divelab/Sys2Bench huggingface dataset:
Question: After increasing 20% in the price of a Book costs Rs.240. Find the actual cost of an article?
Options:
A)100
B)150
C)200
D)220
E)250
Rationale: CP* (120/100) = 240
CP= 2* 100 => CP = 200
Answer:C
Answer: C
Question: The owner of a furniture shop charges his customer 20% more than the cost price. If a customer paid Rs. 8400 for a computer table, then what was the cost price of the computer table?
Options:
A)Rs.6725
B)Rs.6727
C)Rs.6908
D)Rs.6725
E)Rs.7000
Rationale: CP = SP * (100/(100 + profit%))
= 8400(100/120) = Rs.7000.
Answer: E
Answer: E
Question: 6^5 is equal to which of the following?
Options:
A)7,776
B)8,752
C)9,660
D)2,662
E)1,608
Rationale: A is the correct answer. We do not really need to calculate it. The last digit for 6^5 will be 6. Hence A is the correct answer
Answer: A
Question: Set A contains all the even numbers between 42 and 90 inclusive. Set B contains all the even numbers between 62 and 110 inclusive. What is the difference between the sum of elements of set B and the sum of the elements of set A?
Options:
A)300
B)350
C)400
D)450
E)500
Rationale: Each term in set B is 20 more than the corresponding term in set A.
The difference of the sums = 25 * 20 = 500.
The answer is E.
Answer: E
Question: Bus 1 and Bus 2 run between cities A and B. Bus 1 leaves city A at the same time Bus 2 leaves city B,
each at a constant rate of speed. Their first meeting is 80 miles from city A. After reaching their
respective destinations and immediately turning around, their second meeting is 30 miles from city B.
What is the distance in miles between city A and city B?
Options:
A)90
B)120
C)125
D)150
E)210
Rationale: let d=distance between cities A and B
bus 1 distance to first meeting=80 miles
bus 1 distance from first to second meeting=(d-80)+30=d-50 miles
bus 2 distance to first meeting=d-80 miles
bus 2 distance from first to second meeting=80+(d-30)=d+50 miles
because both buses take the same time for the pre-meeting leg, and
the same time for the inter-meeting leg, and their speeds remain
constant, their ratios between pre-meeting distance and inter-meeting
distance should be the same
therefore, 80/(d-50)=(d-80)/(d+50)
d^2=210d
d=210 miles
E
Answer: E
Question: Two trains are moving in the same direction at 72 kmph and 36 kmph. The faster train crosses a man in the slower train in 27 seconds. Find the length of the faster train?
Options:
A)270 m
B)870 m
C)290 m
D)278 m
E)275 m
Rationale: Relative speed = (72 - 36) * 5/18 = 2 * 5 = 10 mps.
Distance covered in 27 sec = 27 * 10 = 270 m.
The length of the faster train = 270 m.
Answer:A
Answer: A
Question: How many of the factors of 1400 are divisible by 2?
Options:
A)12
B)15
C)18
D)21
E)24
Rationale: 1400 = 2^3*5^2*7
The total number of factors is (3+1)(2+1)(1+1) = 24
The odd factors are 1, 5, 7, 25, 35, and 175.
The number of even factors is 24-6 = 18
The answer is C.
Answer: C
Question: Andrew travelling to 7 cities. Gasoline prices varied from city to city. $1.75, $1.61, $1.79, $2.11, $1.96, $2.09, $1.85. What is the median gasoline price?
Options:
A)$1
B)$1.85
C)$1.92
D)$2.13
E)$2.15
Rationale: Ordering the data from least to greatest, we get:
$1.61, $1.75, $1.79, $1.85, $1.96, $2.09, $2.11
The median gasoline price is $1.85. (There were 3 states with higher gasoline prices and 3 with lower prices.)
B
Answer: B
Question: Simplify: 7005-7000+10
Options:
A)6505
B)6305
C)6501
D)6508
E)None of them
Rationale: 7005-7000+10
=7005-(7000/10)
=7005-700
=6305.
Answer is B
Answer: B
Question: a bag of blue and green colored marbles has 50% green marbles, with a total of 40 marbles. how many green marbles need to be added to bring their percentage in the bag up to 80%?
Options:
A)60
B)58
C)42
D)24
E)80
Rationale: A bag of 40 marbles which is 50% green marbles means we have 20 green marbles. To get a bag that is 80% green marbles, let's use this expression: (20 + x)/(40 + x) = 0.8 with x representing the amount of green marbles that must be added to get to 80%. The expression thus represents the new number of green marbles over the total new number of marbles and this fraction should equal 80%. 60 marbles need to be added, as 80/100=0.8 or 80%
choose A
Answer: A
Question: A Bullock cart can travel at 10 kmph uphill road and downhill road 20 kmph. He takes 3 hours to row from lower point A to upper point B and back to A. What is the total distance traveled by bullock card in 3 hours? He must return back to base point A in remaining time.
Options:
A)80
B)20
C)30
D)40
E)60
Rationale: Upward distance traveled per hour - 10 km,
distance traveled after 2 hours = 20 km.
He remained with 1 hour travel downward, i.e. distance traveled in remaining 1 hour downward = 20 km
so total distance traveled from A to B and Back to point A = 20 km upward + 20 km downward = 40 km
(Answer : D)
Answer: D
Question: A and B can do a work in 12 days, B and C in 30 days and C and A in 36 days. In how many days will the work be completed, if all three of them work together?
Options:
A)180/18
B)180/15
C)180/13
D)180/12
E)180/11
Rationale: One day work of A and B = 1/12
One day work of B and C = 1/30
One day work of C and A = 1/36
2(A + B + C) = 1/12 + 1/30 + 1/36
2(A + B + C) = 13/90
(A + B + C) = 13/180
Number of days required = 180/13 days.
Answer: C
Answer: C
Question: A boy has 11 trousers and 11 shirts. In how many different ways can he select a trouser and a shirt?
Options:
A)121ways
B)130ways
C)90ways
D)100ways
E)139ways
Rationale: The boy can select one trouser in nine ways.
The boy can select one shirt in 12 ways.
The number of ways in which he can select one trouser and one shirt is 11 * 11
= 121 ways.
Answer: A
Answer: A
Question: If 3<x<6<y<9, then what is the greatest possible positive integer difference of x and y?
Options:
A)3
B)4
C)5
D)6
E)7
Rationale: 3 < x < 6 < y < 9;
3 < x
y < 9
3 + y < x + 9
y - x < 6.
Positive integer difference is 5 (for example y = 8.5 and x = 3.5)
Answer: C.
Answer: C
Question: If integers ending in zero are excluded, what is the unit digit of the product of the even integers from 302 to 398, inclusive?
Options:
A)2
B)4
C)6
D)8
E)10
Rationale: 302 x 304 x 306 x 308 => unit digit 4
312 x 314 x 316 x 318 => unit digit 4
392 x 394 x 396 x 398 => unit digit 4
So we have 10 numbers ending by 4 to multiply.
Given that 4 x 4 = 16, we have 5 numbers ending by 6 to multiply
Given that 6 x 6 = 36, we have 3 numbers ending by 6 to multiply
So the unit digit is 6
Answer: C
Answer: C
Question: The simple interest on a sum of money will be Rs.1200 after 10 years. If the principal is trebled after 5 years what will be the total interest at the end of the tenth year?
Options:
A)8981
B)3799
C)1200
D)2400
E)1732
Rationale: P --- 10 ---- 1200
P --- 5 ----- 600
3P --- 5 ----- 1800
------
=> 2400
Answer: D
Answer: D
Question: 4, 12, 48, 240, 1440, (...)
Options:
A)7620
B)10080
C)6200
D)10020
E)12500
Rationale: Explanation :
Go on multiplying the given numbers by 3, 4, 5, 6, 7
Answer : Option B
Answer: B
Question: A confectioner decides to sell all of his pastry due to the coming holiday. His pastry goods are equally divided among a group of 50 regular customers. If only 49 customers come to the bakery, each one will receive 6 less pastry goods. How much pastry does the confectioner needs to sell?
Options:
A)500.
B)412.
C)432.
D)502.
E)522.
Rationale: Pastry is divided in 50 customers equally. So,total number of pastry must be a multiple of 50
Only option A satisfies the condition, and hence is the answer
Answer: A
Question: For all positive integers m, [m]=3m when m is odd and [m]=(1/2)*m when m is even. What is [5]*[6] equivalent to?
Options:
A)[45]
B)[30]
C)[15]
D)[5]
E)[50]
Rationale: [5]*[6] = 15*3 = 45 = [15]
The answer is C.
Answer: C
Question: Eight men, ten women and six boys together can complete a piece of work in eight days. In how many days can 20 women complete the same work if 20 men can complete it in 12 days?
Options:
A)33m
B)77m
C)66m
D)22m
E)88m
Rationale: Let the number of units which can be completed by each man, each women and each boy be m/day, w/day and b/day respectively.
Work = 8(8m + 10w +6b)units = 12(20m)
10w +6b = 22m
b is unknown.
We cannot find the relation between m and w.
We cannot answer the question.
Answer: D
Answer: D
Question: In 2005, 25 percent of the math department’s 40 students were female, and in 2007, 40 percent of the math department’s 85 students were female. What was the percent increase from 2005 to 2007 in the number of female students in the department?
Options:
A)15%
B)50%
C)62.5%
D)115%
E)240%
Rationale: Female students in 2005 = 25% of 40 = 10
Female students in 2007 = 40% of 85 = 34
Percentage increase = [(34- 10)/10] *100 % = 240%
Correct Option: E
Answer: E
Question: In a village there are 4 programs organised for the festival. The festival occurs for 2 days. In how many ways that the 2 programs can be divided for 2 days?
Options:
A)4
B)10
C)6
D)5
E)2
Rationale: There are 4 programs for 2 days festival. That is Each day has 2 Programs.
The question is how many ways that the programs is chosen.
4C2=4!/2!*2!=4*3*2!/2!*2*1
4C2=6
There are 6 possible ways to organise the programs
Answer: C
Question: Pipes A and B fill a tank in 5 and 6 hours respectively. Pipe C empties the tank in 12 hours, if you open all the three pipes together this tank will be filled in?
Options:
A)3 9/17 hours
B)4.5 hours
C)1.25 hours
D)4 7/17 hours
E)None of these
Rationale: Explanation:
It is given that Pipe A,B and C work simultaneously.
Pipe A and B are filling the tank and Pipe C is emptying it.
In 1 hour volume of tank filled by Pipe A: 1/5th
In 1 hour volume of tank filled by Pipe B: 1/6th
In 1 hour volume of tank emptied by Pipe C: 1/12th
Working together Pipe A B and C fill x volume of tank.
=>1/x = (1/5)+(1/6)-(1/12)
=>x=60/17 hours
ANSWER A
Answer: A
Question: In a class of 50 students the number of females is 2 more that 5 times the male number .How many males and females are there in the class
Options:
A)42, 8
B)50,8
C)8,42
D)8,50
E)40,10
Rationale: Let the females be x and males be y
x+y=50--1
x=5y+2---2 This can be written as x-5y=2---3
solving 1 and 3 we get y=8
substituting y=8 in (1) we get x=42
Correct ans is A
Answer: A
Question: Jim wishes to buy 3 gifts that cost 15 dollars, 9 dollars, and 8 dollars. He has 1/4 of the money
he needs. How much more money must he earn in order to buy the gifts?
Options:
A)$57
B)$47
C)$7
D)$17
E)$24
Rationale: $15 + $9 + $8 = $32 needed
1/4 x $32 = $8 on hand
$32 - $8 = $24 to earn
correct answer E
Answer: E
Question: A train 240 m in length crosses a telegraph post in 16 seconds. The speed of the train is?
Options:
A)22 kmph
B)77 kmph
C)54 kmph
D)71 kmph
E)88 kmph
Rationale: S = 240/16 * 18/5 = 54 kmph
Answer:C
Answer: C
Question: What percent is 25 gm of 1 kg?
Options:
A)0.5 %
B)1%
C)1.5%
D)2.5%
E)3%
Rationale: 1 kg = 1000 gm
25/1000 × 100 = 2500/1000
=5/2 = 2.5 %
D)
Answer: D
Question: The price of a T.V. set worth Rs. 30000 is to be paid in 20 installments of Rs. 1000 each. If the rate of interest be 6% per annum, and the first installment be paid at the time of purchase, then the value of the last installment covering the interest as well will be?
Options:
A)22678
B)26699
C)29000
D)19000
E)26711
Rationale: Money paid in cash = Rs. 1000
Balance payment = (30000 - 1000) = Rs. 29000
Answer: C
Answer: C
Question: nd the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 12 cm?
Options:
A)243cm2
B)239cm2
C)267cm2
D)230cm2
E)228cm2
Rationale: Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them)
= 1/2 (20 + 18) * (12)
= 228 cm2
Answer:E
Answer: E
Question: Yesterday, a pie shop sold 6 apple pies, 5 peach pies, and 6 pumpkin pies. How many pies did the shop sell in all?
Options:
A)21
B)37
C)13
D)17
E)32
Rationale: Add the number of pies
6+5+6 =17.
Answer is D.
Answer: D
Question: From a pack of cards two cards are drawn one after the other, with replacement. The probability that the first is a red card and the second is a king is?
Options:
A)1/26
B)1/27
C)1/29
D)1/37
E)1/97
Rationale: Let E1 be the event of drawing a red card.
Let E2 be the event of drawing a king .
P(E1 ∩ E2) = P(E1) . P(E2)
(As E1 and E2 are independent)
= 1/2 * 1/13
= 1/26
Answer: A
Answer: A
Question: A and B starts a business with Rs.8000 each, and after 3 months, B withdraws half of his capital . How should they share the profits at the end of the 18 months?
Options:
A)18:11
B)22:13
C)23:12
D)12:7
E)11:10
Rationale: A invests Rs.8000 for 18 months, but B invests Rs.8000 for the first 3 months and then withdraws Rs.4000. So, the investment of B for remaining 15 months is Rs.4000 only.
A : B
8000*18 : (8000*3) + (4000*15)
144000 : 84000
A:B = 12:7
ANSWER:D
Answer: D
Question: The ratio of number of boys and girls in a school is 2:5. If there are 210 students in the school, find the number of girls in the school?
Options:
A)150
B)250
C)300
D)370
E)280
Rationale: Let the number of boys and girls be 2x and 5x
Total students = 210
Number of girls in the school = 5*210/7 = 150
Answer is A
Answer: A
Question: Hector runs a one-product widget shop. Each month, his fixed costs, whether he produces anything or not, amount to x dollars. The marginal cost of producing and selling one widget is a constant y dollars, regardless of the number he produces. He generally can sell widgets for t dollars each. What is the minimum number of widgets he must produce, assuming he sells every widget he produces, to break even (neither make a profit nor incur a loss) in a given month?
Options:
A)x/(y - t)
B)x/(t - y)
C)x
D)x/t
E)x/y
Rationale: Two ways...
1) let a be the total number, so cost = a*y+x and selling price=a*t..
No profit no loss....a*y+x=a*t.....a=x/(t-y)...
B
Answer: B
Question: Five drainage pipes, each draining water from a pool at the same constant rate, together can drain a certain pool in 12 days. How many additional pipes, each draining water at the same constant rate, will be needed to drain the pool in 6 days?
Options:
A) 5
B) 9
C) 10
D) 12
E) 15
Rationale: This is an inverse proportional problem......
5 pipes in 12 days; So for 6 days, it will be = 12 x 5 / 6 = 10
So, 10-5 = 5
Answer A
Answer: A
Question: The salaries of A and B together amount to $2000. A spends 95% of his salary and B, 85% of his. If now, their savings are the same, what is A's salary?
Options:
A)$1000
B)$1250
C)$2500
D)$1500
E)$1200
Rationale: Let A's salary is x
B's salary = 2000-x
(100-95)% of x = (100-85)% of (2000-x)
x = $1500
Answer is D
Answer: D
Question: If 213 × 16 = 3408, then 0.016 × 2.13 is equal to :
Options:
A)0.3408
B)3.408
C)34.08
D)340.8
E)0.03408
Rationale: Solution
0.016 × 2.13 = (16/1000x 213/100)= (16/1000x 213/100) = 3408/100000 = 0.03408. Answer E
Answer: E
Question: A, B and C work on a task. To complete the task alone, B takes twice the time that A would take to complete the task alone and 1/5rd the time that C would take to complete the task alone. If B actually worked for half the number of days that A worked and 3/2 times the number of days that C worked, what proportion of the total work was completed by B?
Options:
A)1/3
B)2/9
C)9/49
D)7/19
E)25/126
Rationale: The first thing to notice is that A is faster than B and B is faster than C.
Since work is proportional to time, in 1 day lets say if A does 2 works, B does 1 work and C does 2/3rd of a work.
If A works for 2 days, B works for 1 day and C works for only 2/3 of the day.
Therefore total work done = (2*2) + (1*1) + (1/5*1/5) = 126/25
Proportion of work done by B = (1*1) / (126/25) = 25/126
Hence Answer E.
Answer: E
Question: A part-time employee whose hourly wage was decreased by 20 percent decided to increase the number of hours worked per week so that the employee's total income did not change. By what percent T should the number of hours worked be increased?
Options:
A)12.5%
B)20%
C)25%
D)50%
E)100%
Rationale: Correct Answer: C
Solution: C. We can set up equations for income before and after the wage reduction. Initially, the employee earns W wage and works H hours per week. After the reduction, the employee earns .8W wage and works X hours. By setting these equations equal to each other, we can determine the increase in hours worked: WH = .8WX (divide both sides by .8W) 1.25H = X We know that the new number of hours worked T will be 25% greater than the original number. The answer is C.
Answer: C
Question: A survey was conducted to find out how many people in a housing colony of 140 residents could swim, dance and drive a car. It was found that the number of people who could not swim was 89, the number of people who could not dance was 100 and that the number of people who could not drive a car was 91. If the number of people who could do at least two of these things, was found to be 37 and the number of people who could do all these things was found to be 6, how many people could not do any of these things?
Options:
A)43
B)44
C)45
D)46
E)48
Rationale: T=n(A)+n(B)+n(C)-n(Exactly two of the events)-2*n(All 3 Events)+n(None of the events)
T=140
n(A)=T-n(A')=140-89=51
n(B)=T-n(B')=140-100=40
n(C)=T-n(C')=140-91=49
n(Exactly two of the events)=n(At least 2 Events)-n(All 3 Events)=37-6=31
n(All 3 Events)=6
140=51+40+49 -31-2*6+n(None of the events)
n(None of the events)=140-51-40-49+31+12=43
Ans:A
Answer: A
Question: If m > 0, which of the following could be true?
I. m^3 > m^2
II. m^2 = m
III. m^2 > m^3
Options:
A)I only
B)III
C)IIIII
D)All of the above
E)None of the above
Rationale: 1. m^3 > m^2 Well this is true for all m > +1 (but not for values 0 < m < 1)
2. m^2 = m This is true for only one positive number, 1
3. m^2 > m^3 Again this is true only for values of m such that 0 < m < 1
So the answer is D, since there exists at least one value of m which satisfies the equation/ in equation(s)
Answer: D
Question: Find the odd man out. 187, 264, 336, 473, 682, 781
Options:
A)187
B)336
C)781
D)682
E)473
Rationale: In all numbers except 336, the middle digit is the sum of other two digits.Answer is B
Answer: B
Question: Two friends A and B apply for a job in the same company. The chances of A getting selected is 2/5 and that of B is 4/7. What is the probability that both of them get selected?
Options:
A)8/35
B)34/35
C)27/35
D)30/35
E)None of these
Rationale: EXPLANATION:
P(A) = 2/5
P(B) = 4/7
E = {A and B both get selected}
P(E) = P(A)*P(B)
= 2/5 * 4/7
= 8/35
ANSWER IS A
Answer: A
Question: 4 boys and 2 girls are to be seated in a row in such a way that the two girls are always together. In how many different ways can they be seated?
Options:
A)120
B)720
C)148
D)240
E)None of these
Rationale: Assume the 2 given students to be together (i.e one).
Now there are five students.
Possible ways of arranging them are = 5! = 120
Now, they (two girls) can arrange themselves in 2! ways.
Hence total ways = 120 × 2 = 240
Answer D
Answer: D
Question: If each permutation of the digits 1,2,3,4,5,6 are listed in the increasing order of magnitude, then 289thterm will be
Options:
A)361452
B)321546
C)321456
D)341256
E)None of these
Rationale: Explanation :
Let's see how many numbers can be formed with the left most digit as 1
The digit '1' is placed at the 1st position (only 1 way of doing this)
Since one digit is placed at the 1st position,
any of the remaining 5 digits can be placed at 2nd position.
Since one digit is placed at the 1st position and another digit is placed
at the 2nd position, any of the remaining 4 digits can be placed at the
3rd position.
So on ...
1	5	4	3	2	1
i.e., total number of ways = (1)( 5)( 4)( 3)( 2)( 1) = 120
i.e., total count of numbers which can be formed
with the left most digit as 1 = 120
Similarly, total count of numbers which can be formed
with the left most digit as 2 = 120
Similarly, total count of numbers which can be formed
with the left most digit as 3 = 120
i.e., 240 numbers (=120 + 120) can be formed
(with left most digit as 1) or (with left most digit as 2)
Similarly, 360 numbers (=120 + 120 + 120) can be formed
(with left most digit as 1) or (with left most digit as 2)
or (with left most digit as 3)
Hence, the left most digit of the 249th number = 3
Now, let's find out how many numbers can be formed
with the left most digit as 3 and next digit as 1
The digit '3' is placed at the 1st position (only 1 way of doing this)
The digit '1' is placed at the 2nd position (only 1 way of doing this)
Any of the remaining 4 digits can be placed at 3rd position.
Since 3 digits are placed in the first three positions, any of the remaining 3 digits
can be placed at the 4th position.
Since 4 digits are placed in the first four positions, any of the remaining 2 digits
can be placed at the 5th position.
Since 5 digits are placed in the first five positions, the remaining 1 digit
can be placed at the 6th position.
1	1	4	3	2	1
i.e., total number of ways = (1)(1)(4)(3)(2)(1) = 24
i.e., Total count of numbers which can be formed
(with the left most digit as 3) and (next digit as 1) = 24
Similarly, total count of numbers which can be formed
(with the left most digit as 3) and (next digit as 2) = 24
Hence, 120 + 120 + 24 + 24 = 288 numbers can be formed
(with left most digit as 1) or (with left most digit as 2)
or (with left most digit as 3 and next digit as 1)
or (with left most digit as 3 and next digit as 2)
Hence, the 289th number is the minimum value number which is formed
with the left most digit as 3 and next digit as 4.
i.e., the number is 341256
Answer : D
Answer: D
Question: Siddharth wants to borrow Rs.5000 at rate of interest 6% p.a. at S.I and lend the same amount at C.I at same rate of interest for two years. What would be his income in the above transaction?
Options:
A)s.21.60
B)s.21.29
C)s.22.29
D)s.18.00
E)s.21.12
Rationale: Amount of money Siddharth borrowed at S.I at 6% p.a. for two years = Rs.5,000
He lend the same amount for C.I at 6% p.a. for two years.
=> Siddharth's income = C.I - S.I
= p[1 + r/ 100]n - p - pnr/100
= p{ [1 + r/ 100]2 - 1 - nr/100
= 5,000{ [1 + 6/100]2 - 1 - 12/100}
= 5,000 {(1.06)2- 1 - 0.12} = 5,000(1.1236 - 1 - 0.12)
= 5,000 (0.0036) = 5 * 3.6 = Rs.18.00
Answer: D
Answer: D
Question: If the price of a certain computer increased 30 percent from a dollars to 351 dollars, then 2a =
Options:
A)540
B)570
C)619
D)649
E)700
Rationale: Before Price increase Price = a
After 30% Price increase Price = a+(30/100)*a= 1.3a = 351 (Given)
i.e. a= 351/1.3 = $270
i.e. 2a = 2*270 = 540
Answer: option A
Answer: A
Question: A man has a certain number of small boxes to pack into parcles. If he packs 3, 4, 5 or 6 in a parcel, he is left with one over; if he packs 7 in a parcle, none is left over. What is the number of boxes, he may have to pack?
Options:
A)106
B)301
C)309
D)400
E)450
Rationale: Explanation:
Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3, 4, 5, or 6 and no remainder when divided by 7. Thus, the number must be of the form (L.C.M of 3, 4, 5, 6) x + 1 i.e., (60x + 1 ) and a multiple of 7. Clearly, for x = 5, the number is a multiple of 7. So the number is 301.
Answer: B) 301
Answer: B
Question: A completes a work in 12 days and B complete the same work in 36 days. If both of them work together, then the number of days required to complete the work will be?
Options:
A)9 days
B)11 days
C)21 days
D)22 days
E)24 days
Rationale: If A can complete a work in x days and B can complete the same work in y days, then, both
of them together can complete the work in x y/ x+ y days
Therefore, here, the required number of days = 12 Ã— 36/ 48 = 9 days.
Option 'A'
Answer: A
Question: Average of five numbers is 26. If one number is excluded the average becomes 28. The excluded number is
Options:
A)10
B)12
C)16
D)20
E)18
Rationale: Explanation:
Number is (5*26) - (4*28) = 130-112 = 18
Answer: Option E
Answer: E
Question: All the faces of cubes are painted with red colour. It is then cut into 64 equal small cubes.Find How many small cubes are there whose two adjacent faces are coloured red ?
Options:
A)0
B)8
C)16
D)24
E)4
Rationale: There are 64 small cubes,Hence one side side of the big cube is 3√64=4 cm
Number of small cubes having two adjacent faces coloured red = (x - 2) x No. of edges
= (4 - 2) x 12
= 24
Answer :D
Answer: D