divelab/Sys2Bench huggingface dataset:
Question: Find the sum The difference between the compound and S.I. on a certain sum of money for 2 years at 10% per annum is Rs.15of money?
Options:
A)1500
B)2999
C)2772
D)2987
E)2661
Rationale: P = 15(100/10)2 => P = 1500.Answer:A
Answer: A
Question: A single discount equivalent to the discount series of 20%, 10% and 8% is?
Options:
A)31.5
B)33.8
C)31.6
D)31.3
E)31.1
Rationale: 100*(80/100)*(90/100)*(92/100)
= 66.2
100 - 66.2 = 33.8
Answer:B
Answer: B
Question: B completes a work in 6 days. A alone can do it in 10 days. If both work together, the work can be completed in how many days?
Options:
A)3.33 days
B)4.33 days
C)5.33 days
D)6.33 days
E)7.33 days
Rationale: 1/5 + 1/10 = 3/10
10/3 = 3.33 days
ANSWER:A
Answer: A
Question: The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:
Options:
A)276
B)299
C)322
D)345
E)365
Rationale: Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322. Answer: Option C
Answer: C
Question: Which of the following reaches the largest value over the range -1 <= R <= 1?
Options:
A)(R+3)^100
B)(R+3)/(R^100)
C)(R+1000)
D)97^97
E)(R/2R)
Rationale: (R+3)/(R^100) can be equal any arbitrarily large number. Some of the other options are large, but they all have a definite maximum. (R+3)/(R^100) does not, so the correct answer is B.
Answer: B
Question: Two circles, one with radius 10 inches and the other with radius 4 inches, are tangent at point A. Two insects start crawling at the same time from point A: one along the larger circle at 3π inches per minute, the other along the smaller circle at 2.5π inches per minute. How much time has elapsed when the two insects meet again at point A?
Options:
A) 15 minutes
B) 30 minutes
C) 40 minutes
D) 1 hour
E) 1 hour, 20 minutes
Rationale: First we figure out the circumference of the circles so we can determine how long it takes to complete one revolution from point A to point A .
C (large) = 20pi
C (small) = 8pi
R (large) = 3pi
R (small) = 5/2pi
It takes 20/3 minutes to make one revolution around the larger circle
It takes 8/(5/2) = 16/5 minutes to make one revolution around the smaller circle
The time it takes them to intersect can be determined by finding the LCM of both times. Get the denominators of both numbers equal one another.
20/3 = 100/15
16/5 = 48/15
LCM of 48 and 100:
48 = 2*2*2*2*3
100 = 2*2*5*5
LCM = 2*2*2*2*3*5*5 = 1200
1200/15 = 80 minutes.
(E) 1 hour, 20 minutes
Answer: E
Question: In how many ways can 3 postcards can be posted in 5 postboxes?
Options:
A)125
B)286
C)296
D)199
E)167
Rationale: Ans: First card can go into any of the five boxes, Second can go into any of the five boxes, Third can go into any of the five boxes = 5×5×5=1255×5×5=125
Answer:A
Answer: A
Question: Which of the following integers is a divisor of(15!+11)?
Options:
A)15
B)14
C)11
D)7
E)2
Rationale: When we say 15!, it just means 1*2*3*4*5*...*13*14*15
So you need to find the integer which is a divisor of
(1*2*3*4*5*...*13*14*15 + 11) = 11 (1*2*3*4*5*...10*12*13*14*15 + 1) We obtain this by taking 11 common.
Hence 11 is definitely a divisor of (15! + 11).
C
Answer: C
Question: Pipes A and B can fill a tank in 6 and 4 hours. Pipe C can empty it in 12 hours. If all pipes are opened together, then the tank will be filled in?
Options:
A)3hr
B)2hr
C)1hr
D)9/7hr
E)5/3hr
Rationale: Net part filled in 1 hour = 1/6 + 1/4 - 1/12 = 1/3
The tank will be full in 3 hr
Answer is A
Answer: A
Question: A number with an interesting property:
When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.
It's not a small number, but it's not really big, either.
When I looked for a smaller number with this property I couldn't find one.
Can you find it?
Options:
A)2588
B)2474
C)2254
D)2519
E)8725
Rationale: D
2519
The number has to end in 9.
Looked brute force for small numbers.
59 and 119 were promising, but no cigar.
Then looked for agreement among
39 + multiples of 40,
69 + multiples of 70 and
89 + multiples of 90
Smallest one was 2519.
Answer: D
Question: In a certain warehouse, 90 percent of the packages weigh less than 75 pounds, and a total of 48 packages weigh less than 25 pounds. If 80 percent of the packages weigh at least 25 pounds, how many of the packages weigh at least 25 pounds but less than 75 pounds?
Options:
A)8
B)64
C)168
D)102
E)144
Rationale: If 80% of the packages weigh at least 25 pounds
This means that 20% of the packages weigh LESS THAN 25 pounds
Let T = TOTAL number of packages
So, 20% of T = # of packages that weigh LESS THAN 25 pounds
48 packages weigh LESS THAN 25 pounds
GREAT. So, 20% of T = 48
Rewrite to get: 0.2T = 48
Solve:T = 240
90% of the packages weigh less than 75 pounds
So, 90% ofT= number of packages that weigh less than 75 pounds
90% of240=216, so216packages weigh less than 75 pounds
OF THOSE216packages that weigh less than 75 pounds,48packages weigh less than 25 pounds.
So, the number of packages that weight BETWEEN 25 and 75 pounds =216-48= 168 = C
Answer: C
Question: Paul and Jack are 2 mechanics who work 8 hours per day changing oil in cars at The Oil Spot. If Paul can change oil in 2 cars per hour, and Jack can change oil in 3 cars per hour. What is the least number of cars per work day the 2 men can finish?
Options:
A)50
B)40
C)35
D)10
E)90
Rationale: The proportion equation can be used. Let x = number cars and C = least number of cars per day the 2 men can finish (C = x/(8 hours), P = cars per hour Paul can change, and J = cars per hour Jack can change. Note: 8 hours = a work day
P + J = C
(2 cars/hour) + (3 cars/hour) = x/(8 hours)
Proportion: 2/1 + 3/1/ = x/300
x = 8(2 + 3) = 40. At least 40 cars per work day.
Answer is B
Answer: B
Question: If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the
average marks of all the students is
Options:
A)54.48
B)54.68
C)54.6
D)54.58
E)None of these
Rationale: Explanation:
(55×50)+(60×55)+(45×60)55+60+45
8750160=54.68
Answer: Option B
Answer: B
Question: The total of 324 of 20 paise and 25 paise make a sum of Rs. 70. The no of 20 paise coins is
Options:
A)238
B)277
C)220
D)200
E)288
Rationale: Explanation:
Let the number of 20 paise coins be x. Then the no of 25 paise coins = (324 - x).
0.20 * (x) + 0.25 (324 - x) = 70 => x = 220..
Answer: C) 220
Answer: C
Question: In the interior of a forest, a certain number of apes equal to the square of one-eighth of the total number are playing and having great fun. The remaining sixteen apes are on a hill and the echo of their shrieks by the adjoining hills frightens them. They came and join the apes in the forest and play with enthusiasm. What is the total number of apes?
Options:
A)48
B)16
C)64
D)80
E)32
Rationale: Let total number be x
No in the interior = (x/8)^2
No outside = 16
So : x - (x/8)^2=16
x^2-64x+32^2=0
(x-32)
So either x=32
E
Answer: E
Question: Find compound interest on Rs. 8000 at 15% per annum for 2 years 4 months, compounded annually.
Options:
A)Rs 309
B)Rs 319
C)Rs 3109
D)Rs 109
E)Rs 209
Rationale: Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs'. [8000 X (1+¬(15/100))2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109.	.
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.
ANSWER C 3109
Answer: C
Question: A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?
Options:
A)22
B)77
C)10
D)27
E)18
Rationale: Number of liters of water in150 liters of the mixture = 20% of 150 = 20/100 * 150 = 30 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).
(30 + P) = 25/100 * (150 + P)
120 + 4P = 150 + P => P = 10 liters.
Answer: C
Answer: C
Question: Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 300 as interest at the end of the loan period, what was the rate of interest?
Options:
A)3.6
B)6
C)5
D)24
E)None of these
Rationale: Let rate = R% and time = R years.
Then, (1200 x R x R)/100 = 300
12R2 = 300
R2 = 25
R = 5.
Answer: Option C
Answer: C
Question: The integer x is divisible by both 7 and 21. Which of the following must be an integer?
Options:
A)x/25
B)x/34
C)x/52
D)x/45
E)x/21
Rationale: Prime factorization of 7 = 1*7
Prime factorization of 21= 3*7
LCM of 7 and 21 = 7 * 3 = 21
Therefore x/21 must be an integer
Answer E
Answer: E
Question: Tanks A and E are each in the shape of a right circular cylinder. The interior of tank A has a height of 10 meters and a circumference of 8 meters, and the interior of tank E has a height of 8 meters and a circumference of 10 meters. The capacity of tank A is what percent of the capacity of tank E?
Options:
A)75%
B)80%
C)100%
D)120%
E)125%
Rationale: B.
For A, r=8/2pi. Its capacity = (4pi)^2 * 10= 160pi
For E, r=10/pi. Its capacity = (5pi)^2 *8 = 200pi
A/E = 160pi/200pi = 0.8=B.
Answer: B
Question: 15 years hence, Rohan will be just four times as old as he was 15 years ago.How old is Rohan at present?
Options:
A)20
B)25
C)30
D)35
E)40
Rationale: Let the present age of Rohan be x years
Then, given: x + 15 = 4 (x - 15)
x = 25
ANSWER:B
Answer: B
Question: The average age of a group of 10 students was 20. The average age increased by 2 years when two new students joined the group. What is the average age of the two new students who joined the group?
Options:
A)22 years
B)30 years
C)44 years
D)32 years
E)None of these
Rationale: Answer
The average age of a group of 10 students is 20.
Therefore, the sum of the ages of all 10 of them = 10 * 20 = 200
When two new students join the group, the average age increases by 2. New average = 22.
Now, there are 12 students.
Therefore, the sum of the ages of all 12 of them = 12 * 22 = 264
Therefore, the sum of the ages of the two new students who joined = 264 - 200 = 64
And the average age of each of the two new students = 64/2 = 32 years.
Answer D
Answer: D
Question: If a car had traveled 30 kmh faster than it actually did, the trip would have lasted 30 minutes less. If the car went exactly 90 km, at what speed did it travel?
Options:
A)35 kmh
B)45 kmh
C)60 kmh
D)75 kmh
E)90 kmh
Rationale: Time = Distance/Speed
Difference in time = 1/2 hrs
90/x - 90/(x + 30) = 1/2
Substitute the value of x from the options. --> x = 60 --> 90/60 - 90/90 = 3/2 - 1 = 1/2
Answer: C
Answer: C
Question: Find the smallest number which when divided by 11 and 13 leaves respective remainders of 4 and 6.
Options:
A)187
B)197
C)207
D)136
E)227
Rationale: Let 'N' is the smallest number which divided by 11 and 13 leaves respective remainders of 4 and 6.
Required number = (LCM of 11 and 13) - (common difference of divisors and remainders) = (143) - (7) = 136.
ANSWER:D
Answer: D
Question: The area of a square is 4096 sq cm. Find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square?
Options:
A)5:16
B)5:12
C)5:19
D)5:18
E)5:14
Rationale: Let the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm.
a2 = 4096 = 212
a = (212)1/2 = 26 = 64
L = 2a and b = a - 24
b : l = a - 24 : 2a = 40 : 128 =
5:16
Answer:E
Answer: E
Question: The number of degrees that the hour hand of a clock moves through between noon and 2.30 in the afternoon of the same day is?
Options:
A)38
B)27
C)75
D)28
E)21
Rationale: Explanation:
The hour hand moves from pointing to 12 to pointing to half way between 2 and 3. The angle covered between each hour marking on the clock is 360/12 = 30. Since the hand has covered 2.5 of these divisions the angle moved through is 75.
Answer: C
Answer: C
Question: 3034 - (1002 / 200.4) = ?
Options:
A)2984
B)3029
C)2982
D)2981
E)None of these
Rationale: 3034-5 = 3029
Answer : B
Answer: B
Question: After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, running north along the same path. After running 3.25 miles south, he decides to run for only 50 minutes more. If Bob runs at a constant rate of 8 minutes per mile, how many miles farther south can he run and still be able to return to the parking lot in 50 minutes?
Options:
A) 1.5
B) 2.25
C) 3.0
D) 3.25
E) 4.75
Rationale: For the first 3.25 miles, Bob takes 3.25x8=26minutes
Say he runs 'x' miles further south. So the total distance he covers is (3.25+2x+3.25) in (50+26) minutes
So we have 6.5+2x=76/8 => x=1.5
Answer A
Answer: A
Question: There is 60% increase in an amount in 6 years at S.I. What will be the C.I. of Rs. 12,000 after 3 years at the same rate?
Options:
A)Rs.3389
B)Rs.3923
C)Rs.3972
D)Rs.3928
E)Rs.3929
Rationale: Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.
R = (100 * 60)/(100 * 6) = 10% p.a.
Now, P = Rs. 12000, T = 3 years and R = 10% p.a.
C.I. = [12000 * {(1 + 10/100)3 - 1}]
= 12000 * 331/1000 =Rs.3972
Answer:C
Answer: C
Question: A garrison of 2000 men has provisions for 54 days. At the end of 15 days, a reinforcement arrives, and it is now found that the provisions will last only for 20 days more. What is the reinforcement?
Options:
A)1898
B)9801
C)1987
D)1900
E)8373
Rationale: 2000 ---- 54
2000 ---- 39
x ----- 20
x*20 = 2000*39
x = 3900
2000
-------
1900
Answer:D
Answer: D
Question: Find the length of the wire required to go 15 times round a square field containing 69696 m2.
Options:
A)15840 m
B)16840 m
C)15820 m
D)15640 m
E)16640 m
Rationale: a2 = 69696 => a = 264
4a = 1056
1056 * 15 = 15840
ANSWER:A
Answer: A
Question: Ram get 450 marks in his exam which is 90% of total marks. What is the total marks?
Options:
A)475
B)600
C)550
D)500
E)525
Rationale: X * (90/100) = 450
X = 5 * 100
X = 500
Answer: D
Answer: D
Question: Jack, Jill, and Sandy each have one try to make a basket from half court. If their individual probabilities of making the basket are 1/6, 1/7, and 1/8 respectively, what is the probability that all three will miss?
Options:
A)3/8
B)5/8
C)7/16
D)9/16
E)23/32
Rationale: The probability that all three will miss is 5/6*6/7*7/8=5/8.
The answer is B.
Answer: B
Question: M and N invested in a shop. The profits were divided in the ratio of 4 : 5 respectively. If M invested Rs. 40,000, the amount invested by N is:
Options:
A)35000
B)40000
C)50000
D)60000
E)None
Rationale: Suppose N invested Rs. x. Then, 40000/x = 4/5 or y = [40000 x 5/ 4] = 50000.
Answer C
Answer: C
Question: 7.51 8.22 7.86 8.36
8.09 7.83 8.30 8.01
7.73 8.25 7.96 8.53
A vending machine is designed to dispense 8 ounces of coffee into a cup. After a test that recorded the number of ounces of coffee in each of 1000 cups dispensed by the vending machine, the 12 listed amounts, in ounces, were selected from the data above. If the 1000 recorded amounts have a mean of 7.5 ounces and a standard deviation of 0.3 ounces, how many of the 12 listed amounts are within 1.7 standard deviation of the mean?
Options:
A)Four
B)Six
C)Nine
D)Ten
E)Eleven
Rationale: Mean = 7.5
Standard Deviation = 0.3
within 1.7 standard deviation of the mean
i.e. 1.7 standard deviation ABOVE the mean = 7.5+1.7*0.3 = 8.01
and 1.7 standard deviation BELOW the mean = 7.5-1.7*0.3 = 6.99
hence, 6 values fall within 1.7 SD from Mean
Answer: Option B
Answer: B
Question: If integer k is equal to the sum of all even multiples of 25 between 200 and 600, what is the greatest prime factor of k?
Options:
A)5
B)7
C)11
D)13
E)17
Rationale: if we break down what the stem is asking what is the sum of all mult of 50 between 200 and 600.
using arithmetic progression to find n : 600 = 200 + (n - 1) 50
400+ 50 = 50n
450 = 50n => n = 9
the sum would be: 9* mean
mean = [600 + 200] / 2 = 400
9*400 =3600
D
Answer: D
Question: If José reads at a constant rate of 2 pages every 6 minutes, how many seconds will it take him to read N pages?
Options:
A)180
B)2N
C)5/2*N
D)24N
E)150
Rationale: Jose would read 1 page in 6/2 min
Jose would read N page in (6/2)*N min i.e. (6/2)*N*60 Seconds = 180N Seconds.
Option A is the correct Answer.
Answer: A
Question: An amount of money is to be divided between P, Q and R in the ratio of 3:7:12. If the difference between the shares of P and Q is Rs.2400, what will be the difference between Q and R's share?
Options:
A)3000
B)2887
C)2666
D)2578
E)2771
Rationale: 4 --- 2000
5 --- ? => 3000
Answer: A
Answer: A
Question: The first three terms of a proportion are 6, 12 and 24. The fourth term is?
Options:
A)22
B)28
C)27
D)36
E)48
Rationale: (12*24)/6 = 48
Answer:E
Answer: E
Question: How much time will take for an amount of Rs. 320 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
Options:
A)8 years
B)4 years
C)3 years
D)5 years 7 months
E)5 years
Rationale: Time = (100 * 81) / (320 * 4.5)
= 5 years 7 months
Answer: D
Answer: D
Question: If 2^5, 3^3, and 11^2 are all factors of the product of 936 and w where w is a positive integer, what is the smallest possible value of w?
Options:
A)132
B)39
C)42
D)65
E)156
Rationale: Here 156 has three two's
two three's
and one 11
rest of them must be in w
so w= 11*3*4 = 132
Smash A
Answer: A
Question: Amy has to go to three different stores this week. How many ways can she go to one or more stores today?
Options:
A)19
B)26
C)24
D)9
E)7
Rationale: Amy has to visit at least one of three stores.
= > 2^3 - 1 = 7
E
Answer: E
Question: A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets $500 more than D, what is A's share?
Options:
A)$1200
B)$1600
C)$2000
D)$2500
E)$3000
Rationale: Let the shares of A, B, C and D be 5x, 2x, 4x and 3x respectively.
Then, 4x - 3x = 500
x = $500
A's share = 5x = 5*$500 = $2500
The answer is D.
Answer: D
Question: For each 6-month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6-month period. If the odds of a light bulb burning out during the first 6-month period following its purchase are 2/3, what are the odds of it burning out during the period from 6 months to 1 year following its purchase?
Options:
A)5/27
B)2/9
C)1/3
D)4/9
E)2/7
Rationale: P(of not burning out in a six mnth period)=1/2 of P(of not burning out in prev 6 mnth period)
P(of burning out in 1st 6 mnth)= 2/3
--->P( of not burning out in 1st 6 mnth)=1-2/3=1/3
---->P(of not burning out in a six mnth period)=1/2 *1/3=1/6--->P(of burning out in a six mnth period)=1-1/3=2/3
now
P( of burning out in 2nd six mnth period)=P( of not burning out in 1st six mnth)*P(of burning out in a six mnth)
=2/3 * 1/6=2/7
Ans E
Answer: E
Question: In how many ways can a cricket eleven be chosen out of a batch of 15 players.
Options:
A)1365
B)2154
C)1245
D)4512
E)7841
Rationale: Required number of ways
= 15C 11 = 15C (15-11)
= 15C4
15C4 = 15 * 14 * 13 * 12 / 4 * 3 * 2 *1
= 1365
Ans: A
Answer: A
Question: For how many values n is 2^n = n^2 ?
Options:
A)None
B)One
C)Two
D)Three
E)More than Three
Rationale: draw a graph with x^2 and 2^x, you'll see 3 intercepts between the 2 curves, but one is in the 2nd quadrant,Thus 3 are valid, and the answer is D
Answer: D
Question: A money lender finds that due to a fall in the annual rate of interest from 8% to 7 1/2 % his yearly income diminishes by Rs.61.50. his capital is
Options:
A)223.78
B)37.78
C)24.6
D)278.88
E)123.0
Rationale: Explanation:
Capital = Rs.x , then
1/2 x=61.5
x = 123.00
Answer: E) Rs.123.00
Answer: E
Question: A train covers a distance of 6 km in 10 min. If it takes 12 sec to pass a telegraph post, then the length of the train is?
Options:
A)120m
B)180m
C)240m
D)220m
E)280m
Rationale: Speed = (6/10 * 60) km/hr = (36 * 5/18) m/sec = 10 m/sec.
Length of the train = 10 * 12 = 120 m.
ANSWER:A
Answer: A
Question: The average (arithmetic mean) of four numbers is 5x + 9. If one of the numbers is 2x, what is the average of the other three numbers?
Options:
A)x + 1
B)3x + 3
C)5x + 1
D)5x + 4
E)6x + 12
Rationale: The sum of all four numbers is 4 * (5x + 9) = 20x + 36.
If one of the numbers is 2x, the sum of the other 3 numbers is 18x+36.
The average of the remaining three numbers is (18x+36)/3 = 6x+12.
The answer is E.
Answer: E
Question: One fourth of a solution that was 15% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight?
Options:
A)34%
B)24%
C)22%
D)19%
E)8.5%
Rationale: Say the second solution (which was 1/4 th of total) was x% sugar, then 3/4*0.15+1/4*x=1*0.16 --> x=0.19. Alternately you can consider total solution to be 100 liters and in this case you'll have: 75*0.15+25*x=100*0.16 --> x=0.19.
Answer: D.
Answer: D
Question: Find compound interest on Rs. 7500 at 4% per annum for 2 years, compounded annually.
Options:
A)Rs 712
B)Rs 812
C)Rs 912
D)Rs 512
E)Rs 612
Rationale: Amount = Rs [7500*(1+(4/100)2] = Rs(7500 * (26/25) * (26/25)) = Rs. 8112.
therefore, C.I. = Rs. (8112 - 7500) = Rs. 612.
ANSWER E 612
Answer: E
Question: An investor bought a 1-year $6,000 certificate of deposit that paid interest at an annual rate of 10 percent compounded semiannually. What was the total amount of interest paid on this certificate at maturity?
Options:
A)$325
B)$600
C)$615
D)$1000
E)$1230
Rationale: 1.05*1.05*6000 = $6615
The answer is C.
Answer: C
Question: A man cheats while buying as well as while selling. While buying he takes 10% more than what he pays for and while selling he gives 20% less than what he claims to. Find the profit percent, if he sells at 9.09% below the cost price of the claimed weight.
Options:
A)19.81%
B)20%
C)37.5%
D)25%
E)37.5%
Rationale: (1 + m1%)(1 + m2%)(1 - d%) = (1 + p%)
11/10 * 5/4 * 10/11 = (1 + p%)
profit % = 25%
ANSWER:D
Answer: D
Question: A bag contains 7 green and 8 white balls. If two balls are drawn simultaneously, the probability that both are of the same colour is?
Options:
A)7/15
B)7/16
C)7/19
D)7/13
E)7/11
Rationale: Drawing two balls of same color from seven green balls can be done in 7C2 ways.
Similarly from eight white balls two can be drawn in ⁸8C2 ways.
= 7/15
Answer: A
Answer: A
Question: During 2005, a company produced an average of 2,500 products per month. How many products will the company need to produce from 2006 through 2008 in order to increase its monthly average for the period from 2005 through 2008 by 300% over its 2005 average?
Options:
A)450,000
B)475,000
C)500,000
D)525,000
E)550,000
Rationale: Company produced 12*2500 = 30,000 products in 2005. If company produces X products from 2006 to 2008, then total amount of product produced in 4 years (2005 through 2008) is X+30,000. The gives the average of (X+30,000)/4.
This average needs to be 200% higher than that in 2005. In math terms, 30,000+300%(30,000) = 120,000. So:
(X+30,000)/4 = 120,000
X+30,000 = 480,000
X = 450,000
The answer is A.
Answer: A
Question: Yearly subscription to professional magazines cost a company $940.00. To Make a 50% cut in the magazine budget, how much less must be spent?
Options:
A)654
B)655
C)656
D)657
E)470
Rationale: total cost 940
940*50/100=470
so the cut in amount is 470
the less amount to be spend is 940-470=470
ANSWER:E
Answer: E
Question: The weight of four dogs is determined to be 42 pounds, 38 pounds, 35 pounds and 41 pounds respectively. The weight of a fifth dog is determined to be y pounds. If the average (arithmetic mean) weight of the first four dogs is the same as that of all five dogs what is the value of y?
Options:
A)31
B)33
C)35
D)37
E)39
Rationale: Total weight of the 4 dogs = (42+38+35+41) = 156
Avg = 156/4 = 39
Total weight of 5 dogs = 156+y or 4(39)+y
Average of 5 dogs as per Question = 39
Equation:
4(39) + y = 5 (39), or y = 39.
Choose E
Answer: E
Question: By selling 150 mangoes, a fruit-seller gains the selling price of 30 mangoes. Find the gain percent?
Options:
A)22
B)25
C)77
D)88
E)99
Rationale: SP = CP + g
150 SP = 150 CP + 30 SP
120 SP = 150 CP
120 --- 30 CP
100 --- ? => 25%
Answer:B
Answer: B