Aptitude Time & Work Problems

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Time & Work Problems

Type I: Calculate Time to Complete Work by 2 or More People

In this type, you have to calculate time taken by 2 or more people to do a job. In the question, you will be given the time required by each member individually. You have to calculate the required time if they work together.

Below is an example question.

Example Question 1: Rahul takes 5 hours to do a job. Benny takes 8 hours to do the same job. How long should it take for Rahul and Benny, working together but independently, to do the same job?

Solution:

From the question, you can write down the below values

Part of work done by Rahul in 1 hour = 1/5 …value 1

Part of work done by Benny in 1 hour = 1/8 …value 2

Part of work done by Rahul and Benny together = value 1 + value 2

= 1/5 + 1/8

= 13/40

Now, you can calculate total hours required by both to complete the work using direct proportion table.

Hours    Work

1    13/40

?    1 (1 represents full work)

Number of hours required by both to complete the work together = 1 x 1/ (13/40) 

40/13 = 3 1/13 hours

 

Type II: Extension to Type 1 When Days and Hours are Given

In this type, you will see time as a measure of days and working hours per day. Below example will help you to understand better.

Example Question 2: Arjun can do a piece of work in 5 days of 8 hours each and Chinna can do it in 4 days of 6 hours each. How long will they take to do it working together 7 ½ hours a day?

Solution:

In question, you can see that Arjun can complete the work in 5 days working 8 hours per day. 

Therefore, Arjun can complete the work in (5 days x 8 hours/day) = 40 hours

Similarly, Chinna can complete the work in (4 days x 6 hours/day) = 24 hours

Now, you have to proceed like type 1.

Part of work that Arjun can do in 1 hour = 1/40 … value 1

Part of work that Chinna can do in 1 hour = 1/24 … value 2

Part of work that Arjun and Chinna can do together in 1 hour = value 1 + value 2

= 1/40 + 1/24 = (3+5)/120 = 8/120

You can calculate total hours required by both to complete the work using direct proportion table

Hours    Work

1    8/120

?    1 (1 represents full work)

Both will finish the work in 1 x 1 / (8/120) = 120/8 hours.

You have to calculate number of days required if they work 7 ½

or 15/2 hours each day. Now, you can calculate the number of days required using direct proportion table method.

Hours    Days

15/2    1

120/8    ?

Therefore, if the friends work 15/2 hours each day, the total number of days to complete the work
= 120/8 x 2/15 = 2 days

 

Type III: Equations Based Time and Work Problems

In this type, you have to form equations based on question data. You have to then solve those equations to arrive at the solution. Now let us see an example.

Example Question 3: A and B can built a wall in 12 days, B and C can do it in 16 days and A and C can do it in 18 days. In how many days will A, B and C finish it separately?

Solution:

You have to assume the following.

Let A’s 1 day of work be X,

B’s 1 day of work be Y,

and C’s 1 day of work be Z.

Part 1: Form Equations Based on Question Data

From the question you know that A and B can build the wall in 12 days.

Part of work completed by A and B together in 1 day = 1/12

Or A’s 1 day of work + B’s 1 day of work = 1/12

Or X + Y = 1/12 … equation 1

You know that B and C can build the wall in 16 days

Part of work completed by B and C together in 1 day = 1/16

Or B’s 1 day of work + C’s 1 day of work = 1/16

Or Y + Z = 1/16 … equation 2

You also know that A and C can build the wall in 18 days

Part of work completed by A and C together in 1 day = 1/18

Or A’s 1 day of work + C’s 1 day of work = 1/18

Or X + Z = 1/18 … equation 3

Part 2: Let Us Solve The Equations

If you add equations 1,2 and 3, you will get the following.

2 (X + Y + Z) = 1/12 + 1/16 + 1/18

2 (X + Y + Z) = (12 + 9 + 8)/144 = 29/144

Or, X + Y + Z = 29/288 … equation 4

a) Subtract equation 1 from equation 4:

(X + Y + Z) – (X + Y) = 29/288 – 1/12

Or, Z = 29-24 / 288

Or, Z = 5/288

b) Subtract equation 2 from equation 4:

(X + Y + Z) – (Y + Z) = 29/288 – 1/16

Or, X = 29-18 / 288

Or, X = 11/288

c) Subtract equation 3 from equation 4:

(X + Y + Z) – (X + Z) = 29/288 – 1/18

Or, Y = 29-16 /288

Or, Y = 13/288

A’s 1 day of work = X = 11/288

Therefore, A can complete the work in 288/11 = 26 2/11 days

B’s 1 day of work = Y = 13/288

Therefore, B can complete the work in 288/13 = 22 2/13 days

C’s 1 day of work = Z = 5/288

Therefore, C can complete the work in 288/5 = 57 3/5 days

 

Type IV: Efficiency Based Time and Work Problems

In this type, efficiency of one worker compared to other worker/workers will be given. You have to use this efficiency data to solve this type. Now let us see an example.

Example Question 4: Sam is twice as good a workman as Raj and together they finish a painting work in 10 days. In how many days will Sam alone finish the work?

Solution:

Let us assume the following.

Sam’s 1 day work = X

and Raj’s 1 day work = Y.

From question you know that Sam is twice as good as workman as Raj. In other words, Sam is 2 times efficient than Raj. Therefore, you can form the below ration.

Sam’s 1 day work : Raj’s 1 day work = 2 : 1

Or X:Y = 2:1

Or, X = 2Y … equation 1

Sam and Raj can finish the work in 10 days.

Part of work completed by Sam and Raj in 1 day = 1/10

Or X + Y = 1/10 …equation 2

If you substitute equation 1 in equation 2, you will get.

2Y + Y = 1/10

3Y = 1/10

Y = 1/30 and

X = 2Y = 2/30 = 1/15

X = Sam’s 1 day work = 1/15

Therefore, Sam can complete the work in 15 days.

 

Type V: Calculate Time When Efficiency is Given in Percentage

This is a slight variation of type 4. In this type efficiency will be given in percentage. Below example will help you to understand this type better.

Example Question 5: Ram alone can fence the garden in 8 days. Bose is 50% more efficient than Ram. How many days does Bose alone take to fence the garden?

Solution:

Let Ram’s 1 day work be X

and let Bose’s 1 day work be Y

From the question, you know that Bose is 50% more efficient than Ram.

This means, if Ram’s does 1 unit of work in 1 day, Bose can do 150/100 x 1 = 1.5 units of work in 1 day

Or, X:Y = 1.5:1

OR X = 1.5Y …equation 1

In question, you can see that Ram can complete the work in 8 days

Or Ram’s 1 day work = X = 1/8

If you substitute X = 1/8 in equation 1, you will get, 

Y = X/1.5 = 1/12

Therefore, Bose’s 1 day work = 1/12

Or, Bose can complete the work in 12 days.

 

Type VI: Calculate Time When Workers Leave in Between

In this type, you will find some workers leave in between and others will complete the work. Below example will help you to understand better.

Example Question 6: Saran can do a piece of work in 50 days. He works for 15 days and then Sanjay alone finishes the remaining work in 35 days. In how many days Sanjay alone can finish the work?

Solution:

Let Saran’s 1 day work be X

and Sanjay’s 1 day work be Y.

Saran can complete the work in 50 days.

Therefore, Saran’s 1 day work = X = 1/50.

Though Saran has the ability to complete the work in 50 days, he leaves in 15 days. From then onwards, Sanjay works to finish the remaining work.

Work done by Saran in 15 days = 15 x X = 15/50 = 3/10

You know that 1 represents full/complete work.

Therefore, remaining work = 1 – 3/10 = 7/10

Sanjay finishes this remaining 7/10 work in 35 days. 

You have to calculate the time that Sanjay will take to finish the whole work. You can use direct proportion table method as shown below.

Days    Work

35    7/10

?    1 (1 represents full work)

Therefore, Sanjay can complete the complete work in 1 x 35 / (7/10) = 10 x 35 / 7 = 50 days.

 

Type VII: Share of Salary Based on Work

In this type, you have to calculate the salary of each working member based on their amount of work. You will find the below example helpful.

Example Question 7: Sakshi and Saranya undertake a typist work for Rs.1000. Sakshi alone can complete it in 8 days while Saranya alone can complete it in 10 days. With the help of Ravi, they finish it in 4 days. Find the share of each.

Solution:

Let Sakshi’s 1 day work be X,

let Saranya’s 1 day work be Y

and Ravi’s 1 day work be Z.

Saskshi can complete the work in 8 days.

Therefore, Sakshi’s 1 day work = X = 1/8

Saranya can complete the work in 10 days.

Therefore, Saranya’s 1 day work = Y = 1/10

When all 3 work together, they complete the work in 4 days.

Therefore, Part of work done by all 3 together = ¼

Or, X + Y + Z = ¼

Z = ¼ – (X + Y) = ¼ – (1/8 + 1/10) = 1/40

To find salary share of each working member, remember the following rule.

Ratio of the salaries between members = Ratio of 1 day (or 1 hour) work of the members.

Based on the above rule,

Salary of Sakshi : Saranya : Ravi = X:Y:Z = 1/8 : 1/10 : 1/40 

= 4 : 5 : 20

Sakshi’s share = 1000 x 4/(4+5+20) = 1000 x 4/29 = 137.93

Saranya’s share = 1000 x 5/(4+5+20) = 1000 x 5/29 = 172.41

Ravi’s share = 1000 x 20/(4+5+20) = 1000 x 20/29 = 689.66

Note: If you have doubt on above calculation, you can refer to type2.

Aptitude Time & Distance Problems

Time & Distance Problems

Type I: Unit Conversion Time And Distance Problems (They Are The Most Basic Type)

This type is very easy to solve. You will get speed in one unit (e.g kilometer per hour or km/h) and you have to convert into another unit (e.g metres per second or m/s).

To solve such problems, you have to remember 2 formulas.

If speed X is given in (kilometres per hour) km/h, then speed in (metres per second) m/s is X x 5/18

If speed X is given in (metres per second) m/s, then speed in (kilometres per hour) km/h is X x 18/5

Example Question 1: A bus moves at a speed of 81 km/h. What is the speed of the bus in metres per second?

Solution:

In question, you can see that the speed is given in km/h. 

You can write, X = 81 km/h

Therefore, speed in m/s (according to our formula) = X x 5/18 

= (81 x 5/18)

= 22.5 m/s

 

Type II: Average Speed When Travelling To A Place And Returning

In this type of questions, a person will move from one place to another at certain speed. Then he will return to the starting place at a different speed. You may be asked to find average speed, distance, etc.

To solve such problems, you have to remember the formula given below.

Let a person move from one place to another at speed X and return to the starting place at a different speed Y.

Then, average speed for the whole journey = 2XY/(X+Y)

Example Question 2: Ram walked from his home to the bank at the rate of 20 km/h and returned back at the rate of 5 km/h. If he took 4 hours and 30 minutes for the whole journey, find the distance of the bank from his home.

Solution:
 

Part 1: Finding Average Speed

From the question, you know the following:

Speed of Ram from home to bank = X = 20 km/h 

Speed of Ram from bank to home = Y = 5 km/h

Also, you know that the average speed for the whole journey = 2XY/X+Y

= (2 x 20 x 5) / (20+5)

= 200 / 25 = 8 kmph.

Part 2: Finding Distance

From the question, you know that the time taken for the whole journey = 4 hrs and 30 min

4 hrs and 30 min can be written as 4 ½ hours or 9/2 hours

In part 1 of the solution, you have found that the average speed = 8 kmph

You know the familiar formula that Speed = Distance / Time

Therefore, total distance travelled by Ram = Average Speed X Time Taken

= 8 x 9/2 = 36 Km

Note: But, this distance is the total distance travelled by Ram from his home to bank plus distance he travelled from bank to home. Therefore, to calculate the distance from home to bank, you have to divide the above value by 2.

Therefore, distance from home to bank = 36/2 = 18 Km

 

Type III: Problems Based On Changing Time And Changing Speed

This type is based on a simple fact that if a person increases his speed he will reach his destination faster and if he decreases his speed he will reach his destination slower.

Though this type looks tough at first, you can solve this type easily if you carefully read and understand the below example.

Example Question 3: If a cyclist rides at a speed of 4 km/h, he reaches the office by 5 minutes late. However, if he rides at a speed of 5 km/h, he reaches 4 minutes earlier. Find the distance covered by him to reach office?

Solution:

Assume that the distance travelled by cyclist (from home to office) to be D km.

Case I: Cyclist’s speed is 4 km/h

Time taken to cover D km at 4 km/h = Distance covered by cyclist / Speed of cyclist= D/4 hours

Note: Above equation is based on the simple formula: Speed = Distance / Time

Case II: Cyclist’s speed is 5 km/h

Time taken to cover D km at 5 km/h = Distance covered by cyclist / Speed of cyclist = D/5 hours.
 

Time difference between case I and II

You know that when cyclist travels at 4 km/h, he reaches 5 minutes LATE but when he travels at 5 kmph, he reaches 4 minutes

EARLIER.

Therefore, difference in time taken between case I and II = (5 minutes + 4 minutes)

= 9 minutes or 3/20 hours

But we know that the time taken in case I is D/4 and that in case II is D/5. Therefore, above equation becomes,

D/4 – D/5 = 3/20

(5D-4D) / 20 = 3/20

D = 3 km

 

Type IV: Time And Speed Problems On Trains

This type is very popular in bank and other government exams. In this type, starting time, speed, etc., of trains will be given. You will asked to find the time of their crossing.

Here is an example question.

Example Question 4: Two trains start from stations A and B which are 400 km apart. First train starts from A at 9.00 am and travels towards B at 50 km/h. Another train starts from B at 10.00 am and travels towards A at 40 km/h. At what time do they meet?

Solution:

Assume that the two trains will meet each other at X hours after 9.00 am.
 

Case I: Distance travelled by train 1 before crossing train 2

First train starts at 9.00 am and travels X hours before meeting. Also, in question you can see that the speed of the first train is 50 km/h.

Let D1 be the distance travelled by the first train in X hours.

You can write the below equation based on the above data:

D1 = Speed of the first train x Time taken by the first train till crossing

D1 = 50X km …equation 1

Case II: Distance travelled by train 2 before crossing train 1

Second train starts at 10.00 am and travels (X – 1) hours before meeting the first train

Do you get a doubt that why we used X – 1 instead of X? Here is your answer:
Our assumption is that the trains meet each other at X hours after 9.00 am. Second train starts at 10.00 am i.e., 1 hour after 9. Therefore, it will take X – 1 hours to meet the first train.

Let D2 be the distance travelled by second train in X-1 hours.

Therefore, D2 = Speed of the second train x Time taken by the second train till crossing

D2 = 40(X-1) km …equation 2

Forming equation for total distance

D1 + D2 = Total distance between A and B

From question, you know that the total distance between A and B is 400 Km

Therefore, D1 + D2 = 400

If you substitute the values from equations 1 and 2 in above equation, you will get,

50X + 40(X-1) = 400

50X + 40X – 40 = 400

90X = 360

X = 4 hours

So the two trains will meet each other after 4 hours after 9.00 am. i.e., at 1.00 pm.

 

Type V: Problems On Bus With Stoppages

Though the bus you are travels appears to travel fast, it will reach its destination very slow if it stops at many places. Type V deals with such cases. Below example will help you to understand clearly.

Example Question 5: A bus without any stoppage can travel a certain distance at an average of 70km/h and with stoppages covers the same distance at an average speed of 50 km/h. How many minutes per hour does the bus stop?

Solution:

Assume that the total distance travelled by the bus to be D km.
 

Case I: Without stoppages

Without stoppages, the average speed of the bus is 70 km/h.

Time taken by bus without stoppages = Distance covered by the bus / Speed of the bus = D/70 hours …equation 1

Case II: With stoppages

With stoppages, the average speed of the bus is 50 km/h.

Time taken by the bus with stoppages = Distance covered by the bus / Speed of the bus = D/50 hours …equation 2

Calculation of Stopping Time

If we subtract the values of time taken by bus with and without stoppages, we can find the total stopping time. In other words,

total stopping time equals the difference in values of equations 1 and 2

Therefore, we can write

Time taken by bus for stoppages (D/50 – D/70) hours = 7D – 5D / 350 

= 2D/350 

= D/175 hours. …. equation 3

Now, have a look at case II. In case II, bus with stoppages travels at 50 km/h to cover the distance D.

Therefore, journey time of bus with stoppages = Distance / Speed = D/50 hours

Also we found that the total stoppage time = D/175 hours (see equation 3)

In D/50 hours of journey, the bus stops for D/175 hours. The total stoppage time for 1 hour of journey can be calculated using the direct

proportion table method as given below.

Journey Time    Stoppage Time

D/50    D/175

1    X

D/50 = D/175X

Or, X = 50/175 = 2/7 hours

= 2/7 x 60 minutes = 17.14 minutes

Alternate Shortcut Method:

You know that the speed of the bus is reduced by 20 km/h due to stoppages

In other words, the time that the bus takes without stopping to drive 20 km/h will be equal to stoppage time per hour.

Time taken to cover 20 km = (20 km / Actual speed of bus without stoppage ) 

= 20/70 hours = 2/7 hours 

= 20/70 x 60 minutes

= 17.14 minute

Aptitude Ratio And Proportion Problems

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Ratio And Proportion Problems

Type 1: Combined Ratio Based On Individual Ratios

You can expect this type of problems not only in bank but also in other government exams. This type is very easy to solve. Below is an example, to understand this type clearly.

Example Question 1: If a:b = 5:8 and b:c = 6:7, Find a:b:c

Solution 1: 

To solve this type, first you have to identify the common term appearing in both the ratios.

In this question, b is common in both the ratios. The value of b in first ratio is 8 and in second ratio is 6.

Now, you have to find the LCM of 8 and 6, which is 24.

Then, you have to transform a:b and b:c so that b becomes 24 in both the cases.
 

Consider first ratio a:b

You know that a:b = 5:8

To transform b to 24, you have to multiply both the terms by 3.

Therefore, a:b = 5×3:8×3 = 15:24

Consider second ratio b:c

You know that b:c = 6:7

To transform b from 6 to 24, you have to multiply both the terms by 4.

Therefore, b:c = 6×4:7×4 = 24:28

After transformation, a:b becomes 15:24 and b:c becomes 24:28

Now, you can spot that b is equal (24) in both the ratios.

Now you to combine both the transformed ratios by writing b value only once.

Therefore, you will get a:b:c = 15:24:28

 

Type 2: Distributing Any Quantity Based On Ratios

In this type, you will find that a particular quantity (e.g .,Amount in rupees, Mixture in litres) is to be shared among individuals based on ratios. You will understand this type after the below example.

Example Question 2: Ram, Gita and Anu shared Rs.5400 among themselves in the ratio 2:3:4. Find the amounts received by each of them.

Solution: 

To solve this type of problems, you have to remember a simple formula shown below:
 

Amount received by a person = (Ratio value of that person / Sum of the ratio values) x Total amount

Based on the above formula, you can easily derive the below 3 formulas:

Amount received by Ram = (Ram’s ratio value / Sum of the ratio values) x Total amount 

Amount received by Gita = (Gita’s ratio value / Sum of the ratio values) x Total amount 

Amount received by Anu = (Anu’s ratio value / Sum of the ratio values) x Total amount

You know that Ram’s ratio value = 2 , Gita’s value = 3 and Anu’s value = 4

Sum of the ratio values = 2+3+4 = 9

And total amount = 5400

Therefore, you can find individual amounts as shown below

Ram’s amount = 2/9 x 5400 = 1200

Gita’s amount = 3/9 x 5400 = 1800

Anu’s amount = 4/9 x 5400 = 2400

 

Type 3: Coins Based Ratio Problems (This type is interesting)

This is a special type of ratio problems is very interesting. If you have not seen this before, below example will help you.

Example Question 3: A bag contains 50p, 20p and 10p coins in the ratio 4 : 8 : 6, amounting to Rs. 210. Find the number of coins of each type.

Solution:

You know that the given ratio of the number 50, 20 and 10 paisa coins is 4:8:6

To make calculations easier, you have to assume number of coins based on their ratio values. For example, the ratio value of 50p coins is 4. Therefore, you have to assume that there are 4X number of 50p coins. (Here X is the unknown quantity, which you will solve)

Similarly, you have to assume that there are 8X number of 20p coins and 6X number of 10p coins. 

You know that in total value of all the coins is Rs. 210.

You also know that, two 50p coins make 1 rupee, five 20p coins make 1 rupee and ten 10 paisa coins make 1 rupee. Therefore, we can write the below 3 equations:

Amount in rupees corresponding to 4x number of 50p coins = 4X x (1/2) 

Amount in rupees corresponding to 8x number of 20p coins = 8X x (1/5) 

Amount in rupees corresponding to 6x number of 10p coins = 6X x (1/10)

Adding all the above three amounts in rupees, you should get Rs. 210. Therefore, you can write,

4X/2 + 8X/5 + 6X/10 = 210

Or (20X + 16X + 6X) / 10 = 210

42X = 2100

X = 50

Number of 50p coins = 4X = 4 x (50) = 200

Number of 20p coins = 8X = 8 x (50) = 400

Number of 10p coins = 6X = 6 x (50) = 300

This type is interesting, isn’t it? Share your views on comments section below. Now let us move on to our final type.

 

Type 4: Mixtures & Replacement Based Ratio Problems

You may see problems that involve replacement of a liquid in a mixture of two different liquids. Now, let us see an example.

Example Question 4: A 15 litres of mixture contains water and milk in the ratio 2 : 4. If 3 litres of this mixture is replaced by 3 litres of water, the ratio of water to milk in the new mixture would be?

Solution:

After 3 litres of mixture is taken out, the remaining mixture will be12 litres.

First you find the amount of water in 12 litres of mixture by using the below formula
 

Amount of water in 12 litres of mixture = (Ratio value of water / Sum of ratios ) x Total Quantity

Note: Above formula is the same as that we used in example 2.

So, Amount of water in 12 litres of mixture = (2/6) x 12 = 4 litres … equation 1

After 3 litres of mixture is taken out, 3 litres of water is added.

Therefore, Amount of water in 15 litres of new mixture = 3 litres of water + Amount of water in 12 litres of mixture

= 3 + 4 = 7 litres of water …. equation 2

(Note: If you doubt from where 4 appeared refer to equation 1)

Therefore, quantity of milk in the mixture = 15 litres of mixture – 7 litres of water

= 8 litres of milk … equation 3

From equations 1 and 2, you can conclude that the ratio of water and milk in the new mixture = 7 : 8

Aptitude Probability Problems

Probability Problems

Type I: Simple Problems Based on Dice, Coins, etc.

You will be very familiar with this type of problems even from school days. This type of problems is very easy to solve. Below is an example.

Example Question 1: In a simultaneous throw of two dice, find the probability of getting a total more than 6.

Solution:
Let S denote the set of all possible outcomes. S is also called sample space.

Note: If you are not clear on what ‘outcome’ means, here is an example. Let us assume that first dice shows 1 and second dice shows 1. Then the outcome is (1,1). If the first dice shows 1 and second dice 2, then the outcome is (1,2) and so on…

When two dice are thrown, you will get the below possible outcomes.

S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),( 4,2),( 4,3),( 4,4),( 4,5),( 4,6),

(5,1),( 5,2),( 5,3),( 5,4),( 5,5),( 5,6),

(6,1),( 6,2),( 6,3),( 6,4),( 6,5),( 6,6)}

Therefore, total number of outcomes, n(S) = 36.

Let E be the event of getting a total more than 6. In other words, sum of the numbers shown on dices should be greater than 6. 

If you closely observe the values in S, you can identify outcomes where the total is greater than 6. Such outcomes will form E.

Therefore, E = {(1,6), (2,5), (2,6), (3,4), (3,5), (3,6), (4,3), (4,4), (4,5), (4,6), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Total number of events, n(E) = 21

Probability of an even E is given by the formula, P(E) = n(E) / n(S)

Probability of getting total more than 6, P(E) = n(E) / n(S) = 21/36 = 7/12

 

Type II: Probability Problems that Require Logical Thinking

This is slightly difficult type you will see in bank exams. You will understand this type after reading the below example.

Example Question 2: In a competitive exam, Arun can answer 75% questions correctly and Vimal can answer 70% questions correctly. Assume that a teacher picks a question and asks both Arun and Vimal to solve that question. what is the probability that at least one of them will solve that question?

Solution:

Here, total percentage of questions, n(S) = 100

Let A be the event that Arun answers correctly and B be the event that Vimal answers correctly.

Percentage of questions Arun answers correctly, n(A) = 75

Percentage of questions Vimal answers correctly, n(B) = 70

Step 1: Calculate Individual Probabilities

Probability of Arun answering a question correctly, P(A) = n(A)/n(S) = 75/100 = ¾

Probability of Vimal answering a question correctly, P(B) = n(B)/n(S) = 70/100 = 7/10

Important Note:

You have found the probabilities of Arun and Vimal answering correctly. To solve this problem, you also have to find the probabilities to answerincorrectly. (You will be using these values in later steps.)

Probability of Arun answering a question incorrectly, P(Aꞌ) = 1 – P(A) = 1 – ¾ = ¼

Probability of Vimal answering a question incorrectly, P(Bꞌ) = 1 – P(B) = 1 – 7/10 = 3/10

Step 2: Think Logically, Read the Question Carefully and Construct the Event

Now its time for you to think logically. Teacher gives a question to both. For the condition that “at least one of them have to answer the question correctly”, you can construct the even as shown below.

Event E:

Arun answers correctly AND Vimal answers incorrectly OR

Arun answers incorrectly AND Vimal answers correctly OR

Both Arun and Vimal answer correctly

Based on the above event, you can write the below probability equation.

P(E) = P(A) AND P(Bꞌ) 

OR P(Aꞌ) AND P(B)

OR P(A) And P(B)

In such probability equations, you can replace “AND” with “X” (multiplication) and “OR” with “+” (addition)

Therefore, P(E) = P(A) X P(Bꞌ) + P(Aꞌ) X P(B) + P(A) X P(B)

= (¾ x 3/10) + (1/4 x 7/10) + (¾ x 1/4) 

= 9/40 + 7/40 + 3/16

= (18+14+15)/80

= 47/80

 

Type III: Probability Problems Based on Drawing Balls at Random

In this type, you will find a collection of balls of 2 or 3 different colours. You have to calculate the probability of drawing (picking) balls based on conditions(which you will find in question). Below example will help you to understand well.

Example Question 3: A box contains 5 yellow and 3 green balls. Two balls are drawn at random. Find the probability that they are of the same color.

Solution:

Step 1: Find n(S)

Let S be the sample space. Then, n(S) = number of ways of selecting any 2 balls. 

Number of ways of selecting 2 balls from 8 balls = 8C2

Therefore, n(s) = 8C2 = 8 x 7 / 1 x 2 = 28

Step 2: Find n(E)

Let E be the event of getting two balls of the same color.

Therefore, n(E) = number of ways of getting (2 balls out of 5 yellow) OR (2 balls out of 3 green)

Or, n(E) = 5C2 + 3C2

= (5×4 / 1×2) + (3×2 / 1×2)

= 10 + 3 = 13

Step 3: Find P(E)

Therefore, required answer = P(E) = n(E) / n(S) 

= 13/28

Aptitude Permutation & Combination Problems

Permutation & Combination Problems

Type I: Make A Word Using Letters Given

In this type of problems, you will be given a word. You have to find the number of words that can be made using the letters in the given word.

Can’t able to understand this type? Consider a word “MATHS”. You can make words like “AMTHS”, “TAMHS” and a lot more words. It is not necessary that the new words made have meaning. In this type of problems, you will be asked to find the total number of words that can be formed.
This type is very easy to solve. Read the below example carefully.

Example Question 1: How many words can be formed by using the letters of the word “MATHEMATICAL”?

Solution:
Step 1:

You have to calculate the number of letters in the given word.

In “MATHEMATICAL” there are 12 letters.

Step 2:
You have to count the number of times each letter appears in the given word. You will get the below table

Letter    Number of times the letter occurs

M    2
A    3
T    2
H    1
E    1
I    1
C    1
L    1

But, you know 1! = 1. Therefore, above equation becomesStep 3:

Now, you can find the number of words that can be made, using the below formula
 

Number of words = factorial of total number of letters / product of factorial of repetitions of each letter

If you substitute the values got in steps 1 and 2, you will get

Number of words = 12! / (2! x 3! x 2! x 1! x 1! x 1! x 1! x 1!)

Number of words = 12! / (2! x 3! x 2!)

= (12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (2 x 1 x 3 x 2 x 1 x 2 x 1)

= (12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / 24

= 19958400

 

Type II: Make A Word When Some Letters Should Be Together

This is actually an extension of type I. This is similar to type I, but additional condition will be given. You will understand this after reading the below example.

Example Question 2: Find number of words that can be made from letters of the word “ABACUS” with a condition that “U” and “S” are always together.

Solution:
You could see a condition that “U” and “S” should always be together. 

In a question, when you find a condition that some letters should always be together, you should treat those letters as 1 (single) letter.

To solve our example question, you have to assume “U” and “S” as 1 single letter. Therefore, the word can be rewritten as
ABAC(US)

As we already did for example question 1, we have to form a table showing number of times each letter appears. The table is as shown below.

Letter    Number of times the letter occurs

A    2
B    1
C    1
US    1

To find the number of words that can be formed from ABAC(US), you can use the formula (as you saw in example question 1),Note: Here, we have written US as 1 single letter

Number of words that can be formed from ABAC(US) = 

factorial of total number of letters / product of factorial of repetitions of each letter
= 5! / 2!.1!.1!.1!
= 5 x 4 x 3 x 2 x 1 / 2 x 1 x 1 x 1 x 1
= 60 words …. value 1

But, this is not your answer. There is one more step

You have assumed (US) as 1 (single) letter right? You have to find number of words that can be formed from the letters “U” and

“S”, and multiply this value with answer of previous step.

If you create repetition table for US, you will get,

Letter    Number of times the letter occurs

U    1
S    1

Now, you have to multiply, value 1 and value 2 to get the answer. Therefore, your answer will be 60 x 2 = 120 wordsNumber of words that can be formed from U and S = factorial of total number of letters / product of factorial of repetitions of each letter

= 2! / 1!.1!

= 2 …. value 2

Hint of practice test question:
You already know there is an online practice test at the end of this tutorial. You will see a question of this type in that test. Read and understand this type (II) carefully to solve the practice test question easily.

 

Type III: Combination Problems Based On Selection From Group

This type is very common in bank exams. In this type, you have to calculate the number of ways a team (short group) can be selected from a large group.

This type is based on an easy formula, which is given below:

Number of ways a team of R members can be selected from a group of N members is given by NCR
 

In the above formula, C represent “Combination”. 
 

The value for NCR can be calculated as N! / [R! x (N-R)!]

Below example will help you to understand clearly.

Example Question 3: In how many ways can 5 students be selected from a group of 10 students?

Solutions:
In the above question, the size of large group N = 10.

From the group of 10 you have to select a team of 5 students.

Therefore, number of members in team = R = 5

You can now calculate the number of ways 5 students can be selected from 10 students using our formula NCR
We have to select 5 students from 10 students.

No of ways of selection = 10C5

= 10! / [(10-5)! x 5!]

= 10! / (5! x 5!)

= (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / [(5 x 4 x 3 x 2 x 1) x (5 x 4 x 3 x 2 x 1)]

= 252

 

Type IV: Problems Based On Selection And Possibilities

This type is an extension to type 3. Here, you have to use the same selection from group formula that you saw in type 3. But, in this type, you have to select from group based on some possibilities (given in question). Below example will help you to understand this type clearly.

Example Question 4: A box contains 3 red balls, 2 yellow balls and 5 green balls. In how many ways can 3 be drawn from the box, if at least 2 green balls are to be included in the draw?

Solution:

You have to select 3 balls. Among the 3 balls at least 2 need to be green.

You can do this in 2 possible ways. 

You can select ( all 3 green balls) OR ( 2 green balls AND 1 non green ball)

In such problems, you have to change OR to addition (+) and AND to multiplication (X)

Therefore, your final answer will be,

Number of ways of selecting 3 green balls + (Number of ways of selecting 2 green balls X Number of ways of selecting 1 non green ball) …. equation 1

Case1: Number of ways of selecting all 3 green balls 

You know that there are 5 green balls. Therefore, size of group, N = 5

You have to select 3 green balls from 5. Therefore, team size, R = 3

Therefore, number of ways 3 green balls can be selected = NCR = 5C3

= 5! / [(5-3)! x 3!]

= 5! / 2!x3!

= 5 x 4 x 3 x 2 x 1 / 2 x 1 x 3 x 2 x 1

= 10

Case 2: Number of ways of selecting 2 green balls 

In this case, size of group, N = 5 and size of team, R = 2

Therefore, number of ways of selecting 2 green balls = NCR = 5C2

= 5! / [(5-2)! x 2!]

= 5! / 3! x 2!

= 5 x 4 x 3 x 2 x 1 / 3 x 2 x 1 x 2 x 1

= 10

Case 3: Number of ways of selecting 1 non green ball

There are 3 red balls and 2 yellow balls. Therefore, there are 5 non green balls in total. Therefore, group size N =5.

You have to select 1 non green ball from 5 non green balls. Therefore, team size, R = 1. 

Therefore, number of ways of selecting 1 non green ball = NCR = 5C1

= 5

We simplified 5C1 to 5 because NC1 is always equal to N

Hint: In combinations, you have to always remember these two rules: NCN = 1 and NC1 = N

Final Substitution:

Now if you substitute the values of cases 1,2 and 3 in equation 1, you will get

Answer = 5C3 + (5C2 X 5C1) = 10 + (10 x 5)

Answer = 60

Therefore, number of ways of selecting 3 balls so that there are at least 2 green balls = 60 ways.

Aptitude Partnership Problems

Partnership Problems

Type I: Profit Share Of Partners When Investment Durations Are Same. (Why This Is The Easiest And Basic Type?)

This type is very easy to solve. You will understand why after reading the below example. In this type, you will find amounts invested by 2 or more partners forsame duration of time. You will be given the profit amount. You have to find the profit share of each member in the group.

Let us see an example.

Example Question 1: Ravi, Kiran and Raj start a business by investing Rs.1,00,000, Rs.1,50,000 and 2,00,000 respectively. Find the share of each out of an annual profit of Rs.50,000.

Solution:
To solve this question you have to remember the below formula:

If duration (months/years) of investment of partners is same, the profit will be shared in the ratio of their investments.

In other words, if duration of investment is same,
 

Ratio of profit between partners = Ratio of investment between partners

In our example, the ratio of investments by Ravi, Kiran and Raj is as follows

1,00,000 : 1,50,000 : 2,00,000

Therefore, based on our formula,

The ratio of their profits is also equal to 1,00,000 : 1,50,000 : 2,00,000

If we simplify the above ratio, we will get

Ratio of profits = 2 : 3 : 4

You know that the total profit earned = Rs. 50,000

Therefore, the profit of each person can be calculated as shown below. (If you don’t understand the below calculations, please refer to formula in example question 2 of ratio problems tutorial. )

Ravi’s profit share = 50,000 x 2/9 = Rs.11,111.11

Kiran’s profit share = 50,000 x 3/9 = Rs.16,666.67

Raj’s profit share = 50,000 x 4/9 = Rs.22,222.22

 

Type II: Profit Share Of Partners When Investment Durations Are Different (How This Is Different From Type 1?)

This type is same as type 1, but here the investment duration of partners will be different. To solve this you have to apply a slightly advanced formula as shown in the example below:

Example Question 2: Sam and Rita are partners in a business. Rita invests Rs.40,000 for 6 months and Sam invests Rs.45,000 for 3 months. Out of a profit of Rs.30,000, Rita’s share is ____ .

Solution:
You have to apply the below formula to solve this problem:

Assume that three partners P1, P2 and P3 have invested amounts A1, A2 and A3 (in Rupees) for durations T1, T2 and T3 (months or years) respectively. Then,
 

Ratio of profit between P1, P2 and P3 = A1 x T1 : A2 x T2 : A3 x T3 

Note: You have to extend the above formula if there are more than 3 partners. For example, for 4 partners the above formula will

become A1 x T1 : A2 x T2 : A3 x T3 : A4 X T4

You know that the total profit earned = Rs. 30,000

From the question, you know that Rita invests Rs. 40,000 for 6 months and Sam invests Rs. 45,000 for 3 months.

In this example, there only 2 partners. Therefore, you have to use the below formula:

Ratio of profit shared between Rita and Sam = Investment of Rita x Duration of investment of Rita : Investment of Sam x Duration of investment of Sam

You know the below values from question:

Investment of Rita = Rs. 40,000

Duration of investment of Rita = 6 months

Investment of Sam = Rs. 45,000

Duration of investment of Sam = 3 months

If you substitute the above values in the formula, you will get

Ratio of profit between Rita and Sam = 40,000 x 6 : 45,000 x 3

= 2,40,000 : 1,35,000

= 16 : 9

Therefore, you can find Rita’s profit as shown below:

Rita’s share = 30,000 x 16/25 = Rs. 19,200

 

Type III: Profit Between Partners When Amounts Are Invested/Withdrawn In Parts (How To Solve This Difficult Type?)

This is the most difficult type of partnership problems. But you can easily solve if you carefully read the below example.

Here is our example.

Example Question 3: Anu, Mahesh and Naren enter into a partnership and invest Rs. 30,000, Rs. 40,000 and Rs.50,000 respectively. Mahesh withdraws Rs.20,000 at the end of first year and Naren withdraws Rs.25,000 at the end of second year. At what ratio will they share their profit at the end of 3 years.

Solution:
In this case, you know Anu has invested Rs 30000 for 3 years (for years I, II and III).

Mahesh invests 40,000 at start but withdraws Rs 20,000 a the end of first year. Therefore, his remaining investment in business will be 40,000 – 20,000 = 20,000.

In other words, Mahesh’s investment is 40,000 for year I and 20,000 for years II & III

And, Naren invests 55,000 at start but withdraws 25,000 at the end of second year. 

Therefore, Naren’s investment is 50,000 for years I & II and 25,000 for year III.

Now, let us see the formula you should use to solve such problems. (This is actually an extension to formula in example question 2)

Let partner P1 has invested amount A1a for T1a months, A1b for T1b months and so on…

Let partner P2 has invested amount A2a for T2a months, A2b for T2b months and so on…

Let partner P3 has invested amount A3a for T3a months, A3b for T3b months and so on… 

Then, ratio of profits between P1, P2 and P3 = (A1a x T1a + A1b x T1b …) : (A2a x T2a + A2b x T2b…) … : (A3a x T3a + A3b x T3b …)

Is the above formula very difficult to understand? It will become easy after you finish reading this example.

To solve our example using the above formula, first let us convert years into months. So, you will get

Anu’s investment = 30000 for 36 months

Mahesh’s investment = 40000 for 12 months and 20000 for 24 months

Naren’s investment = 50000 for 24 months and 25000 for 12 months

Now, if you apply the formula, you will get

Ratio of their profits = (30,000 x 36) : (40,000 x 12 + 20,000 x 24) : (50,000 x 24 + 25,000 x 12)

= 10,80,000 : 96,000 : 15,00,000

= 54 : 48 : 75

Aptitude Boats And Streams Problems

Aptitude Boats And Streams Problems

Boats And Streams: How To Solve 5 Types Of Problems

Type I: Finding Speed Of Boat Using Direct Formula

In this type, you will be finding speed of boat in still water (i.e., when water is not flowing/running). You have to remember a very simple formula as shown below.

Speed of the boat in still water = ½ (Downstream speed + Upstream speed)

Here, downstream speed denotes the speed of the boat in the direction of the stream, and, upstream speed denotes the speed of the

boat against the direction of the stream.

2 more basic formulas that will help you are given below.
 

Downstream speed = Speed of boat in still water + Speed of stream
 

Upstream speed = Speed of boat in still water – Speed of stream

Let us see an example to understand this type.

Example Question 1: A boat travels at 9 km/h along the stream and 6 km/h against the stream. Find the speed of the boat in still water.

Solution:
From the question, you can write down the below values.

Downstream speed of the boat = 9 km/h

Upstream speed of the boat = 6 km/h

You have to substitute the above values in the below formula.

Speed of the boat in still water = ½ (Downstream speed + Upstream speed)

= ½ (9 + 6)

=7.5 km/h

 

Type II: Finding Speed Of Stream Using Direct Formula

This type is similar to type 1. But there is one difference. Here you have to find speed of stream and not the speed of the boat.
You have to use the below formula to find speed of stream.

Speed of stream = ½ ( Downstream speed – upstream speed)

Below is your example.

Example Question 2: A man rows downstream 30 km and upstream 12 km. If he takes 4 hours to cover each distance, then the velocity of the current is:

Solution:
In this question, downstream and upstream speeds are not given directly. Hence you have to calculate them first.

Step 1: Calculation of downstream speed

You know that the man rows 30 Km in 4 hours downstream

You know the familiar formula that Speed = Distance/Time

Therefore, Downstream speed = Distance travelled downstream / Time taken

= 30/4 Km/h

Downstream speed = Distance travelled in downstream / Time taken in Downstream travel

= 30/4 … value 1

Step 2: Calculation of upstream speed

You know that the man rows 12 Km in 4 hours upstream

So, Upstream speed = Distance travelled in upstream / Time taken

=12/4 … value 2

Step 3: Calculation of speed of stream

You have to substitute values got in steps 1 and 2 in below formula to find the speed of the stream.

Speed of the stream = ½ ( Downstream speed – upstream speed)

= ½ (30/4 – 12/4)

= ½(18/4)

= 2.25 km/h

 

Type III: Find Distance Of Places

In this type, you have to find distance of places based on given conditions. Below example will help you to understand better.

Example Question 3: A man can row 5 km/h in still water. If in a river running at 2 km an hour, it takes him 40 minutes to row to a place and return back, how far off is the place ?

Solution:

From the question, you can write down the below values.

Speed of the man in still water = 5 km/h 

And speed of the river = 2 km/h

Using the above data, you have to first calculate downstream and upstream speeds.

Downstream speed = Speed of man in still water + Speed of the river

= 5 + 2 = 7 km/h … value 1

And, Upstream speed = Speed of man in still water – Speed of the river

= 5- 2 = 3 km/h … value 2

The man rows to a particular place and comes back. You have to calculate the distance of this place. Let this distance be X. See the below diagram to understand clearly. (Man starts from A, travels to B and comes back. Therefore distance between A and B = X)

You have to use the below equation to find the value of X,

Total time to travel from A to B and come back to A= Time taken from A to B (downstream) + Time taken from B to A (upstream)

You know the familiar formula, Speed = Distance/Time. Therefore, Time = Distance / Speed. Therefore, above equation becomes,

Total time to travel from A to B and come back to A = Distance from A to B/downstream speed + Distance from B to A/upstream speed

But, as per our assumption, distance from A to B = distance from B to A = X. 

Also we have calculated downstream and upstream speeds at the start (see values 1 and 2). 

So the above equation becomes,

Total time to travel from A to B and come back to A = X/7 + X/3

In question, you can see that the man takes 40 minutes to travel to B and come back to A. You have to convert this to hours and apply

in above equation. (We are converting from minutes to hours because we are using speed values in km perhour units.)

40 minutes = 40/60 hours = 2/3 hours

Our equation becomes,

2/3 = X/7 + X/3

2/3 = (3X + 7X)/21

(21 X 2) / 3 = 10X

X = 42/30

= 1.4 Km

 

Type IV: Using Man’s Still Water Speed Calculate Stream’s Speed

In this type, you have to follow two steps.

1. Using man’s still water speed, you have to calculate upstream and downstream speeds.

2. Using upstream and downstream speeds, you have to find the speed of the stream.

Below example will help you understand better.

Example Question 4: A man can row 9 km/h in still water. It takes him twice as long to row up as to row down the river. Find the rate of the stream.

Solution:
 

Step 1: Calculate upstream and downstream speeds.

Assume that the man’s speed in upstream be X km/h

From the question, you know that his downstream speed is twice of upstream speed.

Then, his downstream speed = 2X km/h

You know the formula that, Man’s speed in still water = ½ (Upstream speed + Downstream speed)

=1/2 (X + 2X) 

= 3X/2

But, in question, the man’s speed in still water is given to be 9 km/h

Therefore, 3x/2 = 9

X = 6 km/h.

Based on our assumptions, you can easily calculate upstream and downstream speeds as shown below. 

Upstream speed = X = 6 km/h

Downstream speed = 2X = 12 km/h

Step 2: Calculate Speed Of The Stream

You already know the basic formula shown below.

Speed of the stream = ½ (Downstream speed – Upstream speed)

If you substitute the downstream and upstream speeds of step 1 in the above formula, you will get,

Speed of the stream = ½(12 – 6)

= 3 km/h.

 

Type V: Equations Based Boats And Stream Problems

In this type, you have to form linear equations based on conditions given. You have to solve those equations to find the answer.

Below example will help you to understand this type clearly.

Example Question 5: Kavin can row 10 km upstream and 20 km downstream in 6 hours. Also, he can row 20 km upstream and 15 km downstream in 9 hours. Find the rate of the current and the speed of the man in still water.

Solution:

You have to make below assumptions to form equations.

Let the upstream speed be X km/h

And downstream speed be Y km/h.

You already know the below equation. (If you are not clear about this, refer to theequation in type 3.)

Time for downstream travel + Time for upstream travel = Total Time for upstream and downstream travel

Using the familiar Speed = Distance / Time formula, the above equation can be simplified as shown below.

Distance travelled in downstream/downstream speed + Distance travelled in upstream/upstream speed = Total Time for upstream and downstream travel

If you substitute the values in question in above equation, you will get the below 2 equations.

10/x + 20/y = 6 …equation 1

20/x + 15/y = 9 …equation 2

Assume that 1/x = u and 1/y = v, Now you rewrite the above equations as given below.

10u + 20v = 6 …equation 3

20u + 15v = 9 …equation 4

If you multiply equation 3 by 2 , you will get, 20u + 40v = 12 …equation 5

If you subtract equation 4 from equation 5, you will get

By cancelling out u, we get, v = 3/25

If you substitute v = 3/25 in equation 3, you will get,

10u + 20(3/25) = 6

10u + 12/5 = 6

10u = 18/5

u = 9/25

From the values of u and v, you can find the downstream and upstream speeds as shown below.

Upstream speed = X = 1/u = 25/9 km

and Downstream speed = Y = 1/v = 25/3 km

You can now calculate the speed of the man in still water, using our familiar formula.

Speed of the man in still water = ½ (downstream speed + upstream speed)

= ½(25/3 + 25/9)

=½(100/9) = 50/9 = 5.6 kmph

Also, you know the formula for speed of the current. 

Speed of the current = ½(downstream speed – upward stream)

= ½(25/3 – 25/9)

=1/2(50/9) = 25/9 = 2.8 km/h

Aptitude Train Problems

Aptitude Train Problems

3 Types Of Train Problems For Bank Exams

Type I: Time Taken by a Train to Cross a Platform or a Man or a Pole

In this type, you have to find the time taken by a train to cross a platform or man or pole. You have to learn an easy formula to solve this type of questions. See the below example.

Example Question 1: A train is moving at a speed of 120 km/hr. If the length of the train is 130 metres, how long will it take to cross a railway platform 170 metres long?

Solution:

From the question, you can write down the below values.

Length of the train = 130 m

Length of the platform = 170 m

Speed of the train = 120 km/h

You have to convert the speed in km/h to m/s. This is because the lengths of the train and platform are given in metres and not in kilometres.

To convert speed from km/h to m/s, you have to multiply it by 5/18.
Speed of the train in m/s = 120 x 5/18 = 100/3 m/s.

Let Lt be the length of the train and Lp be the length of the platform.

Then, you can find the time taken by the train to cross the platform using the below formula.

Time to cross the platform = Lt + Lp / Speed of the train

You know Lt = 130 m, Lp = 170 m and Speed of the train = 100/3 m/s

Therefore, Answer = (130 + 170) / (100/3)

= 300 x 3 /100 = 9 sec

 

Type II: Time Taken for 2 Trains to Cross Each Other

In this type, lengths of 2 trains and their speeds will be given. You have to find the time taken for the first train to cross the second.You have to remember 2 simple formulas to solve this type.

Assume below values:

L1 = Length of the first train

L2 = Length of the second train

S1 = Speed of the first train

S2 = Speed of the second train

Case I: Trains travelling in opposite directions:

Time taken for 2 trains to cross each other = (L1 + L2) / (S1 + S2)

Case II: Trains travelling in same direction:

Time taken for 2 trains to cross each other =

(L1 + L2) / (S1 – S2) when S1 > S2
 

Or

(L1 + L2) / (S2 – S1) when S2 > S1

Here is your example question.

Example Question 2: Two trains of lengths 120 metres and 150 metres respectively are running towards each other on parallel lines. Speed of the first train is 40 km/h and that of the second is 46 km/h. In how much time the first train will cross the second?

Solution:
From the question, you can write down the below values

L1 = 120 m, L2 = 150 m, S1 = 40 km/h and S2 = 46 km/h

As you can see, the speeds are in km/h units but lengths are in metres. Therefore, you have to convert speeds to m/s units.

S1 = 40 x 5/18 m/s

S2 = 46 x 5/18 m/s

According to the formula (that we saw above),

Time taken for the trains to cross each other = 120 + 150 / (40 x 5/18 + 46 x 5/18)

= 120 + 150 / (215/9 )

= 170 x 9 / 215 = 7.12 sec

= 7 sec (approx)

2 Important Points to Note:

1. Instead of 2 trains, if you find a train and a man in question, you have to assume the length of man to be zero.

2. If pole is given in question and length of the pole is not given, you can assume its length to be zero as well.

 

Type III: Equations Based Train Problems

This type can be an extension to type 1 or type 2 or both. Based on data given, you have to form equations to solve this type. Below is an example.

Example Question 3: A train running at 60 km/h takes 10 seconds to pass a platform. Next it takes 5 seconds to pass a man walking at 6 km/h in the same direction. Find the length of the train and the length of the platform.

Solution:
From the question, you know that the train at 60 km/h takes 10 seconds to cross the platform. (Speed of train in m/s = 60 x 5/18 = 50/3 m/s)

Part 1:

In type 1, you saw the below formula.

Time taken for a train to cross a platform = Lt + Lp / Speed of the train

You have to substitute the values (from question) in above equation and simplify as shown below.

10 = Lt + Lp / (50/3)

500 = 3(Lt + Lp)

Or Lt + Lp = 500/3 … equation 1

Part 2:

From the second half of the question, you know that the train takes 5 seconds to cross a man walking 6 km/h in the same direction.

(Speed of man in m/s = 6 x 5/18 = 5/3 m/s)

You know the formula for time taken for 2 trains travelling in same direction to cross each other.

Time taken = (L1 + L2) / (S1 – S2)

If instead of 2 trains, 1 train and 1 man is given, you have to assume his length to be 0.

Therefore L2 = 0. 

S1 = Speed of the train = 50/3 m/s,

S2 = man’s speed = 5/3 m/s

and time taken for train to cross the man = 5 seconds

You have to substitute the values in the formula, Time taken = (L1 + L2) / (S1 – S2).

You will get,

5 = L1 + 0 / (50/3 – 5/3)

5 = L1 / 15

L1 = 75 m

Part 3: (Finding Lp)

In part 2, you found L1 = 75m. L1 is nothing but the length of the train Lt.

If you substitute Lt = 75 m in equation 1, you will get,

75 + Lp = 500/3 

Lp = 500/3 – 75 = 91.6m

Your answer is as follows.

Length of the train = 75 m and

Length of the platform = 91.6 m

Aptitude Simple And Compound Interest Problems

Simple And Compound Interest Problems

Type 1: Simple Interest Formula Based Direct Problems. (Easy But They Can Be Twisted. See Why)

You will find this type to the easiest of all the 4 types we are going to discuss here. In this type, you will be applying the direct simple interest formula.

Here is your familiar simple interest formula
SI = PNR/ 100

The four variables in the above formula are : 

SI = Simple Interest

P = Principal Amount (This the amount invested)

N = Number of years 

R = Rate of interest (per year) in percentage

Note: Amount A you will get by investing an amount P for N number of years at R percent rate per annum will be
 

A = SI +P

Can they be twisted? Yes. See Why

A questioner can give you all values for the variables except one variable. For example, the questioner may give you SI, P and R and ask you to find N. But, nowadays these type of simple questions are hard to find. This type can be twisted with so many variations but the basics is still the same. Below example question is a slightly twisted version.

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

Example Question 1: A certain sum of money amounts to Rs.2000 in 2 years and to Rs.2500 in 3 years. Find the sum and rate of interest.

Solution:
Let P be the amount invested.

You know that the amount becomes 2000 in 2 years and 2500 in 3 years. You can see that the amount increases by Rs. 500 between 2nd and 3rd years.

Therefore, you can easily say that the simple interest for 1 year = 2500 – 2000 = 500

So, simple interest for 2 years = 500 x 2 = 1000

From the question, you know that the amount A after 2 years = 2000

Now using the formula A = P + SI,

you can write P = A – SI = 2000 – 1000 = 1000

Now, you know P = Rs.1000, N = 2 years and Simple interest SI = Rs.1000.

If you substitute above values in the formula SI = PNR / 100, you will get R as shown below:

R = (100 x 1000) / (1000 x 2)

R = 50%

Therefore, the sum invested P = 1000 and rate of interest R = 50%.

 

Type 2. Compound Interest Formula Based Direct Problems. (Are They Similar To Type 1?)

In this type, you will be getting problems dealing directly with the below compound interest formula.
 

CI = [P(1 + R/100)N] – P

The variables in the above formula are as follows:

CI = Compound Interest

P = Principal (Amount invested)

R = Rate of interest in percentage per year

N = Number of years

Note: There is a special case when the interest is compounded (calculated and added to the amount invested) half yearly instead of yearly basis. In that case, CI formula becomes
 

half yearly CI = [P(1 + (R/2)/100)2n] – P

Like type 1 (simple interest), this type can be twisted. But here we are going to see a straightforward example.

Example Question 2: Find compound interest on Rs.6000 at 12% per annum for 2 years compounded annually.

Solution:
From the question, you know that N = 2 years, R = 12%,P = Rs.6000

If you substitute the above values in the formula CI = [P(1 + R/100)n] – P, you will get

CI = [6000(1 + 12/100)2] – 6000

= (6000 x 28/25 x 28/25) – 6000

= 7526.40 – 6000

= Rs.1526.40

 

Type 3: Difference Between Compound And Simple Interests

This type is based o the difference between simple and compound interests. For example, in the below example, you will be given the difference between SI and CI and you have to calculate principal:

Example Question 3: The difference between the compound interest and simple interest on a certain sum at 12% per annum for 2 years is Rs.700. Find the sum.

Solution:
You have to start by assuming principal to be X.

From the question, you know that N = 2 years, R = 12%

Case 1: Let us start with Compound Interest:

You have to apply P = X, N = 2 and R = 12 in compound interest formula as shown below:

CI = [P(1 + R/100)n] – P

= (X[1 + 12/100]2) – X

= (X[112/100]2) – X

= (X[28/25 x 28/25] ) – X

CI = 159X / 625 …. equation 1

Case 2: Now let us move on to Simple Interest

Now you have to substitute P = X, N = 2 and R = 12 in simple interest formula.

SI = PNR/100

= (X x 2 x 12) / 100

= 24X/100 

SI = 12X/50 … equation 2

Case 3: Now let us find P

You know from the question that the difference between simple and compound interests is 700
Therefore,

CI – SI = 700 (We are writing CI first because CI will be higher than SI same rate of interest and same number of years)

If you substitute CI and SI values from equations 1 and 2, you will get

159X/625 – 12X/50 = 700

(318X – 300X)/1250 = 700

18X/1250 = 700

700 x 1250 = 18X

Or, X = 48611.11

Therefore, Principal (Sum invested) = Rs.48611.11

 

Type 4: Direct Problems With Both SI And CI (How Type 3 Is Different From Types 1 and 2?)

This type is a combined type of types 1 and 2. In this type, you have to apply both simple and compound interest formulas according to the question.

You will understand this after reading the below example.

Example Question 4: If the simple interest on a sum of money at 6% per annum for 4 years is Rs.1600, then find the compound interest on the same sum for the same period at the same rate.

Solution:
From the question, you know that R = 6%, N = 4 years, SI = Rs.1600

If you apply the above values in the simple interest formula SI = PNR/100, you will get

1600 = P x 4 x 6 / 100

Or P = (1600 x 100) / 6 x 4

P = 6333.33

Using the above value of P, you have to now calculate CI as shown below: 

CI = [P(1 + R/100)n] – P

= [6333.33(1 + 6/100)4] – 6333.33

= [6333.33 (106/100)4] – 6333.33

= [6333.33 x 53/50 x 53/50 x 53/50 x 53/50] – 6333.33

= 7995.68 – 6333.33
= Rs.1662.35