RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: A can do a piece of work in 12 days. He worked for 15 days and then B completed the remaining work in 10 days. Both of them together will finish it in.
A.12 1/2 days
B.10 1/2 days
C.12 1/3 days
D.10 1/3 days
Answer: A.12 1/2 days
Explanation: Explanation:
15/25 + 10/x = 1 => x = 25
1/25 + 1/25 = 2/25
25/2 = 12 1/2 days
Answer A
Question: A man took loan from a bank at the rate of 5% p.a. S.I. After 13 years he had to pay Rs. 1300 interest only for the period. The principal amount borrowed by him was?
A.1750
B.4000
C.3500
D.2000
Answer: D.2000
Explanation: Principal = (100 * 1300) / (5 * 13)
= Rs.2000
Answer: D
Question: When x is even, [x] = x/2 + 1, when x is odd [x] = 2x + 1 then [7]*[8] = ?
A.[22]
B.[44]
C.[75]
D.[88]
Answer: C.[75]
Explanation: [7]*[8]=(7*2+1)(8/2+1)
=[75].
Ans -C
Question: there are two cubes with ratio between volume is 125:64. then what is the value of length of each cube?
A.25:16
B.16:25
C.4:5
D.5:4
Answer: D.5:4
Explanation: given
6a12:a22=150:96
then a12:a22=25:16
a1:a2=5:4
so the answer is D
Question: Mark purchases a dozen items for $240.00. If Mark sells each item for $30.00 what is his markup%?
A.10
B.50
C.75
D.100
Answer: B.50
Explanation: C.P.=240/12=20
S.P.=30
markup%=((S.P.-C.P.)/C.P.)*100=((30-20)/20)*100=50%
ANSWER:B
Question: 5 percent of 255 =
A.5
B.12.5
C.14.5
D.12.75
Answer: D.12.75
Explanation: 255/100=2.551% =2.555%= 2.55*5 = 12.75 ANSWER:D
Question: A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half. The number of pupils in the class is :
A.45
B.40
C.39
D.37
Answer: B.40
Explanation: Explanation:
Let there be x pupils in the class.
Total increase in marks = (x * 1/2) = x/2.
x/2 = (83 - 63) => x/2 = 20 => x = 40. Answer: B
Question: Find out the C.I on Rs.4000 at 4% p.a. compound half-yearly for 1 1/2 years
A.244.83
B.306.07
C.306.04
D.306.03
Answer: A.244.83
Explanation: A = 4000(51/50)3
= 4244.832
4000
-----------
244.83
Answer:A
Question: In a mixed college 160 students are there in one class. Out of this 160 students 3/5 students are girls. How many boys are there ?
A.A)40
B.B)64
C.C)80
D.D)120
Answer: B.B)64
Explanation: Total Number of Students :160
Total Girls : 160*3/5 =96
Total Boys : 160-96= 64
Answer is B
Question: The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?
A.5m
B.3m
C.9m
D.2m
Answer: D.2m
Explanation: Let the sides of the rectangle be l and b respectively.
From the given data,
(√l2 + b2) = (1 + 108 1/3 %)lb
=> l2 + b2 = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l2 + b2)/lb = 25/12
12(l2 + b2) = 25lb
Adding 24lb on both sides
12l2 + 12b2 + 24lb = 49lb
12(l2 + b2 + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)2 = 49lb
=> 12(14)2 = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l - b = 8 - 6 = 2m.
Answer: D
Question: 100 + 50 * 2 = ?
A.200
B.150
C.100
D.300
Answer: A.200
Explanation: Explanation:
After applying BOMAS rule,
First Multiplication will be done then addition
So it will be 100 + 100 = 200
Option A
Question: The G.C.D. of 1.08, 0.36 and 0.8 is:
A.0.04
B.0.9
C.0.18
D.0.108
Answer: A.0.04
Explanation: Given numbers are 1.08, 0.36 and 0.80. H.C.F. of 108, 36 and 80 is 18,
H.C.F. of given numbers = 0.04.
Answer: Option A
Question: If X = Y/Z, -1 < Z < 0 and 4 < Y, which of the following is correct?
A.X > 4.
B.0 < X <4.
C.-4 < X < 0
D.X < -4.
Answer: D.X < -4.
Explanation: Y is +ve & Z is -ve
Eliminate A & B
For min. value max. numerator and minimize denominator
take Y=4.1 & X=-0.9
X=4.1/-0.9
X<-4
Ans D
Question: A project scheduled to be carried out over a single fiscal year has a budget of $12,000, divided into 12 equal monthly allocations. At the end of the fourth month of that fiscal year, the total amount actually spent on the project was $4,540. By how much was the project over its budget?
A. $380
B. $540
C. $1,050
D. $1,380
Answer: B. $540
Explanation: Difficulty level: 600
Each month's budget = 12000/12 = 1000
Budget for 4 months = 4*1000 = 4000
Actual amount spent = 4540
Amount spent over the budget = 4540 - 4000 = 540
Answer (B),
Regards,
Question: Two trains travel in the same direction at the speeds of 56km/hr. and 29km/hr. respectively. The faster train passes a man in the slower trains in 10 seconds. The length of the faster train (in metres) is
A.100
B.80
C.75
D.120
Answer: C.75
Explanation: Explanation:
The relative speed is 56-29 Km/hr =(27/18) X 5= 7.5 m/s2718×5=7.5m/s
The faster train travels only the length of it’s own = 7.5 x 10 = 75 m
Answer C
Question: One pump drains one-half of a pond in 6 hours, and then a second pump starts draining the pond. The two pumps working together finish emptying the pond in one-half hour. How long would it take the second pump to drain the pond if it had to do the job alone?
A.1 hour
B.1.09 hour
C.3 hours
D.5 hours
Answer: B.1.09 hour
Explanation: The tricky part here, I believed is one half hour = 1/2. Then everything would be easy.
We have the 1st pump working rate/hour = 1/2 :6 = 1/12
Working rate of 2 pumps: 1/2 : 1/2 = 1.
Working rate of 2nd pump: 1 -1/12= 11/12 --> Time taken for the 2nd pump to finish: 1 : 11/12 = 12/11 = 1.09 hours.
B
Question: A person got Rs.40 more when he invested a certain sum at compound interest instead of simple interest for two years at 10% p.a. Find the sum?
A.7500
B.4000
C.5000
D.4500
Answer: B.4000
Explanation: P = (d * 1002) / R2
=> (40 * 100 * 100) / 10 * 10 = Rs.4000
Answer:B
Question: A farmer had d animals, all of which the farmer bought for the same price. The farmer decided to sell the animals. 14 of the animals were sold at a profit of 10% each while the rest were sold at a loss of 20% each. Overall, the farmer made neither a profit nor a loss. Which of the following is equal to d?
A.21
B.24
C.27
D.33
Answer: A.21
Explanation: Let P be the price the farmer paid for each animal.
The total profit on the 14 animals is 14*0.1*P.
The total loss on the other (d-14) animals is (d-14)*0.2*P.
(d-14)*0.2*P = 14*0.1*P
0.2*d = 1.4 + 2.8
d = 21
The answer is A.
Question: The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet?
A.12
B.10
C.11
D.09
Answer: C.11
Explanation: Suppose they meet x hrs after 8 a.m. Then,
(Distance moved by first in x hrs) + [Distance moved by second in (x - 1) hrs] = 330
60x + 75(x - 1) = 330 => x = 3
So, they meet at (8 + 3) i.e., 11 a.m.
Answer:C
Question: A chair is bought for Rs.800/- and sold at Rs.720/- find the loss percentage
A.10% Loss
B.10% Gain
C.12% Loss
D.12% Gain
Answer: A.10% Loss
Explanation: Explanation:
Formula = (Selling price ~ Cost price)/Cost price * 100
= (720 ~ 800)/800 = 10% Loss
Answer: Option A
Question: Thabo owns exactly 180 books, and each book is either paperback fiction, paperback nonfiction, or hardcover nonfiction. If he owns 20 more paperback nonfiction books than hardcover nonfiction books, and twice as many paperback fiction books as paperback nonfiction books, how many hardcover books nonfiction books does Thabo own?
A.10
B.20
C.30
D.40
Answer: C.30
Explanation: I think we can use double-matrix method and solve using only one variable.
Our goal is to find the number of hardcover nonfiction books. Let that number be x. We are given that all 140 books are either paperback fiction, paperback nonfiction, or hardcover nonfiction. This implies that number of hardcover fiction books is 0.
Double-matrix: P = paperback; H = hardcover; F = fiction; NF = nonfiction
P H Total
F 2x+40 0
NF x+20 x
Total 3x+60 x 180
3x + 60 + x = 180
x = 30
Answer (C.)
Question: By selling an article at Rs.600, a profit of 25% is made. Find its cost price?
A.Rs.486
B.Rs.455
C.Rs.487
D.Rs.480
Answer: D.Rs.480
Explanation: SP = 600
CP = (SP)*[100/(100+P)]
= 600 * [100/(100+25)]
= 600 * [100/125]
= Rs.480
Answer: D
Question: Out of 12 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is at least one graduate among them?
A.A)5/11
B.B)5/11
C.C)5/1
D.D)9/11
Answer: D.D)9/11
Explanation: P(at least one graduate) = 1 - P(no graduates)
= 1 - ⁶C₃/¹⁰C₃
= 1 - (6 * 5 * 4)/(10 * 11 * 12)
= 9/11
Answer:D
Question: The H.C.F. of two numbers is 12 and their L.C.M. is 600. If one of the number is 12, find the other?
A.100
B.600
C.120
D.200
Answer: B.600
Explanation: Other number = 12*600/12= 600
Answer is B
Question: Find the odd man out. 8,40,42,60,96
A.8
B.40
C.42
D.60
Answer: C.42
Explanation: All except 42 are multiples of 4
Answer is C
Question: 40 is divided into two parts in such a way that seventh part of first and ninth part of second are equal. Find the smallest part?
A.13.2
B.17.5
C.12.8
D.34.25
Answer: B.17.5
Explanation: x/7 = y/9 => x:y = 7:9
7/16 * 40 = 17.5
ANSWER:B
Question: A part-time employee whose hourly wage was increased by 15 percent decided to reduce the number of hours worked per week so that the employee's total weekly income would remain unchanged. By what percent should the number of hours worked be reduced?
A.9
B.13.04
C.10
D.11
Answer: B.13.04
Explanation: Let original hourly wage be x and let the no of hours worked be y
Total wage will be =x*y
After the increment the wage will be=1.15 x
Now we need to find number of hours worked so that
x*y=1.15x*z i.e z=1/1.15 y
%decrease = (y-1/1.15y)/y *100=13.04%
Thus my answer is B.
Question: 81*82*83*84*85*86*87*89. What should be in the unit place in this product?
A.0
B.1
C.2
D.3
Answer: A.0
Explanation: It should be zero in the unit place in this product.
85*82 will give zero in unit place.
ANSWER:A
Question: At an upscale fast-food restaurant, Shin can buy 3 burgers, 7 shakes, and one cola for $120. At the same place it would cost $162.50 for 4 burgers, 10 shakes, and one cola. How much would it cost for a meal of one burger, one shake, and one cola?
A.$21
B.$35
C.$31
D.$41
Answer: B.$35
Explanation: let's suppose that the price of a burger is BB, of a shake - SS and that of a cola is CC. We can then construct these equations:
3B+7S+C=120
4B+10S+C=162.5
Subtracting the first equation from the second gives us B+3S=42.5
Now if we subtract the new equation two times from first or 3 times from second we will get B+S+C=35. In any case, there is no necessity to know each item's price, just the sum.
Answer: B
Question: A number is selected at random from the first 40 natural numbers. What is the probability that the number is a multiple of either 2 or 21?
A.21/40
B.2/5
C.7/15
D.4/15
Answer: A.21/40
Explanation: Number of Multiples of 2 from 1 through 40 = 40/2 = 20
Number of Multiples of 21 from 1 through 40 = 40/21 = 1
Number of Multiples of 2 and 21 both from 1 through 40 = Number of Multiples of 21*2(=42) = 0
Total favourable cases = 20 + 1 - 0 = 21
Probability = 21 / 40
Answer: option A
Question: A train running at the speed of 60 km/hr crosses a pole in 27 seconds. Find the length of the train?
A.150 meter
B.876 meter
C.167 meter
D.450 meter
Answer: D.450 meter
Explanation: Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 27
= 450 meter
Answer: D
Question: At what time between 7 and 8 o’clock will the hands of a clock be in the same straight line but, not together?
A.5 min. past 7
B.5 2⁄11 min. past 7
C.5 3⁄11 min. past 7
D.5 5⁄11 min. past 7
Answer: D.5 5⁄11 min. past 7
Explanation: When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.
At 7 o’clock, they are 25 min. spaces apart.
∴ Minute hand will have to gain only 5 in. spaces.
55 min.spaces are gained in 60 min.
5 min. spaces are gained in (60⁄55 × 5)min. = 5 5⁄11 min.
∴ Required Time = 5 5⁄11 min. past 7
Answer D
Question: The average marks of 25 students in a class is 100. But a student mark is wrongly noted as 60 instead of 10 then find the correct average marks?
A.A)78
B.B)82
C.C)98
D.D)91
Answer: C.C)98
Explanation: correct avg marks = 100+(10-60)/25
avg = 100-2 = 98
Answer is C
Question: The average age of a college is 21.8 years. If the average age of students of college is 24.2 years and Average age of lecturers of college is 20.6 years. Find the ratio of the number of students to that of Lecturers?
A.3:1
B.1:3
C.2:1
D.1:2
Answer: D.1:2
Explanation: 1:2
Option 'D'
Question: If v = 4p, where p is a prime number greater than 2, how many different positive even divisors does v have, including v ?
A. Two
B. Three
C. Four
D. Six
Answer: C. Four
Explanation: v=4p
since v = 4p thenevendivisors are:
4p/4p= 1
4p/2p= 2
4p/4= p
4p/2= 2p
so the answer is (C)
Question: A car is running at a speed of 108kmph. What distance will it cover in 15sec?
A.300m
B.150m
C.200m
D.450m
Answer: D.450m
Explanation: Speed = 108kmph = 108*5/18 = 30 m/s
Distance covered in 15sec = 30*15 = 450m
Answer is D
Question: Last year Manfred received 26 paychecks. Each of his first 6 paychecks was $750; each of his remaining paychecks was $40 more than each of his first 6 paychecks. To the nearest dollar, what was the average (arithmetic mean) amount of his pay checks for the year?
A. $781
B. $755
C. $765
D. $773
Answer: A. $781
Explanation: = (750*6+ 790 *20)/26
= 781
Answer is A.
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Question: Eleven animals of a circus has to be placed inside in eleven cages one in each cage. If 4 of the cage are too small for 6 of the animal then find the number of ways of caging the animal.
A.604800
B.808250
C.502450
D.784200
Answer: A.604800
Explanation: first we will fill the smaller cages-
these 4 cages will require to be filled by 11-6 = 5 animals => 5C4∗4!=120.
remaining 6 and 1 out of the above 5 can be filled in7∗6∗5∗4∗3∗2=5040
total ways = 120∗5040=604800
ANSWER:A
Question: Let n~ be defined for all positive integers n as the remainder when (n - 1)! is divided by n.
What is the value of 35~ ?
A.0
B.1
C.2
D.8
Answer: A.0
Explanation: n~ = (n-1)!
so 35~ = (35-1)! = 34!
when 34!/35 we have 16*2 inside 34!
hence 35 gets cancelled and we get remainder as 0
A
Question: The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
A.288
B.277
C.245
D.997
Answer: C.245
Explanation: Speed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30 = 25/2 => 2(130 + x) = 750 => x = 245 m
Answer: C
Question: The last digit in the product (784 x 618 x 917 x 463) is?
A.5
B.4
C.2
D.3
Answer: C.2
Explanation: Unit digit in the given product = Unit digit in (4 x 8 x 7 x 3) = (672) = 2
Answer is C
Question: A trader sells 85 meters of cloth for Rs. 8925 at the profit of Rs. 15 per metre of cloth. What is the cost price of one metre of cloth?
A.22
B.77
C.90
D.66
Answer: C.90
Explanation: SP of 1m of cloth = 8925/85 = Rs. 105
CP of 1m of cloth = SP of 1m of cloth - profit on 1m of cloth
= Rs. 105 - Rs. 15 = Rs. 90.
Answer: C
Question: If 6x^2 + x - 12 = (fx + b)(cx + d), then |f| + |b| + |c| + |d| =
For a complete solution and more practice problems, see this blog:
http://magoosh.com/gmat/2012/algebra-on ... to-factor/
A. 10
B. 12
C. 15
D. 18
Answer: B. 12
Explanation: 6x^2+x-12 = 6x^2 + 9x -8x -12
=> 3x(2x+3) -4(2x+3)
=> (2x+3)(3x-4) = (fx+b)(cx+d)
Hence f=2, b=c=3, d=-4
So, 2+3+3+ |-4| = 2+3+3+4 = 12
Answer B.
Question: A mobile battery in 1 hour charges to 20 percent.How much time (in minute) will it require more to charge to 55 percent.
A.145
B.150
C.175
D.165
Answer: D.165
Explanation: 1 hr =20 percent.Thus 15 min=5 percent .Now to charge 55 percent 165 min.Answer:D
Question: The citizens of planet nigiet are 8 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 buildings numbered 1 to 1000 in base 8. How many 3s are used in numbering these buildings?Express your answer in base 10
A.192
B.200
C.196
D.198
Answer: A.192
Explanation: Explanation :
Number 3 in 1-100 comes 20 times.
101-200-> 20 times
201-300-> 20 times
301-400->120 times
401-500-> 20 times
501-600-> 20 times
601-700-> 20 times
701-800-> 20 times
801-900-> 20 times
901-1000-> 20 times
Total times: 300 times (any no. from 1 to 9 except 0 will occur 300 times)
8 based planet.. then
3*8(2) + 0*8(1) + 0*8(0) = 192
Answer : A
Question: In a rectangular coordinate system, what is the area of a triangle whose vertices have the coordinates (2, 0), (6, 3), and (6, -3)?
A.11
B.12
C.13
D.14
Answer: B.12
Explanation: The triangle is symmetric about the x-axis.
The part above the x-axis forms a triangle with a base of 4 and a height of 3.
The area of this part is (1/2)(4)(3).
We can double this to find the area of the whole triangle.
The total area is (2)(1/2)(4)(3) = 12.
The answer is B.
Question: If Rs.7500 are borrowed at C.I at the rate of 4% per annum, then after 2 years the amount to be paid is?
A.2662
B.1286
C.1987
D.8112
Answer: D.8112
Explanation: Explanation:
A = 7500(26/25)2 = 8112
Answer:D
Question: One-third less than 35% of 180 is equal to:
A.15
B.30
C.35
D.42
Answer: D.42
Explanation: Lots of ways to tackle this.
35% of 180 =63
1/3 of 63= 21
So, 1/3less than 63 is equal to 63- 21 = 42
Answer: D
Question: A hotel pays a food distributor d dollars for 6 pounds of Maine lobster. Each pound can make v vats of lobster bisque, and each vat makes b bowls of lobster bisque. If the cost of the lobster per bowl is an integer, and if v and b are different prime integers, then which of the following is the smallest possible value of d?
A.36
B.40
C.45
D.50
Answer: A.36
Explanation: d dollars = 6 pounds = 6v vats = 6vb bowls
bowl = d/(6vb).=> d/(6vb) is an integer
=> v and b are different primes
d/(6vb)=d/(6*2*3)=d/36
integer (d min is for min values of v and b, so for 2 and 3). Hence min value => 36.
Answer: A
Question: If the numbers 1 to 98 are written on 98 pieces of paper, (one on each) and one piece is picked at random, then What is the probability that the number drawn is neither prime nor composite?
A.1/50
B.1/25
C.1/98
D.1
Answer: C.1/98
Explanation: There are 25 primes, 72 composite numbers from 1 to 98. The number
which is neither prime nor composite is 1.
Therefore, required probability = 1/98.
ANSWER:C
Question: The captain of a cricket team of 11 members is 25 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?
A.20 years
B.21 years
C.22 years
D.23 years
Answer: C.22 years
Explanation: Explanation:
Let the average age of the whole team by x years.
11x â€“ (25 + 28) = 9(x -1)
11x â€“ 9x = 44
2x = 44
x = 22.
So, average age of the team is 22 years.
Answer C
Question: A boat can move upstream at 25 kmph and downstream at 40 kmph, then the speed of the current is?
A.5
B.6
C.7
D.7.5
Answer: D.7.5
Explanation: US = 25
DS = 40
M = (40 - 25)/2 = 7.5
Answer: D
Question: Listed below are the mean temperatures measured in the Tokharian desert in a certain month. What is the median temperature R in that particular month?
Mean Temp (C°) No. of days
29 --------------------> 5
30 --------------------> 11
31 --------------------> 8
32 --------------------> 6
A.29
B.30
C.30.5
D.31
Answer: B.30
Explanation: to find out median===>they must be arranged in ascending order.
29 29 29...(5 times)30 30 30 30.....(11 times)31 31 31.....(8 times)32 32 32 .....(6 times)
total days=5+11+8+6=30
therefore 30 numbers ...median R will be==>mean of 15 and 16 th place temperature===>(30+30)/2=30
hence B
Question: In store A there are 10 pairs of pants for every 40 store B has. The price ratio between the pants in store B and the pants in store A is 3:4. If all the pants were sold in both places until the stock ran out, what is the ratio between the total amount stores B earned to the total amount store A earned?
A.3:1
B.2:3.
C.1:3.
D.3:4.
Answer: A.3:1
Explanation: 1st statement : ratio of pants
Store A : Store B
10x : 40x
X:4X
Price :
4y:3y
Total revenue
4xy : 12xy
3:1
Answer : A
Question: The cost of 10 kg of apples 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 Rs.20.50. Find the total cost of 4 kg of apples, 3 kg of rice and 5 kg of flour?
A.Rs.849.40
B.Rs.877.40
C.Rs.901.60
D.Rs.815.20
Answer: B.Rs.877.40
Explanation: Let the costs of each kg of apples and each kg of rice be Rs.a and Rs.r respectively.
10a = 24r and 6 * 20.50 = 2r
a = 12/5 r and r = 61.5
a = 147.6
Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5
= 590.4 + 184.5 + 102.5 = Rs.877.40
ANSWER:B
Question: A customer using a certain telephone calling plan pays a fee of $25 per month, and then receives a discount of 40% on the regular charge for all calls made to country Z. If calls to country Z are regularly charged at $1.60 per minute for the first 3 minutes, and $0.80 per minute for each minute thereafter, what is the maximum the customer could have saved over regular prices if he was charged for 1 hour of calls made to country Z in a certain month?
A.$8.75
B.$12
C.$13.40
D.$17.40
Answer: C.$13.40
Explanation: Discount = 40% of spending
We want to maximise discount so we will need to maximise spending. We do that by assuming that 60 calls were made of 1 min each because the first minute is the most expensive.
Max discount = (40/100)* 60 * 1.6 = (40/100)*96
This would be slightly less than 40. About 38 to 39.
Saving = (Slightly less than 40) - 25 = Slightly less than 15 charged in Z country
Answer (C)
Question: A circular rim 28 inches in a diameter rotates the same number of inches per second as a circular rim 35 inches in diameter. If the smaller rim makes E revilutions per second, how many revilutions per minute does the larger rim makes in terms of E?
A.a) 48pi/E
B.b) 75m
C.c) 48E
D.d) 24E
Answer: C.c) 48E
Explanation: distance covered by a circular wheel per sec* = 2*pi* r * n
where n is number of revolution
So we know the smaller rim makes E rev per sec so the inches coverd SR per sec = 28 * pi* E
the problem the say SR rotates the same number of inches as the LR.
The inches covered by the LR per sec = 35 * pi*Y
So equate both u will get Y = 4/5 E
Theis is in sec we need the answer in min so 48E
answer is C
Question: Let the polynomials be:
P1(x)=3x2−7x+6,
P2(x)=−5x2−4x+2,
P1(x)+P2(x)=
A.2x2 + 11x - 7
B.-2x2 - 11x + 8
C.-2x2 - 11x + 6
D.2x2 - 9x + 7
Answer: B.-2x2 - 11x + 8
Explanation: Solution:
P1(x)+P2(x)=(3x2−7x+6)+(−5x2−4x+2)=P1(x)+P2(x)=(3x2−7x+6)+(−5x2−4x+2)=
3x2−7x+6−5x2−4x+2=3x2−7x+6−5x2−4x+2=
−2x2−11x+8
Answer B
Question: At a certain supplier, a machine of type A costs $30,000 and a machine of type B costs $65,000. Each machine can be purchased by making a 20 percent down payment and repaying the remainder of the cost and the finance charges over a period of time. If the finance charges are equal to 40 percent of the remainder of the cost, how much less would 2 machines of type A cost than 1 machine of type B under this arrangement?
A.$6,600
B.$11,200
C.$12,000
D.$12,800
Answer: A.$6,600
Explanation: Total Cost of 2 Machines of Type A
= 20% of (Cost of 2 machine A) + remainder + 40 % remainder
= 20% of 30000 + (30000 - 20% of 30000) + 40% of (30000 - 20% of 30000)
= 79200
Total Cost of 1 Machine of Type B
= 20% of (Cost of 1 machine B) + remainder + 40 % remainder
= 20% of 65000 + (65000 - 20% of 65000) + 40% of (65000 - 20% of 65000)
= 85800
Diff = 85800 - 79200 = 6600
Hence, A.