RikoteMaster/aqua-rat-mcqa huggingface dataset:
Question: A certain league has four divisions. The respective divisions had 9, 10, 11, and 12 teams qualify for the playoffs. Each division held its own double-elimination tournament -- where a team is eliminated from the tournament upon losing two games -- in order to determine its champion. The four division champions then played in a single-elimination tournament -- where a team is eliminated upon losing one game -- in order to determine the overall league champion. Assuming that there were no ties and no forfeits, what is the maximum number of games that could have been played in order to determine the overall league champion?
A.79
B.83
C.85
D.87
Answer: A.79
Explanation: Consider 4 teams.
Max possible games if a team is eliminated after 2 loses = 3 + 2 = 5
Similarly, if 5 teams are there,
Max possible games = 4 + 3 = 7.
So going forth we can play 15,17,19,21 max possible games for 9,10,11,12 teams respectively.
Adding them up gives 72. Now there are two teams left in each group still. So four more games to give a winner in each team.
So 72+ 4 = 76.
Now of the four teams we can have max 3 games if its a knock out round.
So 76 + 3 = 79.
Answer A = 79.
Question: For the symbol, m”n=n^2−m for all values of m and n. What is the value of 6”3?
A.5
B.3
C.2
D.1
Answer: B.3
Explanation: 6”3= 9 - 6=3
Answer: B
Question: A brick measures 20 cm * 10 cm * 7.5 cm how many bricks will be required for a wall 25 m * 2 m * 0.75 m?
A.33381
B.27991
C.92211
D.25000
Answer: D.25000
Explanation: 25 * 2 * 0.75 = 20/100 * 10/100 * 7.5/100 * x
25 = 1/100 * x => x = 25000
Answer: D
Question: A fraudulent dairy guy declares to market his dairy at price range but he combined it with water and therefore benefits 25%. The percentage of water in the combination is:
A.20%
B.30%
C.50%
D.35%
Answer: A.20%
Explanation: Explanation :
-> If Capability Price of 1 L of milk = Rs.1
And S.Price of 1 L of combination = Rs.1
Benefits = 20%
Capability Price of 1 L of combination =[ (100 / 100+25)*1 ]
= Rs. 100/125 = Rs. 4/5
-> From the rule of allegation
i. Capability Price of 1 L of water = 0
ii. Capability Price of 1 L of milk = 1
iii. (p) = 4/5
iv. d – m = 1 – 4/5 = 1/5
v. m – c = 4/5 – 0 = 4/5
Ratio of milk to water = 4/5 : 1/5 = 4:1
% of water in the combination = (1/5 x 100)% = 20%
Answer : A
Question: Today John is twice as old as Sam and Sam is 3 years younger than Mike. If John, Sam, and Mike are all alive 4 years from today, which of the following must be true on that day. I. John is twice as old as Sam II. Sam is 3 years younger than Mike III. John is older than Mike
A.I, III
B.I
C.II
D.I,II & III
Answer: D.I,II & III
Explanation: Say Mike age is 10
Sam age is 7
John age is 14
After 4 years
Mike age is 14
Sam age is 11
John age is 28
i.e ) after 4 years
Sam is 3 years younger than Mike
John is twice as old as Sam
John is older than Mike
Answer : D
Question: Kate and Danny each have $10. Together, they flip a fair coin 5 times. Every time the coin lands on heads, Kate gives Danny $1. Every time the coin lands on tails, Danny gives Kate $1. After the five coin flips, what is the probability W that Kate has more than $10 but less than $15?
A.5/16
B.1/2
C.12/30
D.W=15/32
Answer: D.W=15/32
Explanation: For 21 heads Kate will end up with < $10 and we want her to win. Therefore, only possibilities W are 3 or 4 heads.
I made an educated guess and it worked fine.
Ans- 'D'
MGMAT's anagram helped here as well.
HHHHT = 5!/4!*1! = 5
HHHTT = 5!/3!*2! = 10
Total acceptable cases = 15
Total cases = 32
P = 15/32.D
Question: Find unit digit of{(6374)^1793 x (625)^317 x (341)^491}.
A.5
B.15
C.0
D.20
Answer: C.0
Explanation: Explanation:
Unit digit of 6374 is 4.
Unit digit of 625 is 5.
Unit digit of 341 is 1.
Just multiply 4*5*1 = 20
So unit digit of 20 is 0.
Answer C.
Question: The ratio by weight, measured in pounds, of books to clothes to electronics in a suitcase initially stands at 5:4:2. Someone removes 7 pounds of clothing from the suitcase, thereby doubling the ratio of books to clothes. How many pounds do the electronics in the suitcase weigh?
A.6
B.7
C.8
D.9
Answer: B.7
Explanation: The weights of the items in the suitcase are 5k, 4k, and 2k.
If removing 7 pounds of clothes doubles the ratio of books to clothes, then 7 pounds represents half the weight of the clothes.
2k = 7 pounds and then k = 3.5 pounds.
The electronics weigh 2(3.5) = 7 pounds.
The answer is B.
Question: A woman has two blouses of different colors, three skirts of different colors, and two different pairs of shoes. She refuses to wear her pink blouse with her green skirt. How many different blouseskirt-shoe combinations could she wear?
A.8
B.10
C.16
D.17
Answer: B.10
Explanation: total ways = 2*3*2=12..
now its better to find the ways in which the pink blose and green skirty are together and subtract from total to get the answer..
the ways in which the pink blose and green skirty are together=1*1*2=2..
total ways =12-2=10..
B
Question: A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm^3, then the weight of the pipe is:
A.3.6 kg
B.3.696 kg
C.36 kg
D.36.9 kg
Answer: B.3.696 kg
Explanation: External radius = 4 cm,
Internal radius = 3 cm.
Volume of iron=(22/7	x [(4)^2 - (3)^2] x 21) cm^3
(22/7x 7 x 1 x 21)cm^3
462 cm^3.
Weight of iron = (462 x 8) gm = 3696 gm = 3.696 kg.
ANSWER B
Question: A train passes a station platform in 34 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
A.228
B.210
C.188
D.266
Answer: B.210
Explanation: The speed of the train is 54 * 5/18 m/s = 15 m/s
The length of the train is 15 x 20 = 300 m
platform + 300 = 15 * 34
platform = 510 - 300 = 210
The answer is B.
Question: One computer can upload 120 megabytes worth of data in 6 seconds. Two computers, including this one, working together, can upload 1300 megabytes worth of data in 42 seconds. How long would it take for the second computer, working on its own, to upload 120 megabytes of data?
A. 6
B. 7
C. 9.13
D. 11
Answer: C. 9.13
Explanation: Since the first computer can upload 120 megabytes worth of data in 6 seconds then in 6*7=42 seconds it can upload 7*120=840 megabytes worth of data, hence the second compute in 42 seconds uploads 1300-840=460 megabytes worth of data. The second computer can upload 120 megabytes of data in 9.13 seconds.
Answer: C.
Question: How many two-digit numbers yield a remainder of 1 when divided by both 6 and 15?
A.0
B.1
C.2
D.3
Answer: C.2
Explanation: Easier to start with numbers that are of the form 15p+1 ---> 16,31,46,61,76,91.
Out of these 2 (31,61) are also of the form 6q+1. Thus 2 is the answer. C is the correct answer.
Question: At a small company, 76 percent of the employees are women, and 60 percent of the employees are married. If 2/3 of the men are single, what fraction of the women are married?
A.13/19
B.1/3
C.9/20
D.7/10
Answer: A.13/19
Explanation: Lets take total employees are 100.
Given that,Total Women= 76 and Total Married = 60.
Total Men = 100 - 76 = 24 and Single men = 2/3*24 = 16.
Married men = total men - single men = 24 - 16 = 8.
Married women= Total married - Married men = 60 - 8 = 52.
Fraction of women are married = Married women / Total Women = 52 / 76 = 13/19. ANS A
Question: If CD represents a positive two digit number, where C and D are single digit integers, which of the following cannot be true?
A.C+D =9
B.(C)(D)=9
C.C-D = 9
D.D-C= 9
Answer: D.D-C= 9
Explanation: (A) CD = 45 or 63 or 27
(B) CD = 33 or 19
(C) 90
(D) impossible
(E) CD = 91
(D) is impossible because 09 is not a valid two digit number --- if there's a zero in the ten's place, that's just 9, a single digit number. Therefore, no valid two digit number would satisfy (D).
ans D
Question: 8^100 is divisible by 17. Find the remainder for this?
A.10
B.12
C.16
D.17
Answer: C.16
Explanation: This is an extremely difficult problem to solve with out Fermat's little theorem. By applying Fermat's little theorem , We know that 816 when divided by 17, the remainder is 1.
So divide 100 by 16 and find the remainder. Remainder = 4
Therefore, 100 = (16 × 6) + 4
Now this problem can be written as 810017 = 816×6+417 = (816)6×8417
Now this problem simply boils down to (1)6×8417 = 8417
84 = 82×82, we need to find the remainder when 64 × 64 is divisible by 17. Or 13 × 13 = 169. When 169 is divided by 17, remainder is 16.
C
Question: The speed of a boat in upstream is 60 kmph and the speed of the boat downstream is 80 kmph. Find the speed of the boat in still water and the speed of the stream?
A.10 kmph
B.22 kmph
C.77 kmph
D.66 kmph
Answer: A.10 kmph
Explanation: Speed of the boat in still water = (60+80)/2 = 70 kmph. Speed of the stream
= (80-60)/2
= 10 kmph.
Answer:A
Question: A single letter is selected at random from the word 'SIM'. The probability that it is a vowel is..??
A.2/11
B.3/11
C.4/11
D.1/3
Answer: D.1/3
Explanation: total vowels = 1, total number of letters = 3
so probability = 1/3
ANSWER:D
Question: A train 600 meters long is running with a speed of 50 kmph. The time taken by it to cross a tunnel 170 meters long is?
A.287 sec
B.288 sec
C.55 sec
D.16 sec
Answer: C.55 sec
Explanation: D = 600 + 170 = 770
S = 50 * 5/18 = 14 mps
T = 770/14 = 55 sec
Answer: C
Question: Which one the following fractions has the smallest value?
A.123/847
B.123/849
C.123/874
D.123/877
Answer: D.123/877
Explanation: Among (A) , (B) , (C) & (D)
(D) is the smallest...
Now compare (D) Vs (E)
(D) 123/877 ~ 14%
(E) 246/1700 ~ 14.47%
So (D) < (E)
Hence correct answer must be (D)
Question: If a is a positive integer, then 5^a+1^a =?
A.Cannot be determined
B.1+(5^a)
C.2a
D.2a^(a + 1)
Answer: B.1+(5^a)
Explanation: 5^a+1^a=5^a+1
Answer is option B
Question: If P gets 25% more than Q, then by what percent does Q get less than P?
A.60%
B.22%
C.12%
D.20%
Answer: C.12%
Explanation: Let Q get 1 unit.
25% of 1 unit = 1/4 = 0.25
P gets (1 + 0.25) i.e., 1.25 units.
Difference = (1.25 - 1) = 0.25
Required percentage = 0.25/1.25 * 100 = 1/5 * 100 = 20%
Answer: C
Question: The present population of a town is 2000. Population increase rate is 20% p.a. Find the population of town after 1 years?
A.1000
B.2400
C.1500
D.1600
Answer: B.2400
Explanation: P = 2000
R = 20%
Required population of town = P(1 + R/100)^T
= 2000(1 + 20/100)
= 2000(6/5)
= 2400
Answer is B
Question: A straight fence is to be constructed from posts 6 inches wide and separated by lengths of chain 5 feet long. If a certain fence begins and ends with a post, which of the following could be the length of the fence in feet? (12 inches = 1 foot).
A.17
B.88
C.266
D.2777
Answer: A.17
Explanation: Explanation:
The fence will consist of one more post than there are chains. (e.g. P-c-P-c-P).
Therefore, a total length has to be a multiple of the length of the chain plus one post (5.5) plus one post extra.We have length = (5.5n + 0.5), where n can be any positive whole number. If n= 3, length =17
Answer: A) 17
Question: If Yury has been writing programming code for 15 hours at a rate of 5 lines of code per 20 seconds, how many lines of code has he written?
A.126(10^2)
B.135(10^2)
C.12.6(10^5)
D.1.26(10^7)
Answer: B.135(10^2)
Explanation: 5 LINES IN 20 SECONDS
SO FOR 1 MINUTE ( 5 X 3 = 15 LINES )
I HOUR = 60 MINUTES
15 X 60 = 900 LINES IN AN HOUR
15 HOURS = 900 X 15 = 135(10^2)
ANSWER B
Question: A sum of money at simple interest amounts ti $2240 in 2 years and to $ 2600 in 5 years. What is the principal amount?
A.$1520
B.$1880
C.$2120
D.$2000
Answer: D.$2000
Explanation: S.I for 3 years = $ (2600-2240) = $ 360
S.I for 2 years = $ ((360/3) x 2) = $ 240
Principal =$ (2240-240) = $ 2000
Answer is D
Question: By selling an shirt for $300, a shop keeper gains 20%. During a clearance sale, the shopkeeper allows a discount of 10% on the marked price. His gain percent during the sale is:
A.2
B.14
C.8%
D.12
Answer: C.8%
Explanation: C
8%
Marked price = $300
C.P. = 100/120 * 300 = $250
Sale price = 90% of $300 = $270
Required gain % = 20/250 * 100 = 8%.
Question: All windows are doors. No door is wall
Conclusions:
1. No window is wall
2. No Wall is door
A.Both 1 and 4 follow
B.Both 1 and 3 follow
C.Both 4 and 1 follow
D.Both 1 and 2 follow
Answer: D.Both 1 and 2 follow
Explanation: Explanation:
Since both the premises are universal and one premise is negative, the conclusion must be universal negative. Also, the conclusion should not contain the middle term. So, 1 follows.
However, 2 is the converse of the second premise and thus it also holds.
Answer: D
Question: If 1 + 9 + 11 = 1, Then what is the value of
12 + 11 + 9 = ?
A.one
B.Ten
C.six
D.five
Answer: B.Ten
Explanation: Solution:
10
Explanation:
Equation 1 + 9 + 11 = 1 can be derived from
One (o) + nine (n) + eleven (e) = one => 1
Similarly for equation,
12 + 11 + 9
Twelve (t) + eleven (e) + nine (n) => ten (10)
Answer B
Question: A delegation of 5 members has to be formed from 3 ladies and 5 gentlemen. In how many ways the delegation can be formed, if 2 particular ladies are always included in the delegation?
A.20
B.34
C.37
D.27
Answer: A.20
Explanation: There are three ladies and five gentlemen and a committee of 5 members to be formed.
Number of ways such that two ladies are always included in the committee
= ⁶C₃ = (6 * 5 * 4)/6
= 20.
Answer: A
Question: There are 3 states and 3 students representing each state. In how many ways can 5 students be chosen such that at Least one student is chosen from each state.
A.108.
B.118
C.128
D.138
Answer: A.108.
Explanation: To choose 5 students so that atleast one student will represent each state can be done in two ways:
A. 3-1-1 (3 students from 1 state and 1 student from other two states)
3C1*3C3*3C1*3C1=27
3C1 - # of ways to choose 3-student state;
3C3 - # of ways to choose 3 students from 3-student state;
3C1 - # of ways to choose 1 student from the first 1-student state;
3C1 - # of ways to choose 1 student from the second 1-student state.
OR
B. 1-2-2 (1 student from 1 state and 2 students from other two states)
3C1*3C1*3C2*3C2=81
3C1 - # of ways to choose 1-student state;
3C1 - # of ways to choose 1 student from the 1-student state;
3C2 - # of ways to choose 2 students from the first 2-student state;
3C2 - # of ways to choose 2 students from the second 2-student state.
27+81=108.
Answer: 108 A
Question: 2 oranges, 3 bananas and 4 apples cost Rs.15. 3 oranges, 2 bananas and 1 apple costs Rs 10. What is the cost of 3 oranges, 3 bananas and 3 apples
A.22
B.28
C.15
D.77
Answer: C.15
Explanation: Explanation:
2 O + 3 B + 4 A = 15 - - - - (1)
3 O + 2 B + 1 A = 10 - - - - (2)
Where A,B and O are number of apple, bananas, and oranges respectively.
Adding 1 and 2,
5 O + 5 B + 5 A = 25 ⇒ 1 O + 1 A + 1 B = 5
now,
3O + 3A + 3B = 5 × 3 = 15
Answer:C
Question: A certain sum amounts to Rs.7350 in 2 years and to Rs.8575 in 3 years.find the sum and rate percent
A.5400
B.3400
C.5200
D.6500
Answer: A.5400
Explanation: Interest for 1 year is the same whether it's simple interest or the compound interest.Now interest of third year = 8575 - 7350 = 1225; means principal for this interest is 7350 if comound interest is takenIf 7350 is the principal interest = 1225if 100 is the principal interest =(1225/7350) * 100 = 50/3 %When a thing increases for two successive times the overall increase on initial amount = a + b + (a * b)/100Therefore overall interest for two years = 50/3 + 50/3 + [(50/3)*(50/3)]/100 = 325/9 %Therefore amount after 2 years = 100 + 325/9 = 1225/9If 1225/9 is the amount principal =100if 7350 is the amount principal =(900/1225)*7350 =5400So sum = 5400; Rate = 50/3 % Answer : A
Question: Two cards are drawn together at random from a pack of 52 cards. What is the probability of both the cards being Queens?
A.1/221
B.1/712
C.1/224
D.1/212
Answer: A.1/221
Explanation: Total number of Queens = 4
Total number of cards = 52
P(Queen in first draw) = 4/52
P(Queen in second draw) = 3/51
=4/52*3/51=1/13*1/17=1/221
Answer A
Question: A printer print a page in 4 second,approx how long time it will take to print 190 page report?
A.12 Min 40 Sec
B.14 Min 40 Sec
C.16 Min 40 Sec
D.18 Min 40 Sec
Answer: A.12 Min 40 Sec
Explanation: 1 Page = 4 Sec
30 Page = 120 Sec (2 Min)
190 Page = 12 Min 40 Sec
ANSWER:A
Question: The prime minister of a country and 4 other dignitaries are scheduled to sit in a row on the 5 chairs represented above. If the prime minister must sit in the center chair, how many different seating arrangements are possible for the 5 people?
A. 4
B.24
C. 20
D. 21
Answer: B.24
Explanation: IMO B
Given that one chair is taken, i think the remaining 4 dignitaries can be arranged 4!. That is 4*3*2*1=24.
Question: A metallic sphere of radius 12 cm is melted and drawn into a wire, whose radius of cross section is 16 cm. What is the length of the wire?
A.9
B.8
C.7
D.6
Answer: A.9
Explanation: Volume of the wire (in Cylindrical shape) is equal to the volume of the sphere.
π(16)2 * h = (4/3)π (12)3 => h = 9 cm
Answer:A
Question: The first three terms of a proportion are 4, 9 and 12. The fourth term is?
A.24
B.25
C.26
D.27
Answer: D.27
Explanation: (9*12)/4 = 27
ANSWER:D
Question: The perimeters of 2 squares are 60cm and 48cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares?
A.20cm
B.28cm
C.36cm
D.48cm
Answer: C.36cm
Explanation: Side of the 1st square = 60/4 = 15cm
Side of the 2nd square = 48/4 = 12cm
Area of the 3rd square = 15^2 - 12^2 = 81cm^2
Side of the 3rd square = 9cm
Required perimeter = 9*4 = 36cm
Answer is C
Question: Compound interest of Rs.4000 at 10% per annum for 1 1/2years will be (interest compounded half yearly).
A.Rs.473
B.Rs.374
C.Rs.495
D.Rs.630
Answer: D.Rs.630
Explanation: 10% interest per annum will be 5% interest half yearly for 3 terms( 1 1/2 years)
So compound interest=4000[1+(5/100)]^3 - 4000=4000[(21/20)^3 -1]
=4000(9261-8000)/8000
=4*1261/8
=630
ANSWER:D
Question: This topic is locked. If you want to discuss this question please re-post it in the respective forum.
If y ≠ 3 and 3t/y is a prime integer greater than 2, which of the following must be true?
I. t = y
II. y = 1
III. t and y are prime integers.
A. None
B. I only
C. II only
D. III only
Answer: A. None
Explanation: 3t/y and if t=y then ans is 3 and 3 is prime number, which is greater than 2. so ans must be B. Please post OA.
Question: Vijay bought 160 shirts at the rate of Rs. 225 per shirt. The transport expenditure was Rs. 1400. He paid an octroi at the rate of Rs. 1.75 per shirt and labour charges were Rs. 320. What should be the selling price of one shirt, if he wants a profit of 50%?
A.282
B.229
C.356.25
D.285
Answer: C.356.25
Explanation: Total CP per shirt = 225 + 1400/160 + 1.75 + 320/160 = Rs. 237.5
SP = CP[(100 + profit%)/100]
= 237.5 * [(100 + 50)/100] = Rs. 356.25.
Answer:C
Question: A man can row his boat with the stream at 6 km/h and against the stream in 4 km/h. The man's rate is?
A.1 kmph
B.7 kmph
C.4 kmph
D.9 kmph
Answer: A.1 kmph
Explanation: DS = 6
US = 4
S = ?
S = (6 - 4)/2
= 1 kmph
Answer:A
Question: If a man walks to his office at ¾ of his usual rate, he reaches office 1/3 of an hour late than usual. What is his usual time to reach office?
A.1 hour
B.9 hour
C.7 hour
D.6 hour
Answer: A.1 hour
Explanation: Speed Ratio = 1:3/4 = 4:3
Time Ratio = 3:4
1 -------- 1/3
3 --------- ? è 1 hour
Answer:A
Question: If P * Q implies (P2 + Q2), 8 * (3 * 2) = ?
A.646
B.86
C.233
D.104
Answer: C.233
Explanation: Explanation:
We have, P * Q = (P2 + Q2) ......... Given
8 * (3 * 2) = ?
8 * 13 = (64 +169) = 233
ANSWER C
Question: At the end of the month, a certain ocean desalination plant’s reservoir contained 14 million gallons of water. This amount is twice the normal level. If this amount represents 70% of the reservoir’s total capacity, how many million gallons short of total capacity is the normal level?
A.2
B.5
C.10
D.12
Answer: C.10
Explanation: the Q talks of total capacity, normal level, present level, shortage etc..
so it is all about not going wrong in these terms
14 MG= 70% of Total..
total = 14/.7=20MG..
Normal level= 1/2 of 20=10MG..
shortage of normal level=20-10=10MG..
C
Question: The probability of a lottery ticket being a prized ticket is 0.2. When 4 tickets are purchased, the probability of winning a prize on atleast one ticket is
A.0.5928
B.0.5904
C.0.2788
D.0.2699
Answer: B.0.5904
Explanation: P(winning prize atleast on one ticket)
= 1 - P("Losing on all tickets")
= 1 - (0.8)4 = (1 + (0.8)2)(1 - (0.8)2)
= (1.64)(0.36) = 0.5904
Answer:B
Question: Right now, the ratio between the ages of Sandy and Molly is 4:3. After 6 years, Sandy’s age will be 66 years. What is Molly's age right now?
A.15 years
B.35 years
C.25 years
D.45 years
Answer: D.45 years
Explanation: Now, Sandy is 66 - 6 = 60
Molly's age is (3/4) * 60 = 45
The answer is D.
Question: Point X is located on line segment AB and point Y is located on line segment CD. If AB = CD and AX < CY, then
A.XB > YD
B.XB < YD
C.AX > XB
D.AX < XB
Answer: A.XB > YD
Explanation: A--X------B
C-----Y---D
XB < YD.
Answer: A.
Question: A bucket full of worms kept for Mike's next fishing trip was discovered by the rat that resides in the basement. The rat eats a sixth of the total number of worms in 5 hours. How much time will it take the rat to finish a quarter of the worms?
A.4 hours
B.6 hours
C.7.5 hours
D.8 hours
Answer: C.7.5 hours
Explanation: In 5 hrs, the rat eats 1/6 of the worms.
In 1 hr, it eats (1/6)/5 = 1/30 of the worms. This is his rate of eating.
Worms to be eaten = 1/4. This is the work that has to be done.
Work = Rate * Time
1/4 = 1/30 * Time
Time = 15/2 = 7.5 hrs
Answer (C)
Question: The area of sector of a circle whose radius is 12 metro and whose angle at the center is 41° is?
A.49m2
B.50m2
C.51.5m2
D.53m2
Answer: C.51.5m2
Explanation: 41/360 * 22/7 * 12 * 12
= 51.5 m2
Answer:C
Question: A man can row upstream at 25 kmph and downstream at 43 kmph, and then find the speed of the man in still water?
A.86
B.34
C.30
D.15
Answer: B.34
Explanation: US = 25
DS = 43
M = (43 + 25)/2
= 34
Answer: B
Question: If logx 4 = 1/4, then x is equal to
A.16
B.64
C.128
D.256
Answer: D.256
Explanation: Solution
log x 4	= 1/4
‹=› x 1/4
= 4
‹=›x= 44
= 256.
Answer D
Question: Siva Reddy walked 2 km west of his house and then turned south covering 4 km. Finally, He moved 3 km towards east and then again 1 km west. How far is he from his initial position?
A.10 km
B.9 km
C.2 km
D.4 km
Answer: D.4 km
Explanation: 4 km
ANSWER D
Question: Every year an amount increases by 1/8th of itself. How much will it be after two years if its present value is Rs.6400?
A.8100
B.3388
C.7767
D.2009
Answer: A.8100
Explanation: 6400* 9/8 * 9/8
= 8100
Answer: A
Question: A Jar contains a mixture of AB in the ratio 4:1. When 40 L of Mixture is replaced with liquid B, ratio becomes 2:3. How many liters of liquid A was present in mixture initially.
A.12
B.15
C.16
D.64
Answer: D.64
Explanation: 40 litres of mixture that is replaced will contain 32 litres of A and 8 litres of B (as A:B = 4:1)
Let the initial volume of the mixture be 4K + 1K = 5K
So by condition ,
[ 4K-32 ]/ [ K-8+40 ] = 2/3 => 12K -96= 2K-16+80 => 10K = 160
Solve for K which is K = 16
So initial volume of liquid A = 4K = 64 litres
Answer : D
Question: The average of 11 numbers is 10.8. If the average of first six is 10.5 and that of the last six is 11.4 the sixth number is?
A.9.5
B.12.6
C.10.5
D.11.5
Answer: B.12.6
Explanation: Explanation:
1 to 11 = 11 * 10.8 = 118.8
1 to 6 = 6 * 10.5 = 63
6 to 11 = 6 * 11.4 = 68.4
63 + 68.4 = 131.4 – 118.8 = 12.6
6th number = 12.6
Option B
Question: John have a 630ml mixture of milk and water in the ratio 7:2. How much water must be added to make the ratio 7:3?
A.40 m
B.50 m
C.70 m
D.90 m
Answer: C.70 m
Explanation: concentration of water in mixture1 =29 (since the ratio of milk and water = 7:2) ...(1)
concentration of water in pure water= 1 ...(2)
Now the above mentioned items are mixed to form mixture2 where milk and water ratio = 7 : 3
=> concentration of water in mixture2 =310
By rule of alligation,
concentration of water in mixture1 (29)
concentration of water in pure water (1)
Mean concentration
(310)
1−310=710
310−29=790
=> Quantity of mixture1 : Quantity of water
=710:790=110:190=1:19
Given that Quantity of mixture1 = 630 ml
=> 630 : Quantity of water =1:19
=> Quantity of water =630×19=70 ml
C