mathewhe/aqua_rat huggingface dataset:
Question: Four dice are thrown simultaneously. Find the probability that all of them show the same face.
(A) 1/216
(B) 1/212
(C) 1/272
(D) 1/220
(E) 8/216
Rationale: Explanation:
The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:
=
n(S) =
Let X be the event that all dice show the same face.
X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)}
n(X) = 6
Hence required probability
Answer: A) 1/216
Answer: (A) 1/216
Question: How many liters of 20% alcohol solution should be added to 40 liters of a 50% alcohol solution to make a 30% solution?
(A) 60 liters
(B) 70 liters
(C) 80 liters
(D) 90 liters
(E) 50 liters
Rationale: Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hence
x + 40 = y
We shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage term.
20% x + 50% * 40 = 30% y
Substitute y by x + 40 in the last equation to obtain.
20% x + 50% * 40 = 30% (x + 40)
Change percentages into fractions.
20 x / 100 + 50 * 40 / 100= 30 x / 100 + 30 * 40 / 100
Multiply all terms by 100 to simplify.
20 x + 50 * 40 = 30 x + 30 * 40
Solve for x.
x = 80 liters
80 liters of 20% alcohol is be added to 40 liters of a 50% alcohol solution to make a 30% solution
Answer C
Answer: (C) 80 liters
Question: Peter walks along Sunrise Boulevard daily. He starts walking at 07:00 from block 10 and walks to block 90 where he turns around and walks back to block 70, where he stops at 07:30. The blocks along the boulevard are numbered sequentially (1,2,3), and each block measures 30 meters. What is Peter's speed in meters per minute?
(A) 100
(B) 180
(C) 198
(D) 216
(E) 252
Rationale: Total distance from 10 to 90 =80 + from 90 to 70 =20
So the dist is 100×30(per block dist)
Speed = 3000mts/30min =100m/min
A is the answer
Answer: (A) 100
Question: A clock gains 15 minutes per day. It is set right at 12 noon. What time will it show at 4.00 am, the next day?
(A) 4 : 10 am
(B) 4 : 45 am
(C) 4 : 20 am
(D) 5 : 00 am
(E) None of these
Rationale: The clock gains 15 min in 24 hours.
Therefore, in 16 hours, it will gain 10 minutes.
Hence, the time shown by the clock will be 4.10 am.
Answer A
Answer: (A) 4 : 10 am
Question: How much more would Rs.40000 fetch, after two years, if it is put at 20% p.a. compound interest payable half yearly than if is put at 20% p.a. compound interest payable yearly?
(A) 482
(B) 725
(C) 964
(D) 837
(E) 923
Rationale: 40000(11/10)4 - 40000(6/5)2 = 964
Answer:C
Answer: (C) 964
Question: A boat running up stram takes 6 hours to cover a certain distance, while it takes 8 hours to cover the same distance running down stream. what is the ratio between the speed of the boat and the speed of water current respectively?
(A) 2 : 3
(B) 5 : 6
(C) 4 : 5
(D) 7 : 1
(E) 8 : 1
Rationale: Explanation:
Let speed of boat is x km/h and speed stream is y km/hr
6(x+y) = 8(x-y)
6x+6y = 8x-8y
14y = 2x
7y = x
x/y = 7/1
7 : 1
Answer: Option D
Answer: (D) 7 : 1
Question: 50% of the apartments in a certain building have windows and hardwood floors. 25% of the apartments without windows have hardwood floors. If 40% of the apartments do not have hardwood floors, what M percent of the apartments with windows have hardwood floors?
(A) 10
(B) 16 2/3
(C) 40
(D) 50
(E) 83 1/3
Rationale: yup you need to construct a matrix
X********| Windows'|No Windows'''| total
hardFl****'|50*****''|10*********'|60
Not hardFl*|10******|30*********'|40
Total*****|60******'|40*********'|100
so M percent of the apartments with windows have hardwood floors = 50/60 *100
= 83 1/3
so E is the answer.[/u]
Answer: (E) 83 1/3
Question: A coin is tossed 4 times. What is the probability a that the number of Heads is equal to the number of Tails?
(A) 1/8
(B) 1/4
(C) 3/8
(D) 1/2
(E) 9/16
Rationale: We have to find the probability that we get 2 heads and 2 tails in 4 tosses of the coin.
Using the binary formula and defining a head as a success,
P(2 heads in 4 tosses) = 4C2* (1/2)^2 * (1/2)^2
= 6/16
a= 3/8
The answer is therefore (C)
Answer: (C) 3/8
Question: If x + (1/x) = 5, what is the value of W= x^2 + (1/x)^2?
(A) W=21
(B) W=22
(C) W=23
(D) W=24
(E) 27
Rationale: Squaring on both sides,
x^2 + (1/x)^2 + 2(x)(1/x) = 5^2
x^2 + (1 /x)^2 = 23
Answer: C
Answer: (C) W=23
Question: In Goshawk-Eurasian nature reserve 30 percent of the birds are Hawks, and 40 percent of the non-Hawks are Paddyfield-Warblers. If there are 25 percent as many Kingfishers as Paddyfield-Warblers in the reserve, then what percent of the birds X in the nature reserve are not Hawks, Paddyfield-Warblers, or Kingfishers?
(A) 25%
(B) 35%
(C) 45%
(D) 70%
(E) 80%
Rationale: 1. We are given the following percentages: 30 (70), 40 (60), 25 (75). There are two threads from here. First starts at 30% and finishes there. Second one starts at 70, then 40, and then 25. We need a value that is divisible by 7, 2, and 5 at least once. lets pick a number now, say 700. So say if Non Hawks are 700 (this is 70% of the total, so total = 1000), then Paddy Warbs are 2/5 x 700 = 1400/5 = 280. Kingfishers, therefore, are 280/4 = 70. Lets add them up. 300 hawks + 280 Peddy Warbs + 70 kingsifhers = 650. So all others are 1000 - 650= 350 or 35% of total birds.
The main job here to to identify the smart number to start the question with. This can be time consuming, but once identified, this question can be solved fairly quickly.
2. Another method: If X is total --> Non Hawks = 0.7X --> Warbs = 0.4(0.7X) --> KFs = 0.25(0.4(0.7X)).
Our job is to find out X: (0.3X + 0.28X + 0.07X)/ X. OR 0.65X/X = 0.65. We need to find 1-0.65 = 0.35 or 35%.B
Answer: (B) 35%
Question: How many seconds will a train 60 meters long take to cross a bridge 120 meters long if the speed of the train is 30 m/sec?
(A) 6 sec
(B) 8 sec
(C) 15 sec
(D) 3 sec
(E) 5 sec
Rationale: D = 60+120 = 180
S = 30 mps
T = 180/30 = 6 sec
A) 6 sec
Answer: (A) 6 sec
Question: Two trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 4 hours and 10 hours respectively. The ratio of their speeds is
(A) 4 : 9
(B) 4 : 3
(C) 4 : 12
(D) 4 : 8
(E) 5 : 2
Rationale: Let us name the trains as A and B. Then, (A's speed) : (B's speed) = b : a = 10 : 4 = 5 : 2.Answer: E
Answer: (E) 5 : 2
Question: The present worth of a sum due sometime hence is Rs. 576 and the banker’s gain is Rs. 16. The true discount is :
(A) 36
(B) 52
(C) 66
(D) 96
(E) None
Rationale: Sol.
T.D. = √P.W. * B.G. = √576 * 16 = 96.
Answer D
Answer: (D) 96
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) 400
(B) 500
(C) 550
(D) 600
(E) 650
Rationale: C's 1 day's work =	1	- 1	+	1 =	1	-	7	=	1	.
3	6	8	3	24	24
A's wages : B's wages : C's wages =	1	:	1	:	1	= 4 : 3 : 1.
6	8	24
C's share (for 3 days) = Rs. 3 x	1	x 3200 = Rs. 400.
Option A
Answer: (A) 400
Question: What is 15% of 2/3 of 0.7?
(A) 0.07
(B) 0.9
(C) 9
(D) 90
(E) None of the above
Rationale: The best way to solve these questions is to convert every term into fraction
(15/100)*(2/3)*(7/10) = 210/3000 = 0.07
Option A
Answer: (A) 0.07
Question: In Arun's opinion, his weight is greater than 62 kg but leas than 72 kg. His brother does not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 65 kg. If all of them are correct in their estimation, what is the average of diferent probable weights of Arun ?
(A) 86.5 kg
(B) 63.5 kg
(C) 46.5 kg
(D) 26.5 kg
(E) 16.5 kg
Rationale: Let Arun's weight be X kg.
According to Arun, 62 < X < 72.
According to Arun's brother, 60 < X < 70.
According to Arun's mother, X < 65.
The values satisfying all the above conditions are 63 and 64.
Required average = (63 + 64) / 2 = 63.5 kg
Answer: B
Answer: (B) 63.5 kg
Question: What is the minimum value of |x-4| + |x+6| + |x-5| ?
(A) -3
(B) 3
(C) 5
(D) 7
(E) 11
Rationale: A cannot be the answer as all the three terms are in modulus and hence the answer will be non negative.
|x-4| >= 0 --> Minimum occurs at x = 4
|x+6| >= 0 --> Minimum occurs at x = -6
|x-5| >= 0 --> Minimum occurs at x = 5
x = -6 --> Result = 10 + 0 + 11 =21. Also any negative value will push the combined value of |x-4| + |x-5| to a value > 9.
x = 4 --> Result = 0 + 10 + 1 = 11
x = 5 --> Result = 1 + 11+ 0 = 12
x = 6 --> Result = 2 + 12 +1 = 15
So minimum value of the expression occurs at x = 4 and the resultant value = 11
Answer: E
Answer: (E) 11
Question: A mixture of sand and cement contains, 3 parts of sand and 5 parts of cement. How much of the mixture must be substituted with sand to make the mixture half sand and half cement?
(A) 1/3
(B) 1/4
(C) 1/5
(D) 1/7
(E) 1/8
Rationale: We have total of 8 parts: 3 parts of sand and 5 parts of cement.
In order there to be half sand and half cement (4 parts of sand and 4 parts of cement), we should remove 1 part of cement. With 1 part of cement comes 3/5 parts of sand, so we should remove 1 + 3/5 = 8/5 part of the mixture, which is (8/5)/8 = 1/5 of the mixture.
Answer: C.
Answer: (C) 1/5
Question: Daniel went to a shop and bought things worth Rs. 40, out of which 30 Paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
(A) A)34.7
(B) B)20
(C) C)21.3
(D) D)21.5
(E) E)22
Rationale: Total cost of the items he purchased = Rs.40
Given that out of this Rs.40, 30 Paise is given as tax
=> Total tax incurred = 30 Paise = Rs.30/100
Let the cost of the tax free items = x
Given that tax rate = 6%
∴ (40−30/100−x)6/100 = 30/100
⇒ 6(40 −0.3 −x) = 30
⇒ (40 − 0.3 − x) = 5
⇒ x = 40 − 0.3 − 5
= 34.7
Ans - A
Answer: (A) A)34.7
Question: What is the smallest positive integer x such that 520-x is the cube of a positive integer
(A) 5
(B) 7
(C) 8
(D) 9
(E) 12
Rationale: Given 520-x is a perfect cube
so we will take 512= 8*8*8
520-x = 512
x =520-512 = 8
correct option is C
Answer: (C) 8
Question: If k, (k+200), (k+350), and 15*K are all multiples of D, then D could equal which of the following ?
(A) 20
(B) D=25
(C) D=75
(D) 100
(E) 150
Rationale: {Multiple of D} - {Multiple of D} = {Multiple of D}.
(K+200) - k = 200 = {Multiple of D}.
(K+350) - k = 350 = {Multiple of D}.
Both 200 and 350 are multiples of only option B.
Answer: B.
Answer: (B) D=25
Question: A question paper consists of five problems, each problem having three internal choices. In how many ways can a candidate attempt one or more problems?
(A) 1278
(B) 2799
(C) 1023
(D) 1782
(E) 2781
Rationale: Given that, the question paper consists of five problems. For each problem, one or two or three or none of the choices can be attempted.
Hence, the required number of ways = 45 - 1.
= 210 - 1 = 1024 - 1 = 1023
Answer: C
Answer: (C) 1023
Question: A plane was originally flying at an altitude of x feet when it ascended 2,000 feet and then descended 5,000 feet. If the plane's altitude after these two changes was 1/7 its original altitude, then the solution of which of the following equations gives the plane's original altitude, in feet ?
(A) x + 2000 = 1/3*(x - 3000)
(B) 1/3*(x - 3000) = x
(C) x + 3000 = 1/3*x
(D) x - 7000 = 1/3*x
(E) x - 3000 = 1/7*x
Rationale: Plane's original altitude = x
Plane's new altitude after ascending 2000 ft = x+2000
Plane's new altitude after descending 5000 ft from previous altitude = x+2000-5000 = x-3000
So, after two changes plane is at 1/7 its original altitude => x-3000 = x/7
Answer (E)
Answer: (E) x - 3000 = 1/7*x
Question: Two employees A and B are paid a total of Rs. 560 per week by their employer. If A is paid 150 percent of the sum paid to B, how much is B paid per week?
(A) s.130
(B) s.140
(C) s.150
(D) s.224
(E) s.280
Rationale: Let the amount paid to A per week = x
and the amount paid to B per week = y
Then x + y = 560
But x = 150% of y = 150y/100 = 15y/10
∴15y/10 + y = 560
⇒ y[15/10 + 1] = 560
⇒ 25y/10 = 560
⇒ 25y = 5600
⇒ y = 5600/25 =Rs.224
D)
Answer: (D) s.224
Question: A bag contains 20 red and 30 white balls. One ball is drawn at random. What is the probability that the ball drawn is red?
(A) 11/5
(B) 2/5
(C) 3/5
(D) 4/5
(E) 1/2
Rationale: Total number of balls = 20+30 = 50
number of red balls = 20
Probability = 20/50 = 2/5
Answer is B
Answer: (B) 2/5
Question: A shopkeeper loses 15%,if an article is sold for Rs. 102. What should be the selling price of the article to gain 20%?
(A) 128
(B) 144
(C) 176
(D) 186
(E) 111
Rationale: Explanation:
Given that SP = Rs. 102 and loss = 15%
CP = [100(SP)]/(100 - l%) = (100 * 102)/85 = 20 * 6 = Rs. 120.
To get 20% profit, New SP = [(100 + p%)CP]/100 = (120 * 120)/100 = Rs. 144
Answer:B
Answer: (B) 144
Question: Total 60 cows 20 cow gives each 2 liter milk 20 cow gives each 3/4 liter milk 20 cow gives each 1/4 liter milk this is split into 3 son per each 20 cows & 20 liter milk how?
(A) 20
(B) 10
(C) 15
(D) 16
(E) 18
Rationale: 20 cow 2 liter each =40 liter
20cow 3/4 liter each= 3/4=0.75*20=15
20 cow 1/4 liter each = 1/4=0.25*20=5
add 40+15+5=60
milk split into 3 son each 20 liter then 60/3=20
ANSWER:A
Answer: (A) 20
Question: Two trains of length 120 m and 210 m are 80 m apart. They start moving towards each other on parallel tracks, at speeds 69 kmph and 82 kmph. After how much time will the trains meet?
(A) 15/10
(B) 19/10
(C) 20/19
(D) 16/10
(E) 27/10
Rationale: They are moving in opposite directions, relative speed is equal to the sum of their speeds.
Relative speed = (69 + 82)*5/18 =41.9 mps.
The time required = d/s = 80/41.9 = 19/10 sec.
Answer:B
Answer: (B) 19/10
Question: If the price of an article went up by 30%, then by what percent should it be brought down to bring it back to its original price?
(A) 2/3%
(B) 2/8%
(C) 2/1%
(D) 1/3%
(E) 23.07%
Rationale: Let the price of the article be Rs. 100.
30% of 100 = 30.
New price = 100 + 30 = Rs. 130
Required percentage = (130 - 100)/130 * 100
= 30/130 * 100 = 23.07%.
Answer:E
Answer: (E) 23.07%
Question: There are 57 people in a certain community who belong to the local golf club, tennis club or both clubs. If twice as many people belong to the tennis club as belong to the golf club, what could be the number of people who belong to both clubs?
(A) 1) 4
(B) 2) 7
(C) 3) 17
(D) 4) 21
(E) 5) 27
Rationale: x -> Golf Club
2x -> Tennis Club
y -> both
x+2x-y = 51
3x - y =51
multiples of 3 greater than 51
60
63
66
69
72 - 21 = 51
IMO (D)
Answer: (D) 4) 21
Question: The size of the wooden block is 5 x 10 x 20 cm .How many such blocks will be required to construct a solid wooden cube of minimum size?
(A) 6
(B) 8
(C) 12
(D) 16
(E) 20
Rationale: Side of smallest cube = L.C.M of 5,10,20 = 20cm
Volume of the cube = (20x20x20)= 8000 cm3
Volume of the block= (5x10x20cm3 = 1000cm3
Number of blocks = (8000/1000) = 8 Answer: B
Answer: (B) 8
Question: If the charge of staying in a student youth hostel $18.00/day for the first week, and $13.00/day for each additional week, How much does it cost to stay for 23 days?
(A) $160
(B) $289
(C) $334
(D) $274
(E) $286
Rationale: Total number of days of stay = 23
Charge of staying in first week = 18*7 = 126 $
Charge of staying for additional days =(23-7)*13 = 16*13 = 208 $
Total charge =126 +208 = 334$
Answer C
Answer: (C) $334
Question: If 'Pink' is called 'Orange', 'Orange' is called 'Blue', 'Blue' is called 'Red', 'Red' is called 'Green', 'Green' is called 'Black' and 'Black' is called 'white', what would be the color of human blood ?
(A) Green
(B) BLUE
(C) PINK
(D) ORANCE
(E) BLACK
Rationale: Explanation:
Pink → Orange
Orange → Blue
Blue → Red
Red → Green
Green → Black
Black → White
The color of human blood is 'Red' but in coding language, 'Red' is called 'Green'. So, the color of human blood is 'Green'.
ANSWER: A
Answer: (A) Green
Question: A group of people participate in some curriculum, 35 of them practice Yoga, 20 study cooking, 15 study weaving, 7 of them study cooking only, 5 of them study both the cooking and yoga, 3 of them participate all curriculums. How many people study both cooking and weaving?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Rationale: Both cooking and weaving = 20 - (7+5+3) = 5
So, the correct answer is E.
Answer: (E) 5
Question: The markup on a necklace is 80 percent of the cost. The markup is what percent of the selling price?
(markup = selling price - cost)
(A)  8%
(B)  10%
(C)  44.44%
(D)  55%
(E)  66 2/3%
Rationale: MP = 0.8 CP
SP = CP + 0.8 CP = 1.8 CP
hence MP = 0.8/1.8 SP = 4/9 SP. Hence MP is 44.44% of SP
Answer C
Answer: (C)  44.44%
Question: Shawn has $150 to buy a new guitar. he found a guitar strap for $20, a guitar amplifier
for $80, and some guitar picks for $8. How much does he have left to buy the guitar?
(A) $52
(B) $42
(C) $32
(D) $22
(E) $12
Rationale: we know that shawn had 150 dollars
then we have
$150all-$20strap-$80 amplifier-$8picks
we take
$52 dollars
correct answer A
Answer: (A) $52
Question: How many even number in the range between 10 to 120 inclusive are not divisible by 3
(A) 15
(B) 30
(C) 31
(D) 37
(E) 46
Rationale: We have to find the number of terms that are divisible by 2 but not by 6( as the question asks for the even numbers only which are not divisible by 3)
For 2,
10,12,14...120
using AP formula, we can say 120 = 10 + (n-1) *2
or n=56.
For 6,
12,18,...120
using AP formula, we can say 120 = 12 + (n-1) *6
or n=19.
Hence, only divisible by 2 but not 3 =56-19 = 37. hence, Answer D
Answer: (D) 37
Question: In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1:2:3. If there are Rs. 30 in all, how many 5 p coins are there?
(A) 165
(B) 716
(C) 150
(D) 143
(E) 145
Rationale: Let the number of 25 p, 10 p and 5 p coins be x, 2x and 3x respectively.
Then, sum of their values = [25x/100 + (10 * 2x)/100 + (5 * 3x)/100] = Rs. 60x/100
60x/100 = 30 => x = 50.
Hence, the number of 5 p coins = 3 * 50 = 150.
Answer:C
Answer: (C) 150
Question: A person bought an article and sold it at a loss of 10%. If he had bought it for 20% less and sold it for Rs.75 less, he could have gained 25%. What is the cost price?
(A) 250
(B) 256
(C) 259
(D) 255
(E) 251
Rationale: CP1 = 100 SP1 = 90
CP2 = 80 SP2 = 80 * (140/100) = 112
22 ----- 100
55 ----- ? => Rs.250
Answer: A
Answer: (A) 250
Question: How many positive number less than 1000 have sum of there digit 17?
(A) 43
(B) 53
(C) 63
(D) 75
(E) 76
Rationale: digits between 1-100 sum 17 =only 2 (89,98)
digits between 101-200 sum 17 =3 (188,179,197)
digits between 201-300 sum 17 = 4 (296,287,278,269)
and so on.....
digits between 301-400 sum 17 =5
digits between 401-500 sum 17 =6
digits between 501-600 sum 17 =7
digits between 601-700 sum 17 =8
digits between 701-800 sum 17 =9
digits between 801-900 sum 17 =10
but in
digits between 901-999 sum 17 =9 (980,971,962,953,944,935,926,917,908)
so total number is
2+3+4+5+6+7+8+9+10+9=63
ANSWER:C
Answer: (C) 63
Question: Solve 7x – 2(x+5) = 4x + 11
(A) 16
(B) 25
(C) 21
(D) 15
(E) 12
Rationale: Sol.
7x – 2(x+5) = 4x + 11
7x - 2x -10 = 4x + 11
=> 7x – 2x - 4x = 11 + 10
=> x = 21
Answer C
Answer: (C) 21
Question: A dishonest dealer professes to sell goods at the cost price but uses a false weight and gains 25%. Find his false weight age?
(A) 298
(B) 268
(C) 800
(D) 266
(E) 226
Rationale: 25 = E/(1000 - E) * 100
1000 - E = 4E
1000 = 5E => E = 200
1000 - 200 = 800
Answer:C
Answer: (C) 800
Question: If |5x-5| = 100, then find the sum of the values of x?
(A) 1
(B) -2
(C) 2
(D) -3
(E) 4
Rationale: |5x-5| = 100
5x-5 = 100 or 5x-5 = -100
5x = 105 or 5x = -95
x = 21 or x = -19
sum = 21-19 = 2
Answer is C
Answer: (C) 2
Question: Bob finishes the first half of an exam in four sixth the time it takes him to finish the second half. If the whole exam takes him an hour, how many minutes does he spend on the first half of the exam?
(A) 12
(B) 24
(C) 27
(D) 36
(E) 40
Rationale: Many times, it is easiest to think of problems like this conceptually (as opposed to formulaically). Conceptually, if the first half of the exam takes 4/6's the time of the second half (6/6's), we can see that the hour the entire exam took to finish can be broken down into tenth's. (Another way to look at this problem is via ratio's - first half is 4:10 and the second half is 6:10). With each tenth of an hour being 6 minutes (60/10), the first half of the exam would have taken 12 minutes. Correct answer is A.
Answer: (A) 12
Question: Excluding stoppages, the speed of a bus is 54 km/hr and including stoppages, it is 45 km/hr. For how many minutes does the bus stop per hour?
(A) 11 min
(B) 10 min
(C) 37 min
(D) 17 min
(E) 16 min
Rationale: Due to stoppages, it covers 9 km less.
Time taken to cover 9 km
= 9/54 * 60 = 10 min.
Answer: B
Answer: (B) 10 min
Question: The sum of 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
(B) 5
(C) 6
(D) 7
(E) 8
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) = 50
5x = 20
x = 4.
Age of the youngest child = x = 4 years.
Answer:A
Answer: (A) 4
Question: A is thrice as efficient as B and is, therefore, able to finish a piece of work 10 days earlier than B. In how many days A and B will finish it together?
(A) 6
(B) 7
(C) 87
(D) 5
(E) 34
Rationale: WC = 3:1
WT = 1:3
x 3x
1/x – 1/3x = 1/10
x = 20/3
3/20 + 1/20 = 1/5 => 5 days
Answer:D
Answer: (D) 5
Question: How long does a train 110 m long running at the speed of 90 km/hr takes to cross a bridge 132 m length?
(A) 82.1 sec
(B) 9.68 sec
(C) 16.1 sec
(D) 13.1 sec
(E) 12.15sec
Rationale: Speed = 90 * 5/18 = 25 m/sec
Total distance covered = 110 + 132 = 242 m.
Required time = 242/25 = 9.68 sec.
Answer:B
Answer: (B) 9.68 sec
Question: A person buys an article at Rs.500. At what price should he sell the article so as to make a profit of 20%?
(A) 600
(B) 882
(C) 772
(D) 652
(E) 271
Rationale: Cost price = Rs.500
profit = 20% of 500 = Rs.100
Selling price = Cost price + Profit
= 500 + 100 = 600
Answer:A
Answer: (A) 600
Question: A computer simulation of a 12 hour analog clock keeps perfect time while it is running, and
has two hands—an hour hand and a minute hand—both of which move continuously around
the 12 hour dial. (For example, at 2:30, the hour hand will be exactly halfway between 2 and
3.) Because of careless programming, the minute hand looks exactly like the hour hand, so
that the two are indistinguishable.
On one day, the clock stops at some time after 12am and before 12pm. How many times could
it have stopped without it being possible after 12pm to tell what time it stopped?
(A) 11
(B) 12
(C) 66
(D) 132
(E) 708
Rationale: A configuration of the hands is called “ambiguous” if you cannot use this broken clock to
distinguish the minute hand from the hour hand.
In the above diagram it must be the minute hand that is pointing to 6 and the hour hand that
is pointing midway between 2 and 3. (If the hour hand were to point to 6, the minute hand
would have to point to 12.) 2:30 is not an ambiguous configuration.
Let m 2 (0; 1) represent the fraction of the way around the clock face for the minute hand and
h 2 (0; 1) the corresponding fraction for the hour hand. For example, 2:30 hasm = 1=2 = 0:5
and h = 2:5=12 = 0:2083.
As the minute hand moves 12 times as fast as the hour hand, a configuration is ambiguous if
m and 12h have the same fractional part. Because the hands are indistinguishable, a time is
also ambiguous when h and 12m have the same fractional part. We express these conditions
as: 12h m( mod 1) and 12m h( mod 1). Thus 144h h( mod 1), and so 143h 0(
mod 1). There are 142 solutions to this equation in the open interval (0; 1): h 2 H1 := f k
143 :
k = 1; 2; : : : ; 142g.
Times when the hour and minute hand perfectly overlap (h = m) are in H1, but these are not
ambiguous times. These configurations occur when 12h h( mod 1). That is, 11h 0(
mod 1). There are 10 such times: h 2 H2 :=
k
11 : k = 1; 2; : : : ; 10
The number of ambiguous times is the cardinality of H1nH2. As H2 H1 the cardinality of
H1nH2 is 142 􀀀 10 = 132, which is answer (d).
correct answer D
Answer: (D) 132