mathewhe/aqua_rat huggingface dataset:
Question: The calendar for the year 2005 is the same as for the year :
(A) 2010
(B) 2011
(C) 2012
(D) 2013
(E) None of these
Rationale: Count the number of days from 2005 onwards to get 0 odd day.
= 7 or 0 odd day.
∴ Calender for the year 2005 is the same as that for the year 2012
Answer C
Answer: (C) 2012
Question: A train with 120 wagons crosses Arun who is going in the same direction, in 36 seconds. It travels for half an hour from the time it starts overtaking the Arun ( he is riding on the horse) before it starts overtaking the Sriram( who is also riding on his horse) coming from the opposite direction in 24 seconds. In how much time (in seconds) after the train has crossed the Sriram do the Arun meets to Sriram?
(A) 2276 sec
(B) 1276 sec
(C) 3576 sec
(D) 2176 sec
(E) 1676 sec
Rationale: Explanation:
Let the length of the train be L metres and speeds of the train Arun and Sriram be R, A and S respectively, then
---------- (i)
and ---------(ii)
From eq.(i) and (ii)
3(R - A ) = 2 (R + K)
R = 3A + 2K
In 30 minutes (i.e 1800 seconds), the train covers 1800R (distance) but the Arun also covers 1800 A (distance) in the same time. Therefore distance between Arun and Sriram, when the train has just crossed Sriram
= 1800 ( R - A) - 24 ( A + K)
Time required =
= (3600 - 24) = 3576 s
Answer: C
Answer: (C) 3576 sec
Question: The number 61 can be written as the sum of the squares of 3 different positive integers. What is the sum of these 3 integers?
(A) 17
(B) 16
(C) 15
(D) 14
(E) 13
Rationale: I think brute force with some common sense should be used to solve this problem.
Write down all perfect squares less than 61: 1, 4, 9, 16, 25, 36, 49.
Now, 61 should be the sum of 3 of those 7 numbers. Also to simplify a little bit trial and error, we can notice that as 61 is an odd numbers then either all three numbers must be odd (odd+odd+odd=odd) OR two must be even and one odd (even+even+odd=odd).
We can find that 61 equals to 9+16+36=3^2+4^2+6^2=61 --> 3+4+6=13.
Answer: E.
Answer: (E) 13
Question: During 2003, a company produced an average of 2,000 products per month. How many products will the company need to produce from 2004 through 2007 in order to increase its monthly average for the period from 2003 through 2007 by 150% over its 2003 average?
(A) 287,000
(B) 290,000
(C) 284,000
(D) 285,000
(E) 286,000
Rationale: Company produced 12*2000 = 24,000 products in 2003. If company produces X products from 2004 to 2007, then total amount of product produced in 4 years (2003 through 2007) is X+24,000. The gives the average of (X+24,000)/4.
This average needs to be 300% higher than that in 2003. In math terms, 36,000+150%(36,000) = 80,000. So:
(X+36,000)/4 = 80,000
X+36,000 = 320,000
X = 284,000
The answer is C.
Answer: (C) 284,000
Question: Susan is traveling from one end of a forest to the other. In order to find her way back, she is leaving morsels of bread in the following pattern: 2 morsels of wheat, 3 morsels of white, and 1 morsel of bap. The pattern repeats after she leaves the morsel of bap. If Susan drops 2,000 morsels of bread, what are the last 3 morsels of bread that she drops?
(A) bap − wheat − wheat
(B) wheat − wheat − white
(C) white − bap − wheat
(D) white − white − white
(E) white − white − bap
Rationale: In a single round Susan drops 6 morsels.
Remainder (2000/6) = 2
Final action = 2 drops of Wheat --> Options B, C, D and E can be eliminated
Answer: A
Answer: (A) bap − wheat − wheat
Question: If there are 5 seats and 3 students, in how many ways can they be seated?
(A) 40
(B) 45
(C) 50
(D) 60
(E) 80
Rationale: The first student has 5 choices, the second student has 4 choices, and the third student has 3 choices. The number of ways to sit is 5*4*3 = 60.
The answer is D.
Answer: (D) 60
Question: If e, f, and g are distinct positive prime numbers, then (e^3)(f)(g) has how many different positive factors greater than 1?
(A) a. 8
(B) b. 9
(C) c. 12
(D) d. 11
(E) e. 27
Rationale: Let Number is (e^3)(f)(g) = (1^3)(6)(10) = 60
We can write 60 as product of two numbers in following ways
1*60
2*30
3*20
4*15
5*12
6*10
6cases = 6*2 i.e. 12 factors (including 1)
Factors greater than 1 = 11
Answer: Option D
Answer: (D) d. 11
Question: Two trains of length 90 m and 100 m are 200 m apart. They start moving towards each other on parallel tracks, at speeds 71 kmph and 89 kmph. After how much time will the trains meet?
(A) 47/10
(B) 46/10
(C) 45/10
(D) 42/10
(E) 43/10
Rationale: They are moving in opposite directions, relative speed is equal to the sum of their speeds.
Relative speed = (71 + 89)*5/18 = 44.4mps.
The time required = d/s = 200/44.4= 45/10 sec.
Answer:C
Answer: (C) 45/10
Question: Find the fraction which has the same ratio to 2/6 that 3/4 has to 1/2
(A) 1/2
(B) 2/5
(C) 6/8
(D) 9/4
(E) 7/5
Rationale: P : 2/6 = 3/4 : 1/2
As the product of the means is equal to the product of the extremes.
P*1/2 = 2/6 * 3/4
P*1/2 = 6/24
P = 1/2 => P = 1/2
ANSWER:A
Answer: (A) 1/2
Question: If the difference between the length and breadth of a rectangle is 23 m and its perimeter is 206 m, what is its area?
(A) 2510
(B) 2535
(C) 2530
(D) 2515
(E) 2520
Rationale: length = breadth +23. Therefore,
4× breadth +2×23=206 m
⇒ breadth =40 m
length =40+23=63 m
Area =63×40=2520 m2
Answer is E.
Answer: (E) 2520
Question: A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit.
(A) Rs. 1900
(B) Rs. 2660
(C) Rs. 2800
(D) Rs. 2840
(E) Rs. 2440
Rationale: For managing, A received = 5% of Rs. 7400 = Rs. 370.
Balance = Rs. (7400 - 370) = Rs. 7030.
Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3)
= 39000 : 42000 : 30000
= 13 : 14 : 10
B's share = Rs.	7030 x14	= Rs. 2660
ANSWER:B
Answer: (B) Rs. 2660
Question: A classroom has 3 electric lamps. From a collocation of 10 electric bulbs of which 6 are good , 3 are selected at random and put in the lamps. Find the probability that the room is lighted.
(A) 1/6
(B) 4 / 120
(C) 29 / 30
(D) 1/9
(E) None of these
Rationale: Explanation :
Out of 10 electric bulbs, 3 are selected in 10C3 ways.
So, number of possible outcomes = T = 10C3 =120.
The room is lighted i.e at least one fo the 3 bulbs chosen at random is a good bulb.
Wherever 'at least one' is present, it is easier to do the problem using complement of the event(i.e room is dark and all 3 bulbs are bad ).
P(room is dark) = F / T = P(none of the bulbs is good)
=> 4C3/120. [since 4 bulbs are bad ].
P (room is lighted )
= 1 - ( 4C3 / 120 ). [ room is lighted if at least one bulb is good ].
or P (at least one)
= 1 - (4 / 120).
= 29 / 30.
Answer : C
Answer: (C) 29 / 30
Question: It takes 30 days to fill a laboratory dish with bacteria. If the size of the bacteria doubles each day, how long did it take for the bacteria to fill one eighth of the dish?
(A) 10 days
(B) 15 days
(C) 27 days
(D) 28 days
(E) 29 days
Rationale: The bacteria doubles each day, so after 29 days, the dish was half full.
After 28 days, the dish was one quarter full.
After 27 days, the dish was one eighth full.
The answer is C.
Answer: (C) 27 days
Question: Which of the following is closest in value to (9^13)-(9^2)?
(A) 9^9
(B) 9^8
(C) 9^13
(D) 9^6
(E) 9^5
Rationale: 9^13 is much larger number than 9^2 and as we need approximate value then it's pretty much negligible in this case, so: (9^13)-(9^2) = ~9^9.
Answer: C.
Similar questions to practice:
Hope it helps.
Answer: (C) 9^13
Question: 325325 is a six digit number. It is divisible by
(A) 7 only
(B) 11 only
(C) 13 only
(D) all 7,11 and 13
(E) None
Rationale: Sol.
Clearly, 325325 is divisible by all 7, 11 and 13.
Answer D
Answer: (D) all 7,11 and 13
Question: The length and breadth of a rectangle is increased by 10% and 25% respectively. What is the increase in the area?
(A) 37.7%
(B) 37.5%
(C) 87.5%
(D) 37.54%
(E) 32.5%
Rationale: 100 * 100 = 10000
110 * 125 = 13750
-----------
3750
10000 ------ 3750
100 ------- ? => 37.5%
Answer:B
Answer: (B) 37.5%
Question: Tom opened a shop investing Rs.500. Jose joined him 2 months later, investing Rs.350. They earned a profit of Rs.1720 after completion of one year. What will be Jose's share of profit?
(A) 80
(B) 20
(C) 30
(D) 41
(E) 15
Rationale: SoL = ~s- So Anju’s share = [5/9] x 1720 = 80
A
Answer: (A) 80
Question: Which of the following cannot be the sum of 3 different odd numbers?
(A) 10
(B) 13
(C) 17
(D) 15
(E) 19
Rationale: Option A: 10 as odd + odd + odd = odd
Correct Option: A
Answer: (A) 10
Question: A sum of money invested at C.I. amounts to Rs. 800 in 3 years to Rs. 880 in 4 years. The rate of interest per annum is?
(A) 2 1/2 %
(B) 4%
(C) 10%
(D) 6 2/3 %
(E) 6%
Rationale: S.I. on Rs. 800 for 1 year = (880 - 800) = Rs. 80 Rate = (100 * 80)/(800 * 1) = 10%
ANSWER:C
Answer: (C) 10%
Question: The average age 40 teachers of a school is 35 years. Out of these, the average age of 25 teachers is 30 years and that of the other 15 teachers is 39 years. The age of the 40th teacher is?
(A) 65
(B) 69
(C) 61
(D) 62
(E) 72
Rationale: Age of the 40th teacher= 40*35 - (25*30+15*39) = 1400-1335= 65 years
Answer is A
Answer: (A) 65
Question: When a 186 meters long rod is cut down into small pieces of length 2.0 meters each. Then how many pieces are available ?
(A) 52
(B) 68
(C) 62
(D) 67
(E) 69
Rationale: Answer
No. of pieces = Total length / Length of each piece
= 186/3.0
= 62
Option: C
Answer: (C) 62
Question: Abhijit started a business investing 70000. Anuja joined him after six months with an amount of 105000 and Sunil joined them with 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Abhijit, Anuja and Sunil respectively, three years after Abhijit started the business?
(A) 42 : 45 : 56
(B) 7 : 6 : 10
(C) 12 : 15 : 16
(D) Cannot be determined
(E) None of these
Rationale: Ratio of their investments
= 70 × 36 : 105 × 30 : 140 × 24 = 12 : 15 : 16
Answer C
Answer: (C) 12 : 15 : 16
Question: The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% of a mean score of 31. The mean score of remaining 55% is
(A) 40.8 approx
(B) 45.3 approx
(C) 51.4 approx
(D) 53.8 approx
(E) None
Rationale: Sol.
Let the required mean score be x. Then,
20 × 80 + 25 × 31 + 55 × x = 52 × 100
⇔ 1600 + 775 + 55x = 5200
⇔ 55x = 2825
⇔ x = 565 / 11 = 51.4.
Answer C
Answer: (C) 51.4 approx
Question: The sum of a number and its square is 1406. What is the number?
(A) 1409
(B) 1402
(C) 1221
(D) 1406
(E) 8828
Rationale: Explanation:
By option Verification 37 + 37 × 37 = 1406
Answer: D
Answer: (D) 1406
Question: How many seconds does a train 125 meters long, traveling at 54 km/h, take to completely cross a bridge of 205 meters in length?
(A) 18
(B) 20
(C) 22
(D) 24
(E) 26
Rationale: 54 km/h = 54000/3600 = 15 m/s
Time = 330 / 15 = 22 seconds
The answer is C.
Answer: (C) 22
Question: A bag making machine can make 20 bags per hour. If the company adjusts the machine to produce twice the amount per hour to satisfy expected rise in demand. How many bags can the machine make in the next 5 days if it works 10 hours a day?
(A) 2,000
(B) 1,000
(C) 400
(D) 50
(E) 1,500
Rationale: 20*2= 40 bags per hour.
5 days* 10 hours* 40 bags per hour= 2,000 bags.
Answer is option A.
Answer: (A) 2,000
Question: The length of the bridge, which a train 170 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
(A) 205
(B) 210
(C) 215
(D) 250
(E) 260
Rationale: Speed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (170 + x)/30 = 25/2 => 2(170 + x) = 750 => x = 205 m.
Answer: Option A
Answer: (A) 205
Question: A can do a work in 25 days and B can do it in 10 days. In how many days A and B can do the work?
(A) 20 days
(B) 10 days
(C) 6 8/7days
(D) 8 days
(E) 7 days
Rationale: Explanation:
A's 1day's work = 1/25
B's 1day's work = 1/10
They work together = 1/25 + 1/10 = 7/50 = 50/7 = 6 8/7days
Answer: Option C
Answer: (C) 6 8/7days
Question: A train 200m long passes a standing man in 10 seconds. What is the speed of the train?
(A) 30km/hr
(B) 36km/hr
(C) 42km/hr
(D) 50km/hr
(E) 72km/hr
Rationale: Speed of the train = 200/10* 18/5 = 72km/hr
Answer is E
Answer: (E) 72km/hr
Question: If books bought at prices ranging from Rs.200 to Rs.350 are sold at prices ranging from Rs.300 to Rs.425,what is the greatest possible profit that might be made in selling eight books ?
(A) Rs.400
(B) Rs.600
(C) Cannot be determined
(D) None of these
(E) Rs.800
Rationale: Solution
Least C.P. = Rs.(200 8) = Rs.1600. Greatest S.P. = Rs.(425 8) = Rs.3400.
Required profit = Rs.(3400 - 1600) = Rs.1800. Answer D
Answer: (D) None of these
Question: Amar takes as much time in running 18 meters as a car takes in covering 48 meters. What will be the distance covered by Amar during the time the car covers 1.6 km?
(A) 339
(B) 278
(C) 289
(D) 600
(E) 821
Rationale: Distance covered by Amar = 18/4.8 (1.6km)
= 3/8(1600) = 600 m
Answer:D
Answer: (D) 600
Question: Find the value of x from the below equation: x^2−7x+10=0
(A) 5 or 2
(B) 3 or 5
(C) 3 or 7
(D) 4 or 6
(E) None
Rationale: Here we need to find out a and b such that a + b = -7 and ab = +10
a = -5 and b = -2 satisfies the above condition.
Hence
x^2−7x+10=(x−5)(x−2)
x2−7x+10=(x−5)(x−2)
x^2−7x+10
=0
⇒(x−5)(x−2)
=0
x2−7x+10=0⇒(x−5)(x−2)=0
Step 3: Equate each factor to 0 and solve the equations
(x−5)(x−2)
=0
⇒(x−5)
=0or (x−2)=0
⇒x= 5 or 2
A
Answer: (A) 5 or 2
Question: Vishal invested 10% more than Trishul. Trishul invested 10% less than Raghu. If the total sum of their investments is Rs. 7225, how much amount did Raghu invest ?
(A) 1287
(B) 2887
(C) 2500
(D) 1129
(E) 1192
Rationale: Let money invested by Raghu = Rs. x
Money invested by Trishul = 9/10 x = 0.9x
Money invested by Vishal = 9/10x * 110/100 = 0.99x
Also, x+0.9x+0.99x = 7225
= x= 7225/2.89 = 2500
Therefore, amount invested by Raghu is Rs. 2500.
Answer: C
Answer: (C) 2500
Question: In how many ways can the letters of the word RAVED be arranged in a row so that the vowels appear in alphabetic order?
(A) 60
(B) 30
(C) 90
(D) 120
(E) 180
Rationale: Two letters can be arranged in 2! ways.
only one combination EIU is required.
5 letters can be arranged in 5! ways.
thus 5!/ 2! * 1 = 60.
A
Answer: (A) 60
Question: If a man lost 4% by selling oranges at the rate of 72 a rupee at how many a rupee must he sell them to gain 44%?
(A) 41
(B) 48
(C) 49
(D) 44
(E) 43
Rationale: 96% ---- 72
144% ---- ?
96/144 * 72 = 48
Answer: B
Answer: (B) 48
Question: In a meeting of 3 representatives from each of 5 different companies, each person shook hands with every person not from his or her own company. If the representatives did not shake hands with people from their own company, how many handshakes took place?
(A) 125
(B) 120
(C) 110
(D) 90
(E) 100
Rationale: In fact, all 15 people will shake hands with 12 others.
So, it SEEMS like the TOTAL number of handshakes = (15)(12)
HOWEVER, we need to keep in mind that we have counted each handshake TWICE.
To account for this DUPLICATION, we must divide (15)(12) by 2.
So, the TOTAL # of handshakes = (15)(12)/2 = 90
Answer : D
Answer: (D) 90
Question: Twelve men and six women together can complete a piece of work in four days. The work done by a women in one day is half the work done by a man in one day. If 12 men and six women started working and after two days, six men left and six women joined, then in hoe many more days will the work be completed?
(A) 1 (1/2) days
(B) 5 (1/2) days
(C) 3 (1/2) days
(D) 7 (1/2) days
(E) 2 (1/2) days
Rationale: E
2 (1/2) days
Work done by a women in one day = 1/2 (work done by a man/day)
One women's capacity = 1/2(one man's capacity)
One man = 2 women.
12 men = 24 women.
12 men + 6 women = 30 women
30 women can complete the work in four days. In the first 2 days they can complete 1/2 of the work. Remaining part of the work = 1/2. If 6 men leave and 6 new women join, then new work force = 30 women - 12 women + 6 women = 24 women.
Time taken by them to complete the remaining work = 1/2 (Time taken by 24 women to complete the work) = 1/2 * (30 * 4)/24 = 2 (1/2) days.
Answer: (E) 2 (1/2) days
Question: . A car covers a distance of 1028 km in 4 hours. Find its speed?
(A) 104
(B) 255
(C) 266
(D) 257
(E) 276
Rationale: 1028/4= 257 kmph
Answer: D
Answer: (D) 257
Question: At the end of the day, February 14th, a florist had 90 roses left in his shop, all of which were red, white or pink in color and either long or short-stemmed. A third of the roses were short-stemmed, 15 of which were white and 10 of which were pink. The percentage of pink roses that were short-stemmed equaled the percentage of red roses that were short-stemmed. If none of the long-stemmed roses were white, what percentage of the long-stemmed roses were red?
(A) 20%
(B) 75%
(C) 50%
(D) 33.3%
(E) 80%
Rationale: R + W + P = 90
S + L = 90
1/3 * 90 = 30
Short-Stemmed White = 15
Short-Stemmed Pink = 10
=> Short-Stemmed Red = 5
10/P = 5/R
=> R = P/2 , P=2R
So Total Long Stemmed = 60
And Long Stemmed Red + Long Stemmed Pink = 60
So Long Stemmed Red/Long Stemmed = ?
Total White = 15 (As no Long stemmed white)
=> R + 2R + 15 = 90
=> 3R = 75 and R = 25
Long Stemmed R = 25 - 5 = 20
So Long Stemmed Red/R = 20/60 = 33.3%
Answer - D
Answer: (D) 33.3%
Question: In a certain game, a player begins with a bag containing tiles numbered 1 through 11, each of which has an equal probability of being selected. The player draws one tile. If the tile is even, the player stops. If not, the player draws another tile without replacing the first. If this second tile is even, the player stops. If not, the player draws a third tile—without replacing either of the first two tiles—and then stops. What is the probability that at the conclusion of the game, the sum of the tiles that the player has drawn is odd?
(A) 5/18
(B) 13/36
(C) 3/8
(D) 5/8
(E) 23/36
Rationale: player begins with containing tiles 1 to 11 and have equal probability of selecting means
Proabability of selecting one number =1/10. here 5 even numbers and 5 odd numbers are there.
Next , player draws one title , if number is even player stops or otherwise title is odd without replacement
player draws second title.
If second title is even , player stops or title is odd without replacement player draws third title.
in third title , without replacement of first and second title, player draws and stops it.
the sum of tilte probability is odd. here two conditions are possible.
1st condition is
1st title is odd+ 2nd title is even stops= probability of selecting one title is 1/10*5c1.
Here are we are not selecting 1st condition as even stops because sum of tile is odd.
Here 5 odd numbers are there we can select 1out of 5 odd numbers.
without replacement of fist we select second tilte is even. is 5/10*5c1/9c1.
here we are selecting one number out of remaining 9 numbers. so probability is 5/18.
we are selecting 1 even number out of 5.
2nd condition is 1stodd without replacement,2nd odd without replacement and 3rd one also odd to get odd as sum of title.
Then probability is 5/10*4/9*3/8=1/12.
Finally sum of probability of two conditions is 5/18+1/12
=23/36.
so option E is correct.
Answer: (E) 23/36
Question: If 0 is the remainder when m is divided by 4, what is the remainder when 6m is divided by 4?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
Rationale: For any m that is divisible by 4, any multiple of m (such as 6m) will also be divisible by 4.
m = 4*a
6m = 6*4*a
The answer is, therefore, (A).
Answer: (A) 0
Question: Which of the following is a pair of co–primes?
(A) (16, 62)
(B) (18, 25)
(C) (21, 35)
(D) (23, 92)
(E) none
Rationale: Solution
H.C.F. of 18 and 25 is 1. So, they are co-primes.
Answer B
Answer: (B) (18, 25)
Question: Krishan and Nandan jointly started a business. Krishan invested three times as Nandan did and invested his money for double time as compared to Nandan. Nandan earned Rs. 4000. If the gain is proportional to the money invested and the time for which the money is invested then the total gain was?
(A) 28,009
(B) 28,029
(C) 28,019
(D) 28,000
(E) 28,0012
Rationale: 3:1
2:1
------
6:1
1 ----- 4000
7 ----- ? => Rs.28,000
Answer: D
Answer: (D) 28,000
Question: One type of liquid contains 10% of water and the second type of liquid contains 15% of water. A glass is filled with 5 parts of 1st liquid and 2 parts of 2nd liquid. The percentage of water in the new mixture in the glass is?
(A) 5.6%
(B) 11.4%
(C) 15.6%
(D) 21.3%
(E) 18.9%
Rationale: Required percentage = (10% of 5 + 15% of 2)/5+2 *100
= 0.8/7 * 100 = 11.4%
Answer is B
Answer: (B) 11.4%
Question: Which of the following numbers is closest to 100 on the number line?
(A) 98
(B) 95
(C) 103
(D) 107
(E) 110
Rationale: We place each of the five answers and 100 on a number line.
Of the five answers given, the two closest numbers to 100 are 98 and 103.
Since 98 is 2 units away from 100 and 103 is 3 units away from 100, then 98 is closest to 100.
Since the smallest positive dierence is 2, then 98 is the closest to 100 on the number line
correct answer A
Answer: (A) 98
Question: Three numbers are in the ratio 3:5:7. The largest number is 35. What is the difference between smallest and largest number?
(A) 16
(B) 18
(C) 20
(D) 22
(E) 24
Rationale: The three numbers are 3x, 5x, and 7x.
The largest number is 35 = 7*5, so x = 5.
The smallest number is 3*5 = 15.
35 - 15 = 20
The answer is C.
Answer: (C) 20
Question: A bus starts from city X. The number of women in the bus is half of the number of men. In city Y, 8 men leave the bus and four women enter. Now, number of men and women is equal. In the beginning, how many passengers entered the bus ?
(A) 15
(B) 30
(C) 36
(D) 45
(E) 46
Rationale: Explanation:
Originally, let number of women = x. Then, number of men = 2x.
So, in city Y, we have : (2x - 8) = (x + 4) or x = 12.
Therefore Total number of passengers in the beginning = (x + 2x) = 3x = 36.
Answer: C
Answer: (C) 36
Question: Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 13 cm
(A) 178 cm2
(B) 179 cm2
(C) 285 cm2
(D) 247 cm2
(E) 197 cm2
Rationale: Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them)
= 1/2 (20 + 18) * (13)
= 247 cm2
Answer: D
Answer: (D) 247 cm2
Question: It takes Avery 2.5 hours to build a brick wall while Tom can do it in 4 hours. If the two start working together and after an hour Avery leaves, how much time will it take Tom to complete the wall on his own?
(A) 1 hour and 24 minutes.
(B) 50 minutes.
(C) 1 hour and 40 minutes.
(D) 2 hours and 10 minutes
(E) 1 hour and 20 minutes
Rationale: Avery's efficiency is 100/2.5 = 40%
Tom's = 100/4= 25%
they worked together for 1 hour and finished 65% of the job
remaining = 35 %
Tom will complete 25 % in 60 minutes , and
10% in 60*10/25 minutes = 24 minutes
Time taken by Tom to finish the remaining on his own = 60+24 = 84 minutes
ANSWER:A
Answer: (A) 1 hour and 24 minutes.
Question: The circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 45 ft long. What is the distance traveled by the cart, when the front wheel has done five more revolutions than the rear wheel?
(A) 20 ft
(B) 25 ft
(C) 450 ft
(D) 900 ft
(E) 1000 ft
Rationale: Point to note: Both the wheels would have traveled the same distance.
Now consider, no. of revolutions made by back wheel as x, which implies that the number of revolutions made by the front wheel is (x+5).
Equating the distance traveled by front wheel to back wheel: (x+5)*30 = x * 45. (Formula for calculating the distance traveled by each wheel is: # of revolutions * circumference.)
Solving this eqn. gives x=10.
Sub x=10 either in (x+5)*30 or in x * 45 to get the distance, which is 450. So the correct choice is C.
Answer: (C) 450 ft