divelab/Sys2Bench huggingface dataset:
Question: Each of the integers from 0 to 9, inclusive, is written on a separate slip of blank paper and the ten slips are dropped into hat. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 9?
Options:
A)3
B)4
C)5
D)6
E)7
Rationale: i'm not sure i got the question right, if it is what i think it is, then 5 should be the right answer
0,1,2,3,4,will not have 2 slip sum of 9, add another (either 6,7,8 or 9) and we'll have a pair with sum 9
Ans:C
Answer: C
Question: If y exceeds x by 20%, then x is less than y by?
Options:
A)12
B)5 4/7
C)22
D)16 2/3
E)100
Rationale: Explanation:
X=100	y=120
120------20
100-------? => 16 2/3%
D
Answer: D
Question: If you are a genius, you can solve it easily.
Use three 9's in a mathematical expression such that it forms the number one. You are not allowed to divide or multiply them.
Options:
A)Not possible
B)Possible
C)Do know
D)Doubt
E)Can't do
Rationale: B
Yes We can do it.
9^(9-9)
Answer: B
Question: The salary of a typist was at first raised by 10% and then the same was reduced by 5%. If he presently draws Rs.5225.What was his original salary?
Options:
A)2277
B)5000
C)1000
D)2651
E)1971
Rationale: X * (110/100) * (95/100) = 5225
X * (11/10) * (1/100) = 55
X = 5000
Answer: B
Answer: B
Question: John has x dollars more than Peter has, and together they have a total of y dollars. Which of the following represents the number of dollars that Peter has?
Options:
A)(y – x)
B)3(y – x)/2
C)(y – x)/2
D)(y – x)/7
E)(2y – x)/2
Rationale: To solve, we will set up two equations. Let's start by defining two variables.
A = number of dollars Peter has
B= number of dollars John has
We are given that John has x dollars more than Peter. We set up an equation:
B= x + A
We are next given that together they have a total of y dollars. We can set up our second equation:
A + B = y
Since we know that B = x + A, we can substitute x + A for B into the second equation A+ B = y.
Notice that, after the substitution, we will only have variables of A, x, and y. Thus, we have:
A + x + A = y
2A + x = y
2A = y – x
A = (y – x)/2
Answer : C
Answer: C
Question: If n! and (n + 1)! have the same units digit, then n can be which of the following?
I. n = 10
II. n = 1
III. n = 0
Options:
A)I only
B)II only
C)III only
D)I and III
E)I, II, and III
Rationale: Given: n! and (n + 1)! have the same units digits
Let us first write down the factorials of different numbers:
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
If we notice here, all the numbers after 4! will end with a 0.
Here is why:
If we multiply any number by 10, a 0 is added to that number
And a 10 is formed by a 5 and a 2, 10 = 5*2
Till 4!, we do not have a "5" in multiplication. But after that, we will have atleast one "5" in the multiplication.
Hence the units digits will be same after 5! and for 0! and 1!
Now checking the statements
I. n = 10
Yes, the units digits of 10! and 11! are same and = 0. TRUE
II. n = 1
units digit f 1! = 1 and units digit of 2! = 2. FALSE
III. n = 0
Units digit of 1! = 1 and units digit of 0! = 1. TRUE
Correct Option: D
Answer: D
Question: A man rows his boat 85 km downstream and 45 km upstream, taking 2 1/2 hours each time. Find the speed of the stream?
Options:
A)7 kmph
B)9 kmph
C)1 kmph
D)8 kmph
E)2 kmph
Rationale: Speed downstream = d/t = 85/(2 1/2) = 34 kmph
Speed upstream = d/t = 45/(2 1/2) = 18 kmph
The speed of the stream
= (34 - 18)/2
= 8 kmph
Answer: D
Answer: D
Question: Which of the following lines is perpendicular to y = − 2x + 3 and has the same y-intercept as y = 2x − 5?
Options:
A)y = -1/2*x + 3
B)y = 1/2*x - 5
C)y = 1/2*x + 3
D)y = 2x + 3
E)y = 2x − 3
Rationale: Since our line has to perpendicular to y = − 2x + 3
Slope of this line , m = -2
Slope will be required line L = 1/2 ( Product of slope of perpendicular lines = -1)
y-intercept of line L is same as y intercept of line y = 2x − 5
y - intercept of line L = -5
Required line = 1/2* x - 5
Answer B
Answer: B
Question: What is the minimum number of square marbles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm?
Options:
A)176
B)187
C)540
D)748
E)546
Rationale: The marbles used to tile the floor are square marbles.
Therefore, the length of the marble=width of the marble.the length of the marble=width of the marble.
As we have to use whole number of marbles, the side of the square should a factor of both 5 m 78 cm and 3m 74. And it should be the highest factor of 5 m 78 cm and 3m 74.
5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm.
The HCF of 578 and 374 = 34.
Hence, the side of the square is 34.
The number of such square marbles required,
=578×374/34×34
=17×11= 187 marbles
Option(B)
Answer: B
Question: A family consist of a grandfather, 5 sons and daughter and 8 grandchildren. They are to be seated in a row for dinner. The grandchildren wish to occupy the 4 seats at each end and the grandfather refuses to have a grandchild on either side of him. The number of ways in which the family can be made to sit is:
Options:
A)21530
B)8! * 360
C)8! * 480
D)8! * 240
E)None
Rationale: Solution:
Total no. of seats,
= 1 grandfather+ 5 sons and daughters + 8 grandchildren
= 14.
The grandchildren can occupy the 4 seats on either side of the table in 4! = 24 ways.
The grandfather can occupy a seat in (5-1)= 4 ways (4 gaps between 5 sons and daughter).
And, the remaining seats can be occupied in 5!= 120 ways (5 seat for sons and daughter).
Hence total number of required ways,
= 8! � 480
Answer: Option C
Answer: C
Question: What least number must be subtracted from 3830 so that the remaining number is divisible by 15?
Options:
A)3
B)1
C)5
D)11
E)15
Rationale: On dividing 3830 by 15, we get remainder = 5.
Required number be subtracted = 5
ANSWER:C
Answer: C
Question: Find the odd man out
864, 420, 200, 93, 40, 16, 6
Options:
A)420
B)200
C)16
D)93
E)6
Rationale: 6x2+4=16
16x2+8=40
40x2+12=92
92x2+16=200.. and so on. 93
ANSWER:D
Answer: D
Question: Four friends decided to go on a vacation to a hill station. During their vacation, they experienced 13 days of heavy snowfall. When it used to snow in the morning, the afternoon was clear and when it used to snow in the afternoon, the day was followed by a clear morning.
In total, they had eleven clear mornings and twelve clear afternoons. Is there a way you can calculate the number of days they spent in their vacation?
Options:
A)16
B)21
C)18
D)19
E)25
Rationale: C
18
Number of days with snow in morning and clear afternoon = A
Number of days with clear morning and snow in afternoon = B
Number of days with no snowfall in morning or afternoon = C
Now, number of days with snowfall = A + B = 13
Number of days with clear mornings = B + C = 11
Number of days with clear afternoons = A + C = 12
Adding these three equations, we get
2 * (A + B + C) = 36
Or, A + B + C = 18
Therefore, they spent eighteen days in total at their vacation.
Answer: C
Question: If x=2y=z, what is x-y, in terms of z?
Options:
A)z
B)z/2
C)z/3
D)2z/5
E)z/5
Rationale: x=2y=z
x = z
and y = z/2
x-y = z - z/2 = z/2
Answer is B
Answer: B
Question: Seven dierent playing cards, with values from ace to seven, are shued and
placed in a row on a table to form a seven-digit number. What is the probability that this
seven-digit number is divisible by 11?
Note: Each of the possible seven-digit numbers is equally likely to occur.
Options:
A)4/35
B)1/7
C)8/35
D)2/7
E)12/35
Rationale: To be divisible by 11, the digits must be arranged so that the difference between
the sum of one set of alternate digits, and the sum of the other set of alternate digits, is either
0 or a multiple of 11. The sum of all seven digits is 28. It is easy to nd that 28 can be
partitioned in only two ways that meet the 11 test: 14j14, and 25j3. The 25j3 is ruled out
because no sum of three dierent digits can be as low as 3. Therefore, only the 14j14 partition
need to be considered.
There are 35 dierent combinations of three digits that can fall into the B positions in the
number ABABABA. Of those 35, only four sum to 14: 167, 257, 347, and 356. Therefore, the
probability that the number will be divisible by 11 is 4/35.
correct answer A
Answer: A
Question: A certain roller coaster has 7 cars, and a passenger is equally likely to ride in any 1 of the 7 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 7 times, what is the probability that the passenger will ride in each of the 7 cars?
Options:
A)0
B)1/9
C)2/9
D)720/16807
E)1
Rationale: If he is to ride 6 times and since he can choose any of the 6 cars each time, total number of ways is
= 7 * 7 * 7 * 7 * 7 * 7
= 117649
Now the number of ways if he is to choose a different car each time is
= 7 * 6 * 5 * 4 * 3 * 2 * 1
= 5040
So the probability is
= 5040/117649 = 720/16807
Answer : D
Answer: D
Question: In an IT company, there are a total of 100 employees including 50 programmers. The number of male employees is 80, including 35 male programmers. How many employees must be selected to guaranty that we have 3 programmers of the same sex?
Options:
A)10
B)50
C)55
D)35
E)65
Rationale: You could pick 50 non-programmers, 2 male programmers, and 2 female programmers, and still not have 3 programmers of the same sex. But if you pick one more person, you must either pick a male or a female programmer, so the answer is 55.
C
Answer: C
Question: Mixture A is 20% oil and 80% material B by weight. If 2 more kilograms of oil are added to the 8 kilograms mixture A, how many kilograms of mixture A must be added to make a 70% material B in the new mixture?
Options:
A)6
B)7
C)8
D)9
E)10
Rationale: Old mix has 20% oil. You add 2 kgs of 100% oil. You get new mix with 30% oil (70% material B).
w1/w2 = (100 - 30)/(30 - 20) = 70/10 = 7/1
For every 7 parts of mix A, you have added 1 part of oil. Since you have added 2 kgs oil, mix A must be 7*2 = 14 kgs.
You already had 8 kgs of mix A, so extra mix A added must be 14 - 8 = 6 kgs.
Answer (A)
Answer: A
Question: A company has 15 equally qualified applicants for 4 open positions. How many different groups of 4 applicants can be chosen by the company to fill the positions if the order of selection does not matter?
Options:
A)1275
B)1315
C)1365
D)1405
E)1455
Rationale: 15C4=1365
The answer is C.
Answer: C
Question: In a card game, a combination of two queen beats all others. If Jose is the first to draw from a standard deck of 52 cards, what is the probability that he wins the game with the best possible combination?
Options:
A)13/221
B)1/221
C)2/52 D
D)3/51
E)4/51
Rationale: he best possible combination is the case when he wins in his first two draws
There are 52 cards in a deck of cards with 4 queens in it.
Probability of first card to be queen = 4/52
Probability of Second card to be queen = 3/51
Probability of Best possible case = (4/52)*(3/51) = 1/221
B
Answer: B
Question: When R is divided by 9, the remainder is 4. All of the following are possible values of R, except for which of the following?
Options:
A)13
B)31
C)49
D)57
E)67
Rationale: The number can be expressed in the form of 9x+4 where x can 0,1,2,3,4....
by evaluating the answer choices carefully we can clearly observe that 57 is the only number which can't be expressed R in the form of 9x+4
In other words we can also say that the (answer - 4) will not be divisible by 9.57 is the number which doesn't follow this condition
Correct Answer - D
Answer: D
Question: If the wheel is 12 cm then the number of revolutions to cover a distance of 1056 cm is?
Options:
A)18
B)26
C)14
D)12
E)91
Rationale: 2 * 22/7 * 12 * x = 1056 => x
= 14
Answer: C
Answer: C
Question: In a football team, scored 3 goals luis diego more than 2 goals more than samuel who scored three parts of what I write down luis and diego. Diego scored many goals?
Options:
A)2
B)4
C)0
D)1
E)3
Rationale: Realizamos un sistemas de tres ecuaciones, Luis=3+Diego; Luis=2+Samuel, Samuel=Luis+Diego/3
Resolvemos
Diego=0
Answer: C
Answer: C
Question: A can do a piece of work in 10 days and B can do the same work in 12 days. A and B worked together for 1 days. How many more days are required to complete the remaining work if they work together?
Options:
A)3 (5/18)
B)3 (5/16)
C)3 (5/11)
D)3 (5/13)
E)4 (5/11)
Rationale: A can do 1/10 of the work in a day.
B can do 1/12 of the work in a 1 day.
Both of them together can do (1/10 + 1/12) part of work in 1 day = (6 + 5)/60 = 11/60
They take 60/11 days to complete the work together.
Given that they already worked for 1 days.
The number of days required to complete remaining work => 60/11 - 1 = 49/11 = 4 (5/11) days.
Answer:E
Answer: E
Question: If (x-13)(x-3) < 0, which of the following must be true?
Options:
A)x^2 + 5x + 6 < 0
B)x^2 + 5x + 6 > 0
C)5 - x < 0
D)x - 5 < 0
E)13 - x > 0
Rationale: (x-13)(x-3)<0 can be written as (13-x)(x-3)>0
So,to inequality hold true multiple of (13-x) and (x-3) must be positive and if I let both elements positive/negative then the range becomes 3<x<13
Answer Choice E matched which says 13-x>0 or x<13,which must be true (I started to look from E to A for making easy decision first)
Correct Answer E
Answer: E
Question: How many numbers from 29 to 119 are exactly divisible by 11?
Options:
A)5
B)7
C)9
D)8
E)12
Rationale: 29/11 = 2 and 119/11 = 10 ==> 10 - 2 = 8 Numbers
Answer : D
Answer: D
Question: In a certain large company, the ratio of college graduates with a graduate degree to non-college graduates is 1:8, and ratio of college graduates without a graduate degree to non-college graduates is 2:3. If one picks a random college graduate at this large company, what is the probability Q this college graduate has a graduate degree?
Options:
A)1/11
B)1/12
C)1/13
D)3/19
E)3/43
Rationale: In believe the answer is D. Please see below for explanation.
0) we are told the following ratios
CGD - College Graduate with Degree
NCG - Non College Graduate
CGN - College Graduate no Degree
CGD NCG CGN
1 8
3 2
In order to make CGD and CGN comparable we need to find the least common multiple of 8 and 3 and that is 24 multiplying the first ratio by 3 and the second ratio by 8 we get
CGD NCG CGN
3 24 16
If one picks a random college graduate at this large company, what is the probability this college graduate has a graduate degree?
Nr of CGD = 3
Nr of CG = 3+ 16 = 19
Probability of CGD / (CG) Q-> 3/19
Answer D
Answer: D
Question: What Is The Next Number
12 13 15 17 111 113 117 119 123 ?
Options:
A)154
B)220
C)129
D)428
E)224
Rationale: C
129.
These are the first 10 prime numbers (2, 3, 5...) prefixed with a 1
Answer: C
Question: 1(1!)+2(2!)+3(3!)....2012(2012!) = ?
Options:
A)1
B)5
C)6
D)7
E)8
Rationale: 1(1!)=1 ⇒⇒ 2!-1
1(1!)+2(2!)=1+4=5 ⇒⇒ 3!-1
1(1!)+2(2!)+3(3!)=1+4+18=23 ⇒⇒ 4!-1
Answer:A
Answer: A
Question: If x > 1, which of the following decreases as x decreases?
I. x + x^3
II. 2x^2 − x
III. 1/(x + 1)
Options:
A)I only
B)II only
C)III only
D)I and III
E)II and II
Rationale: let x= 4 and 3 b old and new values
ST1: x= 4 -> 4+64 = 68
x=3-> 3+27= 30 ...decreased
ST2:x= 4-> 2*16-4 = 28
x=3->2*9-3 = 15
decreased
St3: x=4-> 1/4 = .25
x=3->1/3 = .33
Ans: E
Answer: E
Question: Half of 5 percent written as decimal is
Options:
A)5
B)0.5
C)0.05
D)0.025
E)None of these
Rationale: Explanation:
It will be 1/2(5%) = 1/2(5/100) = 5/200 = 0.025
Answer: Option D
Answer: D
Question: An empty wooden vessel weighs 12% of its total weight when filled with paint. If the weight of a partially filled vessel is one half that of a completely filled vessel, what fraction of the vessel is filled.
Options:
A)3/5
B)6/11
C)1/24
D)4/9
E)2/5
Rationale: An empty wooden vessel weighs 12% of its total weight when filled with paint:
Vessel = 0.12(Vessel + Paint);
12V = V + P (so the weight of completely filled vessel is 10V)
P = 11V (so the weight of the paint when the vessels is completely filled is 9V).
The weight of a partially filled vessel is one half that of a completely filled vessel:
V + P' = 1/2*12V;
P' = 6V (so the weight of the paint when the vessels is partially filled is 6V).
What fraction of the vessel is filled?
So, we need to find the ratio of the weight of the paint when the vessel iscompletely filledto the weight of the paint when the vessel ispartially filled:
P'/P = 6V/11V = 6/11.
Answer: B.
Answer: B
Question: By selling an article at Rs.800, a profit of 25% is made. Find its cost price?
Options:
A)s.486
B)s.455
C)s.640
D)s.480
E)s.489
Rationale: SP = 800
CP = (SP)*[100/(100+P)]
= 800* [100/(100+25)]
= 800* [100/125]
= Rs.640
Answer: C
Answer: C
Question: If a trader sold two cars each at Rs. 325475 and gains 10% on the first and loses 10% on the second, then his profit or loss percent on the whole is?
Options:
A)1.44%
B)1.74%
C)1.84%
D)1.47%
E)1.00%
Rationale: SP of each car is Rs. 325475, he gains 10% on first car and losses 10% on second car.
In this case, there will be loss and percentage of loss is given by
= [(profit%)(loss%)]/100
= (10)(10)/100 % = 1.00%
Answer:E
Answer: E
Question: Two tailors x and y are paid a total of $550 per week by their employer. If x is paid 120% of the sum paid to y. How much is y paid per week?
Options:
A)150
B)100
C)240
D)190
E)550
Rationale: Let the sum paid to y per week be $z
then, z+120% of z = 550
z+(120/100)z = 550
z = 250
Answer is E
Answer: E
Question: Envelopes can be purchased for $1.50 per pack of 100, $1.00 per pack of 50, or $0.03 each. What is the greatest number of envelopes that can be purchased for $8.10?
Options:
A)426
B)430
C)443
D)520
E)486
Rationale: I think first we should see in what group envelopes value less, so that we can buy as many as we can with a specific amount of money:
-the 100 pack costs 1.5$, meaning 0.015$ per envelope
-the 50 pack costs 1$, meaning 0.02$ per envelope
-the single envelope costs 0.03$ per envelope
Thus, we have to buy as many 100 packs as we can, then as meny of 50 packs as we can and the remaining envelopes are single ones.
Having 8.1$, we can buy as many as 5 packs of 100 (5 x 1.5$ = 7.5$). We then have 0.6 $ left. so we buy 20 individual envelopes. If we add up the numbers, we get 500+20=520 envelopes. Answer D
Answer: D
Question: If 23^3*19^4*14^2 = x, what is the units digit of x ?
Options:
A)2
B)3
C)6
D)7
E)9
Rationale: 23^3*19^4*14^2 = x
Units digit of x will be
3^3 * 9^4 * 4^2
=7* 1 *6
=42
Therefore , units digit of x = 2
Answer A
Answer: A
Question: A car takes 6 hours to cover a distance of 270 Km. how much should the speed in Kmph be maintained to cover the same direction in 3/2th of the previous time?
Options:
A)10Kmph
B)20Kmph
C)25Kmph
D)30Kmph
E)27Kmph
Rationale: Time = 6
Distence = 270
3/2 of 6 hours = 6 * 3/2 = 9 Hours
Required speed = 270/9 = 30 Kmph
D)
Answer: D
Question: What sum of money will produce Rs.70 as simple interest in 5 years at 3 1/2 percent?
Options:
A)400
B)500
C)367
D)368
E)339
Rationale: 70 = (P*5*7/2)/100
P = 400
Answer: A
Answer: A
Question: Jasmine has a toaster with two slots that toasts one side of each piece of bread at a time, and it takes one minute to do so.
If she wants to make 3 pieces of toast, what is the least amount of time she needs to toast them on both sides?
Options:
A)1
B)2
C)3 min
D)4
E)5
Rationale: 3 minutes. She puts two pieces in the toaster, toasting one side of each. Then she flips one of them, takes one out, and puts the completely untoasted piece into the toaster. Finally, she takes out the toasted piece and puts the two half-toasted pieces of bread into the toaster for a minute and she's done. C
Answer: C
Question: Adding two non multiples of Q always results in a multiple of Q; which of the following must be the value of Q ?
Options:
A)[A] two
B)[B] three
C)[C] five
D)[D] ten
E)[E] hundred
Rationale: NON- multiples of 2 would always be ODD and sum of two odd would be EVEN.
so irrespective of value of non-multiples, sum of two them would always be even and thus multiple of 2..
A
Answer: A
Question: A towel, when bleached, lost 30% of its length and 40% of its breadth. What is the percentage decrease in area?
Options:
A)24%
B)30 %
C)44%
D)58%
E)64%
Rationale: percentage change in area
=(−30−40+(30×40)/100)%=−58%
i.e., area is decreased by 58%
ANSWER:D
Answer: D
Question: A boat can move upstream at 25 kmph and downstream at 35 kmph, then the speed of the current is?
Options:
A)5
B)6
C)8
D)4
E)2
Rationale: US = 25
DS = 35
M = (35 - 25)/2 = 5 Answer:A
Answer: A
Question: in how many different ways can the letters of the word OPTICAL be arranged so that the vowels always come together
Options:
A)520
B)620
C)720
D)750
E)800
Rationale: The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
ANSWER:C
Answer: C
Question: The price of rice falls by 20%. How much rice can be bought now with the money that was sufficient to buy 40 kg of rice previously?
Options:
A)5 kg
B)15 kg
C)25 kg
D)50 kg
E)None
Rationale: Solution: Let Rs. 100 be spend on rice initially for 40 kg.
As the price falls by 20%, new price for 40 kg rice,
= (100 -20% of 100) = 80
New price of rice = 80/40 = Rs. 2 per kg.
Rice can bought now at = 100/2 = 50 kg.
Answer: Option D
Answer: D
Question: John has taken four (4) tests that have an average of 80. In order to bring his course grade up to a ‘B’, he will need to have a final average of 87. What will he need to average on his final two tests to achieve this grade?
Options:
A)87
B)101
C)92
D)94
E)97
Rationale: 87-80=7*4=28 points
2*87=174
174+28=202 points to be scored in the 2 tests.
Avg of 2 tests = 202/2=101
Answer: B
Traditional Method:
Total scored till now 80*4=320
Total score to avg 87 in 6 tests = 87*6=522
Total to be scored on 2 tests = 522-320=202
Avg on 2 tests = 202/2 = 101
Answer B
Answer: B
Question: Find the C.I. on Rs. 15,625 for 9 months at 16% per annum compounded quarterly?
Options:
A)19587
B)1987
C)1951
D)1998
E)1912
Rationale: P = Rs. 15625, n = 9 months = 3 quarters, R = 16% p.a. per quarter.
Amount = [15625 * (1 + 4/100)3]
= (15625 * 26/25 * 26/25 * 26/25) = Rs. 17576 C.I. = 17576 - 15625 = Rs. 1951.
Answer: C
Answer: C
Question: A large research project was scheduled over the course of a 30-month period, and a budget of $99,000 was allotted for the course of the entire project. After ten months, the project had spent $30,000. Was the project at this point over budget or under budget, and by how much?
Options:
A)under, $4200
B)under, $1200
C)under, $700
D)on bugdget
E)over $5800
Rationale: cost per month = 99,000/30 = 3,000.
Now, multiply by seven months: 10*($3000) = $30,000. That would be the budgeted amount, the amount theoretically supplied to the project. The real amount spent is equal to this, so it is on budget.
$30,000 â€“ #30,000 = $0
Answer = (D)
Answer: D
Question: Directions : In questions no. 1 to 8, select the one which is different from other three alternatives.
Options:
A)Delegate
B)Person
C)Manager
D)Deputy
E)Representative
Rationale: B
Except (B) all are designations.
Answer: B
Question: What percentage of numbers from 1 to 20 have squares that end in the digit 9?
Options:
A)1
B)14
C)20
D)21
E)22
Rationale: Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 18 are 3,7,13,17.
Number of such numbers = 4.
Required percentage = (4/18 * 100) = 22%
ANSWER:E
Answer: E
Question: Mariah has decided to hire three workers. To determine whom she will hire, she has selected a group of 10 candidates. She plans to have one working interview with 3 of the 10 candidates every day to see how well they work together. How many days will it take her to have working interviews with all the different combinations of job candidates?
Options:
A)720
B)120
C)300
D)30
E)333
Rationale: There are C310=120 different groups of three possible out of 10 candidates and since each of these groups is interviewed every day then total of 120 days are needed.
Answer: B
Answer: B
Question: A constructor estimates that 3 people can paint Mr khans house in 4 days. If he uses 4 people instead of 3,how long will they take to complete the job?
Options:
A)2
B)8
C)1
D)3
E)4
Rationale: Explanation:
Use formula For a work Members × days = constant
3 × 4 = 4 × a
a = 3
so answer is 3 days
Answer: D
Answer: D