mathewhe/aqua_rat huggingface dataset:
Question: In a soccer bet,we play 4 teams.first team odd 1.30,second 5.23,third 3.25,fourth 2.05.we place the bet with 5.00 euros.
how much money we expect to win?
(A) 219.5
(B) 226.49
(C) 235.01
(D) 266.74
(E) 669.0
Rationale: in the case we won the bet,we have:
1.30* 5.23* 3.25* 2.0*5.00=219.50
we will win 226.49
so correct answer is B
Answer: (B) 226.49
Question: Joe’s average (arithmetic mean) test score across 4 equally weighted tests was 70. He was allowed to drop his lowest score. After doing so, his average test score improved to 75. What is the lowest test score that was dropped?
(A) 20
(B) 25
(C) 55
(D) 65
(E) 80
Rationale: The arithmetic mean of 4 equally weighted tests was 70. So what we can assume is that we have 4 test scores, each 70.
He dropped his lowest score and the avg went to 75. This means that the lowest score was not 70 and other three scores had given the lowest score 5 each to make it up to 70 too. When the lowest score was removed, the other 3 scores got their 5 back. So the lowest score was 3 *5 = 15 less than 70.
So the lowest score = 70 - 15 = 55
Answer (C)
Answer: (C) 55
Question: 103,100,97,..,91,88,85
Insert the missing number
(A) 88
(B) 78
(C) 94
(D) 98
(E) 108
Rationale: Missing number : 94
Answer : C
Answer: (C) 94
Question: In a series of consecutive numbers, 27 is the eighth smallest number and 20 is the tenth largest number. What is the range of the series?
(A)  32
(B)  30
(C)  28
(D)  16
(E)  25
Rationale: Imp:All are consecutive odd numbers.
27 is the 8th smallest ,there are 7 numbers smaller then 27 ,smallest number :27-14=13
20 is the 10th largest ,so there are 9 numbers larger then 20, largest number:20+18=38
Range:38-13=25
Ans : E
Answer: (E)  25
Question: The area of a triangle is with base 4m and height 5m?
(A) 88 m2
(B) 10 m2
(C) 66 m2
(D) 77 m2
(E) 31 m2
Rationale: 1/2 * 4 * 5 = 10 m2
Answer: B
Answer: (B) 10 m2
Question: A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?
(A) Rs. 375
(B) Rs. 400
(C) Rs. 600
(D) Rs. 800
(E) Rs. 850
Rationale: C's 1 day's work = 1/3 - (1/6 + 1/8) = 1/3 - 7/24 = 1/24
A's wages : B's wages : C's wages = 1/6 : 1/8 : 1/24 = 4: 3 : 1
C's share (for 3 days) = Rs.	(3 * 1/24 * 3200) =Rs.400
Answer = B
Answer: (B) Rs. 400
Question: If the height of an equilateral triangle GBC is equal in length to the hypotenuse of an isosceles right triangle DEF, a side of ΔGBC is how many times as long as a leg of the right triangle DEF?
(A) (√2)/(√3)
(B) (√3)/(√2)
(C) √2
(D) (2√2)/(√3)
(E) 2/(√3)
Rationale: Side of the equilateral triangle GBC = 2m ----> height = m * sqtr(2)
Hipotenuse of the right triangle GBC = m * sqrt(3) ----> side = m * sqtr(3) / sqtr(2)
Question is 2m/m * sqtr(3) / sqtr(2) = 2*sqrt(2)/sqrt(3)=D
Answer: (D) (2√2)/(√3)
Question: An amount of money is to be distributed among Faruk, Vasim and Ranjith in the ratio 3 : 5 : 7. If Vasims share is Rs. 1500, what is the difference between Faruk's and Ranjith's shares?
(A) Rs 1200
(B) Rs 1500
(C) Rs 1600
(D) Rs 1900
(E) Rs 1700
Rationale: Explanation :
Let P = Faruk, Q = Vasim, R = Ranjith
Let P = 3x, Q = 5x and R = 7x.
Then, 5x = 1500 ? x = 300. P=900,Q=1500 and R = 2100.
Hence, (R - p) = (2100 - 900) = 1200
Answer : A
Answer: (A) Rs 1200
Question: In a partnership between A, B and C. A's capital is Rs.5000. If his share of a profit of Rs.800 is Rs.200 and C's share is Rs.130, what is B's capital?
(A) 19878
(B) 12798
(C) 26908
(D) 11750
(E) 22073
Rationale: 200 + 130 = 330
800 - 330 = 470
200 ---- 5000
470 ---- ? => 11750
Answer: D
Answer: (D) 11750
Question: A family has two children. find the probability that both the children are girls given that at least
one of them is a girl?
(A) 1/2
(B) 1/5
(C) 1/3
(D) 1/7
(E) 1/9
Rationale: Let b stand for boy and g for girl. The sample space of the experiment is
S = {(g, g), (g, b), (b, g), (b, b)}
Let E and F denote the following events :
E : ‘both the children are girls’
F : ‘at least one of the child is a girl’
Then E = {(g,g)} and F = {(g,g), (g,b), (b,g)}
Now E n F = {(g,g)}
Thus P(F) = 3/4
and P (E n F )= 1/4
Therefore P(E|F) = P(E ∩ F)/P(F) = (1/4)/(3/4) = 1/3
C
Answer: (C) 1/3
Question: If books bought at prices ranging from Rs. 125 to Rs. 350 are sold at prices ranging from Rs. 170 to Rs 420, what is the greatest possible profit that might be made in selling 12 books ?
(A) Rs. 3500
(B) Rs. 3540
(C) Rs. 3520
(D) Rs. 3200
(E) Rs. 3600
Rationale: The greatest profit is possible only if the cost price of the books are minimum and selling prices are maximum.
Let lowest cost price of the 12 books = 125*12 = Rs. 1,500
Maximum selling price of 12 books = 420 *12 = Rs. 5,040
So, maximum profit = 5040 - 1500 = Rs. 3,540
ANSWER : OPTION B
Answer: (B) Rs. 3540
Question: What is the remainder when 12346 is divided by 9?
(A) 7
(B) 6
(C) 5
(D) 4
(E) 3
Rationale: If the sum of the digits of a number is divisible by 9, the number will be divisible by 9.
1+2+3+4+6 = 16
The nearest multiple of 9 is 9.
16-9= 7.
ANSWER:A
Answer: (A) 7
Question: A’s speed is 30/17 times that of B. If A and B run a race, what part of the length of the race should A give B as a head start, so that the race ends in a dead heat?
(A) 1/17
(B) 13/30
(C) 17/30
(D) 11/30
(E) 11/13
Rationale: We have the ratio of A’s speed and B’s speed. This means, we know how much distance A covers compared with B in the same time.
This is what the beginning of the race will look like:
(Start) A_________B______________________________
If A covers 30 meters, B covers 17 meters in that time. So if the race is 30 meters long, when A reaches the finish line, B would be 13 meters behind him. If we want the race to end in a dead heat, we want B to be at the finish line too at the same time. This means B should get a head start of 13 meters so that he doesn’t need to cover that. In that case, the time required by A (to cover 30 meters) would be the same as the time required by B (to cover 17 meters) to reach the finish line.
So B should get a head start of 13/30th of the race.
Answer (B)
Answer: (B) 13/30
Question: Rs. 700 is divided among A, B, C so that A receives half as much as B and B half as much as C. Then C's share is:
(A) Rs 200
(B) Rs 300
(C) Rs 400
(D) Rs 500
(E) Rs 600
Rationale: Let C = x.
Then B = x/2
and A = x/4
A:B:C = 1:2:4.
C's share Rs.[(4/7)x700) = 400
answer :C
Answer: (C) Rs 400
Question: In a garden, there are four blue flowers, three red flowers, four green flowers, and four pink flowers. What is the probability that a florist will choose three flowers of the same color when randomly picking three flowers?
(A) 11/10
(B) 1/55
(C) 31/10
(D) 3/55
(E) 1/16
Rationale: P(all the same color) = P(1st flower is ANY colorAND2nd flower is same as firstAND3rd flower is also the same color)
= P(1st flower is ANY color)xP(2nd flower is same as 1st)ANDP(3rd flower is the same color)
= 1x3/11x2/10
= 3/55
Answer: D
Answer: (D) 3/55
Question: The current of a stream runs at the rate of 4 kmph. A boat goes 6 km and back to the starting point in 2 hours, then find the speed of the boat in still water?
(A) 9
(B) 7
(C) 8
(D) 6
(E) 4
Rationale: S = 4
M = x
DS = x + 4
US = x - 4
6/(x + 4) + 6/(x - 4) = 2
x = 8
Answer: C
Answer: (C) 8
Question: How many different positive integers are factors of 64 ?
(A)  10
(B)  12
(C)  14
(D)  16
(E)  18
Rationale: 8×8=4^2 × 2^2
So total factors = (3+1)(3+1) = 16
Answer: D
Answer: (D)  16
Question: If a fair coin is tossed six times, what is the probability of getting exactly three heads in a row?
(A) 1/16
(B) 3/16
(C) 1/8
(D) 3/8
(E) 1/2
Rationale: A) 3 heads..
all three heads can be taken as one, so total are 4- 3H,T,T,T...
3H can take any of four places= 4 ways..
B) 4 Heads..
again 3H are one but it cannot have fourth one next to it..
(i)so if we have 3H in edges.. 3H,T,_,_
the last two can be H,T or T,H--- so 2 ways..
similarily if 3H in the rightmost position-- _,_,T,3H---- 2 ways..
(ii) if we have 3H in middle... T,3H,T,H and H,T,3H,T--- 2ways..
TOTAL=2+2+2=6 ways..
C) 5 heads..
Again 3H in edges-- 3H,T,H,H or H,H,T,3H--2 ways.
we cannot have 3H in middle, as it will surely have a H adjacent to it, resulting in 4 H in a row..
D) 6 Heads--
NO possiblity..
TOTAL ways to have exactly 3H in a row=4+6+2=12 ways..
total combinations of 6 tosses= 2^6..
Probability= 12/2^6= 3/16
Answer: B.
Answer: (B) 3/16
Question: At a school, two-fifths of the male students and two-thirds of the female students speak a foreign language. If the number of males is three-fourths the number of females, what fraction of the students speak a foreign language?
(A) 7/15
(B) 8/15
(C) 18/35
(D) 22/35
(E) 58/105
Rationale: Let x be the number of students in the school.
The number of males who speak a foreign language is (2/5)(3/7)x = (6/35)x
The number of females who speak a foreign language is (2/3)(4/7)x = (8/21)x
The total number of students who speak a foreign language is (6/35)x + (8/21)x = (58/105)x
The answer is E.
Answer: (E) 58/105
Question: There is a building of 100 floors
-If an egg drops from the Nth floor or above it will break.
-If it’s dropped from any floor below, it will not break.
You’re given 2 eggs.
Find N
How many drops you need to make?
(A) 14
(B) 22
(C) 18
(D) 25 sec
(E) 38
Rationale: A
14
This follows the below logic.
Say, the egg breaks at floor n we try to find out by going (N-1) till the first floor by doing linear search.
Say for example, I throw the egg from 10th floor, and it breaks, I wíll go to floor 1 to 9 to find out the floor..
Then I would try the same logic for every 10 floors thereby setting a worst case scenario of 19 chances.. I.e. 10,20,30,40,50,60,70,80,90,100,91,92,93,94,95,96,97,98,99
To find optimum solution, let’s try this:
If for every n, egg doesnt break, instead of going to next n, go to N-1, this would save us one drop as we are doing a linear search with second egg when egg1 breaks…
So the series would look something like this..
N + (N-1) + (N-2) + (N-3) +…+ 1
Now this is a series which is equal to N(N+1)/2
Now since it is given that the egg may or may not break from 100th floor..
We can write it as..
N(N+1)/2>=100
And n=14(approx)
So we should start from 14 then move up N-1 to 13 floor I.e. 27,39…
So the floors from where the drop needs to be done are: 14,27,39,50,60,69,77,84,90,95,99,100
So the answer is 14
Answer: (A) 14
Question: A pipe is 30 m long and is 45% longer than another pipe. Find the length of the other pipe
(A) 20.68
(B) 20
(C) 20.12
(D) 20.5
(E) None of these
Rationale: Explanation:
Let length of other pipe be X
According to question,
30 = 45/100 X + X
30 = 0.45X + X
30 = 1.45 X
X= 30/1.45
X = 20.68m
Thus the length of the other pipe is 20.68 metres.
ANSWER A
Answer: (A) 20.68
Question: If 3y + 7 = x^2 + P = 7x + 5, what is the value of P ?
(A) 8(1/4)
(B) 8(1/2)
(C) 8(1/3)
(D) 8(1/5)
(E) None of these
Rationale: Explanation:
3x + 7 = 7x + 5 => 7x - 3x = 2 => 4x = 2 => x = 1/2.
Now, 3x + 7 = x^2 + p => 3/2 + 7 = 1/4 + p => p = 8 * 1/4.
Answer: A
Answer: (A) 8(1/4)
Question: If b is an integer, what is the greatest possible value for b that would still make the following statement true: 11*10^b < 1/10 ?
(A)  –4
(B)  –3
(C)  –2
(D)  –1
(E)  0
Rationale: Just putting values, the LHS becomes
b=0 --> 11
b=-1 --> 11/10
b=-2 --> 11/100
b=-3 --> 11/1000
Anything lower will be smaller than 11/1000. b=-2, equality does not hold but it does for b=-3.
Answer is B.
Answer: (B)  –3
Question: Ram singh goes to his office in the city, every day from his suburban house. His driver Mangaram drops him at the railway station in the morning and picks him up in the evening. Every evening Ram singh reaches the station at 5 o'clock. Mangaram also reaches at the same time. one day Ramsingh started early from his office and came to the station at 4 o'clock. Not wanting to wait for the car he starts walking home. Mangaram starts at normal time, picks him up on the way and takes him back house, half an hour early. How much time did Ram singh walk.
(A) 1/4 hours
(B) 2/4 hours
(C) 3/4 hours
(D) 5/4 hours
(E) 7/4 hours
Rationale: Let the time required for walk is t hours then Ram Singh will reach at the car in time (4+t), due to this walking of Ram Singh time saving = [5-(4+t)]*2 , here 2 is due to forward motion upto railway station and backward from railway station to the point where the car met to Ram Singh. It is given that time saving = 1/2 hour
so 2[1-t] = 1/2 on solving we get t = 3/4 hours
ANSWER:C
Answer: (C) 3/4 hours
Question: By selling an article for 3200, a dealer suffers a loss equal to 1/5 of his investment. Find the cost price. At what price should he had sold it to make a gain of 10% ?
(A) Rs. 3600, Rs. 3960
(B) Rs. 4000, Rs. 4400
(C) Rs. 4800, Rs. 5280
(D) Rs. 4200, Rs. 5080
(E) None of these
Rationale: Let his CP was X.
Loss = X/5
Selling price = CP - loss
3200 = X - (X/5)
X = 4000.
CP = Rs. 4000.
To get 10% gain,
SP = 4000 + 10% of 4000 = 4000 + 400 = Rs. 4400.
ANSWER : B
Answer: (B) Rs. 4000, Rs. 4400
Question: How many words can be formed by using all the letters of the word “DAUGHTR” so that vowels always come together & vowels are never
together?
(A) 4320
(B) 760
(C) 720
(D) 4000
(E) 700
Rationale: Given word contains 8 different letters
When the vowels AUE are always together we may suppose
them to form an entity ,treated as one letter
then the letter to be arranged are DAHTR(AUE)
these 6 letters can be arranged in 6p6 = 6!
= 720 ways
The vowels in the group (AUE) may be arranged in 3! = 6 ways
Required number of words = 760 * 6 =4320
ANSWER A 4320
Answer: (A) 4320
Question: A and B start from house at 10am. They travel on the MG road at 20kmph and 40 kmph. there is a Junction T on their path. A turns left at T junction at 12:00 noon, B reaches T earlier, and turns right. Both of them continue to travel till 2pm. What is the distance between A and B at 2 pm
(A) 160km
(B) 170km
(C) 180km
(D) 190km
(E) 150km
Rationale: at 12 A will travel=20*2=40km
B will travel this 40 km in 40/40=1 hr i.e. by 11am.
after T junction for A- distance travelled =2*20=40 km
for b distance travelled =40*3=120
so distance between A & B is=120+40=160km
ANSWER:A
Answer: (A) 160km
Question: The average (arithmetic mean) of five numbers is 5. If 2 is subtracted from each of three of the numbers, what is the new average?
(A) 3.8
(B) 8.9
(C) 4.5
(D) 7.5
(E) 2.9
Rationale: Sum of 5 numbers = 5*5 =25
If 2 is subtracted from each of three of the numbers , we subtract 2*3=6 from the total sum
Sum of 5 number after subtracting 2 from each of three of the numbers =25-6=19
New average = 19/5=3.8
Answer A
Answer: (A) 3.8
Question: Light glows for every 30 seconds . How many max. times did it glow between 1:57:58 and 3:20:47 am.
(A) 380 times
(B) 142 times
(C) 382 times
(D) 392 times
(E) 165 times
Rationale: Time difference is 1hr,22min,49sec=4969sec. So, light glows floor(4969/30)=165 times.
ANSWER:E
Answer: (E) 165 times
Question: A military commander has to choose 3 soldiers out of a company of 5 for guarding the headquarters. How many different group combinations of 3 does he have to choose from?
(A) 4
(B) 15
(C) 16
(D) 21
(E) 10
Rationale: A military commander has to choose 3 soldiers out of a company of 5.
different group combinations of 3
He can do this in 5C3 ways
5*4*3/3*2*1= 10 ways
ANSWER:E
Answer: (E) 10
Question: The Average of 100, 200, 300, 400, ... , 1000 is 550. If each number is divided by 5, then the new average will be equal to
(A) 450
(B) 45
(C) 55
(D) None of these
(E) Cannot be determined
Rationale: Answer
Actual average
=( 100 + 220 + 300 + ... + 1000 ) / 10
=550
New Average = 550 / 5
=110
Correct Option: D
Answer: (D) None of these
Question: Robert is travelling on his cycle andhas calculated to reach point A at 2 PM. if he travels at 10 kmph, he will reach there at 12Pm if he travels at 15 kmph. At what speed musthe travel to reach A at 1 PM?
(A) 8 kmph
(B) 10 kmph
(C) 12 kmph
(D) 16 kmph
(E) 18 kmph
Rationale: Let the distance travelled by x km.
Then, x - x = 2
10 15
3x - 2x = 60
x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60 hrs = 6 hrs.
10
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
Required speed = 60 kmph. = 12 kmph.
5
C
Answer: (C) 12 kmph
Question: Three runners R, B and C run a race, with runner R finishing 12m ahead of runner B and 18m ahead of runner C, while runner B finishes 8m ahead of runner C. Each runner travels entire distance at a constant speed. What was the length of the race?
(A) 36m
(B) 48m
(C) 60m
(D) 72m
(E) 84m
Rationale: Let distance of race be x mtrs. Then when R finishes x m , B has run (x- 12)mtrs and C has run x-18 mtrs. so at this point B is 6 m ahead of C. Now to finish race b needs to run another 12 m, so he runs another 12 m. when B finishes race he is 8 m ahead of C. so last 12 m B has run, C has run 10 m.
as speeds are constant, we have equation,
x-12/ x-18 = 12/10 > x = 48.
Answer B
Answer: (B) 48m
Question: Jake remembers only the last four letters of his five-letter Klingon name. If he is sure that the first letter is neither "N" nor "Z", and assuming that there are only 10 letters in the Klingon alphabet, what is the probability that he will give the correct name when asked for it by the space attendant?
(A) A)8/100
(B) B)1/8
(C) C)1/9
(D) D)4/5
(E) E)9/10
Rationale: Total letters 10, out of which the first letter is anything of 10 but N or X...
So there are 8 possible letters for first place..
And only one of them is correct..
So 1/8..
Answer: B
Answer: (B) B)1/8
Question: Five years ago, John’s age was half of the age he will be in 2 years. How old is he now?
(A) 14years
(B) 12years
(C) 16years
(D) 17years
(E) 18years
Rationale: Let John=j,j-5=(j+2)/2,then j=12years
ANSWER:B
Answer: (B) 12years
Question: A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant. If the lowest frequency is 110 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?
(A) 110 * sqrt 2
(B) 110 * sqrt (2^7)
(C) 110 * sqrt (2^12)
(D) 110 * the twelfth root of (2^7)
(E) 110 * the seventh root of (2^12)
Rationale: Let the constant be k.
F1 = 110
F2 = 110k
F3 = 110 k * k = 110 * k^2
F13= 110 * k^12
we know F13 = 2 *F1 = 2 * 110 = 220
220/110 = k^12
k = twelfth root of 2
for F7...
F7 = 110 * k^6 ( as we wrote for F2 and F3)
F7 = 110 * (twelfth root of 2) ^ 6
F7 = 110 * sqrt (2)
The answer is A.
Answer: (A) 110 * sqrt 2
Question: In a fort, there are 1200 soldiers. If each soldier consumes 3 kg per day, the provisions available in the fort will last for 30 days. If some more soldiers join, the provisions available will last for 25 days given each soldier consumes 2.5 kg per day. Find the number of soldiers joining the fort in that case.
(A) 627
(B) 528 soldiers
(C) 626
(D) 657
(E) 673
Rationale: Assume x soldiers join the fort. 1200 soldiers have provision for 1200 (days for which provisions last them)(rate of consumption of each soldier)
= (1200)(30)(3) kg.
Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) k
As the same provisions are available
=> (1200)(30)(3) = (1200 + x)(25)(2.5)
x = [(1200)(30)(3)] / (25)(2.5) - 1200 => x = 528.
Answer:B
Answer: (B) 528 soldiers
Question: Danny is sitting on a rectangular box. The area of the front face of the box is half the area of the top face, and the area of the top face is 1.5 times the area of the side face. If the volume of the box is 648, what is the area of the side face of the box?
(A) 32
(B) 78
(C) 84
(D) 90
(E) 72
Rationale: Lets suppose length= l, breadth= b, depth= d
Front face area= l*w = 1/2 w*d (l=1/2 d or d=2l)
top face area= w*d
side face area= w*d = 1.5 d*l (w=1.5l)
Volume = l*w*d= 648
l*1.5l*2l= 648
l=6
Side face area= l*d= l*2l= 6*2*6=72
E is the answer
Answer: (E) 72
Question: If 3x+4y=88 and 3x-4y=14, what is the value of x subs y?
(A) 17/4
(B) 25/4
(C) 27/4
(D) 25/4
(E) 19/4
Rationale: Given 3x+4y=88 ---Eq 1
3x -4y =14 -- Eq 2
Sum both eqns we get 6x = 102 => x = 17
Sum 3x in eq 2 => 51 - 4y = 14.
=> y= 37/4
Now XY = 14 - 37/4 = 19/4.
Option E is correct answer.
Answer: (E) 19/4
Question: The operation T is defined by x T y=1/x+1/y for all nonzero numbers x and y. If z is a number greater than 1, which of the following must be true.
I. z T z/z-1=1
II.x T (-z)=0
III. 2/z T 2/z=z
(A) a.) I
(B) b.) I and II only
(C) c) I and III only
(D) d) II and III only
(E) e) I, II and III only
Rationale: Statement I is TRUE
Statement II is not TRUE
Statement III is TRUE
Answer : C
Answer: (C) c) I and III only
Question: In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the rate in the remaining 40 overs to reach the target of 262 runs?
(A) 6.25
(B) 6.22
(C) 6.29
(D) 5.75
(E) 6.13
Rationale: Required run rate
= [262 - (3.2 * 10)]/40
= 230/40
= 5.75
Answer: D
Answer: (D) 5.75
Question: A trader bought a car at 20% discount on its original price. He sold it at a 40% increase on the price he bought it. What percent of profit did he make on the original price?
(A) 19%
(B) 72%
(C) 12%
(D) 16%
(E) 32%
Rationale: Original price = 100
CP = 80
S = 80*(140/100) = 112
100 - 112 = 12%
Answer:C
Answer: (C) 12%
Question: The total of the ages of Amar, Akbar and Anthony is 62 years. What was the total of their ages three years ago ?
(A) 51 years
(B) 55 years
(C) 74 years
(D) 53 years
(E) 78 years
Rationale: Explanation:
Required sum = (62 - 3 x 3) years = (62 - 9) years = 53 years.
Answer: Option D
Answer: (D) 53 years
Question: What is the greater of the two numbers whose product is 2560, given that the sum of the two numbers exceeds their difference by 64?
(A) 90
(B) 100
(C) 110
(D) 70
(E) None of these
Rationale: Let the greater and the smaller number be g and s respectively.
gs = 2560
g + s exceeds g - s by 64 i.e., g + s - (g - s) = 64
i.e., 2s = 64 => s = 32.
g = 2560/s = 80.
ANSWER:E
Answer: (E) None of these
Question: A and B can do a work in 4days, B and C in 5days, C and A in 7days. If A,B and C work together , they will complete the work in?
(A) 3days
(B) 5days
(C) 7days
(D) 1days
(E) 2dasy
Rationale: A+B 1day work = 1/4
B+C 1day work = 1/5
C+A 1day work = 1/7
Adding we get 2(A+B+C) = 1/4 + 1/5 + 1/7 = 83/140
A+B+C 1 day work = 83/280
A,B,C can finish the work in 280/83 days = 3days approximately
Answer is A
Answer: (A) 3days
Question: What is the greatest possible straight line distance, in inches, between any two points on a rectangular box that is 5 inches wide, 3 inches long, and 4 inches high?
(A) 2√5
(B) 5
(C) 5√2
(D) 10
(E) 50
Rationale: The maximum possible distance in a cuboid is the diagonal.
Diagonal length = sqrt(length^2 + breadth^2 + height^2) = sqrt(50)
= 5√2
(Option C)
Answer: (C) 5√2
Question: Suresh can complete a job in 15 hours. Ashutosh alone can complete the same job in 30 hours. Suresh works for 9 hours and then the remaining job is completed by Ashutosh. How many hours will it take Ashutosh to complete the remaining job alone?
(A) 4
(B) 5
(C) 6
(D) 12
(E) None of these
Rationale: The part of job that Suresh completes in 9 hours
= 9â„15 = 3â„5
Remaining job = 1 - 3â„5 = 2â„5
Remaining job can be done by Ashutosh in 2â„5 Ã— 30 = 12 hours
Answer D
Answer: (D) 12