
E) 7/2
Rationale: Slope = (y2-y1)/(x2-x1)
=> k = (k+4)/(3+2)
=> 5k = k+4
=> k =1
Ans B it is!
Answer: B
Question: The average age of 30students in a class is 5years. If teacher's age is also included then average increases 1year then find the teacher's age?
Options:
A)59
B)55
C)61
D)45
E)36
Rationale: Total age of 50 students = 30*5 = 150
Total age of 51 persons = 31*6 = 186
Age of teacher = 186-150 = 36years
Answer is E
Answer: E
Question: If x + (1/x) = 5, what is the value of R=x^2 + (1/x)^2?
Options:
A)R=21
B)R=22
C)R=23
D)24
E)27
Rationale: Squaring on both sides,
x^2 + (1/x)^2 + 2(x)(1/x) = 5^2
x^2 + (1/x)^2 = 23
Answer: C
Answer: C
Question: An error 5% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square?
Options:
A)4.05%
B)4.02%
C)4%
D)10.28%
E)2%
Rationale: Percentage error in calculated area
=(5+5+(5Ã—5)/100)%
=10.25%
ANSWER:D
Answer: D
Question: Solve the equation for x : 19(x + y) + 17 = 19(-x + y) - 21
Options:
A)1
B)-1
C)2
D)-2
E)3
Rationale: B
-1
19x + 19y + 17 = -19x + 19y - 21
38x = -38 => x = -1
Answer: B
Question: A person bought an article and sold it at a loss of 10%. If he had bought it for 20% less and sold it for Rs.75 less, he could have gained 25%. What is the cost price?
Options:
A)Rs.250
B)Rs.225
C)Rs.275
D)Rs.200
E)Rs.100
Rationale: CP1 = 100 SP1 = 90
CP2 = 80 SP2 = 80 * (140/100) = 112
22 ----- 100
55 ----- ? => Rs.250
ANSWER:A
Answer: A
Question: At Didymus Corporation, there are just two classes of employees: silver and gold. The average salary of gold employees is $70,000 higher than that of silver employees. If there are 120 silver employees and 160 gold employees, then the average salary for the company is how much higher than the average salary for the silver employees?
Options:
A)$24,000
B)$28,000
C)$32,000
D)$36,000
E)$40,000
Rationale: The answer should be -> E. $40,000
Answer: E
Question: A train sets off at 2 p.m. at the speed of 50 kmph. Another train starts at 3:30 p.m. in the same direction at the rate of 65 kmph. At what time the trains will meet?
Options:
A)8.37 p.m
B)8.20 p.m
C)8.30 p.m
D)8.38 p.m
E)8.10 p.m
Rationale: D = 50 * 1 Â½ = 75 km
RS = 65 â€“ 50 = 15
T = 75/15 = 5 h
3.30 + 7 h = 8.30 p.m.
Answer: C
Answer: C
Question: What percent is 4 gm of 1 kg?
Options:
A)1.5 %
B)0.4 %
C)2.5 %
D)3.5 %
E)4 %
Rationale: 1 kg = 1000 gm
4/1000 × 100 = 400/1000
=2/5 = 0.4 %
B)
Answer: B
Question: A card game called “high-low” divides a deck of 52 playing cards into 2 types, “high” cards and “low” cards. There are an equal number of “high” cards and “low” cards in the deck and “high” cards are worth 2 points, while “low” cards are worth 1 point. If you draw cards one at a time, how many ways can you draw “high” and “low” cards to earn 6 points if you must draw exactly 4 “low” cards?
Options:
A)1
B)2
C)3
D)4
E)5
Rationale: Great question Ravih. This is a permutations problem (order matters) with repeating elements. Given thatlowcards are worth 1 pt andhigh cards2 pts, and you must draw 3 low cards, we know that you must also draw 1 high card. The formula for permutations problems with repeating elements isN!/A!B!...where N represents the number of elements in the group and A, B, etc. represent the number of times that repeating elements are repeated. Here there are 4 elements and thelowcard is repeated 3 times. As a result, the formula is:
5!/4! which represents (5*4*3*2*1)/(4*3*2*1) which simplifies to just 5, giving you answer E.
Answer: E
Question: Running at the same constant rate, 100 identical machines can produce a total of 500 coffee bar per minute. At this rate, how many bottles could 20 such machines produce in 2 minutes?
Options:
A) 110
B) 220
C) 330
D) 200
E) 789
Rationale: Let's take the approach that uses the answer choices to eliminate wasted time.
500/100 = 5 coffee bar per minute per machine. 20 machines = 100 per minute. 2 minutes worth = 200 coffe bar.
Looking at the answers it is clear...we can only choose (D)
THE CORRECT ANSWER IS D.
Answer: D
Question: Two workers completed the painting of a building in 10 days. If they were to paint the building separately, how many days will it take each worker, if one of them can complete it in approximately 2 days earlier than the other.
Options:
A)9 and 11
B)19 and 21
C)8 and 10
D)10 and 15
E)10 and 12
Rationale: Work = (A)(B)/(A+B) where A and B are the individual times of each entity.
Here, we're told that (working together) the two workers would complete a job in 10 days. This means that (individually) each of them would take MORE than 10 days to do the job.
Answers E, A and C are illogical, since the individual times must both be greater than 10 days.
so we can TEST the values for Answers B and D.
Using the values for Answers B and D...
Answer B: (19)(21)/(19+21) = 399/40 = 9.97 This is a match (complete it in approximately 2 days earlier than the other)
This IS a match
Final Answer:
B
Answer: B
Question: On Sunday, the original price of a shirt was discounted by 20%. On Monday, the price of the shirt was marked up L percent so that the new price on Monday was 20% more than the original price before the discount. What is the value of L?
Options:
A)20%
B)30%
C)33 1⁄3%
D)40%
E)50%
Rationale: Let the original price be 100
discounted Price = 80
New Price = 20 +100 =120
L=(40/80)*100
Ans:E
Answer: E