divelab/Sys2Bench huggingface dataset:
Question: Three candidates contested an election and received 45990, 20008 and 34884 votes respectively. What percentage of the total votes did the winning candidate get ?
Options:
A)57.25%
B)45.54%
C)75.20%
D)45.59%
E)58%
Rationale: Explanation :
Total number of votes polled = (45990 + 20008 + 34884 )
= 61200.
Required percentage = (45990/ 100882) x 100%
= 45.59%.
Answer : D
Answer: D
Question: The number of stamps that P and Q had were in the ratio of 5:3 respectively. After P gave Q 10 stamps, the ratio of the number of P's stamps to the number of Q's stamps was 9:7. As a result of the gift, P had how many more stamps than Q?
Options:
A)20
B)30
C)40
D)60
E)90
Rationale: P started with 5k stamps and Q started with 3k stamps.
(5k-10)/(3k+10) = 9/7
35k - 27k = 160
k = 20
P has 5(20) - 10 = 90 stamps and Q has 3(20)+10 = 70 stamps.
The answer is A.
Answer: A
Question: the first flight out of Phoenix airport had a late departure. If the next three flights departed on-time, how many subsequent flights need to depart from Phoenix on-time, for the airport's on-time departure rate to be higher than 90%?
Options:
A)6
B)7
C)9
D)10
E)11
Rationale: We need on-time departure rate to be higher than 9/10, so it should be at least 10/11 which means that 10 out of 11 flights must depart on time. Since for now 3 out of 4 flights departed on time then 10−3=7 subsequent flights need to depart on-time.
Answer: B
Answer: B
Question: If the sum of a number and its square is 20, what is the number?
Options:
A)15
B)26
C)28
D)4
E)none of these
Rationale: Let the number be x. Then,
x + x2 = 20
(x + 5)(x - 4) = 0
x = 4
ANSWER:D
Answer: D
Question: If population of certain city increases at the rate of 5%. If population in 1981 was 370440, then population in 1978 was?
Options:
A)320000
B)120029
C)187899
D)178968
E)789698
Rationale: X * (105/100) * (105/100) * (105/100) = 370440
X = 370440/1.157625
X = 320000
Answer:A
Answer: A
Question: Set Q consists of the following five numbers: Q = {5, 8, 13, 21, 32}. Which of the following sets has the same standard deviation as Set Q?
I. {35, 38, 43, 51, 64}
II. {10, 16, 26, 42, 68}
III. {46, 59, 67, 72, 75}
Options:
A) I only
B) III
C) IIII
D) IIIII
E) I, II,III
Rationale: (1)	Multiplying all the numbers change Standard Deviation
(2)	Adding and subtracting all the numbers with the same number keeps the standard deviation same.
If you observe Set I and III are added versions of Set Q .
Set I: 5 has been added to the Set Q
Set III: Subtract each element from 80 and you would find a number there in the Set Q.
Set II: elements are multiplied by 2 and standard deviation changes.
Hence the answer is D.
Answer: D
Question: If x and y are integers and x=50y + 2, which of the following must be odd?
Options:
A)none of the following
B)x + y
C)x + 2y
D)3x-1
E)xy
Rationale: x = 50y + 2, since y is integer, 50y will always be even hence 50y + 2 will be even hence x = even
you don't know whether Y is even or odd, so you need to try.
E: even * even = even ; ->not true
B: even + even = even ; > not true
C: even + 2*even = even ;
A
Answer: A
Question: During the past week, a local medical clinic tested N individuals for two infections. If 1/2 of those tested had infection A and, of those with infection A, 1/6 also had infection B, how many individuals did not have both infection A and B?
Options:
A)N/12
B)4N/15
C)14N/12
D)11N/12
E)4N/5
Rationale: 1/3 of tested had infection A, thus N/2 had infection A;
Of those with infection A, 1/6 also had infection B, thus 1/2*N/6 = N/12 had both infections A and B.
Therefore, N - N/12 = 11N/12 did not have both infection A and B.
Answer: D.
Answer: D
Question: A and B invests Rs.8000 and Rs.9000 in a business. After 4 months, A withdraws half of his capital and 2 months later, B withdraws one-third of his capital. In what ratio should they share the profits at the end of the year?
Options:
A)32:45
B)32:37
C)32:41
D)32:47
E)32:48
Rationale: A : B
(8000*4)+(4000*8) : (9000*6)+(6000*6)
64000 : 90000
32 : 45
Answer: A
Answer: A
Question: If 4<x<8<y<12, then what is the greatest possible positive integer difference of x and y?
Options:
A)3
B)4
C)5
D)6
E)7
Rationale: Let x=4.1 and y=11.1
Greatest possible difference=11.1-4.1=7
Answer E
Answer: E
Question: If a = 105 and a^3= 21 * 35 * 45 * b, what is the value of b?
Options:
A)35
B)42
C)45
D)49
E)54
Rationale: First step will be to break down all the numbers into their prime factors.
105 = 3 * 5 * 7
21 = 7 * 3
35 = 7 * 5
45 = 3 * 3 * 5
so, (105)^3 = 3 * 7 * 7 * 5 * 3 *3 *5 *b
Therefore (3 * 5 * 7) ^ 3 = 3^3 * 5^2 * 7^2 *b
Therefore, b = 3^3 * 5^3 * 7 ^3/3 ^3 * 5 ^2 * 7^2
b = 5*7 = 35
Correct answer A.
Answer: A
Question: Two pipes A and B can separately fill a cistern in 60 minutes and 75 minutes respectively. There is a third pipe in the bottom of the cistern to empty it. If all the three pipes are simultaneously opened, then the cistern is full in 50 minutes. In how much time, the third pipe alone can empty the cistern?
Options:
A)90 min
B)100 min
C)110 min
D)120 min
E)None
Rationale: Solution
Work done by the third pipe in 1 min.	= 1/50 - (1/60+1/75)
= (1/50 - 3/100)
= -1/100.
Therefore, the third pipe alone can empty the cistern in 100 min.
Answer C
Answer: C
Question: A shop has 3000 cycles. It sells 10% of the cycles daily. How many cycles remain after 2 days?
Options:
A)2500
B)2600
C)2450
D)2430
E)2250
Rationale: Formula :
( After =100 denominator
Ago = 100 numerator)
3000 Ã— 90/100 Ã— 90/100 = 2430
D
Answer: D
Question: By selling 33 metres of wire, a shopkeeper gains the price of 11 metres of wire. His gain per cent is
Options:
A)7%
B)50%
C)20%
D)22%
E)32%
Rationale: Suppose S.P of 33 meters of cloth = Rs 33
Therefore, Gain = Rs 11 => C.P = Rs 22
Gain % = 11/22 x 100 = 50%
Gain = 50%
ANSWER:B
Answer: B
Question: For each month of a given year except December, a worker earned the same monthly salary and donated one-tenth of that salary to charity. In December, the worker earned N times his usual monthly salary and donated one-fourth of his earnings to charity. If the worker's charitable contributions totaled one-eighth of his earnings for the entire year, what is the value of N?
Options:
A)8/5
B)5/2
C)3
D)22/10
E)4
Rationale: Let monthly salary for each of the 11 months except December was x, then
11x*1/10 + Nx*1/4 = 1/8(11x + Nx);
11/10 + N/4 = 1/8(11 + N) => 44+10N/40= 11+N/8
352+80N = 440+40N => 40N = 88
N =88/40 =22/10
Answer: D.
Answer: D
Question: A clock is set right at 5 a.m. The clock loses 16 minutes in 24 hours.What will be the true time when the clock indicates 10 p.m. on 4th day?
Options:
A)11
B)66
C)88
D)55
E)18
Rationale: Time from 5 am. on a day to 10 pm. on 4th day = 89 hours.
Now 23 hrs 44 min. of this clock = 24 hours of correct clock.
356/15 hrs of this clock = 24 hours of correct clock
89 hrs of this clock = (24 x 31556 x 89) hrs of correct clock.
= 90 hrs of correct clock.
So, the correct time is 11 p.m.
Answer: A) 11pm
Answer: A
Question: A parking garage rents parking spaces for $15 per week or $30 per month. How much does a person save in a year by renting by the month rather than by the week?
Options:
A) $140
B) $160
C) $220
D) $420
E) $260
Rationale: 10$ per week!
An year has 52 weeks.
Annual charges per year = 52* 15 = 780$
30$ per month!
An year has 12 months.
Annual charges per year = 12 * 30 = 360$
780 - 360 = 420
Ans D
Answer: D
Question: 3, k, 20, m, 4, n
In the list above, k, m, and n are three distinct positive integers and the average (arithmetic mean) of the six numbers in the list is 8. If the median of the list is 6.5, which of the following CANNOT be the value of k, m, or n ?
Options:
A)9
B)8
C)7
D)6
E)5
Rationale: Since median is 6.5, the first options I will try are 6 and 7 - two numbers will be eliminated if I can find a case where this works.
3, 4, 6, 7, 20 - the avg of these numbers is 8. If the last number is also 8, the mean will remain 8 as desired. So in fact, we eliminated 3 options here. k, m and n can be 6, 7 and 8.
Let's try 5 now, not 9 because 9 is more complicated. 9 gives us two number less than 6.5 and 2 more than 6.5. So there will be many different options. If instead 5 is in the list, we now have 3 numbers less than 6.5, so the other 3 numbers must be greater than 6.5 and the average of one of those numbers with 5 must be 6.5. So the fourth number should be 8 on the list to give the median 6.5. These 5 numbers (3, 4, 5, 8, 20) give an average of 8. The sixth number must be 8 to keep the average 8 but numbers must be distinct. So this is not possible. Hence none of k, m and n can be 5.
Answer (E)
Answer: E
Question: The difference of two numbers is 11 and one fifth of their sum is 9. The numbers are :
Options:
A)31, 20
B)30, 19
C)29, 18
D)28, 17
E)none
Rationale: x − y = 11, x + y = 5 × 9 x − y = 11, x + y = 45, y = 17, x = 28
Answer: D
Answer: D
Question: Henry eats X scones in X percent of the time it takes Rachel to eat Y scones. If Rachel eats four scones in ten minutes, then the number of minutes it takes Henry to eat 12 scones must be equal to which of the following?
Options:
A)Y/5
B)3Y/10
C)100Y/(15X)
D)XY/250
E)Y/(5X)
Rationale: Rachel eats 4 scones/10 minutes = 2/5 scones per minute
The time for Rachel to eat 1 scone is 5/2 minutes.
The time for Rachel to eat Y scones is 5Y/2 minutes.
The time for Henry to eat X scones is 5YX/200 minutes.
The time for Henry to eat 1 scone is 5Y/200 minutes.
The time for Henry to eat 12 scones is 60Y/200 = 3Y/10 minutes.
The answer is B.
Answer: B
Question: The length of a rectangular field is 7/5 its width. If the perimeter of the field is 384 meters, what is the width of the field?
Options:
A)50
B)60
C)11
D)80
E)90
Rationale: Let L be the length and W be the width. L = (7/5)W
Perimeter: 2L + 2W = 384, 2(7/5)W + 2W = 384
Solve the above equation to find: W = 80 m and L = 112 m.
correct answer D)80
Answer: D
Question: The simple interest on Rs.12000 at a certain rate of interest in five years is Rs.7200. Find the compound interest on the same amount for five years at the same rate of interest.
Options:
A)3052.82
B)3052.89
C)3052.85
D)3052.8
E)3052.82
Rationale: R = 100 I / PT
=> R = (100 * 7200)/ (12000 * 5) = 12%
CI = P{ [1 + R /100]n - 1}
= 12000 { [ 1 + 12 / 100]2 - 1} = Rs.3052.80
Answer:D
Answer:D
Answer: D
Question: P and Q started a business with respective investments of Rs. 4 lakhs and Rs. 10 lakhs. As P runs the business, his salary is Rs. 5000 per month. If they earned a profit of Rs. 2 lakhs at the end of the year, then find the ratio of their earnings?
Options:
A)1 : 8
B)1 : 1
C)1 : 6
D)1 : 5
E)1 : 3
Rationale: Ratio of investments of P and Q is 2 : 5
Total salary claimed by P = 12 * 5000 = Rs. 60000
Total profit = Rs. 2 lakhs.
Profit is to be shared = Rs. 140000
Share of P = (2/7) * 140000 = Rs. 400000
Share of Q = Rs. 100000
Total earnings of P = (60000 + 40000) = Rs. 100000
Ratio of their earnings = 1 : 1.Answer: D
Answer: D
Question: How many numbers from 10 to 50 are exactly divisible by 3.
Options:
A)13
B)14
C)16
D)17
E)19
Rationale: 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45,48.
13 Numbers.
10/3 = 3 and 50/3 = 16 ==> 16 - 3 = 13. Therefore 13 digits.
A)
Answer: A
Question: If 102 books need to be placed in boxes and each box can contain at most 9 books. How many books will be in the last unfilled box?
Options:
A)12 books
B)10 books
C)0 books
D)3 books
E)9 books
Rationale: The number of books that can be placed in boxes of 9 books is the first number that is divisible by 9 that occurs before 102 .
In order to divide the sum in 9 parts, the amount must be divisible by 9
Divisibility rule of 9: The sum of the digits must be divisible by 9
Sum of digits of 102 =3 and 0 is divisible by 9.
Hence, we need to remove 3 to this number for it to be divisible by 9
Correct Option:D
Answer: D
Question: The cost of paint is 60 per kilograme. A kilogram paint covers 20 square feet. How much will it cost to paint the outside of a cube having each side 10 feet?
Options:
A)3000
B)900
C)1800
D)360
E)None of these
Rationale: Area of cube
= 6× (side)2 = 6 × 10 × 10 = 600 square feet.
Cost to paint outside of the cube = 600⁄20 × 60
= 1800
Answer C
Answer: C
Question: What is the total number of positive integers that are less than 400 and that have no positive factor in common with 400 other than 1?
Options:
A)100
B)120
C)150
D)160
E)180
Rationale: Since 400=2^4*5^2 then a number can not have 2 and/or 5 as a factor.
The odd numbers do not have 2 as a factor and there are 200 odd numbers from 1 to 400.
We then need to eliminate the 40 numbers that end with 5, that is 5, 15, 25,...,395.
There are a total of 200 - 40 = 160 such numbers between 1 and 400.
The answer is D.
Answer: D
Question: Raju age after 15 years will be 5 times his age 5 years back, What is the present age of Raju
Options:
A)15
B)14
C)10
D)8
E)7
Rationale: Explanation:
Clearly,
x+15 = 5(x-5)
<=> 4x = 40 => x = 10
Answer: Option C
Answer: C
Question: Somesh bought a microwave oven and paid 10% less than its original price. He sold it at 30% profit on the price he had paid. What percentage of profit did Somesh earn on the original price?
Options:
A)32%
B)11%
C)20%
D)17%
E)None of these
Rationale: Quicker Method:
% profit which Somesh gets
= –10 +30 – 10×30/100 = +17%
Answer D
Answer: D
Question: In how many ways can 3 boys and 3 girls be arranged at a table if 3 girls are never together?
Options:
A)96
B)92
C)82
D)84
E)90
Rationale: Total ways of arranging 6 person in a circle = 5! = 120 ways
If we consider three women as one, then we have 3+1=4 persons. Arranging these four persons in circle will take = 3! = 6 ways
And two women can arrange themselves in 3! ways
Total ways in which three women sit together in circle = 3!*3! = 36 ways
Total ways in which three women doesn’t sit together in circle = 5! - 3!*3! = 120 - 36 = 84 ways
D
Answer: D
Question: Seven people are on an elevator that stops at exactly 7 floors. What is the probability that exactly one person will push the button for each floor?
Options:
A)7!/7^7
B)7^7/7!
C)7/7!
D)7/7^7
E)1/7^7
Rationale: Each person out of 7 has 7 options, hence total # of outcomes is 7^7;
Favorable outcomes will be 7!, which is # of ways to assign 7 different buttons to 7 people
So basically # of arrangements of 7 distinct objects: 7!.
P=favorable/total=7!/7^7
Answer: A.
Answer: A
Question: Rose is two years older than Bruce who is twice as old as Chris. If the total of the age of Rose, B and Chris be 27 years, then how old is Bruce ?
Options:
A)8 years
B)10 yrs
C)12 years
D)14 years
E)16 years
Rationale: Let Chris's age be x years. Then, Bruce's age = 2x years.Rose's age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25
x = 5.
Hence, Bruce's age = 2x = 10 years.
B
Answer: B
Question: Ravi started walking from his house east direction on Bus stop which is 3km.away. Then he set off in the bus straight towards his right to the school 4 km away. what is the crow flight distance from his house to the school?
Options:
A)1 km
B)5 km
C)7 km
D)12 km
E)None of these
Rationale: Explanation:
The walked will be of in triangular shape and from starting point is 5km
ANSWER IS B
Answer: B
Question: A person goes to his office at 1/3rd of the speed at which he returns from his office. If the avg speed during the whole trip is 21m/h. what is the speedof the person while he was going to his office?
Options:
A)8km/h
B)10km/h
C)12km/h
D)13km/h
E)14km/h
Rationale: u = k , v= 3k
\inline \therefore \frac{2uv}{u+v}\: \: \Rightarrow \frac{2\times k\times 3k}{(k+3k)}=21
\inline \Rightarrow 1.5k = 21
\inline \Rightarrow k=14km/h
E
Answer: E
Question: What will come in place of the x in the following Number series? 2, 7, 27, 107, 427, x
Options:
A)1427
B)1607
C)1707
D)2047
E)2147
Rationale: 22 × 4 - 1 = 77 × 4 - 1 = 2727 × 4 - 1 = 107107 × 4 - 1 = 427427 × 4 - 1 = 1707 C
Answer: C
Question: Two cars start from the opposite places of a main road, 113 km apart. First car runs for 25 km and takes a right turn and then runs 15 km. It then turns left and then runs for another 25 km and then takes the direction back to reach the main road. In the mean time, due to minor break down the other car has run only 35 km along the main road. What would be the distance between two cars at this point?
Options:
A)65
B)38
C)20
D)28
E)21
Rationale: Answer: D) 28 km
Answer: D
Question: If b is the sum of consecutive even integers w, x, y, and z, where w < x < y < z, all of the following must be true EXCEPT
Options:
A)z - w = 3(y - x)
B)b is divisible by 8
C)The average of w, x, y, and z is odd
D)b is divisible by 4
E)w + x + 8 = y + z
Rationale: Just assume that the numbers are 2a, 2a+2, 2a+4 and 2a+6.
b = 2a + 2a+2 + 2a+4 + 2a+6 = 8a + 12
Focus on the easiest options first. You see that b is not divisible by 8. Answer (B)
Answer: B
Question: On Monday, Lou drives his vehicle with 26-inch tires, averaging x miles per hour. On Tuesday, Lou switches the tires on his vehicle to 32-inch tires yet drives to work at the same average speed as on Monday. What is the percent change from Monday to Tuesday in the average number of revolutions that Lou's tires make per second?
Options:
A)Decrease by 14.3%
B)Decrease by 19%
C)Increase by 14.3%
D)Increase by 12.5%
E)Cannot be determined with the given information.
Rationale: 416 is the LCM
On Tuesday 416/32 =13
On Monday 416/26 = 16
Therefore, 13-16/16 = -1/5=-19% decrease
Answer is B
Answer: B
Question: A certain protective additive increases from 30 days to 60 days the time between required maintenance checks on an industrial vehicle. By what percent is the time between maintenance checks increased by using the additive?
Options:
A)25%
B)33 1/3%
C)100%
D)66 2/3%
E)75%
Rationale: General formula for percent increase or decrease, (percent change):
percent=Change/Original∗100
So, the time between maintenance checks increased by (60−30)/30∗100=100
Answer: C
Answer: C
Question: In a decade, a town’s population increases from 175000 to 262500. What is the average % increase of population each year?
Options:
A)9%
B)2%
C)5%
D)7%
E)1%
Rationale: Explanation:
Increase in population in 10 years= 262500-175000= 87500
Now, % increase will be increase in (population /original population) * 100
=> 87500/175000*100= 50% increase in population over 10 years
Therefore, Average % increase each year= 50/10= 5%
ANSWER: C
Answer: C
Question: Three-twentieths of the members of a social club are retirees who are also bridge players, five-twentieths of the members are retirees, and one-half of the members are bridge players. If 120 of the members are neither retirees nor bridge players, what is the total number of members in the social club?
Options:
A)240
B)300
C)200
D)400
E)480
Rationale: {Total} = {Retirees} + {Bridge players} - {Both} + {Neither}
x = 5/20*x + x/2 - 3/20*x + 120
20x = 5x + 10x - 3x + 120*20 (multiply by 20)
12x = 120*20
x = 200.
Answer: C
Answer: C
Question: If you have 4 New Year greeting cards and you want to send them to 2 of your friends, in how many ways can this be done?
Options:
A)360
B)12
C)16
D)24
E)None
Rationale: Solution
We have to find number of permutations of 2 objects out of 4 objects. This number is 4P2=4(4-1)=4*3=12
Answer B
Answer: B
Question: Find the probability of getting a multiple of 3,when an unbiased die is tossed?
Options:
A)3/4
B)2/8
C)1/7
D)1/3
E)1/6
Rationale: Here S={1,2,3,4,5,6}
Let E be the event of getting the multiple of 3
then ,E={3,6}
P(E)=n(E)/n(S)=2/6=1/3
Option D
Answer: D
Question: What percent is 1 gm of 1 kg?
Options:
A)0.1 %
B)1%
C)1.5%
D)2%
E)3%
Rationale: 1 kg = 1000 gm
1/1000 × 100 = 100/1000
=1/10 = 0.1 %
A)
Answer: A
Question: If the cost price of 12 pens is equal to the selling price of 10 pens, the gain percent is:
Options:
A)80%
B)90%
C)20%
D)40%
E)10%
Rationale: Explanation:
Let C.P. of each pen be Re. 1.
Then, C.P. of 10 pens = Rs. 10; S.P. of 10 pens = Rs. 12.
Gain % = 2/10 * 100 = 20%
Answer:C
Answer: C
Question: The average mark obtained by 22 candidates in an examination is 46. The average of the first ten is 55 while the last eleven is 40 .The marks obtained by the 11th candidate is ?
Options:
A)22
B)0
C)19
D)18
E)16
Rationale: It is clear that 22 x 46 = 10 x 55 + K + 11 x 40 ⇒⇒ K = 22
Answer:A
Answer: A
Question: Team / No of games won
A 4
B 7
C 9
D 2
E 2
X ?
According to the incomplete table above, if each of the 6 teams in the league played each of the other teams exactly twice and there were no ties, how many games did team X win? (Only 2 teams play in a game.)
Options:
A)4
B)5
C)6
D)8
E)10
Rationale: 6C2= 15 single games, & 30 in total.
won by others= 24
30-24= 6
ANSWER:C
Answer: C
Question: On a scale that measures the intensity of a certain phenomenon, a reading of j+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of j. On that scale, the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3?
Options:
A) 5
B) 50
C) 10^5
D) 5^10
E) 8^10 - 3^10
Rationale: To solve this problem we need to examine the information in the first sentence. We are told that “a reading of j + 1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of j.”
Let’s practice this idea with some real numbers. Let’s say j is 2. This means that j + 1 = 3. With the information we were given we can say that a reading of 3 is ten times as great as the intensity of a reading of 2.
Furthermore, we can say that a reading of 4 is actually 10 x 10 = 10^2 times as great as the intensity of a reading of 2.
Increasing one more unit, we can say that a reading of 5 is 10 x 10 x 10 = 10^3 times as great as the intensity of a reading of 2.
We have found a pattern, which can be applied to the problem presented in the stem:
3 is “one” unit away from 2, and thus a reading of 3 is 10^1 times as great as the intensity of a reading of 2.
4 is “two” units away from 2, and thus a reading of 4 is 10^2 times as great as the intensity of a reading of 2.
5 is “three” units away from 2, and thus a reading of 5 is 10^3 times as great as the intensity of a measure of 2.
We can use this pattern to easily answer the question. Here we are being asked for the number of times the intensity corresponding to a reading of 8 is as great as the intensity corresponding to a reading of 3. Because 8 is 5 units greater than 3, a reading of 8 is 10^5 times as great as the intensity corresponding to a reading of 3.
Answer C.
Answer: C
Question: How many seconds will a 800 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?
Options:
A)11
B)30
C)99
D)48
E)61
Rationale: Speed of train relative to man = 63 - 3 = 60 km/hr.
= 60 * 5/18 = 50/3 m/sec.
Time taken to pass the man = 800 * 3/50 = 48 sec.
Answer: D
Answer: D
Question: A clock loses 1% time during the first week and then gains 2% time during the next one week. If the clock was set right at 12 noon on a Sunday, what will be the time shown by the clock exactly 14 days from the time it was set right?
Options:
A)1: 36: 48
B)1: 40: 48
C)1: 41: 24
D)10: 19: 12
E)None
Rationale: Solution:
The clock loses 1% time during the first week.
In a day there are 24 hours and in a week there are 7 days. Therefore, there are 7 * 24 = 168 hours in a week.
If the clock loses 1% time during the first week, then it will show a time which is 1% of 168 hours less than 12 Noon at the end of the first week = 1.68 hours less.
Subsequently, the clock gains 2% during the next week. The second week has 168 hours and the clock gains 2% time = 2% of 168 hours = 3.36 hours more than the actual time.
As it lost 1.68 hours during the first week and then gained 3.36 hours during the next week, the net result will be a -1.68 + 3.36 = 1.68 hour net gain in time.
So the clock will show a time which is 1.68 hours more than 12 Noon two weeks from the time it was set right.
1.68 hours = 1 hour and 40.8 minutes = 1 hour + 40 minutes + 48 seconds.
i.e. 1 : 40 : 48 P.M.
Answer B
Answer: B
Question: If 20 men can build a wall 112 metres long in 6 days, what length of a similar wall can be built by 30 men in 3 days?
Options:
A)65mtr.
B)84mtr
C)70mtr.
D)78mtr.
E)17mtr.
Rationale: 20 men is 6 days can build 112 metres
30 men in 3 days can build = 112*(30/20)x(3/6)
= 84 meters
Answer: B.
Answer: B