divelab/Sys2Bench huggingface dataset:
Question: To reach her destination, Jeanette must drive 90 miles. If she drives 6 miles every 7 minutes, how much time will it take her to reach her destination?
Options:
A)2 hours and 2 minutes
B)hours and 45 minutes
C)2 hours and 10 minutes
D)2 hours and 12 minutes
E)2 hours and 15 minutes
Rationale: 7 minute for 6 miles how much for 90? minutes for 1(7/5) mile multiplied by 90 miles.
7*90/6= 105> 1 hours 45 min answer is B
Answer: B
Question: When the number 1y92355 is exactly divisible by 11, then the smallest whole number in place of y is?
Options:
A)1
B)3
C)7
D)9
E)0
Rationale: The given number =1y92355
Sum of the odd places =5+3+9+1=18
Sum of the even places = 5+2+y
(Sum of the odd places)- (Sum of even places) = Number (exactly divisible by 11)
18-(7+y) = divisible by 11
11 � y = divisible by 11.
Y must be 0, to make given number divisible by 11.
E
Answer: E
Question: Sonika deposited Rs.8000 which amounted to Rs.9200 after 3 years at simple interest. Had the interest been 2% more. She would get how much?
Options:
A)9680
B)9687
C)96809
D)9683
E)9682
Rationale: (8000*3*2)/100 = 480
9200
--------
9680
Answer: A
Answer: A
Question: In the faculty of Reverse-Engineering, 226 second year students study numeric methods, 423 second year students study automatic control of airborne vehicles and 134 second year students study them both. How many students are there in the faculty if the second year students are approximately 70% of the total?
Options:
A)515.
B)545.
C)618.
D)644.
E)736.
Rationale: Answer is E :736
Solution : Total number of students studying both are 423 + 226 - 134 = 515 (Subtracting the 134 since they were included in the both the other numbers already).
So 70% of total is 515, so 100% is approx. 736.
Answer: E
Question: Eighty percent of the lights at Hotel California are on at 8 p.m. a certain evening. However, forty percent of the lights that are supposed to be off are actually on and ten percent of the lights that are supposed to be on are actually off. What percent of the lights that are on are supposed to be off?
Options:
A)22(2/9)%
B)16(2/3)%
C)11(1/9)%
D)10%
E)5%
Rationale: a= number of lights supposed to be on
b= number of lights supposed to be off
total = a+b
no of lights that are on = 0.8(a+yb
Light that are supposed to be on but are actually off = 10% of a
Light that are supposed to be on are on = 90% of a
Light that aare supposed to be but are on = 40% of b
Light that aare supposed to be off are off = 60% of b
0.9a + 0.4b = 0.8 (a+b)
a = 4b
Suppose total Lights = 100
so, a = 80 and b = 20
No of lights that are on are on but are supposed to be off = 40% of b = 0.4 (20) = 5
So, the % of lights that are on are supposed to be off = 5/100 = 5%
ANSWER:E
Answer: E
Question: If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p ?
Options:
A)10
B)12
C)14
D)16
E)18
Rationale: Thus proceeding with the "300 + k" assumption, we get:
P = 30!
Option A: 310
=> 310 = 2 * 5 * 31
=> 31 is a prime and is not present in 30!
Thus 310 is not a factor of P
=> A is Eliminated
Option B: 312
=> 312 = 2^3 * 3 * 13
Thus 312 is a factor of P
=> B is the correct answer !
Option C: 314
=> 314 = 2 * 157
=> 157 is a prime and is not present in 30!
Thus 314 is not a factor of P
=> C is Eliminated
Option D: 316
=> 316 = 2^2 * 79
=> 79 is a prime and is not present in 30!
Thus 316 is not a factor of P
=> D is Eliminated
Option E: 318
=> 310 = 2 * 3 * 53
=> 53 is a prime and is not present in 30!
Thus 318 is not a factor of P
=> E is Eliminated
ANS: B
Answer: B
Question: A can give B 100 meters start and C 200 meters start in a kilometer race. How much start can B give C in a kilometer race?
Options:
A)111.12
B)888.55
C)200.05
D)400.04
E)250.5
Rationale: A runs 1000 m while B runs 900 m and C runs 800 m.
The number of meters that C runs when B runs 1000 m,
= (1000 * 800)/900 = 8000/9 = 888.88 m.
B can give C = 1000 - 888.88 = 111.12 m.
ANSWER A
Answer: A
Question: A train 2500 m long can cross an electric pole in 50 sec and then find the speed of the train?
Options:
A)178
B)180
C)182
D)184
E)186
Rationale: Length = Speed * time
Speed = L/T
S = 2500/50
S = 50 M/Sec
Speed= 50*18/5 (To convert M/Sec in to Kmph multiply by 18/5)
Speed = 180 Kmph
Answer:B
Answer: B
Question: How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per kg?
Options:
A)60
B)50
C)55
D)63
E)70
Rationale: Selling price of 1kg = 924
profit = 10%
cost = 100/(100+profit%)*SP
100/(100+10)*9.24= 8.4
Mean price = (8.4-7=1.4) (9-8.4=0.6)
=>1.4:0.6=14:6 = 7:3
3x=27*7=63
ANSWER D
Answer: D
Question: A train traveling at 100 kmph overtakes a motorbike traveling at 64 kmph in 40 seconds. What is the length of the train in meters?
Options:
A)1777 meters
B)1822 meters
C)400 meters
D)1111 meters
E)None of these
Rationale: Explanation :
When a train overtakes another object such as a motorbike, whose length is insignificant compared to the length of the train, the distance traveled by the train is equal to the length of the train.
Because the motorbike is also moving, we have to take the relative speed between the train and the motorbike and not just the speed of the train.
The length of the train = distance traveled by the train while overtaking the motorbike = relative speed between the train and the motorbike * time taken
In this case, as both the objects i.e., the train and the motorbike are moving in the same direction, the relative speed between them = difference between their respective speeds = 100 - 64 = 36 kmph.
Distance traveled by the train while overtaking the motorbike = 36 kmph * 40 seconds.
The final answer is in meters and the speed is given in kmph and the time in seconds.
So let us convert the given speed from kmph to m/sec.
1 kmph = 518518 m/sec
Therefore, 36 kmph = 36 * 518518 = 10 m/sec.
Relative speed = 10 m/sec. Time taken = 40 seconds.
Therefore, distance traveled = 10 * 40 = 400 meters.
Answer: B
Question: A man can row 9 kmph in still water. When the river is running at 1.2 kmph, it takes him 1 hour to row to a place and back. What is the total distance traveled by the man?
Options:
A)6.24 km
B)6 km
C)8.84 km
D)5.66 km
E)10 km
Rationale: M = 9
S = 1.2
DS = 10.2
US = 7.8
x/10.2 + x/7.8 = 1
x = 4.42
D = 4.42 * 2 = 8.84
ANSWER:C
Answer: C
Question: Each year for 4 years, a farmer increased the number of trees in a certain orchard by 1/4 of the number of trees in the orchard of the preceding year. If all of the trees thrived and there were 3125 trees in the orchard at the end of 4 year period, how many trees were in the orchard at the beginning of the 4 year period.
Options:
A)1631
B)1531
C)1533
D)1535
E)1536
Rationale: Trees increase by 1/4 the number of trees in preceding year. Hence, correct answer must be divisible by 4. Based on divisibility rules, if last 2 digits are divisible by 4 then the number is divisible by 4. Thus, we can eliminate A, B, C,D The answer to be E
Again, trees increase by 1/4 the number of trees in preceding year. Hence, the number of trees increase by 5/4 times the number of trees the preceding year.
If x = initial number of trees = 1536
Year 1 = 5/4x
Year 2 = (5/4)(5/4)x
Year 3 = (5/4)(5/4)(5/4)x
Year 4 = (5/4)(5/4)(5/4)(5/4)x
Only for Answer D: (5/4)(5/4)(5/4)(5/4)1536 = 3750
Hence, correct answer = E
Answer: E
Question: 32% of 5/8 =
Options:
A)0.2
B)0.5
C)0.6
D)0.75
E)1.0
Rationale: Should be simple.
0.32*5/8 = 1.6/8 = 0.2
Correct Option : A
Answer: A
Question: The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:
Options:
A)17 kg
B)20 kg
C)26 kg
D)31 kg
E)32 kg
Rationale: Explanation:
Let A, B, C represent their respective weights. Then, we have:
A + B + C = (45 x 3) = 135 .... (i)
A + B = (40 x 2) = 80 .... (ii)
B + C = (43 x 2) = 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31.
B's weight = 31 kg.
ANSWER IS D
Answer: D
Question: A piece of work can finish by a certain number of men in 100 days. If however, there were 10 men less,
it would take 10 days more for the work to be finished. How many men were there originally?
Options:
A)75
B)82
C)100
D)110
E)120
Rationale: Expl: Originally let there be x men.
Less men, more days (Indirect)
:. (x-10): x : : 100 :110 or x-10/x = 100/110
or 11x-110 = 10x or x =110
So, originally there were 110 men.
Answer: D
Answer: D
Question: There are 24 students in Chad’s class. He brought 29 cookies to pass out
for his birthday. How many cookies will each student get? Will there be
any cookies left over?
Options:
A)1-9
B)1-7
C)2-8
D)1-5
E)3-10
Rationale: 29/24 = 1 R 5 Chad will give each student 1 cookie and there will be 5
cookies left over
correct answer D
Answer: D
Question: A magician holds one six-sided die in his left hand and two in his right. What is the probability the number on the dice in his left hand is greater than the sum of the dice in his right?
Options:
A)7/108
B)5/54
C)1/9
D)2/17
E)1/4
Rationale: The least sum of the numbers of the two dice in the right hand is 2. The maximum sum of the numbers on these two dice can be 5, for the left hand dice to be able to have a greater number.
Therefore the number on the dice in the left hand must be 3 or more.
Number of ways to get a sum of 2 with the two dice in the right hand= 1
Number of ways to get a sum of 3 with the two dice in the right hand= 2
Number of ways to get a sum of 4 with the two dice in the right hand= 3
Number of ways to get a sum of 5 with the two dice in the right hand= 4
Therefore required probability = P(two right dice have a sum of 2)*P(left dice has 3) + P(right dice have sum of 2 or 3)*P(left dice has 4) + P(two right dice have a sum of 2,3, or 4)*P(left dice has 5) + P(two right dice have a sum of 2,3,4,or 5)*P(left dice has 6)
= (1/36)*(1/6) + (3/36)*(1/6) + (6/36)*(1/6) + (10/36)*(1/6)
= (1/216) * (1+3+6+10)
= 20/216
= 5/54
Option B
Answer: B
Question: There are 6 ladies and 4 gents to dine at a round table. In how many ways can they seat themselves so that no two gents are seated together.
Options:
A)85400
B)36400
C)86400
D)14400
E)88600
Rationale: First arrange the ladies. This is a circular arrangment. 6 Gents can be arranged in 5! ways.
Now there are 6 spaces (space in between two ladies) and we have to seat 5 gents there.
First select 4 places out of 5 available places. can be done in 5C4ways. Then arrange 5 gents there in 5! ways.
Total: 5!*5C4*4! = 14400
D is the answer.
Answer: D
Question: A man covered a certain distance at some speed. If he had moved
3 kmph faster, he would have taken 40 minutes less. If he had moved
2 kmph slower, he would have taken 40 minutes more. What is the distance in km?
Options:
A)40
B)30
C)35
D)60
E)55
Rationale: speed= 2v1v2/v1-v2=2*3*2/3-2=12 km/hr
dist=12*40/16(1+12/3)=40 km
ANSWER A
Answer: A
Question: The average score of a cricketer for twelve matches is 48 runs. If the average for first eight matches is 40, then average for last four matches is
Options:
A)33.25
B)89
C)64
D)68
E)90
Rationale: Explanation:
=(48×12)−(40×8)/4
=(576−320)/4=64
Answer: Option C
Answer: C
Question: Last year’s receipts from the sale of greeting cards during the week before Mother’s Day totaled $239 million, which represented 10 percent of total greeting card sales for the year. Total greeting card sales for the year totaled how many million dollars?
Options:
A)2,390
B)2,100
C)1,890
D)1,701
E)210
Rationale: 10% ---- 239 millions
for 100% => ( 239 * 100% )/10% = 2390.
Option A.
Answer: A
Question: If A^5 + B^5 = 2500, then the greatest possible value of A is between
Options:
A)0 and 3
B)3 and 6
C)6 and 9
D)9 and 12
E)12 and 15
Rationale: for the greatest possible value of A^5, we must minimize the value of B^5 i.e. lets say B^5 = 0
then we need to find a number A such that A^5 < 2500. 4^5 = 1024 and 5^5 = 3125 so we can say that the maximum possible value of A can be a little more than 4 hence answer = between 3 and 6
hence B
Answer: B
Question: Two trains of equal are running on parallel lines in the same direction at 52 km/hr and 36 km/hr. The faster train passes the slower train in 36 sec. The length of each train is?
Options:
A)50 m
B)72 m
C)80 m
D)82 m
E)84 m
Rationale: Let the length of each train be x m.
Then, distance covered = 2x m.
Relative speed = 52 - 36 = 16 km/hr.
= 16 * 5/18 = 40/9 m/sec.
2x/36 = 40/9 => x = 80.
ANSWER:C
Answer: C
Question: At Deluxe paint store, Fuchsia paint is made by mixing 5 parts of red paint with 3 parts of blue paint. Mauve paint is made by mixing 3 parts of red paint with 5 parts blue paint. How many liters of blue paint must be added to 40 liters of Fuchsia to change it to Mauve paint?
Options:
A)9
B)12
C)15
D)16
E)26.6
Rationale: in 40 liters, Red = 5/8 * 40 = 25 and Blue = 15
so, 25/(15+x) = 3/5
or, x = 26.6 (Answer E)
Answer: E
Question: If 6 (A's capital) = 8 (B's capital) = 10 (C's capital). Then the ratio of their capitals is:
Options:
A)3:4:5
B)12:15:20
C)20:15:12
D)6:8:10
E)8:4:12
Rationale: Explanation:
6A = 8B = 10 C
A:B:C = 1/6:1/8:1/10
= 20:15:12
ANSWER IS C
Answer: C
Question: Find the probability that a leap year selected at random will not have 53 Sundays
Options:
A)6/7
B)5/7
C)4/7
D)3/7
E)2/7
Rationale: There are 366 days in a leap year: 52 weeks and 2 more days. So, 52 Sundays and 2 days.
These 2 days can be: {Mon, Tue}, {Tue, Wed}, {Wed, Thu}, {Thu, Fri}, {Fri, Sat}, {Sat,Sun} and {Sun, Mon} (7 cases). In order to have 53 Sundays we should have either {Sat,Sun} or {Sun, Mon} case.
The probability of that is 2/7.
So, probability of not having 53 sundays is 1 - 2/7 = 5/7
Answer: B.
Answer: B
Question: Seven children â€” A, B, C, D, E, F, and G â€” are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?
Options:
A)600
B)720
C)1440
D)4320
E)4800
Rationale: This problem is a notch harder than anything you are likely to see on the GMAT. If you can master the principles in this problem, you certainly will be able to handle almost any problem the GMAT could concoct.
First, consider the restriction of A & B. As with problem #1 above, there are 12 possibilities for A & B, counting both position and order.
Now, put the other five in any order â€” thatâ€™s 120 possibilities, for a total number of configurations of 1440. That number does not take into account the restriction with C.
Think about those 1440 configurations. In exactly half of them, C will be to the right of A & B, and exactly half, C will be to the left of A & B. Therefore, in (1/2)*1440 = 720 configurations, C will be to the right of A & B. Answer = B
Answer: B
Question: On an order of 4 dozen boxes of a consumer product, a retailer receives an extra dozen free. This is equivalent to allowing him a discount of:
Options:
A)10%
B)20%
C)30%
D)40%
E)50%
Rationale: Clearly, the retailer gets 1 dozen out of 5 dozens free.
Equivalent discount = 1/5 * 100 = 20%.
Answer B) 20%
Answer: B
Question: Pipe A runs 60 liters of water per minute into a pool that has a total volume of 3,600 liters. Pipe B fills a third of the pool in 12 hours.
Working together how long will it take both pipes to fill the pool?
Options:
A)2 hours and 36 minutes
B)36/37 hours
C)9 hours and 12 minutes
D)15 hours
E)1 hour and 48 minutes.
Rationale: Pipe A has a rate of 60 liters per minute which equals a rate of 3600 liters per hour.
Pipe B has a rate of 100 liters per hour.
100 + 3600 (A+B) = Together they have a rate of 3700 liters per hour.
3600 (The pool) / 3700 = 36/37 hours
Answer -B
Answer: B
Question: At what rate percent per annum will the simple interest on a sum of money be 6/5 of the amount in 10 years?
Options:
A)4%
B)7%
C)9%
D)3%
E)12%
Rationale: Let sum = x. Then, S.I. = 6x/5, Time
= 10 years.
Rate = (100 * 6x) / (x * 5 * 10)
= 12%
Answer:E
Answer: E
Question: Angela, Bernie, and Colleen can complete a job, all working together, in 3 hours. Angela and Bernie, working together at their respective rates, can complete the same job in 5 hours. How long would it take Colleen, working alone, to complete the entire job?
Options:
A)7.5 hours
B)10 hours
C)12 hours
D)16 hours
E)20 hours
Rationale: A+B+C complete the job in 3 hours.
A+B complete the job in 5 hours.
A+B and C complete the job in 3 hours -->1/(A+B) + 1/C = 1/3-->1/5+1/C=1/3 ----> C=7.5 hours. ANS A
Answer: A
Question: HCF and LCM two numbers are 16 and 396 respectively. If one of the numbers is 36, then the other number is?
Options:
A)36
B)66
C)132
D)176
E)364
Rationale: 16 * 396 = 36 * x
x = 176
ANSWER:D
Answer: D
Question: A man has Rs. 312 in the denominations of one-rupee notes, five-rupee notes and twenty-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
Options:
A)36
B)24
C)28
D)32
E)38
Rationale: Explanation :
Let the number of notes of each denomination be x
Then, x + 5x + 20x = 312
=> 26x = 312
=> x = 312/26 = 12
the total number of notes that he has = 3x = 3 × 12 = 36
Answer : A
Answer: A
Question: If shares of two persons in profits are Rs.400 and Rs.500 then ratio of their capitals is
Options:
A)3:4
B)4:5
C)4:3
D)1:3
E)1:2
Rationale: profit=time*capital
so 400:500=4:5
ANSWER:B
Answer: B
Question: Brian is analyzing the performance of certain stocks. He finds that for the first fifteen days of a 30 day period, the daily closing price per share of Phoenix Pharma Inc. conformed to the function f(x)=0.4x+2f(x)=0.4x+2 , where x represents the day in the period. He found that for the last 15 days in the same period, the stock followed the function f(x)=(−1/7)x+9f(x)=(−1/7)x+9, where x represents the day in the period. Approximately what is the median closing price per share of Phoenix Pharma Inc. for the entire 30 day period?
Options:
A)4.7
B)5.55
C)6.71
D)8
E)11.1
Rationale: First 15 days: F1, x= 1 to 15 and f(x)= 2.4 to 8
Last 15 days: F2, x= 16 to 30 and f(x) = ~4.7 to ~6.7 (30th to 16th)
Median is average of 15th and 16th value when all of the above values are arranged in order
Do not try to calculate all the values. Go for the choices.
11.1 - outside the range of values
8- last value
6.71 - last value for the F2, so already we have 15 values less than 6.71. Additionally F1 has some values less.
4.7 - First value for the F2, so already we have 15 values greater than 4.7. Additionally F1 has some values greater.
Hence 5.55 must be the answer.
ANSWER:B
Answer: B
Question: Set J consists of 10 consecutive even numbers. If the smallest term in the set is -4, what is the range of the positive integers in set J?
Options:
A)8
B)12
C)14
D)16
E)18
Rationale: Since there are only 10 integers, another approach is the just list all 10.
We get: -4, -2,0,2, 4, 6, 8, 10, 12, 14
Range of POSITIVE INTEGERS =14-2= 12
Answer: B
Answer: B
Question: Someone on a skateboard is traveling 15 miles per hour. How many feet does she travel in 30 seconds? (1 mile = 5280 feet)
Options:
A)600 ft
B)660 ft
C)670 ft
D)680 ft
E)700 ft
Rationale: per second => 15*5280 ft/60*60 =22 ft
30 seconds => 22 * 30 = 660 ft
Answer : B
Answer: B
Question: A ratio between two numbers is 3 : 4 and their L.C.M. is 84. The first number is
Options:
A)60
B)45
C)21
D)15
E)None
Rationale: Sol.
Let the required numbers be 3x and 4x. Then, their L.C.M. is 12x.
∴ 12x = 84⇔ x = 7. Hence, the first number is 21.
Answer C
Answer: C
Question: The number x of cars sold each week varies with the price y in dollars according to the equation x = 800,000 – 50y. What would be the total weekly revenue T, in dollars, from the sale of cars priced at $ 15,000?
Options:
A)50,000
B)750,000
C)850,000
D)7,500,000
E)T=750,000,000
Rationale: number of cars sold = x=800000-50y
y=15000
x=800000-750000=50000
revenue from 50000 cars=15000*50000=750000000 E
Answer: E
Question: On Saturday morning, Malachi will begin a camping vacation and he will return home at the end of the first day on which it rains. If on the first three days of the vacation the probability of rain on each day is 0.2, what is the probability that Malachi will return home at the end of the day on the following Monday?
Options:
A)0.008
B)0.128
C)0.488
D)0.512
E)0.64
Rationale: Probability of rain on Saturday OR Probability of rain on Sunday (given that it didn't rain on Saturday) OR Probability of rain on Monday (given that it didn't rain on Sunday)
= 0.2 + (0.8 x 0.2) + (0.8 x 0.8 x 0.2)
= 0.2 + 0.16 + (0.128 <- The actual answer!)
=0.488
ANSWER:C
Answer: C
Question: What will come in place of the x in the following Number series? 6, 12, 21, x , 48
Options:
A)33
B)45
C)75
D)56
E)85
Rationale: (A)
The pattern is + 6, + 9, + 12, +15 ………..
So the missing term is = 21 + 12 = 33
Answer: A
Question: A person sells 36 oranges per rupee and suffers a loss of 4%. Find how many oranges per rupee to be sold to have again of 8%?
Options:
A)30
B)31
C)32
D)33
E)None of these
Rationale: 1⁄36 : (100 - 4) :: x : (100 + 8)
⇒ x = 108/96×36=132
He sells 32 oranges per rupee.
Answer C
Answer: C
Question: A can do a job in 20 days and B can do it in 30 days. A and B working together will finish twice the amount of work in days?
Options:
A)22 1/6
B)24
C)22 1/0
D)22 1/9
E)22 1/1
Rationale: 1/20 + 1/30 = 5/60 = 1/12
12 = 12 *2 = 24 days
Answer: B
Answer: B
Question: A walks at 10 kmph and 7 hours after his start, B cycles after him at 20 kmph. How far from the start does B catch up with A?
Options:
A)100km
B)140km
C)50km
D)120km
E)200km
Rationale: Suppose after x km from the start B catches up with A. Then, the difference in the time taken by A to cover x km and that taken by B to cover x km is 7 hours.
x/10 - x/20 = 7
x = 140 km
Answer is B
Answer: B
Question: A car travells the first 1/2 of certain distance with a speed of 10km/hr the next 1/4 distance witha a speed of 20 km/hr, and the last 1/4 distance with a speed of 60 km/hr. the average speed of the carfor the whole journey is?
Options:
A)12km/hr
B)15km/hr
C)89km/hr
D)52km/hr
E)79km/hr
Rationale: Let the whole distance travelled be x km
and the average speed of the car for he whole journey be y km/hr
then (x/2)/10+(x/4)/20+(x/4)/60=x/y
x/20+x/80+x/240=x/y
1/15y=1
y=15km/hr
Answer(B)
Answer: B
Question: Two carpenters, working in the same pace, can build 2 desks in two hours and a half. How many desks can 4 carpenters build in 3 hours?
Options:
A)2.4.
B)3.6.
C)4.8.
D)5.5.
E)6.4
Rationale: W = 2 desks
T = 2.5 hrs
Rate of 2 carpenters = 2×R
Rate = work done/ time
2xR = 2/2.5
R = 1/2.5 = 2/5 (this is the rate of each carpenter)
Work done by 4 carpenters in 3 hrs = 4 × rate of each carpenter x time = 4× 2/5 × 3 =4.8 desks
C is the correct answer.
Answer: C
Question: A bag contains a certain number of 50 paise coins, 20 paise coins and 10 paise coins inthe ratio 3:4:5. If the total value of all the coins in the bag is Rs.630, find the number of 50 paise coins ?
Options:
A)498
B)488
C)428
D)528
E)675
Rationale: 50*3k + 20*4k + 10*5k = 63000
280k = 63000 => k =225
50p coins = 3k = 3*225 = 675
ANSWER:E
Answer: E
Question: The average of four results is 222 and that of the first three is 219. The fourth result is?
Options:
A)231
B)201
C)85
D)192
E)215
Rationale: 4 * 222 â€“ 3 * 219 = 231
Answer: A
Answer: A
Question: A train passes a man standing on a platform in 8 seconds and also crosses the platform which is 267 metres long in 20 seconds. The length of the train (in metres) is:
Options:
A)178
B)176
C)175
D)96
E)None of these
Rationale: Explanation:
Let the length of train be L m.
Acc. to question
(267+L)/20 = L/8
2136+8L=20L
L= 2136/12 = 178 m
Answer A
Answer: A
Question: A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is?
Options:
A)76 km/hr
B)50 km/hr
C)17 km/hr
D)18 km/hr
E)14 km/hr
Rationale: Speed of the train relative to man
= (125/10) m/sec = (25/2) m/sec. [(25/2) * (18/5)] km/hr
= 45 km/hr. Let the speed of the train be x km/hr. Then, relative speed
= (x - 5) km/hr. x - 5
= 45 ==> x = 50 km/hr.
Answer: B
Answer: B