RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: How many seconds will a 500 meter 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?
A.228
B.277
C.500
D.288
Answer: C.500
Explanation: Let length of tunnel is x meter
Distance = 800+x meter
Time = 1 minute = 60 seconds
Speed = 78 km/hr = 78*5/18 m/s = 65/3 m/s
Distance = Speed*Time
800+x = (65/3) * 60
800+x = 20 * 65 = 1300
x = 1300 - 800 = 500 meters.Answer:C
Question: Out of eight crew members three particular members can sit only on the left side. Another two particular members can sit only on the right side. Find the number of ways in which the crew can be arranged so that four men can sit on each side.
A.864
B.863
C.865
D.1728
Answer: D.1728
Explanation: Explanation :
The required number of ways :-
=>3C2×4!×4!.
=>1728.
Answer : D
Question: The average earning of a mechanic for the first-four days of a week is Rs. 18 and for the last four days is Rs. 22. If he earns Rs . 41 on the fourth day, his average earning for the whole week is ?
A.Rs . 18.95
B.Rs 16
C.Rs. 17
D.Rs. 25.71
Answer: C.Rs. 17
Explanation: Answer
Total earning for the week = Sum of earning of first four days + Sum of earning of last four days - Earning of 4th day
= 4 x18 + 4 x 22 -41
= Rs. 119
âˆ´ Average earning = 119 / 7
= Rs. 17
Correct Option: C
Question: FOG + FOG + FOG + FOG = 1444.
If F,O and G are digits in a three-digit number in the preceding equation the F + O + G = ?
A.11
B.12
C.13
D.10
Answer: D.10
Explanation: given,FOG + FOG + FOG + FOG = 1444
4(FOG)=1444
=>FOG=361
Face value of F is:3
Face value of O is:6
Face value of G is:1
so F+O+G=3+6+4=10
ANSWER:D
Question: A fruit wholesale dealer has been paid $329,864 in his savings account. What is the least amount of money (in whole number of dollars) that he must add to his account if if he wants to be sure that all 9 of his salesmen have paid evenly?
A.$1
B.$2
C.$3
D.$4
Answer: D.$4
Explanation: To find the least amount the fruit wholesale dealer should add to his account to ensure all 9 of his salemen have paid evenly,
he needs to make the total divisible by 9
Simply add the individual digits of the total = 3+2+9+8+6+4 = 32
If you add 4 , the number is divisible by 9 (32+ 4)
Correct Option : D
Question: If m,n,o,p are four consecutive even numbers, their average is
A.(m + 3)
B.5(m+4)
C.5(m+n+o+p)
D.mnop
Answer: A.(m + 3)
Explanation: Explanation :
Clearly,n=m+2,o=m+4,p=m+6.
Average (m + (m + 2) + (m + 4) + (m + 6)) / 4
= (4m + 12) / 4
= (m + 3)
Answer : A
Question: Six lights begin to flash together at 12:00 noon and flash respectively at intervals of 3, 4, 5, 6, 7, and 8 seconds. Not including the flash at 12:00 noon, how many more times will all six lights flash together before 1:00 pm (i.e. one hour later)?
A.4
B.6
C.8
D.10
Answer: A.4
Explanation: The least common multiple is 2*2*2*3*5*7=840.
3600 seconds / 840 = 4 + remainder.
The answer is A.
Question: A caterer ordered 125 ice-cream bars and 125 sundaes. If the total price was $225.00 and the price of each ice-cream bar was $0.60, what was the price of each sundae?
A.$0.60
B.$0.80
C.$1.00
D.$1.20
Answer: D.$1.20
Explanation: Let price of a sundae = S
Price of ice cream bar = .6 $
125 * .6 + 125*S = 225
=> 125*S = 150
=> S = 1.2
Answer D
Question: If a can do a work in 10 days and a and b can do a same piece of job in 5 days. IN how days can do a work b alone?
A.10
B.20
C.5
D.15
Answer: A.10
Explanation: a's one day work=1/10
a+b=1/5(one day work)
b=(1/5)-a=(1/5)-(1/10)
=1/10(one day work)
b can done a work in 10 days.
Answer option A
Question: A certain factory produces buttons and buckles at a uniform weight. If the total weight of 2 buttons and 2 buckles is one third of 11 buckles and 3 buttons, then the weight of 3 buttons and 2 buckles is how many times that of 4 buckles and 6 buttons?
A.7/15.
B.4/9.
C.6/11.
D.1/2.
Answer: D.1/2.
Explanation: 2buttons + 2buckles = 1/3(11buckles + 3buttons)
6buttons + 6buckles = 11buckles + 3buttons
3buttons = 5buckles
Weight of 3buttons and 2buckles = 5buckles + 2buckles = 7buckles
Weight of 6buttons and 5buckles = 10buckles + 4buckles = 14buckles
(3buttons and 2buckles) = x * (6buttons and 5buckles)
7 = x*14
x = 7/14=1/2
Answer: D
Question: A 4-letter code word consists of letters L, M, AND N. If the code includes all the three letters, how many such codes are possible?
A.72
B.36
C.24
D.44
Answer: B.36
Explanation: Note that the correct answer to this question is 36, not 72.
L-LMN can be arranged in 4!/2!=12 ways;
M-LMN can be arranged in 4!/2!=12 ways;
N-LMN can be arranged in 4!/2!=12 ways;
Total: 12+12+12=36.
Answer: B
Question: A candidate got 35% of the votes and lost to the rival by 1350 votes. How many votes were cast?
A.4000
B.4500
C.5000
D.5500
Answer: B.4500
Explanation: 35%-----------L
65%-----------W
------------------
30% = 1350
10% = 450
100% of the votes = 4500
Answer: B
Question: A and B invest in a business in the ratio 3: 2. Assume that 5% of the total profit goes to charity. If A's share is Rs. 855, what is the total profit?
A.1200
B.1300
C.1500
D.1400
Answer: C.1500
Explanation: Assume that the total profit is x
Since 5% goes for charity, 95% of x will be divided between A and B in the ratio 3:2
Therefore, A's profit =(95x/100)×(3/5)
Given that A's share is Rs. 855. Therefore,
(95x/100)×(3/5)=855
⇒95x/100=855×5/3=285×5=1425
⇒x=1425×100/95
=285×100/19=1500
Answer is C.
Hence the total profit = 1500
Question: How many positive integers less than 100 have a remainder of 2 when divided by 13?
A.5
B.7
C.8
D.9
Answer: C.8
Explanation: take the multiples of 13 and add 2
0X13+2=2 ....13X7+2=93
there are 14 numbers
answer C
Question: There are 6periods in each working day of a school. In how many ways can 1 arrange 5subjects such that each subject is allowed at least 1period?
A.1200
B.1800
C.2000
D.2500
Answer: B.1800
Explanation: 5 subjects can be arranged in 6 periods in 6P5 ways.
Remaining 1 period can be arranged in 5P1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements = (6P5 x 5P1)/2! = 1800
Alternatively this can be derived using the following approach.
5 subjects can be selected in 5C5 ways.
Remaining 1 subject can be selected in 5C1 ways.
These 6 subjects can be arranged themselves in 6! ways.
Since two subjects are same, we need to divide by 2!
Total number of arrangements = (5C5 × 5C1 × 6!)/2! = 1800
B
Question: A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in:
A.14 hrs
B.15 hrs
C.10 hrs
D.20 hrs
Answer: A.14 hrs
Explanation: work done in 1 hr = 1/2-3/7 = 1/14
==>14 hrs
ANSWER A
Question: A train 240 m in length crosses a telegraph post in 16 seconds. The speed of the train is?
A.50
B.54
C.52
D.56
Answer: B.54
Explanation: Explanation:
speed= Distance/ Time
S = 240/16 * 18/5 = 54 kmph
Answer:B
Question: In a certain accounting class of 100 students, 70% of the students took the final exam on the assigned day while the rest of the students took the exam on a make-up date. If the students on the assigned day had an average score of 60%, and the students on the make-up date had an average score of 80%, what was the average score for the entire class?
A.64%
B.66%
C.68%
D.70%
Answer: B.66%
Explanation: 70% of the class scored 60% and 30% of the class scored 80%.
The difference between 60% and 80% is 20%.
The average will be 60% + 0.3(20%) = 66%.
The answer is B.
Question: Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same Job.How long should it take both A and B, working together but independently, to do the same job?
A.40/9 days
B.9 days
C.40 days
D.94 days
Answer: A.40/9 days
Explanation: A’s 1 hour's work = 1/8
B's 1 hour's work = 1/10
(A + B)'s 1 hour's work = (1/8) +(1/10)=9/40
Both A and B will finish the work in 40/9 days.
Ans: A
Question: A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?
A.316/439
B.316/438
C.316/435
D.316/299
Answer: C.316/435
Explanation: Let A be the event of getting two oranges and
B be the event of getting two non-defective fruits.
and be the event of getting two non-defective oranges
Answer: C) 316/435
Question: Find the least number must be subtracted from 427398 so that remaining no.is divisible by 15?
A.2
B.3
C.5
D.7
Answer: B.3
Explanation: On dividing 427398 by 15 we get the remainder 3, so 3 should be subtracted
B
Question: The cost of registration at a professional association meeting was $50 per person; a lunch for registrants only was available for an additional $22 per person. If the number of registrants who paid for lunch was 50 more than the number who did not, and if receipts for registration and lunch totaled $82,900, how many people paid just for registration at the meeting?
A.700
B.650
C.1,300
D.1,500
Answer: B.650
Explanation: Hope this might be useful to you.
Let the number of people who have opted only to register = x
Now since the registration cost is 50$ per person, the total amount sums to = 50x $
As per the information given in the question, the number of registrants who paid for lunch was 50 more than the number who did not. That means, total number of people who registered and paid for lunch = 50 + x.
For the people who registered for lunch the cost is 50 $ (for the event registration) + 22 $ (for lunch) = 72 $.
Total amount in this case sums to = 72(50 + x) = 3600 + 72x
Now, total amount received was 82900.
Thus, from the above data,
50x + 3600 + 72x = 82900
122x = 82900 - 3600
122x = 79300
x = 650
Hence the correct Ans is B
Question: A bullock cart has to cover a distance of 80km in 10hrs. If it covers half of the journey in 3/5th time. what should be its speed to cover the remaining distance in the time left.
A.10 kmph
B.20 kmph
C.30 kmph
D.40 kmph
Answer: A.10 kmph
Explanation: A
10 kmph
Time left = 10 - 3/5*10
= 4 hr 10 km/h
speed =40 km /4 hr
=10 kmph
Question: If 30 oxen can plough 1/7th of a field in 2 days, how many days 18 oxen will take to do the remaining work?
A.30 days
B.20 days
C.15 days
D.18 days
Answer: B.20 days
Explanation: Solution: We will use work equivalence method,
30/18 = (1/7)/(6/7)*x/2;
5/3 = (1/6)*x/2;
Or, x = 60/3 = 20 days.
Answer: Option B
Question: A shopkeeper sells 500 metres of cloth for Rs. 15000 at a loss of Rs.10 per metre. Find his cost price for one metre of cloth?
A.s.40
B.s.12
C.s.96
D.s.50
Answer: A.s.40
Explanation: SP per metre = 15000/500
= Rs. 30 Loss per metre
= Rs. 10 CP per metre = 30 + 10
= Rs.40
Answer: A
Question: 15 percent of new students select a science major, 10 percent select an arts major, and 35 percent select an engineering major, what fraction of students remain undecided?
A.4/5
B.3/5
C.3/10
D.2/5
Answer: D.2/5
Explanation: 100%-(15%+10%+35%)=60% remains for other allocations --> 40% = 40/100 = 2/5.
Answer: D.
Question: The present ages of three persons in proportions 5 : 6 : 4. Five years ago, the sum of their ages was 60. Find their present ages (in years).
A.8, 20, 28
B.16, 28, 36
C.25,30,20
D.18, 24, 32
Answer: C.25,30,20
Explanation: Explanation:
Let their present ages be 5x, 6x and 4x years respectively.
Then, (5x - 5) + (6x - 5) + (4x - 5) = 60
15x-15 = 60
x = 5
Their present ages are 5x = 25 years, 6x = 30 years and 4x = 20 years respectively.
Answer: C
Question: The number of new words that can be formed by rearranging the letters of the word 'LIVE' is?
A.25
B.27
C.19
D.23
Answer: D.23
Explanation: Number of words which can be formed
= 4! - 1 = 24- 1
= 23.
Answer:D
Question: Number of ways in which the letters of word AMBI can be arranged with vowels in alphabetical order, is
A.360
B.240
C.12
D.480
Answer: C.12
Explanation: Solution
Order of vowels of fixed
âˆ´ required number of ways are 4!/2!.
Answer C
Question: It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:
A.3:4
B.4:5
C.6:3
D.4:4
Answer: A.3:4
Explanation: A
3:4
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then, 120/x + 480/y = 8 --> 1/x + 4/y = 1/15 ------ (i)
Also, 200/x + 400/y = 25/3 --> 1/x + 2/y = 1/24 ----(ii)
Solving (i) and (ii), we get: x = 60 and y = 80.
--> Ratio of speeds = 60 : 80 = 3 : 4.
Question: Rap is working on a set of Data Sufficiency problems for his December GMAT: a geometry problem, an algebra problem, and a data interpretation problem. He has determined that statement 1 of the geometry problem is insufficient on its own, that both statement 1 and 2 of the algebra problem are insufficient on their own, and that statement 2 of the data interpretation problem is insufficient on its own. If the probabilities are expressed as percents T, approximately how much greater is the probability that all three answers are “C” after Rap figures out that statement 1 of the data interpretation problem is also insufficient on its own?
A.T=2.3%
B.T=2.8%
C.T=3.3%
D.T=5.6%
Answer: B.T=2.8%
Explanation: Step 1: Figure out what we need to know.We have two probabilities T that we need to figure out: the one before Raj figures out that statement 1 of the data interpretation problem is also insufficient, and the one after. The answer is the difference between them, in the form of a percent.
Step 2: The first probability.
If statement 1 of the geometry problem is insufficient, then the only valid answer choices are B, C, and E; the probability that it is C is 1/3.
If both statements of the algebra problem are insufficient on their own, then the only remaining valid answers are C and E; the probability that it is C is 1/2
If statement 2 of the data interpretation problem is insufficient, then the remaining answers are A, C, and E, and the probability that C is correct is 1/3.
The probability of all three occurring is the product of the probability fractions: (1/3)*(1/2)*(1/3) = 1/18.
Step 3: The second probability.
Only the third problem has changed; if Raj now knows that statement 1 is also insufficient, the valid answer choices are only C and E, leaving a 1/2 probability that the answer is C.
The probability of all three occurring is still the product of those fractions, but this time they are (1/3)*(1/2)*(1/2) = 1/12.
Step 4: The answer.Note that here, as will sometimes happen on the GMAT, values that you compute “along the way” appear as wrong answer choices. This problem calls for the difference between our two numbers, (1/12)-(1/18) = 1/36, which is a 2.77777% chance, rounded to 2.8% —the correct answer is Choice B.
Question: If the a radio is sold for rs 490 and sold for rs 465.50.find loss value.
A.24.5
B.12.5
C.17.25
D.6.5
Answer: A.24.5
Explanation: cp=rs490,sp= 465.50.
loss=rs(490-465.50)=rs 24.50.
Ans: A
Question: If rupee one produces rupees nine over a period of 40 years, find the rate of simple interest?
A.22 1/9 %
B.22 2/2 %
C.23 1/2 %
D.22 1/2 %
Answer: D.22 1/2 %
Explanation: 9 = (1*40*R)/100
R = 22 1/2 %
Answer:D
Question: David and Lewis leave Chennai for Tirupati simultaneously at 7 A.M in the morning driving in two cars at speeds of 60 mph and 80 mph respectively. As soon as Lewis reaches Tirupati he returns back to Chennai along the same route and meets David on the way back. If the distance between the two cities is 160 miles, how far from Chennai did David and Lewis meet?
A.100 mlies
B.120 miles
C.90 miles
D.95 miles
Answer: B.120 miles
Explanation: Time taken by Lewis to reach Tirupati = 160/80 = 2 hours
In 2 hours, David travels 60*2 = 120 miles
So distance at which they meet should be greater than 120 miles.
Only B satisfies.
Answer is B.
Question: In how many ways can 4 boys and 2girls be arranged ata table if 2 girls are never together?
A.76
B.74
C.72
D.73
Answer: C.72
Explanation: Total ways of arranging 6 person in a circle = 5! = 120 ways
If we consider two women as one, then we have 4+1=5 persons. Arranging these five persons in circle will take = 4! = 24 ways
And two women can arrange themselves in 2! ways
Total ways in which two women sit together in circle = 4!*2! = 48 ways
Total ways in which two women doesn’t sit together in circle = 5! - 4!*2! =120 - 48 = 72 ways
C
Question: If 0 < c< 1 < d, which of the following must be true?
A.1 < 1/c< 1/d
B.1/c< 1 < 1/d
C.1/c< 1/d< 1
D.1/d< 1 < 1/c
Answer: D.1/d< 1 < 1/c
Explanation: 0<c<1<d. Let c=1/2 and d=2
Substituting values:
A. 1<2<1/2 ---Wrong
B. 2<1< 1/2 ---Wrong
C. 2<1/2<1 ---Wrong
D. 1/2<1<2 ---Correct
E. 1/2<2<1 ---Wrong
Hence answer is D
Question: a farmer have 10%cow,15%buffalo,25%goat,50% chicken.he has 30 goat.then calculate the no. of cow,buffalo and chicken.
A.12,15,45
B.12,18,60
C.10,18,45
D.10,16,60
Answer: B.12,18,60
Explanation: total no. of livestock is x.
x*25/100=30
x=3000/25=120
no.of cow=120*10/100=12
no. of buffalo=120*15/100=18
no. of chicken=120*50/100=60
answer B
Question: The total of a company's assets in 1994 was 200% greater than the total in 1993, which in turn was 300% greater than the total in 1992. If the total of the company's assets in 1992 was N dollars, what were the company's assets in 1994?
A.5N
B.6N
C.9N
D.12N
Answer: D.12N
Explanation: In 1992, the company's assets were N.
In 1993, the company's assets were N + 3N =4N.
In 1994, the company's assets were 4N + 8N = 12N.
The answer is D.
Question: A fill pipe can fill 1/6 of cistern in 90 minutes in how many minutes, it can fill 5/6 of the cistern?
A.450 min
B.360 min
C.240 min
D.306 min
Answer: A.450 min
Explanation: 1/6 of the cistern can fill in 90 min
5/6 of the cistern can fill in = 90*6*5/6 = 450 min
Answer is A
Question: A farmer grows Broccoli in his farm that is in the shape of a square. Each Broccoli takes 1 square foot of area in his garden. This year, he has increased his output by 101 Broccoli when compared to last year. The shape of the area used for growing the Broccoli has remained a square in both these years. How many Broccoli did he produce this year?
A.2400
B.2601
C.1987
D.2000
Answer: B.2601
Explanation: Explanatory Answer
The shape of the area used for growing Broccoli has remained a square in both the years.
Let the side of the square area used for growing Broccoli this year be X ft.
Therefore, the area of the ground used for cultivation this year = X2 sq.ft.
Let the side of the square area used for growing Broccoli last year be Y ft.
Therefore, the area of the ground used for cultivation last year = Y2 sq.ft.
As the number of Broccoli grown has increased by 101, the area would have increased by 101 sq ft because each Broccoli takes 1 sq ft space.
Hence, X2 - Y2 = 101
(X + Y)(X - Y) = 101.
101 is a prime number and hence it will have only two factors. i.e., 101and 1.
Therefore, 101 can be expressed as product of 2 numbers in only way = 101 * 1
i.e., (X + Y)(X - Y) = 101* 1
So, (X + Y) should be 101 and (X - Y) should be 1.
Solving the two equations we get X = 51 and Y = 50.
Therefore, number of Broccoli produced this year = X2 = 512 = 2601.
Alternative Approach : Use answer choices
The area in both the years are squares of two numbers.
That rules out choice A,C and D. as 2400,1987 and 2000 are not the square of any number.
Check Choice B: If this year's produce is 2601, last year's produce would have been 2601 - 101 = 2500
2500 is the square of 50.
So, 2601 is the answer.
Choice B
Question: 2, 3, 5, 7, 11, ?, 17
A.15
B.14
C.13
D.12
Answer: C.13
Explanation: The series consists of prime numbers starting from 2.
Answer : C.
Question: Out of 18 students studying in a class, 7 are from Maharashtra, 5 are from Karnataka and 3 are from Goa. Four students are to be selected at random. What are the chances that at least one is from Karnataka?
A.12â„13
B.11â„13
C.10â„15
D.59/77
Answer: D.59/77
Explanation: Total possible ways of selecting 4 students out of 18 students = 18C4 =3060
The no. of ways of selecting 4 students in which no student belongs to Karnataka = 13C4=715
âˆ´ Hence no. of ways of selecting at least one student from Karnataka = 18C4 â€“ 13C4 = 2345
âˆ´ Probability = 2345â„3060= 59/77
Answer D
Question: John makes $60 a week from his job. He earns a raise and now makes $100 a week. What is the % increase?
A.16%
B.66.67%
C.17%
D.17.61%
Answer: B.66.67%
Explanation: Increase = (40/60)*100 = (4/6)*100 = 66.67%.
B
Question: If the sides of a triangle are 30 cm, 28 cm and 10 cm, what is its area?
A.120
B.110
C.130
D.140
Answer: D.140
Explanation: The triangle with sides 30 cm, 28 cm and 10 cm is right angled, where the hypotenuse is 30 cm.
Area of the triangle = 1/2 * 28 * 10 = 140 cm2
Answer: Option D
Question: Pipe A fills a swimming pool in 4 hours. Pipe B empties the pool in 6 hours. If pipe A was opened at 5:00 am and Pipe B at 6:00 am, at what time will the pool be full?
A.13:00
B.14:00
C.15:00
D.17:00
Answer: C.15:00
Explanation: Pipe A fills the pool in 4 hrs.
1 hour's work : 1/4
Pipe B empties the pool in 6 hrs.
1 hour's work : 1/6
Together if they work, 1 hour's work = 1/4 -1/6 = 1/12
Given : Pipe A started at 5:00 a.m and Pipe B at 6:00 a.m
Pool filled after 1 hour by Pipe A : 1/4 or 3/12
After 6:00 a.m
Pool filled after 1 hour with both the pipes on : 1/12
Pool filled after 9 hours with both pipes on : 9/12
Pool filled in 1 hour + Pool filled in 9 hours = 3/12 +9/12 =1
Therefore, it takes 10 hrs to fill the pool
As Pipe A started at 5:00 a.m, pool is full at 15:00 hrs
Answer : C
Question: we have 7 weights one of 2 kg one of 3 kg one of 5 kg one of 1 kg one of 10 kg one of 15 kg one of 20 kg by using only five weight how many weight u can weight
A.19
B.18
C.24
D.21
Answer: D.21
Explanation: select 5 weight out of 7, 7C5=21
ANSWER:D
Question: Obra drove 140π meters along a circular track. If the area enclosed by the circular track on which she drove is 57,600π square meters, what percentage of the circular track did Obra drive?
A.6.67%
B.12.5%
C.18.75%
D.29.16%
Answer: D.29.16%
Explanation: area enclosed by the circular track on which she drove is 57,600π square meters
So, π(r^2) = 57,600π ---> (r^2)=57,600 ---> r=240
Circumference of the circular track = 2πr = 480π
Therefore, part of circumference covered = 140π/480π =29.16%
Hence, answer is D.
Question: A and B go around a circular track of length 600 m on a cycle at speeds of 36 kmph and 54 kmph. After how much time will they meet for the first time at the starting point?
A.120
B.288
C.277
D.126
Answer: A.120
Explanation: Time taken to meet for the first time at the starting point
= LCM { length of the track / speed of A , length of the track / speed of B}
= LCM { 600/ (36 * 5/18) , 600/ (54 * 5 /18) }
= LCM (60, 40) = 120 sec.
Answer: A
Question: Walking at 5/6th of its usual speed a cab is 8mnts late. Find its usual time to cover the journey?
A.25m
B.45m
C.40m
D.50m
Answer: C.40m
Explanation: New speed = 5/6th of usual speed
New time = 6/5th of usual time
6/5 UT - UT = 8m
UT/5 = 8m
UT = 40m
Answer is C
Question: A rectangular grass field is 70 m * 55 m, it has a path of 2.5 m wide all round it on the outside. Find the area of the path and the cost of constructing it at Rs.2 per sq m?
A.s.1350
B.s.1300
C.s.1328
D.s.1397
Answer: B.s.1300
Explanation: Area = (l + b +2d) 2d
= (70 + 55 +2.5 * 2) 2*2.5 => 650
650 * 2 = Rs.1300
Answer: B
Question: The numerator of a fraction is 1 less than twice the denominator. If the numerator and denominator are both increased by \
1, the fraction becomes 3/5. Find the fraction?
A.A)5/3
B.B)3/5
C.C)4/7
D.D)5/9
Answer: A.A)5/3
Explanation: Let the numerator and denominator of the fraction be 'n' and 'd' respectively.
n = 2d - 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n - 1) + 3 => n = 5
n = 2d - 1 => d = 3
Hence the fraction is : 5/3
ANSWER:A
Question: The product of three consecutive numbers is 336. Then the sum of the smallest two numbers is ?
A.13
B.15
C.20
D.38
Answer: A.13
Explanation: Product of three numbers = 336
336 = 6*7*8.
So, the three numbers are 6, 7 and 8.
And Sum of smallest of these two = 6+7 = 13.
ANSWER : OPTION A
Question: The edges of a cuboid are respectively 3cm, 4cm and 12 cm. Find the length of the diagonal of cuboid?
A.28 m3
B.27 m3
C.60 m3
D.18 m3
Answer: C.60 m3
Explanation: lb = 20 ; lh = 15 ; fh
= 12
(lbh)2 = 20 * 15 * 12 => lbh
= 60 m3
Answer: C