divelab/Sys2Bench huggingface dataset:
Question: There are some chickens and some chicken field. If we sell 75 chickens Then the feed will last 20 days longer.If we buy 100 chickens more, then the feed will get over before 15 days.What is the present number of chickens?
Options:
A)100
B)200
C)300
D)400
E)500
Rationale: let x=number of chicken,y=no of days,z= each day chicken's feed....so total feed=xyz then eqn 1 will be xyz=(x-75)(y+20)z.....
and eqn 2 will be xyz=(x+100)(y-15)z if we solve this two eqn we get x=300
ANSWER:C
Answer: C
Question: Insert the missing number.
2, 6, 12, 20, 30, 42, 56,72 (....)
Options:
A)22
B)38
C)72
D)66
E)90
Rationale: Explanation:
The pattern is 1 x 2, 2 x 3, 3 x 4, 4 x 5, 5 x 6, 6 x 7, 7 x 8, 8 x 9
So, the next number is 9 x 10 = 90.
Answer: E) 90
Answer: E
Question: The ratio of efficiency of A is to C is 5:3.The ratio of no.of days taken by B is to C is 2:3.A takes 6 days less than C,when A and C completes the work individually.B and C left the work after 2 days.the no of days taken by A to finish the remaining work?
Options:
A)4.5
B)5
C)6
D)7
E)8
Rationale: A : B : C
Efficiency 10 : 9 : 6
No of days 9x : 10x : 15x
given=> 15x-9x = 6
hence, x = 1
Number of days taken by A = 9 and daily work done=10
Number of days taken by B= 10 and daily work done=9 (total work =90)
Number of days taken by C = 15 and daily work done=6
work done by B and C in initial 2 days =30
rest work =60
Number of days required by A to finish = 60/10=6 days
ANSWER:C
Answer: C
Question: When tossing two coins once, what is the probability of heads on both the coins?
Options:
A)3/4
B)2/4
C)1/3
D)1/4
E)1/2
Rationale: Since two coins are tossed, sample space = 4
Getting heads on both the coins=1
p(E) = 1/4 = 1/4
ANSWER:D
Answer: D
Question: The number of years of service of the eight employees in a production department are 15, 10, 9, 17, 6, 5, 14 and 16. What is the range in the number of years of service of the eight employees?
Options:
A)10
B)11
C)12
D)13
E)14
Rationale: = 17-5
= 12
Answer C
Answer: C
Question: L.C.M of two numbers is 192 and there H.C.F is 16. If one of them is 48. Find the other
Options:
A)32
B)64
C)48
D)68
E)78
Rationale: Explanation:
Product of two numbers = product of their L.C.M and H.C.F
48 x a = 192 x 16
a = (192 x 16)/48 = 64
Answer: Option B
Answer: B
Question: A Cistern is filled by pipe A in 8 hrs and the full Cistern can be leaked out by an exhaust pipe B in 12 hrs. If both the pipes are opened in what time the Cistern is full?
Options:
A)12 hrs
B)24 hrs
C)16 hrs
D)32 hrs
E)None of these
Rationale: Explanation :
Pipe A can fill 1⁄8 of the cistern in 1 hour.
Pipe B can empty 1⁄12 of the cistern in 1 hour
Both Pipe A and B together can effectively fill 1⁄8-1⁄12= 1⁄24 of the cistern in 1 hour
i.e, the cistern will be full in 24 hrs. Answer : Option B
Answer: B
Question: Among the employees of a certain company, 52 percent of the employees are male and 48 percent are female. In this company 60 percent of the male employees are married and 50 percent of the female employees are married. If one employee in the company is randomly selected, approximately what is the probability that he or she is NOT married?
Options:
A) 0.4
B) 0.3
C) 0.5
D) 0.6
E) 0.7
Rationale: Much faster: draw 4 quadrants
______________Male (52%)_____Female (48%)
Married :__________60%____________40%
Not Married:_______30%____________50%
Therefore: the probability of picking one random person Not Married (he or she) is: 0,52 x 0,30 + 0,50 x 0,50 = 0,15 + 0,25 (approx.)
Solution: approx. 0,4 (answer A)
Answer: A
Question: There are 50 employees in the office of ABC Company. Of these, 22 have taken an accounting course, 15 have taken a course in finance and 14 have taken a marketing course. Nine of the employees have taken exactly two of the courses and 3 employee has taken all three of the courses. How many of the 50 employees have taken none of the courses?
Options:
A)9
B)8
C)14
D)10
E)11
Rationale: 50 Employees. Counting every different attendand to the courses we have:
Accounting: 22
Finance: 15
Marketing: 14
Which would add up to 51 different attendands, which is not possible.
Now 9 have taken exactly 2 courses, which means that there are 9 less different attendands. Say that 9 of the Finance attentands also attended Accounting.
51-9= 42
3 Person has taken all three courses. As above, we subtract him from the number of different attendands. Since this time the person took all three courses, we have to substract him two times.
42-6= 36.
Answer : C
Answer: C
Question: Define r* by the equation r* = π-r. Then ((−π)*)* =
Options:
A) −2π
B) -1
C) −π
D) 2π
E) 4π
Rationale: for r* f(f(−π)) = f(π − (−π)) = f(π + π) = f(2π) = π − 2π = −π=C
Answer: C
Question: In how many ways the word "DARKEST" can be arranged?
Options:
A)5040
B)4506
C)3450
D)8900
E)Cannot be determined
Rationale: Explanation :
The required number of ways = 7!
= 5040
Answer : A
Answer: A
Question: A boat goes 100 km downstream in 10 hours, and 75 m upstream in 15 hours. The speed of the stream is?
Options:
A)1
B)2
C)3
D)4
E)5
Rationale: DS = 6
US = 4
S = ?
S = (6 - 4)/2 = 1 kmph Answer: A
Answer: A
Question: A chartered bank offers a five-year Escalator Guaranteed Investment Certificate.In successive years it pays annual interest rates of 4%, 4.5%, 5%, 5.5%, and 6%, respectively, compounded at the end of each year. The bank also offers regular five-year GICs paying a fixed rate of 5% compounded annually. Calculate and compare the maturity values of $1000 invested in each type of GIC. (Note that 5% is the average of the five successive one-year rates paid on the Escalator GIC.)
Options:
A)1276.28
B)1276.22
C)1276.29
D)1276.21
E)1276.23
Rationale: Explanation:
FV = $1000(1.04)(1.045)(1.05)(1.055)(1.06) = $1276.14
the maturity value of the regular GIC is
FV = $= $1276.28
Answer: A) 1276.28
Answer: A
Question: Each side of a certain parallelogram has length 5. If the area of the parallelogram is 20. Which of the following is the measure of one of its angles?
Options:
A)30
B)45
C)53.13
D)90
E)120
Rationale: area of a parallelogram = b*h
b*h=20
h=20/5=4
sin theta =opp/hyp= 4/5
theta = sin inv of 4/5=53.13 deg
C
Answer: C
Question: January 1, 2008 is Wednesday. What day of the week lies on Jan 1, 2009?
Options:
A)Saturday
B)Wednesday
C)Thursday
D)Tuesday
E)Friday
Rationale: Explanation:
Number of odd days in 2008 = 2 (since it is a leap year)
(we have taken the complete year 2008 because we need to find out the odd days from 01-Jan-2008 to 31-Dec-2008, that is the whole year 2008)
Given that January 1, 2008 is Wednesday.
Hence January 1, 2009 = (Wednesday + 2 odd days) = Friday
Answer: Option E
Answer: E
Question: Find the Value of X :log(x+1) + (log(x-1) = log8
Options:
A)5
B)3
C)6
D)9
E)8
Rationale: Explanation:
log(x+1)(x-1) = log8
log(x^2-1) = log8
x^2-1 = 8
x^2 = 9
x = {-3, 3}
Testing for extraneous solutions eliminates -3 leaving you with:
x = 3
Answer B
Answer: B
Question: The sum of the ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child ?
Options:
A)4 years
B)8 years
C)10 years
D)12 years
E)None
Rationale: Solution
Let the ages of the children be x,(x+3),(x+6),(x+9) and (x+12) years.
Then,	x+(x+3)+(x+6)+(x+9)+(x+12)=50
5x=20
x=4.
Age of the youngest child	=4 years.
Answer A
Answer: A
Question: A plane flies 420 miles with the wind and 350 miles against the wind in the same length of time. If the speed of the wind is 23 mph, what is the speed of the plain in still air?
Options:
A)196
B)210
C)216
D)235
E)253
Rationale: The speed of the plane in still air = x miles/hour
The speed of the wind is 23 mph
Speed with the wind = (x + 23) mph
Speed against the wind = (x – 23) mph
Time = Distance/ Speed
According to the problem,
420/(x + 23) = 350/(x – 23)
420(x – 23) = 350(x + 23)
420x – 9660 = 350x + 805
420x – 350x = 8050 + 9660
70x = 17710
x = 17710/70
x = 253 ANSWER : E
Answer: E
Question: If the time is currently 1:30 pm, what time will it be in exactly 643 hours?
Options:
A)5:30 am
B)6:30 am
C)7:30 am
D)8:30 am
E)9:30 am
Rationale: 643 = 26(24) + 19/24
The time will be 19 hours later than 1:30 pm which is 8:30 am.
The answer is D.
Answer: D
Question: The ratio of the cost price and selling price is 2:3. The profit percent is?
Options:
A)20%
B)25%
C)50%
D)75%
E)100%
Rationale: Let the C.P. = $2x
Then S.P. = $3x
Then, Gain = 3x-2x = $x
Gain% = x/2x *100 = 50%
Answer is C
Answer: C
Question: In a function they are distributing noble prize. In how many ways can 3 prizes be distributed among 4 boys when No boy gets more than one prize?
Options:
A)12
B)15
C)18
D)24
E)26
Rationale: Sol. In this case, repetitions are not allowed. So, the first prize can be given in 4 ways. The second in 3 ways and the third in 2 ways. But fundamental principle, (4 x 3 x 2) ways = 24 ways
4: Or 4p = — 4:- 4X3X2X1- 24 ways
D
Answer: D
Question: A boy traveled from the village to the post-office at the rate of 12.5 kmph and walked back at the rate of 2 kmph. If the whole journey took 5 hours 48 minutes, find the distance of the post-office from the village.
Options:
A)5 km
B)7 km
C)9 km
D)10 km
E)12 km
Rationale: Explanation :
Solution: Average speed = 2xy/(x+y) km/hr
= (2*12.5*2)/(12.5+2) km/hr
= 50/14.5 km/hr.
Total distance = (50/14.5 * 29/5) km.
= 20 km.
Required distance = 20/2 = 10 km.
Answer : D
Answer: D
Question: The first flight out of Phoenix airport had a late departure. If the next three flights departed on-time, how many subsequent flights need to depart from Phoenix on-time, for the airport's on-time departure rate to be higher than 40%?
I will see what is the quickest way to solve it then I will provide the explanation
Options:
A)6
B)2
C)9
D)10
E)11
Rationale: The following approach might be the easiest one and less error prone.
We need on-time departure rate to be higher than 4/10, so it should be at least 5/11, which means that 5 out of 11 flights must depart on time. Since for now 3 out of 4 flights departed on time then 5-3=2 subsequent flights need to depart on-time.
Answer: B.
Answer: B
Question: There are 5 pairs of socks and 2 socks are worn from that such that the pair of socks worn are not of the same pair. what is the number of pair that can be formed.
Options:
A)19
B)20
C)30
D)32
E)25
Rationale: First of all you should remember that there is a difference in left and right sock.
now no. of way to select any of the sock = 5
and for second = 4
so total methods = 5*4 = 20
ANSWER:B
Answer: B
Question: Due to construction, the speed limit along an 4-mile section of highway is reduced from 55 miles per hour to 35 miles per hour. Approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ?
Options:
A)A) 3.12
B)B) 8
C)C) 2.49
D)D) 15
E)E) 24
Rationale: Old time in minutes to cross 4 miles stretch = 4*60/55 = 4*12/11 = 4.36
New time in minutes to cross 4 miles stretch = 4*60/35 = 4*12/7 = 6.85
Time difference = 2.49
Ans:C
Answer: C
Question: What least number must be subtracted from 427398 so that remaining no.is divisible by 10
Options:
A)3
B)5
C)6
D)7
E)8
Rationale: Explanation:
On dividing 427398 by 10 we get the remainder 8, so 8 should be subtracted
Answer: Option E
Answer: E
Question: A batsman in his 20th innings makes a score of 90, and thereby increases his average by 2. What is his average after the 20th innings? He had never been ’not out’.
Options:
A)52
B)37
C)39
D)43
E)42
Rationale: Average score before 20th innings
= 90 - 2 × 20= 50
Average score after 20th innings
=> 50 + 2 = 52
ANSWER:A
Answer: A
Question: The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle?
Options:
A)6m
B)5m
C)2m
D)8m
E)3m
Rationale: Let the sides of the rectangle be l and b respectively.
From the given data,
(√l2 + b2) = (1 + 108 1/3 %)lb
=> l2 + b2 = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l2 + b2)/lb = 25/12
12(l2 + b2) = 25lb
Adding 24lb on both sides
12l2 + 12b2 + 24lb = 49lb
12(l2 + b2 + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)2 = 49lb
=> 12(14)2 = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l - b = 8 - 6
= 2m.
Answer:D
Answer: D
Question: Four milkmen rented a pasture. A grazed 24 cows for 3 months; B 10 for 5 months; C 35 cows for 4 months and D 21 cows for 3 months. If A's share of rent is Rs. 2880, find the total rent of the field.
Options:
A)3750
B)3550
C)4250
D)2250
E)13000
Rationale: Ratio of shares of A, B, C, D = (24 x 3) : (10 x 5) : (35 x 4) : (21 x 3) = 72 : 50 : 140 : 63.
Let total rent be Rs. x. Then, A’s share = Rs. (72x)/325
(72x)/325=2880 = x=(2880 x 325)/72 = 13000
Hence, total rent of the field is Rs. 13000.
Answer is E.
Answer: E
Question: What is the remainder when 3^30 is divided by 5?
Options:
A)0
B)1
C)2
D)3
E)4
Rationale: The units digit of powers of 3 follow a repeating cycle of four: {3, 9, 7, 1}
30 has the form 4k+2, so the units digit of 3^30 is 9.
The remainder when dividing by 5 is 4.
The answer is E.
Answer: E
Question: The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is?
Options:
A)8.09%
B)6.07%
C)1.09%
D)6.09%
E)2.09%
Rationale: Explanation:
Amount of Rs. 100 for 1 year when compounded half-yearly = [100 * (1 + 3/100)2] = Rs. 106.09 Effective rate = (106.09 - 100) = 6.09%
Answer: D
Answer: D
Question: The perimeter of an isosceles right triangle is 8 + 8 sq rt 2. What is the length of the hypotenuse of the triangle?
Options:
A)4
B)5.656
C)5
D)6
E)7
Rationale: side of triangle is a then perimeter = a+ a +a.sqrt2 ( right angle and pythagorus) =2a +a .sqrt 2 = 8 + 8 Sqrt2 or,
a. ( 2+ sqrt2) = 8(1 + sqrt2), a= 8*(1+sqrt2)/2+sqrt2 =8*2.414/3.414 =0.707*8 then hypotenuse = 5.656
B
Answer: B
Question: Shekar scored 76, 65, 82, 67 and 85 marks in Mathematics, Science, Social studies, English and Biology respectively. What are his average marks?
Options:
A)65
B)69
C)75
D)85
E)90
Rationale: Explanation :
Average= (76+65+82+67+85)/5 = 375/5 =75
Hence average=75
Answer : C
Answer: C
Question: If cost of 15 eggs be 75 rupees, then find out the cost of 5 dozen eggs.
Options:
A)300
B)400
C)500
D)600
E)700
Rationale: Cost of 15 eggs = 75 rupees
Cost of 1 egg = 75/15 = 5 rupees,
Cost of 5 dozens, i.e. 5 x 12 = 60 x 5 = 300
Answer : A
Answer: A
Question: If 5 < x < 9 and y = x + 3, what is the greatest possible integer value of x + y?
Options:
A)19
B)20
C)21
D)22
E)23
Rationale: x+y = x+x+3 = 2x+3
We need to maximize this value and it needs to be an integer.
2x is an integer when the decimal of x is .0 or .5
The largest such value is 8.5
Then x+y = 8.5 + 11.5 = 20.
The answer is B.
Answer: B
Question: A student was asked to find the arithmetic mean of the numbers 3,11,7,9,15,13,8,19,17,21,14 and x. He found the mean to be 12. What should be the remainder in the place of x?
Options:
A)5
B)6
C)7
D)8
E)9
Rationale: We have (3+11+7+9+15+13+8+19+17+21+14+x) / 12 = 12
137+x = 144
x = 7
Answer is C
Answer: C
Question: Find the minimum value of 5cosA + 12sinA + 12
Options:
A)-1
B)1
C)2
D)3
E)None of these
Rationale: Solution:
5cosA +12sinA + 12 = 13(5/13 cosA +12/13sinA) + 12
Now, for any values of B we can get sinB =5/13 and we can replace cosB = 12/13.
We see that our assumption is right because we satisfy the condition sin^2B + cos^2B=1 so we get 13(sinBcosA +cosBsinA) + 12 =13(sin(A+B))+12. Therefore we know that minimum value of sinx=-1 and greatest is 1 so the least value of the expression becomes when sin(A+B) =-1 then the value of the whole term becomes 13.(-1) + 12 = -13 +12 = -1
Answer A
Answer: A
Question: The average (arithmetic mean) of 4 different integers is 70. If the largest integer is 90, what is the least possible value of the smallest integer?
Options:
A)1
B)19
C)29
D)13
E)33
Rationale: Total of Integers=70*4=280
Lowest of the least possible integer is when the middle 2 intergers are at the maximum or equal to the highest possible integer.
But all integers are distinct. So if the largest integer is 90, then the middle 2 will be 88 and 89
Lowest of least possible integer = 280-(90+89+88)=280-267=13
Answer: D
Answer: D
Question: In a school with 620 students, the average age of the boys is 12 years and that of the girls is 11 years. If the average age of the school is 11 years 9 months, then the number of girls in the school is
Options:
A)150
B)200
C)155
D)350
E)None
Rationale: Sol.
Let the number of grils be x.
Then, number of boys = (620 - x).
Then, (11 3/4 × 620)
⇔ 11x + 12(620 - x) ⇔ x = 7440 - 7285 ⇔ 155.
Answer C
Answer: C
Question: An article costing rs.300 is sold at 10% discount on a mark-up price. What is the selling price after
discount?
Options:
A)270
B)116
C)126
D)136
E)146
Rationale: 300*90/100=270
ANSWER:A
Answer: A
Question: A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?
Options:
A)9:1
B)9:8
C)9:2
D)9:7
E)9:6
Rationale: Explanation:
Milk = 3/5 * 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 - 6 = 6 liters
Remaining water = 8 - 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed, then amount of milk removed = 4/5 * 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 - 8) = 8 liters.
Remaining water = (4 -2) = 2 liters.
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1
Answer: A
Answer: A
Question: The average age of three boys is 15 years. If their ages are in the ratio 3:5:7 the average age of the young boy is ______ years?
Options:
A)18
B)8
C)9
D)19
E)17
Rationale: avg of three boys is 15 years given ratio of their ages are 3:5:7
let x be the age
(3x+5x+7x)/3 =15 implies x=3
then the age of the young boy is 3x3=9
ANSWER:C
Answer: C
Question: The number of ways in which six men and six women can be seated in a row, so that no two girls sit together is?
Options:
A)6! * ⁷P₆
B)5! * 7
C)6! * 7
D)4! * 4
E)None
Rationale: We can initially arrange the six boys in 6! ways.
Having done this, now three are seven places and six women to be arranged. This can be done in ⁷P₆ ways.
Hence required number of ways = 6! * ⁷P₆
A
Answer: A
Question: Medals are to be awarded to three teams in a 10-team competition. If one medal is gold, one medal is silver, and one medal is bronze, how many different ways are there to award the three medals to teams in the competition?
Options:
A)10!/7!
B)10!/(3!7!)
C)10!/3!
D)7!/3!
E)7!/94!3!)
Rationale: Choosing 3 teams out of 10 when order of the teams matters - P310=10!7!
Or: choosing which 3 teams out of 10 will get the medals - C310C103 and arranging them - 3!3!, so total - C310∗3!=10!7!
Or:
1-2-3-4-5-6-7-8-9-10 (teams);
G-S-B-N-N-N-N-N-N-N (GSB - medals, N - no medal);
Permutation of 10 letters out of which 7 N's are identical is 10!7!10!7! (so you'll get 10!7!10!7! different ways of assigning the medals to the teams).
Answer: A.
Answer: A
Question: In a 1000 m race, A beats B by 40 m and B beats C by 100 m. In the same race, by how many meters does A beat C?
Options:
A)145 m
B)136 m
C)268 m
D)129 m
E)218 m
Rationale: By the time A covers 1000 m, B covers (1000 - 40) = 960 m.
By the time B covers 1000 m, C covers (1000 - 100) = 900 m.
So, the ratio of speeds of A and C =
1000/960 * 1000/900 = 1000/864. So, by the time A covers 1000 m, C covers 864 m.
So in 1000 m race A beats C by 1000 - 864 = 136 m.
Answer:B
Answer: B
Question: In 1950, Richard was 4 times as old as Robert. In 1955, Richard was 3 times as old as Robert. In which year was Richard 1.75 times as old as Robert?
Options:
A)1960
B)1965
C)1970
D)1975
E)1980
Rationale: In 1950: Ri = 4 Ro -------------- eq1
In 1955: Ri +5 = 3(Ro+5)---------eq2
Thus in 1950, solving eq1 and eq2
Ro= 10 , Ri =40
Now for each year we can calculate:
1960: Ri = 50 , Ro =20
1965: Ri= 55, Ro=25
1980: Ri= 40, Ro = 70
Thus8Ans:E
Answer: E
Question: A train covers a distance of 12 km in 10 min. If it takes 10 sec to pass a telegraph post, then the length of the train is?
Options:
A)178 m
B)168 m
C)120 m
D)168 m
E)200 m
Rationale: Speed = (12/10 * 60) km/hr = (72 * 5/18) m/sec
= 20 m/sec.
Length of the train = 20 * 10
= 200 m.
Answer: E
Answer: E
Question: find how many 4 digits no. is less then 1239 such that all 4 digits are distinct.
Options:
A)68
B)69
C)70
D)71
E)72
Rationale: case 1- 1000-1099- there are 56 distinct digits
case 2- 1200-1209- there are 7 distinct digits
case 3- 1230-1239(excluding 1239)- there are 6 distinct digits
hence, 56+7+6=69
ANSWER:B
Answer: B
Question: A fruit seller sells mangoes at the rate of Rs.14 per kg and thereby loses 15%. At what price per kg, he should have sold them to make a profit of 15%?
Options:
A)Rs.11.81
B)Rs.12
C)Rs.18.94
D)Rs.12.31
E)None
Rationale: Solution
85 : 14 = 115 : x
x= (14Ã—115/85)
= Rs.18.94
Hence, S.p per Kg	= Rs.18.94
Answer C
Answer: C
Question: The average of six number is 3.95. The average of two of them is 3.4, while the average of the other two is 3.85. what is the average of the remaining three number ?
Options:
A)4.5
B)4.6
C)4.7
D)4.8
E)None of these
Rationale: solution:
Sum of the remaining three numbers = (3.95 × 6) - [(3.4 × 2) + (3.85 ×2)]
=23.70 -(6.8 + 7.7) = 23.70 - 14.5 = 9.20.
∴ Required average =9.2/3 = 3.06 Answer E
Answer: E
Question: How many four digit numbers have no repeat digits, do not contain zero, and have a sum of digits F equal to 28?
Options:
A)14
B)24
C)28
D)48
E)96
Rationale: First, look for all 4 digits without repeat that add up to 28. To avoid repetition, start with the highest numbers first.
Start from the largest number possible 9874.
Then the next largest number possible is 9865.
After this, you'll realize no other solution. Clearly the solution needs to start with a 9 (cuz otherwise 8765 is the largest possible, but only equals 26). With a 9, you also need an 8 (cuz otherwise 9765 is the largest possible, but only equals 27). With 98__ only 74 and 65 work.
So you have two solutions. Each can be rearranged in 4!=24 ways. So F=24+24=48.D
Answer: D
Question: How many unique positive odd integers less than 80 are equal to the product of a positive multiple of 5 and an odd number?
Options:
A)4
B)8
C)11
D)12
E)15
Rationale: The Question basically asks how many positive odd integers less than 80 are odd multiples of 5
So we have 5,15,25,35,45,55, 65,75
=8
B
Answer: B
Question: A man can row downstream at 18 kmph and upstream at 10 kmph. Find the speed of the man in still water and the speed of stream respectively?
Options:
A)13,3
B)12,6
C)15,3
D)14,4
E)None of these
Rationale: Let the speed of the man in still water and speed of stream be x kmph and y kmph respectively.
Given x + y = 18 --- (1)
and x - y = 10 --- (2)
From (1) & (2) 2x = 28 => x = 14, y = 4.
ANSWER:D
Answer: D
Question: Factor: 6x4y3 â€“ 96y3
Options:
A)A) 3y3(x2 + 4)(x + 2)(x -2)
B)B) 6y3(x2 + 4)(x + 2)(x -2)
C)C) 3y3(x2 + 4)(x + 2)(x -3)
D)D) 3y3(x2 + 4)(x + 3)(x -2)
E)E) 3y3(x2 + 4)(x + 2)(x -4)
Rationale: 6x4y3â€“ 963.
= 6y3(x4 â€“ 16).
= 6y3[(x2)2 - 42].
= 6y3(x2 + 4)(x2 - 4).
= 6y3(x2 + 4)(x2 - 22).
= 6y3(x2 + 4)(x + 2)(x -2).
Answer: (B) 6y3(x2 + 4)(x + 2)(x -2)
Answer: B
Question: Seven children — A, B, C, D, E, F, G and H— are going to sit in seven chairs in a row. The children CF have to sit next to each other, and the others can sit in any order in any remaining chairs. How many possible configurations are there for the children?
Options:
A)600
B)720
C)1440
D)10080
E)4800
Rationale: for such questions , we can treat both as one then total person=7..
these 7 can be arranged in 7! ways.. but within these 7, one consists of two people who can be arranged in 2 ways CF or FC.. so ans =7!*2=10080
ans D
Answer: D