mathewhe/aqua_rat huggingface dataset:
Question: The original price of a car was $25,200. Because the car owner thought he could get more money for the car, he increased the price of the car to 110% of its original price. After a week, the car had not sold, so the owner then discount ed the price by 10%, and the car was finally sold. What price was the car sold for?
(A) $25,200
(B) $25,000
(C) $24,948
(D) $24,542
(E) $23,658
Rationale: The car is sold for 1.1*0.9*25200
=0.99 * 25200
=(1 - 0.01) * 25200
=25200 - 252
=24948
C is the answer
Answer: (C) $24,948
Question: Each student at a certain university is given a four-character identification code, the rest two characters of which are digits between 0 and 9, inclusive, and the last two characters of which are selected from the 26 letters of the alphabet. If characters may be repeated and the same characters used in a different order constitute a different code, how many different identification codes can be generated following these rules?
(A) 135,200
(B) 67,600
(C) 64,000
(D) 60,840
(E) 58,500
Rationale: First two parts for the code, are digits between 0-9, therefore, 10 options for the first part of the code,
and as characters may be repeated, 10 options for the second part as well
Therefore, we have 10 X 10 possibilities for the first and second part of the code
Last two parts of the code, are characters selected from the 26 letters of the alphabet, therefore, 26 options for the third part of the code,
and as characters may be repeated, 26 options for the fourth part as well
Therefore, we have 26 X 26 possibilities for the third and fourth part of the code
so, in all total no. of different identification codes generated following these rules
= 10 X 10 X 26 X 26 = 67600
Answer B
Answer: (B) 67,600
Question: Pipe A can fill a tank in 2 hr, pipe B in 4hr and pipe C in 8hr. If all the pipes are open in how many hours will tank be filled?
(A) 1hr 27min
(B) 2hr 19min
(C) 3hr 14min
(D) 3hr
(E) 5hr 45min
Rationale: Part filled by (A+B+C) in 1hour = 1/2 + 1/4 + 1/8 = 7/8
All the 3 pipes together will fill the tank in 1hour 27min
Answer is A
Answer: (A) 1hr 27min
Question: Find large number from below question The difference of two numbers is 1375. On dividing the larger number by the smaller, we get 6 as quotient and the 15 as remainder
(A) 1235
(B) 1345
(C) 1678
(D) 1767
(E) 1647
Rationale: Let the smaller number be x. Then larger number = (x + 1375).
x + 1375 = 6x + 15
5x = 1360
x = 272
Large number = 272+1375 = 1647
E
Answer: (E) 1647
Question: The H.C.F of two numbers is 12 and their L.C.M is 6600. If one of the numbers is 288, then the other is?
(A) 255
(B) 260
(C) 265
(D) 270
(E) 275
Rationale: Other number = (12 * 6600)/288
= 275.
Answer: E
Answer: (E) 275
Question: Varma can read a book in k minutes. What part of the book can he read in 8 minutes?(k>8)
(A) (k - 8)/k
(B) k/8
(C) 8 + k
(D) k-8
(E) 8/k
Rationale: Option E
Explanation:
Part of the book he can read in 1 minute = 1/k
Part of the book he can read in 8 minutes =8/k.
Answer: (E) 8/k
Question: If Rs. 527 be divided among A, B, C in such a way that A gets 2/3 of what B gets and B gets 1/4 of what C gets, then their shares are respectively?
(A) s.300
(B) s.360
(C) s.389
(D) s.368
(E) s.372
Rationale: (A = 2/3 B and B = 1/4 C) = A/B = 2/3 and B/C = 1/4
A:B = 2:3 and B:C = 1:4 = 3:12
A:B:C = 2:3:12
A;s share = 527 * 2/17 = Rs. 62
B's share = 527 * 3/17 = Rs. 93
C's share = 527 * 12/17 =Rs.372.
Answer:E
Answer: (E) s.372
Question: If six numbers are randomly selected without replacement from set {1, 2, 3, 4, 5, 6}, what is the probability that the numbers are selected in ascending order?
(A) 1/280
(B) 1/640
(C) 1/720
(D) 1/240
(E) 1/120
Rationale: The total number of ways to arrange 6 numbers is 6! = 720.
There is only one way to arrange the numbers in ascending order.
P(ascending order) = 1/720
The answer is C.
Answer: (C) 1/720
Question: A man took loan from a bank at the rate of 12% p.a. S.I. After 20 years he had to pay Rs. 2100 interest only for the period. The principal amount borrowed by him was?
(A) 600
(B) 715
(C) 600
(D) 875
(E) 900
Rationale: Principal = (100 * 2100) / (12 * 20)
= Rs.875
Answer: D
Answer: (D) 875
Question: In a camp,there is a meal for 120 men or 200 children.If 150 children have taken the meal,how many men will be catered to with the remaining meal ?
(A) 20
(B) 30
(C) 40
(D) 50
(E) 60
Rationale: Solution
There is a metal for 200 children. 150 children have taken the meal.
Remaining meal is to be catered to 50 children.
Now, 200 children = 120 men
50 children =( 120/200x50)men= 30 men. Answer B
Answer: (B) 30
Question: The ratio of men to women in a class is 6 to 5. If 3 men drop out of the class and the ratio of men to women after the drop out is 1 to 1, then how many men are currently in the class?
(A) 13
(B) 14
(C) 15
(D) 18
(E) 22
Rationale: Equation on the basis of question
6x-3/5x=1/1
x=3
Original number of men = 6*3=18
Original Number of Women = 5*3=15
Current number of men = 18-3=15 (Answer)
Current number of women = 15
Ratio = 15/15= 1/1 ( Just to verify)
Ans C
Answer: (C) 15
Question: 40% of a number is added to 120, the result is the same number. Find the number?
(A) 300
(B) 277
(C) 266
(D) 200
(E) 121
Rationale: :
(40/100) * X + 120 = X
2X = 400
X = 200
Answer: D
Answer: (D) 200
Question: In a shop, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, find out approximately what percentage of the selling price is the profit?
(A) 180 %
(B) 120 %
(C) 90 %
(D) 80 %
(E) 70 %
Rationale: Let the CP = 100
Profit=(320 / 100) ×100=320
SP=CP+Profit=100+320=420
If the cost increases by 25%, New CP=(125/100) ×100=125
Selling Price is constant, hence New SP=420
Profit=SP–CP=420–125=295
Required Percentage =(295/420)×100
=2950 / 42
=1475 / 21 ≈ 70
ANSWER:E
Answer: (E) 70 %
Question: Two cars are driving toward each other. The first car is traveling at a speed of 60 km/h, which is 25% slower than the second car's speed. If the distance between the cars is 770 km, how many hours will it take until the two cars meet?
(A) 4
(B) 4.5
(C) 5
(D) 5.5
(E) 6
Rationale: The speed of the first car is 60 km/h.
The speed of the second car is 60 / 0.75 = 80 km/h.
The two cars complete a total of 140 km each hour.
The time it takes the cars to meet is 770 / 140 = 5.5 hours.
The answer is D.
Answer: (D) 5.5
Question: Aishwarya Rai walks into a bank to cash out her check.
By mistake the bank teller gives her dollar amount in change, and her cent amount in dollars.
On the way home she spends 5 cent, and then suddenly she notices that she has twice the amount of her check.
How much was her check amount ?
(A) 21.55
(B) 25.89
(C) 30.56
(D) 31.55
(E) 31.63
Rationale: E
The check was for dollars 31.63.
The bank teller gave her dollars 63.31
She spent .05, and then she had dollars 63.26, which is twice the check.
Let x be the dollars of the check, and y be the cent.
The check was for 100x + y cent
He was given 100y + x cent
Also
100y + x - 5 = 2(100x + y)
Expanding this out and rearranging, we find:
98y = 199x + 5
or 199x ≡ -5 (mod 98)
or 98*2*x + 3x ≡ -5 (mod 98)
3x ≡ -5 ≡ 93 (mod 98)
this quickly leads to x = 31
Answer: (E) 31.63
Question: Insert the missing number. 2, 7, 10, 22, 18, 37, 26
(A) 42
(B) 52
(C) 46
(D) 62
(E) 68
Rationale: Explanation :
There are two series here
2, 10, 18, 26, ... (Increase by 8)
7, 22, 37, ... (Increase by 15)
Hence, next term is 37+15 = 52
Answer : Option B
Answer: (B) 52
Question: Working at a constant rate, P can finish a job in 4 hours. Q, also working at a constant rate, can finish the same job in 15 hours. If they work together for 3 hours, how many more minutes will it take P to finish the job, working alone at his constant rate?
(A) 5
(B) 8
(C) 10
(D) 12
(E) 15
Rationale: Each hour they complete 1/4 + 1/15 = 19/60 of the job.
In 3 hours, they complete 3(19/60) = 19/20 of the job.
The time for P to finish is (1/20) / (1/4) = (1/5) hour = 12 minutes
The answer is D.
Answer: (D) 12
Question: 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter. The length if the wire in meters will be:
(A) 76 m
(B) 80 m
(C) 84 m
(D) 88 m
(E) None of these
Rationale: Explanation:
Let the length of the wire be h
Radius=1/2mm=120cm
πr2h=66
22/7∗1/20∗1/20∗h=66
=>h=66∗20∗20∗7/22=8400cm=84m
Option C
Answer: (C) 84 m
Question: What is the sum of the first 23 natural numbers (starting from 1)?
(A) 256
(B) 266
(C) 276
(D) 286
(E) 296
Rationale: The sum of n natural numbers = n*(n+1) / 2 = 23*24/2 = 276
The answer is C.
Answer: (C) 276
Question: If the price of sugar rises from Rs. 6 per kg to Rs. 7.50 per kg, a person, to have no increase in his expenditure on sugar, will have to reduce his consumption of sugar by
(A) 10%
(B) 20%
(C) 25%
(D) 30%
(E) None
Rationale: Solution
Let the original consumption = 100 kg and new consumption = x kg.
So, 100 x 6 = x × 7.50
x = 80 kg.
Reduction in consumption = 20%.
Answer B
Answer: (B) 20%
Question: Of 40 applicants for a job, 32 had at least 5 years of prior work experience, 24 had advanced degrees, and 16 had at least 5 years of prior work experience and advanced degrees. How many of the applicants had neither 5 years of prior work experience nor advanced degrees?
(A) 0
(B) 2
(C) 4
(D) 8
(E) 16
Rationale: soln:
The 16 with both experience and degrees are the first group;
we can subtract the 16 from the 24 with advanced degrees to determine that 8 have advanced degrees but not 5 years of experience,
we can subtract the 16 from the 32 with experience to determine that there are 16 applicants with 5 years of experience but no advanced degrees
We can then subtract these three groups from the total pool of 40 applicants to determine how many lack both
experience and advanced degrees; 40-16-16-8=0,
so there were no such applicants among the 40
ANSWER:A
Answer: (A) 0
Question: What is the ratio whose term differ by 30 and the measure of which is 2/7?
(A) A)12:42
(B) B)16:56
(C) C)71:85
(D) D)32:39
(E) E)41:52
Rationale: Let the ratio be x:(x+30)
Then, x/(x+30) = 2/7
x=12
required ratio = 12:42
Answer is A
Answer: (A) A)12:42
Question: There are some pigeons and hares in a zoo. If heads are counted, there are 200. If legs are counted, there are 530. The number of hares in the zoo is?
(A) 90
(B) 99
(C) 77
(D) 65
(E) 221
Rationale: 200*2 = 400
530
-----
130
1----2
?----130 = 65
Answer:D
Answer: (D) 65
Question: On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
(A) Rs. 20
(B) Rs. 60.
(C) Rs. 40
(D) Rs. 30
(E) Rs. 90
Rationale: (C.P. of 17 balls) - (S.P. of 17 balls) = (C.P. of 5 balls)
C.P. of 12 balls = S.P. of 17 balls = Rs.720.
C.P. of 1 ball = Rs. 720/12 = Rs. 60.
Answer:B
Answer: (B) Rs. 60.
Question: What sum of money put at C.I amounts in 2 years to Rs.8840 and in 3 years to Rs.9261?
(A) 8054
(B) 8877
(C) 2877
(D) 2678
(E) 1011
Rationale: 8840 ---- 421
100 ---- ? => 4.76%
x *104.76/100 * 104.76/100 = 8840
x*1.0975=8840
x=8840/1.0975 => 8054.54
Answer: A
Answer: (A) 8054
Question: The population of a bacteria colony doubles every day. If it was started 3 days ago with 3 bacteria and each bacteria lives for 12 days, how large is the colony today?
(A) 512
(B) 1024
(C) 768
(D) 2048
(E) 4096
Rationale: 9 days ago- 2
8 days ago- 4
7 days ago- 8
6 days ago- 16
5 days ago- 32
4 days ago- 64
3 days ago- 128
2 days ago- 256
Yesterday- 512
Today- 1024
Answer:B
Answer: (B) 1024
Question: Two travelers set out on a long odyssey. The first traveler starts from city X and travels north on a certain day and covers 1 km on the first day and on subsequent days, he travels 2 km more than the previous day. After 3 days, a second traveler sets out from city X in the same direction as the first traveler and on his first day he travels 12 km and on subsequent days he travels 1 km more than the previous day. On how many days will the second traveler be ahead of the first?
(A) 2 days
(B) 6 days
(C) From the 2nd day after the 2nd traveler starts
(D) From the 3rd day after the 2nd traveler starts
(E) None
Rationale: Explanatory Answer
Travel Pattern of Traveller 2
Day 1: 1 km; Day 2: (1 + 2) = 3 km; Day 3: (1 + 2 + 2) = 5km and so on
i.e. on each of these days he covers the following distance Day 1: 1km, Day 2: 3km, Day 3: 5km and so on.
The distances covered each day by the first traveller is in an AP with the first term being 1 and the common difference being 2
After 3 days, the first traveler would have traveled 9 kms. He is ahead of the second traveler at the end of 3 days by 9 kms.
Travel Pattern of Traveller 2
The second traveler travels as follows: On his first day he travels 12 kms, his second day he travels 13 kms and so on.
The distances covered each day by the second traveller is in an AP with the first term being 12 and the common difference being 1
Step 2
Compute total distance covered by each of the travellers
Let the number of days travelled by the second traveller be n.
Therefore, the first traveller would have travelled for (n + 3) days. Traveler 1 has covered a total distance of 1+ 3 + 5 + 7 + ..... + (n + 3) km.
The total distance covered by traveler 1 is the summation of the AP.
Sum of an AP = (n/2)(2a1 + (n-1)d)
Total distance travelled by traveller 1 in (n + 3) days = n+3/2[2(1)+(n+3−1)(2)]
The above term simplifies as (n + 3)2
The total distance covered by the second traveler = 12 + 13 + 14 + ... + n terms
Total distance travelled by traveller 2 in n days = n2[2(12)+(n−1)(1)] = n(n+23)2
On the days that the second traveller is ahead of the first one, the total distance covered by the second traveller till that day will be more than the total distance covered by the first traveller.
Note: All values calculated are with respect to the day from which the second traveller started travelling
Therefore, let us solve the inequality n(n+23)/2 > (n + 3)2
n2 + 23n > 2(n2 + 6n + 9)
Or n2 - 11n + 18 < 0
The quadratic factorizes as (n - 2)(n - 9) < 0
Possibility 1: (n - 2) > 0 AND (n - 9) < 0
i.e., n > 2 AND n < 9
Therefore, 2 < n < 9
Possibility 2: (n - 2) < 0 AND (n - 9) > 0
i.e., n < 2 AND n > 9
Infeasible solution
Therefore, starting the 3rd day and up to the 8th day the second traveller was ahead of the first traveller
The correct answer is Choice (2). 6 days
Answer B
Answer: (B) 6 days
Question: After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member?
(A) 2 years
(B) 4 years
(C) 8 years
(D) 15 years
(E) 17 years
Rationale: Solution
Age decreased	= (5 x 3) years
15 years
So, required difference =15years.
Answer D
Answer: (D) 15 years
Question: At a certain resort, each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants. The 3 restaurants are the family buffet, the dining room, and the snack bar. Exactly 19 employees are trained to work in the family buffet, 16 are trained to work in the dining room, and 12 are trained to work in the snack bar. If 4 employees are trained to work in exactly 2 restaurants, how many employees are trained to work in all 3 restaurants?
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6
Rationale: 39 = 19 + 16 + 12 - 4 - 2x
2x = 19 + 16 + 12 - 4 - 39
= 43- 39
=4
x = 2
A
Answer: (A) 2
Question: If log1087.5 = 5.9421, then the number of digits in (875)10 is?
(A) 30
(B) 70
(C) 27
(D) 26
(E) 25
Rationale: X = (875)10 = (87.5 x 10)10
Therefore, log10X = 10(log1087.5 + 1)
= 10(5.9421 + 1)
= 10(6.9421) = 69.421
X = antilog(69.421)
Therefore, number of digits in X = 70.
ANSWER:B
Answer: (B) 70
Question: How many integers between 323,700 and 458,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,349
Rationale: There is one number in hundred with 1 in the tens digit and 3 in the units digit:13, 113, 213, 313, ...
The difference between 323,700 and 458,600 is 458,600-323,700=134,900 - one number per each hundred gives 134,900/100=1,349 numbers.
Answer: E.
Answer: (E) 1,349
Question: The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm. Find the ratio of the breadth and the area of the rectangle?
(A) 1:96
(B) 1:98
(C) 1:95
(D) 1:45
(E) 1:92
Rationale: Let the length and the breadth of the rectangle be 4x cm and 3x respectively.
(4x)(3x) = 6912
12x2 = 6912
x2 = 576 = 4 * 144 = 22 * 122 (x > 0)
=> x = 2 * 12 = 2
Ratio of the breadth and the areas = 3x : 12x2 = 1 : 4x =1:96.
Answer: A
Answer: (A) 1:96
Question: A man invested Rs. 4940 in Rs. 10 shares quoted at Rs. 9.50. If the rate of dividend be 14%, his annual income is
(A) 708
(B) 718
(C) 738
(D) 748
(E) 728
Rationale: Market Value of a share = Rs.9.50
Investment = Rs.4940
Number of shares = 4940/9.50 = 520
Face Value of a share = Rs.10
dividend = 14%
dividend per share = (10×14)/100 = Rs. 1.4
His annual income = 520 × 1.4 = Rs.728
Answer is E.
Answer: (E) 728
Question: In how many ways can the letters of the word REGULATION be arranged such that all the vowels always appear together?
(A) 10!*5!
(B) 10!*6!
(C) 10!
(D) 6!*5!
(E) 5!*5!
Rationale: The 10 letters can be grouped into 5 consonants and one set of 5 vowels.
The number of ways to arrange 6 units is 6!
Then, for each arrangement, the 5 vowels can be arranged in 5! ways.
The total number of arrangements is 6!*5!
The answer is D.
Answer: (D) 6!*5!
Question: Each digit 1 through 5 is used exactly once to create a 5-digit integer. If the 3 and the24 cannot be adjacent digits in the integer, how many 5-digit integers are possible?
(A)  48
(B)  72
(C)  76
(D)  78
(E)  90
Rationale: Number of arrangements using 5 distinct digits = 5!
Number of arrangements in which 3 and 4 are adjacent - consider 3 and 2 together as one group. Now you have 4 numbers/groups to arrange which can be done in 4! ways. In each of these arrangements, 3 and 2 can be arranged as 32 or 23.
Number of arrangements in which 3 and 4 are not adjacent = 5! - 2*4! = 72
Answer : B
Answer: (B)  72
Question: A girl has exactly 6 quarters, 8 dimes, and 10 nickels in her pocket. If she withdraws a number of these coins at random, how many coins would she have to withdraw to ensure that she has at least a 50 percent chance of withdrawing at least one quarter?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: When the probability of NOT drawing a quarter is 50% or less, then the probability of drawing a quarter is at least 50%.
P(not drawing a quarter with 2 coins) = 18/24 * 17/23 = 51/92 > 50%
P(not drawing a quarter with 3 coins) = 51/92 * 16/22 = 102/253 < 50%
The answer is C.
Answer: (C) 3
Question: The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
(A) 20 years
(B) 30 years
(C) 60 years
(D) 40 years
(E) 35 years
Rationale: D
40 years
Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.
Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years.
Husband's present age = (90 - 50) years = 40 years.
Answer: (D) 40 years
Question: Two trains of equal lengths are running at speeds of 30 kmph and 60 kmph. The two trains crossed each other in 30 seconds when travelling in opposite direction. In what time will they cross each other when travelling in the same direction?
(A) 90
(B) 82
(C) 16
(D) 67
(E) 12
Rationale: Let the length of each train be x m.
(x + x) / (60 + 30)5/18 = (750 * 18) / (30 * 5)
= 90 sec.
Answer: A
Answer: (A) 90
Question: It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)),
when p=(a+b+c)/2, such that a, b, c are the lengths of sides of the triangle. If the triangle has 300, 300, and 360 as the side’s lengths, what is the triangle’s area?
(A) 34,200
(B) 36,200
(C) 43,200
(D) 42,200
(E) 38,200
Rationale: It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)),
when p=(a+b+c)/2,
-> P=(300+300+360)/2=480, area=√(480(480-300)(480-300)(480-360))=43,200. Therefore, the answer is C.
Answer: (C) 43,200
Question: My brother is 3 years elder to me.My father was 28 years of age when my sister was born while my mother was 26 years of age when I was born.If my sister was 4 years of age when my brother was born,then,what was the age of my father and mother respectively when my brother was born ?
(A) 32 yrs,23 yrs
(B) 32 yrs,29 yrs
(C) 35 yrs,29 yrs
(D) 35 yrs,33 yrs
(E) None of these
Rationale: Solution
Clealy,my brother was born 3 years before I was born and 4 years after my sister was born.
So,father's age when brother was born =(28+4) years =32 years,
mother's age when brother was born =(26-3) years =23 years . Answer A
Answer: (A) 32 yrs,23 yrs
Question: Club X has more than 50 but fewer than 80 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at the table that has fewer than 6 members?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: The number of members is 4j+3 = 5k+3.
Let's start at 4(12)+3 = 51 and keep adding 4 until we find a number in the form 5k+3.
51, 55, 59, 63 = 5(12) + 3
The club has 63 members.
63/6 = 6(10) + 3
The answer is C.
Answer: (C) 3
Question: What is the smallest positive integer x, such that 7000x is a perfect cube?
(A) 4
(B) 6
(C) 8
(D) 12
(E) 18
Rationale: Take out the factors of 7000x that will come 10^3*7. for perfect cube you need every no. raise to the power 3. for 7000x to be a perfect cube, need two 7 that means 49.
E is the answer.
Answer: (D) 12
Question: A train moves fast a telegraph post and a bridge 264 m long in 8 sec and 20 sec respectively. What is the speed of the train?
(A) 79.5
(B) 39.5
(C) 79.9
(D) 79.2
(E) 79.1
Rationale: Let the length of the train be x m and its speed be y m/sec.
Then, x/y = 8 => x = 8y
(x + 264)/20 = y
y = 22
Speed = 22 m/sec = 22 * 18/5 = 79.2 km/hr.
Answer:D
Answer: (D) 79.2
Question: In a college 1/5th of the girls and 1/8th of the boys took part in a social camp. What of the total number of student in the college took part in the camp ?
(A) 13/40
(B) 13/80
(C) 2/13
(D) Data inadequate
(E) None
Rationale: Answer
Out of the 5 girls, 1 took part in the camp. Out of the 8 boys, 1 took part in the camp.
∴ 2/13 of total number of students took part in the camp.
Correct Option: C
Answer: (C) 2/13
Question: A farming field can be ploughed by 6 tractors in 4 days. When 6 tractors work together, each of them ploughs 120 hectares a day. If two of the tractors were moved to another field, then the remaining 4 tractors could plough the same field in 5 days. How many hectares a day would one tractor plough then?
(A) 144
(B) 255
(C) 148
(D) 265
(E) 320
Rationale: If each of 66 tractors ploughed 120120 hectares a day and they finished the work in 44 days, then the whole field is: 120⋅6⋅4=720⋅4=2880120⋅6⋅4=720⋅4=2880 hectares. Let's suppose that each of the four tractors ploughed xx hectares a day. Therefore in 5 days they ploughed
5⋅4⋅x=20⋅x5⋅4⋅x=20⋅x hectares, which equals the area of the whole field, 2880 hectares.
So, we get 20x=288020x=2880
x=288020=144x=288020=144. Hence, each of the four tractors would plough 144 hectares a day.
correct answer is A)144
Answer: (A) 144
Question: Using all the letters of the word "NOKIA", how many words can be formed, which begin with N and end with A?
(A) 7
(B) 6
(C) 8
(D) 9
(E) 3
Rationale: There are five letters in the given word.
Consider 5 blanks ....
The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.
The number of words = 3! = 6.
Answer:B
Answer: (B) 6
Question: Water consists of hydrogen and oxygen, and the approximate ratio, by mass, of hydrogen to oxygen is 2:16. Approximately how many grams of oxygen are there in 153 grams of water?
(A)  16
(B)  72
(C)  112
(D)  136
(E)  142
Rationale: Since the ratio, by mass, of hydrogen to oxygen is 2:16, then oxygen is 16/(2+16)=8/9 of mass of water.
Therefore there are 153*8/9=136 grams of oxygen in 153 grams of water.
Answer: D.
Answer: (D)  136