Problem: 

Normal Method:

Add 2 on both sides we get

Shortcut Method:

then

that means from the given problem

---

Problem: 

Normal Method:

Subtract 2 from both sides we get

Shortcut Method:

then

that means from the given problem

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Short cut Method:

Solution:

Example:

Solution:

---

Problem: 

Normal Method:

According to Standard Algebraic Identities:

then by substituting the given values we get

Shortcut Method:

Given that 

Let a = 1 and b = 3

then  and 

Let's pick 

then  and 

then

or

therefore

---

Normal Method:

Given that 

Squaring on both sides

According to Standard Algebraic Identities:

then by substituting the given values we get

Shortcut Method:

Given that 

Let's pick 

then  and

therefore

---

Solving Linear Equations without Quadratic Form:

Problem: Solve the below linear equation for x.

Solution:

The value of x can be solved by rewriting the R.H.S into the form of L.H.S

Then,

and

Example 1:

Solve the below linear equations for x.

Solution:

Example 2:

Solve the below linear equations for x.

Solution:

Then,

and

---

Solving Linear Equations in One Variable Orally:

Type 1:

Then the value of x will get using the below shortcut method:

Example 1:

Solve the below linear equations for x.

Solution:

Example 2:

Solve the below linear equations for x.

Solution:

Example 3:

Solve the below linear equations for x.

Solution:

---

Solving Linear Equations in One Variable Orally:

Type 2:

Then the value of x will get using the below shortcut method:

Example 1:

Solve the below linear equations for x.

Solution:

Example 2:

Solve the below linear equations for x.

Solution:

Example 3:

Solve the below linear equations for x.

Solution:

---

Solving Linear Equations in One Variable Orally:

Type 3:

Then the value of x will get using the below shortcut method:

Example 1:

Solve the below linear equations for x.

Solution:

Example 2:

Solve the below linear equations for x.

Solution:

Example 3:

Solve the below linear equations for x.

Solution:

---

Solving Linear Equations in One Variable Orally:

Type 4:

Then the value of x will get using the below shortcut method:

Example 1:

Solve the below linear equations for x.

Solution:

Example 2:

Solve the below linear equations for x.

Solution:

Example 3:

Solve the below linear equations for x.

Solution:

Example 4:

Solve the below linear equations for x.

Solution:

---

Solving Linear Equations directly by Cross Multiplying:

Problem: Solve the below linear equations for x and y.

6x + 5y = 8

2x – 3y = 12

Solution:

Method:

Then,

By substituting the value of “x” in any given linear equation we can get the value of “y”

By applying above method for given equations we get

Let the given equations be

Example 1:

Solve the below linear equations for x and y.

Example 2:

Solve the below linear equations for x and y.

&

---

Trick #26

Translating Coin Problems into Specific Types of Equations:

Typical Problem in English:

Adam has 72 coins in his piggy bank. The piggy bank contains only dimes and quarters. If he has $14.70 in his piggy bank, write an equation that can be used to determine q, the number of quarters he has?

Mathematical Translation:

The total value of all coins is 1470 cents.

Let the number of quarters be represented by q and the value of quarters

be represented by 25q.

Let the number of dimes be represented by 72 – q and the value of dimes

be represented by 10(75 – q)

Write:

25q +10(72 − q) =1470

Solve for q.

q = 30

Key Points:

Work with cents as units. Remember that each coin has a specific value in cents

---

Translating Consecutive Integer Problems into Specific Types of Equations:

Typical Problem in English:

The sum of three consecutive odd integers is 18 less than five times the middle number. Find the three integers.

Mathematical Translation:

Let x represent the first integer.

Let x + 2 represent the middle integer.

Let x + 4 represent the 3^rd integer.

Write:

Solve for x, x +2, and x + 4.

7, 9, 11

Key Points:

For consecutive integer problems, define your variables as x, x + 1, and

x + 2

For consecutive even or odd integer problems, define your variables as x,

x +2, and x + 4.

---

Translating Age Problems into Specific Types of Equations:

Typical Problem in English:

Jaqueline has two sisters. One of the sisters is 7 years older than Jaqueline. The other sister is 3 years younger than Jaqueline. The product of Jaqueline’s sisters' ages is 24. How old is Jaqueline?

Mathematical Translation:

Let x represent Jaqueline’s age.

Let x+7 represent the older sister’s age.

Let x - 3 represent the younger sister’s age.

Write:

Solve for x.

x = 5

Key Points:

Define your variables. Check your answers.

Remember than “is” means “=”.

---

Translating Missing Number in the Average Problems into Specific Types of Equations:

Typical Problem in English:

FINE Medicals is a small business with five employees. The mean (average) weekly salary for the five employees is $360. If the weekly salaries of four of the employees are $340, $340, $345, and $425, what is the salary of the fifth employee?

Mathematical Translation:

Let  represent the missing salary Write:

Solve for 

 = $350

Key Points:

Substitute given values into the following formula for finding the average.

then solve for the missing value.

---

Translating Number Problems into Specific Types of Equations:

Typical Problem in English:

Twice the larger of two numbers is ten more than five times the smaller, and the sum of four times the larger and three times the smaller is 39. What are the numbers?

Mathematical Translation:

Let x represent the larger.

Let y represent the smaller.

Write two equations:

2x =10 + 5y

And

4x + 3y = 46

Solve as a system of equations.

x = 10 and y = 2 

Key Points:

Define your variables. Check your answers.

Remember than “is” means “=”.

---

Translating Area, Volume and Perimeter Problems into Specific Types of Equations:

Typical Problem in English:

If the length of a rectangular prism is doubled, its width is tripled, and its height remains the same, what is the volume of the new rectangular prism in relation to the volume of the original rectangular prism?

Mathematical Translation:

Use the formula 

Let the volume of the original rectangular prism be represented by lwh.

Let the volume of the new rectangular prism be represented by 

which simplifies to 6 times lwh.

The new rectangular prism has six times the volume of the original rectangular prism.

Key Points:

Use a geometric formula as a guide.

---

Write equations or expressions out of word  problems: 

Example 1:

Three times the sum of a number and four is equal to five times the number, decreased by two. If x represents the number, write an equation that is a correct translation of the statement?

Equation:

3(x − 4) = 5x − 2

Example 2:

The product of a number and 3, increased by 5, is 7 less than twice the number. Write an equation that can be used to find this number, n?

Equation:

3n + 5 = 2n − 7

Example 3:

The width of a rectangle is 3 less than twice the length, x. If the area of the rectangle is 43 square feet, write an equation that can be used to find the length, in feet?

Equation:

x (2x − 3) = 43

Example 4:

If n is an odd integer, write an equation that can be used to find three consecutive odd integers whose sum is -3?

Equation:

Example 5:

The length of a rectangular window is 5 feet more than its width, w. The area of the window is 36 square feet. Write an equation that could be used to find the dimensions of the window?

Equation:

or

Example 6:

Mary has $1.35 in nickels and dimes in her pocket. If she has six more dimes than nickels, write an equation that can be used to determine x, the number of nickels she has?

Equation:

or

Example 7:

If h represents a number, write an equation that is a correct translation of "Sixty more than 9 times a number is 375"?

Equation:

9h + 60 = 375

Example 8:

The ages of three brothers are consecutive even integers. Three times the age of the youngest brother exceeds the oldest brother's age by 48 years.

Write an equation that could be used to find the age of the youngest brother?

Equation:

3x = 48 + (x + 4)

or

3x − (x + 4) = 48 

Example 9:

The width of a rectangle is 4 less than half the length. If l represents the length, write an equation that could be used to find the width, w?

Equation:

---

Solving a special type of linear equation problem:

Problem: Solve the below linear equations for x and y.

Solution:

This type of equations can be solved by adding and subtracting both equations so that we can get two new equations by solving the resultant equations we can get the values of x and y.

Adding:

Subtracting:

can be written as  and   can be written as 

By solving new equations, we get

then x = 2, therefore y = 1

Example: Solve the below linear equations for x and y.

Solution:

Adding:

Subtracting:

 can be written as  and   can be written as 

By solving new equations, we get

then y = 1, therefore x = 3

---

Solving complex linear equations:

Problem: Solve the below linear equation for x.

Solution:

We can solve this type of equations by taking the LCM of the denominators of both sides and cross multiplying both sides with LCM.

Given equation can be written as

Example: Solve the below linear equation for x.

Solution:

Given equation can be written as

---

Solving a special type of linear equation problem:

Problem: Solve the below linear equation for x.

Solution:

We can solve this type of equations by cross multiplying.

Example: Solve the below linear equation for x.

Solution:

---

If an amount is to be divided among A, B and C in the ratio l : m : n and the difference between A and B is ‘R’, then

---

If there are notes of ‘x’ dollars, ‘y’ dollars and ‘z’ dollars in a box in the ratio m:n:r and the total value of notes is ‘R’, then

---

If adding/subtracting a certain quantity gives new ratio, then multiplier 

---

If the ratio of alligation of milk and water in a glass is m:n and in other glass alligation is p:q, then the ratio of milk and water in third glass which contains alligation of both glasses is

---

If the ratio of milk and water in the alligation of A litre is p:q then water must be added in it so that ratio of milk and water would be r:s is

---

The ratio of income of two persons A and B is p:q. If the ratio of their expenditures are r:s, then the monthly income of A and B, when each one of them saves ‘R’ dollars will be

---

Let ‘x’ be a number which is subtracted from a, b, c and d to make them proportional, then

Let ‘x’ be a number which is added to a, b, c and d to make them proportional, then

Here, a, b, c and d should always be in ascending order.

Example-1:

When a particular number is subtracted from each of 7, 9, 11 and 15, the resulting numbers are in proportion. The number to be subtracted is?

Solution:

By Applying Trick 92,

The number will be x

Example-2:

Which number when added to each of the numbers 6, 7, 15, 17 will make the resulting numbers proportional?

Solution:

By Applying Trick 92,

Required number

---

Example:

 

Solution:

By Applying Trick 93,

Here,

---

Two numbers are in the ratio a:b and if each number is increased by x, the ratio becomes c:d.

Then the two numbers will be

Example:

If two numbers are in the ratio 2 : 3 and the ratio becomes 3 : 4 when 8 is

added to both the numbers, then the sum of the two numbers is?

Solution:

By Applying Trick 94,

Here, a = 2, b = 3, x= 8, c = 3, d = 4

Sum of numbers = 16 + 24 = 40

---

Two numbers are in the ratio a:b and if x is subtracted from each number the ratio becomes c:d.

The two numbers will be

Example:

Two numbers are in the ratio 5 : 7. If 9 is subtracted from each of them, their ratio becomes 7 : 11. The difference of the numbers is?

Solution:

By Applying Trick 95,

Here, a = 5, b = 7, x = 9, c = 7, d = 11

---

If the ratio of present age and the ratio of age after ‘n’ years is given then present age factor is given by

---

If x is the present age factor, and the difference in cross product of ratio is zero then,

---

If the ratio of ‘some years ago’ and ‘after some years’ is given. And Before  years, the ratio of ages of A and B was a : b.

after  years, the ratio of their ages will be c: d.

When, the difference in ratios is equal, then

---

If the product of present ages is given, then,

---

If sum of present age and ratio of the ages is given then, present age factor,

---

If the ratio of ages and difference in ages is given then,

---

The ratio of ages of A and B was x: y ‘n’ years ago.

---

 

---

---

The cost of cheap object is Rs. C/kg and the cost of costly object is Rs. D/kg.

If the mixture of both object costs Rs. M/kg. then

---

Quantity of x in mixture

---

If from x litre of liquid A, p litre is withdrawn and same quantity of liquid B is added. Again, from mixture q litre mixture is withdrawn and same quantity of liquid B is added. Again, from mixture, r litre is withdrawn and same quantity of liquid B is added, then In final mixture, liquid A is

If only one process is repeated n times, then liquid A in final mixture is

---

If x is initial amount of liquid, p is the amount which is drawn, and this process is repeated n-times such that the resultant mixture is in the ratio a: b then,

---

There are two pots of same volume. Both the pots contain mixture of milk and water in the ratio m:n and p:q respectively. If both the mixtures are mixed together in a big pot, then what will be the final ratio of milk and water?

---

 

If ‘d’ unit water is added to it then

---

There is x% milk in ‘a’ unit mixture of milk and water.

The amount of milk that should be added to increase the percentage of milk from x% to y% is given by

---

There is x% water in ‘a’ unit the mixture of sugar and water.

The quantity of water vapourised such that decrease in the percentage of water is from x% to y% is given by

---

Principal or Sum:

The money borrowed or lent out for a certain period is called the principal or the sum.

Interest:

Extra money paid for using other’s money is called interest.

Simple Interest: 

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

Formulae:

Let Principal = P, Rate = R% per annum and time = T years.

Example-1:

At some rate of simple interest, A lent $6000 to B for 2 years and $1500 to C for 4 years and received $900 as interest from both of them together. The rate of interest per annum was?

Solution:

If rate of interest be R% p.a. then,

Example-2:

$500 was invested at 12% per annum simple interest and a certain sum of money invested at 10% per annum simple interest. If the sum of the interest on both the sum after 4 years is $480, the latter sum of money is?

Solution:

Simple interest gained from $500

Let the other Principal be x.

S.I. gained = $(480 – 240) = $240

Example-3:

A lends $2500 to B and a certain sum to C at the same time at 7% annual simple interest. If after 4 years, A altogether receives $1120 as interest from B and C, the sum lent to C is?

Solution:

Let the sum lent to C be x According to the question,

or 2500 × 28 + 28x = 112000 or

2500 + x = 4000 or

x = 4000 – 2500 = 1500

---

If there are distinct rates of interest for distinct time periods i.e.

---

If a certain sum becomes ‘n’ times of itself in T years on Simple Interest, then the rate per cent per annum is.

Example-1:

A sum of money becomes 7/6 of itself in 3 years at a certain rate of simple interest. The rate per annum is?

Solution:

By Applying Trick 114,

Example-2:

In how many years will a sum of money double itself at  simple interest per annum?

Solution:

By Applying Trick 114,

---

 

---

If Simple Interest (S.I.) becomes ‘n’ times of principal i.e.

S.I. = P × n then.

RT = n × 100

 

Example-1:

The simple interest on a sum after 4 years is 1/5 of the sum. The rate of interest per annum is?

Solution:

By Applying Trick 116,

Example-2:

Simple interest on a certain sum for 6 years is 9/25 of the sum. The rate of interest is?

Solution:

By Applying Trick 116,

---

If an Amount (A) becomes ‘n’ times of certain sum (P) i.e.

---

If the difference between two simple interests is ‘x’ calculated at different annual rates and times, then principal (P) is

Example:

The difference between the simple interest received from two different

banks on $500 for 2 years is $2.50. The difference between their (per

annum) rate of interest is?

Solution:

By Applying Trick 118,

Here, P = $500, x = $2.50, Difference in time = 2 years.

Difference in rate =?

Different in rate = 0.25%

---

---

 

and

---

If a sum is to be deposited in equal instalments, then,

Example-1:

What annual instalment will discharge a debt of $6450 due in 4 years at 5% simple interest?

Solution:

By Applying Trick 121,

Example-2:

What equal instalment of annual payment will discharge a debt which is due as $848 at the end of 4 years at 4% per annum simple interest?

Solution:

By Applying Trick 121,

Here, A = $848, T = 4 years, r = 4%

---

To find the rate of interest under current deposit plan,

where n = no. of months

---

   

Example-1:

The simple interest on a certain sum at 5% per annum for 3 years and 4

years differ by $42. The sum is?

Solution:

By Applying Trick 124,

Example-2:

The difference between the simple interest received from two different sources on $1500 for 3 years is 413.50. The difference between their rates of interest is?

Solution:

By Applying Trick 124

---

Example:

A sum of money lent at simple interest amounts to $880 in 2 years and to $920 in 3 years. The sum of money (in dollars) is?

Solution:

---

Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, say yearly or half yearly or quarterly to settle the previous account. In such case, the amount after first unit of time becomes the principal for the second unit, the amount after second unit becomes the principal for the third unit and so on. After a specified period, the difference between the amount and the money borrowed is called the Compound Interest (abbreviated as C.I.) for that period.

Let Principal = P, Rate = R% per annum, Time = n years.

Example-1:

The compound interest on $10,000 in 2 years at 4% per annum, the interest being compounded half-yearly, is?

Solution:

Example-2:

In what time will 1000 becomes 1331 at 10% per annum compounded annually?

Solution:

Let the required time be n years. Then,

Example-3:

The principal, which will amount to $270.40 in 2 years at the rate of 4% per annum compound interest, is?

Solution:

Let the principal be $P

---

If there are distinct ‘rates of interest’ for distinct time periods i.e.,

C.I. = A – P

Example:

If the rate of interest be 4% per annum for first year, 5% per annum for second year and 6% per annum for third year, then the compound interest of $10,000 for 3 years will be?

Solution:

By Applying Trick 125,

A = $11575.2

C.I. = $(11575.2–10000)

= $1575.2

---

If the time is in fractional form i.e., t = nF, then

Example:

Find compound interest on 10,000 for  years at 10% per annum, compounded yearly?

Solution:

By Applying Trick 126,

A = $13975.5

CI = $(13975.5 – 10,000)

CI = $3975.5

---

A certain sum becomes ‘m’ times of itself in ‘t’ years on compound interest then the time it will take to become mn times of itself is t × n years.

 

Example:

A sum of money placed at compound interest doubles itself in 5 years. In how many years, it would amount to eight times of itself at the same rate of interest?

Solution:

By Applying Trick 127,

---

The difference between C.I. and S.I. on a sum ‘P’ in 2 years at the rate of R% rate of compound interest will be

Example:

If the difference between the compound interest and simple interest on a sum at 5% rate of interest per annum for three years is 36.60, then the sum is?

Solution:

By Applying Trick 128,

---

If on compound interest, a sum becomes A in ‘a’ years and B in ‘b’ years then,

where n is a whole number

 

Example:

A sum of money amounts to $4,840 in 2 years and to $5,324 in 3 years at compound interest compounded annually. The rate of interest per annum is?

Solution:

By Applying Trick 129,

Here, b – a = 3 – 2 = 1 B = $5,324, A = $4,840

---

If a sum becomes ‘n’ times of itself in ‘t’ years on compound interest, then

Example:

If the amount is 2.25 times of the sum after 2 years at compound interest (compound annually), the rate of interest per annum is?

Solution:

By Applying Trick 130,

Here, n = 2.25, t = 2 years

---

If a sum ‘P’ is borrowed at r% annual compound interest which is to be paid in ‘n’ equal annual installments including interest, then

Example:

A builder borrows $2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 years in two equal yearly instalments.

How much will each instalment be?

Solution:

By Applying Trick 131,

Here, P = $2550, n = 2, r = 4%

---

The simple interest for a certain sum for 2 years at an annual rate interest R% is S.I., then

Example:

If the compound interest on a sum for 2 years at  per annum is $510, the simple interest on the same sum at the same rate for the same period of time is?

Solution:

By Applying Trick 132,

Here, C.I. = $510

---

Example-1:

A sum of $12,000, deposited at compound interest becomes double after 5 years. How much will it be after 20 years?

Solution:

By Applying Trick 133,

Example-2:

A sum of money becomes double in 3 years at compound interest compounded annually. At the same rate, in how many years will it become four times of itself?

Solution:

Example-3:

A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to four times itself?

Solution:

---

Work Word Problem:

  It involves a number of people or machines working together to complete a task.

  We are usually given individual rates of completion.

   We are asked to find out how long it would take if they work together.

• The Work Problem Concept:

STEP 1: Calculate how much work each person/machine does in one unit of time (could be days, hours, minutes, etc.)

  If we are given that A completes a certain amount of work in X hours, simply reciprocate the number of hours to get the per hour work. Thus, in one hour, A would complete  of the work. But what is the logic behind this?

  Let me explain with the help of an example.

  Assume we are given that Jack paints a wall in 5 hours. This means that in every hour, he completes a fraction of the work so that at the end of 5 hours, the fraction of work he has completed will become 1 (that means he has completed the task).

  Thus, if in 5 hours the fraction of work completed is 1, then in 1 hour, the fraction of work completed will be (1*1)/5

STEP 2: Add up the amount of work done by each person/machine in that one unit of time.

  This would give us the total amount of work completed by both of them in one hour. For example, if A completes  of the work in one hour and B completes  of the work in one hour, then TOGETHER, they can complete  of the work in one hour.

STEP 3: Calculate total amount of time taken for work to be completed when all persons/machines are working together.

  The logic is similar to one we used in STEP 1, the only difference being that we use it in reverse order.

  Suppose  

  This means that in one hour, A and B working together will complete  of the work. Therefore, working together, they will complete the work in Z hours.

 

Example-1:

  Jack can paint a wall in 3 hours. John can do the same job in 5 hours. How long will it take if they work together?

Solution:

  This is a simple straightforward question wherein we must just follow steps 1 to 3 in order to obtain the answer.

  STEP 1: Calculate how much work each person does in one hour.

  Jack → (1/3) of the work

  John → (1/5) of the work

  STEP 2: Add up the amount of work done by each person in one hour.

  Work done in one hour when both are working together  

  STEP 3: Calculate total amount of time taken when both work together.

  If they complete 8/15 of the work in 1 hour, then they would complete 1 job in 15/8 hours.

 

Example-2:

  Working, independently X takes 12 hours to finish a certain work. He finishes 2/3 of the work. The rest of the work is finished by Y whose rate is 1/10 of X. In how much time does Y finish his work?

Solution:

  Now the only reason this is trickier than the first problem is because the sequence of events is slightly more complicated. The concept however is the same. So, if our understanding of the concept is clear, we should have no trouble at all dealing with this.

   ‘Working, independently X takes 12 hours to finish a certain work’ This statement tells us that in one hour, X will finish 1/12 of the work.

   ‘He finishes 2/3 of the work’, This tells us that 1/3 of the work still remains.

   ‘The rest of the work is finished by Y whose rate is (1/10) of X’ Y has to complete 1/3 of the work.

    ‘Y's rate is (1/10) that of X‘. We have already calculated rate at which X works to be 1/12. Therefore, rate at which Y works is 

    ‘In how much time does Y finish his work?’ If Y completes 1/120 of the work in 1 hour, then he will complete 1/3 of the work in 40 hours.

 

  So, as you can see, even though the question might have been a little difficult to follow at first reading, the solution was in fact quite simple. We didn’t use any new concepts. All we did was apply our knowledge of theconcept we learnt earlier to the information in the question in order to answer what was being asked.

 

Example-3:

  Working together, printer A and printer B would finish a task in 24 minutes. Printer A alone would finish the task in 60 minutes. How many pages does the task contain if printer B prints 5 pages a minute more than printer A?

Solution:

  This problem is interesting because it tests not only our knowledge of the concept of word problems, but also our ability to ‘translate English to Math’ ‘Working together, printer A and printer B would finish a task in 24 minutes’ This tells us that A and B combined would work at the rate of 1/24 per minute.

   ‘Printer A alone would finish the task in 60 minutes’ This tells us that A works at a rate of 1/60 per minute.

  At this point, it should strike you that with just this much information, it is possible to calculate the rate at which B works:

  ‘B prints 5 pages a minute more than printer A’ This means that the difference between the amount of work B and

  A complete in one minute corresponds to 5 pages. So, let us calculate that difference.

  ‘How many pages does the task contain?’ If 1/120 of the job consists of 5 pages, then the 1 job will consist of 

 

Example-4:

Machine A and Machine B are used to manufacture 660 sprockets. It takes machine A ten hours longer to produce 660 sprockets than machine B.

Machine B produces 10% more sprockets per hour than machine A. How many sprockets per hour does machine A produce?

Solution:

We are told that B produces 10% more sprockets per hour than A, thus 

---

If M₁ men can finish W₁ work in D₁ days and M₂ men can finish M₂ work in D₂ days then, Relation is

If M₁ men finish W₁ work in D₁ days, working T₁ time each day and M₂ men finish W₂ work in D₂ days, working T₂ time each day, then

---

If A completes a piece of work in ‘x’ days, and B completes the same work in ‘y’ days, then,

Work done by A in 1 day = 1/x, Work done by B in 1 day = 1/y

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If A can do a work in ‘x’ days, B can do the same work in ‘y’ days, C can do the same work in ‘z’ days then, total time taken by A, B and C to complete

If workers are more than 3 then total time taken by A, B, C ...... so on to 

Example:

A, B and C individually can do a work in 10 days, 12 days and 15 days respectively. If they start working together, then the number of days required to finish the work is?

Solution:

By Applying Trick 136,

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If A alone can do a certain work in ‘x’ days and A and B together can do the same work in ‘y’ days, then B alone can do the same work in

Example:

A and B together can dig a trench in 12 days, which A alone can dig in 28 days; B alone can dig it in

Solution:

By Applying Trick 137,

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If A and B can do a work in ‘x’ days, B and C can do the same work in ‘y’ days, C and A can do the same work in ‘z’ days.

Then total time taken, when A, B and C work together

Example:

A and B can do a given piece of work in 8 days, B and C can do the same work in 12 days and A, B, C complete it in 6 days. Number of days required to finish the work by A and C is?

Solution:

By Applying Trick 138,

Let the time taken by A and C is x days

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Total work = (work of one day) × (total no. of working days) Remaining

Work = 1 – (work done)

Work done by A = (Work done in 1 day by A) × (total no. of days worked by A, B and C and so on

Where A can complete work in x days, B in y days, C in z days and so on.

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If A can finish m/n part of the work in D days. Then, Total time taken to finish the work by

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(i) If A can do a work in ‘x’ days and B can do the same work in ‘y’ days and when they started working together, B left the work ‘m’ days before

(ii) A leaves the work ‘m’ days before its completion then total time taken

Example-1:

A can do a piece of work in 18 days and B in 12 days. They began the work together, but B left the work 3 days before its completion. In how many days, in all, was the work completed?

Solution:

By Applying Trick 141(i),

Here, x = 18, y = 12, m = 3

Example-2:

A and B alone can complete work in 9 days and18 days respectively. They worked together; however, 3 days before the completion of the work A left. In how many days was the work completed?

Solution:

By Applying Trick 141(ii),

Here, x = 9, y = 18, m = 3

Total time taken

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If A and B together can finish a certain work in ‘a’ days. They worked together for ‘b’ days and then ‘B’ (or A) left the work. A (or B) finished the rest work in ‘d’ days, then Total time taken by A (or B) alone to complete the work

Example:

A and B can together finish a work in 30 days. They worked at it for 20 days and then B left. The remaining work was done by A alone in 20 more days.

A alone can finish the work in?

Solution:

By Applying Trick 142,

Here, a = 30, b = 20, d = 20

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Example:

4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?

Solution:

By Applying Trick 144,

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If food is available for ‘a’ days for ‘A’ men at a certain place and after ‘b’ days. ‘B’ men join, then the remaining food will serve total men for

If food is available for ‘a’ days for ‘A’ men at a certain place, and after ‘b’ days ‘B’ men leave then the remaining food will serve remaining men for

Example:

40 men can complete a work in 18 days. Eight days after they started working together, 10 more men joined them. How many days will they now take to complete the remaining work?

Solution:

By Applying Trick 143,

Here, A = 40, a = 18

b = 8, B = 10

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Example:

If 8 men or 12 boys can do a piece of work in 16 days, the number of days required to complete the work by 20 men and 6 boys is?

Solution:

Here, A = 8, B = 12, a = 16

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Example:

If 1 man or 2 women or 3 boys can complete a piece of work in 88 days,then 1 man, 1 woman and 1 boy together will complete it in?

Solution:

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If ‘A’ men can do a certain work in ‘a’ days and ‘B’ women can do the same work in ‘b’ days, then the total time taken when A₁ men and  B₁ women work together is

Example-1:

5 men can do a piece of work in 6 days while 10 women can do it in 5 days.

In how many days can 5 women and 3 men do it?

Solution:

Example-2:

A man, a woman and a boy can complete a work in 20 days, 30 days and 60 days respectively. How many boys must assist 2 men and 8 women so as to complete the work in 2 days?

Solution:

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The comparison of rate of work done is called efficiency of doing work.

Example:

Mike can do a work in 15 days. John is 50 per cent more efficient than

Mike in doing the work. In how many days will John do that work?

Solution:

Efficiency and time taken are inversely proportional

Hence, John completes the work in 10 days.

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If the efficiency to work of A is twice the efficiency to work of B, then,

Example:

A man and a boy received $800 as wages for 5 days for the work they did

together. The man’s efficiency in the work was three times that of the boy.

What are the daily wages of the boy?

Solution:

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If A can do a work in ‘x’ days and B is R% more efficient than A, then ‘B’ alone will do the same work in

Example:

A can do a piece of work in 70 days and B is 40% more efficient than A. The number of days taken by B to do the same work is?

Solution:

Here, x = 70, r = 40%

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A, B and C can do a certain work together within ‘x’ days. While, any two of them can do the same work separately in ‘y’ and ‘z’ days, then in how many days can 3rd do the same work

Example:

A, B and C can complete a work in 8 days. B alone can do it in 18 days and C alone can do it in 24 days. In how many days can A alone do the same work?

Solution:

Here, x = 8, y = 18, z = 24

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A and B can do a work in ‘x’ days, B and C can do the same work in ‘y’ days. C and A can do the same work in ‘z’ days.

Then, all can do alone the work as following:

Example-1:

A and B can do a piece of work in 10 days. B and C can do it in 12 days. A and C can do it in 15 days. How long will A take to do it alone?

Solution:

Example-2:

If A and B together can complete a work in 18 days, A and C together in 12 days and B and C together in 9 days, then B alone can do the work in?

Solution:

Example-3:

A and B can do a piece of work in 10 days. B and C can do it in 12 days. C and A in 15 days. In how many days will C finish it alone?

Solution:

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A can do a certain work in ‘m’ days and B can do the same work in ‘n’ days.

They worked together for ‘P’ days and after this A left the work, then in how many days did B alone do the rest of work

when after ‘P’ days B left the work, then in how many days did A alone do the rest of work

Example-1:

A can do a piece of work in 12 days and B in 15 days. They work together for 5 days and then B left. The days taken by A to finish the remaining work is?

Solution:

Here, m = 12, n = 15, p = 5

Example-2:

A can do a piece of work in 12 days and B can do it in 18 days. They work together for 2 days and then A leaves. How long will B take to finish the remaining work?

Solution:

Here, m = 12, n = 18, p = 2

Time taken by B

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Example:

While working 7 hours a day, A alone can complete a piece of work in 6 days and B alone in 8 days. In what time would they complete it together, working 8 hours a day?

Solution:

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The efficiency of A to work is ‘n’ times more than that of B, both start to work together and finish it in ‘D’ days. Then, A and B will separately complete, the work in

Example:

A can do in one day three times the work done by B in one day. They together finish 2/5 of the work in 9 days. The number of days by which B can do the work alone is?

Solution:

Here, n = 3 and D

(Time taken to finish whole work)

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Some people finish a certain work in ‘D’ days. If there were ‘a’ less people, then the work would be completed in ‘d’ days more, what was the number of people initially

Example:

A certain number of persons can complete a piece of work in 55 days. If there were 6 persons more, the work could be finished in 11 days less. How many persons were originally there?

Solution:

Here, D = 55

a = 6, d = 11

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If A:B = x:y, B:C = p:q and C:D = m:n then,

Example-1:

If A : B = 3 : 4, B : C = 5 : 7 and C : D = 8 : 9 then A : D is equal to?

Solution:

By Applying Trick 79,

Example-2:

Solution:

By Applying Trick 79,

Thus,

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Example:

 

Solution:

By Applying Trick 80,

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If an amount R is to be divided between A and B in the ratio m:n then

where m > n

Example-1:

If an amount $250 is to be divided between P and Q in the ratio 3:2 then

how much will Q get on his part?

Solution:

By Applying Trick 81,

Therefore, Part of Q = 100

Example-2:

If $1000 is divided between A and B in the ratio 3 : 2, then A will receive?

Solution:

By Applying Trick 82,

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If the ratio of A and B is m:n and the difference in their share is ‘R’ units then,

where m > n

 

Example:

If the ratio of A and B is 5:2 and the difference in their share is $75

then how much will A get?

Solution:

By Applying Trick 82,

Here, m:n = 5:2 and R = 75,

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If the ratio of A and B is m:n and the part of A is ‘R’, then

where m > n

 

Example:

Marks of two candidates P and Q are in the ratio of 2 : 5. If the marks of P is

120, marks of Q will be?

Solution:

By Applying Trick 83,

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where m>n

 

Example:

Solution:

By Applying Trick 84,

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Complex Radical:

Example: Find the value of the 

Solution:

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Associative:

Example: Find the value of the 

Solution:

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Simple Product:

Example: Find the value of the 

Solution: 

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Simple Quotient:

 Example: Find the value of the 

Solution:

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Complex Product:

Example: Find the value of the 

Solution:

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Complex Quotient:

Example: Find the value of the 

Solution:

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Nesting:

Example: Find the value of the 

Solution:

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Last Digit: Repeated Multiplication

The problems that deal with the last digit of a number resulting from repeated multiplication can be solved easily by observing the repeating patterns of the unit’s digits of consecutive integral powers of numbers from 0 to 9.

For ‘2’:

The repeating pattern of the unit’s digit of consecutive integral powers of 2 are {2, 4, 8, 6}.

Example:

Solution:

As per the trick 31,

The repeating pattern of the unit’s digit of consecutive integral powers of 2 are {2, 4, 8, 6}

Then the unit’s digit of  is 6.

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For ‘3’:

The repeating pattern of the unit’s digit of consecutive integral powers of 3 are {3, 9, 7, 1}.

Example:

Solution:

As per the trick 32,

The repeating pattern of the unit’s digit of consecutive integral powers of 3 are {3, 9, 7, 1}

Then the unit’s digit of  is 9.

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For ‘4’:

The repeating pattern of the unit’s digit of consecutive integral powers of 4 are {4, 6}.

Example:

Solution:

As per the trick 33,

The repeating pattern of the unit’s digit of consecutive integral powers of 4 are {4, 6}.

Then the unit’s digit of  is 4.

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For ‘7’:

The repeating pattern of the unit’s digit of consecutive integral powers of 7are {7, 9, 3, 1}.

Example:

Solution:

As per the trick 34,

The repeating pattern of the unit’s digit of consecutive integral powers of 7 are {7, 9, 3, 1}.

Then the unit’s digit of  is 3.

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For ‘8’:

The repeating pattern of the unit’s digit of consecutive integral powers of 8 are {8, 4, 2, 6}.

Example:

Solution:

As per the trick 35,

The repeating pattern of the unit’s digit of consecutive integral powers of 8 are {8, 4, 2, 6}.

Then the unit’s digit of  is 6.

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For ‘9’:

The repeating pattern of the unit’s digit of consecutive integral powers of 9 are {9,1}. 

Example:

Solution:

As per the trick 36,

The repeating pattern of the unit’s digit of consecutive integral powers of 9 are {9,1}.

Then the unit’s digit of  is 1.

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The last digit of any number that ends in 0, 1, 5, or 6, will always remain unchanged on repeated multiplication with itself.

Example:

Solution:

As per the trick 37,

The unit’s digit will remain itself for consecutive integral powers of 5

Then the unit’s digit of  is 5.

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When a number is divided by 10, the remainder is the same as the last digit of that number.

Example:

Solution:

The last digit of 7 repeats in a cycle of 4, and there are 19 full cycles, with one remainder, therefore, the last digit of  is 7, which is also the remainder when  is divided by 10.

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The units digit of the fifth power of a number is the same as the unit’s digit of the original number. 

Example-1:

Solution:

As per the trick 39,

The units digit of the fifth power of a number is the same as the unit’s digit of the original number.

Then the unit’s digit of  is 7.

Example-2:

Solution:

As per the trick 39,

The units digit of the fifth power of a number is the same as the unit’s digit of the original number.

Then the unit’s digit of  is 8.

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The number  has a units digit of zero, in other words when  is divided by 10, the remainder is always zero.

Example:

Solution:

As per the trick 40,

The number  has a units digit of zero

Then the unit’s digit of  is 0.

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If a and b are consecutive integers and a > b, then the units digit of  is always 1.

Example:

Solution:

As per the trick 41,

If a and b are consecutive integers and a > b, then the units digit of  is always 1.

Then the unit’s digit of `is 1.

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Last two digits of numbers ending in 1:

Example-1:

Solution:

Multiply the tens digit of the number (4 here) with the last digit of the exponent (9 here) to get the tens digit. The unit digit will always equal to one.

In no time at all you can calculate the answer to be 61 (4 × 9 = 36).

Therefore, 6 will be the tens digit and one will be the unit digit. 

Example-2:

Solution:

The last two digits of  will be 01 and the last two digits of  will be 21.

Therefore, the last two digits of  will be the last two digits of 

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Last two digits of numbers which end in 3, 7 and 9:

Convert the number till the number gives 1 as the last digit and then find the last two digits according to the previous method. 

Example-1:

Solution:

 therefore, we need to find the last two digits of 

Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method.

The last two digits are  so the tens digit will be 8 and last digit will be 1)

Example-2:

Solution:

 therefore, we need to find the last two digits of 

By the previous method, the last two digits of 

Example-3:

Solution:

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Last two digits of numbers which end in 2,4,6,8:

There is only one even two-digit number which always ends in itself (last two digits) i.e, 76 raised to any power gives the last two digits as 76.

Therefore, our purpose is to get 76 as last two digits for even numbers. 

We know that  ends in 76 and  ends in 24. Also, 24 raised to an even power always ends with 76 and 24 raised to an odd power always ends with 24. Therefore,   will end in 76 and   will end in 24.

Example-1:

Solution:

(NOTE: Here if you need to multiply 76 with  then you can straightaway write the last two digits of  because when 76 is multiplied with  the last two digits remain the same as the last two digits of . Therefore, the last two digits of  will be the last two digits of 

Same method we can use for any number which is of the form 

Example-2:

Solution:

Example-3:

Solution:

Example-4:

Solution:

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Fractional Equality:

Example: If x/3 = 5/2 then find the value of x.

Solution:

Given: x/3 = 5/2

2x = 15 

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Fractional Equivalency:

Example: The given fractions 5/16 and x/12 are equivalent fractions, then find the value of x.

Solution:

Given: 5/16 = x/12

x = (5 x 12)/16

x = 60/16

x =15/4

Therefore, the value of x is 15/4.

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Addition (like denominators):

Example: If   then find the value of x+y?

Solution:

Given: 

x + y = 6

Therefore, the value of x + y is 6.

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Addition (unlike denominators):

Note: bd is the common denominator

Example 1: If   then find the value of 2p + 3q when pq = 1?

Solution:

Given:  

Therefore, the value of 2q + 3q is 4.

Example 2: 

Solution:

First, let’s separate out the denominators and simplify them.

and

We are adding the reciprocals of these two fractions:

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Example: If  then find the value of m - n?

Solution:

Therefore, the value of m - n is 20.

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Example: If  then find the value of  when xy = 2?

Solution:

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Example: If  then find the value of xy?

Solution:

Given: 

Therefore, the value of xy 100.

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Example:

Solution:

Let’s think about this is in stages.  First, call the entire denominator D; then (0.2)/D = 4.  From this, we must recognize that D must be 1/4 of 0.2, or D = 0.05.

Now, set that denominator equal to 0.05.

0.3 – x = 0.05

x = 0.3 – 0.05 = 0.25 = 1/4

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Example:

If  then find the value of x?

Solution:

Given that 

then  ( from the above formula)

Therefore x = 100

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Example:

If  then find the value of ab?

Solution:

Given that 

then  ( from the above formula)

Therefore ab = 50

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Example:

If  then find the value of 

Solution:

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Converting Improper Fractions to Mixed Fractions:

  Divide the numerator by the denominator

  Write down the whole number answer

  Then write down any remainder above the denominator

Example:

Convert 11/4 to a mixed fraction.

Solution:

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Converting Mixed Fractions to Improper Fractions:

 

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