Aptitude Profit and Loss Problems

Aptitude Profit and Loss Problems

7 Types Of Profit And Loss Problems To Gain Easy Marks In Bank Exams

Type I: Direct Formula Based Profit And Loss Percentages (Why This Is Easy?)

This type is very straightforward and is formula based. This is very easy because, you have to remember just 4 very simple formulas to solve this type.

Let CP be cost price of an item and SP be its selling price.
If SP is greater than CP, then there is profit in the transaction. Profit value and percentage can be calculated using below two formulas:

Profit = SP – CP
Profit Percentage = (Profit / CP) x 100%

If SP is lesser than CP, then there will be loss in the transaction. Loss value and percentage can be calculated using the below formulas.

Loss = CP – SP
Loss Percentage = (Loss / CP) x 100%

Now, let us see an example question based on type 1.

Example Question 1: Ram buys a book for Rs.100 and sells it for Rs.150. Find his gain or loss percentage.

Solution:
You can write down the CP and SP values from the question as follows:

Cost Price CP =Rs.100 and Selling Price SP = Rs.150

Here, SP is greater than CP. Therefore, there is profit in the transaction.

Based on formula, you know that Profit = SP – CP = 150 – 100 = Rs. 50

You also know the formula that ​Profit Percentage = (Profit / CP) x 100%

Therefore, Profit Percentage = (50 / 100) x 100% = (1/2)x100 % = 50%

Type II: Profit And Loss When Selling Different Varieties Of Same Item (Is This New To You?)

In this type, a seller will buy two (or more) varieties of an item at two different cost prices. Then he will sell them together (by mixing them) at common selling price.

You will understand this type clearly after reading the below example.

Example Question 2: Uma bought a number of roses at 4 for a rupee and an equal number at 2 for a rupee. At what price per dozen should she sell them to make a profit of 25%?

Solution:
Uma buys two varieties of roses. Type I at 4 roses per rupee and type II at 2 roses per rupee.

CP of 4 roses of type I = 1 and 

CP of 2 roses of type II = 1

Therefore, CP of 1 rose of type I = ¼ and

CP of 1 rose of type II = 1/2

Now assume that Uma had bought 1 dozen (12) roses of each variety.

Therefore, CP of 1 dozen roses of type I = ¼ x12 = 3 and

CP of 1 dozen roses of type II = 1/2 x 12 = 6

If Uma mixes 1 dozen of type I and 1 dozen of type II together,

CP of 2 dozen mixed roses = CP of 1 dozen roses of type I + CP of 1 dozen roses of type II

= 3 + 6 = Rs. 9

So, CP of 1 dozen mixed roses = 9/2 = Rs. 4.5

Let SP of 1 dozen mixed roses be X

You know that the Profit = SP – CP = X – 4.5

And Profit Percentage = Profit / CP x 100%

= (X – 4.5) / 4.5 x 100%

To answer the question, you have to find X value when profit percentage is 25. Therefore,

(X – 4.5) / 4.5 x 100 = 25

Or X – 4.5 = 25 x 4.5 / 100

Or X – 4.5 = 1.125

Or X = 5.625

Therefore, to make a profit of 25%, Uma has to sell the mixture at Rs. 5.625 per dozen

Type III: Same Selling Prize, Equal Profit And Loss Percentages (Why This Is Interesting?)

This is a very interesting type. Though this looks hard to solve, you can solve this type easily by using a super simple formula. Read on…

Assume that a vendor sells 2 items at same selling price. Also assume he makes profit in one transaction and loss in the other. Let the profit percentage in the first transaction be equal to the loss percentage in second transaction. In such case, overall there will be a loss. Type III deals with such problems.

You will understand this type after reading the below example.

Example Question 3: A man sold two bicycles at Rs.1500 each. He sold one at a loss of 23% and other at a profit of 23%. Find his profit or loss percentage.

Solution:
Whenever you see such problems where one is sold at x% loss and another at an equal x% profit, you can be sure that there will always be loss.

To calculate loss %, you can use the below shortcut formula:

If one item is sold at X% profit and other at X% loss and selling prices in both the transactions are equal, then

Loss % = (X/10)2 

In our example, the value of X is 23

Therefore, Loss percentage = (23/10)2 = 2.3 x 2.3 = 5.29

 

Type IV: Profit When Seller Is Not Honest And Uses False Weighing Stone Or Scale

If a seller (e.g., vegetable seller) uses false weighing stone (for example, 750 gram instead of 1 kilogram weighing stone), he will make higher profit compared to an honest seller, right?

Type IV is all about such dishonest sellers.
(Like type III, you can solve type IV questions using simple formula.)
Here is your example question:

Example Question 4: A seller uses a weighing stone of 900gms instead of 1 Kg. Find his real profit percent.

Solution:
You have to use below formula in such problems:

Real Profit % = Error / (True value – Error) x 100

Here, Error is the difference between weights of true weighing stone and the seller’s false weighing stone.
True Value denotes the correct weight of the stone (which an honest seller will use).

In question, you will see that the seller uses 900g weight instead of 1000g or 1Kg weight.

Therefore, Error = 1000 – 900 = 100

But a true weighing stone will be 1 Kg or 1000g.

Therefore, True value = 1000

If you apply above values in our Real Profit % formula, you will get

Real Profit % = 100 / (1000 – 100) x 100

= 100/900 x 100 = 11.11%

Type V: Multiple Transactions Based Profit And Loss Problems

In all the above types, you saw only one transaction. In the below example, you will find two or more continuous transactions. Now let us directly go to our example.

Example Question 5: Rahul sells a bicycle to Banu at a profit of 15%. Banu sells it to Sona at a profit of 20%. If sona pays Rs.3000 for it, then the cost price of the bicycle for Rahul is.

Solution:
First, assume CP of the bicycle when Rahul bought be Rs.X.

He sells it to Banu at profit of 15%. In other words, Banu buys the bicycle from Rahul by giving 15% more than Rahul’s CP.

Therefore, CP of bicycle to Banu = 15% more than CP of bicycle to Rahul

CP of bicycle to Banu = CP of bicycle to Rahul + 15/100 x CP of bicycle to Rahul 

= X + 15/100 x X

= X x (115/100) …equation 1

Banu sells it to Sona at a profit of 20%. In other words, Sona buys the bicycle from Banu by giving 20% more than Banu’s CP.

Therefore, CP of bicycle to Sona = 20% more than CP of bicycle to Banu

= CP of bicycle to Banu + 20/100 x CP of bicycle to Banu

= CP of bicycle to Banu x (120/100)

But you know from equation 1 that CP of bicycle to Banu = X x (115/100). If you substitute this in above equation, you will get:

CP of bicycle to Sona = X x (115/100) x (120/100) … equation 2

In question, you can see that Sona pays Rs 3000 for the bicycle.

Or, CP of bicycle to Sona = 3000

If you substitute above value in equation 2, you will get,

3000 = 120/100 x 115/100 x X

Or X = 3000 x 100/120 x 100/115 

Or X = Rs. 2173.91

Therefore, CP of bicycle to Rahul is Rs. Rs. 2173.91

Type VI: Marked Price And Discounts

In shops, you can see products with price mentioned on labels. This is called marked price (or printed price). If a seller gives discount on marked price, you will get this type VI problems.

Let us see an example for type VI.

Example Question 6: A vendor buys 30 pencils at the marked price of 25 pencils from a wholesaler. If he sells these pencils giving a discount of 2%, then what is his profit percentage ?

Solution:
First, let us assume that the marked price of each pencil be Rs.1

In question, you can see that the Vendor buys 30 pencils at the marked price of 25 pencils.

Therefore, CP of 30 pencils = Marked price of 25 pencils = 25 x 1 = Rs.25

Without discount, SP of 30 pencils = Marked price of 30 pencils = 30 x 1 = Rs. 30

But, the vendor sells these pencils at a discount of 2%.

Therefore, SP of 30 pencils = Marked price of 30 pencils – (2/100) of Marked price of 30 pencils

= Marked price of 30 pencils (1 – 2/100)

= Marked price of 30 pencils x 98/100

= 30 x 98/100

= Rs. 29.40

Therefore, his Profit = SP – CP

= 29.40 – 25 = Rs. 4.40

Profit % = Profit /CP x 100

= 4.40/25 x 100

= 17.6%

Type VII: Profit And Loss Problems With Ratio Calculations

If any of the above types is combined with ratio calculation, you will get this type VII. Truely, this type is an extension to any of the above types. Let us see an example, which is an extension to type VI with ratio calculation.

Example Question 7: A vendor earns a profit of 10% on selling a book at 15% discount on the printed price (marked price). The ratio of the cost price to the printed price of the book is.

Solution:

Let the CP be Rs.100.

Vendor earns a profit of 10%. Therefore his SP will be CP + 10% of CP

SP = 100 + 10% of 100

Or SP = Rs. 110 … equation 1

Let the printed price be Rs.X

From the question, you know that SP is printed price with 15% discount.

Or SP = Printed Price – 15% Printed Price

Or SP = X – (15/100) x X

Or SP = .85X … equation 2

From equations 1 and 2, you can write,

.85X = 110

or X = 110/.85 = 11000/85 = 2200/17

Based on our assumption that CP is 100, we have found that printed price will be 2200/17

Therefore, required ratio = CP : Printed price

=100 : 2200/17

To simplify the above ratio, you can multiply both the terms by 17. So the above ratio becomes,

1700 : 2200 

=17 : 22

Aptitude Pipes & Cisterns Problems

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Pipes & Cisterns Problems

Type I: Calculate Time Taken to Fill a Tank By 2 or More Pipes

This is the easiest type of pipes and cisterns problems. In this type, you will be given time taken by each pipe to individually fill a tank. You then have to find the time taken to fill the tank when all the pipes are opened together. Below is your example.

Example Question 1: Two pipes P and Q can fill a tank in 30 and 42 hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?

Solution:

You know that the pipe P takes 30 hours to fill the tank. You have to first calculate the portion of the tank filled in 1 hour. You can use the direct proportion table as shown below.

Hours    Portion of the tank

30    1 (1 represents full tank)

1    ?

In 1 hour, the portion of the tank filled by pipe P = 1 x 1/30 = 1/30 … value 1

By the same logic, in 1 hour, the portion of the tank filled by pipe Q =

1/42 … value 2

In 1 hour, the portion of the tank filled by both pipes P and Q together = value 1 + value 2

= 1/30 + 1/42 = 12/210 = 2/35

You can find the time taken by both the pipes together to fill the entire tank using direct proportion table as shown below.

Hours    Portion of the tank

1    2/35

?    1 (1 represents full tank)

Therefore, pipes P and Q can fill the tank in 1 x 1 / (2/35) = 35/2 hrs = 17 hours and 30 minutes.

Type II: Calculate Time Taken to Fill a Tank With Leakage

This is an extension to type 1. You have to find the time taken to fill tanks with leakages. Below example will help you to understand better.

Example Question 2: In a school, to fill a cistern, the authorities use two taps namely A and B. A can fill the tank in 20 hours and B can fill in 28 hours. The pipes are opened simultaneously and it is found that due to leakage in the cistern, it takes 20 minutes more to fill the cistern. If the cistern is full, then find the time taken by the leak to empty it.

Solution:

You have to solve this problem in 3 parts.

Part 1: Assume That There is No Leakage

Portion of the tank filled by pipe A in 1 hour = 1/20 … value 1

Portion of the tank filled by pipe B in 1 hour = 1/28 … value 2

Portion of the tank filled by pipes A and B together in 1 hour = value 1 + value 2

= (1/20 + 1/28)

= 12/140

= 6/70

Therefore, time taken to fill full tank by pipes A and B together = 1/(6/70)

= 70/6 hours … value 3

Note: Till the last step, every step is the same as in example 1.

Part 2: Include Leakage

In the question, it is given that due to leakage, the pipes will take extra 20 minutes (or 1/3 hours) to fill the tank.

(20 minutes = 20/60 hours = 1/3 hours)

Therefore, time taken to fill full tank by the pipes A and B together when there is leakage

= value 3 got in part 1 + 1/3 hours

= 70/6 + 1/3

= 72/6

= 12 hours

Part 3: Calculate Portion of Tank Emptied by Leakage Alone

Portion of tank filled by 2 pipes together with leakage in 1 hour = 1/12 … value 4

From part 1, you know that

Portion of the tank filled by 2 pipes together without leakage 1 hour = 

6/70 … value 5

Portion of the tank emptied by leakage alone in 1 hour can be found by subtracting value 4 from value 5

Portion of the tank emptied by leakage alone in 1 hour = 6/70 – 1/12

= (36 – 35) /420 

= 1/420

Therefore, leakage alone will empty the cistern in 420 hours.

Type III: Equations Based Pipes and Cistern Problems

In this type, you have to form equations based on conditions given in the question. Solving those equations will help you to find the answer. Here is your example.

Example Question 3: A tank can be filled in 10 hours by two taps Tap1 and Tap2. Tap2 is twice as fast as tap1. How much time will tap1 alone take to fill the tank?

Solution:

Assume that the tank can be filled by tap1 in X hours.

tap2 is twice as fast as tap1. Therefore, Then Tap2 can fill it in X/2 hours.

(Some candidates may make a mistake and write 2X hours instead of X/2 hours. If you have the same doubt, here is your explanation. When speed increases time decreases. Therefore, if tap2 is 2 times faster than tap1, then the time taken by tap2 will be half of that of tap1.)

Portion of the tank filled by tap1 in 1 hour = 1/X … value 1

Portion of the tank filled by tap2 in 1 hour = 2/X … value 2

Portion of the tank filled by both the taps in 1 hour = value 1 + value 2

= 1/X + 2/X … value 3

In the question, you can see that the taps together take 10 hours to fill the tank.

Therefore,

Portion of the tank filled by both the taps in 1 hour = 1/10 … value 4

As you can see, both the values 3 and 4 are same. Therefore, you can equate these values to find X.

i.e., 1/X + 2/X = 1/10

3/X = 1/10

X = 30 hours

So, tap1 alone will take 30 hours to fill the tank.

Type IV: Calculate Time Taken When Pipes Are Opened For Different Periods

In all the above types, all the pipes were open for the full duration. But what if a pipe is closed when other pipes are still open? For scenarios like this, you have to learn this type. Below example will help you.

Example Question 4: In a five-star hotel, there are three inlets namely P, Q and R, which can fill the tank in 10 hours. After working together for 5 hours, R is closed and inlets P and Q fill the remaining part in 12 hours. Find the time taken by R alone to fill the tank.

Solution:

From the question, you know that the 3 pipes (if opened together) will take 10 hours to fill the entire tank.

Portion of the tank filled by the pipes P, Q and R in 1 hour = 1/10

Part 1: First 5 Hours When All 3 Pipes Are Open

Portion of the tank filled by the pipes P, Q and R in 5 hours = 5 x 1/10 = ½

Remaining portion of tank to be filled = 1 – ½ = ½

Part 2: Second 5 Hours When Only P And Q Are Open 

P and Q take 12 hours to fill the remaining part

So, portion of the tank filled by the pipes P and Q in second 5 hours = 1/2

Portion of the tank filled by P and Q in 1 hour can be found using the direct proportion table shown below.

Hours    Portion of the tank

12    ½

1    ?

Portion of the tank filled by P and Q in 1 hour = 1 x ½ / 12 = 1/24

Part 3: Calculate Time Taken by R Alone to Fill the Tank

At the start of the solution, we found the below value.

Portion of the tank filled by the pipes P, Q and R in 1 hour = 1/10 …value 1

In part 2, we found the below value.

Portion of the tank filled by P and Q in 1 hour = 1/24 … value 2

If you subtract value 2 from value 1, you will get the portion of the tank filled by the pipe R in an hour.

Therefore, portion of the tank filled by R in 1 hour = 1/10 – 1/24 = 7/120

Therefore, time is taken by R to fill the entire tank = 1/(7/120) = 120/7 hours

= 17 hours 8 minutes (approximately)

Type V: Calculate Number of Pipes

Below is an example of this type.

Example Question 5: In a college hostel, the water tank is fitted with 6 pipes, in which some of them are used to fill the tank and others to drain the tank. Each of the filling pipes can fill the tank in 6 hours and each of the draining pipes can drain the tank in 4 hours. If all the pipes are opened together, the tank will drain in 4 hours. How many of the pipes are fill pipes?

Solution:

Assume that there are X fill pipes.

In the question, you can see that the total number of pipes is 6.

Therefore, will be (6-X) drain pipes.

Part 1: Calculate Values for Fill Pipes

Each fill pipe (alone) will take 6 hours to fill the tank.

So, portion of the tank filled by each fill pipe in 1 hour = 1/6

Therefore, portion of the tank filled by X fill pipes in 1 hour = X x 1/6 = 

X/6 … value 1

Part 2: Calculate Values for Drain Pipes

Each drain pipe (alone) will take 4 hours to drain the tank.

So, portion of the tank drained by each drain pipe in 1 hour = 1/4

Therefore, portion of the tank drained by (6-X) drain pipes in 1 hour = 

(6-X)/4 … value 2

Part 3: Final Solution

If all the pipes are opened together, then in 1 hour 1/4 of the tank gets drained.

Therefore, you can form the below equation.

Portion of the tank filled by X fill pipes in 1 hour – Portion of the tank drained by (6-X) drain pipes in 1 hour = -1/4

Note: Since the water is draining out you have to use “-” (minus) sign before 1/4

If you substitute value 1 (from part 1) and value 2 (from part 2) in above equation, you will get

X/6 – (6-X)/4 = -1/4 

4X – 36X +6X / 24 = -1/4

10x -36 = -6

10X = 30

X =3

Therefore, there are 3 fill pipes.

Aptitude Percentage Problems

Percentage Problems

Type I: Direct Simplification Type (You Can Skip If You Think This Type Is Easy)

This type is the easiest of all types of percentage problems. This is the same type of problems you would have seen in your school days. If you find this too easy, you can skip to next type.

Below is an example for type 1.

Example Question 1: 20% of 180 + ? = 10% of 350 + 5% of 120

Solution:

In above question, you have to find the missing value. Let us substitute X in place of question mark and start simplification as given below.

20% of 180 + X = 10% of 350 + 5% of 120

20/100 x 180 + X = 10/100 x 350 + 5/100 x 120

X = 10/100 x 350 + 5/100 x 120 – 20/100 x 180

X = 5

Type II: Salary Comparison Percentage Problems:

You will see this type often in bank exams. Though it looks tough, it is very easy if you use simple formula. You should remember the below two formulas.

Assume that there are two persons A and B and A’s salary is X% more than B’s salary.

Then B’s salary is lesser than A’s salary by X/(100+X) x 100%

Variant of the above formula:

If A’s salary is X% lesser than B’s salary,

then B’s salary will be more than A’s salary by X/(100-X) x 100%

Below is an example question using the above formula.

Example Question 2:

Ram’s salary is 30% more than Renu’s salary, by how much percent is Renu’s salary less than Ram?

Solution:

From the question you know that Ram’s salary is 30% more than that of Renu.

Therefore our value of X (to use in formula) = 30

Now, Renu’s formula is lesser than that of Renu by X/(100+X) x 100%

= 30/(100 + 30) x 100%

= 30/130 x 100%

= 23.07%

Type III: Appreciation And Depreciation Based Percentage Problems

As you know, appreciation refers to increase in value and depreciation means decrease in value. When appreciation or depreciation is given in percentage, then we get this type III.

There are two simple formulas you have to remember to solve these problems. They are as follows.

When a value Voriginal increases (i.e., appreciates) by R% per annum (i.e., per year), then the final value Vfinal after N number of years is given by

Vfinal = Voriginal(1 + R/100)N

When a value Voriginal decreases (i.e., depreciates) by R% per annum (i.e., per year), then the final value Vfinal after N years is given by

Vfinal = Voriginal(1 – R/100)N

You will understand this type clearly after seeing the below example.

Example Question 3: The population of the Chennai city in the year 2015 is 3,48,600. If it increases at the rate of 5% per annum, what will be its population in 2017?

Solution:

From the question, you can write down the below values to apply in our formula

Chennai’s population in 2015 = Voriginal = 3,48,600

Rate of increase of population = R = 5%

Now you have to find Vfinal after 2 years.

If you substitue the above values in the Vfinal = Voriginal(1 + R/100)N, you will get

Vfinal = 3,48,600(1 + 5/100)2

= 3,48,600(1 + 5/100)2

= 3,48,600 x 105/100 x 105/100

= 3,84,331.5

But you know population cannot be a fractional number. So we can approximate the above value to 3,84,332.

So you have now find out that the population of Chennai after 2 years (i.e in 2017) will be 3,84,332

Type IV: Price And Consumption Based Percentage Problems

Assume that every month you buy 10 Kgs of onions at 50 per Kg (i.e., you will be paying Rs. 500 in total). Due to lack of rainfall, assume that shopkeeper raises the price of onion to 60 per Kg.

Now, to buy 10 Kgs of onion you have to spend Rs. 600. But if you don’t want to spend the extra 100 rupees, you have to buy lesser than 10 Kg of onions. In other words, you have reduce your consumption (usage). If you are asked to calculate the reduction in consumption, you will get this type IV.

Below are two simple formulas to solve these problems. Though the concept is slightly different, the below formulas look exactly the same as that of type II formulas.

If price of an item increases by X%, then the reduction in consumption (use) so that expenditure (money spent) will not increase is X/(100+X) x 100%

Variant of the above formula:

If price of an item decreases by X%, then the increase in consumption (use) so that expenditure (money spent) will not decrease is X/(100-X) x 100%

Hint: In practice test (at the end of the tutorial), you may get problem using the second formula. But the example here uses first formula.

Let us now see an example:

Example Question 4: The price of oil increases by 10%. By how much percent must a person reduce his consumption so that his expenditure on it does not increase?

Solution:

From the question, you know that the increase in price of oil = X = 10%

If you apply the above value in the formula (that we saw above), you will get:

Reduction in consumption = [X / (100+X)] x 100%

= [10/(100+10)] x 100%

= 10/110 x 100%

= 9.09%

Type V: Set Theory Formula Based Percentage Problems

In school days when learning Venn Diagrams, you would have come across a diagram and formula as shown below:

n(A∪B) = n(A) + n(B)- n(A∩B)

Now, there can be some percentage problems which use the above formula. Below is an example:

Example Question 5: In an examination 30% of total students failed in Maths, 15% in Hindi and 5% in both. Find the percentage of those who failed in both the subjects.

Solution:

If you draw a simple venn diagram and substitute the values in question, you will get

n(A) = percentage of students who failed in maths = 30%

n(B) = percentage of students who failed in hindi = 15%

The shaded region represents the number of students failed in both.

Therefore, n(A∩B) = percentage of students who failed in both = 5%

n(A∪B) represents the number of students who failed at least in 1 subject. (i.e., n(A∪B) denotes students who failed in maths or hindi or both)

You know the formula that n(A∪B) = n(A) + n(B)- n(A∩B)

If you substitute values for n(A), n(B) and n(A∩B), you will get 

n(A∪B) = 30 + 15 -5 = 40%

Therefore, you have found that 40% of students have failed in at least 1 subject. 

Therefore 100 – 40 = 60% students would have passed in both the subjects.

Aptitude Clock Problems

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Clock Problems

Type 1: Finding Angle Between Minute And Hour Hands

This is the easiest type in clock problems. Two simple facts you should know to solve these problems are:

The hour hand finishes full rotation in 12 hours. That is, the hour hand traces 360 degrees in 12 hours.

The minute hand finishes full rotation in 60 minutes. That is, the minute hand traces 360 degrees in 60 minutes.

Now, lets us move to our example question…

Example Question 1: If the time is 5.30 am, what is the angle between minute and hour hands.

To solve this problem, you have to individually find the angles traced by hour hand and minute hand, and find the difference between the two.

Step 1. Angle for hour hand:

You know that the angle traced by hour hand in 12 hours is 360 degrees.

Now, as first step, you have to find the angle traced by hour hand at 5 hours 30 minutes or 5 1/2 hours or 11/2 hours.

Using direct proportion table below, you can easily find this angle. (You may know this simple method already)

Hours    Angle

12        360

11/2    x

Since, hours and angle are in direct proportion (i.e angle increases as hour increases and vice versa), we can write…

12 / (11/2) = 360 / x

Or x = (360 / 12) x (11 / 2)

x = 165 degrees

Therefore, angle traced by hour hand at 5.30 am will be 165 degrees

Step 2. Angle for minute hand:

You know that the angle traced by minute hand in 60 minutes is 360 degrees

Now, you have to find the angle traced by the minute hand in 30 minutes (This is because, we are finding the angle for 5.30 am and minute part in 5.30 am is 30 minutes)

Now, you have to use the direct proportion table for minute hand

Minutes      Angle

60          360

30           y

Since minutes and angle are in direct proportion, we can write,

60/30 = 360/y

Or y = 360 x 30 / 60 = 180 degrees

Therefore, angle traced by minute hand in 5.30 am will be 180 degrees

Step 3: Finding the difference

Therefore, angle between hour and minute hands in 5 hour 30 minutes will be the

difference between 165 and 180 degrees, which is 15 degrees.

Type 2: Finding Time If Angle Is Given

This type is the straight opposite of type 1 problems. In type 1, you have to find the angle if time is given. In this type, you will be given the angle and you will be finding the time.

Below 4 simple facts will help you to solve these types of problems:

If the hour and minute hands are in straight line and in same direction, the angle between them is 0 degrees or 360 degrees. And they are 0 minute spaces apart (they coincide).

If the hour and minute hands are in straight line and in opposite directions, the angle between them is 180 degrees. And they are 30 minute spaces apart.

If the angle between minute and hour hands is 90 degrees (right angle), then they will be 15 minute spaces apart.

Below data is based on measurement in clockwise direction:

Angle Between Minute And Hour Hand (in Degrees)    Minute , Space Between Minute And Hour Hands (in Minutes)

30 , 5

60 , 10

90 ,15

120 , 20

150 , 25

180    , 30

210    , 35

240    , 40

270    , 45

300    , 50

330    , 55

360    , 60

Minute hand gains 55 minute spaces on hour hand in 60 minutes. This point confuses many youngsters. If you are not able to understand this, here is your explanation for this 4th point:

Whenever minute hand completes one full rotation, the hour hand will be moving by 5 minute spaces. For example, if the time is 2 o clock. Now minute hand will be at 12 and hour hand will be at 2.

When the minute hand completes one full rotation and reaches number 12 again, the hour hand will be pointing at 3.

In other words, whenever small hand (hour hand) moves 5 minute spaces, the long hand (minute hand) moves 60 minute spaces. That is, it will gain or be ahead of hour hand by 60 – 5 = 55 minutes.

You will understand type 2 after reading the below example:

Example Question 2: Assume that the time now is 5 pm. At what time from now, minute hand will be ahead of hour hand by 90 degrees?

At 5 pm, the minute and hour hands will be 25 minute spaces apart. If you do not understand this, please check the below diagram.

For minute hand to move ahead of the hour hand by 90 degrees, it should gain 15 minute spaces on hour hand. (If you want to know why 90 degrees corresponds to 15 minute spaces, please read again the facts we discussed at the start of this type 2)

But the minute hand is behind the hour hand by 25 minute spaces. Therefore, first the minute hand must coincide with hour hand by covering 25 minute spaces and then it has to move ahead by 15 minute spaces.

So, it has to gain a total of 25 + 15 = 40 minute spaces.

We already know, minute hand gains 55 minute spaces on hour hand in 60 minutes. (check the 4th fact if not clear)

Using direct proportion table, we can calculate the minutes required for minute hand to gain 40 minute spaces

Minute Space Gain    Minutes

55    60

40    x

Here x is the number of minutes required for minute hand to move ahead of hour hand by 90 degrees.

55/40 = 60/x

Or x = 60 x 40 / 55

x = 2400/55 = 480/11 = 43 7/11 minutes.

Therefore, at 5 hour and 43 7/11 minutes, the minute hand will be ahead of hour hand by 90 degrees.

Note: In this question, we found the time when minute hand is ahead of hour hand by 90 degrees. But, if the question asks us to find the time when minute hand is behind hour hand by 90 degrees, we can repeat the same solution by calculating the time taken for minute hand to gain 25 – 15 = 10 minute spaces.

Type 3: Correct Time On Incorrect (Fast or Slow) Clocks

When some clocks are not perfect, they become either faster or slower than regular clocks. You may get problems based on incorrect clocks.

Below example will help you to solve these types of problems…

Example Question 3: Rohit buys a new clock and sets time to 5. pm (by seeing the correct time in a regular clock). But the new clock is faulty and it gains 20 minutes in 4 hours. After 3 days, Rohit sees that the faulty new clock is showing 9 pm. But, what will be the actual time in a regular correct clock?

Duration between 5 pm on day one to 9 pm on day 3 can be calculated as follows:

Time between 5 pm on day 1 to 5 pm on day 2 = 24 hours

Time between 5 pm on day 2 to 5 pm on day 3 = 24 hours

Time between 5 pm on day 3 to 9 pm on day 3 = 4 hours

Therefore, time between 5 pm on day 1 to 11 pm on day 3 = 24 + 24 + 4

= 52 hours or

52 x 60 = 3120 minutes

From the question, you know that hours and 20 minutes on faulty clock is same as 4 hours on regular clock.

To make calculations easier, we are converting above two times into minutes:

4 hours = 4 x 60 = 240 minutes and

4 hour 20 minutes = 4 x 60 + 20 = 260 minutes

Now you can calculate the minutes on a regular correct clock corresponding to 3120 minutes on the faulty clock using direct proportion table as shown below:

Minutes on faulty clock    Minutes on regular clock

260    240

3120    x

260/3120 = 240/x

x = 240 x 3120 / 260 = 2880 minutes

Therefore, 3120 minutes on faulty clock will be equal to 2880 minutes on regular clock

Hence, difference in minutes between regular clock and faulty clock at 9 pm on third day, will be

3120 – 2880 = 240 minutes or 4 hours

Therefore, regular clock will be 4 hours behind faulty clock and hence the correct time will be 9 – 4 = 5 pm.

Aptitude Average Problems

Average Problems

Type 1: Number Series Summation Based Averages: What Are They?

In number system topic you will see questions like, “find sum of first 25 natural numbers“, “find sum of squares of first 30 natural numbers“, etc. If these types of questions are combined with averages, you will get type 1 of average problems.

For example, below question falls under Type 1:

Example Question 1: Find the average of first 30 natural numbers:

To solve this, you have to know the formula for first n natural numbers, which is n(n+1)/2

Therefore, sum of first 30 natural numbers = 30 x (31)/2 = 465

Also, you know that average = sum of observations/ number of observations

Therefore, average of first 30 natural numbers = Sum of first 30 natural numbers / 30

= 465/30

= 15.5

View more number series formulas given below:

Sum of first n natural numbers = n(n+1)/2

Sum of squares of first n natural numbers = n(n+1)(2n+1)/6

Sum of cubes of first n natural numbers = (n(n+1)/2) 2

Sum of first n even natural numbers = n(n+1)

Sum of first n odd natural numbers = n2

Sum of n number of terms of a natural number series in which the difference between any two consecutive terms is same = n/2 (first term + last term)

Type 2: Consecutive Even/Odd Type Problems:

You can expect problems based on average of consecutive odd or even numbers. To be clear, let us see an example.

Example Question 2: Average of 4 consecutive odd numbers is 16. Find the second odd number in the series.

Let us assume four odd numbers to be x,x+2,x+4,x+6

Note: Here we are assuming the numbers to be x, x+2, x+4 and x+6 because, difference between any two consecutive odd numbers is 2. If the question is on consecutive even numbers, again you can assume in similar way because, difference between two consecutive even numbers is also 2. Therefore, assumption is same for both odd and even number series.

Average of the above four odd numbers = Sum of the 4 odd numbers / 4 = 16 …(16 is given in question)

Therefore, (x)+(x+2)+(x+4)+(x+6)/4 = 16

(4x+12)/4 = 16

4x+12=64

4x=52

x=13

Second odd number = x+2 = 13+2 = 15

Type 3 : Change In Average Based Problems

You will find problems based on change in average when a new person joins or leaves an existing group.

Below example will help you to understand this type fully.

Example Question 3: In a classroom at a cbse school in Jaipur, average weight of 5 students is 70 Kgs. The average increased to 73, after Rohit joined as a new student. What is the weight of Rohit?

Case I: Before Rohit Joined

Average weight of 5 students = Sum of weights of the 5 students / 5

70=Sum of weights of the 5 students/5

Or Sum of weights of the 5 students = 70 x 5 = 350 Kg

Case 2: After Rohit Joined

If you assume x to be Rohit’s weight, new average = Sum of 6 weights of students / 6

= (Sum of weights of old 5 students + Weight of Rohit) / 6

= (350 + x)/6 = 73 (In question, you will find 73 is the new average)

Or, 350 + x = 438

x = 88 Kgs = Rohit’s weight

Type 4: Multiple Groups Based Average Problems

You will get problems involving 2 or more groups, their individual averages and combined average of all the groups.

You will understand type 4 after reading the below example…

Example Question 4: A total of 35 soccer players is divided into two teams of 15 and 20 players. The average weight of the first team is 60 Kg and that of the second team is 70 Kg. The team manager wants to know the average weight of the whole team. Help him to find the answer.

You know the formula for average = Sum of observations / Number of observations

Therefore, Sum of observations = Average x Number of observations

You will be using the above formula for this example.

Now consider group 1:

Number of members in group 1 = 15

Average weight for group 1 = 60

Sum of weights of all 15 members in group 1 = 15 x 60 = 900

Now consider group 2:

Number of members in group 2 = 20

Average weight for group 2 = 70

Sum of weights of all 20 members in group 2 = 20 x 70 = 1400

Now, average of the entire group = Sum of weights of all 35 members / 35

= Sum of weights of all members of group 1 + Sum of weights of all members of group 2 / 35

= 900 + 1400 / 35

= 2300/35

= 65.71

Type 5: Distance And Speed Based Averages

In bank exams, you may also get average questions based on distance and speed.

Example Question 5: A man travels by motorcycle from his home to office. He covers his first half of journey at 40 km/hr and realizes he is late. He then increases his speed by 50% for his remaining journey. Find his average speed for the whole distance (from home to office).

The man covers first half of his journey at 40 km/hr.

He increases his speed by 50% for remaining journey. Therefore, speed for his remaining journey = (50% of 40 + 40)

= 60km/hr.

You can solve this problem using two methods: Direct method and Shortcut method.

Here is your direct method:

Average speed = Total distance / Total time

Let the total distance for the whole journey is 2d

Then, first half distance = d and Second half distance = d

Time taken to travel first half = First half distance / Speed for first half journey

= d/40

Time taken to travel second half = Second half distance / Speed for second half journey

= d/60

Total time taken = d/40 + d/60

Now you can apply the above values in the below formula:

Average speed = Total distance / Total time

= 2d / (d/40 + d/60)

= 2 / (1/40 + 1/60)

= 2 / (10/240)

= 480/10

= 48 Kmph

Now, here is your shortcut method:

For such problems, you can use the below shortcut formula:

If x is the speed for one half of the journey and y is the speed for remaining half

The average speed for the whole journey = 2xy/x+y

In our case, x = 40 kmph and y = 60 kmph

Therefore, average speed = 2xy/x+y

= 2 x 40 x 60 / (40 + 60)

= 4800/100 = 48 kmph.

Aptitude Age Problems

Aptitude Age Problems

3 Easy Types Of Age Problems In Which You Should Not Miss Even One Mark

Type 1: Ratio Based Age Problems

In this type, you will be given ratios between ages in question. You then have to then find the present ages of the people. In some cases, you may also be asked to find past or future ages.

This the easiest type of age problems. You will know why after reading the below example.

Example Question1: Ratio between ages of Rahul and Ravi is 3/5. After 10 years, the ratio will become 2/3. Find the present ages of Rahul and Ravi.

To solve this, let us assume Rahul’s age to be x and Ravi’s age to be y.

You can then write, x/y = 3/5

Or, 5x=3y

Or, x=3y/5 ….equation 1

In question, you can find that after 10 years the ratio becomes 2/3.

(After 10 years, Rahul’s age will become x+10

and Ravi’s age will become y+10)

Therefore You can write, (x+10)/(y+10)=2/3

Or 3x+30 = 2y+20

Or 2y-3x=10 … equation 2

But, you already know from equation 1 that x = 3y/5. If you substitute this equation 2, you will get

2y-3(3y/5)=10

Or 2y-9y/5=10

Or 10y-9y/5=10

y=50

If you substitute, y = 50 in equation 1, you will get x = 3 x 50/5 = 30

Therefore, your answer is Rahul’s age = 30 and Ravi’s age = 50

This type is very easy, isn’t it? If you have doubts, please use the comments section at the end of the article.

Type 2: Equation Solving Type Age Problems

This type of age problems is probably the most important. You will find the reason after the solution to the below example.

Example Question 2: In 5 years, Reshma will be 2 times older than Satya. Renuka, who is Reshma’s sister is 5 years younger than Reshma. Before 5 years, Renuka was 3 times older than Satya. Find present age of Satya.

To solve this question, let us assume Satya’s age to be x, Reshma’s age to be y and Renuka’s age to be z.

In 5 years, Reshma will be 2 times older than Satya. Therefore, you can write

y+5=2(x+5)

Or y-2x=5

Or y=2x+5…equation 1

You know that Renuka is 5 years younger than Reshma. Therefore, you can write

z=y-5 ….equation 2

Also, from question, you can find that before 5 years, Renuka was 3 times older than Satya. So,

z-5=3(x-5)

Or z-3x=20

Or z=3x-10 …equation 3

If you substitute the value of y from equation 1 in 2, you will get

z=2x

Now you have to substitute z=2x in equation 3. So you will get,

2x=3x-10

Or x=10

Therefore, you have found the answer that Satya’s age = 10

Why This type is very important? (Why You Should Not Lose Concentration)

A questioner can twist these types of questions to any extent. Therefore, they are very important. You have to clearly understand the question to form correct equations. You also should not make any careless mistakes at the start or middle. Or you have to rework the solution from the start.

Type 3: Finding Ratio Between Ages 

If in type 2 (previous type), you are asked to find ratio between ages instead of ages, you will get type 3.

In example question 2, let us assume you are asked to find the ratio between ages of Satya and Reshma. Here is your solution…

From the solution to example question 2, you know x = Satya’s age = 10

If you substitute, x = 10 in equation 1 of example 2, you will get

Reshma’s age = y = 25

Aptitude CI And SI

CI And SI

Simple Interest (SI)

Principal: - The money borrowed or lent out for certain period is called the principal or the Sum.

Interest: - Extra money paid for using other money is called interest.

If the interest on a sum borrowed for certain period is reckoned uniformly, then it is called simple interest.

Let Principal = P, Rate = r % per annum (p.a.), and Time = t years then

Simple Interest(SI)= ((P×r×t))/100

Using this formula we can also find out

P=(100×SI)/(r×t);

r=(100×SI)/(P×t);

t=(100×SI)/(P×r).

Compound Interest:

When compound interest is applied, interest is paid on both the original principal and on earned interest.

So for one year Simple interest and Compound interest both are equal.

Suppose if you make a deposit into a bank account that pays compounded interest, you will receive interest payments on the original amount

that you deposited, as well as additional interest payments.

This allows your investment to grow even more than if you were paid only simple interest.

So Amount at the end of 1st year (or Period) will become the principal for the 2nd year (or Period) and

Amount at the end of 2nd year (or Period) becomes the Principal of 3rd year.

Amount = Principal + Interest

A= P (1+r/100) ^n

A= Amount,

P= Principal,

r= Rate %,

n= no. of years.

So Compound Interest = [P (1+r/100) ^ n - P]

= P [(1+r/100) ^ n – 1]

Condition:-

1.When interest is compounded annually,

Amount = P(1+r/100)^n

2.When interest is compounded half yearly,

Amount = P(1+(r/2)/100)^2n

3.When interest is compounded Quarterly,

Amount =P(1+(r/4)/100)^4n

4.When interest is compounded annually but time is in fraction, say 3 whole 2/5 year

Amount = P(1+r/100)^3×(1+(2r/5)/100)

5.When Rates are different for different years, say r1%, r2%, and r3% for 1st, 2nd and 3rd year respectively.

Then,

Amount = P(1+r1/100)×(1+r2/100)×(1+r3/100).

Present worth of Rs. x due n years hence is given by:

Present Worth = x/(1+r/100)

Difference between Compound Interest & Simple interest Concept For Two years

CI – SI =P(r/100)^2

For Three Year

CI – SI =P(r^2/(100^2 ))×(300+r)/100)

For Two year

CI/SI=(200+r)/200

Quant Quiz for Simple Interest & Compound Interest

1. A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:

A) Rs. 720

B) Rs. 698

C) Rs. 678

D) Rs. 696

E) none of these

2. A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 % p.a. in 5 years. What is the sum?

A) Rs. 8045

B) Rs. 8925

C) Rs. 8900

D) Rs. 8032.45

E) none of these

3. A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is:

A) 12 %

B) 13 %

C) 8 %

D) 12.5 %

4. A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 6.25% p.a. for 2 years.Find his gain in the transaction per year.

A) Rs. 112.50

B) Rs. 175

C) Rs. 150

D) Rs. 125.50

5. A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period.The principal amount borrowed by him was:

A) Rs. 12000

B) Rs.15000

C) Rs. 12500

D) Rs. 22000

6.How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?

A)3 year

B)4 year

C)5 year

D)6 year

7. Bhavika took a loan of Rs. 1200 with simple interest for as many years as the rate of interest.If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

A)3.6

B) 5

C) 6

D)25

8. A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:

A) 5 %

B) 7%

C)10 %

D) 12%

9.A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1st January and 1st July of a year.

At the end of the year, the amount he would have gained by way of interest is:

A)123

B) 122

C)121

D)120

10.The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is:

A)2.5

B) 2

C) 3

D) 4

E) none of these

11.At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?

A)8 %

B) 9%

C) 6 %

D) 8.5 %

E) none of these

12.The difference between simple interest and compound on Rs. 1200 for one year at 10% per annum reckoned half-yearly is:

A)Rs. 3

B) Rs. 4

C) Rs. 3.5

D) Rs. 7.5

E) none of these

13.The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:

A) 4

B) 5

C) 6

D) 2.5

E) none of these

14.What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.?

A) Rs.10123.20

B) Rs. 9000

C) Rs. 12000

D) Rs. 10163.34

E) none of these

15.Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is:

A)Rs. 1650

B)Rs. 2000

C)Rs. 1750

D) Rs.1550

E) none of these

Answers with Explanation:

Anwers:

1.B

2.B

3.A

4.A

5.B

6.B

7.C

8.C

9.C

10.B

Aptitude Time, Speed & Distance

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Time, Speed & Distance

Suppose, Delhi to Agra is 120 km. And my motorcycle covers 40 km in one hour. So, how much time I will take to reach Agra?

Simple! 3 hrs. time.

But my friend's car covers 60 km in an hour. He will take how much time?

Simple! 2 hrs. time.

Means to say, my friend will reach Agra 1 hour before me.

So, keeping the distance constant, we got two times for two speeds. The time taken is inversely proportional to speed.


Basic formula we used here for calculation of time taken is:

Time taken = Distance/Speed

And using this formula, we can calculate speed, or, distance, if two other things are known

Speed = Distance/Time

Distance = Speed * Time


Feel this in mind before we go further...


Let’s now entertain the concept of average speed!

Question: I travel half of my journey by Bus with speed of 60kmph and the rest half on my friend's motorcycle with speed of 80kmph. What is my average speed of total journey?

Average speed is that speed which covers the total distance in the total time (that is, the total time taken to cover the distance if I go by variable speeds)

Average speed = Total distance / Total time taken

Now, here in this question, 'speed' is variable (means changing). Distance is taken constant. So, Time taken will also be variable depending upon the speeds.

Time = Distance/speed, T= D/S

Let total distance be 2D, so that for 1st speed we have half distance 'D', and for second speed we have second half distance 'D'

S1 = 60kmph
S2 = 80kmph

So, we have

T1 = D/60
T2 = D/80

Now, average speed = total distance / total time

Total distance = distance for 1st time + Distance for 2nd time = 2D
Total time = D/60 + D/80 = [7D/240]

Average speed will be = [2D] / ( [7D/240] ) = 480/7 kmph


Let’s derive this formula

Let 1st speed (60) = X
Let 2nd speed (80) = Y

T1 = D/X
T2 = D/Y

Total Distance = 2D
Total time = D/X + D/Y = [Y+X]*D/[XY]

Average Speed will be = [2D] / ( [Y+X]*D/[XY] ) = [2XY]/[X+Y]

Note: This formula we have derived taking the distances for both the speed as equal. So, if in questions, distances varies, this formula will fail to be applicable.


If you can remember the formula, then its fine, but if not, it’s still is fine. Problem is to just find the average speed. Our suggestion is to stick to basic concepts.


Now let's test you:

Quiz Ques 1: Uday travels one third of its journey by train with speed of 60kmph and the rest of journey by car with speed of 80kmph. Find his average speed of his journey?

(Answer to this and all quiz questions later)


The distance of the college and home of Rajeev is 80km. One day, he was late by 1 hour than normal time to leave for college, so he increased his speed by 4kmph and thus he reached to college at the normal time. What is the changed speed of Rajeev?

To solve this question, first feel what the question is saying.

Distance is 80km. It is constant. Only speed is changed.

Now, let’s say his normal speed is X kmph, Then he will reach the college in 80/X hour time. (Equation 1)

With [X+4] speed, he will reach the college in 80/[X+4] hour time. (Equation 2)

Now the question says, he is late 1 hour but with X+4 speed, he reaches the college on time.

That means time in (Equation 1) must be 1 hour more than the time in (Equation 2)

Quiz Ques 2: Can you solve further and find the increased speed of Rajeev?


Let’s now solve a very good question which will clear many concepts in a single run!!


The distance between two places P and Q is 700km. Two persons A and B started towards Q and P from P and Q simultaneously. The speed of A is 30kmph and speed of B is 40kmph. They meet at a point M which lies on the way from P to Q.

(i) How long will they take to meet each other at M?

(ii) What is the ratio of PM : MQ?

(iii) What is the distance MQ?

(iv) What is the extra time needed by A to reach at Q than to reach at P by B?

(V) What is the ratio of time taken by A and B to reach their respective destinations after meeting at M?

(vi) In how many hours will they be separated by only 560 km before meeting each other?

(vii) How long will it take to separate then by 280 km from each other when they cross M (time to be considered after their meeting)?

The concept of relative motion is entertained here.

By relative motion, we means the motion of one thing with respect to another thing. Suppose you're sitting on the pillion seat behind the motorcycle of your friend who is driving the motorcycle at 40kmph, then the relative speed of you with respect to motorcycle (or your friend) will be zero because for your friend, you are not moving an inch. But with respect to a person selling ice cream in the corner shop, your relative speed will be 40kmph, because for him, you're moving with a speed of 40kmph.

Now, you steal his ice cream, and get ahead. He also had a bike and he's now driving his bike behind you with a speed of 50kmph. Will he catch you?

Of course, he will. Coz now, the relative speed of him is 10kmph more with respect to you. He will catch you sooner.

The concepts when put mathematically is this:

If two bodies A and B are moving with speed Sa and Sb, then relative speed will be

Sa - Sb, if they're moving in the same direction, and

Sa + Sb, if they're moving in the opposite direction.


(i) Now apply this concept.

A and B, both are moving in opposite direction with speeds of 30kmph and 40kmph. So, their relative speed will be?

Ans: 30+40 = 70kmph.

They will take how much time to reach at point M?

They will cover total Distance = 700km / with speed of 70 kmph = will take Time = 10 hour to reach at point M


Understand this before you go further to solve the rest of questions!


(ii) It took 10 hour by both of them to reach at M.

With speed 30kmph, A has covered 30*10 = 300km = PM
With speed 40kmph, B has covered 40*10 = 400km = MQ

Ratio of distances PM : MQ = 300 : 400 = 3 : 4

Note: know this that if time is taken constant, the ratio of distances will be equal to the ratio of their speed. (Just because distance = speed * time)

How?

D1 = S1*T1
D2 = S2*T2

T1 = T2

------> D1/D2 = S1/S2


(iii) Distance MQ = 400 km


(iv) Time taken by A to reach Q = distance/speed = 700km/30kmph = 70/3 hour
Time taken by B to reach P = distance/speed = 700km/40kmph = 70/4 hour

Extra time taken by A will be ---- 70/3 - 70/4 = 70/12 hour


Understand this before you go further!!


(v) When A has reached at point M, A has covered 300km (because PM = 300km) and B has covered 400km (because MQ = 400km). Now, A has to cover 400km more and B has to cover 300km more. So,

Time Ta taken by A to cover MQ = distance MQ/speed = 400/30 hour
Time Tb taken by B to cover PM = distance PM/speed = 300/40 hour

Ratio of their time = Ta/Tb = [400/30]/[300/40] = 16/9


If derived (just like we've solved), we will get to know that this ratio [Ta/Tb] of their time is the ratio of the reciprocal of squares of their speed.

Why not we derive this?

Suppose A travels X km with speed Sa and B travels 700-X km with speed Sb and reaches point M in time T.

Time taken to reach point M will be equal.

i.e. [X]/Sa = [700-X]/Sb
i.e. [700-X]/[X] = [Sb/Sa]

Now, for A, rest distance to cover is 700-A with speed Sa, and for B, rest distance to cover is X with speed Sb, they will take time Ta and Tb to reach their destinations.

Ta = [700-X]/Sa
Tb = [X]/Sb

So, ratio of their times will be = Ta/Tb = ([700-X]/Sa) / ([X]/Sb) = ([700-X]/[X]) * ([Sb/Sa])

But we know that [700-X]/[X] = [Sb/Sa]

So, putting this in equation, we gets, Ta/Tb = ([Sb/Sa]) * ([Sb/Sa])

i.e. Ta/Tb = square of [Sb/Sa] ------ (note Sb/Sa and not Sa/Sb)

Sb = 40 kmph, Sa = 30kmph  Ta/Tb = square of [40/30] = square of [4/3] = 16/9


(vi) They will be separated by only 560 Km if they have covered 700-560 = 140 km.

With relative speed of 70kmph, they will cover 140km in 2 hour. So, that means, after 2 hour, they will be separated by 560km


(vii) Again, after crossing at the point M, their relative speed still will be the same. I.e. they will cover 280km in 280/7 = 4 hour time.


Understood? Now, try to solve this question!!

Quiz Ques 3: A lives at P and B lives at Q. A usually goes to meet B at Q. He covers the distance in 3 hour at 150kmph. On a particular day, B started moving away from A While A was moving towards Q, thus A took 5 hours to meet B. What is the speed of B?
 

Concept of Boats and Stream

The concepts of boats and streams is also based on this relative speed.

When boat goes downstream, the speed of flowing water helps the boat to move faster with more speed. When boat goes upstream, the speed of flowing water tries to cancel the speed of boat. The boat moves slower this time.

If, speed of stream = S and speed of boat is B, then

Downstream speed, D = B + S
Upstream speed, U = B - S

B generally means speed of boat in still water.

Hence, B = [D+U]/2 and S = D-B = [D-U]/2


Let’s apply this concept in this question:

Ques: A man can row 9 kmph in still water. It takes him twice as long as to row up as to row down, Find the rate of stream of water.

Let distance covered by boat be 'D'
Speed of stream be 'S'
Speed of boat is 9kmph

• Downstream time Td = distance/speed = D/[9+S]
• Upstream time Tu = distance/speed = D/[9-S]

Tu is twice than Td

• D/[9-S] = 2*D/[9+S]

On solving, we will get S = 3 kmph


Understood? Solve this question now:

Quiz Ques 4: A man can row at 10 kmph in still water. If the river flows at 3 kmph and, it takes 12 hours more in upstream than to go downstream for the same distance. How far is the place?
 

Concept of Races

In races, questions are asked of two or three player race. Questions are like,

Ques: In a race between Ram and Rahim, Ram has won the 1 km race by 100 meters. What is the ratio of their speeds?

The concepts are no different than those that we have already covered. Just some twists in the questions. The objective is to find what the question is saying. Answers will follow.

Now, first step and the most important step is to feel in mind's eye what all is happening in the race. When Ram has just won the race of 1km, Rahim is how far behind him?

Ans is 100 meters behind him.

And Ram has covered 1000 meters, Rahim has covered how much distance?

Ans is 900 meters.

And Ram has covered 1000 meters in the same time it took Rahim to cover 900 meters.

Distance covered are in the ratio of 1000:900

Know the previous concept that when distance is different and time is constant, speed is directly proportional to the distance.

So, speed ratio of Ram : Rahim will also be 1000:900 = 10 : 9


This question will cover the remaining logic.

Ques: in a 1000m race, Ravi gives Vinod a start of 40m and beats him by 19 seconds. If Ravi gives a start of 30 sec to Vinod, then Vinod beats Ravi by 40m. What is the ratio of speed of Ravi to that of Vinod?

Case1: Now, visualize, in 1000m race, Ravi has given Vinod a start of 40m and beats him by 19 seconds. Means to say, Ravi runs 1000m while Vinod runs only 960m.

Second thing, when Ravi completed his 1000m, Vinod is still running and he runs for 19s more.

When putting it mathematically, if Ravi has completed the 1000m in T1 seconds, Vinod took T1+19 seconds to complete the 960m.

Case2: Ravi gives Vinod a start of 30s, then Vinod beat Ravi by 40m. Means to say, When the race is finished, Vinod has run 1000m while Ravi has run only 960m.

Second thing, Vinod has given the start of 30s. When Vinod has completed his 1000m, Ravi is still behind him 40m (i.e. Ravi has completed 960m)

When putting it mathematically, if Vinod has run 1000m in T2+30 seconds, Ravi has run 960m in T2 seconds.


Based on all the above facts, we'll find their speeds.

Ravi's speed = 1000/T1 = 960/T2

T1 = [25/24]*T2

Also, Vinod's Speed = 960/[T1 + 19] == 1000/[T2 + 30]

Solving this by putting T1's value, we gets, T2 = 120s

Required ratio = [960/T2] / [1000/(T2 + 30)] == [960/120] / [1000/150] = 6/5 = Answer.


If you understood all this, try solvingthesequestion using the same basic concepts.

Quiz Ques5: In a 1600m race, A beats B by 80m and C by 60m. If they run at the same time, then by what distance will C beat B in a 400 m race?

Quiz Ques 6: A beats B by 100m in a race of 1200m and B beats C by 200m in a race of 1600m. Approximately by how many meters can A beat C in a race of 9600m?
 

Circular Motion Concept

In circular tracks, the radius should be given or the length of the track is given. If length is given then its okay, if radius is given then put the formula L = 2*pi*radius to find the length of track.

Two or more runners will run this track with unequal speeds. They will run in the same direction or opposite direction.

We've covered that when two people run in same direction, their relative speed = speed of person with more speed - speed of person with less speed.
And when they run in opposite direction, their relative speed becomes = speed of 1st person + speed of 2nd person.

This same concept is to be used here.

Time taken by them to meet for first time will be = length of trace / relative speed.

Means to say, if A runs 1000m race with speed of 25mps and B runs race with speed of 15mps and they both are running in opposite direction, then

Time taken = length of track 1000m / relative speed 25+15mps = 1000/40 = 25seconds

If they run in the same direction, they will take = 1000/10 = 100 seconds.


Sometimes, question is asked of their meeting at the same starting point. For, this, LCM of their time is taken

That is to say, if A runs 1000m with 25mps speed, he will take = 40sec
B runs 1000m with 15mps speed = he will take = 200/3 sec

LCM of 40 and 200/3 = 200

So, they will take 200 sec to meet again at starting point.

This covers the basics of Time and Distance.

Note: These time and speed questions entertains practical imagination of examinee. If one can practically feel how boring it is to drive at 30 kmph and how thrilling the drive becomes if we go at 90 kmph, then he can do more with these questions. The idea is just to get a feel out of these questions. Answers will automatically follow.