mathewhe/aqua_rat huggingface dataset:
Question: A train running at the speed of 60 km/hr crosses a pole in 6 sec. What is the length of the train?
(A) 120 m
(B) 100 m
(C) 324 m
(D) Cannot be determined
(E) None of these
Rationale: Speed = 60* 5/18 = 50/3 m/sec
Length of the train = speed * time = 50/3 * 6 = 100 m
ANSWER:B
Answer: (B) 100 m
Question: A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average(arithmetic mean) of the other 20 numbers in the list, then n is what fraction T of the sum of the 21 numbers in the list?
(A)  1/20
(B)  1/6
(C)  1/5
(D)  4/21
(E)  5/21
Rationale: This is how I used to calculate which I think works pretty well:
if you let the average of the 20 other numbers equal a, can you write this equation for sum of the list (S)
n + 20a = S
the question tells us that
n = 4a
plug this back into the first equation and you get that the sum is 24a
4a + 20a = 24a
therefore fraction T of n to the total would be
4a/24a or 1/6
answer B
Answer: (B)  1/6
Question: The difference between the compound interest and simple interest on a certain sum at 10% per annum for 2years is $631. Find the sum?
(A) $52000
(B) $61000
(C) $71540
(D) $63100
(E) $56210
Rationale: Let the sum be $x
C.I. = x(1+10/100)^2 - x = 21x/100
S.I. = (X*10*2)/100 = X/5
C.I.- S.I. = (21x/100)-(x/5) = x/100
x/100 = 631
x = 63100
Answer is D
Answer: (D) $63100
Question: Thrice the square of a natural number decreased by 4 times the number is equal to 50 more than the number. The number is:
(A) 4
(B) 5
(C) 6
(D) 7
(E) 10
Rationale: Let the number be x. Then,
3x2 - 4x = x + 50
3x2 - 5x - 50 = 0
(3x + 10)(x - 5) = 0
x = 5
ANSWER:B
Answer: (B) 5
Question: The length of a brigade which a train 130 m long and traveling at 45kmph can cross in 30 seconds, is
(A) 196 mts
(B) 220 mts
(C) 230 mts
(D) 245 mts
(E) None of these
Rationale: Explanation:
Let the length of the brigade is x mts
Then Distance = 130 + x mts
given speed = 45 kmph = 45 x5/18 m/s
Time = 30 sec
T = D/S
=> 30 = 130+x/(45x5/18)
=> x = 245 mts.
ANSWER IS D
Answer: (D) 245 mts
Question: A shopkeeper’s price is 50% above the cost price. If he allows his customer a discount of 30% what profit does he make?
(A) 5%
(B) 10%
(C) 15%
(D) 20%
(E) None of these
Rationale: Let C.P. = 100, then M. P. = 150
S.P. = 70% of 150 = 105
∴ % profit = 105−100/100×100=5%
Answer A
Answer: (A) 5%
Question: Rs.8000 become Rs.9261 in a certain interval of time at the rate of 5% per annum of C.I. Find the time?
(A) 6
(B) 5
(C) 6
(D) 3
(E) 2
Rationale: 9261 = 8000(21/20)N
(21/20)3 = (21/20)N => N = 3
Answer:D
Answer: (D) 3
Question: What is the least possible value of x^2-15 where x is a positive number.
(A) -15
(B) -14
(C) -13
(D) -12
(E) -11
Rationale: If x is a positive integer, the lowest value would be x = 1, because 0 is not included in the natural numbers. That's a technicality the GMAT would not expect students to know.
If x = 1, then the expression equals -14.
Answer = (B).
Answer: (B) -14
Question: The probability of a lottery ticket being a prized ticket is 0.2. When 4 tickets are purchased, the probability of winning a prize on atleast one ticket is ?
(A) 123 km/hr
(B) 187 km/hr
(C) 179 km/hr
(D) 120 km/hr
(E) 169 km/hr
Rationale: Let speed of the car be x km/hr.
Then, speed of the train = 150/100 x = 3/2 x km/hr.
75/x - 75/(3/2 x) = 125/(10 * 60)
75/x - 50/x = 5/24 = 120 km/hr.
Answer:D
Answer: (D) 120 km/hr
Question: If a and y are positive integer and a/y is an even integer, which of the following can be an odd integer?
(A) x
(B) xy
(C) x−y
(D) x+2y
(E) a/3
Rationale: C a-Y
Let says a =12 Y=3 iT BECOMES a-Y =9 This is odd True
a = 12 Y = 6 THEN a-Y = 2, Which is even hence this is False.
so both Yes and NO situations are exists for this Option. C can't be the correct answer same logic goes for E. I believe this question is incomplete b'coz of Irrelevant Options.
Answer: (E) a/3
Question: If cost of 15 eggs be 75 rupees, then find out the cost of 4 dozen eggs.
(A) 300
(B) 240
(C) 500
(D) 600
(E) 700
Rationale: Cost of 15 eggs = 75 rupees
Cost of 1 egg = 75/15 = 5 rupees,
Cost of 4 dozens, i.e. 5 x 12 = 60 x 4 = 240
Answer : B
Answer: (B) 240
Question: Convert the 10/36 m/s into kilometers per hour?
(A) 2.9 kmph
(B) 9.9 kmph
(C) 1.3 kmph
(D) 1 kmph
(E) 5.7 kmph
Rationale: 10/36 m/s = 10/36 * 18/5 = 1/1
= 1kmph.
Answer: D
Answer: (D) 1 kmph
Question: Find the total no of 10 digits whose sum is 4.
(A) 210
(B) 220
(C) 230
(D) 240
(E) 250
Rationale: 4=4+0(one 4 & 9 zeroes)=> 1
4=3+1(one 3,one 1 & 8 zeroes)=> 2*9!/8!=18
4=2+2(two 2's & 8 zeroes)=> 9
4=1+1+2(two 1's & one 2)=> 9!/7! + 9!/(7!*2!)=72+36=108
4=1+1+1+1(four 1's)=> 9!/(6!*3!)=84
total no. of 10 digits no. whose sum is 4 =1+18+9+108+84=220
ans 220
ANSWER:B
Answer: (B) 220
Question: If 1.5 x= 0.04 y, then the value of (y-x) (y+x) is:
(A) 1405.25
(B) 1305.25
(C) 1205.25
(D) 1505.25
(E) 1605.25
Rationale: let us assume x=1 and y=37.5
then 1.5(x)=0.04(y)
that is 1.5(1)=0.04(37.5)
= 1.5 =1.5
therfore from given equation (37.5-1)(37.5+1)
=(36.5)(38.5)
=1405.25
ANSWER:A
Answer: (A) 1405.25
Question: A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr. find the time taken by the boat to go 68 km downstream?
(A) 2 hours
(B) 3 hours
(C) 4 hours
(D) 5 hours
(E) None
Rationale: Solution
Speed Downstream	= (13 + 4) km/hr
= 17 km/hr.
Time taken to travel 68 km downstream	=(68 / 17)hrs
= 4 hrs.
Answer C
Answer: (C) 4 hours
Question: 9 men working 7 hours a day can complete a piece of work in 15 days. How many days can 6 men working for 9 hours a day, complete the same piece of work?
(A) 15 ¾ days
(B) 16 days
(C) 16 ¾ days
(D) 17 ½ days
(E) 18 days
Rationale: Explanation:
1 work done = 9 × 7 × 15
9 × 7 × 15 = 6 × 9 × ? days
? days = 9 × 7 × 15/6 × 9
= 35/2 = 17 ½ days.
Answer: Option D
Answer: (D) 17 ½ days
Question: A biker covered half the distance between two towns in 2 hr 30 min. After that he increased his speed by 2 km/hr. He covered the second half of the distance in 2 hr 20 min. Find the distance between the two towns and the initial speed of the biker.
(A) 248
(B) 140
(C) 287
(D) 167
(E) 111
Rationale: Let x km/hr be the initial speed of the biker, then his speed during the second part of the trip is x + 2 km/hr. Half the distance between two cities equals 23060⋅x23060⋅x km and 22060⋅(x+2)22060⋅(x+2) km. From the equation: 23060⋅x=22060⋅(x+2)23060⋅x=22060⋅(x+2) we get x=28x=28 km/hr.
The intial speed of the biker is 28 km/h.
Half the distance between the two towns is
2h30min×28=2.5×28=702h30min×28=2.5×28=70.
So the distance is 2×70=1402×70=140 km.
Answer:B
Answer: (B) 140
Question: A man swims downstream 72 km and upstream 45 km taking 9 hours each time; what is the speed of the current?
(A) 1.9
(B) 1.5
(C) 1.1
(D) 1.3
(E) 1.2
Rationale: 72 --- 9 DS = 8
? ---- 1
45 ---- 9 US = 5
? ---- 1 S = ?
S = (8 - 5)/2 = 1.5
Answer: B
Answer: (B) 1.5
Question: The total surface area of a solid hemisphere of diameter 14 cm, is :
(A) 462 cm²
(B) 530 cm²
(C) 1345 cm²
(D) 1788 cm²
(E) None
Rationale: Sol.
Total surface area = 3∏R² = [3 * 22/7 * 7 * 7] cm² = 462 cm²
Answer A
Answer: (A) 462 cm²
Question: Can you replace the question mark with the correct number, given the pair of numbers exhibits similar relationship?
? : 3839 :: 11 : 1209
(A) 6
(B) 16
(C) 26
(D) 36
(E) 46
Rationale: Formular used :
pow(x,3) - pow(x,2) - 1
11*11*11 - 11*11 - 1
1331 - 121 - 1
1209
Similarly, we can solve
pow($x,3) - pow($x,2) - 1 = 3839
This condition holds true for number-16.
correct answer is B)16
Answer: (B) 16
Question: If the S.P of Rs. 24 results in a 20% discount on the list price, What S.P would result in a 30% discount on the list price?
(A) Rs. 27
(B) Rs. 21
(C) Rs.20
(D) Rs. 9
(E) Rs. 10
Rationale: Let the list price be Rs. X,
80/100*x = 24, x = 24*100/80 = 30
Required S.P = 70% of Rs. 30
= 70*30/100
=21
ANSWER:B
Answer: (B) Rs. 21
Question: What is 10-8+6-4+...+(-18)?
(A)  -8
(B)  10
(C)  12
(D)  -4
(E)  16
Rationale: The expression considers all even numbers between 10 and -18 with alternate addition and subtraction of the numbers.
The numbers to be used are: 10, 8, 6, 4, 2, 0, -2, -4, -6, -8, -10, -12, -14, -16, and -18.
Now, the first term is positive and the next term is subtracted. So, the required expression becomes,
10 - 8 + 6 - 4 + 2 - 0 + (-2) - (-4) + (-6) - (-8) + (-10) - (-12) + (-14) - (-16) + (-18)
= 10 - 8 + 6 - 4 + 2 - 0 - 2 + 4 - 6 + 8 - 10 + 12 - 14 + 16 - 18
= 58 - 62
= -4
Hence the correct answer choice is D.
Answer: (D)  -4
Question: In what ratio mental A at Rs.68 per kg be mixed with another metal at Rs.96 per kg so that cost of alloy (mixture) is Rs.76 per kg?
(A) 5:8
(B) 4:7
(C) 3:7
(D) 9:5
(E) 5:2
Rationale: (96-76)/(76-68)=20/8=5/2
ANSWER:E
Answer: (E) 5:2
Question: The sum of the non-prime numbers between 10 and 20, non-inclusive, is
(A) 60
(B) 75
(C) 90
(D) 105
(E) 120
Rationale: sum of consecutive integers from 11 to 19, inclusive ====> (A1 + An)/2 * # of terms = (11 + 19)/2 * 9 = 15 *9 = 135
sum of non-prime numbers b/w 10 and 20, non inclusive ===> 135 - 60 (i.e., 11 + 13 + 17 + 19, being the prime #s in the range) = 75
Answer: B
Answer: (B) 75
Question: Two persons start running simultaneously around a circular track of length 100 m from the same point at speeds of 5 km/hr and 13 km/hr. When will they meet for the first time any where on the track if they are moving in opposite directions?
(A) 20 sec
(B) 37 sec
(C) 17 sec
(D) 23 sec
(E) 27 sec
Rationale: Time taken to meet for the first time anywhere on the track
= length of the track / relative speed
= 100 / (5 + 13)5/18 = 100* 18 / 18 * 5 = 20 seconds.
Answer : A
Answer: (A) 20 sec
Question: At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but, not together?
(A) 5 min. past 7
(B) 5(2/11) min. past 7
(C) 5(3/11) min. past 7
(D) 5(5/11) min. past 7
(E) 5(5/11) min. past 8
Rationale: When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.
At 7 o'clock, they are 25 min. spaces apart.
Minute hand will have to gain only 5 min. spaces.
55 min. spaces are gained in 60 min.
5 min. spaces are gained in 5(5/11) min.
ANSWER:D
Answer: (D) 5(5/11) min. past 7
Question: A certain sum of money is invested for one year at a certain rate of simple interest. If the rate of interest were 3% higher, then revenue earned would be 20% higher. What is the rate of interest?
(A) 24% p.a.
(B) 4% p.a.
(C) 12% p.a.
(D) 15% p.a.
(E) 18% p.a.
Rationale: Let the rate of interest be x%.
Now it will be (x + 3)%
% increase = (((x + 3) - x)/x) * 100 = 20
=> x = 15
ANSWER:D
Answer: (D) 15% p.a.
Question: In a party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples?
(A) 5/21
(B) 5/22
(C) 5/12
(D) 5/29
(E) 5/21
Rationale: Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ * ⁶C₁)
As remaining person can be any one among three couples left.
Required probability = (⁵C₂ * ⁶C₁)/¹⁰C₅
= (10 * 6)/252 = 5/21
Answer: A
Answer: (A) 5/21
Question: A company’s four cars running 8 hrs a day consume 1200 lts of fuel in 10 days. In the next 6 days, the company will need to run 9 cars for 12 hrs each so it rents 5 more cars which consume 20% less fuel than the company’s four cars. How many lts of fuel will be consumed in the next 6 days?
(A) 1200 lt
(B) 1555 lt
(C) 1664 lt
(D) 1728 lt
(E) 2160 lt
Rationale: Given that
4 cars running 8 hrs a day consume 1200 lts. of fuel in 10 days.
1 car consumption per hour per day = 1200 /4 *8 *10 = 3.75 litre
Now question say new car consume 20% less fuel than the company’s four cars = 80/100 of 3.75 = 3 ( 20 percent less than 3.75)
Hence we calculate total consumption for next 6 days, the company will need to run 5 new cars for 12 hrs = 3*12 *6*5 =1080
similarly = old 4 car consumption for next 6 days for 12 hrs = 3.75*6*12*4 = 1080
hence total is = 1080+1080 = 2160 lt
Ans is E .
Answer: (E) 2160 lt
Question: Sandy walks around a circular field at the rate of one round per hour while Molly runs around it at the rate of three rounds per hour. If they start in the same direction from the same point, how many minutes will it take until they meet again?
(A) 15
(B) 30
(C) 45
(D) 60
(E) 75
Rationale: Since Sandy and Molly move in the same direction along the circle, they will first meet each other when there is a difference of one round between the two.
The relative speed of Sandy and Molly is 3 - 1 = 2 rounds per hour.
The time required to complete one round at this speed is 1/2 hours = 30 minutes.
The answer is B.
Answer: (B) 30
Question: If it takes 10 days to dig a trench 100 m long, 50 m broad and 10 m deep, what length of trench, 25 m broad and 15 m deep can be dug in 30 days ?
(A) 400 m
(B) 200 m
(C) 100 m
(D) 88(8 / 9) m
(E) 89(8 / 9) m
Rationale: We can dig a trench of 100*50*10 in 10 days. The volume of this trench is 50 000m^3. This means that 50 000 m^3 of dirt is moved every 10 days. Thus, it is found that 3(50 000) of dirt, or 150 000 m^3 of dirt is moved in 30 days. Since we know that the dimensions of this hole are 25, 15, and x and the volume is 150 000, we can create the equation 15*25*x=150 000. Solving for x, we find that the missing dimension is 400m
ANSWER:A
Answer: (A) 400 m
Question: A rectangular field is 63 yards long and 21 yards wide. A fence is needed for the perimeter of the
field. Fencing is also needed to divide the field into three square sections. How many feet of fencing
are needed? (It is a good idea to make a drawing for this one.)
(A) 630 ft
(B) 530 ft
(C) 430 ft
(D) 330 ft
(E) 230 ft
Rationale: 63 + 21 + 63 + 21 = 168 yds. for
outside of field.
21 + 21 = 42 yds. for dividing sections
168 + 42 = 210 yds. of fencing
210 x 3 = 630 ft. of fencing
correct answer A
Answer: (A) 630 ft
Question: The equation (M + )/21 = (p – 7)/21 relates two temperature scales, where M is the number of degrees on one scale and P is the number of degrees on the other scale. Which of the following equations can be used to convert temperatures from the P scale to the M scale?
(A) P
(B) P-7
(C) P-8
(D) P-6
(E) P-5
Rationale: M/21 =(P-7)/21
M = (P-7)*21/21
=(P-7)*1 = P-7
B
Answer: (B) P-7
Question: Series E(n) is such that i*E(i) = j*E(j) for any pair of positive integers (i, j). If E(1) is a positive integer, which of the following is possible?
I. 2*E(100) = E(99) + E(98)
II. E(1) is the only integer in the series
III. The series does not contain negative numbers
(A) I only
(B) II only
(C) IIII only
(D) IIIII only
(E) I, IIIII
Rationale: First thing I want to understand is this relation: i*E(i) = j*E(j) for any pair of positive integers. I will take examples to understand it.
When i = 1 and j = 2, E(1) = 2*E(2)
So E(2) = E(1)/2
When i = 1 and j = 3, E(1) = 3*E(3)
So E(3) = E(1)/3
I see it now. The series is: E(1), E(1)/2, E(1)/3, E(1)/4 and so on...
II and III are easily possible. We can see that without any calculations.
II. E(1) is the only integer in the series
If E(1) = 1, then series becomes 1, 1/2, 1/3, 1/4 ... all fractions except E(1)
III. The series does not contain negative numbers
Again, same series as above applies. In fact, since E(1) is a positive integer, this must be true.
I. 2*E(100) = E(99) + E(98)
2*E(1)/100 = E(1)/99 + E(1)/98 (cancel E(1) from both sides)
2/100 = 1/99 + 1/98
Not true hence this is not possible
Answer (D)
Answer: (D) IIIII only
Question: If a man walks to his office at 3/4 of his usual rate, he reaches office 1/3 of an hour later than usual. What is his usual time to reach office.
(A) 1⁄2 hr
(B) 1 hr
(C) 3⁄4 hr
(D) 2 hrs
(E) None of these
Rationale: If new speed is a⁄b of original speed, then usual time × (b⁄a - 1) = change in time
∴ usual time × (4⁄3 - 1) = 1⁄3
⇒ usual time = 1⁄3 × 3 = 1 hr
Answer B
Answer: (B) 1 hr
Question: Two trains are moving in the same direction at 72 kmph and 36 kmph. The faster train crosses a man in the slower train in 27 seconds. Find the length of the faster train?
(A) 250 m
(B) 260 m
(C) 240 m
(D) 270 m
(E) 280 m
Rationale: D
270 m
Relative speed = (72 - 36) * 5/18 = 2 * 5 = 10 mps.
Distance covered in 27 sec = 27 * 10 = 270 m.
The length of the faster train = 270 m.
Answer: (D) 270 m
Question: An accurate clock shows 8 o'clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 11 o'clock in the morning?
(A) 30
(B) 60
(C) 90
(D) 120
(E) 150
Rationale: The hour hand will move 3/12 of a circle.
(3/12)*360 = 90 degrees.
The answer is C.
Answer: (C) 90
Question: There are 18 stations between Hyderabad and Bangalore. How many second class tickets have to be printed, so that a passenger can travel from any station to any other station?
(A) 120
(B) 380
(C) 400
(D) 800
(E) 250
Rationale: Explanation:
The total number of stations = 20
From 20 stations we have to choose any two stations and the direction of travel (i.e., Hyderabad to Bangalore is different from Bangalore to Hyderabad) in ²⁰P₂ ways.
20P2 = 20 * 19 = 380.
Answer: B
Answer: (B) 380
Question: Two cars are traveling in the same direction along the same route. The red car travels at a constant speed of 20 miles per hour, and the black car is traveling at a constant speed of 50 miles per hour. If the red car is 20 miles ahead of the black car, how many hours will it take the black car to overtake the red car?
(A) 0.1
(B) 0.6
(C) 1
(D) 1.5
(E) 2
Rationale: Option B
20+20t=50t
t=0.6
Answer: (B) 0.6
Question: The side of a rhombus is 26 m and length of one of its diagonals is 20 m. The area of the rhombus is?
(A) 288
(B) 208
(C) 200
(D) 480
(E) 261
Rationale: 262 – 102 = 242
d1 = 20 d2 = 48
1/2 * 20 * 48 = 480
Answer: D
Answer: (D) 480
Question: Jane earned enough money by selling seashells at 25 cents each to buy several used paperback books at 55 cents each. If he spent all of the money he earned selling seashells to buy the books, what is the least number of seashells he could have sold ?
(A) 5
(B) 10
(C) 17
(D) 22
(E) 30
Rationale: Let's TEST Answer A: 5 seashells...
With 5 seashells, Jamie would have 5(25) = 125 cents. This would allow him to buy 2 books for 110 cents total, but he would then have 15 cents left over. We're told that he spent ALL of his money on books, so this cannot be the correct answer.
Answer D: 22 seashells....
With 22 seashells, Jamie would have 22(25) = 550 cents. This would allow him to buy 10 books for 550 cents total, with no money left over. This is an exact MATCH for what we were told, so this MUST be the answer.
D
Answer: (D) 22
Question: Evaluate :	136.09 + 43.9
(A) 1000
(B) 3000
(C) 7000
(D) 5000
(E) 2000
Rationale: 25/0.0005 = (25*10000) / (0.0005*10000) = 25000 / 5 = 5000
Answer is D
Answer: (D) 5000
Question: A manufacturer of a certain product can expect that between 0.3 percent and 0.5 percent of the units manufactured will be defective. If the retail price is $1,500 per unit and the manufacturer offers a full refund for defective units, how much money can the manufacturer expect to need to cover the refunds on 20,000 units?
(A)  Between $90,000 and $150,000
(B)  Between $30,000 and $50,000
(C)  Between $60,000 and $100,000
(D)  Between $150,000 and $250,000
(E)  Between $300,000 and $500,000
Rationale: Number of defective units is between = .3 % of 20000 and .5% of 20000
= 60 and 100
Retail Price per unit = 1500 $
Expected price of refund is between = 1500 x 60 and 1500 x 100
=90,000 and 1,50,000 dollars
Answer A
Answer: (A)  Between $90,000 and $150,000
Question: If the height of an equilateral triangle WBC is equal in length to the hypotenuse of an isosceles right triangle DEF, a side of ΔWBC 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 WBC = 2m ----> height = m * sqtr(2)
Hipotenuse of the right triangle WBC = 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: If the length of a rectangle is increased by 10% and the area is unchanged , then its corresponding breadth must be decreased by?
(A) 9 1/19
(B) 9 1/15
(C) 9 1/16
(D) 9 1/11
(E) 9 1/10
Rationale: Suppose length = 100m and breadth = x m
Area = 100 m
New length = 110m and breadth = (x-y% of x)
Then, \inline 110\times (x-\frac{y}{100}x)=100\times x
\inline \Rightarrow \inline 110\times (1-\frac{y}{100})=100
\inline \Rightarrow \inline 1-\frac{y}{100}=\frac{100}{110}
\inline \Rightarrow \frac{y}{100}=1-\frac{100}{110}=\frac{1}{11}
\inline y=\frac{100}{11}=9\frac{1}{10}%
Answer:E
Answer: (E) 9 1/10
Question: Find the remainder when 6^50 is divided by 215
(A) 34
(B) 35
(C) 36
(D) 37
(E) 38
Rationale: (6^3)^16*6^2 /215
(215 + 1)^16 * 6^2/215
1^16 * 6^2/215
= 36
ANSWER:C
Answer: (C) 36
Question: Payal finished 2/3 of a book. She calculated that she finished 20 more pages than she has yet to read. How long is her book?
(A) A)60
(B) B)150
(C) C)130
(D) D)90
(E) E)210
Rationale: Let x be the total number of pages in the book, then she finished 2/3*x pages.
Then she has x−2/3*x=1/3*x pages left.
2/3*x−1/3*x=20
1/3*x=20
x=60
So the book is 270 pages long.
Answer is A.
Answer: (A) A)60
Question: If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :
(A) 603
(B) 604
(C) 611
(D) 511
(E) 601
Rationale: If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways.
If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.
If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.
The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.
ANSWER:E
Answer: (E) 601
Question: A person travels from P to Q at a speed of 40 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trips ?
(A) 44 km/hour
(B) 46 km/hour
(C) 48 km/hour
(D) 50 km/hour
(E) None of these
Rationale: Explanation:
Speed while going = 40 km/hr
Speed while returning = 150% of 40 = 60 km/hr
Average speed =
2xy /x+y
=2∗40∗60 /40+60=4800/100=48Km/hr
Option C
Answer: (C) 48 km/hour
Question: In a 2000 m race, A beats B by 180 meters or 50 seconds. Find the speed of B?
(A) 3.6m/s
(B) 2m/s
(C) 5m/s
(D) 7m/s
(E) 3m/s
Rationale: Since A beats B by 180 m or 50 seconds,
it implies that B covers 180 m in 50 seconds.
Hence speed of B = 180/50 = 3.6 m/s.
Answer: A
Answer: (A) 3.6m/s
Question: A starts a business with a capital of Rs. 8500. B joins in the business with Rs.8500 after some time. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3 : 1?
(A) 4 months
(B) 6 months
(C) 7 months
(D) 8 months
(E) None of these
Rationale: Explanation :
Let B joins for x months. Then
A:B = 8500*12 : x* 8500 = 3 : 1
=> 85*12 : 85x= 3 : 1
=> 85*12/ 85x = 3/1 = 3
=> 85*4 /85x = 1
=> 4/x = 1
=> x = 4. Answer : Option A
Answer: (A) 4 months
Question: How many times does the minute hand to go around the clock in one day?
(A) 1355
(B) 1416
(C) 1432
(D) 1400
(E) 1300
Rationale: The minute hand will go around once. Thus, the second hand will pass the minute hand 59 times in an hour. This means they will meet 59Ã—24=1416 times in a day.
Answer B
Answer: (B) 1416
Question: A person travels equal distances with speeds of 9 km/hr, 10km/hr and 11 km/hr and takes a total time of 47 minutes. The total distance is?
(A) 6 km
(B) 3 km
(C) 8 km
(D) 9 km
(E) 2 km
Rationale: Let the total distance be 3x km.
Then, x/9 + x/10 + x/11 = 47/60
29x/96 = 47/60 => x = 2.59
Total distance = 3 *2.59 = 7.78 km.
Answer:C
Answer: (C) 8 km
Question: How many cases do you need if you have to pack 240 pairs of shoes into cases that each hold 24 shoes?
(A) 35
(B) 15
(C) 20
(D) 25
(E) 30
Rationale: C
20
240 pairs of shoes = 480 shoes.
480÷ 24 = 20.
Answer: (C) 20
Question: The average weight of 19 students is 15kg. by the admission of a new student the average weight is reduced to 14.9 kg. the weight of the new student is ?
(A) 10.6 kg
(B) 10.8 kg
(C) 11 kg
(D) 13 kg
(E) None
Rationale: Answer
Weight of new student = Total weight of all 20 students - Total weight of initial 19 students
= (20 x 14.9 - 19 x 15) kg
= 13 kg.
Correct Option: D
Answer: (D) 13 kg