RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: Alicia lives in a town whose streets are on a grid system, with all streets running east-west or north-south without breaks. Her school, located on a corner, lies three blocks south and three blocks east of her home, also located on a corner. If Alicia s equally likely to choose any possible path from home to school, and if she only walks south or east, what is the probability she will walk south for the first two blocks?
A.1/3
B.1/5
C.1/6
D.1/7
Answer: B.1/5
Explanation: Total number of ways to go home, considering that the order order matters =
Options: SSSEEE
Slots : ------
total permutations of 6 options in 6 slots: 6P6 = 6! = 6*5*4*3*2
Total ways to select "South" in the first 2 positions and anything else in the subsequent 4 positions =
Options: SS ????
Slots: -- ----
Permutations of 3 "S" in 2 slots AND Permutations of 4 Choices in 4 slots =
3P2 * 4P4 = 3! * 4! = 3*2*4*3*2
The probability is therefore:
3P2 * 4P4 = 6P6
3*2*4*3*2= 1/5
ANSWER:B
Question: There are 4 more women than men on Centerville's board of education. If there are 4 members on the board, how many are women?
A. 3
B. 4
C. 6
D. 7
Answer: B. 4
Explanation: Given: M=W-4 so, W+(W-4)=4 --> W=4.
Answer: B.
Question: If O is the center of the circle in the figure above and the area of the unshaded sector is 5, what is the area of the shaded region?
A.25/√π
B.30/√π
C.20
D.25
Answer: D.25
Explanation: 60/360 = 1/6
1/6 of total area = 5
5/6 of total area = 5*5 =25
ANSWER:D
Question: What quantity of water should taken out to concentrate 27 liters of 40% acidic liquid to 60% acidic liquid?
A.5liters
B.10liters
C.15liters
D.9liters
Answer: D.9liters
Explanation: Required answer is = 27(60-40)/60 = 9 liters
Answer is D
Question: A rectangular grass field is 75 m * 55 m, it has a path of 3.2 m wide all round it on the outside. Find the area of the path and the cost of constructing it at Rs.2 per sq m?
A.s.1350
B.s.1327
C.s.1328
D.s.1364
Answer: D.s.1364
Explanation: Area = (l + b +2d) 2d
= (75 + 55 +3.2 * 2) 2*2.5 => 682
682 * 2 = Rs.1364
Answer: D
Question: Two sides of a triangle have length 6 and 8. Which of the following are possible areas of the triangle?
I. 6
II. 12
III. 24
A.I, II, and III
B.I and II only
C.II and III only
D.I and III only
Answer: A.I, II, and III
Explanation: Answer = A. I, II, and III
With dimensions of only 2 sides given (for a triangle), this becomes an open ended question
Any area is possible out of the three
Question: A 240 meter long train running at the speed of 120 kmph crosses another train running in the opposite direction at the speed of 80 kmph in 9 seconds.What is the lenght of other train.
A.260m
B.220m
C.230m
D.240m
Answer: A.260m
Explanation: Relative speeds=(120+80)km/hr
=200km/hr=(200*5/18)m/s=(500/9)m/s
let length of train be xm
x+240/9=500/9
x=260
Ans is 260m
ANSWER: A
Question: A rower can row upstream at 7 km/hr and downstream at 15 km/hr. What is the rower's rate in still water?
A.9 km/hr
B.10 km/hr
C.11 km/hr
D.12 km/hr
Answer: C.11 km/hr
Explanation: The rate in still water is 1/2 * (15 + 7) = 11 km/hr
The answer is C.
Question: Two vertical pole of 2 m and 8 m stand apart on a horizontal plane.The height in meters at the point of intersection of the line joining the top of each pole to the bottom of other pole is ....
A.1.8
B.1.6
C.1.4
D.5.4
Answer: C.1.4
Explanation: formula is...(p*q)/p+q
here p=2,q=8
height in meters=(8*2)/8+2=16/10=1.6
ANSWER:C
Question: To be considered for “movie of the year,” a film must appear in at least 1/4 of the top-10-movies lists submitted by the Cinematic Academy’s 760 members. What is the smallest number of top-10 lists a film can appear on and still be considered for “movie of the year”?
A.191
B.190
C.193
D.212
Answer: B.190
Explanation: Total movies submitted are 760.
As per question we need to take 1/4 of 760 to be considered for top 10 movies = 190
approximate the value we 190.
IMO option B is the correct answer...
Question: If 4 people are selected from a group of 8 married couples, what is the probability that none of them would be married to each other?
A.1/33
B.2/33
C.8/13
D.16/33
Answer: C.8/13
Explanation: If we are to select 4 people from 8 couples WITHOUT any restriction, how many ways can we make the selection? 16!/4!12! = 1820
If we are to select 4 people from 8 couples WITH restriction that no married couple can both make it to the group, only a representative?
8!/4!4! = 70
But we know that to select a person from each couple, take 2 possibilities
70*2*2*2*2 = 1120
Probability = Desired/All Possibilities = 1120/1820 = 8/13
Answer: C
Question: The cost of 3 chairs and 10 tables is Rs 9856. What is the cost of 6 chairs and 20
tables?
A.Rs 17227
B.Rs 18712
C.Rs 19172
D.Rs 19712
Answer: D.Rs 19712
Explanation: Let the cost of one chair be Rs x and that of a table be = Rs y
According to the question,
3x + 10y = Rs 9856
⇒ 2 × (3x + 10y) = 2 × 9856
∴ 6x + 20y = Rs 19712
Answer D
Question: A certain hall contains two cuckoo clocks. If the first clock chirps 20 times per hour and the second clock chirps 7 times per hour, and both clocks chirp for the first time at 2:03 pm, at what time will the first clock have chirped three times as many times as the second clock can chirp per hour?
A.2:53 pm
B.3:03 pm
C.3:13 pm
D.3:23 pm
Answer: B.3:03 pm
Explanation: The question is asking when the first clock will have chirped 21 times (3*7 chirps).
The first clock chirps once every 3 minutes.
If the first chirp is at 2:03, then the 21st chirp is at 63 minutes after 2:00.
The answer is B.
Question: How many cubes of 10cm edge can be put in a cubical box of 1m edge
A.10
B.100
C.1000
D.10000
Answer: C.1000
Explanation: Number of cubes=(100x100x100)/(10x10x10)=1000 Answer: C) 1000
Question: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x,y}, {w, x, y}, and { } (the empty subset).
How many subsets of the set {w, x, y, z} contain w?
A.Four
B.Five
C.Seven
D.Eight
Answer: A.Four
Explanation: The number of ways of selecting 0 or more from n elements is 2^n.
Hence here, once you select w , the number of ways of selecting 0 or more from x,y and z ( 3 elements) is 2^3 = 8.
The same applies for {w, x, y}. If you select w here, the number of ways of selecting 0 or more from x and y = 2^2 = 4
ANSWER:A
Question: The greatest common factor of two numbers is 5!. Which of the following can be the second number, if one of the numbers is 6!?
A.3(5!)
B.5(5!)
C.4(5!)
D.6(5!)
Answer: B.5(5!)
Explanation: GCF is the product of common factors of the numbers involved.
GCF = 5!
a = 6! = 6*5!
b will certainly have 5! and cannot have any more common factors with a (as this will increase the GCF)
Looking at the answers only 5 (5!) and 6! will have GCF as 5!
Ans B
Question: There are twelve consecutive flags at an equal interval of distance. A man passes the 8th flag in 8 seconds. How many more seconds will he take to pass the remaining 4 flags ?
A.2.57 seconds
B.3.57 seconds
C.4.57 seconds
D.5.57 seconds
Answer: C.4.57 seconds
Explanation: let the 1st flag i placed at the origin and the remaining on the X-axis.......then the coordinate on 8th flag is (7,0)..therefore distance travelled is 7 units in 4 second ......and coordinate of 12th flag is (11,0)..there r 4 units ....so time taken is (8*4)/7=4.57 seconds
ANSWER:C
Question: A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that atleast one bulb is good?
A.125/129
B.125/122
C.125/126
D.125/128
Answer: C.125/126
Explanation: Required probability
= 1 - 1/126
= 125/126
Answer: C
Question: If a number is exactly divisible by 11 and 13, which of the following types the number must be?
A.Divisible by (11 + 13)
B.Divisible by (13 - 11)
C.Divisible by (11 x 13)
D.Divisible by (13 ÷ 11)
Answer: C.Divisible by (11 x 13)
Explanation: Answer
LCM of 11 and 13 will be (11 x 13). Hence, if a number is exactly divisible by 11 x 13, then the same number must be exactly divisible by their LCM or by (11 x 13).
Correct Option: C
Question: If x = 1- 2t and y = 2t - 2, then for what value of t does x = y?
A. 5/2
B. 3/2
C. 3/4
D. 2/5
Answer: C. 3/4
Explanation: We are given x = 1 – 2t and y = 2t – 2, and we need to determine the value for t when x = y. We should notice that both x and y are already in terms of t. Thus, we can substitute 1 – 2t for x and 2t – 2 for y in the equation x = y. This gives us:
1 – 2t = 2t – 2
3 = 4t
3/4 = t
The answer is C.
Question: If x and y are positive integers, and 4x^2=2y, then which of the following must be a multiple of 9?
I. x^2
II. y^2
III. xy
A.I only
B.II only
C.III only
D.I and II only
Answer: B.II only
Explanation: 4x^2 = 3y
since x,y are positive integers, x^2 = x*x is divisible by 3 -> x is divisible by 3 and y is divisible by x^2
-> x^2 and y is divisible by 9 -> y^2 is divisible by 9
(1),(2), and (3) must be true
Answer is B
Question: The average (arithmetic mean) of 20, 40, and 60 is 8 more than the average of 10, 70, and what number?
A.16
B.25
C.35
D.45
Answer: A.16
Explanation: a1 = 120/3 = 40
a2 = a1 -8 = 32
sum of second list = 32*3 = 96
therefore the number = 96 -80 = 16
ANSWER:A
Question: Anil had to do a multiplication. Instead of taking 35 as one of the multipliers, he took 53. As a result, the product went up by 540. What is the new product?
A.1590
B.1600
C.1950
D.1630
Answer: A.1590
Explanation: Explanation :
Let the number that Anil wanted to multiply be 'Y'.
He was expected to find the value of 35Y.
Instead, he found the value of 53Y.
The difference between the value that he got (53Y) and what he was expected to get (35Y) is 540.
=> 53Y - 35Y = 540
=> (53 - 35) * Y = 540
=> Y = 30
Therefore, the correct product = 53 * 30 = 1590
Answer : A
Question: Murali travelled from city A to city B at a speed of 50 kmph and from city B to city C at 50 kmph. What is the average speed of Murali from A to C given that the ratio of distances between A to B and B to C is 2 : 3?
A.kmph
B.kmph
C.kmph
D.kmph
Answer: B.kmph
Explanation: Let the distances between city A to B and B to C be 2x km and 3x km respectively.
Total time taken to cover from A to C
= (2x)/50 + (3x)/50
= (2x + 3x)/50 = 5x/100
= x/20 Average speed
= (2x + 3x)/(x/20) = 100 kmph.
Answer: B
Question: A city council will select 2 of 9 available firefighters and 1 of 4 available police officers to serve on an advisory panel. How many different groups of 3 could serve on the panel?
A.36
B.72
C.144
D.216
Answer: C.144
Explanation: OA D
9C2 * 4C1
using the combination formula to choose 2 from 9 and 1 from 4
C
Question: you need to pict any number from 1, 3,5,7,9,11,13 and 15 to make below equation true.
()+()+()=30
can you solve it?
A.7+15+9
B.6+15+9
C.6+14+9
D.6+15+8
Answer: B.6+15+9
Explanation: Correct Answer : B
3x+15+9=30
3x =3*2*1 =6
therefore :- 6+15+9= 30
Question: A number is said to be a “digifac” if each of its digits is a factor of the number itself. What is the sum Z of the missing digits of the following five-digit digifac: 9, 5, 3 _ _ ?
A. 5
B. 7
C. 9
D. 10
Answer: D. 10
Explanation: Here, the term “digifac” should look intimidating. You probably haven’t studied digifacs before, so how should you approach this problem? Well, keep in mind that digifacs aren’t being tested; in fact, the author of this question just made that term up, and then defined it for you. What makes this question hard is that the non-challenge-seeker (I think I just made that term up, too…) will see the unfamiliar term “digifac” and lose faith immediately. “I don’t know what that is!” She who finds the challenge in the GMAT fun, however, will read the definition and think “got it – I need to find the two digits that ensure that 9, 5, and 3 are both factors of the overall number, and that the remaining two digits are also factors”. And work from there. The number must be divisible by 5, so the only units digits that work are 0 or 5. And the number must be divisible by 9 (and also 3), so we need the sum Z of all digits to be a multiple of 9. 9 + 5 + 3 = 17, so our only options are to get the sum to 18 (by adding 1) or to 27 (by adding 10). A quick glance at the answer choices shows that 0 1 isn’t an option. Why not? That would require 0 to be one of the digits…and 0 isn’t a factor of anything. So the units digit must be 5, making the tens digit 5, and we have 95,355. That number is a multiple of 5, 3, and 9, so it works:the correct answer is D, and more importantly this fun challenge required no “trivial” information about digifacs…that term only existed to obscure the link between the given information and the path to the answer.D
Question: If $0.80 is the commission for sales of $2,000, what percent of the sales amount is the commission?
A.4%
B.0.4%
C.0.04%
D.0.004%
Answer: C.0.04%
Explanation: % of sales amount of commission
= (Commission/Total Value) *100
= (0.8/2000) * 100
= 0.04%
The answer is C.
Question: A bookseller sells his books at a 10% markup in price. If he sells a book for $11.00, how much did he pay for it?
A.$14.40
B.$14.00
C.$10.00
D.$9.60
Answer: C.$10.00
Explanation: Let the cost price of book = x
Selling price of book = 11$
Markup % = 10
(110/100)x = 11
=> x = 10
Answer C
Question: Which of the following could be the sides of an obtuse angled triangle? all values are in degree
I. 19, 29, 39
II. 29, 39, 49
III. 10,110, 60
A.III only
B.II only
C.III only
D.I and II only
Answer: A.III only
Explanation: obtuse angled triangle means one angle more than 90deg and total 180 deg.. III
A
Question: With what number must 6.2587 be multiplied in order to obtain the number 625.87?
A.1000
B.1,0000
C.100
D.100,000
Answer: C.100
Explanation: Explanation: The decimal must be moved two places to the right. To do this, we must multiply by a number with two zeroes. The correct answer is 100.The decimal must be moved two places to the right. To do this, we must multiply by a number with two zeroes. The correct answer is 100.
Answer : (C)
Question: Students of three different classes appeared in common examination. Pass average of 10 students of first class was 60%, pass average of 15 students of second class was 60% and pass average of 25 students of third class was 80% then what will be the pass average of all students of three classes?
A.74%
B.75%
C.69%
D.70%
Answer: D.70%
Explanation: Solution:
Sum of pass students of first, second and third class,
= (60% of 10) + (60% of 15)+ (80% of 25)
= 6+9+20 = 35
Total students appeared,
= 10+15+25 = 50
Pass average,
= 35*100/50 = 70%.
Answer: Option D
Question: A car crosses a 600 m long bridge in 5 minutes. Whatis the speed of car in km per hour?
A.7 km/hr
B.7.2 km/hr
C.7.4 km/hr
D.7.8 km/hr
Answer: B.7.2 km/hr
Explanation: Speed = 600 m/sec.
5 x 60
= 2 m/sec.
Converting m/sec to km/hr (see important formulas section)
= 2 x 18 km/hr
5
= 7.2 km/hr
B
Question: In a certain sequence, the first term is 2, and each successive term is 2 more than the reciprocal of the term that immediately precedes it. What is the fifth term in this sequence?
A.13/8
B.21/13
C.58/5
D.70/29
Answer: D.70/29
Explanation: let five terms in the sequence be a,b,c,d,e
a=2
b= 2+1/2=5/2
c=2+2/5= 12/5
d= 2+5/12 = 29/12
e= 2+12/29 = 70/29
hence answer should be D.
Question: The number of roots common between the two equations x3 + 3x2 + 4x + 5 = 0 and x3 + 2x2 + 7x + 3 = 0 is
A.0
B.1
C.2
D.3
Answer: A.0
Explanation: Explanation :
Let α is the common root.
∴ α3 + 3α2 + 4α + 5 = 0. ............................(i)
α3 + 2α2 + 7α + 3 = 0. .............................(ii)
On subtracting (ii) from (i), we get
α2 - 3α + 2 = 0.
α = 2, α = 1.
But the above values of α do not satisfy any of the equations. Thus, no root is common.
Answer : A
Question: If 4XZ + YW = 4 and XW + YZ = 8, what is the value of the expression (2X + Y)(2Z + W)?
A.22
B.20
C.24
D.26
Answer: B.20
Explanation: (2x+y)*(2z+w) = 4+2(8) =20
Answer : B
Question: B completes a work in 7 days. A alone can do it in 10 days. If both work together, the work can be completed in how many days?
A.3.75 days
B.4.11 days
C.5.11days
D.6.75 days
Answer: B.4.11 days
Explanation: 1/7 + 1/10 = 17/70
70/17 = 4.11 days
ANSWER:B
Question: In one hour, a boat goes 8 km along the stream and 2 km against the stream. the sped of the boat in still water (in km/hr) is :
A.3
B.5
C.8
D.9
Answer: B.5
Explanation: Solution
Speed in still water= 1/2(8+2) km/hr= 5kmph. Answer B
Question: There are three departments having students 62,58,24 .In an exam they have to be seated in rooms such that each room has equal number of students and each room has students of one type only (No mixing of departments). Find the minimum number of rooms required?
A.72
B.74
C.75
D.76
Answer: A.72
Explanation: WE NEED TO TAKE GCD WHICH IS 2
THUS ALL THE ROOMS WILL HAVE 2 STUDENTS OF THE SAME DEPT
1) 62/2 = 31
2) 58/2 = 29
3) 24/2 = 12
TOTAL NO. OF MIN ROOMS REQD = 31+12+29 = 72
ANSWER:A
Question: P runs 25% faster than Q so P gives Q a 60 meter head start. If the race ends in a tie, how far (in meters) did P run in the race?
A.100
B.200
C.300
D.400
Answer: C.300
Explanation: Let D be the race distance that P ran.
Let T be the time it took to complete the race.
Let V be Q's speed.
T = D/1.25V = (D-60) / V
D=1.25D-75
0.25D=75
D=300 meters.
The answer is C.
Question: Ramu bought an old car for Rs. 42000. He spent Rs. 13000 on repairs and sold it for Rs. 64500. What is his profit percent?
A.22
B.77
C.17.3
D.99
Answer: C.17.3
Explanation: Total CP = Rs. 42000 + Rs. 13000 = Rs. 55000 and SP = Rs. 64500
Profit(%) = (64500 - 55000)/55000 * 100 = 17.3%
Answer: C
Question: Indu gave Bindu Rs.1250 on compound interest for 2 years at 4% per annum. How much loss would Indu has suffered had she given it to Bindu for 2 years at 4% per annum simple interest?
A.5
B.2
C.9
D.5
Answer: B.2
Explanation: 1250
= D(100/4)2
D = 2
Answer: B
Question: Let x be a positive integer, what is the remainder when 2^(3x+3)+3 is divided by 3?
A.0
B.1
C.2
D.3
Answer: B.1
Explanation: For: (2^(3x+3)+3)/3
Let n = 3x + 3
2^n
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32, last digit pattern repeats 2, 4, 8, 6, 2, ....
Rule: x^(m+n) = (x^m)(x^n)
(2^(3x))(2^3)
Rule: x^(mn) = (x^m)^n
((2^3)^x)(8) => Now, (8)(8) + 3 = (64) + 3 => and, 4 + 2 + 1 => Finally, (6 + 1)/3, and remainder is 1.
Answer: B
Question: A polling company surveyed a certain country, and it found that 35% of that country’s registered voters had an unfavorable impression of both of that state’s major political parties and that 20% had a favorable impression only of Party W. If one registered voter has a favorable impression of both parties for every two registered voters who have a favorable impression only of Party B, then what percentage of the country’s registered voters have a favorable impression of both parties (assuming that respondents to the poll were given a choice between favorable and unfavorable impressions only)?
A.15
B.20
C.30
D.35
Answer: A.15
Explanation: S=100
not( W and B ) =35
only W=20
(W and B)/B=1/2
let ( W and B ) =x
only B =2x
so now, 20+35+x+2x=100
x= 15
A ans
Question: If 25% of a number is equal to one-third of another number, what is the ratio of first number to the second number?
A.4:3
B.1:4
C.3:7
D.6:11
Answer: A.4:3
Explanation: Let 25% of A = 1/3 B
then 25A/100 = 1B/3
A/4 = B/3
A/B = 4/3
A:B = 4:3
Answer is A
Question: If 7a - 3b = 10b + 50 = -12b - 2a, what is the value of 9a + 9b?
A.-9
B.-6
C.0
D.6
Answer: C.0
Explanation: (i) 7a - 13b = 50
(ii) 2a + 22b = -50
Adding (i) and (ii):
9a + 9b = 0
The answer is C.
Question: A pupil's marks were wrongly entered as 73 instead of 40. Due to the average marks for the class got increased by half. The number of pupils in the class is?
A.18
B.82
C.66
D.27
Answer: C.66
Explanation: Let there be x pupils in the class.
Total increase in marks
= (x * 1/2) = x/2
x/2 = (73 - 40) => x/2
= 33 => x = 66.
Answer: C
Question: On a scale that measures the intensity of a certain phenomenon, a reading of a+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of a. On that scale, the intensity corresponding to a reading of 8 is how many times as great as the intensity corresponding to a reading of 3?
A. 5
B. 50
C. 10^5
D. 5^10
Answer: C. 10^5
Explanation: To solve this problem we need to examine the information in the first sentence. We are told that “a reading of a + 1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of a.”
Let’s practice this idea with some real numbers. Let’s say a is 2. This means that a + 1 = 3. With the information we were given we can say that a reading of 3 is ten times as great as the intensity of a reading of 2.
Furthermore, we can say that a reading of 4 is actually 10 x 10 = 10^2 times as great as the intensity of a reading of 2.
Increasing one more unit, we can say that a reading of 5 is 10 x 10 x 10 = 10^3 times as great as the intensity of a reading of 2.
We have found a pattern, which can be applied to the problem presented in the stem:
3 is “one” unit away from 2, and thus a reading of 3 is 10^1 times as great as the intensity of a reading of 2.
4 is “two” units away from 2, and thus a reading of 4 is 10^2 times as great as the intensity of a reading of 2.
5 is “three” units away from 2, and thus a reading of 5 is 10^3 times as great as the intensity of a measure of 2.
We can use this pattern to easily answer the question. Here we are being asked for the number of times the intensity corresponding to a reading of 8 is as great as the intensity corresponding to a reading of 3. Because 8 is 5 units greater than 3, a reading of 8 is 10^5 times as great as the intensity corresponding to a reading of 3.
Answer C.
Question: Two family reunions are happening at the same hotel, the Oates reunion and the Yellow reunion. All 100 guests at the hotel attend at least one of the reunions. If 42 people attend the Oates reunion and 65 people attend the Yellow reunion, how many people attend both reunions?
A. 2
B. 7
C. 10
D. 16
Answer: B. 7
Explanation: No of people in Oates reunion=42
No of people in Yellow reunion=65
Attending both=x
All guests attend at least one.
Therefore, 100=42+65-(Both)
Both=7
Answer B
Question: Harriet planned to complete a certain task on Wednesday, January 1, but because of illness the completion date was postponed 45 days. On which day of the week in February was the task completed?
A.Tuesday
B.Wednesday
C.Saturday
D.Sunday
Answer: C.Saturday
Explanation: This type of story problem is essentially just about a repeating pattern and a 'remainder.'
We're told that Harriet planned to complete a certain task on WEDNESDAY, January 1, but because of illness the completion date was postponed 45 DAYS. We're asked on what day of the week the task was actually completed on.
Since the task was supposed to be completed on a Wednesday....
A 1 day postponement would be Thursday
A 2 day postponement would be Friday
A 3 day postponement would be Saturday
A 4 day postponement would be Sunday
A 5 day postponement would be Monday
A 6 day postponement would be Tuesday
A 7 day postponement would be Wednesday
A 8 day postponement would be Thursday
Etc.
So this question is ultimately asking how many 'groups of 7' are there, and how many 'leftover days' are there, when you postpone 45 days.
45/7 = 6 remainder 3
This means that there are 6 full weeks AND 3 extra days after a Wednesday....Thus, the task was completed on a Saturday.
C
Question: If abc = ab^3 = 180 where a, b and c are positive integers, what is the value of `c’?
A.0
B.1
C.2
D.3
Answer: B.1
Explanation: abc = a(b^3) = 180
on solving abc = a(b^3) we get c = b^2
and for a(b^3) = 180 , there is only one possibility a = 180 and b = 1 so c = 1^2 = 1
ANSWER:B
Question: Find the number of ways in which 8064 can be resolved as the product of two factors?
A.22
B.24
C.21
D.20
Answer: B.24
Explanation: Solution:
Total number of ways in which 8064 can be resolved as the product of two factors is 24 as below:
(1,8064), (2,4032), (3,2688), (4,2016), (6,1344), (7,1152),(8,1008), (9,896), (12,672), (14,576), (16,504), (18,448), (21,884), (24,336), (28,288), (32,252), (36,224), (42,192), (48,168), (56,144), (63,128), (68,126), (72,112), (84,96).
Answer: Option B
Question: How many paying stones, each measuring 4 m * 2 m are required to pave a rectangular court yard 40 m long and 20 m board?
A.99
B.18
C.100
D.17
Answer: C.100
Explanation: 40 * 20 = 4 * 2 * x => x
= 100
Answer: C