divelab/Sys2Bench huggingface dataset:
Question: In an effort to plan out expenses, the Roberts family is representing its annual budget as a circle graph. Each sector of the graph is proportional to the amount of the budget it represents. If “clothes and shoes” takes up 54° of the chart, how much of the Roberts’s $30,000 annual budget is dedicated to clothes and shoes?
Options:
A)$1,500
B)$3,000
C)$4,500
D)$5,000
E)$5,400
Rationale: Given for clothes and shoes” takes up 54° of the circular chart.
Then clothes and shoes = 54/360 * 30000.
=> 4500.
IMO option C is correct answer...
Answer: C
Question: The owner of a furniture shop charges his customer 24% more than the cost price. If a customer paid Rs. 8215 for a computer table, then what was the cost price of the computer table?
Options:
A)7297
B)6625
C)2871
D)6725
E)2981
Rationale: Explanation:
CP = SP * (100/(100 + profit%))
= 8215(100/124) = Rs. 6625.
Answer: B
Answer: B
Question: A rectangular table seats 2 people on each of two sides, with every person directly facing another person across the table. If four people choose their seats at random, what is probability that any two of them directly face other?
Options:
A)1/3
B)1/8
C)1/7
D)15/56
E)4/7
Rationale: A B
------------
| TABLE |
------------
C D
A to D are 4 people.
Prob to select any 1 person = 1
Prob to select the person opposite to the chosen person = 1/3
For ex. If we select A as the person than prob of choosing C is 1/3.
Hence, answer will be A.
Answer: A
Question: The average of first 25 prime numbers is?
Options:
A)9.9
B)8.8
C)5.6
D)11.9
E)11.11
Rationale: Explanation:
Average = (2+3+5+7+11+13+17+19+23/9 = 100/9= 11.11(Approx)
Answer is E
Answer: E
Question: A train A leaves Mysore at 6 a.m. and reaches Bangalore at 10 a.m. Another train leaves B Bangalore at 8 a.m. and reaches Mysore at 11.30 a.m. At what time do the two trains cross each other ?
Options:
A)8.30 a.m.
B)8.45 a.m.
C)8.56 a.m.
D)9 a.m.
E)None of these
Rationale: Explanation :
Solution: Let the distance between mysore and bangalore be x km and let the trains meet y hours after 8 a.m.
Clearly, A covers x km in 4 hours and B covers x km in 4.5 hrs = (7/2) hrs
.'. Speed of A = x/4 kmph, speed of B = 2x/7 kmph.
Distance covered by A in (y+2) hrs + distance covered by B in y hrs = x.
.'. x/4(y+2) + 2x/7 * y = x
=> (y+2)/4 + 2y/7 = 1
=> y = 14/15 hrs =14*60/15 = 56 min.
Hence, the trains meet at 8.56 a.m.
Answer : C
Answer: C
Question: The smallest number when increased by "2 " is exactly divisible by 12, 30, 48, 74 and 100 is:
Options:
A)14439
B)44398
C)44400
D)44402
E)15005
Rationale: LCM = 44400
44400 - 2 = 44398
ANSWER:B
Answer: B
Question: Four different books (A, B, C and D) are to be arranged on a shelf. Books A and B are to be arranged first and second starting from the right of the shelf. The number of different orders in which books C and D may be arranged is
Options:
A)16!
B)12!
C)8!
D)11!
E)2!
Rationale: Solution
Since books A and B are arranged first and second, only books C and D will change order. Therefore it an arrangement problem involving 2 items and the number of different order is given by
2!
Answer E
Answer: E
Question: The cost of a one-family home was $ 120,000 in 1980. In 1988, the price had increased to $ 174,000. What was the percent increase in the cost of the home?
Options:
A)60%
B)50%
C)55%
D)40%
E)45%
Rationale: Increase = 174000-120000 = 54000
% increase = 54000*100/120000 = 45%
Answer: option E
Answer: E
Question: Find the greatest number of four digits which is perfectly divisible by 3 and when divided by 5, 7 and 9 leaves a remainder 3 in each case.
Options:
A)1128
B)1659
C)2687
D)1667
E)1891
Rationale: LCM of 3, 5, 7, 9 = 315. Greatest number of four digits which is divisible by 315 is 9765. The required number = 9765 + 3 = 9768. Ask doubt with Question Id: 1658 13) Find greatest number of four digits which when increased by 3568 is exactly divisible by 6, 8, 12, 20. a) 9992 b) 9785 c) 9840 d) None Explanation: Greatest number of 4 digits = 9999. 9999 + 3568 = 13567. LCM of 6, 8, 12 and 20 is 120.
∴ Required number of 4 digits = 9999 – 7 = 9992 i.e. 9992 is the number to which if 3568 is added, then it is exactly divisible by 6, 8, 12, 20. Ask doubt with Question Id: 1659
Answer:B
Answer: B
Question: Three friends are buying a gift for a friend. Declan contributes 9 dollars more than 1/4 the cost of the gift, Ed contributes 1 dollar less than 1/3 the cost of the gift, and Frank contributes the remaining 22 dollars. What is the cost of the gift?
Options:
A)48
B)54
C)60
D)66
E)72
Rationale: Declan = D, Ed = E, Frank = F
t = total
D + E + F = t
(t/4 + 9) + (t/3 - 1) + 22 = t
t = 30 + (7t/12)
12t = 30(12) + 7t
5t = 5*6(12)
t = 72
The correct answer is E.
Answer: E
Question: In the exam the average was found to be p mark. After deducting typographical error the average marks of 85 student reduced from 75 to 50 . Due to this the average came down by 25 marks. What was the number of students who took the exam?
Options:
A)100
B)85
C)110
D)130
E)160
Rationale: Explanation :
Number of student = (75-50) * 85/25 = 85
Answer : B
Answer: B
Question: A coin has two sides. One side has the number 1 on it and the other side has the number 2 on it. If the coin is flipped three times what is the probability that the sum of the numbers on the landing side of the coin will be greater than 3?
Options:
A)3/8
B)1/16
C)7/8
D)1/2
E)1/4
Rationale: One approach to solve the problem is to list the different possibilities for a toss of coin three times. Because there are two outcomes and the coin is tossed three times, the table will have 2*2*2 or 8 rows.
Next add the resulting rows together to find the sum (the fourth column in the table below).
Toss 1 | Toss 2 | Toss 3 | Sum
1 ---------- 1 -------- 1 ------ 3
1 ---------- 1 -------- 2 ------ 4
1 ---------- 2 -------- 1 ------ 4
1 ---------- 2 -------- 2 ------ 5
2 ---------- 1 -------- 1 ------ 4
2 ---------- 1 -------- 2 ------ 5
2 ---------- 2 -------- 1 ------ 5
2 ---------- 2 -------- 2 ------ 6
From the table we see that there are 7 situations where the sum of the tosses will be greater than 3. And there are 8 possible combinations resulting in a probability of
7/8 or a probability of 7/8.
SO the correct answer is C.
Answer: C
Question: Three partners A, B, C in a business invested money such that 4(Aâ€™s capital) = 2(Bâ€™s capital) = 12(Câ€™s capital) then, The ratio of their capitals is
Options:
A)63 : 45 : 34
B)63 : 54 : 34
C)36 : 54 : 28
D)3 : 6 : 1
E)None of these
Rationale: Explanation:
Let 4(Aâ€™s capital) = 2(Bâ€™s capital) = 12(Câ€™s capital) = Rs.x
Then, Aâ€™s capital = Rs x/4, Bâ€™s capital = Rs. x/2 and Câ€™s capital = Rs. x/12
A : B : C = x/4 : x/2 : x/12
3 : 6 : 1
Answer: Option D
Answer: D
Question: A man saves 20% of his monthly salary. If an account of dearness of things he is to increase his monthly expenses by 20%, he is only able to save Rs. 230 per month. What is his monthly salary?
Options:
A)5000
B)5750
C)2878
D)2990
E)2771
Rationale: Income = Rs. 100
Expenditure = Rs. 80
Savings = Rs. 20
Present Expenditure 80*(20/100) = Rs. 96
Present Savings = 100 â€“ 96 =Rs. 4
100 ------ 4
? --------- 230 => 5750
Answer: B
Answer: B
Question: Find the average of the first 9 natural numbers?
Options:
A)8
B)5
C)10
D)11
E)12
Rationale: Average of the first 'n' natural numbers = (n + 1)/2
We get (9 + 1)/2 = 5
ANSWER:B
Answer: B
Question: If a and b are positive integers, which of the following cannot be an integer?
Options:
A)(2b)/(4a+1)
B)(2b)/(3a+1)
C)(2a+1)/(4b)
D)(2a+1)/(4b+1)
E)(2a·b)/(4b+2)
Rationale: I'd first look for an option which gives odd/even for any values of a and b, because odd/even is not an integer.
Option C fits: 2a+1/4b=even+odd/even=odd/even≠integer
To discard other A consider a=5 and b=1.
To discard other B consider a=2 and b=1.
To discard other D consider a=2 and b=1.
To discard other E consider a=3 and b=1.
Answer: C.
Answer: C
Question: Matt is touring a nation in which coins are issued in two amounts, 2¢ and 5¢, which are made of iron and copper, respectively. If Matt has ten iron coins and ten copper coins, how many different sums from 1¢ to 70¢ can he make with a combination of his coins?
Options:
A)66
B)67
C)68
D)69
E)70
Rationale: The total sum is 10*2 + 10*5 = 70¢. If you can make each sum from 1 to 70 (1¢, 2¢, 3¢, ..., 70¢), then the answer would be 70 (maximum possible).
Now, with 2¢ and 5¢ we cannot make 1¢ and 3¢. We also cannot make 69¢ and 67¢ (since total sum is 70¢ we cannot remove 1¢ or 3¢ to get 69¢ or 67¢).
So, out of 70 sums 4 are for sure not possible, so the answer must be 70 - 4 = 66 sums or less. Only A fits.
Answer: A.
Answer: A
Question: What is the greatest prime factor of 12!11! + 11!10!?
Options:
A)7
B)11
C)13
D)17
E)19
Rationale: Soln:
12!11!+11!10!= 12*11*10!*11*10!+11*10!*10!=10!*10!*11(11*12+1)=10!*10!*11*133=10!*10!*11*19*7
greatest prime factor in 10! is 7 which is less than 19.
So Ans is E
Answer: E
Question: An empty wooden vessel weighs 12% of its total weight when filled with paint. If the weight of a partially filled vessel is one half that of a completely filled vessel, what fraction of the vessel is filled.
Options:
A)3/5
B)5/11
C)1/24
D)4/9
E)2/5
Rationale: An empty wooden vessel weighs 12% of its total weight when filled with paint:
Vessel = 0.12(Vessel + Paint);
12V = V + P (so the weight of completely filled vessel is 12V)
P = 11V (so the weight of the paint when the vessels is completely filled is 11V).
The weight of a partially filled vessel is one half that of a completely filled vessel:
V + P' = 1/2*12V;
P' = 5V (so the weight of the paint when the vessels is partially filled is 5V).
What fraction of the vessel is filled?
So, we need to find the ratio of the weight of the paint when the vessel iscompletely filledto the weight of the paint when the vessel ispartially filled:
P'/P = 5V/11V = 5/11.
Answer: B.
Answer: B
Question: A sum fetched a total simple interest of Rs. 750 at the rate of 6 p.c.p.a. in 5 years. What is the sum?
Options:
A)2600
B)2500
C)2900
D)2800
E)2700
Rationale: Sol.
Principal = Rs. [100 * 750 / 6 * 5] = Rs. [75000 / 30] = Rs. 2500.
Answer B
Answer: B
Question: A salesman sold twice as much pears in the afternoon than in the morning. If he sold $420 kilograms of pears that day, how many kilograms did he sell in the afternoon?
Options:
A)120
B)180
C)240
D)280
E)320
Rationale: 3x=420
x=140
Therefore, the salesman sold 140kg in the morning and 2⋅140=280 kg in the afternoon. So answer is D.
Answer: D
Question: If x = 5/8 and y = -1/2, what is the value of the expression -2x – y^2?
Options:
A)-3/2
B)-1
C)1
D)3/2
E)7/4
Rationale: -2X- y^2
-2x5/8 - (1/2)^2
-5/4-1/4
-3/2
A correct
Answer: A
Question: A train left station X at A hour B minutes. It reached station Y at B hour C minutes on the same day, after travelling C hours A minutes (clock shows time from 0 hours to 24 hours). Number of possible value(s) of A is:
Options:
A)0
B)1
C)2
D)3
E)4
Rationale: Let's try to get hours and minutes values separately..
Starting time + time taken = Reaching time..
A hours + C hours = B hours -------- (i)
A, C and B cannot have values greater than or equal to 24
B minutes + A minutes = C minutes -------- (ii)
Looking at the two equations, we get that no value of A satisfies both the equations.
Hence, option A
Answer: A
Question: From January 1, 1991, to January 1, 1993, the number of people enrolled in health maintenance organizations increased by 20 percent. The enrollment on January 1, 1993, was 50 million. How many million people, to the nearest million, were enrolled in health maintenance organizations on January 1, 1991 ?
Options:
A) 38
B) 39
C) 40
D) 41
E) 42
Rationale: 1.20x=50 --> 6/5*x=50 --> x=50*5/6=250/6=~42.
Answer: E.
Answer: E
Question: A = {2, 3, 5, 7, 11}
B = {2, 4, 6, 17}
Two integers will be randomly selected from sets A and B, one integer from set A and one from set B, and then multiplied together. How many different products can be obtained?
Options:
A)15
B)16
C)25
D)20
E)36
Rationale: By looking at Set A, we can see that it's all primes. Thus, we should immediately break down the elements in Set B to their prime factors. That gives :
Set A = {2,3,5,7,11}
Set B = {2, 2x2, 3x2, 13}
Apart from 2x3x2 (taking 2 from set A) which is the same as 3x2x2(taking 3 from set A); there is nothing which can be repeated. Thus, the total unique product = 20-1 = 25.
C
Answer: C
Question: In the first hour of a two-hour trip, a car traveled 3d kilometers, and in the second hour of the trip, the car traveled one-half that distance. What is the average rate at which the car traveled during the trip, in kilometers per hour?
Options:
A) d
B) 1/3*d
C) 1/2*d
D) 5/4*d
E) 3/2*d
Rationale: Solution:
Lets look at the information given in the question...
Total time travelled = 2 hrs
Distance travelled first hour = 3d
Distance travelled second hour = 3d/2
The question is asking for the avg speed at which the car travels for 2 hrs
We know that avg speed = total distance/total time
Lets place the information given in the question to the formula above = (d+3d/2)/2
= 5/4*d
We find that option D is the answer as it gives the same expression as found in our calculation above. Answer option D.
Answer: D
Question: Walking across campus, a student interviewed a group of students. 25% of the students took a finance class last semester, 53% took a marketing class last semester, and 40% took neither a finance nor a marketing class last semester. What percent of the students in the group took both a finance and a marketing class?
Options:
A)60%
B)50%
C)25%
D)18%
E)10%
Rationale: {Total} = {finance} + {marketing} - {both} + {neither}
100 = 25 + 53 - {both} + 40
{both} = 18%.
Answer: D.
Answer: D
Question: Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 sec. What is the length of the fast train?
Options:
A)27 7/6
B)27 7/5
C)27 7/3
D)27 7/9
E)27 7/1
Rationale: Relative speed = (40 - 20) = 20 km/hr.
= 20 * 5/ 18 = 50/9 m/sec.
Length of faster train = 50/9 * 5 = 250/9 = 27 7/9 m.
Answer: D
Answer: D
Question: What is the least positive integer X for which (X^2) + 164 is divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11?
Options:
A)27556
B)800
C)796
D)87
E)166
Rationale: 2 = 2
3 = 3
4 = 2 * 2
5 = 5
6 = 2 * 3
7 = 7
8 = 2 * 2 * 2
9 = 3 * 3
10 = 2 * 5
11 = 11
So (X ^ 2) + 164 = (2 * 2 * 2 * 3 * 3 * 5 * 7 * 11) = 27720.
Then (X^2) = 27556.
Then X = 166, which is option E.
Answer: E
Question: The sum of three consecutive odd natural numbers, each divisible by 3 is 72. What is the largest among them?
Options:
A)21
B)24
C)27
D)36
E)45
Rationale: Let the 3 consecutive odd numbers each divisible by 3 be X, X+3, X+6
Given, X + X+3 + X+6 = 72
3X = 63
X = 21
The numbers are 21, 24, 27
Therefore the largest number is 27
Answer C.
Answer: C
Question: Which of the following statements must be true?
1) If x/y is a fraction and y has 17 as one of its prime factors then the resulting decimal will be terminating
2) If x/y is a fraction where 2 and 5 are the only two prime factors of y then the resulting decimal will be terminating.
3) If x/y is a fraction such that y=(10^n -1) for n≥1 then the repeating decimals in the result would just be the digits of the numerator.
4) If the interest is accounted annually then for the first compounded interest would be equal to the simple interest.
Options:
A)2,3
B)2,3,4
C)1,2,3
D)2,4
E)1,3
Rationale: Statement 3: Let's try some examples:
1/9 = 0.11111.....
23/99 = 0.2323232323.....
61/999 = 0.061061061061061.....
Statement 3 is always true.
Statement 4: I think first year is implied by the word "first".
Let's take principal = 100 and interest rate = 20%
Compound interest after 1 year = 100*(1+1/5) - 100 = 20
Simple interest after 1 year = (100*20*1)/100 = 20
Let's take principal = 300 and interest rate = 50%
Compound interest after 1 year = 300*(1+1/2) - 300 = 150
Simple interest after 1 year = (300*50*1)/100 = 150
Statement 4 is always true.
Answer (B).
Answer: B
Question: There are 5 chess amateurs playing in Villa's chess club tournament. If each chess amateur plays with exactly 16 other amateurs, what is the total number of chess games possible to be played in the tournament?
Options:
A)10
B)20
C)40
D)60
E)120
Rationale: Each person is one participant of 4 games. So there are in all 16*5 = 80 instances of one participant games. But each game has 2 participants so total number of games = 80/2 = 40
C
Answer: C
Question: The unit digit in the product (293 * 567 * 917 * 343) is:
Options:
A)2
B)1
C)6
D)8
E)10
Rationale: Explanation:
Unit digit in the given product = Unit Digit in (3*7*7*4) = 1
ANSWER: B
Answer: B
Question: Three people are playing cards. What is the probability that a card drawn from a pack of 52 cards will be a diamond or a king?
Options:
A)2/11
B)2/9
C)4/13
D)3/14
E)3/17
Rationale: Here, n(S) = 52
There are 13 cards of diamond (including one king) and there are 3 more kings.
Let E = event of getting a diamond or a king.
Then, n(E) = (13 + 3) = 16
P (E) = 16 / 52 = 4/13
C
Answer: C
Question: What is the sum of the prime factors of1549 and the prime factors of 5049?
Options:
A)73
B)63
C)65
D)74
E)75
Rationale: The prime factors of 5049 are 3*3*3*11*17
The prime factors of 1549 are 3*3*7*23
3+3+3+3+3+11+17+7+23 = 73. A is the correct answer
Answer: A
Question: Look at this series: 14, 28, 20, 40, 32, 64, ... What number should come next?
Options:
A)52
B)56
C)96
D)128
E)None
Rationale: Explanation: This is an alternating multiplication and subtracting series: First, multiply by 2 and then subtract 8.
Answer: Option B
Answer: B
Question: which number need to add to 859622 to get a number exactly divisible by 456?
Options:
A)678342
B)546546
C)576765
D)645235
E)859622
Rationale: Dividend = Quotient * Divisor + Reminder
859622 / 456 gives quotient = 1885 and reminder = 62.
So, the next number divisible by 456 is 456 places infront of 456 * 1885
Which means 456 – 62 = 394 should be added to 859622.
E
Answer: E
Question: 96 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?
Options:
A)76
B)76
C)42
D)58
E)45
Rationale: x/7 = y/9
=> x:y
= 7:9
7/16 * 96
= 42
Answer: C
Answer: C
Question: How many 1/6s are there in 37 1/2?
Options:
A)225
B)425
C)520
D)600
E)700
Rationale: Required number = (75/2)/(1/6)
= (75/2 x 6/1)
= 225.
ANSWER:A
Answer: A
Question: P works 25% more efficiently than Q and Q works 50% more efficiently than R. To complete a certain project, P alone takes 50 days less than Q alone. If, in this project P alone works for 60 days and then Q alone works for 135 days, in how many days can R alone complete the remaining work?
Options:
A)50 days
B)92.4 days
C)100 days
D)150 days
E)80 days
Rationale: P works 25% more efficiently than Q: Something that takes Q 5 days, takes P 4 days
Q works 50% more efficiently than R: Something that takes R 7.5 days, takes Q 5 days
P alone takes 50 days less than Q: For every 4 days P works, Q has to work an extra day. Hence P alone can do it in 200 days and Q alone in 250 days and hence R alone in 385 days
P works for 60 days--> 60/200 work done => 30%
Q works for 135 days--> 135/250 work done => 54%
24% work left ... R alone will take 24% * 385 = 92.4 days
Answer is (B)
Answer: B
Question: Carina has 115 ounces of coffee divided into 5- and 10-ounce packages. If she has 2 more 5-ounce packages than 10-ounce packages, how many 10-ounce packages does she have?
Options:
A)5
B)7
C)6
D)4
E)3
Rationale: Lets say 5 and 10 ounce packages be x and y respectively.
Given that, 5x+10y=115 and x=y+2. What is the value of y.
Substituting the x in first equation, 5y+10+10y=85 ->y=105/15.=7
B
Answer: B
Question: If n + 16 is a positive odd integer, which of the following must be true about the sum of the next four integers?
Options:
A)It is a multiple of 3
B)It is a multiple of 4
C)It is a prime number
D)It is a multiple of 5
E)It is a multiple of 2
Rationale: n+16 is a positive odd integer. This means n will be an odd integer
The next 4 integers would be: n + 16, n + 17, n + 18, n + 19
Sum = 4n + 70
This sum will always be even.
Option E
Answer: E
Question: A can complete a work in 12 days with a working of 8 hours per day. B can complete the same work in 8 days when working 10 hours a day. If A and B work together, working 8 hours a day, the work can be completed in --- days.
Options:
A)5 5⁄11
B)4 5⁄11
C)6 4⁄11
D)6 5⁄11
E)7 5⁄11
Rationale: Explanation :
A can complete the work in 12 days working 8 hours a day
=> Number of hours A can complete the work = 12×8 = 96 hours
=> Work done by A in 1 hour = 1/96
B can complete the work in 8 days working 10 hours a day
=> Number of hours B can complete the work = 8×10 = 80 hours
=> Work done by B in 1 hour = 1/80
Work done by A and B in 1 hour = 1/96 + 1/80 = 11/480
=> A and B can complete the work in 480/11 hours
A and B works 8 hours a day
Hence total days to complete the work with A and B working together
= (480/11)/ (8) = 60/11 days = 5 5⁄11 days
Answer : Option A
Answer: A
Question: If a certain sample of data has a mean of 20.0 and a standard deviation of 4.0, which of the following values is more than 2.5 standard deviations from the mean?
Options:
A) 12.0
B) 13.5
C) 17.0
D) 23.5
E) 26.5
Rationale: The standard deviation is 4. 2.5 standard deviations, which is just 2.5 x 4.0 is 10.0. The only answer more than 10.0 away from the mean is A.
Answer: A
Question: A group of 7 teenagers recently visited the Japanese Restaurant. The total bill for the meal, including 20% gratuity, came to $840. On average, how much did the meal of each individual cost before gratuity?
Options:
A)$160
B)$96
C)$90
D)$10
E)$55
Rationale: 7 teenagers - so total 7 people
The bill $840 includes 20% gratuity... so the actual cost of dinner was $700
Now, the cost per person will be $700/7 which is $10
Option D
Answer: D
Question: If 48 machine can finish a job in 80 days, then how many more machines would be needed to finish the job in one -fifth less time?
Options:
A)a. 4
B)b. 8
C)c. 10
D)d. 12
E)e. 16
Rationale: You might think of this in a management context - we can use the principle of 'person-hours' to solve any problem where we have identical workers. So, using simpler numbers, suppose you know that 6 identical employees, working simultaneously, would finish a job in 5 hours. Then that job requires 6*5 = 30 total hours of person-work. If instead you wanted the job done in 3 hours, you'd assign 30/3 = 10 employees to do the job, because you want to get a total of 30 hours of work from the employees.
We can solve this problem identically. If 48 machines (identical ones, I assume) work simultaneously for 80 days, they will do a total of 48*80 machine-days of work. So the job requires 48*80 days of machine work in total. We instead want the job done in 1/5 less time, so in 64 days. So we'll need 48*80/64 = 60 machines, or 12 additional machines.
D
Answer: D
Question: The speed of a boat in upstream is 60 kmph and the speed of the boat downstream is 100 kmph. Find the speed of the boat in still water and the speed of the stream?
Options:
A)10 kmph
B)20 kmph
C)16 kmph
D)18 kmph
E)19 kmph
Rationale: Speed of the boat in still water = (60+100)/2 = 80 kmph. Speed of the stream = (100-60)/2 = 20 kmph.Answer: B
Answer: B
Question: By travelling at 40 kmph, a person reaches his destination on time. He covered two-third the total distance in one-third of the total time. What speed should he maintain for the remaining distance to reach his destination on time?
Options:
A)20
B)99
C)88
D)77
E)61
Rationale: Let the time taken to reach the destination be 3x hours. Total distance = 40 * 3x = 120x km
He covered 2/3 * 120x = 80x km in 1/3 * 3x = x hours So, the remaining 40x km, he has to cover in 2x hours. Required speed = 40x/2x = 20 kmph.
Answer: A
Answer: A
Question: How many numbers between 100 and 600 are divisible by 2, 3, and 7 together?
Options:
A)112
B)77
C)267
D)12
E)99
Rationale: Explanation:
As the division is by 2, 3, 7 together, the numbers are to be divisible by: 2*3*7 = 42
The limits are 100 and 600
The first number divisible is 42*3 = 126
To find out the last number divisible by 42 within 600:
600/42 = 14.28
Hence, 42*14 = 588 is the last number divisible by 42 within 600
Hence, total numbers divisible by 2, 3, 7 together are (14 – 2) = 12
ANSWER: D
Answer: D