Question -1
There are 200 fishes in an aquarium out of which 99% fishes are Red,
How many fishes do we have to remove to make their percentage 98%.
Constraint:
Total Fishes = 200
Red Fishes = 99%
Blue Fishes = 1%
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Constraint:
Total Fishes = 200
Red Fishes = 99%
Blue Fishes = 1%
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Answer
Remove 100 red fishes
Explanation:
Red Fishes = 200 * 99% = 198
Blue Fishes = 200 * 1% = 2
Remove 100 Red Fishes
Now, Remaining Fishes = 100 out of which 98% are red.
Explanation:
Red Fishes = 200 * 99% = 198
Blue Fishes = 200 * 1% = 2
Remove 100 Red Fishes
Now, Remaining Fishes = 100 out of which 98% are red.
Question -2
You have given two ropes these ropes have special property that they will take 1 hour in burning.
Can you measure 45 minutes using these ropes?
Constraint:
Ropes are identical and each rope take 1 hour to burn completely. You can't cut them.
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Can you measure 45 minutes using these ropes?
Constraint:
Ropes are identical and each rope take 1 hour to burn completely. You can't cut them.
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Answer
Yes.
Explanation:
Burn rope 1 from one end.
Burn rope 2 from both the end, it burns completely in 30 minutes, when rope 2 is completely burned rope one will be half burned.
Now, set fire on other end of rope 1, this will burn rope 1 in 15 minutes.
Explanation:
Burn rope 1 from one end.
Burn rope 2 from both the end, it burns completely in 30 minutes, when rope 2 is completely burned rope one will be half burned.
Now, set fire on other end of rope 1, this will burn rope 1 in 15 minutes.
Question -3
An Arab Sheikh told his two Sons to Race their Camels to a distant city to see who will inherit his fortune.
The son whose camel will reaches the gate second will win.
After wandering aimlessly for days, Brothers asked a wise man for guidance.
Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man said to them?
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Upon receiving the advice, they jump on the camels and race to the city as fast as they can. What did the wise man said to them?
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Answer
The wise man told them to switch camels.
Explanation:
The great sage told them to swap camels. It is the second camel, not the person, to reach the city gates that wins fortune for its owner.
Explanation:
The great sage told them to swap camels. It is the second camel, not the person, to reach the city gates that wins fortune for its owner.
Question -4
There is a room with three identical light bulbs,
Their identical buttons are outside the room.
You have to find out which button belongs to which bulb, when you can visit the room only once?
Constraint:
You can go in room only once.
You can toggle buttons On/Off as many times as you want
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You have to find out which button belongs to which bulb, when you can visit the room only once?
Constraint:
You can go in room only once.
You can toggle buttons On/Off as many times as you want
Click to view the answer.
Answer
Switch on button one for 2 min now switch it off.
Turn the second button on and go to room.
Second button is on so lightening bulb belong to second button.
Now, touch remaining two bulbs one would be hotter that belong to first button that you have switched on first time. Now, you know third button belongs to remaining bulb.
Turn the second button on and go to room.
Second button is on so lightening bulb belong to second button.
Now, touch remaining two bulbs one would be hotter that belong to first button that you have switched on first time. Now, you know third button belongs to remaining bulb.
Question -5
Question:
You are in a dark room and there is a table in front of you on which 10 coins are placed.
You can touch the coins but can't tell, which way up are they, by feel.
You are told that there are 5 coins head up, and 5 coins tails up, but not which ones.
You are asked to divide coins in 2 groups such that both groups of coins will have same number of heads up?
Constraint:
Total 10 coins are placed on table.
5 are Heads up.
5 are Tails up.
You can flip the coins any number of times.
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Constraint:
Total 10 coins are placed on table.
5 are Heads up.
5 are Tails up.
You can flip the coins any number of times.
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Answer
Divide coins in a group of 5 coins and flip one of the group of coins once.
Explanation:
G1 = (H, H, T, T, T)
G2 = (T, T, H, H, H) => Flip (H, H, T, T, T)
G1 = (H, T, H, T, H)
G2 = (T, H, T, H, T) => Flip (H, T, H, T, H)
G1 = (T, T, T, T, T)
G2 = (H, H, H, H, H) => Flip (T, T, T, T, T) G1 = (H, H, H, H, H)
G2 = (T, T, T, T, T) => Flip (H, H, H, H, H)
Explanation:
G1 = (H, H, T, T, T)
G2 = (T, T, H, H, H) => Flip (H, H, T, T, T)
G1 = (H, T, H, T, H)
G2 = (T, H, T, H, T) => Flip (H, T, H, T, H)
G1 = (T, T, T, T, T)
G2 = (H, H, H, H, H) => Flip (T, T, T, T, T) G1 = (H, H, H, H, H)
G2 = (T, T, T, T, T) => Flip (H, H, H, H, H)
Question - 6
Question:
There was a pond with lotus, pond was magical.
Everyday pond has double lotus than previous day.
On day 15, pond was full of lotus.
How many days will it take to fill 25% of pond with lotus?
Constraint:
Each day, twice lotus grows.
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Constraint:
Each day, twice lotus grows.
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Answer
13
Explanation:
On 15th day, pond was full. Everyday pond has double lotus than previous day, Means, previous day pond has half lotus than today. So, on 14th day, pond would be half filled i.e. 50% And, on 13th day, pond would be half the 14th day i.e. 25% filled.
Explanation:
On 15th day, pond was full. Everyday pond has double lotus than previous day, Means, previous day pond has half lotus than today. So, on 14th day, pond would be half filled i.e. 50% And, on 13th day, pond would be half the 14th day i.e. 25% filled.
Question - 7
Question:
Maximize probability of White Ball.
There are two empty bowls in a room. You have 50 white balls and 50 black balls. After you place the balls in the bowls, a random ball will be picked from a random bowl. Distribute the balls (all of them) into the bowls to maximize the chance of picking a white ball.
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There are two empty bowls in a room. You have 50 white balls and 50 black balls. After you place the balls in the bowls, a random ball will be picked from a random bowl. Distribute the balls (all of them) into the bowls to maximize the chance of picking a white ball.
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Answer
First, let us assume that we divided the balls into jars equally so each jar will contain 50 balls.
So the probability of selecting a white ball will be=probability of selecting the first jar*probability of white ball in the first jar + probability of selecting the second jar*probability of white ball in the second jar
= (1/2)*(0/50)+(1/2)*(50/50)= 0.5
Since we have to maximize the probability so we will increase the probability of white ball in the first jar and keep the second probability same mean equal to 1
so we add 49 white balls with 50 black balls in the first jar and only one white ball in the second jar
so the probability will be now = (1/2)*(49/99)+(1/2)*(1/1) = 0.747
Therefore, probability of getting white ball becomes 1/2*1 + 1/2*49/99 which is approximately 3/4.
Therefore, probability of getting white ball becomes 1/2*1 + 1/2*49/99 which is approximately 3/4.
Question - 8
Question:
A snail was trying to climb a 30 feet wall.
every hour snail climbs up 3 feet then slide back 2 feet.
In how many hours snail will be on top of the wall?
Constraint:
Snail walks at constant speed
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Constraint:
Snail walks at constant speed
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Answer
28 hours.
Explanation:
Each hour snail, Climb 3 Feet Slide 2 Feet So, Total Climb = 3 - 2 = 1 feet / hour In 27 hours snail will be at 3 feet away from top. In next hour (28th hour) , Snail will be at top of wall. (Does not slide down from top)
Explanation:
Each hour snail, Climb 3 Feet Slide 2 Feet So, Total Climb = 3 - 2 = 1 feet / hour In 27 hours snail will be at 3 feet away from top. In next hour (28th hour) , Snail will be at top of wall. (Does not slide down from top)
Question - 9
Question:
In a country, all families want a boy. They keep having babies till a boy is born. What is the expected ratio of boys and girls in the country?
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Answer
Assumptions: Probability of having a boy or girl is same. Also, the probability of next kid being a boy doesn’t depend on history.
The problem can be solved by counting expected number of girls before a baby boy is born.
Let N be the expected no. of girls before a boy is born Let p be the probability that a child is girl and (1-p) be probability that a child is boy. N can be written as sum of following infinite series. N = 0*(1-p) + 1*p*(1-p) + 2*p*p*(1-p) + 3*p*p*p*(1-p) + 4*p*p*p*p*(1-p) +..... Putting p = 1/2 and (1-p) = 1/2 in above formula. N = 0*(1/2) + 1*(1/2)2 + 2*(1/2)3 + 3*(1/2)4 + 4*(1/2)5 + ... 1/2*N = 0*(1/2)2 + 1*(1/2)3 + 2*(1/2)4 + 3*(1/2)5 + 4*(1/2)6 + ... N - N/2 = 1*(1/2)2 + 1*(1/2)3 + 1*(1/2)4 + 1*(1/2)5 + 1*(1/2)6 + ... Using sum formula of infinite geometrical progression with ratio less than 1 N/2 = (1/4)/(1-1/2) = 1/2 N = 1
So Expected Number of number of girls = 1 Since the expected number of girls is 1 and as mentioned there is always a baby boy, the expected ratio of boys and girls is 50:50.
The problem can be solved by counting expected number of girls before a baby boy is born.
Let N be the expected no. of girls before a boy is born Let p be the probability that a child is girl and (1-p) be probability that a child is boy. N can be written as sum of following infinite series. N = 0*(1-p) + 1*p*(1-p) + 2*p*p*(1-p) + 3*p*p*p*(1-p) + 4*p*p*p*p*(1-p) +..... Putting p = 1/2 and (1-p) = 1/2 in above formula. N = 0*(1/2) + 1*(1/2)2 + 2*(1/2)3 + 3*(1/2)4 + 4*(1/2)5 + ... 1/2*N = 0*(1/2)2 + 1*(1/2)3 + 2*(1/2)4 + 3*(1/2)5 + 4*(1/2)6 + ... N - N/2 = 1*(1/2)2 + 1*(1/2)3 + 1*(1/2)4 + 1*(1/2)5 + 1*(1/2)6 + ... Using sum formula of infinite geometrical progression with ratio less than 1 N/2 = (1/4)/(1-1/2) = 1/2 N = 1
So Expected Number of number of girls = 1 Since the expected number of girls is 1 and as mentioned there is always a baby boy, the expected ratio of boys and girls is 50:50.
Question - 10
Question:
Look at this series: 7, 10, 8, 11, 9, 12, ... What number should come next?
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Click to view the answer.
Answer
10
Explanation:
This is a simple alternating addition and subtraction series. In the first pattern, 3 is added; in the second, 2 is subtracted.
Explanation:
This is a simple alternating addition and subtraction series. In the first pattern, 3 is added; in the second, 2 is subtracted.
Question - 11
Question:
What was the day of the week on 28th May, 2006?
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Click to view the answer.
Answer
Sunday
Explanation:
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days
Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days
148 days = (21 weeks + 1 day) 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.
Given day is Sunday.
Explanation:
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 6 odd days
Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days
148 days = (21 weeks + 1 day) 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.
Given day is Sunday.
Question - 12
Question:
Kick odd one out from following:
396, 462, 572, 427, 671, 264
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396, 462, 572, 427, 671, 264
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Answer
427
Explanation:
In each number except 427, the middle digit is the sum of other two.
Explanation:
In each number except 427, the middle digit is the sum of other two.
Question - 13
Question:
A Man bought an item of Rs 100 from the Shopkeeper(S1). He paid him a 500 Rs Note. Realizing that he did not have change, the shopkeeper got change for that note from another shopkeeper(S2) and paid Rs 400 to the Man.
After a few days, S2 realized that the note is fake, And this railed at S1 and took 500 Rs back from him.
So in this whole process how much money did S1 loose in the end?
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Click to view the answer.
Answer
500
Explanation:
The total loss for shopkeeper(S1) = 500
Consider a transaction, the Man came with a counterfeit 500 Rs note which can be considered of 0 value. Now the Man took the item (cost of the item 100 Rs ) and 400 Rs (the change given by shopkeeper(S1) to the Man) from the transaction, total of 500 Rs. Now the equivalent amount should be lost by someone, thus shopkeeper(S1) lost 500 Rs. Another shopkeeper(S2) gave 500 Rs and took back the same amount hence no loss for him.
Explanation:
The total loss for shopkeeper(S1) = 500
Consider a transaction, the Man came with a counterfeit 500 Rs note which can be considered of 0 value. Now the Man took the item (cost of the item 100 Rs ) and 400 Rs (the change given by shopkeeper(S1) to the Man) from the transaction, total of 500 Rs. Now the equivalent amount should be lost by someone, thus shopkeeper(S1) lost 500 Rs. Another shopkeeper(S2) gave 500 Rs and took back the same amount hence no loss for him.
Question - 14
Question:
Look at this series: 53, 53, 40, 40, 27, 27, ...
What number should come next?
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What number should come next?
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Answer
14
Explanation:
In this series, each number is repeated, then 13 is subtracted to arrive at the next number.
Explanation:
In this series, each number is repeated, then 13 is subtracted to arrive at the next number.
Question - 15
Question: This time look at both the letter pattern and the number pattern.
B2CD, _____, BCD4, B5CD, BC6D
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B2CD, _____, BCD4, B5CD, BC6D
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Answer
BC3D
Explanation: Because the letters are the same, concentrate on the number series, which is a simple 2, 3, 4, 5, 6 series, and follows each letter in order.
Explanation: Because the letters are the same, concentrate on the number series, which is a simple 2, 3, 4, 5, 6 series, and follows each letter in order.
Question - 16
Question: The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then find the value of x?
Answer
16
Explanation: Let Cost Price of each article be Rs.1
Cost Price of x articles = Rs. x.
Selling Price of x articles = Rs. 20.
Profit = Rs. (20 - x).
i.e. ((20 - x)/x * 100) = 25
2000 - 100x = 25x
125x = 2000
x = 16.
Explanation: Let Cost Price of each article be Rs.1
Cost Price of x articles = Rs. x.
Selling Price of x articles = Rs. 20.
Profit = Rs. (20 - x).
i.e. ((20 - x)/x * 100) = 25
2000 - 100x = 25x
125x = 2000
x = 16.
Question - 17
Question: Here are some words translated from an artificial language.
agnoscrenia means poisonous spider
delanocrenia means poisonous snake
agnosdeery means brown spider
Which word could mean "black widow spider"?
Answer
agnosvitriblunin
Explanation:
In this language, the noun appears first and the adjectives follow. Since agnos means spider and should appear first, choices a and d can be ruled out. Choice b can be ruled out because delano means snake.
Explanation:
In this language, the noun appears first and the adjectives follow. Since agnos means spider and should appear first, choices a and d can be ruled out. Choice b can be ruled out because delano means snake.
Question - 18
Question:
Here are some words translated from an artificial language.
gorblflur means fan belt
pixngorbl means ceiling fan
arthtusl means tile roof
Which word could mean "ceiling tile"?
Answer
pixnarth
Explanation:
Gorbl means fan; flur means belt; pixn means ceiling; arth means tile; and tusl means roof. Therefore, pixnarth is the correct choice.
Explanation:
Gorbl means fan; flur means belt; pixn means ceiling; arth means tile; and tusl means roof. Therefore, pixnarth is the correct choice.
Question - 19
Question:
Look at this series: 664, 332, 340, 170, ____, 89, ... What number should fill the blank?
Answer
178
Explanation:
This is an alternating division and addition series: First, divide by 2, and then add 8.
Explanation:
This is an alternating division and addition series: First, divide by 2, and then add 8.
Question - 20
Question:
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is?
Answer
21/46
Explanation:
Let S be the sample space and E be the event of selecting 1 girl and 2 boys. Then, n(S)= Number ways of selecting 3 students out of 25 = 25C3
= (25 x 24 x 23)/(3 x 2 x 1)
= 2300.
n(E) = (10C1 x 15C2)
= 10 x (15 x 14)/(2 x 1)
= 1050.
P(E) = n(E)/n(S)=1050/2300= 21/46
Explanation:
Let S be the sample space and E be the event of selecting 1 girl and 2 boys. Then, n(S)= Number ways of selecting 3 students out of 25 = 25C3
= (25 x 24 x 23)/(3 x 2 x 1)
= 2300.
n(E) = (10C1 x 15C2)
= 10 x (15 x 14)/(2 x 1)
= 1050.
P(E) = n(E)/n(S)=1050/2300= 21/46
