RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: If in a cricket one day match, there were no wide balls, no wides, no extras and no overthrows.
What is the maximum number of runs that a batsman can score in an ideal case scenario?
A.1652
B.1651
C.1653
D.1654
Answer: C.1653
Explanation: Solution:
1653
Explanation:
For an ideal case, the batsman will hit a six on each ball. But if he hits six on the last ball of the over, the strike will change in the next over. Thus, the best he can do in the last ball is run 3 runs so that he retains the strike even in the next over. Thus the total runs that he can score in each over:
6 * 5 + 3 = 33
But this will have to go like it only till the 49th over. In the last over, he can hit a six in the last ball as well as that will be the last ball of the match.
Thus runs for the last over will be 6 * 6 = 36.
Hence the maximum runs = 33 * 49 + 36 = 1653
Answer C
Question: Joseph bought two varieties of rice, costing 5 cents per ounce and 6 cents per ounce each, and mixed them in some ratio. Then he sold the mixture at 7 cents per ounce,making a profit of 20 percent. What was the ratio of the mixture?
A.1:10
B.1:5
C.2:7
D.3:8
Answer: B.1:5
Explanation: Let x ounce of ist be mixed with y ounce of 2nd
thus total CP = 5x+6y
Sly SP = 7x+7y
Now profit % = 20, its very easy to solve profit n loss problems using Fraction rule.
SP = CP * 120/100 ( since profit so SP >CP n hence multiplied by 120/100 which is > 1
thus 7x+ 7y= (5x+6y) * 120/100
or, x:y = 1:5
ANSWER:B
Question: There are 8 books in a shelf that consist of 2 paperback books and 6 hardback books. If 4 books are selected at random without replacement, how many different combinations are there that at least one paperback book is selected?
A.40
B.45
C.50
D.55
Answer: D.55
Explanation: It's almost always better to solve "at least one" combinations/probability questions with {at least one}={total}-{none}.
Total # of selections of 4 books out of 8: C48=70C84=70;
Total # of selections of 4 hardback books out of 8 (so none paperback): C46=15C64=15;
{at least one}={total}-{none}=70-15=55.
Answer: D
Question: If n is the greatest positive integer for which 4^n is a factor of 10!, then n =?
A.2
B.4
C.6
D.8
Answer: B.4
Explanation: 10! = 3628800
E. 4^10 = 1048576 (3628800/1048576) - this is not a factor of 10!
D. 4^8 = 65536 (3628800/65536) - this is not a factor of 10!
C. 4^6 = 4096 (3628800/4096) - this is not a factor of 10!
B. 4^4 = 256 (3628800/256) - this is a factor of 10!
B is answer
Question: Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is
A.17/25
B.17/29
C.17/21
D.17/20
Answer: D.17/20
Explanation: n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
P(EᴜP) = 10/20 + 8/20 - 1/20
=17/20
Answer: D
Question: √530493 ,493 is between
A.300 and 600
B.200 and 400
C.600 and 800
D.100 and 300
Answer: C.600 and 800
Explanation: 530493 is greater than 49,0000 (700^2) and smaller than 64,0000 (800^2)
700 < √530493 < 800. Only option satisfying this condition is 600 < √530493 < 800.
Answer:C
Question: How many figures are required to number the pages the pages of a book containing 1000 pages?
A.2793
B.2333
C.2993
D.2893
Answer: D.2893
Explanation: 1 to 9 = 9 * 1 = 9
10 to 99 = 90 * 2 = 180
100 to 999 = 900 * 3 = 2700
1000 = 4
-----------
2893
ANSWER:D
Question: A, B and C can do a work in 6 days, 8 days and 12 days respectively. In how many days can all three of them working together, complete the work?
A.2 2/3 days
B.2 6/3 days
C.2 2/7 days
D.6 2/3 days
Answer: A.2 2/3 days
Explanation: Work done by all three of them in one day
= 1/6 + 1/8 + 1/12 = 3/8.
The number of days required
= 8/3
= 2 2/3 days
Answer:A
Question: A train is 360 meter long is running at a speed of 45 km/hour. In what time will it pass a bridge of 140 meter length?
A.20 sec
B.30 sec
C.40 sec
D.50 sec
Answer: C.40 sec
Explanation: Speed = 45 Km/hr = 45*(5/18) m/sec = 25/2 m/sec
Total distance = 360+140 = 500 meter
Time = Distance/speed
= 500 * (2/25) = 40 seconds
ANSWER:C
Question: A man has some hens and cows. If the number of heads be 46 and the number of feet equals 136, then the number of hens will be
A.22
B.24
C.26
D.20
Answer: B.24
Explanation: Explanation :
Let number of hens = h and number of cows = c
number of heads = 46
=> h + c = 46 ---(Equation 1)
number of feet = 136
=> 2h + 4c = 136
=> h + 2c = 68 ---(Equation 2)
(Equation 2) - (Equation 1) gives
2c - c = 68 - 46
=> c = 22
Substituting the value of c in Equation 1, we get
h + 22 = 46
=> h = 46 - 22 = 24
i.e., number of hens = 24
Answer : B
Question: The mall charges 60 cents for the first hour of parking and $4 for each additional hour until the customer reaches 4 hours, after that the parking fee is $2 per hour. If Edward parked his car in the mall for 7 hours and 30 minutes, how much is he going to pay?
A.$25.2
B.$22.
C.$13.
D.$14.5.
Answer: A.$25.2
Explanation: 0.60+4*4+2*4+0.60= $25.2
Answer :$25.2
Question: Suppose you work for a manufacturing plant that pays you $12.50 an hour plus $0.16 for each widget you make. How many widgets must you produce in a 40 hour week to earn $660 (before payroll deductions)?
A.800
B.900
C.1000
D.1100
Answer: C.1000
Explanation: total pay = 40*$12.50 + $0.16*x = 660
x = 160 / 0.16 = 1000
The answer is C.
Question: Pat, Kate, and Mark charged a total of 135 hours to a certain project. If Pat charged twice as much time to the project as Kate and 1/3 as much time as Mark, how many more hours did Mark charge to the project than Kate?
A.18
B.36
C.75
D.90
Answer: C.75
Explanation: 75
all u do is do
2:1:6=>2x+x+6x=135=>x= 15
30:15:90
90-15=75
Answer C
Question: A group of store managers must assemble 280 displays for an upcoming sale. If they assemble 25 percent of the displays during the first hour and 40 percent of the remaining displays during the second hour, how many of the displays will not have been assembled by the end of the second hour?
A.70
B.98
C.126
D.168
Answer: C.126
Explanation: During the first hour 0.25*280 = 70 displays were assembled.
During the second hour 0.4*(280-70) = 84 displays were assembled.
Thus, 280 - (70+84) = 126 displays had not been assembled by the end of the second hour.
Answer: C.
Question: A certain ball team has an equal number of right- and left-handed players. On a certain day, one-third of the players were absent from practice. Of the players at practice that day, one-third were right handed. What is the ratio of the number of right-handed players who were not at practice that day to the number of left handed players who were not at practice?
A.1/3
B.5/1
C.5/7
D.7/5
Answer: B.5/1
Explanation: Say the total number of players is 18, 9 right-handed and 9 left-handed.
On a certain day, two-thirds of the players were absent from practice --> 6 absent and 12 present.
Of the players at practice that day, one-third were right-handed --> 12*1/3=4 were right-handed and 8 left-handed.
The number of right-handed players who were not at practice that day is 9-4=5.
The number of left-handed players who were not at practice that days is 9-8=1.
The ratio = 5/1.
Answer: B
Question: If a person walks at 15 km/hr instead of 10 km/hr, he would have walked 10km more. The actual distance traveled by him is?
A.50 km
B.30 km
C.18 km
D.16 km
Answer: B.30 km
Explanation: Let the actual distance traveled be x km. Then,
x/10 = (x + 10)/15
5x - 150 =>x = 30 km.
Answer: B
Question: If Q is an odd number and the median of Q consecutive integers is 150, what is the largest of these integers?
A.Q/2 + 150
B.Q/2 + 149
C.(Q - 1)/2 + 150
D.(Q + 149)/2
Answer: C.(Q - 1)/2 + 150
Explanation: Consider the easiest case, say Q=3, then;
Set = {149, 150, 151};
The largest integer = 151.
Now, plug Q=3 into the answers to see which yields 151.
Answer: C.
Question: If the number 5 * 2 is divisible by 6, then * = ?
A.2
B.5
C.3
D.1
Answer: A.2
Explanation: 6 = 3 x 2. Clearly, 5 * 2 is divisible by 2. Replace * by x.
Then, (5 + x + 2) must be divisible by 3. So, x = 2.
ANSWER A
Question: If p and q are both odd integers, which of the following must be an odd integer?
A.p+q
B.2p+q
C.p+3q
D.2p+q+1
Answer: B.2p+q
Explanation: p is odd; for this assume p=1
q is odd; for this assume q = 1
1+1 = 2 Even
(2*1)+1 = 3 Odd
1 + (3*1) = 4 Even
(2*1)+1+1 = 4 Even
1+(1^2) = 1 Even
Answer:B
Question: P and Q started a business with respective investments of Rs. 4 lakhs and Rs. 10 lakhs. As P runs the business, his salary is Rs. 5000 per month. If they earned a profit of Rs. 2 lakhs at the end of the year, then find the ratio of their earnings?
A.1 : 5
B.1 : 4
C.1 : 2
D.1 : 1
Answer: D.1 : 1
Explanation: Ratio of investments of P and Q is 2 : 5
Total salary claimed by P = 12 * 5000 = Rs. 60000
Total profit = Rs. 2 lakhs.
Profit is to be shared = Rs. 140000
Share of P = (2/7) * 140000 = Rs. 400000
Share of Q = Rs. 100000
Total earnings of P = (60000 + 40000) = Rs. 100000
Ratio of their earnings = 1 : 1.Answer: D
Question: A can do a work in 6 days , B can do a work in 8 days and C can do it in 12 days. B left work after 3 days. For how many number of days should A and C should work together to complete the remaining work ?
A.0.1
B.0.2
C.0.3
D.0.4
Answer: D.0.4
Explanation: b work 1/8 *3=3/8
remaining work =1-3/8=5/8
a and c work together =1/6+1/12=3/12=1/4
remaining work is done in=8/5 *1/4=2/5=0.4
ANSWER:D
Question: Jane and Thomas are among the 8 people from which a committee of 4 people is to be selected. How many different possible committees of 4 people can be selected from these 8 people if at least one of either Jane or Thomas is to be selected?
A.28
B.46
C.55
D.63
Answer: C.55
Explanation: Solution:
J+T+6 = total 8 ,
we need to select at least one of the J and T or both= total 4 out of 8 - 4 out of 6
8C4 - 6C4
=55
Ans:C
Question: A reduction of 20% in the price of salt enables a lady to obtain 5 kgs more for Rs.100, find the original price per kg?
A.6
B.1
C.5
D.4
Answer: C.5
Explanation: 100*(20/100) = 20 --- 5
? --- 1 => Rs.4
100 --- 80
? ---4 => Rs.5
Answer: C
Question: If A is the product of all positive integers less than 31, than what is the greatest integer k for which A/18^k is an integer?
A.3
B.6
C.7
D.14
Answer: C.7
Explanation: 30! A must be divisible by (3*3*2)^k.
As I will find a lot of 2's among 30!'s factors I will focus on the 3's (that also occurs twice instead of only once as a factor in 18).
Every multiple of 3 will provide me with atleast one factor of 3. There aren't that many so I list them quickly.
3 6 9*
12 15 18*
21 24 27**
30
The stars are marked as they are multiples of 9* or 27** and thus provides two or three 3's respectively.
3=3 (occurs 10 times) +10
3*3=9 (occurs 3 times) +3(adding just one additional three each as they were not counted for in the multiples of 3)
3*3*3=27 (occurs 1 time) +1 (again, adding just one as they were previously not counted)
Total: 14.
For every k we add two 3's. So we must divide 14 by 2 to get the highest integer K before we receive a non-integer result.
Answer: 7 -> Answer choice: C.
Question: The car A has 1.5 times the price of car B which is Rs.250000 more. What is the price of car B?
A.100000
B.200000
C.500000
D.800000
Answer: C.500000
Explanation: Let the price of car B be x, (x+250000/x)=3/2
Up on cross multiplication and solving for x, we get x=500000
Answer:C
Question: A began a business with Rs. 85,000. He was joined afterwards by B with Ks. 40,000. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3 : 1 ?
A.6.5 months
B.3.5 months
C.2.5 months
D.8.5 months
Answer: D.8.5 months
Explanation: Suppose B joined for x months . Then, ( 85000 * 12 )/(40000 * x) = 3. or x = (85000 * 12) / (40000 * 3) = 8.5
So, B joined for 8.5 months.
Answer: D
Question: In the xy-plane the point (-2 -3) is the centre of a circle, the point (-2, 1) lies inside the circle and the point (4, -3) lies outside the circle. If the radius r of the circle r is an integer then diameter ?
A.6
B.10
C.4
D.3
Answer: B.10
Explanation: Can be solved without much calculations.
You are given that (-2,-3) is the center of the circle. Point (4,-3) lies inside the circle ---> the radius is lesser than distance of (-2,-3) from (4,-3) ---> lesser than 6 units but the radius will also be greater than the distance of (-2,-3) from (-2,1) ----> greater than 4 units.
Thus the radius is >4 but <6 and as it is an integer, the only possible value of radius = 5 units.
Diameter =2r= 10
B is the correct answer.
Question: Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 10 cm.
A.190 cm2
B.885 cm2
C.285 cm2
D.825 cm2
Answer: A.190 cm2
Explanation: Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them)
= 1/2 (20 + 18) * (10)
= 190 cm2
Answer: A
Question: If 0 < x < 5 and -2 < y < 9, then the lowest possible value for xy is
A.between 60 and 70
B.between 0 and 60
C.between - 50 and 50
D.between 0 and -70
Answer: C.between - 50 and 50
Explanation: If 0 < x < 5 and -2 < y < 9,
Say y = -1 (-2<y<9)
x = 4
So xy = -4 which is more than -70.
Answer is C.
Question: A boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 5 km/hr, find the time taken by the boat to go 70 km downstream.
A.4.25 hr
B.3.8 hr
C.8.25 hr
D.2.25 hr
Answer: B.3.8 hr
Explanation: speed of boat in still water = 13 km/hr
speed of the stream = 5 km/hr
Speed downstream = (13+5) = 18 km/hr
Time taken to travel 70 km downstream =70â„18 = 3.8 hours
Answer is B
Question: There is a clock which shows false timing ie it shows 10 minutes less for each 1 hour....if the clock shows 12 noon now how many minutes will be lost when it shows 3 pm ?
A.16 minutes
B.26 minutes
C.36 minutes
D.46 minutes
Answer: C.36 minutes
Explanation: For 1 hour, 10minutes is slower, then for
06 miuntes - 1 minutes is slower
30 minutes - 5 minutes
Actual time false time
12 12
01 PM 12.50 PM (10 minutes slower)
02 PM 01.40 PM (12.50 + 60 min - 10 minutes slower)
03 PM 02.30 PM (01.40 + 60 min - 10 minutes slower)
03.36 PM 03.00 PM (02.30 + 36 min - 6 minutes slower )
So, 36 minutes will be lost when it is showing 3.00 PM. and the actual time is 3.36 PM
ANSWER:C
Question: A hat company ships its hats, individually wrapped, in 8-inch by 10-inch by 12-inch boxes. Each hat is valued at $7.50. If the company’s latest order required a truck with at least 336,000 cubic inches of storage space in which to ship the hats in their boxes, what was the minimum value of the order?
A.$960
B.$1,350
C.$1,725
D.$2,625
Answer: D.$2,625
Explanation: Number of boxes= Total volume/volume of one box= 336,000/ (8*10*12)= 350
One box costs 7.50, so 350 box will cost= 350*7.5= 2625
D is the answer
Question: In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded medals, then how many groups of medal winners are possible?
A.20
B.56
C.120
D.560
Answer: B.56
Explanation: 1) Is it important here that how many go into finals... NO....
what is important is HOW many get medals..3
finally these 3 can be any of the 8 - 8C3=8!5!3!=8∗7∗63∗2=568C3=8!5!3!=8∗7∗63∗2=56
2) Is order important...NO
we are looking for groups only..
ans 56 ANSWER:B
Question: In how much time would the simple interest on a certain sum be 0.125 times the principal at 10% per annum?
A.1 1/4
B.1 1/8
C.1 1/2
D.1 1/3
Answer: A.1 1/4
Explanation: Let sum = x. Then, S.I. = 0.125x = 1/8 x, R = 10%
Time = (100 * x) / (x * 8 * 10) = 5/4 = 1 1/4 years.
Answer: A
Question: Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
A.(2/8)∗(3/8)^2∗(3/8)^2∗5!/(2!∗2!)
B.(2/8)∗(3/8)^2∗(3/8)^2∗5!
C.(2/8)∗(3/8)^2∗(3/8)2
D.(28)∗(38)∗(38)^2(28)∗(38)∗(38)^2
Answer: A.(2/8)∗(3/8)^2∗(3/8)^2∗5!/(2!∗2!)
Explanation: CASE I.
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?
The probability would be: 2/8*3/8*3/8*3/8*3/8*5!/2!2!=14.8%
CORRECT ANSWER:A
Question: A satellite is composed of 30 modular units, each of which is equipped with a set of sensors, some of which have been upgraded. Each unit contains the same number of non-upgraded sensors. If the number of non-upgraded sensors on one unit is 1/6 the total number of upgraded sensors on the entire satellite, what fraction of the sensors on the satellite have been upgraded?
A.5/6
B.1/5
C.1/6
D.1/7
Answer: C.1/6
Explanation: Let x be the number of upgraded sensors on the satellite.
The number of non-upgraded sensors per unit is x/6.
The number of non-upgraded sensors on the whole satellite is 30(x/6) = 5x.
The fraction of sensors which have been upgraded is x / (x+5x) = x/6x = 1/6
The answer is C.
Question: Sakshi can do a piece of work in 30 days.Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work :
A.15
B.16
C.14
D.25
Answer: C.14
Explanation: Solution
Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4 .
Suppose Tanya taken x days to do the work.
5 : 4 : : 30 : x ⇒ x = (30x4/5) ⇒ x = 24 days.
Hence,Tanya takes 16 days is complete the work.
Answer C
Question: If n = 8^8 – 8, what is the units digit of n?
A.0
B.1
C.8
D.2
Answer: C.8
Explanation: 8 ^ 8 - 8 = 8 (8 ^ 7 - 1)
==> 8 (2 ^ 21 - 1)
Last digit of 2 ^ 21 is 2 based on what explanation LiveStronger is saying. 2 ^ 24 - 1 yields 2 - 1 = 1 as the unit digit. Now on multiply this with 8, we get unit digit as 8
Answer : C
Question: By mixing two qualities of pulses in the ratio 2: 3 and selling the mixture at the rate of Rs 22 per kilogram, a shopkeeper makes a profit of 10 %. If the cost of the smaller quantity be Rs 14 per kg, the cost per kg of the larger quantity is:
A.33
B.88
C.24
D.77
Answer: C.24
Explanation: Explanation:
Cost Price of 5 kg = Rs.(14*2 + x*3) = (28 + 3x).
Sell price of 5 kg = Rs. (22x5) = Rs. 110.
[{110 - (28 + 3x)}/(28 + 3x) ]* 100 =10
[82-3x/28 + 3x]= 1 / 10
820 - 30x = 28 +3x
33x = 792
x = 24
Answer: C) Rs 24
Question: In a stream running at 2kmph,a motor boat goes 6km upstream and back again to the starting point in 33 minutes.find the speed of the motorboat in still water.
A.22
B.24
C.26
D.28
Answer: A.22
Explanation: let the speed of the motarboat in still water be x kmph.then,
6/x+2 +6/x-2=33/60
11x2-240x-44=0
11x2-242x+2x-44=0
(x-22)(11x+2)=0
x=22.
Answer is A
Question: Four years ago,the father’s age was three times the age of his son.The total of the ages of the father and the son after four years,will be 64 years.What is the father’s age at present ?
A.32 years
B.36 years
C.44 years
D.None of these
Answer: D.None of these
Explanation: Solution
Let the ages of father and son 4 years ago be 3x and x years respectively.
Then,[(3x + 4) + 4] + [(x + 4)+ 4]= 64 ⇔ 4x = 48 ⇔ x =12.
∴ Father's present age = 3x = 36 years. Answer D
Question: In a certain game, you pick a card from a standard deck of 52 cards. If the card is a spade, you win. If the card is not a spade, the person replaces the card to the deck, reshuffles, and draws again. The person keeps repeating that process until he picks a spade, and the point is to measure how many draws did it take before the person picked a spade and won. What is the probability that there will be at least four draws involved in a win, i.e. someone picking her first spade on the fourth draw or later?
A. 1/2
B. 9/16
C. 27/64
D. 13/16
Answer: C. 27/64
Explanation: Probability of picking a spade on any draw = 1/4
Probability of NOT picking a heart on the first draw AND on the second draw And 3rd draw = [1-(1/4)] X [1-(1/4)] X [1-(1/4)]= 3/4 X 3/4 X 3/4 = 27/64
Answer : C
Question: The population of a town is 15000. It decreases annually at the rate of 20% p.a. What will be its population after 2 years?
A.4300
B.4500
C.5120
D.9600
Answer: D.9600
Explanation: 15000 × 80/100 × 80/100 = 9600
Answer : D
Question: Virat buys an old car for Rs. 50000 and spends Rs. 6000 on its repairs. If he sells the scooter for Rs. 70000, his gain percent is:
A.25%
B.11 (1/9) %
C.12 (2/3) %
D.12%
Answer: A.25%
Explanation: Explanation:
C.P. = Rs. (50000 + 6000) = Rs. 56000
S.P. = Rs. 70000.
Gain% = ( (14000 / 56000) * 100 ) % = 25%
ANSWER A
Question: A pipe can fill a cistern in 88 minutes. Find the time in which 1/11 part of the cistern will be filled?
A.5min
B.2min
C.8min
D.1min
Answer: C.8min
Explanation: Full cistern filled in = 88 minutes
1/11 part filled in = 88*1/11 = 8minutes
Answer is C
Question: A grocer has a sale of Rs. 5420, Rs. 5660, Rs. 6200, Rs. 6350 and Rs. 6500 for 5 consecutive months. Find the sale he should have in the sixth month, so that he gets an average sale of Rs. 6000?
A.Rs. 5879
B.Rs. 5872
C.Rs. 5871
D.Rs. 5870
Answer: D.Rs. 5870
Explanation: Explanation:
Total sale for 5 months = Rs. (5420 + 5660 + 6200 + 6350 + 6500) = Rs. 30,130
Therefore, required sale = Rs. [(6000 * 6) – 30,130] = Rs. (36000 – 30,130) = Rs. 5870
ANSWER: D
Question: How many seconds will a 800 meter long train moving with a speed of 63 km/hr take to cross a man walking with a speed of 3 km/hr in the direction of the train ?
A.48
B.52
C.38
D.42
Answer: A.48
Explanation: Explanation:
Here distance d = 800 mts
speed s = 63 - 3 = 60 kmph = 60 x 5/18 m/s
time t = 800x 18 / 60x 5= 48 sec.
ANSWER IS A
Question: City A's population is 68000, decreasing at a rate of 80 people per year. City B having population 42000 is increasing at a rate of 120 people per year. In how many years both the cities will have same population?
A.130years
B.140years
C.150years
D.160years
Answer: A.130years
Explanation: city A population = 68000
city B population = 42000
68000 decreasing at a rate 0f 80p = 42000 increasing at a rate of 120p
68000-80p=42000+120p
200p=26000
p=130years
ANSWER:A
Question: Alex deposited x dollars into a new account that earned 7 percent annual interest, compounded annually. One year later Alex deposited an additional x dollars into the account. If there were no other transactions and if the account contained w dollars at the end of two years, which of the following expresses x in terms of w ?
A.w/(1+1.08)
B.w/(1.08+1.16)
C.w/(1.16+1.24)
D.w/(1.07+1.07^2)
Answer: D.w/(1.07+1.07^2)
Explanation: Account at the end of the first year would be 1.07x dollars. At this time x dollars was deposited, hence the account at the beginning of the second year would be (1.07x+x) dollars. Account at the end of the second year would be (1.07x+x)*1.07=w --> x(1.07^2+1.07)=w --> x=w/(1.07+1.07^2).
Answer: D.
Question: For each 6-month period during a light bulb's life span, the odds of it not burning out from over-use are half what they were in the previous 6-month period. If the odds of a light bulb burning out during the first 6-month period following its purchase are 1/5, what are the odds of it burning out during the period from 6months to 1 year following its purchase?
A.5/27
B.2/9
C.1/2
D.4/9
Answer: C.1/2
Explanation: P(of not burning out in a six mnth period)=1/2 of P(of not burning out in prev 6 mnth period)
P(of burning out in 1st 6 mnth)= 1/5
--->P( of not burning out in 1st 6 mnth)=1-1/5=4/5
---->P(of not burning out in a six mnth period)=1/2 *4/5=4/10=2/5
--->P(of burning out in a six mnth period)=1-2/5=3/5
now
P( of burning out in 2nd six mnth period)=P( of not burning out in 1st six mnth)*P(of burning out in a six mnth)
=4/5 * 3/5=1/2
Ans C
Question: Arrange the following in a logical order:
1. Birth
2. College
3. Marriage
4. School
5. Employment
A.1, 3, 5, 4, 2
B.1, 4, 2, 5, 3
C.2, 5, 3, 1, 4
D.2, 4, 1, 5, 3
Answer: B.1, 4, 2, 5, 3
Explanation: Explanation:
The given words when arranged in the order of various events as they occur in man's life, the sequence is : Birth, School, College, Employment, Marriage.
Sequence is : 1, 4, 2, 5, 3.
ANSWER B
Question: A basket has 5 apples and 4 oranges. Three fruits are picked at random. The probability that at least 2 apples are picked is
A.25/42
B.25/49
C.25/46
D.25/40
Answer: A.25/42
Explanation: Total fruits = 9
Since there must be at least two apples,
(5C2 * 4C1)/9C3 + 5C2/3C3 = 25/42.
Answer: A