Quantitative Aptitude Part – 2 is developed with a view to the latest examination pattern of various banking services and government  jobs. Every care is taken to solve questions through Shot – Cut  Methods  instead of traditional methods.

Quantitative Aptitude Part – 2 to Ensure Your Success

SIMPLE INTEREST & COMPOUND INTEREST

The money paid by the borrower to the leader for the use of money lent is called the Interest and the sum lent is called the Principal.

The sum of Principal and interest is called Amount .

Rate  =  Interest on Rs. 100 for 1 year.

Interest is of two kinds : 

• Simple Interest and

• Compound Interest

When interest is calculated on the original principal for any length of time it is called Simple Interest (S.I.) .

Where, 

P stands for Principle ; I stands for Simple Interest.

T stand for Time ( in years ).

R stands for Rate per cent per annum ( ie. per year )  &  A stands for Amount.

If the Interest is reckoned in 2nd year on amount of 1st year, in 3rd year on amount of second year and so on, then such an interest is called Compound Interest (C.I.)

Example 2. The simple interest on a certain sum of money at 5% per annum for 8 years is Rs. 840. At What rate of interest the same  amount of interest can be received on the same sum after 5 years ?

Solution :      Sum =   100 x 840 / 5 x 8   =   Rs. 2100/- .   

Now, the new rate per cent =   100 x 840 / 2100 x 5  =  8 % .

( Here, S.I. is Rs. 840, sum is  Rs. 2100 and time is 5 years ) .

Explanation:

Let the principal be Rs.P

S.I.=Px5x8/100=Rs.840

P=840×100/5×8=Rs2100

Required rate = (840×100)/(2100×5)=8%

The answer is 8%.

Example 3. At what rate % per annum will a sum of money double itself in 20 years ?

Solution :      Suppose the sum is Rs.100. So Rs.100 amounts to Rs. 200  in 20 years .

P   =    100, I = 200 – 100  = 100, T = 20 .           

R   =   100 x 100 / 100 x 20   =  5%

Example 4. A sum of money lent out at simple interest doubled itself in 5  years . In how many years will it become 6 times itself ?

Solution :    Let the sum of money be Rs.P.

Simple Interest for 5 years = Principal Amount

S.I.=PxRx5/100=P R=100/5=20%

The amount after 5 years is 2P.

Simple Interest for the next 5 years on the amount 2P = 4P

S.I.=2PxRx5/100=4P

R=100/2=50%

Therefore, the sum will become 6 times itself in (5 + 5) = 10 years.

Example 5. A Sum of money , placed at Compound Interest , doubles itself in three years.  In how many years will it amount to eight times itself ?

Solution ( SHORT – CUT )  :

Suppose the sum is X. Now it becomes  2X in 3 years.

If we reinvest 2X for another 3 years, it becomes 4X in another 3 years i.e. X become 4X in 6 years.

Reinvest 4X for another 3 years, it becomes 8 X  i.e. X becomes 8X in 9 years.

Example 7. A lent Rs. 1400 to B for 2 years and Rs. 120 to C for 5 years and received  from both Rs. 102 as interest. Find the rate per cent .

Solution :     Let the rate of interest be R% p.a.

Interest received from B = 1400 x R x 2/100

= 28R Interest received from C = 120 x R x 5/100 = 6R

Total interest received = 28R + 6R = 34R = 102

R = 3

Therefore, the rate of interest = 3% p.a.

Example 8. In what time will the simple interest on Rs. 9000 at 5% be  equal to the  interest on Rs. 6000 for 10 years at   3% ?

Solution :      Let T be the time,

As interest on both the sums is equal,

So, ( 9000 x 5 x T ) / ( 6000 x 3 x 10 / 100 )         

or    T = 4 years.

Explanation:

Simple Interest = PRT/100

Given: Principal (P1) = 9000

Rate of Interest(R1) = 5%

Time (T1) = ?

Principal (P2) = 6000

Rate of Interest(R2) = 3%

Time (T2) = 10 years

Simple Interest(SI1) = Simple Interest(SI2) (P1R1T1)/100 = (P2R2T2)/100

=> (90005T1)/100 = (6000310)/100

=> T1 = 4 years.

Hence, the simple interest on Rs. 9000 at 5% will be equal to the interest on Rs. 6000 for 10 years at 3% in 4 years.

Example 9.  A sum of Rs. 2000 is lent in two parts so that interest on the first part for 2 years at 4% may be equal to the interest on the second part for 4 years at 2% . Find the first sum .

Solution :      Let P be the first sum.

So, the second sum would be ( 2000 – P ).

Now,

( P x 2 x 4 / 100 ) = [ ( 2000 – P ) x 4 x 2 / 100 ]

or    P  =  Rs. 1000/-

Example 10. Find the difference between S. I. and C.I. on Rs. 1000 at 10% for 2 years .

Solution ( SHORT – CUT ) 

S.I. on Rs. 1000 for 1 years at 10 % =  Rs. 100

S.I. on Rs. 100 for 1 year at 10% = Rs. 10

Hence , the difference between S.I. and C.I. = Rs. 10

Example 11. The S.I. on a sum of money lent out for 2 years at 10% is Rs.100. What is  C.I.?

Solution :   

S.I. for 1 year = Rs. 50

S.I. on Rs. 50 for 1 year  = 50 x 10 x 1 / 100  =  Rs.5

C.I. for 2 years = 50 + ( 50 + 5 )  =  Rs. 105

Example 12. The C.I. on a certain sum of money for 2 years is Rs. 104 and the S.I. for two years at the same rate is Rs. 100. Find the rate per cent .

R = [(100*C.I.)/(S.I.*(1 + R/100))]^0.5

Where R is the rate of interest in % C.I. is the compound interest S.I. is the simple interest

In this case,

R = [(100*104)/(100*100*(1 + R/100))]^0.5

= [(104/10000)]^0.5 = 9%

Hence, the rate of interest is 9%.

SPECIAL NOTE :

The ^ symbol in the shortcut formula for finding the rate of interest means “to the power of”. So, in the formula, (1 + R/100)^0.5 means (1 + R/100) to the power of 0.5.

The ^ symbol can also be used to mean other things in mathematics, such as exponentiation and Cartesian product. It is important to be aware of the context in which the ^ symbol is being used to determine its meaning.

Example 13. Sanjay invested a sum of money at C.I. It amounted to Rs. 3000 in 2 years and to Rs. 3600 in 3 years. Find  the rate per cent.

Solution :      Suppose the sum was Rs. X

Now , X  +  C.I.  in 3 years  =  Rs. 3600

X +   C.I. in 2 years  =  Rs. 3000

The difference of Rs. 600 ( Rs. 3600 – 3000 ) is the S.I. for 1 year on Rs. 3000

Rate %  =  600 x 100 / 3000 x 1   =  20%

AVERAGE

( or  Mean )

  Average = Sum of Quantities / Number of Quantities

The average of a set of quantities is calculated by adding all the quantities and then dividing by the number of quantities. For example, if you have the quantities 2, 3, 4, and 5, the average would be calculated as follows:

Average = (2 + 3 + 4 + 5)/4 = 3.5

The average is a measure of central tendency, which means that it indicates the center of the distribution of the quantities. In this case, the average of 2, 3, 4, and 5 is 3.5, which means that half of the quantities are greater than 3.5 and half are less than 3.5.

The average is a useful tool for comparing sets of quantities. For example, if you wanted to compare the average test scores of two different classes, you could calculate the average test score for each class and then compare the two averages.

The average is also a useful tool for making predictions. For example, if you knew the average height of a group of people, you could predict that a randomly selected person from that group would be about that height.

Case 1 a :      When a new person joins a group and average age (or weight etc.) of the group is increased .

Example  1. The average of a class of 50 boys, which is 16.9 years, is raised to 17 years by coming of a new boy. Find the age of the new boy.

Solution :  

Sum of the ages of 50 boys   = 50 x 16.9  = 845 years 

Sum of the ages of 51 boys  = 17 x 51     = 867 year

Age of the new boy = 867 – 845  = 22 years .

SHORT – CUT :

The average age of boys increased by 1 years.

Their total age is increased by 51 x 0.1 =  5.1 years 

The age of the new boy = 16.9 + 5.1  =  22  years

Example  2. The average age of 30 boys is 15 years. If the teacher’s age in  included,  the average age increases by 1. What is the teacher’s age ?

Solution ( SHORT – CUT ) :

The average age is increased by 1 year

Hence,  their total age is increased by  31  x  1  =  31 .  

Teacher’s age = 15 + 31 = 46 years.

CASE 1b : When a new person joins a group and average age ( or weight etc. ) of  the group is decreased.

Example  3.  The average age of 18 girls of a class is 22 years 2 months .By admission of one more girl, the average age is reduced to 22 years. Find the age of new girl.

Solution :     

The average age of 18 girls of a class is 22 years 2 months = 22(2/12) years = 22.5 years

The total age of 18 girls is 1822.5 = 405 years

The total age of 19 girls after the admission of one new girl is 1922 = 428 years

The age of the new girl is 428-405 = 23 years

So the answer is 23.

Example  4. The average of the runs made by 10 players in a cricket team of 11 players is 43. If the 11th player’s runs are also considered, the average of 11 players’ runs decreases by 1. How many runs did the eleventh player make ?

Solution :      Total runs of 10 players = 43 x 10 = 430

Total runs of 11 players = 42 x 11 =  462

Hence , the 11th player made 462 – 430  = 32 runs.

SHORT – CUT :

Since ,the average of the runs decreased by 1 run.

Hence , total runs decreased by 11 x 1 = 11 runs 11th player’s runs =  43 – 11  = 32 runs.

CASE 2 a :    When a person leaves a group and average ( age, weight etc.) of the group is increased .

Note  :  In this case, the person who leaves the group would always be of  lesser weight ( or age etc. ) than  the “ average” of group.

Example  5. The average age of a class of 50 boys, which is 16.9 yrs., is increased to 17 years when a boy left the class.  Find the age of the boy who left the class.

Solution ( SHORT – CUT ) :

Since , the average age of 50 boys increased by 0.1 .

Hence , their total age is increased by 49 x 0.1 = 4.9 years .

The age of the boy leaving the class = 16.9 – 4.9 = 12 year .

Example  6 .The average of Shreya’s marks in 7 subjects is 68. His average in 6 subjects, excluding history, is 70. How many marks did he get in History ?

Solution :     

The average increased by 2 marks .

Hence , Total marks increased by 6 x 2 = 12

His marks in History = 68 – 12   =  56 marks .

CASE 2 b. When a person leaves the group and the average ( age, weight etc. ) of the group is decreased .

Note :  In this case the person who leaves the group would always be of greater weight (or age etc.) than the ‘average’ of the group.

Example  7. The average weight of 10 girls and a boy of a Class is 32 Kg. When the boy’s weight is not considered the average weight of the class decreases by 2 kg. Find the weight of the boy .

Solution : 

The total weight of 10 girls and a boy is 1032 = 320 Kg

When the boy’s weight is not considered, the total weight of 10 girls is 1030 = 300 Kg

Therefore, the weight of the boy is 320 – 300 = 20 Kg

So the answer is 20.

CASE 3a.      When a person belonging to a group is replaced by a new person and the average ( age,  weight etc.) of the group is increased .

Note : In this case , the person who replaces any person of a group would be of greater weight ( or age etc.) than the weight ( or age etc. ) of the person replaced .

Example  8. The average weight of 8 players of a team is increased by 1 kg. when one of the player, who weights 60 kgs., is replaced by a new player. What is the weight of the new player ?

Solution : 

The average weight of 8 players is increases by 1 kg.

Their total weight is increased by 8 x 1 = 8 Kgs .

The weight of the new player = 60 + 8   = 68 Kgs.

Example  9. The average age of 12 dancers is increased by 2 years, when two of them, whose ages are 22 and 26 years , are replaced by two new dancers. What is the average age of the new dancers ?

Solution :  

The average age of 12 dancers is increased by 2 years.

There total age increased by 12 x 2  = 24 years.

The total age of the two new dancers = 22 + 26 + 24 = 72

Hence , Average age of new dancers = 72 / 2  =  36 years.

Example 10. The average temperature for Monday, Tuesday and Wednesday was 40 ֹ C . The average for Tuesday , Wednesday and Thursday was 41 C  and that for Thursday being 45 ֹ C . What was the  temperature on  Monday ?

Solution: Let the temperatures on Monday, Tuesday, Wednesday and Thursday be M, T, W and TH respectively.

Given:

• (M + T + W)/3 = 40
• (T + W + TH)/3 = 41

Solving the two equations, we get:

• M = 39
• TH = 42

Therefore, the temperature on Monday was 39 degrees Celsius.

Example 11. The average temperature from the 6th to the 13th of January ( both days  Inclusive ) was 12.8 ֹ C . And from 7^th to 14^th January it was 13.4 ֹ  C . The  temperature on the 6th was 9.8 ֹ C .  Find the temperature on 14^th .

Solution : Given that average temperature from 6th to 13th January = 12.8°C

Total temperature from 6th to 13th January = 12.8 * 7 = 89.6°C

Given that average temperature from 7th to 14th January = 13.4°C

Total temperature from 7th to 14th January = 13.4 x 7 = 93.8°C

But the temperature on 6th was 9.8°C

Hence, the temperature on 14th = 93.8 – 89.6 – 9.8 = 4.4°C

So the temperature on 14th is 4.4°C.

Case 3 b :     When a person belonging to a group is replaced by a new person and the average ( age,  weight etc.) of the group is decreased.

Note : In this case, the person who replaces any person of a group would be of lesser weight ( age etc.) than the weight ( or age etc.) of the person replaced .

Example 12. The average weight of 20 girls in a Class is decreased by 500 grams when one of the girls who weights 35  kgs is replaced by a new girl. Find  the weight of the new girl .

Solution :      Average weight of 20 girls is decreased by 500 gms ( or ½ kg )  .

Hence ,their total weight is decreased by 20 x ½ =10 kgs.

Weight of the new girl = 35 – 10 = 25 kgs.

Some other Important Questions

Example  1. A cycle dealer sold 8 bicycles at  Rs. 1350 each and 7 bicycles at Rs.1200  each. Find the average price at which he sold them .

Solution :

Average price = (8 x 1350 + 7 x 1200) / (8 + 7)

= 10400/15 = Rs. 693.33

So the average price at which he sold the bicycles is Rs.693.33.

Example  2. The average age of three boys is 15 years and their ages are in proportion  3 : 5 : 7 . What is the age in   years of the youngest boy ?

Solution :     

Total age of three boys = 15 x 3 = 45 years.

Their age ratio    =    3 : 5 : 7

Hence , by Unitary Method ,  age of the youngest = 3/15 x 45  =  9 years .

Example  3. The average expenditure of a man for the first four months is Rs. 16000 and for the next eight months is Rs. 18000. Find his monthly average income if he saves Rs. 12000 during the year .

Solution : Total expenditure for the first four months = Rs.16000 x 4   =  Rs.64,000/- .

Total expenditure for the next eight months = Rs. 18,000 x 8  =  Rs. 1,44,000/- .

Total expenditure during the year = Rs.64,000 + Rs.1,44,000 =  Rs. 2,08,000/- .

Hence , his monthly average income =   20,8000 + 12000 / 12   =  Rs. 18,333.33/- .

Example  4. The age of three children in a family is 20% of the average age of father and the eldest child. The total age of the mother and the youngest child is 39 years. If the father’s age is 26 years, What is the age of the second child ?

Solution : 

Given that father’s age is 26 years.

Average age of father and eldest child = (26 + eldest child_age)/2

20% of the average age of father and eldest child = (20/100) x ((26 + eldest_child_age) / 2)

Total age of three children = 20% of the average age of father and eldest child.

The total age of the mother and the youngest child is 39 years.

Let the age of the youngest child be youngest child age. So, the age of the mother is 39 – youngest child age.

The total age of all the members of the family is 26 + eldest_child_age + 39 – youngest_child_age + 3 x (20/100) x ((26 + eldest_child_age) / 2) = 91

Solving for eldest child age, you get eldest child age = 18 years.

Time , Work & Wages , Pipe & Cistern

• Time: Time is the duration for which an activity is performed. It can be measured in seconds, minutes, hours, days, weeks, months, years, etc.
• Work: Work is the exertion of physical or mental effort in order to achieve a desired result. It can be measured in terms of the amount of effort required, the amount of time taken, or the amount of output produced.
• Wages: Wages are the payment given to someone for their work. They can be paid in terms of money, goods, or services.
• Pipes and Cisterns: Pipes and cisterns are used to transport and store water. They can be used to fill a cistern from a river or stream, or to empty a cistern into a field or garden.

Here are some of the basic formulas for Time, Work & Wages, and Pipes and Cisterns:

• Formula for calculating the amount of work done: Work = Force x Distance
• Formula for calculating the amount of time taken to complete a task: Time = Work / Rate
• Formula for calculating the hourly wage: Hourly wage = Total wage / Number of hours worked
• Formula for calculating the rate of water flow through a pipe: Rate of water flow = Volume of water / Time
• Formula for calculating the time taken to fill a cistern with water: Time taken to fill a cistern = Volume of water in the cistern / Rate of water flow

Time is inversely proportional to speed. This means that as speed increases, time decreases. And as speed decreases, time increases.

This can be explained by the formula:

Speed = Distance / Time

Where:

• Speed is the rate at which an object is moving.
• Distance is the length of the path that an object travels.
• Time is the duration for which an object travels a certain distance.

If we increase the speed, the distance will be covered in less time. And if we decrease the speed, the distance will be covered in more time.

For example, if we drive a car at 60 miles per hour, we will cover 60 miles in 1 hour. But if we drive the car at 30 miles per hour, we will cover 60 miles in 2 hours.

This is because the car will travel a longer distance in the same amount of time if it is going slower.

So, if a person takes 5 days to finish a work, he will do 1/5 of the work in one day .

If A can do a piece of work in 7 days and B in 6 days, the ratio of the work done by A and B in the same time is  6 : 7.  

It can be shown like this :

Some Solved Examples

Example  1. A can do a piece of work in 10 days and B can do it in 15 days. What time will they require to do it working together ?

Solution : A and B can do 1/10 + 1/15 or 1/6 of the work in 1 day.

Hence , A and B can do the whole work in 6 days.

Example  2.  A can do a piece of work in 12 days and B in 20 days. With the help of C,they finish the work in 5 days. How long will it take C to finish the work ?

Solution :       

Suppose, C can finish the work in C days.

Then, in one day, C can do 1/C of the work.

In one day, A can do 1/12 of the work.

In one day, B can do 1/20 of the work.

In one day, all three of them together can do 1/C + 1/12 + 1/20 of the work.

We are given that all three of them together can finish the work in 5 days.

Hence, in one day, all three of them together can do 1/5 of the work.

Therefore, we have 1/C + 1/12 + 1/20 = 1/5

Solving for C, we get C = 10 days.

Therefore, it will take C 10 days to finish the work.

Short-Cut :

1/C = 1/5 – 1/12 – 1/20 = 1/10

=> C = 10 days

Therefore, it will take C 10 days to finish the work.

Example  3.  A can do a piece of work in 12 days. If he is joined by B who is 50 % more efficient, in what time will A and B together finish the work ?

Solution :     

If A does X work in 12 days .

B does 150% of  X  or 1.5X work in 12 days.     

B does X work in 12/1.5  =  8 days

(A+B)’ s 1 day’s work  = 1/12 + 1/8  =  5/24

Hence , A and B can do the whole work in 24/5 days or  4.8   days .

Example  4. A is twice as good a workman as B and together they finish a  work in 18 days. In how many days can A alone do it .

Solution :     

A + A/2 or 3A/2 finish the work in 18 days  as B = A/2.                                               

Hence, A will finish the work in 3/2 x 18 =  27 days .

Example  5. A can do a piece of work in 25 days. He worked at it for 5 days , then B finished it in 12 days . In how many days can A and B together finish the work ?

Solution :     

A can finish the work in A days and B can finish the work in B days.

Work done by A in 1 day = 1/A Work done by B in 1 day = 1/B

We are given that A can do a piece of work in 25 days. He worked at it for 5 days, then B finished it in 12 days.

Work done by A in 5 days = 5 × (1/25) = 1/5 Remaining work = 1 – 1/5 = 4/5

This work is done by B in 12 days. Hence, work done by B in 1 day = (4/5)/12 = 1/15

Work done by A and B in 1 day = 1/A + 1/B = (1/25) + (1/15) = 8/75

=> A and B together can finish the work in 75/8 days.

Short-Cut :

Work done by A in 5 days = 5 x 1/25 = 1/5

Remaining work = 1 – 1/5 = 4/5

This work is done by B in 12 days.

Hence, work done by B in 1 day = 4/60 = 1/15

Work done by A and B in 1 day = 1/25 + 1/15 = 8/75

=> A and B together can finish the work in 75/8 days.

Special Note :

The symbol => is called the implication symbol. It means “implies” or “if…then”.

In the above equation, it means that if A and B work together, they can finish the work in 75/8 days.

Here is an example of how the implication symbol can be used in a sentence:

• If it is raining, then the ground is wet.

In this sentence, the implication symbol is used to show that the two statements are related. In other words, if the first statement is true, then the second statement must also be true.

Example  6.  A certain number of men can do a work in 100 days . If there were 18 men  more it could be finished in 10 days less . How many men are there ?

Solution : 

Men 1 x  Time 1  =  Men 2 x Time 2

Let the number of men initially be X.

We know that, more men, less days (indirect proportion)

Hence, we can write as,

X ∝ 100 X × 100 = (X + 18) × 90 100X

= 90X + 1620 10X = 1620 X = 162

Therefore, there were 162 men initially.

Example  7.  If 4 men or 6 women can do a work in 20 days . How long will 4 men and 9 women take to do it ?

Solution ( SHORT – CUT ) :

4 men is equivalent to 6 women in terms of completing the work. So, 1 man is equivalent to 6/4 = 1.5 women.

Hence, 4 men + 9 women = 4 + 9 × 1.5 = 13.5 women.

6 women can do the work in 20 days. So, 13.5 women can do the work in 6 × 20 / 13.5 = 9 1/3 days.

Therefore, 4 men and 9 women can do the work in 9 1/3 days.

Example  8. Pipe A can fill a tank in 8 hours and pipe B can fill in 10 hours .If they are opened on alternate hours and if pipe A is opened first, in how many hours the tank shall be full ?

Solution :     

The time it takes to fill the tank when the two pipes are opened on alternate hours can be calculated using the following steps:

1. Calculate the part of the tank filled by each pipe in one hour.
2. Find the least common multiple of the two rates.
3. Divide the least common multiple by the sum of the two rates.
4. Multiply the result by 2, since the pipes are opened on alternate hours.

In this case, we have:

• The part of the tank filled by pipe A in one hour is 1/8.
• The part of the tank filled by pipe B in one hour is 1/10.
• The least common multiple of 8 and 10 is 40.
• The sum of the rates of pipe A and pipe B is 1/8 + 1/10 = 9/40.

So, the time it takes to fill the tank is

time = (2 * 40) / (9/40) = 8 8/9 hours.

Therefore, the tank will be full in 8 8/9 hours.

Solution ( SHORT – CUT ) :

1. Find the rate at which the two pipes fill the tank together. This is the sum of their individual rates, which is 1/8 + 1/10 = 9/40.
2. Divide the capacity of the tank by the combined rate to find the number of hours it takes to fill the tank with both pipes open. This is 40/9 hours.
3. Since the pipes are opened on alternate hours, the actual time it takes to fill the tank is twice this amount, or 8 8/9 hours.

Example  9. Two pipes A and B can fill a cistern in 10 and 15 minutes respectively. Both are opened together, but at the end of 5 minutes the first  is turned off. How much longer will the cistern take to fill ?

Given that pipe A can fill a cistern in 10 minutes and pipe B can fill in 15 minutes.

Part of the cistern filled by pipe A and pipe B together in one minute = (1/10) + (1/15) = 1/6

In 5 minutes, the part of the cistern filled by pipe A and pipe B together = 5 x (1/6) = 5/6

Remaining part of the cistern = 1 – 5/6 = 1/6

Since pipe A is turned off, only pipe B will fill the remaining part of the cistern.

Time taken by pipe B to fill the remaining part of the cistern = (1/6) / (1/15) = 5 minutes

So, the cistern will take 5 minutes more to fill.

Example  10.  4 men can earn as much as 5 women and 3 women earn as much as 8   boys.  If 6 boys earn Rs. 3600 daily what are the daily wage of a man ?

Solution ( SHORT – CUT )  

1. Let the daily wage of a man be m.
2. The daily wage of a woman is 800/3 rupees.
3. The daily wage of a boy is 600 rupees.

4m = 5 × (800/3) => m = 125 rupees .

Explanation:

Given:

• 4 men can earn as much as 5 women.
• 3 women earn as much as 8 boys.
• 6 boys earn Rs. 3600 daily.

To find:

• Daily wage of a man.

Solution:

The daily wage of a boy = 3600/6 = 600 rupees.

The daily wage of a woman = 600 × 8/3 = 1600/3 rupees.

The daily wage of a man = 1600/3 × 5/4 = 125 rupees.

Therefore, the daily wage of a man is 125 rupees.

Example  11 . A alone can do a piece of work in 15 days. B alone can do it in 25 days. If the total wages for work are Rs. 12,000 . How much A and B is paid if  they both finish the work together ?

Solution :     

Rule  :  Wages are paid in proportional to units of job done by each in the same time .

The wages of A and B will be distributed in the ratio of their individual efficiencies.

The efficiency of A = 1/15

The efficiency of B = 1/25

The ratio of their efficiencies = 5 : 4

So, A will get 5/9 of the total wages and B will get 4/9 of the total wages.

The amount paid to A is 5/9 × 12000 = Rs. 6666.66

The amount paid to B is 4/9 × 12000 = Rs. 4000.00

Therefore, A will be paid Rs. 6666.66 and B will be paid Rs. 4000.00 if they both finish the work together.

Short – Cut:

1. Let the total wages for A be a and the total wages for B be b.
2. The ratio of their wages is 5 : 4.
3. So, a/b = 5/4.
4. The total wages for the work is a + b = 12000.
5. Substitute the value of a/b in the equation a + b = 12000.

a + b = 12000

5/4a + 5/4b = 12000

a + b = 9600

6. The amount paid to A is a = 9600 × 5/9 = 6666.66
7. The amount paid to B is b = 9600 × 4/9 = 4000.00

Therefore, A will be paid Rs. 6666.66 and B will be paid Rs. 4000.00 if they both finish the work together.

 TIME & DISTANCE

RULE – 1

  DISTANCE   =  SPEED  x  TIME  

From the above equation we can conclude that :

(i)         If speed is constant, distance increases or decreases with time .

(ii)        If time is fixed, distance increases or decreases with speed.

(iii)       If distance is constant, speed increases or decreases with decrease or increase of time .

RULE – 2

CONCEPTS OF RELATIVE SPEED

• If two bodies move in opposite direction, their relative speed can be obtained by adding their speeds .

• If two bodies move in same direction, their relative speed can be obtained by subtracting their speeds .

OTHER CONCEPTS

When two bodies ( or trains ) Cross each other, add their lengths .

• When faster train overtakes the slower train , add their lengths .

• Only length of  the faster train is considered when a faster train crosses or overtakes a person sitting in the slower train.

• Only length of the slower train is considered when a slower train crosses a person sitting in a faster train.

• If a train crosses a platform or bridge or a tunnel, the length of the train is added to the length of platform, bridge or a tunnel .

• If a train crosses a man or an electric pole, the length of only the train is considered .

RULE – 3

Memorize The Following Chart

18 Km/Sec    =         5 meter/sec.

36 Km/Sec     =         10 meter/sec.

54 Km/Sec     =         15 meter/sec.

72 Km/Sec     =         20  meter/sec.

15 Km/Sec     =         25/6 meter/sec.

30 Km/Sec     =         25/3 meter/sec

45 Km/Sec     =         25/2 meter/Sec

60 Km/Sec     =         50/3 meter/sec

90 Km/Sec     =         25 meter/sec.

 CONVERSION  FACTORS

To convert kilometer/hour into meter/second	Multiply by  5/18
To convert Meter/second into kilometer/hour	Multiply by  18/5
To convert kilometer/hour into meter/minute	Multiply by  50/3
To convert Meter/minute into Kilometer/hour	Multiply by  3/50

Some Solved Examples

Example 1 :  A train traveling with a speed of 75 km/hour speeds past a pole in 12 seconds. What is its length in meters?

Solution :  Length of the train = distance covered in 12 seconds at the rate of 75  Kms/hour. Hence , the required length = 12 X 75 X 5/18 = 250 meters.

Explanation:

Given:

• Speed of the train = 75 km/hour
• Time taken to cross the pole = 12 seconds

To find:

• Length of the train in meters

Solution:

First convert the speed of the train from km/hour to m/s.

Speed = 75 km/hour

= 75000 m / 3600 s

= 20.833 m/s

The length of the train is equal to the distance traveled by the train in 12 seconds.

Distance = Speed × Time

= 20.833 m/s × 12 s

= 250 meters

Therefore, the length of the train is 250 meters.

Example 2 :  Two trains are running in opposite directions with speeds of 62 Km/hr and 40 Km/hr respectively. If the length of one train is 250 meters and they cross each other in 18 seconds, what will be the length of the other train ?

Solution :      

Given that:

• Speed of the first train = 62 Km/hr
• Speed of the second train = 40 Km/hr
• Length of the first train = 250 meters
• Time taken to cross each other = 18 seconds

The relative speed of the two trains = 62 + 40 = 102 Km/hr.

102 Km/hr = 102000 m / 3600 s = 28.6 m/s

The total distance traveled by the two trains to cross each other is equal to the sum of their lengths.

Distance = Speed × Time

= 28.6 m/s × 18 s = 514 meters

Therefore, the length of the second train is 514 meters – 250 meters = 264 meters

So the answer is 264.

 Example 3 : A 160 meter long train crosses another 180 meter long train running at a speed of 80 Km/hr in the opposite direction in 6 seconds. What is the speed of the first train in Km/hr.

Given that:

• The length of the first train = 160 meters
• The length of the second train = 180 meters
• The relative speed of the two trains = 80 Km/hr
• Time taken to cross each other = 6 seconds

The relative speed of the two trains = (80 + 0) km/hr = 80 km/hr

80 Km/hr = 80000 m / 3600 s = 22.22 m/s

The total distance traveled by the two trains to cross each other is equal to the sum of their lengths.

Distance = Speed × Time

= 22.22 m/s × 6 s = 133.32 meters

Therefore, the speed of the first train is (160 + 180)/6 = 44 m/s

Converting m/s to km/hr

44 m/s × 3600/1000 = 158.4 km/hr

So the answer is 158.4 .

Example 4 :  A man covers a distance between A and B in 45 minutes. If the speed is reduced by 5 Km/hr, he will cover it in 48 minutes. What is the speed of  man ?

Solution :

1. Let the original speed be X km/hr.
2. The new speed is (X – 5) km/hr.
3. The time taken to cover the distance is increased by 3 minutes = 3/60 hour = 1/20 hour.
4. The total distance traveled by the man is the same in both cases.

X × 45/60 = (X – 5) × 48/60

9X = 8(X – 5)

X = 40 km/hr

Therefore, the original speed of the man is 40 km/hr.

Example 5 :   Two trains travel in the same direction at 66 Km/hr and 30 Km/hr. The faster train passes a man in the slower train in 12 Sec. Find the length of the faster train .

Solution :  The faster train will pass the man when it has gained a distance equal to its own length .

Now, in 1 hour the faster train gains 66 – 30  0r  36 Km.

Hence , in 1 second the faster train gains  36×5/18 = 10 m  

In 12 seconds the faster train gains 10 X 12  = 120 m. 

The required length is 120 meters .

Example 6 :  A train starts from station ‘X’ at 60 Km/hr. and reaches station ‘Y’ in 45 minutes. If the speed is reduced by 6 Km/hr, how much more time will the train take to return from station ‘Y’ to station ‘X’ ?

Solution ( SHORT CUT )

          SPEED 1  x Time 1    =   Speed 2  x  Time 2  

Hence , 60 X 45  =  ( 60 – 6 ) x Time 2          

or              

 Time 2 = 60 x 45 / 54   =   50 minutes

Hence , extra  time taken = 50 – 45  =  5 minutes.

Explanation:

Given that:

• The speed of the train from station X to Y = 60 Km/hr
• The time taken to reach station Y from station X = 45 minutes

The distance between station X and station Y is equal to the product of the speed of the train and the time taken.

Distance = Speed × Time

= 60 × 45/60 Km = 45 Km

If the speed of the train is reduced by 6 Km/hr, the new speed of the train = 60 – 6 = 54 Km/hr.

The time taken by the train to return from station Y to station X = 45/54 hours

= 45 × 60/54 minutes = 40 minutes.

The extra time taken by the train to return from station Y to station X = 40 – 45 = 5 minutes.

So the answer is 5.

Example 7 :  The distance between two cities ‘X’ and ‘Y’ is 300 Kms. A train leaves city ‘X’ and travels toward city ‘Y’ at a speed of 60 Km/hr. At the same time another train starts from city ‘Y’ and travels towards city X at 90 Km/hr. After how many hours do the two trains meet ?

Solution :      The trains are traveling in opposite direction, hence their relative speed = 60 + 90 = 150 Km/hr.

Distance   =   300 Kms.

Hence , time after which the trains meet  =   Distance / Speed = 300 / 150 =  2 hours.

Explanation :

Given that:

• The distance between city X and city Y = 300 Kms
• The speed of the train leaving city X = 60 Km/hr
• The speed of the train leaving city Y = 90 Km/hr

The two trains are moving in opposite directions, so their relative speed is equal to the sum of their speeds.

Relative Speed = 60 + 90 = 150 Km/hr

The time taken by the two trains to meet = Distance/Relative Speed

Time = 300 / 150 hours = 2 hours .

So the answer is 2 .

Example 8 :  A train traveling at 60 Km/hr leaves city ‘X’ at 6 a.m. and another train traveling at 80 Km/hr starts at 9 a.m. in the same direction. At what time are two trains together .

Solution :      In 3 hours ( between 6 a.m. and 9 a.m.) the first train gains 180 Kms. over the second train .

The second train gains 80 – 60 = 20 Km/hr over the first train.

Hence , the two trains are together after  180 / 20  = 9 hours from  9 a.m. i.e. at  6 a.m.

Example 9 : Ram has to be at a certain place at a certain time and finds that he shall be 15 minutes late if he walks at 6 Km/hr and 10 minutes  soon if he walks at 9 Km/ hr. How far he has to walk ?

Solution :     

Suppose Ram has to walk X Km.

Now, to walk X Km. He requires X/6 hours in first  case and X/9  hours in the second case.

Hence, he saves X/6  – X/9  i.e. X/18 hours in the second case which is equal to 15 + 10 or 25 minutes ( or 25/60 hrs. )

Hence , X/18 = 25/60      or      X  = 7.5 Kms.

Example 10 :  Walking 3/4 of his usual speed, a man is two hours late. Find his usual  time.

Solution :  

Since the man walks at 3/4 of his usual rate, the time that he takes is 4/3 of his usual time . ( as Time ∝  )

Hence, 4/3 of usual time = usual time + 2 hours.

1/3 of usual time   =   2  hours

Hence , Usual time   =  2 x 3 hours   =   6 hours.

Example 11 : A man takes 6 hours and 15 minutes to walk a certain distance and riding back. He could walk both ways in 7 hours and 45 minutes. How long could it take him to ride both ways ?

Solution :  One way walking time  + One way riding time = 6 hours 15 minutes     

or    2 ways walking time + 2 ways riding time  = 12 hours 30 minutes .

Hence , 2 ways riding time   =  12 hrs. 30 minutes – 2 ways walking time

                                          = 12 hrs. 30 minutes – 7 hrs. 45 minutes  =  5 hrs. 15 minutes .

Example 12 : A man goes to a certain place at 20  Kms./hr and returns from that place at 30 Kms/hr. What was his  average speed ?

Solution (SHORT – CUT)  

RULE : In this kind of questions, when distance is same :

         Average Speed  =    2XY / X + Y

Where, X and Y are first and second speeds respectively.

In the above question, X is 25 Kms./ hr. and Y is 30 Kms/hr.  Average Speed   =   2 x 20 x 30 / 20 + 30  =   24 Kms/hr.

STREAMS

If U is the speed of a boat in still water and V is the speed of the river ( or current or stream )

U + V = boats speed down stream ( i.e. with the stream )

U – V =  boats speed up stream ( i.e. against the stream )

U = Down Rate + Uprate / 2

V = Down Rate – Uprate / 2

Some Solved Examples

Example 1 :  The speed of a boat in standing water is 20 Kms./hour and that of the current is 4 Kms./hour. Find the distance traveled in 12 minutes (1) down stream and (2) up stream .

Solution :      U   =  20 Kms./hr.        V   =   4 Kms./hr.

Hence , Speed of boat down stream   =  U + V  or  20 + 4    =   24 Kms./hr.

Hence , distance covered in 12 minutes  =  24 x 1/5 = 4.8 Kms. ( as 12 minute is 1/5th of 60 minutes )

Similarly, Speed of boat up stream    =     U – V =  20 – 4    =     16 kms./hr.

Hence , 16 x 1/5    =    3.2 Kms.

Example 2 : A man rows up stream 18 Kms. and down stream 24 Kms. taking 6 hours  each time. Find the velocity of current .

Solution ( SHORT – CUT ) 

Down rate = 24/6  =  3 Kms./hr. , Up rate = 18/6 = 3 Kms./hr.

Hence , velocity of the current ( V )  =   4 – 3 / 2  =  0.5 Km/hr.

Example 3 : A boat moves down stream at the rate of 2 Kms in 10 minutes and up  stream at the rate of 8 Km/hr. What is the speed of boat in still water ?

Solution :      Speed of boat down stream = 2 x 6 = 12 Km/hr.

Speed of boat up stream  =  8 Km/hr .

Hence ,   U   =   12 + 8 / 2   =  10 Km/hr.

Example 4 : A person can row 18 Kms./hr in still water and he finds that it takes him twice as long as to row up as to row down the river. Find the ratio of the stream .

Solution  (  SHORT – CUT ) 

According to the question ,   U + V   =  2 ( U – V )

( As Speed ∝  1 / Time )

Hence , U =   3 V        

or          V =   18/3  =  6 Km/hrs

Quantitative Aptitude Part – 1

Quantitative Aptitude Part – 3