divelab/Sys2Bench huggingface dataset:
Question: A can do a piece of work in 15 days and B in 20 days. They began the work together but 5 days before the completion of the work, A leaves. The work was completed in?
Options:
A)8 days
B)10 days
C)15 days
D)11 3/7 days
E)None of these
Rationale: Explanation:
(x â€“ 5)/15 + x/20 = 1
x = 11 3/7 days
ANSWER IS D
Answer: D
Question: If log 2 = 0.30103, the number of digits in 520 is
Options:
A)14
B)16
C)18
D)25
E)26
Rationale: Solution
Log 520	=20 log 5
=20 ×[log(10/2)]
=20 (log 10 - log 2)
=20 (1 - 0.3010)
=20×0.6990
=13.9800.
Characteristics	= 13.
Hence, the number of digits in Log 5(20) is 14.
Answer A
Answer: A
Question: 60 litres of diesel is required to travel 600 km using a 800 cc engine. If the volume of diesel required to cover a distance varies directly as the capacity of the engine, then how many litres of diesel is required to travel 800 km using 1200 cc engine?
Options:
A)80 litres
B)90 litres
C)120 litres
D)170 litres
E)None of these
Rationale: Explanatory Answer
To cover a distance of 800 kms using a 800 cc engine, the amount of diesel required = 800/600*60 = 80 litres.
However, the vehicle uses a 1200 cc engine and the question states that the amount of diesel required varies directly as the engine capacity.
i.e., for instance, if the capacity of engine doubles, the diesel requirement will double too.
Therefore, with a 1200 cc engine, quantity of diesel required = 1200/800*80 = 120 litres.
Answer C
Answer: C
Question: If 2^z is a factor of 26!, which of the following could be the value of z?
I. 22
II. 23
III. 24
Options:
A)I ​ only
B)II only
C)I​ and II
D)I, II, and III
E)None
Rationale: 26! contains all numbers from 1 to 26. To find the factor with the highest power of 2, we count the number of multiples of 2, 4, 8 and 16.
Multiples of 2 = 13
Multiples of 4 = 6
Multiples of 8 = 3
Multiples of 16 = 1
Total count = 23
The answer is C.
Answer: C
Question: Find the missing number
5 : 24 :: 8 : ?
Options:
A)61
B)62
C)63
D)64
E)65
Rationale: 5^2 -1= 24
so 8^2 -1 =63
ANSWER:C
Answer: C
Question: What should be the least number to be added to the 3198 number to make it divisible by 8?
Options:
A)12
B)17
C)2
D)77
E)26
Rationale: Answer : 2
Option : C
Answer: C
Question: In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the rate in the remaining 40 overs to reach the target of 292 runs?
Options:
A)6.5
B)6.22
C)6.29
D)6.39
E)6.13
Rationale: Required run rate
= [292 - (3.2 * 10)]/40
= 260/40
= 6.5
Answer: A
Answer: A
Question: A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
Options:
A)Rs. 650
B)Rs. 690
C)Rs. 698
D)Rs. 700
E)Rs. 740
Rationale: S.I. for 1 year = Rs. (854 - 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 - 117) = Rs. 698.
Answer: Option C
Answer: C
Question: In how many years will a sum of money doubles itself at 25% per annum on simple interest?
Options:
A)4%
B)8%
C)10%
D)12%
E)15%
Rationale: P = (P*25*R)/100
R = 4%
Answer:A
Answer: A
Question: Entry fee in an exhibition was Rs. 1. Later, this was reduced by 25% which increased the sale by 20%. The percentage increase in the number of visitors is :
Options:
A)33
B)37
C)60
D)28
E)27
Rationale: Explanation:
Let the total original sale be Rs. 100. Then, original number of visitors = 100.
New number of visitors = 120/0.75 = 160.
Increase % = 60 %.
Answer: C) 60 %
Answer: C
Question: If 8 persons can do 8 times of a particular work in 8 days, then, 9 persons can do 9 times of that work in?
Options:
A)3 days
B)4 days
C)6 days
D)7 days
E)8 days
Rationale: That is, 1 person can do one time of the work in 8 days.
Therefore, 9 persons can do 9 times work in the same 8 days itself
E)
Answer: E
Question: Consider two statements
Statement 1 => If R is an integer and R=2p+1 where p is not an integer.
Statement 2 => If S is an integer and S=2q where q is not an integer
Which of the following correctly specifies the Even/Odd nature of R and S respectively
Options:
A)1) Odd,Even
B)2) Odd,Odd
C)3) Even,Odd
D)4) Even,Even
E)5) Cannot be determined
Rationale: answer will be E
R = integer
R = 2p+1 = integer here p can be any n/2 and n can be even or odd , matters alot 2 get cancelled out
if n were even R would odd
if n odd R would be even
ANSWER IS E
Answer: E
Question: A die is thrown twice and the sum of the numbers appearing is observed to be 6.
find the conditional probability that the number 4 has appeared at least once?
Options:
A)1/4
B)1/5
C)2/5
D)3/5
E)3/7
Rationale: Let E be the event that ‘number 4 appears at least once’ and F be
the event
that ‘the sum of the numbers appearing is 6’.
Then, E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4),
(5,4), (6,4)}
and F = {(1,5), (2,4), (3,3), (4,2), (5,1)}
We have P(E) = 11/36
and P (F) = 5/36
Also EnF = {(2,4), (4,2)}
Therefore P(EnF) = 2/36
the required probability
P(E|F) = P(EnF)/P(F) = (2/36)/(5/36) = 2/5.
C)
Answer: C
Question: If a card is drawn from a well shuffled pack of cards, the probability of drawing a spade or a king is -
Options:
A)8/13
B)6/13
C)3/13
D)4/13
E)1/13
Rationale: Explanation:
P(Sá´œK) = P(S) + P(K) - P(Sâˆ©K), where S denotes spade and K denotes king.
P(Sá´œK) = 13/52 + 4/52 - 1/52 = 4/13
Answer: D
Answer: D
Question: Ben
ips a fair coin repeatedly. Given that Ben's first coin
ip is heads, compute the probability
Ben
ips two heads in a row before Ben
ips two tails in a row.
Options:
A)2/3
B)3/3
C)1/3
D)1/1
E)2/2
Rationale: If Ben successfully
ips two heads in a row, then Ben must have
ipped the coin an
even number of times. If Ben successfully
ips two tails in a row, then he must have
ipped
the coin an odd number of times. However, the coin must have been
ipped at least twice. The
desired probability is therefore
(1/3)/(1/2)=2/3
correct answer A
Answer: A
Question: Suppose 5 monkeys take 5 minutes to eat 5 bananas. How many monkeys would it take to eat 15 bananas in 15 minutes?
Options:
A)9
B)10
C)11
D)5
E)13
Rationale: One monkey takes 5 min to eat 1 banana , so in 15 mins 1 monkey will eat 3 bananas , so for 15 bananas in 15 min we need 15/3 =5 monkeys
ANSWER:D
Answer: D
Question: The average of 11 numbers is 10.9. If the average of first six is 10.5 and that of the last six is 11.6 the sixth number is?
Options:
A)11.4
B)11.3
C)11.8
D)11.5
E)12.7
Rationale: 1 to 11 = 11 * 10.9 = 119.9
1 to 6 = 6 * 10.5 = 63
6 to 11 = 6 * 11.6 = 69.6
63 + 69.6 = 132.6 – 119.9 = 12.7
6th number = 12.7
Answer: E
Answer: E
Question: A cask is filled with alcohol and water in the ratio 5”3 Sixteen gallons of this are drawn off and the cask is filled with water and then the proportion of alcohol to water becomes 3:5 How many gallons does the cask hold ?
Options:
A)32
B)36
C)38
D)40
E)42
Rationale: ratio of alcohal and water is 25:15
Initially when Sixteen gallons of this are drawn off
ratio of alcohal and water is 15:9
and the cask is filled with water 15:25 which is 3:5
ANSWER:D
Answer: D
Question: The average weight of 19 students is 15kg. by the admission of a new student the average weight is reduced to 14.7 kg. the weight of the new student is ?
Options:
A)10.6 kg
B)10.8 kg
C)11 kg
D)14.9 kg
E)None
Rationale: Answer
Weight of new student = Total weight of all 20 students - Total weight of initial 19 students
= (20 x 14.7 - 19 x 15) kg
= 9 kg.
Correct Option: E
Answer: E
Question: There are 5 men and 11 women. In how many ways one can select a group of 6 with at-least 3 men?
Options:
A)1932
B)1933
C)1934
D)1935
E)1936
Rationale: (5c3*11c3)+(5c4*11c2)+(5c5*11c1)
1936
ANSWER:E
Answer: E
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 22 years. The present age of the husband is
Options:
A)40
B)36
C)45
D)55
E)65
Rationale: Explanation:
Sum of the present ages of husband, wife and child = (27 * 3 + 3 * 3) years = 90 years.
Sum of the present ages of wife and child = (22 * 2 + 5 * 2) years = 54 years.
Husband's present age = (90 - 54) years = 36 years
Option B
Answer: B
Question: Sand is poured into a box so that the box is being filled at the rate of 3 cubic feet per hour. If the empty rectangular box is 5 feet long, 4 feet wide, and 3 feet deep, approximately how many hours does it take to fill the box?
Options:
A)14
B)16
C)18
D)20
E)22
Rationale: The volume the box is: length*width*depth = 5*4*3= 60 cubic feet.
60 cubic feet / 3 cubic feet per hour = 20 hours.
It will take 20 hours to fill the box.
The answer is D.
Answer: D
Question: The present worth of a certain sum due sometime hence is Rs. 3400 and the true discount is Rs. 340. The banker's gain is:
Options:
A)Rs. 21
B)Rs. 17
C)Rs. 18
D)Rs. 34
E)None of these
Rationale: Explanation :
BG = (TD)2/PW=(340)2/3400=340 × 340/3400 = 340/10 = Rs. 34 Answer : Option D
Answer: D
Question: A woman complete a journey in 5 hours. She travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
Options:
A)334 km.
B)216 km.
C)112 km.
D)224 km.
E)544 km.
Rationale: 0.5x/21 + 0.5x/24 = 5
--> x/21 + x/24 = 10
--> x = 112 km.
C
Answer: C
Question: 10, 25, 45, 54, 60, 75, 80
Options:
A)10
B)45
C)49
D)54
E)75
Rationale: Each of the numbers except 54 is multiple of 5.
answer : C
Answer: C
Question: P and Q are the only two applicants qualified for a short-term research project that pays 480 dollars in total. Candidate P has more experience and, if hired, would be paid 50 percent more per hour than candidate Q would be paid. Candidate Q, if hired, would require 10 hours more than candidate P to do the job. Candidate P’s hourly wage is how many dollars greater than candidate Q’s hourly wage?
Options:
A)$5
B)$8
C)$11
D)$14
E)$17
Rationale: Let Q's hourly wage be x, then P's hourly wage is 1.5x
Let T be the number of hours that Q needs, then P needs T-10 hours to do the job.
Since they both are paid an equal total amount of $480:
x*T=1.5x*(T-10)
T = 30 hours and Q's hourly wage is 480/30 = $16
P's hourly wage is 480/(T-10) = $24 which is $8 per hour more.
The answer is B.
Answer: B
Question: A semicircular shaped window has diameter of 63cm. Its perimeter equals
Options:
A)162
B)766
C)298
D)277
E)198
Rationale: Explanation:
perimeter of window = {\color{Blue} \Pi }r+2r
= [{(22/7) * (63/2)} +63]
= 99+63
= 162 cm
Answer: A) 162
Answer: A
Question: Among 700people, 56% like strawberry jam, 44% like apple jam, and 40% like raspberry jam. If 30% of the people like both strawberry and apple jam, what is the largest possible number of people who like raspberry jam but do not like either strawberry or apple jam?
Options:
A)20
B)60
C)80
D)186
E)210
Rationale: 56%+44%-30%=70% who like either Strawberry, Apple, or both.
That means 100%-70% = 30% do not like either Apple of Strawberry jam.
30% is the most possible that like raspberry jam but do not like either strawberry or apple (assuming nobody in the group does not like any)
E
Answer: E
Question: Arithmetic mean and standard deviation of a certain normal distribution are 13.5 and 0.5. What value is exactly 2 standard deviations less than the mean?
Options:
A) 10.5
B) 11
C) 11.5
D) 12
E) 12.5
Rationale: The value which is exactly two SD below the mean is: mean-2*SD=13.5-2*0.5=12.5.
Answer: E.
Answer: E
Question: The average age of the mother and her six children is 10 years which is reduced by 5 years if the age of the mother is excluded. How old is the mother?
Options:
A)40
B)41
C)42
D)43
E)44
Rationale: Total age of mother and 6 children = avg x n = 10 x (6+1)= 70
If mother is excluded new average is (10 - 5) = 5
so for 6 children = 6 x 5 =30
so mother's age is 70 - 30 =40
Answer: Option A
Answer: A
Question: Today Jim is four times as old as Fred, and Sam is 2 years younger than Fred. Four years ago Jim was 9 times as old as Sam. How old is Jim now?
Options:
A)8
B)12
C)16
D)20
E)40
Rationale: We're asked how old Jim is NOW. We're given three facts to work with:
1) Today, Jim is 4 times as old as Fred
2) Today, Sam is 2 years younger than Fred
3) Four years ago, Jim was 9 times as old as Sam.
let's TEST Answer E: 40
IF....Jim is currently 40 years old....
Fred is 10 years old
Sam is 8 years old
4 years ago, Jim was 36 and Sam was 4, so Jim WAS 9 times Sam's age. This is an exact MATCH for what we were told, so this MUST be the answer.
E
Answer: E
Question: The positive two-digit integers m and n have the same digits, but in reverse order. Which of the following must be a factor of m + n?
Options:
A)33
B)9
C)10
D)12
E)14
Rationale: Remember: When you take the difference between the two, it will always be 9. e.g 23-32=9, 89-98=9
and when you add both integers, the sum will always be a multiple of 11 e.g 23+32=55, 89+98= 187
33 is multiple of 11.
Answer : A
Answer: A
Question: If a train, travelling at a speed of 90 kmph, crosses a pole in 5 sec, then the length of train is?
Options:
A)281
B)125
C)288
D)266
E)121
Rationale: D = 90 * 5/18 * 5
= 125 m
Answer: B
Answer: B
Question: Aish started a business investing Rs. 25,000 in 2010, In 2011, he invested an additional amount of Rs. 10,000 and Roshan joined him with an amount of Rs. 35,000. In 2012, Aish invested another additional amount of Rs. 10,000 and Atul joined them with an amount of Rs. 35,000. What will be Roshan’s share in the profit of Rs. 1,50,000 earned at the end of 3 years from the start of the business in 2010?
Options:
A)Rs.45000
B)Rs.50000
C)Rs.60000
D)Rs.70000
E)None of these
Rationale: Explanation :
Aish : Roshan : Atul
= (25000 x 12 + 35000 x 12 + 45000 x 12) : (35000 x 24) : (35000 x 12)
= 1260000 : 840000 : 420000 = 3 : 2 : 1
Roshan's share = Rs. 150000 x 2/6
= Rs. 50000.
Answer : B
Answer: B
Question: Find the sum of all odd numbers upto 100.
Options:
A)200
B)1000
C)2500
D)3140
E)3200
Rationale: Sol. The given numbers are 1, 3, 5, 7, ..., 99.
This is an A.P. with a = 1 and d = 2.
Let it contain n terms. Then,
1 + (n - 1) x 2 = 99 or n = 50.
Required sum = (n/2) (first term + last term)
= 50(1 + 99)/2 = 2500.
ANSWER C 2500
Answer: C
Question: How long does a train 110 m long traveling at 60 kmph takes to cross a bridge of 170 m in length?
Options:
A)16.6
B)16.0
C)16.3
D)16.8
E)16.1
Rationale: D = 110 + 170 = 280 m
S = 60 * 5/18 = 50/3
T = 280 * 3/50 = 16.8 sec
Answer: D
Answer: D
Question: The area of sector of a circle whose radius is 12 metro and whose angle at the center is 40° is?
Options:
A)53m2
B)51m2
C)50.2m2
D)45m2
E)40m2
Rationale: 40/360 * 22/7 * 12 * 12
= 50.2 m2
Answer:C
Answer: C
Question: If the average (arithmetic mean) of the four numbers K, 2K + 3, 4K – 4 and 5K + 1 is 63, what is the value of K?
Options:
A) 11
B) 15 3/4
C) 21
D) 23
E) 25 3/10
Rationale: K + 2K +3 + 4K - 4 + 5K +1 = 12K
(12K)/4 = 63
12K = 63 * 4 = 252
K = 252 / 12 = 21
Answer C.
Answer: C
Question: The average amount with a group of seven numbers is Rs. 30. If the newly joined member has Rs. 75 with him, what was the average amount with the group before his joining the group?
Options:
A)s. 25.6
B)s. 45/2
C)s. 16.6
D)s. 26
E)s. 25/2
Rationale: Total members in the group = 7
Average amount = Rs. 30
Total amount with them = 7 * 30 = Rs. 210
One number has Rs. 50. So, the amount with remaining 6 people = 210 - 75 = Rs. 135
The average amount with them = 135/6 = Rs. 45/2
ANSWER:B
Answer: B
Question: With a uniform speed a car covers the distance in 8 hours. Had the speed been increased by 3 km/hr, the same distance could have been covered in 7 1/2 hours. What is the distance covered?
Options:
A)360 km
B)480 km
C)278 km
D)297 km
E)671 km
Rationale: Let the distance be x km. Then,
x/(7 1/2) - x/8 = 3
2x/15 - x/8 = 3 => x = 360 km.
Answer:A
Answer: A
Question: The sum of the first 50 positive even integers is 2,550. What is the sum of the odd integers from 101 to 200, inclusive?
Options:
A)5,050
B)7,500
C)10,500
D)15,000
E)19,600
Rationale: 101+103+.......199
If we remove 100 from each of these it will be sum of 1st 100 odd numbers.
so 101+103+.......199 = 50 * 100 + (1+3+5+7+......)
sum of 1st 100 natural numbers = (100 * 101) / 2 = 5050
Sum of 1st 50 positive even integers = 2550
sum of 1st 100 odd numbers = 5050 - 2550 = 2500
so 101+103+.......199 = 50 * 100 + (1+3+5+7+......) = 5000 + 2500 = 7500
B is the answer.
Answer: B
Question: The measures of angles A, B and C of a triangle are in the ratio 3:4:5. What is the measure, in degrees, of the largest angle?
Options:
A)35°
B)65°
C)85°
D)75°
E)55°
Rationale: Let the sizes of angles A, B and C be written as follows
A = 3 x, B = 4 x and C = 5 x , where x is a common factor.
The sum of all three angles is equal to 180°. hence
3 x + 4 x + 5 x = 180°
Solve the above equation for x
x = 15
The size of the largest angle is equal to
5 x = 5 * 15 = 75°
correct answer D
Answer: D
Question: X does a work in 30 days. Y does the same work in 45 days. In how many days they together will do the same work?
Options:
A)18
B)12
C)20
D)30
E)15
Rationale: X's 1day's work= 1/30
Y's 1 day's work=1/45
(X+Y)'s 1day's work= (1/30+1/45)=1/18
both together will finish the work in 18 days.
correct option is A
Answer: A
Question: The average of five consecutive odd numbers is 61. What is the difference between the highest and lowest numbers
Options:
A)7
B)8
C)6
D)4
E)2
Rationale: Explanation:
Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.
Then [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.
or 5x + 20 = 305 or x = 57.
So, required difference = (57 + 8) - 57 = 8
Answer: B
Answer: B
Question: The amount of time that three people worked on a special project was in the ratio of 1 to 2 to 3. If the project took 120 hours, how many more hours did the hardest working person work than the person who worked the least?
Options:
A)35 hours
B)40 hours
C)36 hours
D)38 hours
E)42 hours
Rationale: Let the persons be A, B, C.
Hours worked:
A = 1*120/6 = 20 hours
B = 2*120/6 = 40 hours
C = 3*120/6 = 60 hours
C is the hardest worker and A worked for the least number of hours. So the difference is 60-20 = 40 hours.
Answer : B
Answer: B
Question: The simple interest in 14 months on a certain sum at the rate of 6 per cent per annum is 250 more than the interest on the same sum at the rate of 8 per cent in 8 months. How much amount was borrowed?
Options:
A)15000
B)25000
C)7500
D)14500
E)None of these
Rationale: Let the amount be x.
From the question, x×14×6/1200−x×8×8/1200=250
∴ x = 15000
Answer A
Answer: A
Question: The surface area of a sphere is 4πR2, where R is the radius of the sphere. If the area of the base of a hemisphere is 3, what is the surface area W of that hemisphere?
Options:
A) 6/π
B) 9/π
C) 6
D) 9
E) 12
Rationale: Given Area of the base of a hemisphere is 3 = PI * R^2
Thus R = Sqrt ( 3 / PI ) .
Surface area of whole sphere = 4*PI*R^2 .
= 4 * PI * 3 / PI
= 12 .
Since the hemisphere is half of a sphere the Surface area of the hemisphere = 12 / 2
= 6 ( curved part , not including the flat rounded base ) .
But the total surface area = 6 + Area of the base of a hemisphere .
= 6 + 3
= 9.
Answer is D !!
Answer: D
Question: If P is a prime number greater than 3, find the remainder when P^2 + 12 is divided by 12.
Options:
A)6
B)1
C)0
D)8
E)7
Rationale: Every prime number greater than 3 can be written 6N+1 or 6N-1.
If P = 6N+1, then P^2 + 13 = 36N^2 + 12N + 1 + 12 = 36N^2 + 12N + 12 + 1
If P = 6N-1, then P^2 + 13 = 36N^2 - 12N + 1 + 12 = 36N^2 - 12N + 12 + 1
When divided by 12, it must leave a remainder of 1.
The answer is B.
Answer: B
Question: A person has to cover a distance of 6 km in 45 minutes. If he covers one-half of the distance in two-thirds of the total time; to cover the remaining distance in the remaining time, his speed (in km/hr) must be :
Options:
A)7 km/hr
B)8 km/hr
C)11 km/hr
D)12 km/hr
E)None of these
Rationale: Sol.
Remaing distance = 3 km and
Remaining time = [1/3 * 45] min = 15 min = 1/4 hour.
∴ Required speed = (3*4) km/hr = 12 km/hr.
Answer D
Answer: D
Question: What is the value of x if (64^2)-(36^2)=20*x?
Options:
A)120
B)100
C)140
D)130
E)115
Rationale: 20*x=(64+36)(64-36)
20*x=100*28
x=(100*28)/20
x=5*28
x=140
Option C
Answer: C
Question: Two members of a certain club are selected to speak at the next club meeting. If there are 78 different possible selections of the 2 club members, how many members does the club have?
Options:
A)5
B)6
C)7
D)8
E)13
Rationale: Method-1
Total No. of Selection of r out of n objects are defined by nCr = n! / [(r!)(n-r)!]
i.e. If total member = n
then nC2 = n! / [(2!)(n-2)!] = 78
i.e. n*(n-1)*n-2!/ [(2!)n-2!] = 78
i.e. n*(n-1) = 156
(for Positive Values of n)
therefore, n*(n-1) = 156
i.e. n= 13
Answer: Option E
Answer: E
Question: A present value of a machine is $128000. Its value depletiation rate is 25% per annum then find the machine value after 3 years
Options:
A)A)$48000
B)B)$54000
C)C)$36520
D)D)$32450
E)E)$42500
Rationale: P= $128000
R = 25%
T = 3 years
Machine value after 3years = P/(1-R/100)^T
=128000*3/4 * 3/4 * 3/4
=$54000
Answer is B
Answer: B
Question: There are 18 balls in a box: 10 balls are green, 6 are blue and 2 are white. Then 1 green and 1 blue balls are taken from the box and put away. What is the probability that a blue ball is selected at random from the box?
Options:
A)5/16
B)4/13
C)3/13
D)5/13
E)6/13
Rationale: If 1 green and 1 blue ball are taken from the box, then there are 9 green, 5 blue and 2 white balls, a total of 16 balls. If one ball is selected at random, the probability that it is blue is given by 5 / 16
Answer : A
Answer: A
Question: A, B and C enter into a partnership. They invest Rs. 40,000, Rs. 80,000 and Rs. 1,20,000 respectively. At the end of the first year, B withdraws Rs. 40,000, while at the end of the second year, C withdraws Rs. 80,000. IN what ratio will the profit be shared at the end of 3 years?
Options:
A)3:4:8
B)3:4:7
C)3:4:5
D)3:4:1
E)3:4:0
Rationale: A:B:C = (40000 * 36) : (80000 * 12 + 40000 * 24) : (120000 * 24 + 40000 * 12)
= 144:192:336 = 3:4:7
Answer: B
Answer: B
Question: In how many ways can three boys can be seated on five chairs?
Options:
A)72
B)28
C)60
D)58
E)39
Rationale: There are three boys.
The first boy can sit in any of the five chairs (5 ways).
5
Now there are 4 chairs remaining. The second boy can sit in any of the four chairs (4 ways).
5	4
Now there are 3 chairs remaining. The third boy can sit in any of the three chairs (3 ways).
5	4	3
Hence, total number of ways in which 3 boys can be seated on 5 chairs
=5×4×3=60
Ans: C
Answer: C
Question: How many paying stones, each measuring 2 1/2 m * 2 m are required to pave a rectangular court yard 30 m long and 16 1/2 m board?
Options:
A)99
B)88
C)77
D)665
E)41
Rationale: 30 * 33/2 = 5/2 * 2 * x => x = 99
Answer:A
Answer: A
Question: Tough and Tricky questions: Number Properties.
What is the smallest positive integer x such that 320x is the cube of a positive integer?
Options:
A)2
B)3
C)5
D)15
E)25
Rationale: We want to know the smallest x that will make 320x a CUBE of some number. Let's call that number y.
Let's first figure out what we're working with. The prime factorization of 320 can be visualized:
...........320
......../.......\
......32.......10
...../..\....../...\
...8.....4...2.....5
.../.\ ../.\
..4..2 2..2
./.\
2..2
So, we have 2 * 2 * 2 * 2 * 2 * 2* 5 that can be multiplied together to get 320. Now we need to figure out what we need to make 320 * x into a cube of y (y^3=320*x).
We have six 2s, and one 5. To arrange these numbers in identical triples (2,2,5), we need at least two more 5s. Each of these triples will give us the value of y (2*2*5=20), which, multiplied by itself three times, gives us 320 * x.
Looking at the factors we need to complete the triples, we get 5 * 5 = 25. We know this is the smallest number possible because prime factors by definition cannot be broken down any further.
Therefore, we can go with answer choice E.
If time permits, we can do a sanity check. We calculated that y should be 2 * 2 * 5, or 20. 20 * 20 * 20 = 8000. Also, 320 * 25 = 8000.
Answer: E
Answer: E
Question: What is the sum of the integers from -150 to 155, inclusive?
Options:
A)865
B)955
C)1050
D)765
E)750
Rationale: In an arithmetic progression,
the nth term is given by tn=a+(n-1)d
here tn=155, a=-150, d=1
hence, 155=-150+(n-1)
or n=306
Sum of n terms can be calculated by sn=n/2(a+l)
a=first term, l=last term, n=no. of terms
sn=306*(-150+155)/2
sn=306*5/2 = 765
Answer : D
Answer: D