RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: A man purchased a grinder for Rs. 120 and sold it at a profit of 10%. The selling price is:
A.Rs. 736
B.Rs. 812
C.Rs. 132
D.None of these
Answer: C.Rs. 132
Explanation: s.p.=110/100 × 120 = 132
Answer: C
Question: What percent of 5 letter combinations that can be made from the letters of the word VERMONT and in which each letter can be used only once are the combinations in which the first letter is a vowel and the last letter is a consonant?
A.
B.
C.
D.
E.
A.20%
B.23.8%
C.25%
D.30.2%
Answer: B.23.8%
Explanation: TOTAL number of 5 letter combinations = 7P5 = 7!/2! = 3*4*5*6*7
There are 2 vowels: e,o and 5 consonants: v,r,m,n,t
If ABCDE is the required combination, A has 2 options, E has 5 options, B has 5, C has 4, and D has 3
Number of required combinations = 2*5*5*4*3
Answer = 2*5*5*4*3 / 3*4*5*6*7 = 5/21 = 23.8%
ANSWER: B
Question: If each data point in a certain set Q is increased by a factor of 6, what happens with the set's standard deviation?
A.Doesn't Change
B.Decreases by Factor 6
C.Increases by 6
D.Increases by Factor 6
Answer: D.Increases by Factor 6
Explanation: Standard deviation questions will always be conceptual in nature, because the GMAT does not require that you be able to calculate standard deviation.Veritasteaches a concept known as themapping strategyfor data sufficiency, which helps make a problem like this easily understandable. The mapping strategy is just looking at data sufficiency with sets of numbers like you would look at a map. Moving left and right on the map does not change the map's scale (which represents standard deviation in this analogy), just as adding or subtracting the same value in all numbers of a set does not change that set's standard deviation. For example, the set: 2, 5, 8, 11 has the same standard deviation of that same set after adding 5 to each number: 7, 10, 13, 16. Multiplying and dividing the entire set by a number other than 1, however, does affect standard deviation, and it does so proportionately. Multiplying by a number with an absolute value greater than 1 is like zooming in on a map...the standard deviation increases. Consider that same set multiplied by 2: 4, 10, 16, 22....this has a larger standard deviation than 2, 5, 8, 11. The opposite goes for division - dividing each number in a set of numbers by a number with an absolute value greater than 1 will reduce standard deviation, like zooming out on a map reduces the map's scale. Here because each number was multiplied by a factor of 6, the standard deviation will increase proportionately, giving us answer choice D.
Question: The length of a rectangular plot is thrice its breadth. If the area of the rectangular plot is 867 sq m, then what is the breadth of the rectangular plot?
A.1
B.17
C.88
D.12
Answer: B.17
Explanation: Let the breadth of the plot be b m.
Length of the plot = 3 b m
(3b)(b) = 867
3b2 = 867
b2 = 289 = 172 (b > 0)
b = 17 m.
Answer: B
Question: On flipping 3 coins together what is the outcomes for at least one head?
A.{HHH, HHT, HTH, THH, TTH, THT, TTT}
B.{HHH, HHT, HTH, THH, TTH, THT, HTT}
C.{HHH, HHT, HTH, TTT, TTH, THT, HTT}
D.{TTT, HHT, HTH, THH, TTH, THT, HTT}
Answer: B.{HHH, HHT, HTH, THH, TTH, THT, HTT}
Explanation: Sample space, S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Favorable outcomes for at least one heads
= favorable outcomes for exactly one heads and favorable outcomes for more than one heads
= {HHH, HHT, HTH, THH, TTH, THT, HTT}
B
Question: One pipe can fill a tank four times as fast as another pipe. If together the two pipes can fill the tank in 40 minutes, then the slower pipe alone will be able to fill the tank in?
A.100min
B.150min
C.200min
D.250min
Answer: C.200min
Explanation: Let the slower pipe alone fill the tank in x minutes
Then, faster pipe will fill it in x/4 minutes
1/x + 4/x = 1/40
5/x = 1/40
x = 200 min
Answer is C
Question: The average age of 36 students in a group is 14 years. when teacher's age is included to it, the average increases by one.What is the teacher's age in years?
A.31
B.36
C.51
D.53
Answer: C.51
Explanation: age of the teacher=(37*15-36*14)
=51 years.
ANSWER:C
Question: A car traveling at a certain constant speed takes 5 seconds longer to travel 1 km than it would take to travel 1 km at 90 km/hour. At what speed, in km/hr, is the car traveling?
A.70
B.80
C.74
D.75
Answer: B.80
Explanation: Time to cover 1 kilometer at 80 kilometers per hour is 1/90 hours = 3,600/90 seconds = 40 seconds;
Time to cover 1 kilometer at regular speed is 40 + 5 = 45 seconds = 45/3,600 hours = 1/80 hours;
So, we get that to cover 1 kilometer 1/80 hours is needed --> regular speed 80 kilometers per hour (rate is a reciprocal of time or rate=distance/time).
Answer: B.
Question: John weighs twice as much as Susan. Susan's weight is 60% of Sam's weight. Mark weighs 50% of Lynn's weight. Lynn weighs 140% of John's weight. Which of these 5 people weighs the least?
A.John
B.Susan
C.Sam
D.Mark
Answer: B.Susan
Explanation: John weighs twice as much as Susan --> J=2S --> John cannot weigh the least;
Susan's weight is 60% of Sam's weight --> S=0.6*Sam --> Sam cannot weigh the least;
Mark weighs 50% of Lynn's weight --> M=0.5*L --> Lynn cannot weigh the least;
Lynn weighs 140% of John's weight --> L=1.9*J --> John cannot weigh the least (we already know that);
As you can see the answer is either Susan (S) or Mark (M). To compare their weights we should express their weights in regards of the same person: M=0.5*L=0.5*(1.4*J)=0.7*J and from J=2S, S=0.5*J --> Susan weights the least: Susan's weight is 50% of that of John and Mark's weight is more than 50%of that of John.
Answer: B.
Question: The sum of ages of 5 girls born at the intervals of 3 years each is 40 years. What is the age of the youngest girl
A.years
B.years
C.years
D.years
Answer: A.years
Explanation: 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) = 40
5x = 10
x = 2.
Age of the youngest child = x = 2 years.
A)
Question: If x^3 < y^2 > z, which of the statements could be true?
I. x < y < z
II. x < z < y
III. y < x < z
A.I only
B.III only
C.I and II only
D.II and III only
Answer: C.I and II only
Explanation: Given : x^3 < y^2 > z
I. x < y < z is true for x = 3, y = 6, z = 7
II. x < z < y is true for x = 3 , y = 5 , z = 4
III. y < x < z is not true
Answer: option C
Question: Wendy begins sanding a kitchen floor by herself and works for 6 hours. She is then joined by Bruce, and together the two of them finish sanding the floor in 2 hours. If Bruce can sand the floor by himself in 20 hours, how long would it take Wendy to sand the floor by herself?
A.9/80 hours
B.3/20 hours
C.20/3 hours
D.80/9 hours
Answer: D.80/9 hours
Explanation: Let Wendy finishes sanding the floor alone in W hours while B be the hours for Bruce.
Thus in 1 hour Wendy finishes 1/w of the work while Bruce finishes 1/B of the work.
If wendy works for 6 hours and is then joined by Bruce to finish the work in 2 more hours,
6/W + 2/W+2/B = 1 (1 denotes the total amount of work)
8/W + 2/B =1 and given B =20 hours.
Thus W = 80/9 hours , D is the correct answer.
Question: An amount of Rs. 100000 is invested in two types of shares. The first yields an interest of 9% p.a and the second, 11% p.a. If the total interest at the end of one year is 9 3/4 %, then the amount invested in each share was?
A.Rs. 52500; Rs. 47500
B.Rs. 62500; Rs. 37500
C.Rs. 72500; Rs. 27500
D.Rs. 82500; Rs. 17500
Answer: B.Rs. 62500; Rs. 37500
Explanation: Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 - x). Then,
(x * 9 * 1)/100 + [(100000 - x) * 11 * 1]/100 = (100000 * 39/4 * 1/100)
(9x + 1100000 - 11x)/100 = 39000/4 = 9750
x = 62500
Sum invested at 9% = Rs. 62500
Sum invested at 11% = Rs. (100000 - 62500) = Rs. 37500.
ANSWER:B
Question: what is summation of [3^k(28Ck)] (summation range 28 to k=0) where 28Ck is the number of ways of choosing k items from 28 items??
A.4^28
B.4^27
C.4^26
D.4^25
Answer: A.4^28
Explanation: The expansion of (1+x)^n=nC0 x^0+nC1 x^1+nC2 x^2+......
putting the value of x=3 and n=28 we get
(1+3)^28=28C0 3^0+28C1 3^1+28C2 3^2+28C3 3^3.......
which is the given summation we need to find hence,
ans= 4^28
ANSWER:A
Question: If an integer is randomly chosen from the numbers 250 to 449, inclusive, what is the probability that all the digits of the number are even?
A.3/5
B.17/50
C.1/8
D.51/199
Answer: C.1/8
Explanation: Another method is to enumerate the hundreds and tens digits which are acceptable:
26
28
40
42
44
In each case, any of the 5 even digits (0, 2, 4, 6, 8) can take the units place.
So acceptable cases = 5*5 = 25
Total number of integers = 200
Probability = 25/200 = 1/8
ANSWER:C
Question: Difference of two numbers is 1660. If 7.5% of the number is 12.5% of the other number , find the number ?
A.1470
B.4150
C.3012
D.4430
Answer: B.4150
Explanation: Let the numbers be x and y. Then , 7.5 % of x =12.5% of y
X = 125*y/75 = 5*y/3.
Now, x-y =1660
5*y/3 –y =1660
2*y/3= 1660
y =[ (1660*3)/2] =2490.
One number = 2490, Second number =5*y/3 =4150.
ANSWER B 4150
Question: A and B together have Sterling 1220. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount does B have?
A.Sterling 460
B.Sterling 488
C.Sterling 550
D.Sterling 664
Answer: B.Sterling 488
Explanation: Explanation: 4/15A = 2/5B
A = (2/5 x 15/4)B
A = 3/2 B
A/B = 3/2
A : B = 3 : 2.
B's share = Sterling (1220 x 2/5) = Sterling 488.
Answer: Option B
Question: A box contains nine slips that are each labeled with one number: 1, 2, 3, 5, 8, 13, 21 and 34 Two of the slips are drawn at random from the box without replacement. What is the probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box?
A.7/72
B.3/14
C.7/36
D.15/36
Answer: B.3/14
Explanation: Probability = No: of desired outcomes / Total No: of outcomes.
You are picking two slips out of 8 slips. So
Total No: of outcomes = 8C2 = 28
Desired outcome : sum of the numbers on the two slips is equal to one of the numbers left in the box. How many such outcomes are there?
If you look at the numbers closely, you will see that the following pair of numbers will give you the desired outcome.
(1,2) (2,3) (3,5) (5,8) (8,13) (13,21) . There are 6 such pairs. If the two numbers which I pick is from any of these 6 pairs, then I get my desired outcome.
So No: of desired outcomes = 6
Probability = 6/28 = 3/14
Answer :B
Question: The probability that a man will be alive for 10more yrs is 3/4 & the probability that his wife will alive for 10 more yrs is 1/5. The probability that none of them will be alive for 10 more yrs, is
A.1/2
B.1
C.2/3
D.1/5
Answer: D.1/5
Explanation: Sol. Required probability = Pg.) x P(B) = (1— D x(1 —I) =: x 1=1/5
Ans.(D)
Question: Grayson, a chemist, has 8 litres of a solution that contains 15% hydrochloric acid(HCI), plus unlimited 5% HCI solution. For a particular experiment, however, he needs to prepare some that contains 12% HCI. How much 5% HCI solution should be added to the 15% HCI solution to get the 12% HCI solution needed for the experiment?
A.7.6 litres
B.10 litres
C.12.8 litres
D.3.4 litres
Answer: D.3.4 litres
Explanation: 0.15(8)+0.05x = 0.12(8 + x)
Solve the equation.
0.15(8) + 0.05x = 0.12(8 + x)
1.2 + 0.05x = 0.96 + 0.12x
1.2 – 0.07x = 0.96
-0.07x = -0.24
x ≈ 3.4
Answer is D.
Question: 3 men or 6 women can do a piece of work in 20 days. In how many days will 12 men and 8 women do the same work?
A.15/9 days
B.15/4 days
C.18/4 days
D.11/4 days
Answer: B.15/4 days
Explanation: 3M = 6W ---- 20 days
12M + 8W -----?
24W + 8 W = 32W ---?
6W ---- 20 32 -----?
6 * 20 = 32 * x => x = 15/4 days
Answer:B
Question: Find the C.I. on Rs. 15,625 for 9 months at 16% per annum compounded quarterly?
A.Rs.1939
B.Rs.1912
C.Rs.1951
D.Rs.1908
Answer: C.Rs.1951
Explanation: P = Rs. 15625, n = 9 months = 3 quarters, R = 16% p.a. per quarter.
Amount = [15625 * (1 + 4/100)3]
= (15625 * 26/25 * 26/25 * 26/25)
= Rs. 17576 C.I.
= 17576 - 15625
= Rs.1951.
Answer:C
Question: A 3-digit number 4a3 is added to another 3-digit number 984 to give a 4-digit number 13b7, which is divisible by 11. Then, (a + b) = ?
A.10
B.11
C.12
D.15
Answer: A.10
Explanation: Explanation:
4 a 3 |
9 8 4 } ==> a + 8 = b ==> b - a = 8
13 b 7 |
Also, 13 b7 is divisible by 11 (7 + 3) - (b + 1) = (9 - b)
(9 - b) = 0
b = 9
(b = 9 and a = 1) (a + b) = 10.
Answer: Option A
Question: In the following series how many times the sum of two consecutive digits is even?
1 2 3 4 6 3 4 2 5 9 3 6 7 4 1 2 3 6 7 6 5 4 3
A.5
B.2
C.3
D.4
Answer: D.4
Explanation: 4+6
4+2
5+9
9+3
so 4 times
ANSWER:D
Question: (4x - 7)(4x + 7) + (2x - 9)(2x + 9) =
A.20x2 - 2x - 100
B.20x2 - 130
C.12x2 - 32
D.12x2 + 32
Answer: B.20x2 - 130
Explanation: Solution:
(4x - 7)(4x + 7) + (2x - 9)(2x + 9) =
[(4x)2 - 72] + [(2x)2 - 92] =
(16x2 - 49) + (4x2 - 81) =
16x2 - 49 + 4x2 - 81 =
20x2 - 130
Answer B
Question: Solve: 18800+470+20
A.2
B.5
C.4
D.0
Answer: A.2
Explanation: 18800+470+20=(18800/470)+20=40/20=2.
Answer is A.
Question: 45 men can complete a work in 18 days. Six days after they started working, 30 more men joined them. How many days will they now take to complete the remaining work?
A.6.6
B.10
C.15
D.7.2
Answer: D.7.2
Explanation: 45*18 men can complete the work in 1 day
1 man's 1 day work = 1/810
45 men 6 days work = 6*1/18 = 1/3
remaining work = 1-1/3 = 2/3
75 men's 1 day's work = 75/810 = 5/54
2/3 work is done by them in 54/5 * 2/3 = 7.2 days
Answer is D
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 16 hours and 81 hours respectively. The ratio of their speeds is?
A.4:9
B.9:4
C.4:5
D.4:1
Answer: B.9:4
Explanation: Let us name the trains A and B.
Then, (A's speed) : (B's speed)
= √b : √a = √81 : √16
= 9:4
Answer:B
Question: A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.
A.7/19
B.7/15
C.7/29
D.7/12
Answer: B.7/15
Explanation: Let S be the sample space
Then n(S) = no of ways of drawing 2 balls out of (6+4) = \inline {\color{Black}10C_{2}} = =45
Let E = event of getting both balls of same colour
Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)
=\inline {\color{Black}6C_{2}+4C_{2}} = = 15+6 = 21
Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15
Answer: B
Question: A train 100 m long crosses a platform 125 m long in 15 sec; find the speed of the train?
A.87 kmph
B.65 kmph
C.54 kmph
D.16 kmph
Answer: C.54 kmph
Explanation: D = 100 + 125 = 225
T = 15
S = 225/15 * 18/5 = 54 kmph
Answer:C
Question: Their are two boxes containing a total number of 60 balls, each box has an equal number of red and blue balls. What is the probability of getting a red ball from one box and a blue ball from the other
A.1/2
B.1/4
C.1/16
D.1/8
Answer: A.1/2
Explanation: The number of possible selections is 60.
Let E be the event of getting an even number on one die and an odd number on the other. Since there exist an equal number of blue and red balls, probability of picking a blue or red ball is = 30/60 = 1/2
P(E) = 1 - 1/2
= 1/2.
Answer:A
Question: Find the simple interest on Rs.500 for 9 months at 6 paisa per month?
A.337
B.270
C.287
D.266
Answer: B.270
Explanation: I = (500*9*6)/100
= 270
Answer: B
Question: Peter invested an amount of Rs. 12,000 at the rate of 10% p.a simple interest and another amount at the rate of 20% p.a. simple interest. The total interest earned at the end of one year on the total amount invested became 14% p.a. Find the total amount invested?
A.20000
B.3888
C.2677
D.29812
Answer: A.20000
Explanation: Let the second amount be Rs. x. Then,
(12000 * 10 * 1)/100 + (x * 20 * 1)/100 = [(12000 + x) * 14 * 1] / 100
120000 + 20x = 16800 + 14x
x = 8000
Total investment = 12000 + 8000 = Rs. 20,000.
Answer: A
Question: A statue is being carved by a sculptor. The original piece of marble weighed 250 kg. In the first week 30 percent is cut away. In the second week 20 percent of the remainder is cut away. In the third week the statue is completed when 25 percent of the remainder is cut away. What is the weight of the final statue?
A.105 kg
B.103 kg
C.108 kg
D.125 kg
Answer: A.105 kg
Explanation: A
105 kg
250 × 0.7 × 0.8 × 0.75 = 105 kg.
Question: A chair is bought for Rs.600/- and sold at a loss of 30% find its selling price
A.s.420/-
B.s.550/-
C.s.560/-
D.s.590/-
Answer: A.s.420/-
Explanation: 100 % ------> 600 (100 * 6 = 600)
70 % ------> 540 (70 * 6 = 420)
Selling price = Rs.420/-
A)
Question: 30 people attend a party. 4 men are single and the rest are there with their wives. There are no children in the party. In all 22 women are present. Then the number of married men at the party is ?
A.5
B.6
C.4
D.8
Answer: C.4
Explanation: Total people = number of men + number of women
30 = 4 + number of married men + 22
number of married men = 30- 22-4 = 4 men
ANSWER: C
Question: The average age of a committee of 8 members is 40 years. A member aged 55 years
retired and his place was taken by another member aged 39 years . The average age of
present committee is;
A.39 years
B.38 years
C.36 years
D.35 years
Answer: B.38 years
Explanation: Exp. Total age of the committee = 40*8 = 320,
Total age when a member is retired
and a new one was joined = 320-55 +39, =304
Average age of present committee = 304/8 = 38.
Answer: B
Question: Three positive integers a, b, and c are such that their average is 22 and a ≤ b ≤ c. If the median is (a + 11), what is the least possible value of c?
A.21
B.23
C.25
D.27
Answer: D.27
Explanation: Solution:
We know that the average value is 22 therefore (a+b+c)/3 = 22
so a+b+c = 66
and b = a + 11
therefore a + (a + 11) + c = 66
The least value of c is when c = b
so take c = a + 11
hence, a + (a + 11) + (a + 11) = 66
i.e., a = 44/3 a non integer therefore c is not equal to b
so take c = b+1 therefore c = a+12
a + (a + 11) + (a + 12) = 66
i.e., a= 43/3 again a non integer therefore c is not equal to b+1
next take c= b+2 therefore c = a+13
a + (a + 11) + (a + 13) = 66
i.e., a=42/3 = 14
hence, a = 14, b = 25, and c = 27
Answer: D
Question: John was thrice as old as Tom 6 years ago. John will be 15/13 times as old as Tom in 6 years. How old is Tom today?
A.5
B.7
C.9
D.11
Answer: B.7
Explanation: J-6 = 3(T-6), so J = 3T-12
J+6 = 15/13 * (T+6)
13J + 78 = 15T + 90
13(3T-12) + 78 = 15T + 90
24T = 168
T = 7
The answer is B.
Question: A computer is programmed to multiply consecutive even integers 2*4*6*8*…*n until the product is divisible by 2211, What is the value of n?
A.22
B.134
C.62
D.122
Answer: B.134
Explanation: Factorise 2211.. 3*11*67..
So n has to be a multiple of largest prime number,61..
So n=2*61=134..
ans :B
Question: The calendar for the year 2007 will be the same for the year
A.2017
B.2018
C.2014
D.2016
Answer: B.2018
Explanation: Explanation:
For a year to have the same calendar with 2007 ,the total odd days from 2007 should be 0.
Year:	2007	2008	2009	2010	2011	2012	2013	2014	2015	2016	2017
Odd Days: 1 2 1 1 1 2 1 1 1 2 1
Take the year 2014 given in the choice.
Total odd days in the period 2007-2013 = 5 normal years + 2 leap year
= 5 x 1 + 2 x 2 = 9 odd days
= 2 odd day (As we can reduce multiples of 7 from odd days which will not change anything)
Take the year 2016 given in the choice.
Number of odd days in the period 2007-2015 = 7 normal years + 2 leap year
= 7 x 1 + 2 x 2 = 11 odd days
= 4 odd days
(Even if the odd days were 0, calendar of 2007 will not be same as the calendar of 2016 because 2007 is not a leap year whereas 2016 is a leap year. In fact, you can straight away ignore this choice due to this fact without even bothering to check the odd days)
Take the year 2017 given in the choice.
Number of odd days in the period 2007-2016 = 7 normal years + 3 leap year
= 7 x 1 + 3 x 2 = 13 odd days
= 6 odd days
Take the year 2018 given in the choice.
Number of odd days in the period 2007-2017 = 8 normal years + 3 leap year
= 8 x 1 + 3 x 2 = 14 odd days
= 0 odd day (As we can reduce multiples of 7 from odd days which will not change anything)
Also, both 2007 and 2018 are not leap years.
Since total odd days in the period 2007-2017 = 0 and both 2007 and 2018 are of same type, 2018 will have the same calendar as that of 2007
Answer: Option B
Question: For an upcoming charity event, a male vocalist has agreed to sing 4 out of 6 “old songs” and 2 out of 5 “new songs.” How many ways can the singer make his selection?
A.25
B.50
C.150
D.480
Answer: C.150
Explanation: Number of ways the vocalist can sing "old songs" = C64=15
Number of ways the vocalist can sing "new songs" = C52=10
Therefore total number of ways the singer can make the selection = (Number of ways the vocalist can sing "old songs")*(Number of ways the vocalist can sing "new songs")= 15*10=150
Answer: C
Question: An gun can take a maximum of four shots at an enemy plane moving away from it. The probability of hitting the plane at the 1st, 2nd, third and 4th shots are 1.4, 1.3, 1.2 & 1.1 respectively. What is the probability that the plane is hit when all the four shots are fired?
A.0.53435
B.0.69766
C.1.34354
D.1.45465
Answer: B.0.69766
Explanation: Required probability:
=(0.4×0.7×0.8×0.9)+(0.6×0.3×0.8×0.9)+(0.6×0.7×0.2×0.9)+(0.6×0.7×0.8×0.1)+(0.4×0.3×0.8×0.9)+(0.4×0.7×0.2×0.9)+(0.4×0.7×0.8×0.1)+(0.6×0.3×0.2×0.9)+(0.6×0.3×0.8×0.1)+(0.6×0.7×0.2×0.1)+(0.4×0.3×0.2×0.9)+(0.6×0.3×0.2×0.1)+(0.4×0.3×0.8×0.1)+(0.4×0.7×0.2×0.1)+(0.4×0.3×0.2×0.1)=(0.4×0.7×0.8×0.9)+(0.6×0.3×0.8×0.9)+(0.6×0.7×0.2×0.9)+(0.6×0.7×0.8×0.1)+(0.4×0.3×0.8×0.9)+(0.4×0.7×0.2×0.9)+(0.4×0.7×0.8×0.1)+(0.6×0.3×0.2×0.9)+(0.6×0.3×0.8×0.1)+(0.6×0.7×0.2×0.1)+(0.4×0.3×0.2×0.9)+(0.6×0.3×0.2×0.1)+(0.4×0.3×0.8×0.1)+(0.4×0.7×0.2×0.1)+(0.4×0.3×0.2×0.1)
=0.2016+0.1296+0.756+0.336+0.864+0.504+0.224+0.324+0.144+0.0084+0.0216+0.0036+0.0096+0.0056+0.002=0.2016+0.1296+0.756+0.336+0.864+0.504+0.224+0.324+0.144+0.0084+0.0216+0.0036+0.0096+0.0056+0.002
=0.6976=0.6976
Edit: Thank you Vaibhav for providing an alternative method.
Alternate Method:
probability that the plane is hit when all the four shots are fired,
P=1−probability of not hitting the targetP=1−probability of not hitting the target
=1−(0.6×0.7×0.8×0.9)=1−(0.6×0.7×0.8×0.9)
=1−0.3024=1−0.3024
=0.6976
B
Question: Find smallest number that you can write using all the vowels(A,E,O,I,U) exactly once?
A.5000
B.6000
C.1000
D.4000
Answer: A.5000
Explanation: A
fIvE thOUsAnd(5000)
Question: A circular path of 13 m radius has marginal walk 2 m wide all round it. Find the cost of leveling the walk at 25p per m2?
A.Rs.43
B.Rs.23
C.Rs.44
D.Rs.20
Answer: C.Rs.44
Explanation: π (152 - 132) = 176
176 * 1/4 = Rs.44
Answer: C
Question: There is food for 760 men for 22 days. How many more men should join after two days so that the same food may last for 4 days more?
A.3022
B.3040
C.3099
D.3088
Answer: B.3040
Explanation: 760 ---- 22
760 ---- 20
x ----- 4
x*4 = 760*20
x = 3800
760
-------
3040
Answer: B
Question: A person crosses a 1000 m long street in 5 minutes. What is his speed in km per hour?
A.5.2 km/hr.
B.8.2 km/hr.
C.12.0 km/hr.
D.3.2 km/hr.
Answer: C.12.0 km/hr.
Explanation: C
12.0 km/hr.
Speed=(1000/5 * 60) m/sec
= 3.33 m/sec.
Converting m/sec to km/hr
=(3.33*18/5) km/hr
= 12.0 km/hr.
Question: Which of the following is closest to the difference between sum Z of all proper fractions (fractions less than 1) in the form 1/x , where x is a positive digit, and the product of all proper fractions in the form y/(y+1), where y is a positive digit?
A.2.82
B.2.72
C.1.82
D.1.72
Answer: D.1.72
Explanation: Sum Z of all proper fractions (fractions less than 1) in the form 1/x, where x is a positive digit:
1/1 + 1/2 + 1/3 +.....+ 1/9
This is a harmonic progression. Harmonic progression is inverse of arithmetic progression.
Approximate sum of a harmonic progression with even number of terms = number of terms * (average of middle 2 terms)
Approximate sum of a harmonic progression with odd number of terms = number of terms * (middle term)
The actual sum will be slightly more than the approximation.
Here we have 9 terms (odd).
Sum = 9 * 1/5
= 9/5
Product of all proper fractions in the form y/(y+1), where y is a positive digit:
1/2 * 2/3 * 3/4 *.....*9/10
We will be left with 1/10.
Required = 9/5 - 1/10
= 1.8 - 0.1
= 1.7
Closest is 1.72
Answer (D).