400 puzzle --2

Math gurus may use series formula to solve it.(series: 1,3,5,7,9,11.....upto 18250 terms)

A worker earns a 5% raise. A year later, the worker receives a 2.5% cut in pay, & now his salary is Rs. 22702.68

What was his salary to begin with?

Answer

Rs.22176

Assume his salary was Rs. X

He earns 5% raise. So his salary is (105*X)/100

A year later he receives 2.5% cut. So his salary is ((105*X)/100)*(97.5/100) which is Rs. 22702.68

Hence, solving equation ((105*X)/100)*(97.5/100) = 22702.68
X = 22176

At 6'o a clock ticks 6 times. The time between first and last ticks is 30 seconds. How long does it tick at 12'o.

Answer

66 seconds

It is given that the time between first and last ticks at 6'o is 30 seconds.
Total time gaps between first and last ticks at 6'o = 5
(i.e. between 1 & 2, 2 & 3, 3 & 4, 4 & 5 and 5 & 6)

So time gap between two ticks = 30/5 = 6 seconds.

Now, total time gaps between first and last ticks at 12'o = 11
Therefore time taken for 12 ticks = 11 * 6 = 66 seconds (and not 60 seconds)

500 men are arranged in an array of 10 rows and 50 columns according to their heights.

Tallest among each row of all are asked to come out. And the shortest among them is A.

Similarly after resuming them to their original positions, the shortest among each column are asked to come out. And the tallest among them is B.

Now who is taller A or B ?

Answer

No one is taller, both are same as A and B are the same person.

As it is mentioned that 500 men are arranged in an array of 10 rows and 50 columns according to their heights. Let's assume that position numbers represent their heights. Hence, the shortest among the 50, 100, 150, ... 450, 500 is person with height 50 i.e. A. Similarly the tallest among 1, 2, 3, 4, 5, ..... 48, 48, 50 is person with height 50 i.e. B

Now, both A and B are the person with height 50. Hence both are same.

In Mr. Mehta's family, there are one grandfather, one grandmother, two fathers, two mothers, one father-in-law, one mother-in-law, four children, three grandchildren, one brother, two sisters, two sons, two daughters and one daughter-in-law.

How many members are there in Mr. Mehta's family? Give minimal possible answer.

Answer

There are 7 members in Mr. Mehta's family. Mother & Father of Mr. Mehta, Mr. & Mrs. Mehta, his son and two daughters.

Mother & Father of Mr. Mehta
|
|
Mr. & Mrs. Mehta
|
|
One Son & Two Daughters

When Alexander the Great attacked the forces of Porus, an Indian soldier was captured by the Greeks. He had displayed such bravery in battle, however, that the enemy offered to let him choose how he wanted to be killed. They told him, "If you tell a lie, you will put to the sword, and if you tell the truth you will be hanged."

The soldier could make only one statement. He made that statement and went free. What did he say?

Answer

The soldier said, "You will put me to the sword."

The soldier has to say a Paradox to save himself. If his statement is true, he will be hanged, which is not the sword and hence false. If his statement is false, he will be put to the sword, which will make it true. A Paradox !!!

A person wanted to withdraw X rupees and Y paise from the bank. But cashier made a mistake and gave him Y rupees and X paise. Neither the person nor the cashier noticed that.

After spending 20 paise, the person counts the money. And to his surprise, he has double the amount he wanted to withdraw.

Find X and Y. (1 Rupee = 100 Paise)

Answer As given, the person wanted to withdraw 100X + Y paise. But he got 100Y + X paise. After spending 20 paise, he has double the amount he wanted to withdraw. Hence, the equation is         2 * (100X + Y) = 100Y + X - 20         200X + 2Y = 100Y +X - 20         199X - 98Y = -20         98Y - 199X = 20 Now, we got one equation; but there are 2 variables. We have to apply little bit of logic over here. We know that if we interchange X & Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y = 2X or Y = 2X+1 Case I : Y=2X Solving two equations simultaneously 98Y - 199X = 20 Y - 2X = 0 We get X = - 20/3 & Y = - 40/2 Case II : Y=2X+1 Solving two equations simultaneously 98Y - 199X = 20 Y - 2X = 1 We get X = 26 & Y = 53 Now, its obvious that he wanted to withdraw Rs. 26.53

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BrainVista Answer

The game of Tic-Tac-Toe is being played between two players. Only the last mark to be placed in the game as shown.

Who will win the game, O or X? Can you tell which was the sixth mark and at which position? Do explain your answer.
At the Party:

1.  There were 9 men and children.

2.  There were 2 more women than children.

3.  The number of different man-woman couples possible was 24. Note that if there were 7 men and 5 women, then there would have been 35 man-woman couples possible.

Also, of the three groups - men, women and children - at the party:

4.  There were 4 of one group.

5.  There were 6 of one group.

6.  There were 8 of one group.

Exactly one of the above 6 statements is false.

Can you tell which one is false? Also, how many men, women and children are there at the party

Assume that both the players are intelligent enough.

Answer

O will win the game. The sixth mark was X in square 9.

The 7th mark must be placed in square 5 which is the win situation for both X and O. Hence, the 6th mark must be placed in a line already containing two of the opponents marks. There are two such possibilities - the 6th mark would have been either O in square 7 or X in square 9.

As we know both the players are intelligent enough, the 6th mark could not be O in square 7. Instead, he would have placed O in square 5 and would have won.

Hence, the sixth mark must be X placed in square 9. And the seventh mark will be O. Thus O will win the game.

Answer

Statement (4) is false. There are 3 men, 8 women and 6 children.

Assume that Statements (4), (5) and (6) are all true. Then, Statement (1) is false. But then Statement (2) and (3) both can not be true. Thus, contradictory to the fact that exactly one statement is false.

So Statement (4) or Statement (5) or Statement (6) is false. Also, Statements (1), (2) and (3) all are true.

From (1) and (2), there are 11 men and women. Then from (3), there are 2 possible cases - either there are 8 men and 3 women or there are 3 men and 8 women.

If there are 8 men and 3 women, then there is 1 child. Then Statements (4) and (5) both are false, which is not possible.

Hence, there are 3 men, 8 women and 6 children. Statement (4) is false.

There is a shortage of tubelights, bulbs and fans in a village - Kharghar. It is found that

·   All houses do not have either tubelight or bulb or fan.

·   exactly 19% of houses do not have just one of these.

·   atleast 67% of houses do not have tubelights.

·   atleast 83% of houses do not have bulbs.

·   atleast 73% of houses do not have fans.

What percentage of houses do not have tubelight, bulb and fan?

Answer

42% houses do not have tubelight, bulb and fan.

Let's assume that there are 100 houses. Hence, there should be total 300 items i.e. 100 tubelights, 100 bulbs and 100 fans.

From the given data, we know that there is shortage of atleast (67+83+73) 223 items in every 100 houses.

Also, exactly 19 houses do not have just one item. It means that remaining 81 houses should account for the shortage of remaining (223-19) 204 items. If those remaining 81 houses do not have 2 items each, there would be a shortage of 162 items. But total of 204 items are short. Hence, atleast (204-162) 42 houses do not have all 3 items - tubelight, bulb and fan.

Thus, 42% houses do not have tubelight, bulb and fan.

Mr. Subramaniam rents a private car for Andheri-Colaba-Andheri trip. It costs him Rs. 300 everyday.

One day the car driver informed Mr. Subramaniam that there were two students from Bandra who wished to go from Bandra to Colaba and back to Bandra. Bandra is halfway between Andheri and Colaba. Mr. Subramaniam asked the driver to let the students travel with him.

On the first day when they came, Mr. Subramaniam said, "If you tell me the mathematically correct price you should pay individually for your portion of the trip, I will let you travel for free."

How much should the individual student pay for their journey?

Answer

The individual student should pay Rs. 50 for their journey.

Note that 3 persons are travelling between Bandra and Colaba.

The entire trip costs Rs. 300 to Mr. Subramanian. Hence, half of the trip costs Rs. 150.

For Andheri-Bandra-Andheri, only one person i.e. Mr. Subramaniam is travelling. Hence, he would pay Rs. 150.

For Bandra-Colaba-Bandra, three persons i.e Mr. Subramaniam and two students, are travelling. Hence, each student would pay Rs. 50.

Substitute digits for the letters to make the following Division true

                   O U T

           -------------

   S T E M | D E M I S E

           | D M O C

           -------------

               T U I S

               S T E M

              ----------

                 Z Z Z E

                 Z U M M

                --------

                   I S T

Note that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be

Answer

C=0, U=1, S=2, T=3, O=4, M=5, I=6, Z=7, E=8, D=9

It is obvious that U=1 (as U*STEM=STEM) and C=0 (as I-C=I).

S*O is a single digit and also S*T is a single digit. Hence, their values (O, S, T) must be 2, 3 or 4 (as they can not be 0 or 1 or greater than 4).

Consider, STEM*O=DMOC, where C=0. It means that M must be 5. Now, its simple. O=4, S=2, T=3, E=8, Z=7, I=6 and D=9.

                   O U T                        4 1 3

           -------------                -------------

   S T E M | D E M I S E        2 3 8 5 | 9 8 5 6 2 8

           | D M O C                    | 9 5 4 0

           -------------                -------------

               T U I S                      3 1 6 2

               S T E M                      2 3 8 5

              ----------                   ----------

                 Z Z Z E                      7 7 7 8

                 Z U M M                      7 1 5 5

                --------                     --------

                   I S T                        6 2 3

Also, when arranged from 0 to 9, it spells CUSTOMIZED.

At what time after 4.00 p.m. is the minutes hand of a clock exactly aligned with the hour hand?

Answer

4:21:49.5

Assume that X minutes after 4.00 PM minute hand exactly aligns with and hour hand.

For every minute, minute hand travels 6 degrees.
Hence, for X minutes it will travel 6 * X degrees.

For every minute, hour hand travels 1/2 degrees.
Hence, for X minutes it will travel X/2 degrees.

At 4.00 PM, the angle between minute hand and hour hand is 120 degrees. Also, after X minutes, minute hand and hour hand are exactly aligned. So the angle with respect to 12 i.e. Vertical Plane will be same. Therefore,

6 * X = 120 + X/2
12 * X = 240 + X
11 * X = 240
X = 21.8182
X = 21 minutes 49.5 seconds

Hence, at 4:21:49.5 minute hand is exactly aligned with the hour hand.

400 puzzle -- 1

Three friends divided some bullets equally. After all of them shot 4 bullets the total number of bullets remaining is equal to the bullets each had after division. Find the original number divided.

Answer

18

Assume that initial there were 3*X bullets.

So they got X bullets each after division.

All of them shot 4 bullets. So now they have (X - 4) bullets each.

But it is given that,after they shot 4 bullets each, total number of bullets remaining is equal to the bullets each had after division i.e. X

Therefore, the equation is
3 * (X - 4) = X
3 * X - 12 = X
2 * X = 12
X = 6

Therefore the total bullets before division is = 3 * X = 18

Find sum of digits of D.

Let
A= 1999¹⁹⁹⁹
B = sum of digits of A
C = sum of digits of B
D = sum of digits of C

(HINT : A = B = C = D (mod 9))

Answer

The sum of the digits od D is 1.

Let E = sum of digits of D.

It follows from the hint that A = E (mod 9)

Consider,

        A = 1999¹⁹⁹⁹

          < 2000²⁰⁰⁰

          = 2²⁰⁰⁰ * 1000²⁰⁰⁰

          = 1024²⁰⁰ * 10⁶⁰⁰⁰

          < 10⁸⁰⁰ * 10⁶⁰⁰⁰

          = 10⁶⁸⁰⁰

       

        i.e. A < 10⁶⁸⁰⁰

        i.e. B < 6800 * 9 = 61200

        i.e. C < 5 * 9 = 45

        i.e. D < 2 * 9 = 18

        i.e. E <= 9

        i.e. E is a single digit number.

        Also,

        1999 = 1 (mod 9)

        so 1999¹⁹⁹⁹ = 1 (mod 9)

Therefore we conclude that E=1.

There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.

In the mean time the whole platoon has moved ahead by 50m.

The question is how much distance did the last person cover in that time. Assuming that he ran the whole distance with uniform speed.
Submitted

Answer

The last person covered 120.71 meters.

It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backword - are equal.

Let's assume that when the last person reached the first person, the platoon moved X meters forward.

Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.

Similarly, while moving back the last person moved [50-(50-X)] X meters whereas the platoon moved (50-X) meters.

Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X

Solving, X=35.355 meters

Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters


Note that at first glance, one might think that the total distance covered by the last person is 100 meters, as he ran the total lenght of the platoon (50 meters) twice. TRUE, but that's the relative distance covered by the last person i.e. assuming that the platoon is stationary.

If you take a marker & start from a corner on a cube, what is the maximum number of edges you can trace across if you never trace across the same edge twice, never remove the marker from the cube, & never trace anywhere on the cube, except for the corners & edges?

Answer

9

To verify this, you can make a drawing of a cube, & number each of its 12 edges. Then, always starting from 1 corner & 1 edge, you can determine all of the possible combinations for tracing along the edges of a cube.

There is no need to start from other corners or edges of the cube, as you will only be repeating the same combinations. The process is a little more involved than this, but is useful for solving many types of spatial puzzles.

One of Mr. Bajaj, his wife, their son and Mr. Bajaj's mother is an Engineer and another is a Doctor.

·   If the Doctor is a male, then the Engineer is a male.

·   If the Engineer is younger than the Doctor, then the Engineer and the Doctor are not blood relatives.

·   If the Engineer is a female, then she and the Doctor are blood relatives.

Can you tell who is the Doctor and the Engineer?

Answer

Mr. Bajaj is the Engineer and either his wife or his son is the Doctor.

Mr. Bajaj's wife and mother are not blood relatives. So from 3, if the Engineer is a female, the Doctor is a male. But from 1, if the Doctor is a male, then the Engineer is a male. Thus, there is a contradiction, if the Engineer is a female. Hence, either Mr. Bajaj or his son is the Engineer.

Mr. Bajaj's son is the youngest of all four and is blood relative of each of them. So from 2, Mr. Bajaj's son is not the Engineer. Hence, Mr. Bajaj is the Engineer.

Now from 2, Mr. Bajaj's mother can not be the Doctor. So the Doctor is either his wife or his son . It is not possible to determine anything further.

Three men - Sam, Cam and Laurie - are married to Carrie, Billy and Tina, but not necessarily in the same order.

Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge. No wife partners her husband and Cam does not play bridge.

Who is married to Cam?

Answer

Carrie is married to Cam.

"Sam's wife and Billy's Husband play Carrie and Tina's husband at bridge."

It means that Sam is not married to either Billy or Carrie. Thus, Sam is married to Tina.

As Cam does not play bridge, Billy's husband must be Laurie.

Hence, Carrie is married to Cam.

There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many as they had. After a month Y gave as many tractors to X and Z as many as they have. After a month Z did the same thing. At the end of this transaction each one of them had 24.

Find the tractors each originally had?

Answer

One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing.

It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & Y as many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)

Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)

Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)

Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.

A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he need?

Answer

The sign-maker will need 192 zeroes.

Divide 1000 building numbers into groups of 100 each as follow:
(1..100), (101..200), (201..300), ....... (901..1000)

For the first group, sign-maker will need 11 zeroes.
For group numbers 2 to 9, he will require 20 zeroes each.
And for group number 10, he will require 21 zeroes.

The total numbers of zeroes required are
= 11 + 8*20 + 21
= 11 + 160 + 21
= 192

There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?

Answer

It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.

1.  Take 8 coins and weigh 4 against 4.

o If both are not equal, goto step 2

o If both are equal, goto step 3

2.  One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.

o If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.

§ If both are equal, L4 is the odd coin and is lighter.

§ If L2 is light, L2 is the odd coin and is lighter.

§ If L3 is light, L3 is the odd coin and is lighter.

o If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2

§ If both are equal, there is some error.

§ If H1 is heavy, H1 is the odd coin and is heavier.

§ If H2 is heavy, H2 is the odd coin and is heavier.

o If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4

§ If both are equal, L1 is the odd coin and is lighter.

§ If H3 is heavy, H3 is the odd coin and is heavier.

§ If H4 is heavy, H4 is the odd coin and is heavier.

3.  The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.

o If both are equal, there is some error.

o If X is heavy, X is the odd coin and is heavier.

o If X is light, X is the odd coin and is lighter.

In a sports contest there were m medals awarded on n successive days (n > 1).

1.  On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.

2.  On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.

3.  On the n^th and last day, the remaining n medals were awarded.

How many days did the contest last, and how many medals were awarded altogether?

Answer

Total 36 medals were awarded and the contest was for 6 days.

On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals
On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals
On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals
On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals
On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals
On day 6: Medals awarded 6

I got this answer by writing small program. If anyone know any other simpler method, do submit it.

A number of 9 digits has the following properties:

·   The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3, the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the first n digits is divisible by n, 2<=n<=9.

·   Each digit in the number is different i.e. no digits are repeated.

·   The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.

Find the number.

Answer

The answer is 381654729

One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.

The other way to solve this problem is by writing a computer program that systematically tries all possibilities

1/3 rd of the contents of a container evaporated on the 1st day. 3/4th of the remaining contents of the container evaporated on the second day. What part of the contents of the container is left at the end of the second day?

Answer

Assume that contents of the container is X

On the first day 1/3rd is evaporated.
(1 - 1/3) of X is remaining i.e. (2/3)X

On the Second day 3/4th is evaporated. Hence,
(1- 3/4) of (2/3)X is remaining
i.e. (1/4)(2/3)X = (1/6) X

Hence 1/6th of the contents of the container is remaining

Vipul was studying for his examinations and the lights went off. It was around 1:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last six hours and the thin one two hours less. When he finally went to sleep, the thick candle was twice as long as the thin one.

For how long did Vipul study in candle light?

Answer

Vipul studied for 3 hours in candle light.

Assume that the initial lenght of both the candle was L and Vipul studied for X hours.

In X hours, total thick candle burnt = XL/6
In X hours, total thin candle burnt = XL/4

After X hours, total thick candle remaining = L - XL/6
After X hours, total thin candle remaining = L - XL/4

Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep.
(L - XL/6) = 2(L - XL/4)
(6 - X)/6 = (4 - X)/2
(6 - X) = 3*(4 - X)
6 - X = 12 - 3X
2X = 6
X = 3

Hence, Vipul studied for 3 hours i.e. 180 minutes in candle light.

If you started a business in which you earned Rs.1 on the first day, Rs.3 on the second day, Rs.5 on the third day, Rs.7 on the fourth day, & so on.

How much would you have earned with this business after 50 years (assuming there are exactly 365 days in every year)?

Answer

Rs.333,062,500

To begin with, you want to know the total number of days: 365 x 50 = 18250.

By experimentation, the following formula can be discovered, & used to determine the amount earned for any particular day: 1 + 2(x-1), with x being the number of the day. Take half of the 18250 days, & pair them up with the other half in the following way: day 1 with day 18250, day 2 with day 18249, & so on, & you will see that if you add these pairs together, they always equal Rs.36500.

Multiply this number by the total number of pairs (9125), & you have the amount you would have earned in 50 years.

Ratio and proportion difficult questions

1.Three containers A,B and C are having mixtures of milk and water in the ratio of 1:5 and 3:5 and 5:7 respectively. If the capacities of the containers are in the ratio of all the three containers are in the ratio 5:4:5, find the ratio of milk to water, if the mixtures of all the three containers are mixed together.
Sol: Assume that there are 500,400 and 500 liters respectively in the 3 containers.
Then ,we have, 83.33, 150 and 208.33 liters of milk in each of the three containers.
Thus, the total milk is 441.66 liters. Hence, the amount of water in the mixture is 1400-441.66=958.33liters.
Hence, the ratio of milk to water is 441.66:958.33 => 53:115(using division by .3333)
The calculation thought process should be
(441*2+2):(958*3+1)=1325:2875
Dividing by 25 => 53:115.

2.A certain number of one rupee,fifty parse and twenty five paise coins are in the ratio of 2:5:3:4, add up to Rs 210. How many 50 paise coins were there?
Sol: the ratio of 2.5:3:4 can be written as 5:6:8
let us assume that there are 5 one rupee coins,6 fifty paise coins and 8 twenty-five paise coins in all.
their value=(5*1)+(6*.50)+(8*.25)=5+3+2=Rs 10
If the total is Rs 10,number of 50 paise coins are 6.
if the total is Rs 210, number of 50 paise coins would be 210*6/10=126.

3.The incomes of A and B are in the ratio of 4:3 and their expenditure are in the ratio of 2:1 . if each one saves Rs 1000,what are their incomes?
Sol: Ratio of incomes of A and B=4:3
Ratio of expenditures of A and B=2:1
Amount of money saved by A=Amount of money saved by B=Rs 1000
let the incomes of A and B be 4x and 3x respectively
let the expense of A and B be 2y and 1yrespectively
Amount of money saved by A=(income-expenditure)=4x-2y=Rs 1000
Amount of money saved by B=3x-y=Rs 1000
this can be even written as 6x-2y=Rs 2000
now solve 1 and 3 to get
x=Rs 500
therefore income of A=4x=4*500=Rs 2000
income of B=3x=3*500=Rs 1500

4.A sum of Rs 1162 is divided among A,B and C. Such that 4 times A's share share is equal to 5 times B's share and 7 times C's share . What is the share of C?
Sol: 4 times of A's share =5 times of B's share=7 times of C's share=1
therefore , the ratio of their share =1/4:1/5:1/7
LCM of 4,5,7=140
therefore, ¼:1/5:1/7=35:28:20
the ratio now can be written as 35:28:20
therefore C's share=(20/83)*1162=20*14=Rs 280.

5.The ratio of the present ages of saritha and her mother is 2:9, mother's age at the time of saritha's birth was 28 years , what is saritha's present age?
Sol: ratio of ages of saritha and her mother =2:9
let the present age of saritha be 2x years. then the mother's present age would be 9x years
Difference in their ages =28 years
9x-2x=28 years
7x=28=>x=4
therefore saritha's age =2*4=8 years

Time and Distance questions

Problems

1)A person covers a certain distance at 7kmph .How many meters
does he cover in 2 minutes.

Solution::
speed=72kmph=72*5/18 = 20m/s
distance covered in 2min =20*2*60 = 2400m

2)If a man runs at 3m/s. How many km does he run in 1hr 40min

Solution::
speed of the man = 3*18/5 kmph
= 54/5kmph
Distance covered in 5/3 hrs=54/5*5/3 = 18km

3)Walking at the rate of 4knph a man covers certain distance
in 2hr 45 min. Running at a speed of 16.5 kmph the man will
cover the same distance in.

Solution::
Distance=Speed* time
4*11/4=11km
New speed =16.5kmph
therefore Time=D/S=11/16.5 = 40min

Complex Problems

1)A train covers a distance in 50 min ,if it runs at a speed
of 48kmph on an average.The speed at which the train must run
to reduce the time of journey to 40min will be.

Solution::
Time=50/60 hr=5/6hr
Speed=48mph
distance=S*T=48*5/6=40km
time=40/60hr=2/3hr
New speed = 40* 3/2 kmph= 60kmph

2)Vikas can cover a distance in 1hr 24min by covering 2/3 of
the distance at 4 kmph and the rest at 5kmph.the total
distance is?

Solution::
Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km


3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr.
the usual time is.

Solution::
Usual speed = S
Usual time = T
Distance = D
New Speed is ¾ S
New time is 4/3 T
4/3 T – T = 5/2
T=15/2 = 7 ½

4)A man covers a distance on scooter .had he moved 3kmph
faster he would have taken 40 min less. If he had moved
2kmph slower he would have taken 40min more.the distance is.

Solution::
Let distance = x m
Usual rate = y kmph
x/y – x/y+3 = 40/60 hr
2y(y+3) = 9x --------------1
x/y-2 – x/y = 40/60 hr y(y-2) = 3x -----------------2

divide 1 & 2 equations
by solving we get x = 40

5)Excluding stoppages,the speed of the bus is 54kmph and
including stoppages,it is 45kmph.for how many min does the bus
stop per hr.

Solution::
Due to stoppages,it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min

6)Two boys starting from the same place walk at a rate of
5kmph and 5.5kmph respectively.wht time will they take to be
8.5km apart, if they walk in the same direction

Solution::
The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmph
Distance between them is 8.5 km
Time= 8.5km / 0.5 kmph = 17 hrs

7)2 trains starting at the same time from 2 stations 200km
apart and going in opposite direction cross each other ata
distance of 110km from one of the stations.what is the ratio of
their speeds.

Solution::
In same time ,they cover 110km & 90 km respectively
so ratio of their speed =110:90 = 11:9

8)Two trains start from A & B and travel towards each other at
speed of 50kmph and 60kmph resp. At the time of the meeting the
second train has traveled 120km more than the first.the distance
between them.

Solution::
Let the distance traveled by the first train be x km
then distance covered by the second train is x + 120km
x/50 = x+120 / 60
x= 600
so the distance between A & B is x + x + 120 = 1320 km

9)A thief steals a ca r at 2.30pm and drives it at 60kmph.the
theft is discovered at 3pm and the owner sets off in another car
at 75kmph when will he overtake the thief

Solution::
Let the thief is overtaken x hrs after 2.30pm
distance covered by the thief in x hrs = distance covered by
the owner in x-1/2 hr
60x = 75 ( x- ½)
x= 5/2 hr
thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm

10)In covering distance,the speed of A & B are in the ratio
of 3:4.A takes 30min more than B to reach the destion.The time
taken by A to reach the destinstion is.

Solution::
Ratio of speed = 3:4
Ratio of time = 4:3
let A takes 4x hrs,B takes 3x hrs
then 4x-3x = 30/60 hr
x = ½ hr
Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr

11)A motorist covers a distance of 39km in 45min by moving at a
speed of xkmph for the first 15min.then moving at double the
speed for the next 20 min and then again moving at his original
speed for the rest of the journey .then x=?

Solution::
Total distance = 39 km
Total time = 45 min
D = S*T
x * 15/60 + 2x * 20/60 + x * 10/60 = 39 km
x = 36 kmph

12)A & B are two towns.Mr.Fara covers the distance from A t0 B
on cycle at 17kmph and returns to A by a tonga running at a
uniform speed of 8kmph.his average speed during the whole
journey is.

Solution::
When same distance is covered with different speeds,then the
average speed = 2xy / x+y
=10.88kmph

13)A car covers 4 successive 3km stretches at speed of
10kmph,20kmph,30kmph&:60kmph resp. Its average speed is.

Solution::
Average speed = total distance / total time
total distance = 4 * 3 = 12 km
total time = 3/10 + 3/20 + 3/30 + 3/60
= 36/60 hr
speed =12/36 * 60 = 20 kmph

14)A person walks at 5kmph for 6hr and at 4kmph for 12hr.
The average speed is.

Solution::
avg speed = total distance/total time
= 5*6 + 4*12 / 18
=4 1/3 mph

15)A bullock cart has to cover a distance of 80km in 10hrs.
If it covers half of the journeyin 3/5th time.wht should be its
speed to cover the remaining distance in the time left.

Solution::
Time left = 10 - 3/5*10
= 4 hr
speed =40 km /4 hr
=10 kmph

16)The ratio between the speeds of the A& B is 2:3 an
therefore A takes 10 min more than the time taken by B to reach
the destination.If A had walked at double the speed ,he would
have covered the distance in ?

Solution::
Ratio of speed = 2:3
Ratio of time = 3:2
A takes 10 min more
3x-2x = 10 min
A's time=30 min
--->A covers the distance in 30 min ,if its speed is x
-> He will cover the same distance in 15 min,if its speed
doubles (i.e 2x)

17)A is twice as fast as B and B is thrice as fast as C is.
The journey covered by B in?

Solution::
speed's ratio
a : b = 2: 1
b : c = 3:1
Time's ratio
b : c = 1:3
b : c = 18:54
(if c covers in 54 min i..e twice to 18 min )

18)A man performed 3/5 of the total journey by ratio 17/20 by
bus and the remaining 65km on foot.wht is his total journey.

Solution::
Let total distance is x
x-(3/5x + 17/20 x) =6.5
x- 19x/20 = 6.5
x=20 * 6.5
=130 km

19)A train M leaves Meerat at 5 am and reaches Delhi at 9am .
Another train N leaves Delhi at 7am and reaches Meerut at 1030am
At what time do the 2 trains cross one another

Solution::
Let the distance between Meerut & Delhi be x
they meet after y hr after 7am
M covers x in 4hr
N covers x in 3 ½ i.e 7/2 hr
speed of M =x/4
speed of N = 2x/7
Distance covered by M in y+2 hr + Distance covered by N in
y hr is x
x/4 (y+2) +2x/7(y)=x
y=14/15hr or 56 min

20)A man takes 5hr 45min in walking to certain place and riding
back. He would have gained 2hrs by riding both ways.The time he
would take to walk both ways is?

Solution::
Let x be the speed of walked
Let y be the speed of ride
Let D be the distance

Then D/x + D/y = 23/4 hr -------1
D/y + D/y = 23/4 – 2 hr
D/y = 15/8 --------2
substitute 2 in 1
D/x + 15/8 = 23/4
D/x = 23/4 -15/8 =46-15/8 =31/8
Time taken for walk one way is 31/8 hr
time taken to walk to and fro is 2*31/8 = 31/4 hr
=7 hr 45 min

Trains

(1) Time taken by a train x mt long in passing a signal post
or a pole or a standing man = time taken by the train to cover x mt

(2) Time taken by a train x mt long in passing a stationary
object of length y mt = time taken by the train to cover x+y mt

(3) Suppose two trains or two bodies are moving in the same
direction at u kmph and v kmph such that u > v then their
relative speed is u-v kmph

(4)If two trains of length x km and y km are moving in opposite
diredtions at u kmph and vmph,then time taken by the train to
cross each other = (x+y)/(u+v) hr

(5) Suppose two trains or two bdies are moving in opposite direction
at u kmph and v kmph then,their relative speed = (u+v) kmph

(6)If two train start at the same time from 2 points A & B towards
each other and after crossing they take a & b hours in reaching B & A
respectively then A's speed : B's speed = (b^1/2 : a^1/2 )