mathewhe/aqua_rat huggingface dataset:
Question: A trader has 80 kg of rice, a part of which he sells at 14% profit and rest at 6% loss. On the whole his loss is 4% . What is the quantity sold at 14% profit and that at 6% loss?
(A) 8 and 72 kg
(B) 10 and 40 kg
(C) 15 and 35 kg
(D) 20 and 30 kg
(E) 40 and 30 kg
Rationale: Alligation Method
I PART II PART
14 (x1) -6 (x2)
\ /
x
/ \
(x2-x) (x-x1)
2 18
1 9
Ratio of quantities sold at 14% profit and 6% loss = 1: 9
Quantity sold at 14% profit = 80/1+9 *1 = 8 kg
Quantity sold at 6% loss = 80/1+9 *9 = 72 kg
ANSWER:A
Answer: (A) 8 and 72 kg
Question: A merchant sells an item at a 20% discount, but still makes a gross profit of 50 percent of the cost. What percent of the cost would the gross profit on the item have been if it had been sold without the discount?
(A) 20%
(B) 40%
(C) 87.5%
(D) 60%
(E) 75%
Rationale: original SP = X
Cost =C
current selling price = .8x ( 20% discount)
.8x = 1.5C (50% profit)
x= 1.5/.8 *C
x= 15/8 C
Original selling price is 1.875C which is 87.5% profit
Answer C
Answer: (C) 87.5%
Question: If the average of a sequence of consecutive multiple of 18 is 153, and the greatest term is 270, how many terms Q in the sequence are smaller than the average?
(A) 5
(B) 6
(C) 7
(D) 8
(E) 9
Rationale: Let the sequence have N terms starting 18*K
--> 18K, 18(K+1), 18(K+2), .... , 18(K+N-1) is the sequence.
The sum of the sequence is
---> 18K + 18(K+1) + 18(K+2) + .... + 18(K+N-1) =18*(K+(K+1)+...+(K+N-1)) = 18*(N*(2K+N-1)/2).
So the average is
153=[18*(N*(2K+N-1)/2)]/N = 18*(2K+N-1)/2 = 9 *(2K+N-1) ---> by cancelling 9 we get 2K+N-1=17 --> 2K+N=18 ---*).
On the other hand, since the greatest term is 270, 18(K+N-1)=270 --> by cancelling 18 we get K+N-1=15 --> K+N=16 ---**).
By *), **) we have K=2, N=14.
Since 153/18 = 8.5, 18*8<153<18*9. So the number of terms Q which are smaller than the average is 18*2, 18*3, ..... ,18*8 ---> 7. The answer is C.
Answer: (C) 7
Question: If a and b are both prime numbers greater than 10, which of the following CANNOT be true?
I. ab is an even number.
II. a-b = 117.
III. ( a + B) is even.
(A)  I, II and III
(B) I only
(C)  I and III only
(D)  II and III only
(E)  I and II only
Rationale: MANHATTAN GMATOFFICIAL SOLUTION:
Since a and b are both prime numbers greater than 10, they must both be odd. Therefore ab must be an odd number, so Statement I cannot be true. Similarly, if a and b are both odd, then a - b cannot equal 117 (an odd number). This difference must be even. Therefore, Statement II cannot be true. Finally, since a and b are both odd, a + b must be even, so Statement III will always be true. Since Statements I and II CANNOT be true, but Statement III IS true,the correct answer is (E).
Answer : E
Answer: (E)  I and II only
Question: A and B together have Rs. 1210. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount does B have?
(A) 290
(B) 570
(C) 500
(D) 484
(E) 421
Rationale: (4/15) * A = (2/5) * B
A / B = 3 / 2
Therefore B's share is = (1210 * (2/5)) = 484
Ans - D
Answer: (D) 484
Question: In how many ways the letters of the word RAIXBON be arranged?
(A) 2535
(B) 2754
(C) 2966
(D) 5040
(E) 6454
Rationale: Total ways of arranging the letters = 7! = 5040 ways.
D)
Answer: (D) 5040
Question: The average weight of 8 person's increases by 2 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
(A) 70 kg
(B) 80 kg
(C) 81 kg
(D) 90 kg
(E) 91 kg
Rationale: Total weight increased = (8 x 2) kg = 16 kg.
Weight of new person = (65 + 16) kg = 81 kg.
C)
Answer: (C) 81 kg
Question: If a store adds 30 chairs to its current inventory, the total number of chairs will be the same as three-halves the current inventory of chairs. If the manager wants to increase the current inventory by 50%, what will the new inventory of chairs be?
(A) 40
(B) 90
(C) 100
(D) 140
(E) 160
Rationale: Explanation:
Letâ€™s say t = total current inventory of chairs. The first sentence states that 30 + t = (3/2)t. First solve for the current inventory:
30 + t = (3/2)t
30 = (3/2)t âˆ’ t
60 = (1/2)t
60 = t
The manager wants to increase this by 50%.
50% of 60 is 30, so the new inventory will be 90
Answer: B
Answer: (B) 90
Question: An isosceles triangle with sides 13cm,13cm and base of 10cm is inscribed in a circle with radius r.what is the value of 'r'?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Rationale: height of triangle = (13^2 - (10/2)^2)^1/2 = 12 cm
area of triangle = abc/4r = (1/2)*base*height
=> 13*13*10/ 4r = 1/2 *10*12
=> r = 169/24 = 7.04
ANSWER:E
Answer: (E) 7
Question: A, B and C shared the profit in a business in the ratio 5 : 7 : 8. They had partnered for 28 months, 8 months and 7 months respectively. What was the ratio of their investments?
(A) 10:12:14
(B) 12:24:28
(C) 20:22:12
(D) 10:49:64
(E) None of these
Rationale: Explanation :
Ratio in which A,B and C shared the profit = 5 : 7 : 8
Assume that the ratio of their investment = a:b:c
Then 28a : 8b : 7c = 5 : 7 : 8
Good to go, we got the equation. Now it's a matter of time to find a, b and c
28a/8b = 5/7
=> 196a = 40b
=> b = 196a/40 = 49a/20 ---(1)
28a/7c =5/8
=> 4a/c =5/8
=>32a = 5c
=> c =32a/5 ---(2)
a : b : c = a : 196a/40 : 32a/5 = 1 : 196/40 : 32/5 = 10 : 49 : 64. Answer : Option D
Answer: (D) 10:49:64
Question: The cost per pound of green tea and coffee were the same in June. In July, the price of coffee shot up by 100% and that of green tea dropped by 90%. If in July, a mixture containing equal quantities of green tea and coffee costs $3.15 for 3lbs, how much did a pound of green tea cost in July?
(A) $4
(B) $0.1
(C) $1
(D) $3
(E) $1.65
Rationale: Lets assume price of Coffee in June =100x
Price of green Tea in June =100x
Price of Coffee in July = 200x (because of 100% increase in price)
Price of green Tea in July = 10x (because of 90% Decrease in price)
Price of 1.5 pound of Coffee 1.5 pound of green Tea in July will be = 300x + 15x = 315x
As per question
315x= 3.15$
x = 0.01S
So the Price of green Tea in July = 10x = 10 x 0.01 = 0.1$/pound
Answer B
Answer: (B) $0.1
Question: Twenty identical machines, running continuously at the same constant rate, take 14 days to complete a shipment. How many additional machines, each running at the same constant rate, would be needed to reduce the time required to complete a shipment by four days?
(A) 8
(B) 9
(C) 12
(D) 14
(E) 16
Rationale: From the question you know that 20R = 1/14. The question asks you (partially) to make the rate from 1/14 to 1/10 (drop from 14 day to 10). So the only thing that you need to do is to find the magic number than can convert 1/14 to 1/10.
So 1/14 * x = 1/10 (1 equation with one unknown). So by solving this you get x = 14/10 or 7/5. Thats it then! Take the magic number 7/5 and multiply BOTH sides of the original equation and you have:
20*(7/5)*R = (7/5) * 1/14
4 * 7 * R = 1/10, Hence 28R = 1/10, therefore 8 more machines!
A
Answer: (A) 8
Question: The integer 6 is the product of two consecutive integers (6 = 2 × 3) and the product of three consecutive integers (6 = 1 × 2 × 3). What is the next integer greater than 6 that is both the product of two consecutive integers and the product of three consecutive integers?
(A) 153
(B) 210
(C) 272
(D) 336
(E) 600
Rationale: Since the question asks the lowest possible so let us start with 153:
Any number divisible by 6 would satisfy that.
153 - 17 x 9
210 - divisible by 6 so it will have the pair 1 x 2 x 3
ANSWER:B
Answer: (B) 210
Question: The distance between Delhi and Mathura is 155 kms. A starts from Delhi with a speed of 20 kmph at 7 a.m. for Mathura and B starts from Mathura with a speed of 25 kmph at 8 p.m. from Delhi. When will they meet?
(A) 10.50 a.m.
(B) 11.00 a.m.
(C) 10.30 a.m.
(D) 11.40 a.m.
(E) 1.40 a.m.
Rationale: D = 155 â€“ 20 = 135
RS = 20 + 25 = 45
T = 135/45 = 3 hours
8 a.m. + 3 hrs = 11.00 a.m.
ANSWER:B
Answer: (B) 11.00 a.m.
Question: How many ways can you arrange 16 people into 4 rows of 4 desks each?
(A) 14!
(B) 15!
(C) 16!
(D) 17!
(E) 18!
Rationale: for 1st person there are 16 places
for 2nd person there are 15 places
for 3rd ..... 14 places
no. of ways will be = 16*15*14......3*2*1
= 16!
ANSWER:C
Answer: (C) 16!
Question: The total marks obtained by a student in Mathematics and Physics is 60 and his score in Chemistry is 20 marks more than that in Physics. Find the average marks scored in Mathamatics and Chemistry together?
(A) 40
(B) 99
(C) 88
(D) 77
(E) 66
Rationale: Let the marks obtained by the student in Mathematics, Physics and Chemistry be M, P and C respectively.
Given , M + C = 60 and C - P = 20 M + C / 2 = [(M + P) + (C - P)] / 2
= (60 + 20) / 2 = 40.
Answer:A
Answer: (A) 40
Question: Employees of a certain company are each to receive a unique 7-digit identification code consisting of the digits 0, 1, 2, 3, 4, 5, and 6 such that no digit is used more than once in any given code. In valid codes, the second digit in the code is exactly twice the first digit. How many valid codes are there?
(A) 42
(B) 120
(C) 210
(D) 360
(E) 840
Rationale: One can use slot method to solve this.
7 digit identification code has 7 slots to fill.
For the first slot there are three possible options. (1,2 and 3 because first digit has to be half of second)
For the second slot there are three possible options. (2,4 and 6 because second digit has to be twice the second)
To fill the first two slots there are only three possible ways. Because if the ,
if the first is 1 the second has to be 2 (only 1 way)
if the first is 2 the second has to be 4 (only 1 way)
if the first is 3 the second has to be 6 (only 1 way) . So total ways to fill first two slots = 3ways
You are now left with 5 digits and 5 slots to fill, which can be filled in 5! ways i.e. 120 ways.
Total no: of valid codes will be therefore : 120 * 3 = 360 ways.
Answer : D
Answer: (D) 360
Question: In what time will a train 50 m long cross an electric pole, it its speed be 360 km/hr?
(A) 0.25
(B) 0.5
(C) 1.5
(D) 2.5
(E) 3.5
Rationale: Speed = 360 * 5/18 = 100 m/sec
Time taken = 50/100 = 0.5 sec.
Answer: B
Answer: (B) 0.5
Question: Two trains travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 8 seconds.The length of the faster train is
(A) 80 m
(B) 100 m
(C) 120 m
(D) 180 m
(E) 160 m
Rationale: Relative speed= 81 km/h
= 81 x 5/18 m/sec
=22.5 m/sec
length of slower train= speed x time
=36 x 8 km/hr
=288 x 5/18 m/sec
=80 m/sec
Let length of faster train= b
time cross to each other 8 = (80+b)x 18/(81 x 5)
=3240-1440=18b
1800/18= b
b=100 m
ANSWER:B
Answer: (B) 100 m
Question: A man buys 12 lts of liquid which contains 20% of the liquid and the rest is water. He then mixes it with 10 lts of another mixture with 30% of liquid. What is the % of water in the new mixture?
(A) 75.45
(B) 76.45
(C) 77.45
(D) 74.45
(E) 73.45
Rationale: 20% in 12 lts is 2.4. So water = 12-2.4 = 9.6 lts. 30% of 10 lts = 3. So water in 2nd mixture = 10-3 = 7 lts. Now total quantity = 12 + 10 = 22 lts. Total water in it will be 9.6 + 7 = 16.6 lts. % of water = (100*16.6)/22 = 75.45
ANSWER:A
Answer: (A) 75.45
Question: In how many ways can 6 arts and 3 science students be arranged in a row so that the arts and the science students are placed alternately?
(A) 2880
(B) 1850
(C) 3260
(D) 5200
(E) 4320
Rationale: The six art students can be arranged in those 6 A-slots in 6! = 720 ways.
The three science students can be arranged in those 3 S-slots in 3! = 6 ways.
Total arrangements = 720*6 = 4320
Answer =(E)
Answer: (E) 4320
Question: For every even positive integer m, f(m) represents the product of all even integers from 2 to m, inclusive. For example, f(12)=2x4x6x8x10x12. What is the greatest prime factor of f(36)?
(A) 2
(B) 5
(C) 11
(D) 17
(E) 23
Rationale: f(36) = 2*4*6*8*10*12*14*16*18*20*22*24*26*28*30*32*34*36
The greatest prime factor in this list is 17.
The answer is D.
Answer: (D) 17
Question: A certain number of horses and an equal number of men are going somewhere. Half of the owners are on their horses' back while the remaining ones are walking along leading their horses. If the number of legs walking on the ground is 80, how many horses are there ?
(A) 10
(B) 12
(C) 14
(D) 16
(E) 18
Rationale: legs 16*4 = 64
now half on their horses so remaining on the walk so 8 men
8 men has 16 legs
so, 16+ 64 = 80 legs walking
ANSWER:D
Answer: (D) 16
Question: This is how Edward’s Lotteries work. First, 9 different numbers are selected. Tickets with exactly 5 of the 9 numbers randomly selected are printed such that no two tickets have the same set of numbers. Finally, the winning ticket is the one containing the 5 numbers drawn from the 9 randomly. There is exactly one winning ticket in the lottery system. How many tickets can the lottery system print?
(A)  9P6
(B)  9P3
(C)  9C9
(D)  9C5
(E)  69
Rationale: since we have to select random 5 numbers from 9 and they all are distinct. i think it should be 9C5 or D.
Answer: (D)  9C5
Question: In a boat 25 persons were sitting. Their average weight increased one kilogram when One man goes and a new man comes in. The weight of the new man is 70kgs. Find the Weight of the man who is going ?
(A) 45
(B) 66
(C) 44
(D) 33
(E) 61
Rationale: Weight increased per person is 1 kg. Total increase in weight = 25 kgs Weight of new man is 70 kgs, (Which means his weight is 25 kgs heavier) The weight of the old man was 70 – 25 = 45 kgs
Answer: A
Answer: (A) 45
Question: If P = Q/R, -1 < R < 0 and 4 < Q, which of the following is correct?
(A) P > 4.
(B) 0 < P <4.
(C) -4 < P < 0.
(D) P < -4.
(E) P < -20.
Rationale: Q is +veR is -ve
Eliminate AB
For min. value max. numerator and minimize dinominator
take Q=4.1P=-0.9
P=4.1/-0.9
P<-4
Ans D
Answer: (D) P < -4.
Question: Find the greatest number which on dividing 1661 and 2045, leaves a reminder of 10 and 13
respectively
(A) 125
(B) 127
(C) 129
(D) 131
(E) None of these
Rationale: Explanation:
In this type of question, its obvious we need to calculate the HCF, trick is
HCF of (1661 - 10) and (2045 -13)
= HCF (1651, 2032) = 127
Answer: Option B
Answer: (B) 127
Question: The average age of three boys is 120 years and their ages are in proportion 2:6:8. What is the age in years of the youngest boy?
(A) 10
(B) 9
(C) 15
(D) 6
(E) 12
Rationale: 2x + 6x + 8x = 120
x =7.5
2x = 15
Answer:C
Answer: (C) 15
Question: Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.64000?
(A) 81000
(B) 81007
(C) 81008
(D) 81066
(E) 81022
Rationale: 64000* 9/8 * 9/8 = 81000
Answer:A
Answer: (A) 81000
Question: A certain car uses 12 gallons of gasoline in traveling 120 miles. In order for the car to travel the same distance using 10 gallons of gasoline, by how many miles per gallon must the car’s gas mileage be increased?
(A) 2
(B) 4
(C) 6
(D) 8
(E) 10
Rationale: 120/10 = 12. The difference is 12- 10 = 2.
Answer A
Answer: (A) 2
Question: A rower can row 6 km/h in still water. When the river is running at 2 km/h, it takes the rower 1 hour to row to Big Rock and back. How many kilometers is it to Big Rock?
(A) 2.21
(B) 2.45
(C) 2.67
(D) 2.83
(E) 2.95
Rationale: Let x be the distance to Big Rock.
time = x/4 + x/8 = 1
x = 32/12 = 2.67 km
The answer is C.
Answer: (C) 2.67
Question: A bag contains 6 white balls, 7 black balls, and 7 red balls. One by one, three balls are drawn out without replacement. What is the probability that the third ball is red?
(A) 0.25
(B) 0.15
(C) 0.35
(D) 0.45
(E) 0.4
Rationale: Each ball has an equal probability of being the third ball drawn.
P(3rd ball is red) = 7/20 = 0.35
The answer is C.
Answer: (C) 0.35
Question: 10,8, x, 6
The median of the list of positive integers above is 7. Which of the following could be the average (arithmetic mean) of the list?
(A) 3
(B) 5.5
(C) 6.25
(D) 7
(E) 7.5
Rationale: GivenThe median of the list of positive integers above is 7
So arranging the numbers from lowest to highest, avg of 2 middle terms needs to be 7
so the sequence will be x 6 8 10
Lets say x = 4, which gives us mean = sum/4 = 7
D is correct
Answer: (D) 7
Question: The avearge score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 41. Then find the average for the last four matches?
(A) 33.25
(B) 35.75
(C) 34.25
(D) 35
(E) none
Rationale: Solution
Required average	=(38.9 x 10)-(41 x 6)/ 4
= 143 / 4.
= 35.75
Answer B
Answer: (B) 35.75
Question: Volume of a hemisphere is 19404 cu. cm. Its radius is :
(A) 19 cm
(B) 20 cm
(C) 21 cm
(D) 30 cm
(E) None
Rationale: Sol.
Let the radius be R cm. Then,
2/3 * 22/7 * R³ = 19404 ⇔ R³
= [19404 * 21/44] = (21)³ ⇔ R = 21 cm.
Answer C
Answer: (C) 21 cm
Question: A driver just filled the car's gas tank with 27 liters of gasohol, a mixture consisting of 5% ethanol and 95% gasoline. If the car runs best on a mixture consisting of 10% ethanol and 90% gasoline, how many liters of ethanol must be added into the gas tank for the car to achieve optimum performance?
(A) 1.2
(B) 1.5
(C) 1.8
(D) 2.1
(E) 2.4
Rationale: Let x be the number of liters of ethanol added to the gas tank.
0.05(27) + x = 0.1(27+x)
0.9x = 2.7 - 1.35 = 1.35
x = 1.5 liters
The answer is B.
Answer: (B) 1.5
Question: An alloy contains only zinc and copper. One such alloy weighing 15-gm contains zinc and copper in the ratio of 2: 3 by weight. If 10 gm of zinc is added then find what amount of copper has to be removed from the alloy such that the final alloy has zinc and copper in the ratio of 4:1 by weight?
(A) 5 gm
(B) 5.5 gm
(C) 6 gm
(D) 4.8 gm
(E) 5.8 gm
Rationale: Weight of zinc in the alloy= 6 gm and weight of copper in the alloy = 9 gms.
Suppose ’x’ gms of copper is removed.
Weight of copper in the alloy=9 - x
Weight of zinc in the alloy=6 + 10= 16.
Now,16 = 4(9 - x)
Therefore x = 5 gm
ANSWER:A
Answer: (A) 5 gm
Question: Let us say that a table tennis tournament was going on with knock out terms which means the one who loses the match is out of the tournament. 100 players took part in that tournament.
How many matches were played?
(A) 99 matches.
(B) 88 matches.
(C) 77 matches.
(D) 66 matches.
(E) 55 matches.
Rationale: Solution:
99 matches.
The number of matches will always sum up to one less than the number of players in a knock out tournament. You may calculate it in any manner. Thus 99 matches were played.
Answer A
Answer: (A) 99 matches.
Question: The sum of ages of 5 children born 3 years different each is 60 yrs. What is the age of the Elder child?
(A) 17
(B) 19
(C) 16
(D) 18
(E) 21
Rationale: Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 60
5x = 30
x = 6.
x+12= 6+12= 18
D
Answer: (D) 18
Question: If two integers x, y (x>y) are selected from -10 to 1 (inclusive), how many possible cases are there?
(A) 150
(B) 66
(C) 190
(D) 210
(E) 240
Rationale: If two integers x, y (x>y) are selected from -10 to 9 (inclusive), how many possible cases are there?
A. 150
B. 180
C. 190
D. 210
E. 240
--> 12C2=12*11/2=66. Therefore, the answer is B.
Answer: (B) 66
Question: A 48 lts container containing liquid A and B contains 25% liquid A. a few lts of the mixture is released and replaced with equal amount of Liquid B. If this process is repeated once, the cylinder is found to contain 16% liquid A .How many lts of the mixture was released each time.?
(A) 4.8
(B) 9.6
(C) 7.68
(D) 9
(E) None of the Above
Rationale: 25% of mixture is liquid A => 12 ltrs of liquid A and 36 ltrs of liquid B together form 48 ltrs of the mixture.
This implies that the ratio of liquids A and B in the mixture is is 1:3
Let the amount of liquid removed i x ltrs => x/4 ltrs of liquid A and 3x/4 ltrs of liquid B is removed.
Given that the x liters of mixture removed is added with same quantity of liquid B =>
12-(x/4) + {36-(3x/4) + x} = 48
=> {12 -(x/4)} + {36 + x/4} = 48
The ratio of liquid A to total mixture after first time the process is done = 16% =>
{12- (x/4)} / 48 = 16/100
12- (x/4) = 7.68
x/4 = 4.32 => x= 17.28
ANSWER:C
Answer: (C) 7.68
Question: When tossing two coins once, what is the probability of heads on both the coins?
(A) 1/4
(B) 1/2
(C) 3/4
(D) 1/3
(E) None of the above
Rationale: Total number of outcomes possible when a coin is tossed = 2 (âˆµ Head or Tail)
Hence, total number of outcomes possible when two coins are tossed, n(S) = 2 Ã— 2 = 4
(âˆµ Here, S = {HH, HT, TH, TT})
E = event of getting heads on both the coins = {HH}
Hence, n(E) = 1
P(E) = n(E)/n(S)=1/4
Answer:A
Answer: (A) 1/4
Question: An art gallery owner is hanging paintings for a new show. Of the seven paintings she has to choose from, she can only hang three on the main wall of the gallery. Assuming that she hangs as many as possible on that wall, in how many ways can she arrange the paintings?
(A) 18
(B) 30
(C) 64
(D) 840
(E) 216
Rationale: Arrangement means permutations questions..
7P3 = 7!/3! = 840
.
ANS option D is correct answer..
Answer: (D) 840
Question: Ali and Faizer are 27m apart. They both start walking and meet 9 hrs later, if they travel in the same direction and meet after 3 hrs if they walk in opposite dirction, Aliâ€™s speed must most likely be ?
(A) 2 m/h
(B) 7 m/h
(C) 6 m/h
(D) 9 m/h
(E) 0 m/h
Rationale: Explanation:
let, speed of Ali=X mph and Faizer=Y mph
Now, relative speed when Ali is walking int the same direction with Faizer= (X-Y)mph
hence, according to question, 27/(X-Y)=9 ; solving X-Y=3; --- (1)
Now relative speed when walking in the opposite direction= (X+Y) mph
hence, according to question, 27/(X+y)=3; solving X+Y=9; --- (2)
solving 1 and 2 we get X=6 and Y=3, where X is the speed of Ali.
Answer: C
Answer: (C) 6 m/h
Question: A began business with Rs.45000 and was joined afterwards by B with Rs.5400. When did B join if the profits at the end of the year were divided in the ratio of 2:1?
(A) 6
(B) 0
(C) 7
(D) 5
(E) 4
Rationale: 45*12 : 54*x = 2: 1
x = 5
12 -5 = 7
Answer:C
Answer: (C) 7
Question: Simplify 586645 * 9999
(A) 5865863355
(B) 5665863355
(C) 4865863355
(D) 4665863355
(E) None of these
Rationale: Explanation:
Although it is a simple question, but the trick is to save time in solving this.
Rather than multiplying it we can do as follows:
586645 * (10000 - 1) = 5866450000 - 586645 = 5865863355
Option A
Answer: (A) 5865863355
Question: In how many no. between 10 and 30 exactly Two of the digits is 3?
(A) 1
(B) 20
(C) 30
(D) 40
(E) 50
Rationale: It's simple can be solved by elimination of answer choices.
Option B and C are too large, not possible. Even DE are large to have correct choice.
Ans : A
Answer: (A) 1
Question: An article is bought for Rs.823 and sold for Rs.1000, find the gain percent?
(A) 21.4%
(B) 21.5%
(C) 21.6%
(D) 21.7%
(E) 21.8%
Rationale: 823 ---- 177
100 ---- ? => 21.5%
Answer:B
Answer: (B) 21.5%
Question: Ann is organizing a two day workshop next month. Exactly 100 people will be attending either one day kje the workshop or both. Out of them, 70% will be there for the first day, and 80% for the second. Help Ann calculate how many people will be present for both days of the workshop?
(A) 30
(B) 40
(C) 50
(D) 60
(E) 70
Rationale: The total number of people attending the workshop is n = 100.
The number of people attending the first day of the workshop is 70% of it, thus
n (A)=70/100×100=70.
The number of people attending the second day of the workshop is 80%, thus
n (B)=80/100×100=80.
In order to calculate the number of people attending both days of the workshop we can use the following equation:
n (AnB)=n (A)+n (B) -n = 70+80-100=150-100=50
So, 50 people will be attending both the first and the second day of Ann's workshop.
ANSWER: C
Answer: (C) 50
Question: Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank?
(A) 12
(B) 55
(C) 77
(D) 82
(E) 22
Rationale: Part filled by A in 1 min. = 1/20
Part filled by B in 1 min. = 1/30
part filled by (A + B) in 1 min. = 1/20 + 1/30 = 1/12.
Both the pipes can fill the tank in 12 minutes.
Answer: A
Answer: (A) 12