mathewhe/aqua_rat huggingface dataset:
Question: In three coloured boxes - Red, Green and Blue, 108 balls are placed. There are twice as many balls in the green and red boxes combined as there are in the blue box and twice as many in the blue box as there are in the red box. How many balls are there in the green box ?
(A) 18
(B) 36
(C) 45
(D) 54
(E) None of these
Rationale: Let R, G and B represent the number of balls in red, green and blue boxes respectively.
Then, .
R + G + B = 108 ...(i),
G + R = 2B ...(ii)
B = 2R ...(iii)
From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.
Putting G = 3R and B = 2R in (i), we get:
R + 3R + 2R = 108 6R = 108 R = 18.
Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.
ANSWER:D
Answer: (D) 54
Question: There are three foam generators in the factory, each of the first two can generate 8 liters of foam in one hour and the third can generate 18 liters in an hour. The three generators start working together at the same time and after one hour and a half one of the first generators stops working and two hours after that the third generator stops working and only one generator is left. If 5 hours after they all started to work the last generator stops working, how many liters of foam were generated?
(A) 120.
(B) 132.
(C) 146.
(D) 154.
(E) 115
Rationale: Let the foam generators capacity be -
A = 8 lit/hr
B = 8 lit/hr
C = 18 lit/hr
Total foam generation in 1 hour will be 34 lit ( 8 + 8 + 18 ) ; since in 1 and 1/2 hr they will generate 34 + 17 => 51 litres...
Now one of the first generators stops working ( Say A stops working ) , so we have -
B = 8 lit/hr
C = 18 lit/hr
Total foam generation in 2 hour will be 52 litres {2 (8+18)}
The third generator stops working and now only B works for the remaining time 1 and 1/2 ( 5 - 1 and 1/2 - 2 )...
Foam generated by B will be 3/2 * 8 => 12 litres
So, total foam generated will be 115 Litres ( 51 + 52 + 12 ) ; hence answer will be (E)
Answer: (E) 115
Question: In a game, one player throws two fair, six-sided die at the same time. If the player receives at least a two or a one on either die, that player wins. What is the probability that a player wins after playing the game once?
(A) 1/3
(B) 4/9
(C) 5/9
(D) 2/3
(E) 3/4
Rationale: We're looking for the number of outcomes that include AT LEAST a 1 or a 2.
#getting a one in 1st die = {(1,1), (1,2), ...(1,6) }= 6
#getting a one in 2nd die {(3,1), ...(5,1),(6,1) } = 4 [exclude 1,1 and 2,1] since it will be counted twice
#getting a two in 1st die {(3,1), (3,2), ...(3,6) }= 6
#getting a two in 1st die {(3,2), (4,2), ...(6,2) } = 4 [exclude 1,2 and 2,2] since it will be counted twice
Total possibities = 20
Probability of rolling at least a 1 or a 2 on two dice: 20/36 = 5/9
C
Answer: (C) 5/9
Question: A rectangular field has a length 10 meters more than it is width. If the area of the field is 144, what is the length (in meters) of the rectangular field?
(A) 16
(B) 18
(C) 20
(D) 22
(E) 24
Rationale: area = l * w = (L) * (L-10) = 171
trial and error:
20*10 = 200 (too high)
19*9 = 171 (too high)
18*8 = 144
The length is 18 meters.
The answer is B.
Answer: (B) 18
Question: Find the smallest number which leaves 22,35, 48 and 61 as remainders when divided by 26, 39, 52 and 65 respectively.
(A) 67
(B) 26
(C) 99
(D) 26
(E) 91
Rationale: Answer: D
Answer: (D) 26
Question: 8, 12, 18, 27, 40.5, (...)
(A) 62
(B) 60.75
(C) 60.5
(D) 60.25
(E) 60
Rationale: 8
(8×3)÷ 2 = 12
(12×3)÷ 2 = 18
(18×3)÷ 2 = 27
(27×3)÷ 2 = 40.5
(40.5×3)÷ 2 = 60.75
Answer is B.
Answer: (B) 60.75
Question: In the faculty of Reverse-Engineering, 100 second year students study numeric methods, 100 second year students study automatic control of airborne vehicles and 25 second year students study them both. How many students are there in the faculty if the second year students are approximately 50% of the total?
(A) 340
(B) 335
(C) 370
(D) 360
(E) 350
Rationale: Total number of students studying both are 100 + 100 - 25 = 175 (Subtracting the 25 since they were included in the both the other numbers already).
So 50% of total is 175, so 100% is 350
Answer is E
Answer: (E) 350
Question: Which of the following list a prime number ?
(A) 1
(B) 4
(C) 8
(D) 7
(E) 10
Rationale: D
7
Clearly, 7 is a prime number.
Answer is D
Answer: (D) 7
Question: The average weight of A, B and C is 65 kg. If the average weight of A and B be 60 kg and that of B and C be 70 kg, then the weight of A is:
(A) 55 kg
(B) 65 kg
(C) 75 kg
(D) 45 kg
(E) 63 kg
Rationale: EXPLANATION
Let A, B, C represent their respective weights. Then, we have:
A + B + C = (65 x 3) = 195 â€¦. (i)
A + B = (60 x 2) = 120 â€¦. (ii)
B + C = (70 x 2) = 140 â€¦.(iii)
Adding (ii) and (iii), we get: A + 2B + C = 260 â€¦. (iv)
Subtracting (i) from (iv), we get : B = 65.
SUBSTITUTING B IN (ii)
A+65=120
A=55
Aâ€™s weight = 55 kg.
Answer A
Answer: (A) 55 kg
Question: 0.002 x 0.5 = ?
(A) 0.0001
(B) 0.001
(C) 0.01
(D) 0.1
(E) NONE OF THESE
Rationale: Explanation:
2 x 5 = 10.
Sum of decimal places = 4
0.002 x 0.5 = 0.001
Answer- B
Answer: (B) 0.001
Question: If 2x = 3y = 10, then 3xy = ?
(A)  100
(B)  200
(C)  120
(D)  50
(E)  20
Rationale: 2x = 10;x=5
3y = 10;y=10/3
Multiply: 3xy=3*5*10/3=50
Answer: D.
Answer: (D)  50
Question: Dany bought a total of 26 game cards some of which cost $0.25 each and some of which cost $0.15 each. If Dany spent $4.2 to buy these cards, how many cards of each type did he buy?
(A) 4
(B) 10
(C) 7
(D) 23
(E) 12
Rationale: Let X be the number of cards that cost $0.25 each and Y the number of cards that cost $0.15 each. The total number of cards is 26. Hence
X + Y = 26
If X is the number of cards at $0.25, then the X cards cost
0.25 X
If Y is the number of cards at $0.15, then the Y cards cost
0.15 Y
The total cost of the X cards and the the Y cards is known to be $4.2 and also given by
0.25 X + 0.15 Y = 4.2
We now need to solve the system of equations
X + Y = 26
0.25 X + 0.15 Y = 4.2
The first equation gives Y = 26 - X. Substitute y by 26 - x in the second equation and solve
0.25 X + 0.15 (26 - X) = 4.2
X(0.25 - 0.15) + 3.9 = 4.2
0.1 X = 0.3
X = 3 and Y = 26 - 3 = 23
correct answer D
Answer: (D) 23
Question: The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is 6912 sq cm. Find the ratio of the breadth and the area of the rectangle?
(A) 1:96
(B) 1:28
(C) 2:28
(D) 1:29
(E) 1:12
Rationale: Let the length and the breadth of the rectangle be 4x cm and 3x respectively.
(4x)(3x) = 6912
12x2 = 6912
x2 = 576 = 4 * 144 = 22 * 122 (x > 0)
=> x = 2 * 12 = 24
Ratio of the breadth and the areas = 3x : 12x2 = 1 : 4x
= 1:96.
Answer:A
Answer: (A) 1:96
Question: If x+y=-10, and x=25/y, what is the value of x^2+y^2?
(A) 55
(B) 65
(C) 50
(D) 75
(E) 85
Rationale: x^2 + y^2 should make you think of these formulas:
(x+y)(x+y) = x^2+y^2+2xy
We already know (x+y) = -10 and x*y = 25
(x+y)(x+y) = (-10)(-10) = x^2+y^2+2*(25)
x^2+y^2 = 100 - 50 = 50
Answer : C
Answer: (C) 50
Question: The cricket team of 11members is 26yrs old & the wicket keeper is 3 yrs older. If the ages ofthese 2are excluded, the average age of theremaining players is 1 year less than the average age of the whole team. What is the average age of the team ?
(A) 19
(B) 21
(C) 23
(D) 25
(E) 27
Rationale: Let the average age of the whole team be x years.
11x - (26 + 29) = 9 (x - 1)
=> 11x - 9x = 46
=> 2x = 46
=> x = 23.
So, average age of the team is 23 years.
C
Answer: (C) 23
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) 237
(B) 267
(C) 298
(D) 480
(E) 291
Rationale: 262 – 102 = 242
d1 = 20 d2 = 48
1/2 * 20 * 48 = 480
Answer: D
Answer: (D) 480
Question: Find the ratio in which rice at Rs.7.00 a kg be mixed with rice at Rs.5.70 a kg to produce a mixture worth Rs.6.30 a kg?
(A) 1 : 3
(B) 2 : 3
(C) 3 : 4
(D) 6 : 7
(E) None of these
Rationale: Solution
Required ratio = 60 : 70 = 6 : 7
Answer D
Answer: (D) 6 : 7
Question: ABCDEFGH is an equiangular octagon with side lengths 2,4√2,1,3√2,2,3√2,3,and 2√2
in that order. Compute the area of the octagon.
(A) 25
(B) 35
(C) 45
(D) 55
(E) 65
Rationale: The octagon can be inscribed in a square with side length 8. The four right triangles
at the vertices of the square have a combined area of 2 + 8 + 9/2 + 9/2 = 19, giving a remaining
area of 64- 19 = 45 .
correct answer C
Answer: (C) 45
Question: Two trains of equal are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 sec. The length of each train is?
(A) 50
(B) 99
(C) 77
(D) 26
(E) 23
Rationale: Let the length of each train be x m.
Then, distance covered = 2x m.
Relative speed = 46 - 36 = 10 km/hr.
= 10 * 5/18 = 25/9 m/sec.
2x/36 = 25/9 => x = 50.Answer: A
Answer: (A) 50
Question: What is the smallest integer that is the sum of 2 distinct prime integers, each greater than 140?
(A) 270
(B) 372
(C) 273
(D) 199
(E) 300
Rationale: A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. Here are the first few prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, etc.
Immediate two prime numbers after 70 are - 71 and 73
Sum = 149 + 151= 300
Answer - E
Answer: (E) 300
Question: In one hour,a boat goes 13 km along the stream and 9 km against the stream.The speed of the boat in still water (in km/hr) is :
(A) 3
(B) 5
(C) 11
(D) 9
(E) 10
Rationale: Solution
Speed in still water =1/2(13+9)kmph.= 11 kmph. Answer C
Answer: (C) 11
Question: A solution contains 8 parts of water for every 7 parts of Lemonade syrup. How many parts of the solution should be removed and replaced with water so that the solution will now contain 25% lemonade syrup?
(A) 1.5
(B) 1.75
(C) 2.14
(D) 2.34
(E) 2.64
Rationale: Let the total solution is 150 L with 80 L water70 L syrup.
To make 25% syrup solution, the result solution must have 112.5 L syrup and 37.5 L syrup.
Therefore we are taking 32.5 L of syrup from initial solution and replacing with water.
using urinary method:
70 L syrup in 150 L solution
32.5 L syrup in 69.6 L solution
We started by multiplying 10
Now to get to the result we need to divide by 32.5 => amount of solution to be replaced with water = (69.6/32.5) = 2.14.
Correct option : C
Answer: (C) 2.14
Question: A bag contains 12 white and 18 black balls. 2balls are drawn in succession. What is the probability that first is white and second is black?
(A) 31/145
(B) 23/145
(C) 36/145
(D) 31/115
(E) 41/215
Rationale: The probability that first ball is white:
=12C130C1=12C130C1
=1230=1230
=25=25
Since, the ball is not replaced; hence the number of balls left in bag is 29.
Hence, the probability the second ball is black:
=18C129C1=18C129C1
=1829=1829
Required probability,
=(25)×(1829)=(25)×(1829)
=36/145
B
Answer: (B) 23/145
Question: The number of people at Ovations Bar in the hour from 12 p.m. to 1 p.m. was 30% greater than the number of people there from 11 a.m. to 12 p.m. The number of the people at the bar from 11 a.m. to 12 p.m. was 20% less than the number there from 10 a.m. to 11 a.m. If 92 people were at the bar from 11 a.m. to 1 p.m., which of the following is the number of people in the bar from 10 a.m. to 11 a.m.?
(A) 10
(B) 20
(C) 30
(D) 40
(E) 50
Rationale: No of people from 10 - 11 is x
No of people from 11 - 12 is 4x/5
No of people from 12 to 1 is (13/10)(4x/5)
Given that 4x/5 + (13/10)(4x/5) = 92x/50 = 92
Then x = 50
E
Answer: (E) 50
Question: A number exceeds 20% of itself by 40. The number is ?
(A) 50
(B) 60
(C) 80
(D) 320
(E) None
Rationale: Answer
∵ A - 20% of A = 40
⇒ A - A/5 = 40
⇒ 4A/5 =40
∴ A = (5 x 40)/4 = 50.
Correct Option: A
Answer: (A) 50
Question: Aish started a business investing Rs. 25,000 in 2010, In 2011, he invested an additional amount of Rs. 10,000 and Roshan joined him with an amount of Rs. 35,000. In 2012, Aish invested another additional amount of Rs. 10,000 and Atul joined them with an amount of Rs. 35,000. What will be Roshan’s share in the profit of Rs. 2,70,000 earned at the end of 3 years from the start of the business in 2010?
(A) Rs.45000
(B) Rs.50000
(C) Rs.60000
(D) Rs.90000
(E) None of these
Rationale: Explanation :
Aish : Roshan : Atul
= (25000 x 12 + 35000 x 12 + 45000 x 12) : (35000 x 24) : (35000 x 12)
= 1260000 : 840000 : 420000 = 3 : 2 : 1
Roshan's share = Rs. 270000 x 2/6
= Rs. 90000.
Answer : D
Answer: (D) Rs.90000
Question: If one-third of one-fourth of a number is 12, then three-tenth of that number is:
A. 35
(A) 43.2
(B) 87
(C) 26
(D) 54
(E) 01
Rationale: Explanation:
The number is 1/3 of 1/4 is =15 then 1/3 Ã— 1/4 = 12
Number is 144 then 144 Ã— 3/10 = 43.2
Answer: A
Answer: (A) 43.2
Question: Three pipes A, B and C can fill a tank from empty to full in 20 minutes, 20 minutes and 30 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P, Q and R respectively. What is the proportion of solution P in the liquid in the tank after 3 minutes?
(A) 3/8
(B) 2/8
(C) 6/7
(D) 3/5
(E) 2/5
Rationale: Part filled by (A + B + C) in 3 minutes = 3(1/20 + 1/20 + 1/30) = 6/15
Part filled by A in 3 minutes = 3/20
Required ratio = 3/20 * 15/6 =3/8
Answer:A
Answer: (A) 3/8
Question: In a certain habitat, animals live in groups of two and groups of three. Each time a new animal
arrives at the habitat, it randomly chooses a group. If the chosen group has two animals in
it, the new animal joins the group forming a group of three animals. If the chosen group has
three animals in it, the new animal takes one of the animals away from the group of three and
forms a separate group of two animals.
If initially there are five animals in the habitat, and new animals arrive one at a time, what is
the probability that the fourth new animal joins a group of two animals?
(A) 1/3
(B) 1/2
(C) 5/9
(D) 5/8
(E) none of the previous choices
Rationale: After the first new animal arrives there will be either three groups of 2 animals or two groups
of 3 animals. After the second new animal arrives, in both cases we will end up with two groups
of 2 animals and one group of 3 animals. The third new animal will either A: join a group of two
animals, or B: join a group of three animals. Event A happens with probability 2=3 and we end up
with one group of 2 animals and two groups of 3 animals; Event B happens with probability 1=3
and we end up with four groups of 2 animals. If event A happens, the fourth new animal joins a
group of 2 animals with probability 1=3. If event B happens, the fourth animal joins a group of 2
animals with probability 1. The probability of the fourth new animal joining a group of 2 animals
is (2=3)(1=3) + (1=3)(1) = 5=9.
correct answer C
Answer: (C) 5/9
Question: If Rs.7500 are borrowed at C.I at the rate of 4% per annum, then after 1 year the amount to be paid is?
(A) 3377
(B) 2688
(C) 2688
(D) 7800
(E) 1268
Rationale: A = 7500(26/25)^1 = 7800
Answer: D
Answer: (D) 7800
Question: A pair of toys was bought for $162 at a discount of 10%. What must be the marked price of each of the toy?
(A) $300
(B) $500
(C) $90
(D) $400
(E) $600
Rationale: S.P. of each of the article = 162/2 = $81
Let M.P = $x
90% of x = 81
x = 81*100/90 =$90
Answer is C
Answer: (C) $90
Question: 33 men can complete a piece of work in 18 days. In how many days will 27 men complete the same work ?
(A) 22
(B) 77
(C) 36
(D) 25
(E) 13
Rationale: Explanation:
Less Men, means more Days {Indirect Proportion}
Let the number of days be x
then,
27 : 33 :: 18 : x
[Please pay attention, we have written 27 : 33 rather than 33 : 27, in indirect proportion, if you get it then chain rule is clear to you :)]
{\color{Blue} x=\frac{33\times 18}{27}}
x = 22
So 22 days will be required to get work done by 27 men.
Answer: A
Answer: (A) 22
Question: Tom can hit a target 3 times in 6 shots, Karan can hit the target 2 times in 6 shots & Ashwini can hit 4 times in 4 shots. What is the probability that at least 2 shots hit the target ?
(A) 1/2
(B) 1/3
(C) 2/3
(D) 3/5
(E) 3/7
Rationale: Ashwini hits the target definitely, hence required probability that atleast 2 shots hit the target is given by
Karan hits tha target and Raja not hit the target.
or
Karan does not hit the target and Raja hits the target.
or.
Karan hits the target and Raja hits the target
= 2/6 x 3/6 + 4/6 x 3/6 + 2/6 x 3/6
= 24/36 = 2/3
C
Answer: (C) 2/3
Question: A can do a piece of work in 3 days of 5 hours each and B alone can do it in 5 days of 3 hours each.how long will they take it to do working together
3/4 hours a day?
(A) 6 days
(B) 5 days
(C) 10 days
(D) 9 days
(E) 8 days
Rationale: A's work per hour=1/15
B's work per hour=1/15
A & B's work per hour together=(1/15)+(1/15)=2/15
so a & B together complete the work in 15/2 hours...
if they work 2 1/2=5/2 hours a day,
it will take (15/2)/(3/4) days=10 days...
ANSWER: C
Answer: (C) 10 days
Question: If (1/2)^18 (1/81)^k = 1/18^18, then k =
(A) 9
(B) 12
(C) 16
(D) 24
(E) 36
Rationale: I'm going to focus on denominator only..
(2^18) . ((3^4)^k = 18^18
(2^18) . ((3^4k) = (2. 3^2)^18
(2^18) . ((3^4k) = (2^24) . (3^2)^18
hence
4k = 36
k = 9
answer A
I hope it's quite clear
Answer: (A) 9
Question: Will must choose a 3-character computer password, consisting of 1 letter from the alphabet and 2 distinct digits, in any order.From how many different passwords can Will choose?
(A) 390
(B) 2,340
(C) 4,680
(D) 7,020
(E) 14,040
Rationale: Take the task of creating a password and break it into stages.
Stage 1: Select the one letter to be used in the code
There are 26 letters from which to choose, so we can complete this stage in 26 ways.
Stage 2: Select the two digits to be used in the code
Since the order in which we select the two digits does not matter, we can use combinations.
We can select 2 digits from 10 women in 10C2 ways (45 ways)
So, we can complete stage 2 in 45 ways
NOTE: We now have the 3 characters to be used in the code. At this point, we need to arrange those 3 characters.
Stage 3: Arrange the 3 selected characters.
RULE: We can arrange n unique objects in n! ways.
So, we can arrange the 3 characters in 3! ways (6 ways)
So we can complete stage 3 in 6 ways.
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create a password ) in (26)(45)(6) ways
(26)(45)(6) = 7020 ANS:D
Answer: (D) 7,020
Question: A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is :
(A) 1/4
(B) 1/10
(C) 7/15
(D) 8/15
(E) 9/15
Rationale: Explanation:
A's 1 day's work =	1/15
B's 1 day's work =	1/20
(A + B)'s 1 day's work =	(1/15 + 1/20)=7/60
(A + B)'s 4 day's work =7/60 X 4 = 7/15
Therefore, Remaining work =	1-7/15 =8/15
ANSWER IS D
Answer: (D) 8/15
Question: The difference of the squares of two consecutive even integers is divisible by which of the following integers ?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Rationale: Let the two consecutive even integers be 2n and (2n + 2). Then,
(2n + 2)2 = (2n + 2 + 2n)(2n + 2 - 2n)
= 2(4n + 2)
= 4(2n + 1), which is divisible by 4.
Answer: Option B
Answer: (B) 4
Question: A train 200 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
(A) 45 km/hr
(B) 50 km/hr
(C) 77 km/hr
(D) 55 km/hr
(E) 56 km/hr
Rationale: Speed of the train relative to man = (200/10) m/sec = 20 m/sec. [20 * (18/5)] km/hr = 72 km/hr. Let the speed of the train be x km/hr. Then, relative speed = (x - 5) km/hr. x - 5 = 72==> x = 77 km/hr.
ANSWER: C
Answer: (C) 77 km/hr
Question: In how many ways can 6 arts and 4 science students be arranged in a row so that the arts and the science students are placed alternately?
(A) 12880
(B) 11850
(C) 13260
(D) 15200
(E) 17280
Rationale: The six art students can be arranged in those 6 A-slots in 6! = 720 ways.
The four science students can be arranged in those 4 S-slots in 4! = 24 ways.
Total arrangements = 720*24 = 17280
Answer =(E)
Answer: (E) 17280
Question: Nina has exactly enough money to purchase 6 widgets. If the cost of each widget were reduced by $1.5, then Nina would have exactly enough money to purchase 8 widgets. How much money does Nina have?
(A) $22
(B) $24
(C) $30
(D) $36
(E) $40
Rationale: D its is.
Let price = X
(x-1.5)8 = 6X
X= 6
hence total money = 6*6 = 36
Answer: (D) $36
Question: 1 blue dice and 2 red Dice are thrown....what is the probability that the no. appeared on blue dice is always greater than both the no. appeared on the 2 red dice..
(A) 55/216
(B) 56/216
(C) 57/216
(D) 54/216
(E) 53/216
Rationale: if the no. on blue dice is 2 then 2 red dice can take (1,1) value = 1^2
if the no. on blue dice is 3 then 2 red dice can take (1,1)(1,2)(2,1)(2,2) = 2^2
if the no. on blue dice is 4 then 2 red dice can take (1,1)(1,2)(2,1)(2,2)(1,3)(2,3)(3,2)(3,1)(3,3) = 3^2
Similarly, if the no. on blue dice is 5 then 2 red dice can take = 4^2
and if the no. on blue dice is 6 then 2 red dice can take = 5^2
therefor, probability = total no. of favourable outcome/total no. of possible outcome
= 1^2+2^2+3^2+4^2+5^2 / 216
= 55/216
ANSWER:A
Answer: (A) 55/216
Question: The sum of the ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
(A) 4 years
(B) 8 years
(C) 10 years
(D) 15 years
(E) None of these
Rationale: Explanation:
Let the ages of the children be x, (x + 3), (x + 6), (x + 9) and (x +12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
5x = 20 => x = 4.
Age of youngest child = x = 4 years.
ANSWER IS A
Answer: (A) 4 years
Question: A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
(A) 80
(B) 85
(C) 90
(D) 92
(E) 98
Rationale: Let number of notes of each denomination be x.
Then x + 5x + 10x = 480
16x = 480
x = 30.
Hence, total number of notes = 3x = 90.
C
Answer: (C) 90
Question: It is required to fix a pipe such that water flowing through it at a speed of 7 metres per minute fills a tank of capacity 440 cubic metres in 5 minutes. The inner radius of the pipe should be :
(A) √2 m
(B) 2 m
(C) 1⁄2 m
(D) 1⁄√2 m
(E) None of these
Rationale: Let inner radius of the pipe be r.
Then, 440 = 22⁄7 × r2 × 7 × 5
or r2=440/22×5=4
or r = √4 = 2 m
Answer B
Answer: (B) 2 m
Question: A and B together can complete a piece of work in 4 days.If A alone can complete the same work in 12 days, in how many days can B alone complete that work?
(A) 5
(B) 7
(C) 6
(D) 4
(E) 8
Rationale: A and B's one day work=1/4
A's 1 day's work=1/12
B's 1 day's work=(1/4)-(1/12)=2/12=1/6
Hence B alone can complete the work in 6 days.
Answer is option C
Answer: (C) 6
Question: A metallic sheet is of rectangular shape with dimensions 48 m x 38 m. From each of its corners, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m3) is:
(A) 4830
(B) 5632
(C) 6420
(D) 7500
(E) 8960
Rationale: l = (48 - 16)m = 32 m,
b = (38 -16)m = 22 m,
h = 8 m.
Volume of the box = (32 x 22 x 8) m3 = 5632 m3.
Answer: Option B
Answer: (B) 5632
Question: If 5^29 * 4^15 = 2 * 10^n, what is the value of n?
(A) 11
(B) 21
(C) 22
(D) 29
(E) 32
Rationale: My attempt:
5^(29) * 4^(15) = 2*10^p
5^(29) * 2 ^ (30) = 2 * (2*5)^p
5^(29) * 2 ^ (30) = 2 * 2^(p) * 5^(p)
5^(29) * 2 ^ (30) = 2^(p+1) * 5^(p)
=> (p+1) = 30
=> p = 29
Answer: 29 (D)
Answer: (D) 29
Question: In a race of length X metres, Johnson beats Lewis by M metres and Greene by N metres, By how many metres does Lewis beat Greene in the same race ? (M<N)
(A) X(X-N) / X-M
(B) X(N-M) / X-M
(C) X-N
(D) M-N
(E) M+N
Rationale: As the time is constant, the ratio of distances will be the same as the ratio of speeds.
If Sj,Sl,Sg are the speeds of Johnson, Lewis, and Greene respectively, then
Sj/Sl = X/(X-M)
and Sj/Sg = X/(X-N)
=> Sg/Sl = (X-N)/(X-M)
Therefore the speeds of Lewis and Greene are in the ratio (X-M)/(X-N)
When Lewis finishes the race, the time run by him and Greene are same
=> The ratio of the speeds of Lewis and Greene will be the same as the ratio of distances run by them.
=> Distance run by Greene when Lewis finishes the race = (X-N)/(X-M) * X
=> Lewis beats Greene by X- X*(X-N)/(X-M) = X [ 1 - (X-N)/(X-M)] = X (N-M) / (X-M)
Option (B) is therefore correct.
Answer: (B) X(N-M) / X-M