Aptitude Solved Problems - Ratios and Proportions

Aptitude Solved Problems :

Ratios and Proportions

1. A sum of Money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B′s share ?

Answer: Rs. 2000

Explanation:

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively.

From the given data,

4x - 3x = 1000 

x = 1000

∴ B′s share = Rs. 2x 

= Rs. (2 * 1000)

= Rs. 2000

2. Iron and copper are melted together in the ratio 9 : 11. What is the weight of melted mixture, if 28.8 kg of iron has been consumed in it ?

Answer: 64 kg

Explanation

For 9 kg Iron, mixture melted 

= ( 9 + 11 ) kg

= 20 kg

For 28.8 kg Iron, mixture melted

 = (( 20 / 9 ) * 28.8 ) kg 

= ( 20 * 3.2 ) kg

= 64 kg

3. The salaries A, B, C are in the ratio 2 : 3 : 5. If the increments of 15%, 10% and 20% are allowed respectively in their salaries, then what will be new ratio of their salaries?

Answer: 23 : 33 : 60

Explanation:

Let the salaries of A, B and C are 2x, 3x and 5x respectively.

A′s new salary = (( 115 / 100) * 2x )

= ( 23 / 20 ) * 2x

= ( 23x / 10 )

B′s new salary = (( 110 / 100 ) * 3x )

= ( 11 / 10 ) * 3x

= ( 33x / 10 )

C′s new salary = (( 120 / 100 ) * 5x )

= ( 12 / 10 ) * 5x

= 6x

∴ New ratio = ( 23x / 10 ) : ( 33x / 10 ) : 6x

= 23x : 33x : 60x

= 23 : 33 : 60

4. A certain amount was divided between A and B in the ratio 4 : 3. If B′s share was Rs. 4800, find the total amount ?

Answer: Rs. 11200

Explanation:

A : B = 4 : 3

Total amount = 7

If B′s share = Rs. 3

If B′s share is Rs. 4800, total amount = Rs. (( 7 / 3 ) * 4800)

= Rs. ( 7 * 1600 )

= Rs. 11200

Numbers Formulae

 Numbers Formulae

Some important formulas:

(a + b)(a - b) = a 2-b 2.

(a + b)2 = (a 2 + b 2 + 2ab).

(a - b)2 = (a 2 + b 2 - 2ab).

(a + b + c)2 = a 2 + b 2 + c 2 + 2(ab+bc+ca).

(a 3 + b 3) = (a+b)(a 2 -ab+ b 2).

(a 3- b 3) = (a-b)(a 2 +ab+ b 2).

(a 3 + b 3+ c 3- 3abc) = (a+b+c)(a 2+b 2+ c 2-ab-bc-ca).

When a + b + c = 0; then a 3 + b 3 + c 3 = 3abc.

Properties of Numbers

A three digit number formed by repeating a one digit number thrice, is always divisible by 111.

Since 111=3 × 37, such numbers are divisible by 3, 37, 111.

Ex:222, 777, 888 etc, are divisible by 3, 37, 111.

A four digit number formed by repeating a two digit number, is always divisible by 101.

Ex: 2525, 3636, 8383 etc, are divisible by 101.

The number zero surrounded by the same two digit number on both sides. Any such five digit number is always divisible by 1001.

Since 1001 = 7 × 11 × 13. Such numbers are divisible by 7, 11, 13 and 1001.

Ex: 52052, 68068, 93093, 85085.

A six digit number formed by repeating a three digit number is always divisible by 1001.

Since 1001 = 7 × 11 × 13. Such numbers are divisible by 7,11,13 also. `

Ex: 436436, 850850, 789789, 123123.

If sum of digits of a natural number is subtracted from the number itself, the resulting number is always divisible by 9.

Ex: 143: Sum of digits = 1 + 4 + 3 = 8

Now 143 - 8 = 135, which is divisible by 9.

Sum of two digit number and number formed by interchanging its digits, is always divisible by 11.

Ex1: 38 + 83 = 121 which is divisible by 11.

Ex2: 98 + 89 = 187 which is divisible by 11.

Digits of a two digit number and reversed. Difference between the original number and new number is always divisible by 9.

Ex: 71 - 17 = 54, which is divisible by 9.

Digits of three digits number are reversed. Difference between the original number and new number is always divisible by 99.

Ex: 731 - 137 = 594, which is divisible by 99.

Difference between square of a two digit number (xy) and square of a number formed by interchanging the digits (yx), is always divisible by 99.

Ex: 21 2 - 12 2 = 441 - 144 = 297 = 99 × 3.

Test of Divisibility:

Divisibility By 2:

A number is divisible by 2 if its unit digit is any of of ,2,4,6,8.

Ex: 58694 is divisible by 2, while 86945 is not divisibleby 2.

Divisibility By 3:

A number is divisible by 3 only when the sum of its digits is divisible by 3.

Ex: In the number 695421, the sum of digits = 27, which is divisible by 3, then 695421 is divisible by 3.

Divisibility By 4:

A number is divisible by 4 is the last two digits is divisible by 4.

Ex1: 6879376 is divisible by 4, since 76 is divisible by 4.

Ex:2 496138 is not divisible by 4, since 38 is not divisible by 4.

Divisibility By 5:

A number is divisible by 5 when its unit digit is 0 or 5.

Ex: 76895 and 68790 are divisible with 5.

Divisibility By 8:

A number is divisible by 8 if the number formed by hundreds tens and units digit of the given number is divisible by 8.

Ex: In the number 16789352 the number formed by last three digits 352 is divisible by 8 then whole number is divisible by 8.

Divisibility By 9:

A number is divisible by 9 only when the sum of its digits is divisible by 9.

Ex: In number 246591, the sum of digits = 27, which is divisible by 9, then whole number is divisible by 9.

Divisibility By 10:

A number is divisible by 10 only when its unit digit is 0.

Ex: Each of the numbers 76890 and 68790 is divisible by 10.

Divisibility By 11:

A number is divisible by 11 if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11.

Ex1:

Take the number 29435417.

= (Sum of its digits at odd places) - (Sum of its digits at even places).

= ( 7 + 4 + 3 + 9 ) - ( 1 + 5 + 4 + 2 )

= ( 23 - 12)

= 11.

Ex2:

Consider the number 57463822.

= (Sum of its digits at odd places) - (Sum of its digits at even places).

= ( 2 + 8 + 6 + 7 ) - ( 2 + 3 + 4 + 5 )

= ( 23 - 14 )

= 9

which is not divisible by 11. Then 57463822 is not divisible by 11.

Tipes of Numbers

 Tipes of Numbers

★Numbers

★ Natural Numbers(N)

★ Whole Numbers(W)

★ Integers(Z)

★ Even Numbers

★ Odd Numbers

★ Rational Numbers

★ Mixed Numbers

★ Irrational Numbers

★ Real Numbers

★ Composite Numbers

★ Prime Numbers

★ Twin-Prime Numbers

★ Co-Prime Numbers

★ Face Value and Base Value

★ BODMAS Rule

Natural Numbers(N):

(Positive Integers) 1,2,3,4,‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

Whole Numbers(W):

(set of all natural numbers and 0) 0,1,2,3,4,5,‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 

Integers(Z):

 ‐‐‐‐‐-3,-2,-1,0,1,2,3,4,5,‐‐‐‐‐‐‐‐‐

Even Numbers:

2,4,6,8,,‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

Numbers are exactly divisible by 2. These numbers are end with even number or zero.

Odd Numbers:

1,3,5,7,8,‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

Numbers are not exactly divisible by 2. These numbers are end with odd number.

Sum of any two odd number is always an even number.

Product of any two odd numbers always an odd number.

Rational Numbers:

Numbers which can be expressed in the form of p⁄q, where p and q both are integers, and q is a non-zero integer.

A number expressed in the form of p ÷ q is also called a fraction, where p is numerator and q is denominator.

A fraction when converted into decimal, gives finite or recurring digits after decimal sign.

Ex: 1⁄3, 1⁄2, 7⁄4

Mixed Numbers:

Numbers which consists of a whole number and a fraction.

Ex: 31⁄4, 52⁄7, 46⁄17, etc.

Irrational Numbers:

Numbers which cannot be expressed in the form of p ÷ q where p and q both are integers i.e., the numbers which are not Rational.

These numbers when converted into decimal, give infinite and non-recurring digits after decimal sign.

Ex: π , √ 2  , √ 3  , √ 27 

Real Numbers:

Real Numbers are combination of rational and irrational numbers.

Complex or imaginary Numbers:

These numbers are written as a+bi where i2=-1.

Composite Numbers:

The number which have more than two factors.

Ex: 4,6,8,9.

Prime Numbers:

A number which is not divisible by any number other than 1 and itself.Such numbers have only two factors,the number itself and 1.

Ex: 2,3,5,7,11,13,17,19.

1 is neither a prime number nor a composite number.

Product of any two prime numbers is always a composite number, since the product is divisible by both the prime numbers.

Twin-Prime Numbers:

Pairs of such prime numbers whose difference is 2.

Ex: 3 and 5, 11 and 13.

Co-Prime Numbers:

Co-Prime Numbers are those numbers which are prime to each other i.e., they don’t have any common factor other than 1.

Since these numbers do not have any common factor, their HCF is 1 and their LCM is equal to product of the numbers.

Co-prime numbers can be prime or composite numbers. Any two prime numbers are always co-prime numbers.

Ex1: 3 and 5 both numbers are prime numbers.

Ex2: 8 and 15 both numbers are composite numbers but they are prime to each other i.e., they don’t have any common factor.

Face Value and Base Value:

Face value is absolute value of a digit in a number.

Place value or Local value is value of a digit in relation to its position in the numbers.

Ex:  Face value and Place value of 9 in 12921 is 9 and 900.

BODMAS Rule:

Any simplification of an expression may involve various operations, these operations must be performed in the following order of preference.

B:Bracket:

If more than one type of brackets are found, they are operated in the following order,

First: (   ) line or bar bracket or vinculum bracket.

Second: (  ) small bracket.

Third: { } Mid bracket.

Forth: [ ] Large Bracket.

O(of): means multiplication but is operated before division.

D: Division.

M: Multiplication.

A: Addition.

S: Subtraction.

Note:

If sum of two numbers a and b is x and difference between the numbers is y.

then a = x+y⁄2 and

b = x-y⁄2

If on adding 9 to a two digit number (say xy), its digits are reversed.; Then units digit of the number exceeds its tens digit by 1. Similarly, if on adding 9a to a two digit number its digits are reversed, then units digit of the number exceeds its tens digit by a i.e., x+a = y

if digits are reversed on subtracting 9a from the number then x-a=y

Difference between two digits of the number is number added/subtracted÷ 9.

if on adding 99a to a three digit number (say xyz), its digits are reversed, then x+a = z

If digits are reversed on subtracting 99a from the number, then x-a = z.

If a composite number C, can be expressed as am × bn × cp‐‐‐‐‐, where a,b,c‐‐‐ are prime factors and m,n,p‐‐‐ are natural numbers, then total number of factors of C

= (m+1)×(n+1)×(p+1)‐‐‐‐

Total number of factors includes unity and the number itself.

Age Problems

 

Age Problems

Type 1: Ratio Based Age Problems (Why These Are Very Easy? Read On…)

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

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

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

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

You can then write, x/y = 3/5
Or, 5x=3y
Or, x=3y/5 ….equation 1

In question, you can find that after 10 years the ratio becomes 2/3.
(After 10 years, Rahul’s age will become x+10
and Ravi’s age will become y+10)

Therefore You can write, (x+10)/(y+10)=2/3
Or 3x+30 = 2y+20
Or 2y-3x=10 … equation 2

But, you already know from equation 1 that x = 3y/5. If you substitute this equation 2, you will get
2y-3(3y/5)=10
Or 2y-9y/5=10
Or 10y-9y/5=10
y=50

If you substitute, y = 50 in equation 1, you will get x = 3 x 50/5 = 30
Therefore, your answer is Rahul’s age = 30 and Ravi’s age = 50

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

Type 2: Equation Solving Type Age Problems (Why This Type Is Really Important?)

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

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

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

In 5 years, Reshma will be 2 times older than Satya. Therefore, you can write
y+5=2(x+5)
Or y-2x=5
Or y=2x+5…equation 1

You know that Renuka is 5 years younger than Reshma. Therefore, you can write
z=y-5 ….equation 2

Also, from question, you can find that before 5 years, Renuka was 3 times older than Satya. So,
z-5=3(x-5)
Or z-3x=20
Or z=3x-10 …equation 3

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

Now you have to substitute z=2x in equation 3. So you will get,
2x=3x-10
Or x=10
Therefore, you have found the answer that Satya’s age = 10

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

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

Type 3: Finding Ratio Between Ages (Why This Is An Extension To Type 2 )

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

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

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

If you substitute, x = 10 in equation 1 of example 2, you will get
Reshma’s age = y = 25

 

 

 

Boats And Streams

 

Boats And Streams

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)

A|-------------------------X-----------------------|B

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 per hour 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

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

Note: To solve such linear equations, there is another simple shortcut. Here is the video link to that shortcut.

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

 

 

 

Profit and Loss

 

Profit and Loss

Type I: Direct Formula Based Profit And Loss Percentages (Why This Is Easy?)
This type is very straightforward and is formula based. This is very easy because, you have to remember just 4 very simple formulas to solve this type.

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

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

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

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

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

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

Solution:
You can write down the CP and SP values from the question as follows:
Cost Price CP =Rs.100 and Selling Price SP = Rs.150
Here, SP is greater than CP. Therefore, there is profit in the transaction.

Based on formula, you know that Profit = SP – CP = 150 – 100 = Rs. 50
You also know the formula that ?Profit Percentage = (Profit / CP) x 100%
Therefore, Profit Percentage = (50 / 100) x 100% = (1/2)x100 % = 50%

Type II: Profit And Loss When Selling Different Varieties Of Same Item (Is This New To You?)
In this type, a seller will buy two (or more) varieties of an item at two different cost prices. Then he will sell them together (by mixing them) at common selling price.

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

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

Solution:
Uma buys two varieties of roses. Type I at 4 roses per rupee and type II at 2 roses per rupee.
CP of 4 roses of type I = 1 and
CP of 2 roses of type II = 1

Therefore, CP of 1 rose of type I = ¼ and
CP of 1 rose of type II = 1/2

Now assume that Uma had bought 1 dozen (12) roses of each variety.
Therefore, CP of 1 dozen roses of type I = ¼ x12 = 3 and
CP of 1 dozen roses of type II = 1/2 x 12 = 6

If Uma mixes 1 dozen of type I and 1 dozen of type II together,
CP of 2 dozen mixed roses = CP of 1 dozen roses of type I + CP of 1 dozen roses of type II
= 3 + 6 = Rs. 9
So, CP of 1 dozen mixed roses = 9/2 = Rs. 4.5

Let SP of 1 dozen mixed roses be X
You know that the Profit = SP – CP = X – 4.5
And Profit Percentage = Profit / CP x 100%
= (X – 4.5) / 4.5 x 100%

To answer the question, you have to find X value when profit percentage is 25. Therefore,
(X – 4.5) / 4.5 x 100 = 25
Or X – 4.5 = 25 x 4.5 / 100
Or X – 4.5 = 1.125
Or X = 5.625
Therefore, to make a profit of 25%, Uma has to sell the mixture at Rs. 5.625 per dozen

Type III: Same Selling Prize, Equal Profit And Loss Percentages (Why This Is Interesting?)
This is a very interesting type. Though this looks hard to solve, you can solve this type easily by using a super simple formula. Read on…

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

You will understand this type after reading the below example.

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

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

If one item is sold at X% profit and other at X% loss and selling prices in both the transactions are equal, then
Loss % = (X/10)2
In our example, the value of X is 23
Therefore, Loss percentage = (23/10)2 = 2.3 x 2.3 = 5.29

If you have doubts regarding this type, please use the comments section below.

Type IV: Profit When Seller Is Not Honest And Uses False Weighing Stone Or Scale
If a seller (e.g., vegetable seller) uses false weighing stone (for example, 750 gram instead of 1 kilogram weighing stone), he will make higher profit compared to an honest seller, right?

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

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

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

Real Profit % = Error / (True value – Error) x 100
Here, Error is the difference between weights of true weighing stone and the seller’s false weighing stone.
True Value denotes the correct weight of the stone (which an honest seller will use).

In question, you will see that the seller uses 900g weight instead of 1000g or 1Kg weight.
Therefore, Error = 1000 – 900 = 100
But a true weighing stone will be 1 Kg or 1000g.
Therefore, True value = 1000

If you apply above values in our Real Profit % formula, you will get
Real Profit % = 100 / (1000 – 100) x 100
= 100/900 x 100 = 11.11%

Type V: Multiple Transactions Based Profit And Loss Problems
In all the above types, you saw only one transaction. In the below example, you will find two or more continuous transactions. Now let us directly go to our example.

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

Solution:
First, assume CP of the bicycle when Rahul bought be Rs.X.
He sells it to Banu at profit of 15%. In other words, Banu buys the bicycle from Rahul by giving 15% more than Rahul’s CP.
Therefore, CP of bicycle to Banu = 15% more than CP of bicycle to Rahul

CP of bicycle to Banu = CP of bicycle to Rahul + 15/100 x CP of bicycle to Rahul
= X + 15/100 x X
= X x (115/100) …equation 1

Banu sells it to Sona at a profit of 20%. In other words, Sona buys the bicycle from Banu by giving 20% more than Banu’s CP.
Therefore, CP of bicycle to Sona = 20% more than CP of bicycle to Banu
= CP of bicycle to Banu + 20/100 x CP of bicycle to Banu
= CP of bicycle to Banu x (120/100)

But you know from equation 1 that CP of bicycle to Banu = X x (115/100). If you substitute this in above equation, you will get:
CP of bicycle to Sona = X x (115/100) x (120/100) … equation 2

In question, you can see that Sona pays Rs 3000 for the bicycle.
Or, CP of bicycle to Sona = 3000
If you substitute above value in equation 2, you will get,
3000 = 120/100 x 115/100 x X
Or X = 3000 x 100/120 x 100/115
Or X = Rs. 2173.91
Therefore, CP of bicycle to Rahul is Rs. Rs. 2173.91

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

Let us see an example for type VI.

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

Solution:
First, let us assume that the marked price of each pencil be Rs.1
In question, you can see that the Vendor buys 30 pencils at the marked price of 25 pencils.
Therefore, CP of 30 pencils = Marked price of 25 pencils = 25 x 1 = Rs.25

Without discount, SP of 30 pencils = Marked price of 30 pencils = 30 x 1 = Rs. 30
But, the vendor sells these pencils at a discount of 2%.

Therefore, SP of 30 pencils = Marked price of 30 pencils – (2/100) of Marked price of 30 pencils
= Marked price of 30 pencils (1 – 2/100)
= Marked price of 30 pencils x 98/100
= 30 x 98/100
= Rs. 29.40

Therefore, his Profit = SP – CP
= 29.40 – 25 = Rs. 4.40
Profit % = Profit /CP x 100
= 4.40/25 x 100
= 17.6%

Type VII: Profit And Loss Problems With Ratio Calculations
If any of the above types is combined with ratio calculation, you will get this type VII. Truely, this type is an extension to any of the above types. Let us see an example, which is an extension to type VI with ratio calculation.

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

Solution:
Let the CP be Rs.100.
Vendor earns a profit of 10%. Therefore his SP will be CP + 10% of CP
SP = 100 + 10% of 100
Or SP = Rs. 110 … equation 1

Let the printed price be Rs.X
From the question, you know that SP is printed price with 15% discount.
Or SP = Printed Price – 15% Printed Price
Or SP = X – (15/100) x X
Or SP = .85X … equation 2

From equations 1 and 2, you can write,
.85X = 110
or X = 110/.85 = 11000/85 = 2200/17

Based on our assumption that CP is 100, we have found that printed price will be 2200/17
Therefore, required ratio = CP : Printed price
=100 : 2200/17
To simplify the above ratio, you can multiply both the terms by 17. So the above ratio becomes,
1700 : 2200
=17 : 22

 

 

 

Time & Work

 

Time & Work

Time and Work is a very easy topic and at least 2 Questions are asked from this topic in Any Exam whether it's Clerk or PO or SSC.What actually the Time and Work Topic is all about? In simple words It's how much time a person will take to Complete a Given Work.
Here we are providing some cases of time and work and efficient method to solve it. Hope this would be helpful to all aspirants.

These Tricks are shared by one of the our ardent BA reader Insomniac.We wish you good luck for your future.

Note -: In conventional Method work is always treated as 1
Example: So if I say that a person can complete a work in 15 days that means he will do 1/15 work in one day, It's simple maths.
Now another person does the same work in 30  days. So he will do 1/30 work in 1 day.
Now if i ask in how many days they will complete the work together. What we gonna do is Add their 1 day of work
 like 1/15 + 1/30 = (2+1)/30 = 3/30 = 1/10
Now this 1/10 we got is actually their 1 day work, So if they do 1/10 work in one day then it's simple they will complete the whole work in 10 days.
Now that was the conventional method and I believe that you all know how to solve Questions through Conventional method.

So now lets move on to the Faster method i.e efficiency method.

In efficiency method the Work is not treated in numerical value, Like in Conventional method the work is 1 but here the work is treated as percentage.
So by common sense the work is always treated as 100%
So when i say a person completes a work in 15 days it means he will do 100/15 % work in 1 day i.e 6.66% work in 1 day
If another person does the work in 30 days that means he will do 3.33% work in 1 day.
And together they will do 6.66 + 3.33 = 9.99 or 10% work in one day So in how many days they will do the complete work, that will be 100/10 = 10 days.
Now it may sound difficult That we have to convert Each value in % but don't worry you don't have to convert each value, You just have to learn all the values till 1/30 and then it will be a cakewalk.

Now we will take Some standard Cases Of time and work and you all are free to ask any problem if you have in any case.

Case 1 - A does a work in X days, B does a Work in Y days In how many days they will complete the work.
Question- A completes the work in 10 days and B completes the work in 15 days In how many days they will complete the work.
Conventional Method
Work done by A in 1 day = 1/10
Work Done by B in 1 day = 1/15
Work done By A & B together in 1 day = 1/10 + 1/15 = (3+2)/30 = 5/30 = 1/6
As A & B Completes 1/6 work in one day So they will complete the whole work in 6 Days.

Efficiency method.
Efficiency of A =100/10 = 10%
Efficiency of B = 100/15 = 6.66%
Efficiency of A & B Together = 10+ 6.66 = 16.66%
So the time taken by A & B together to Complete the work will be 100/16.66 = 6 Days.

Case -2 A can do a work in X days and B can do it Y days, In how many days the work is completed if they work alternatively Started by A.
Now in these type of question the person are not actually working together, what happens in this type of question is that A works for 1 day and then on 2nd day B work and on 3rd again A work and on Fourth again B works and so on till the work is completed.
For example A can do a work in 10 days B can do it 15 days and how many days they will finish it if The work is started by A
So again work done by A in one day = 1/10
 ''     ''          ''        ''      ''  B      ''      ''      = 1/15
Now the work done by Togther will be = 1/10 + 1/15 = 1/6 [ Note now this 1/6 work is not done by them in 1 day but in 2 days Actually, See A worked for 1 day and did 1/10 work on the second day B worked and finished the 1/15 work So in total 2 days they did 1/6 work]
So in 2 days they did 1/6 work so in how many days they will complete the whole work, Simple 12 days.
Efficienecy Method
A's Efficiency = 10%
B's Efficiency = 6.66%
A + B Efficiency = 16.66%
Work done by A and B in 2 days [ remember 2 days because they are not working together but working alternatively] = 16.66%
So time taken by them to complete 100% work = 100/(16.66 = 6 [ but always remember multiply this by 2, Beacause 16.66% work is done by them in 2 days and not in 1 day.
So The answer will be 6*2 = 12 days.

Case 3: A can do a work in X days, B can do the work Y days and A leaves after working Z days.
Question- A can do a work in 10 days and B can do it in 15 days, A works for 2 days and then leaves, In how many days will the work be completed?
Now here we can se that A leaves after 2 days that means A and B only worked for 2 days and the remaining work is done by B alone.
So first we have to calculated the work done by A and B together in 2 days.
So work done be A in 1 day = 1/10
""   "         "        "  B "  "   "     = 1/15
Work done by A & B together is 1 day = 1/10 + 1/15 = 1/6
Work done by A & B together in 2 days = (1/6) * 2 = 1/3
So remaining work = 1 - 1/3 = 2/3
Now this 2/3 work has to be done by B alone.
So it takes 15 days for B to do Complete work, How much time it will be taken by B to do 2/3 work ? So it will be 15*(2/3) = 10 days
So the work will be completed in 2 + 10 days = 12 days

Efficiency method
A's efficiency = 10%
B's Efficiency = 6.66%
Total a+b = 16.66%
work done by A and B together in 2 days = 16.66*2 = 33.33%
Work remaining = 66.66%
time taken by B to complete 66.66% work = 66.66/6.666 = 10 days
Total time = 10+2 = 12 days

Case 4
A can do a piece of Work in 10 days and B can do it in 15 days, In how many days will the work be completed if B leaves 2 days before the completion on work.
Now in this question B leaves before 2 days
Together they can do the work in what = 1/10 + 1/15 = 1/6
That means 6 Days.
That means Together they could have completed the work in 6 days but B works only till 4th day and The remaining work will be done by A alone
So they worked together for 4 days in all So work done by them in 4 days = (1/6)*4 = 4/6 = 2/3
remaining work = 1/3
Now this 1/3 work will be done by A alone
Now A can do the complete work in 10 days, So the time taken by him to do 1/3 work = 10 * (1/3) = 10/3 days or 3.33 days
So total time taken = 4+ 3.33 days = 7.33 days
Efficiency method
A's efficiency = 10%
B's efficiency = 6.66%
Total = 16.66%
Work will be completed in 6 days
Work done in 4 days = 66.66%
remaining = 33.33%
A will complete the remaining in = 33.33/10 = 3.33
Total = 4+3.33 = 7.33

Case 5: A can do a Work in X days B can Do it in Y days, In how many days The work will get completed if B leaves 2 days before the actual completion of work.

Question: A can do a work in 10 days B can do it in 15 days, In how many days The work will get completed if B leaves 2 days before the Actual Completion of Work.what is the difference between this Actual completion of work and Completion of Work?
See in last example the work was supposed to get completed in 6 days, So we just Solved the question taking into consideration that B leaves 2 days before the completion of work i.e B worked for 4 days and the rest 2 days work was don by A alone and Completes that work in 3.33 days Total 7.33 days.
So if i ask In this question If B left 2 days before the actual completion then it means B should have left on 5.33 days Got it ?
Now back to the question.
Now just imagine that the work gets completed in x days.
So A would work for x days[ A works for the whole time ]
And B would work for x-2 days[ because B left 2 days before the actual completion of work]
So now According to Question
x/10 + (x-2)/15 = 1 [ Beacuse work is always one ]
(3x+2x-4)/30 = 1
5x -4 = 30
5x = 34
x = 6.8 days.
So the work will be completed in 6.8 Days.
It can also be asked That after how many days B left, So the answer would be Simple 6.8 - 2 = 4.8 days

Efficiency Method
A's Efficiency = 10%
B's Efficiency = 6.66%
Let the no. of days be x
so Accordring to question
10x + 6.66(x-2) = 100[ Work is always 100% in efficiency method ]
10x + 6.66x - 13.33 = 100
16.66x = 113.33
x = 113.33/16.66 = 6.8
Answer = 6.8 days

 

 

 

Percentage I

Percentage I

Percentage:
The word defines itself Per means 1 upon something and Cent Is like Paise. In India we have 1 rupee = 100 paise
So per cent = 1/100 Part of something or %[ This sign even means 1/100] So if i say 20% of something Just multiply that something by 20/100 or 0.2
Like wise if is say 30% then it simply means i want to know 30/100th or 3/10th part value of something.

Why Do we Use percentage ?
Basically It is used for comparison.
Like If i say i got 400 marks in 10th and the other guy says that he got 600 marks in 10th. So Numerically He has got more marks than me  But does his score is relatively better than me ?
For that purpose we must know that He got 600 marks out of how many marks. Let's say he got 600 out of 1000. So his percentage marks will be 60%
And I got 400 out of 500. So my % marks will be 80%.

Now you can easily say that My marks are better because i am getting 80% and he is getting 60%.
How to calculate Percentage [ You all know it but just say I am telling it to myself :P ]
Well the simple formula is [Value/total value] * 100
For example A  Ring Contains 63 gm of Gold and total weight of ring is 70 gm. Find the percentage of Gold in the ring ?
So By the Formula [Value/Total Value]*100
[63/70]*100
[9/10]*100
90%.

% to Decimal Conversion or Fraction Conversion.

Just remember in Fraction Conversion we leave the fraction as it is without Converting Into Decimal. While in Decimal Conversion we first convert into fraction then Write the Decimal Value of that fraction.
Very easy You just have to dive by 100 nothing else.
For example 30% = 30/100 = 0.3
21% = 21/100 = 0.21
99% = 99/100 = 0.99
60% = 60/100 or 3/5 = 0.6

Case 1 [ Percentage of Quantity ]
Find the no. of male Students i.e boys, If there are 47% male students in the school and Total no. of students in the school is 1000.
As i said If you See anywhere % of something. Just convert the no. into it decimal value and multiply by that Something.
So in the above Question Boys are 47%[ Convert this into Decimal and you will get 0.47]  of 1000 [ Something]
So what we gonna do friends we will just multiply it by 0.47
So the no. of Boys will be 0.47*1000 = 470.

Lets See another Example.
A student scored 85% marks. Total marks are 400. How much marks did he score.
So a student got 85% marks out of 400
So again 85%[ Convert 85% in decimal i.e 0.85] of 400 [something]
So the answer will be 0.85*400 = 340
so 340 is our answer.

Well they can also Change the Final Question.
Like in First Example they asked Find the no. of male students. They could have asked the no. of students that are not male.
So what we should do in that case. Nothing to worry my friend just do the usual job 47% are male that means that the rest 53% are not male now calculate 535 of 1000 that will be your answer i.e 530

Case 2 [ Inverse Case]
Now in the case 1 we were just asked to Find the % value something. But What if % value of something is given and we have to find the Total Value. ?
Now to worry below example will make it clear.
30% of a Number is 150. What is the number.
So after examining the question we can say that 30% of Some number is 150[ But we don't know yet what is the original number]
When we don't know about something Just Assign a variable to that value.
So we say that the Original Number is x
So as mentioned in the question 30% of x = 150
[ Convert 30% into decimal] 0.3*x= 150
0.3x = 150
x = 150/0.3
x = 500
So you see it's Quite easy.
Likewise Many Different Question can be formed on the same logic. Lets discuss 1
There are 200 girls in the class and girls and girls make up 25% of the class. Find the total No. of students in the class.
We don't know the No. of students so assume that no. of students is x
So what is given in the Question.
25% of Total Students in the class are Girls and Total Girls in the class is 200
lets just try convert above [English] Statement into mathematical Form
25% of x = 200
0.25*x = 200x = 200/0.25
x = 800
So total no. of students in the class = 800.

Case 3 - Percentage Change.[ Very Important For DI]
The simple Way to put that is [( Change in Quantity/ Original Quantity) *100 ] Also [change in quantity = Final Quantity - Initial Quantity]
Note- The quantity in whose respect % change is asked is considered as the base By base i mean the original value in the above formula.
Let me make it clear to you with the help of Some Example.
The height of Nikhil some times ago was 160cm. Now his height is 200cm. Find the % change in his height?
So if we analyse the above question We can say that all we have to calculate is the % change in the height of  Nikhil with respect to his Earlier Height.
So now lets apply the formula here [( Change in quantity/ Original Quantity) * 100]
Which will be [{( 200-160)/160} * 100 ]
[{40/160}*100]
[(1/4)*100]
25%
So the % change in height = 25%.

Case 4- Use of Base Value and With Respect To Cases[ Very very Important For DI]
Suppose Salary of  Raman is 80,000 and Salary of Ved is 1,00,000. The questions are
What per cent is the salary of Ved to that of Raman?
It's a very simple Question If you just know in whose respect you have to find the %.
Now in the above Question we have to find the % of ved salary with respect to ELF's Salary [ remember jiske respect me % nikalna hota hai Wohi base hota hai ]
So here we have to find Ved's Salary with respect to Raman So we use the formula [ (Value/ In whose respect it is asked)*100]
So ved's salary in respect to Raman's salary will be [ (1,00,00/80,000)*100 ] = 125%
So VED's Salary is 125% of Raman's Salary.

If the question was just opposite.
Like What percent is the salary of Raman to that of Ved. (In this Question the Base will be Ved's Salary)
So lets just apply the formula [ Value/ In whose respect it is asked) * 100]
(80,000/1,00,000)*100 = 80%
So elf's salary is 80% of VED's Salary.

Case 5 - Product Constancy [ Most Important Because With it's Application You can also solve Questions related to Time and Work, Speed Time Distance,  Average etc. This Concept has a very huge application]
i - Speed*Time = Distance
ii- efficiency*time = work
ii- Length*breadth= area
iv- Average*No. of elements = Total value
v - rate*quantity = Expenditure

let me make you clear with an example.
The price of sugar is increased by 25% then by how much per cent should a customer reduce the consumption ( i.e quantity used) Of sugar so that he has not increase his expense on Sugar.
Just remember  If one factor of product constancy is increased by P% then the other factor will be decreased by [(p)/(100+p) * 100] To maintain the Product Constancy.

Now in the above Question The rate of sugar is increased by 25% So by how much % we should reduce the quantity to maintain the same expenditure
Just apply the above formula [(p)/(100+p)*100]  = (25/125)*100 = 20%
Now It sound Simple but It is difficult to remind these formulas at the time of Solving Question So let me Give you simple method of learning this Formula.
Just Imagine In Your mind that the Quantity is 100. Ok
if the value is inreased by 25 % how much should the consumption be reduced.  
Now all you have to remember is [(How much % value is Increased/ What it becomes after increase) * 100 ]

No as i said In your mind the Quantity is 100. How much the value is increased in the above Question yeah 25%
And how much it will become after 25% increase if the Quantity was 100 yeah That will be 125. So the answer will be
(25/125)*100 = 20%
lets try again If the price of petrol is increased by 50%. By how much % the consumption be reduced so the expenditure remains same.

Just apply the formula How much increased = 50
What it will become after 50% increase = 150
% redcution required = (50/150)* 100 = 33.33%
In the same way you can also use the same formula for calculating just the opposite.

For Example If the price of Sugar is reduced by 20% by how much should the family increase it's consumption So the expenditure remains same?
How much % decrease ? = yeah it is 20%
What it will become after 20% decrease = Yeah 80
So Increase required = (20/80)*100 = 25%

lets do one more question.
The price of petrol is reduced by 33.33% but how much % should a person increase his consumption so that His expenditure remains constant.

How much decrease = 33.33
What it will become after 33.33% decrease = 66.66
So % increase required = (33.33/66.66)*100 = 50%

The Length of Rectangle is Increased by 25% By what % the breadth be reduced so that are remains Constant?
Try Again How much Increase 25
What it becomes 125
% to be reduced = (25/125)*100 = 20%
Same way the Question of Time Speed Distance can also be solved But i will teach that when i will Explain Time Speed And Distance.

Case 6 - Increase or decrease In value to Get Back the Original Value.
Remember if a value P is increase by x % then we have to decrease the resultant value by
[{ x/(x+100)} *100]% to get back the original value.
For Example Rocky's Salary is 1000rs and it Increased by 10%. How much % His salary must be Decreased So that he Gets His original Salary.
Apply the above Formula [{x/(x+100)}*100 = [ {10/(100+10)}*100] = 100/11 or 9.09090%
In case of Decrease.
The formula will be [ {x/(100-x)}*100]%
Rocky Salary is 1000 and it is decreased by 10%. By how much % his salary must be increased so that he gets His Original Salary.
Apply the formula here [{10/(100-10)}*100]% = 100/9 % or 11.11%
But Instead of Doing All this BS you can Also Apply My previous formula here.

Like Salary Increased 10. what wit will become 110.
How much it should be reduced = (10/110)*100 = 100/11 = 9.09%
Again Salary decreased = 10, What it will become 90.
So how much it should be increased to get the original salary = (10/90)*100 = 100/9 = 11.11%

CASE - 7 Concept of "by" and 'to"
Please note that there is very Big Difference between by and to.
Eg . The income is reduced BY 40% it means the New  Income 60% of the original value.
And If income is to 40% it means The new Income is 40% of the Original Value.

Case 8 - Consecutive Increase in Percentage.

Suppose the Salary of Sumit is first increase by 20% and Then again it's Increased by 20%.
What is the Total Percentage Increase in His Salary.
Now don't try to be smart here and just add 20% and 30%And say That it's 50% Increase-- THAT WILL BE TOTALLY WRONG.
Actuall let me make the Picture a Little bit Clear. What actually Happens in the case of Consecutive Increase and Decrease.
Now Just Suppose that The Salary of Sumit was 1000 Rs. it gets Increased by 20% so What it will become ? Yeah you are right 1200rs.
Now When it is Again Increased by 30% Then we are Calculating that 30% increase on 1200RS ans not on 1000 So the Inrease will be 30% of 1200 which will be 360. So increased salary will be 1200 + 360 = 1560.

Now are two Shortcut Methods Here.

1st Simple Multiplication.
When I say Something is increased by 20% It means It's Value Is increased by 20 % or It's Total value is 120% of the original Value Ok ?
Like 100 is incresed by 20% That means it's final value will be 100 + 20% of 100 = 100 + 20 = 120.
So if  Sumit  Salary is Increased By 20% that the Value will be 100*1.2.
And If it's Again increased by 30% then the value will be 100*1.2*1.3 = 100*1.56 = 156.
Total increase = 56.
Percentage Increase = 56%

2nd Method Formula Approach.

Well the formula is [x + y + (xy)/100]% [Note this formula works only when there is 2 increases]
Now apply the formula in above Question you will get 20 + 30 + (20*30)/100 = 50 + 600/100 = 50 + 6 = 56%

Same Sumit's Salary is 1000rs and if it's asked the Salary of Sumit is Increased First by 20% then 30% and then again by 40% then what will be Total Increase and Final Salary.
It's Pretty simple now 1000*1.2*1.3*14 = 2184
That's the final Salary and % inrease = [(2184-1000)/1000]*100 = (1184/1000)*100 = 118.4%

Some Similar Questions Are like.
The Side of Square is Increased by 10% what will be the increase in Area.
So Just Let The Value of Eaxh side is x, If it's Increased by 10% Then it will become 1.1x
As You know are = Side*side
So Initially The Area without Increase Would Have been x*x = x^2
After Increase the Area will be 1.1x*1.1x = 1.21x^2
So total % increase in area will be 21%.

Case -9 Consecutive Increase and Decrease Simultaneously
In the last case we saw the case of % increase but now we will learn how to solve when there is a consecutive Increase and as well As decrease.
It's same as the last example.
Sumit's Salary is 1000rs Suppose the salary of Megamind was first Increased by 30% and Then decreased by 20%. What will be final Increase or Decrease in His Salary.
Just do The same Thing 30% increase means 1000*1.3
And then 20% Decrease mean 0.8times [ remember we have to decrease here and 20% decrease means 0.2 Point decrease]
So total decease = 1000*1.3*0.8 = 1040.
Final Increase = 40RS
% Increase will be 4%.
With Formula.
Remember the formula x + y + (xy)/1000
Same formula can be used here But remember Increase means +ve Sign and decrease means -ve sign,
So apply here now 30 - 20 + (30)(-20)/100 = 10 - 600/100 = 10 - 6 = 4%.

 

Quiz : Time : 5-6 minutes.

1.  Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. What are the marks obtained by them?
A) 42, 36
B) 44, 39
C) 42, 33
D) 44, 37
E) None of these.

2. A fruit seller had some oranges. He sells 40% oranges and still has 420 oranges. How many oranges he had originally?
A) 690
B) 700
C) 720
D) 745
E) None of these.

3.  A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
A) 45(6/11) %
B) 45(7/11) %
C) 45(5/11) %
D) 46(5/11) %
E) None of these.

4.  In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, what was the number of valid votes that the other candidate got?
A) 3690
B) 2700
C) 5720
D) 4745
E) None of these

5.  A student has to obtain 33% of the total marks to pass. He got 125 marks and failed by 40 marks. The maximum marks are
A) 600
B) 700
C) 520
D) 500
E) None of these.

6. A housewife saved Rs. 2.50 in buying an item on sale. If she spent Rs. 25 for the item, approximately how much percent she saved in the transaction ?
A) 6 %
B) 7 %
C) 10 %
D) 9(1/11) %
E) None of these.

7. A pipe X is 30 meters and 45% longer than another pipe Y. find the length of the pipe Y?
A) 600/29 meter
B) 37.25 meter
C) 20 meter
D) 50 meter
E) None of these.

8. On my sister’s 15th birthday, she was 159 cm in height, having grown 6% since the year  before. How tall was she the previous year ?
A)156 cm
B) 150 cm
C) 155 cm
D) 172 cm
E) None of these

9. Sumit got 30% of the maximum marks in an examination and failed by 10 marks. However, Sujith who took the same examination got 40% of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination?
A) 96
B) 150
C) 75
D) 85
E) None of these

10. 30% of the men are more than 25 years old and 80% of the men are less than or equal to  50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
A) 60 %
B) 50%
C) 80 %
D) 85 %
E) None of these

Answers:

1. C
2. B
3. C
4. B
5. D
6. D
7. A
8. B
9. D
10.C

Ratio And Proportion

Ratio And Proportion

What is Ratio?
A ratio is a relationship between two numbers by division of the same kind. The ration of a to b is written as a : b = a / b In ratio a : b , we can say that a as the first term or antecedent and b, the second term or consequent.

Example :  The ratio 4 : 9 we can represent as  4 / 9 after this 4 is a antecedent and , consequent = 9

Rule of ration :  In ratio multiplication or division of each and every term of a ratio by the same non- zero number does not affect the ratio.

Different type of ratio problem is given in Quantitative Aptitude which is a very essential topic in banking exam. Under below given some more example for your better practice.
Anything we learn in our school days was basics and that is well enough for passing our school exams. Now the time has come to learn for our competitive exams. For this we need our basics but also we have to learn something new. That’s where shortcut tricks and formula are comes into action.

What is Proportion?
The idea of proportions is that two ratio are equal.
If a : b = c : d, we write a : b : : c : d,
Ex. 3 / 15 = 1 / 5
a and d called extremes, where as b and c called mean terms.

Proportion of quantities
the four quantities a, b, c, d said proportion then we can express it
a : b = c : d
Then a : b : : c : d  <–> ( a x d ) = ( b x c )
product of means = product of extremes.

If there is given three quantities like a, b, c of same kind then then we can say it proportion of continued.
a : b = b : c the middle number b is called mean proportion. a and c are called extreme numbers.
So, b2 = ac. ( middle number )2 = ( First number x Last number ).

Quiz on ratio and proportion:-

1 If P : Q : R = 2 : 3 : 4 , Then P / Q : Q / R : R / P = ?
A. 8: 9: 24
B. 7: 9: 24
C. 4: 6: 15
D. 8: 11: 24
E. None of these

2: If 2P = 3Q = 4R, Then P : Q : R = ?
A. 2: 3: 5
B. 2: 3: 4
C. 3: 5: 6
D. 1: 2: 3
E. None of these

3: If P : Q = 2 : 3 , Q : R = 4 : 5 and R : S = 6 : 7 , then P : S = ?
A. 18: 25
B. 17: 35
C. 16: 35
D. 8: 11
E. None of these

4:Rama distributes his pencil among his four friends Rakesh, Rahul, Ranjan, and Rohit in the ratio 1 / 2 : 1 / 3 : 1 / 4 : 1 / 5 . What is the minimum number of pencils that the person should have?
A. 66
B. 64
C. 72
D. 77
E. None of these

5: Two numbers are respectively 40% and 60% more than third number. Find the ration of two numbers ?
A. 8: 7
B. 7: 9
C. 9: 11
D. 8: 13
E. None of these

6: Rs 1210 were divided among three person P, Q, R so that P : Q = 5 : 4 and Q : R = 9 : 10. Then R gets the amount.
A. 450
B. 400
C. 500
D. 375
E. None of these

7: Share Rs.4200 among joy, sanjay and bijoy in the ration 2 : 4 : 6.Find the amount received by sanjay.
A. 1200
B. 1300
C. 1400
D. 1500
E. None of these

8 :Find the mean proportional between given two number that is 64 and 49.
A. 45
B. 52
C. 54
D. 56
E. None of these

9: What number has to be added to each term of 3 : 5 to make the ratio 5 : 6 .
A. 7
B. 6
C. 8
D. 5
E. None of these

10:Rs. 385 were divided among P , Q , R in such a way that P had Rs 20 more than Q and R had Rs 15 more than P . How much was R’s share?
A. 156
B. 145
C. 152
D. 150
E. None of these

Answers with Explanation:-
1. P : Q : R = 2 : 3 : 4 .
Let P = 2k,
Q = 3k,
R = 4k.
Then,
P / Q = 2k / 3k = 2 / 3 ,
Q / R = 3k / 4k = 3 / 4
R / P = 4k / 2k = 2 / 1.
SO, P / Q : Q / R : R / P = 2 / 3 : 3 / 4 : 2 / 1 = 8 : 9 : 24.

2. Let 2P = 3Q = 4R = k ,
Then ,
P = k / 2,
Q = k / 3 ,
R = k / 4.
SO , P : Q : R = k / 2 : k / 3 : k / 4 = 6 : 4 : 3.

3. (C)

4. Rakesh : Rahul : Ranjan : Rohit = 1 / 2 : 1 / 3 : 1 / 4 : 1 / 5
Step 1: At First we need to do is LCM of 2,3,4 and 5 is 60.
Step 2: Then pencil are distributed in ratio among friends,
Rakesh = ( 1 / 2 x 60 ) = 30.
Rahul = ( 1 / 3 x 60 ) = 20.
Ranjan = ( 1 / 4 x 60 ) = 15.
Rohit = ( 1 / 5 x 60 ) = 12.
Step 3: Total number of pencils are ( 30 x + 20 x + 15 x + 12 x) = 77 x.
For minimum number of pencils x = 1 .
The person should have atleast 77 pencils.

5.Step 1: Let the third number is A
Then first number is 140% of A = 140 x A / 100 = 7A / 5 and second number is 160% of B = 160 x B / 100 = 8B /5.
Step 2: now ratio of first and second number is 7A / 5 : 8B / 5 = 35A : 40B = 7 : 8.

6. P : Q = 5 : 4, Q : R = 9 : 10 = ( 9 x 4 / 9 ) : ( 10 x 4 / 9 ) = 4 : 40 / 9.
So, P : Q : R = 5 : 4 : 40 /9 = 45 : 36 : 40
Sum of ratio terms is = ( 45 + 36 + 40 ) =121.
R share of amount is Rs (1210 x 40 / 121) = Rs. 400.

7. Amount received by sanjay.
4 / 12 X 4200 = 1400= ( related ratio / sum of ratio ) x Total amount
So, the Amount received by sanjay is 1400.

8. The mean proportion of two numbers is
Root of 64 and 49 is √8 x √ 7 = 8 x 7 = 56.
So, the mean proportional is 56.

9.Let the number to be added x , Then
3 + x / 5 + x = 5 / 6
6 ( 3 + x ) = 5 ( 5 + x )
x = ( 25 – 18 ) = 7
So , the number to be added is 7 .

10. Let Q gets Rs x. Then We can say P gets Rs (x + 20 ) and R gets Rs ( x + 35) .
x + 20 + x + x + 35 = 385
3x = 330
x = 110 .
R’s share = Rs ( 110 + 35 ) = Rs 145 .

 

 

 

Number Series Rules with solved

 

Number Series Rules with solved

Dear Readers,

Since many Banking exams are in the streak and Maths section is like to be out of depth for many Aspirants. Here we are sharing Rules pertained to Number Series which is prepared and provided by one of our Ardent Readers Stallon. Number series form which at least 4- 5 questions use to come in the Banking Exams. We anticipate these would be viable to all of you.

Number Series Rules with solved Problems

Series is an important chapter from banking examination point of view. Following are some of the important rules or order on which the number series can be made :-
Pure Series
Difference Series
Ratio Series
Mixed Series
Geometric Series
Two Tier Arithmetic Series
Other Types

1. Pure Series

In this type of number series, the number itself obeys certain order so that the character of the series can be found out.
The number itself may be.
 

Perfect Square

Example :
121, 144, 169, 225 ?
Answer - 256
Perfect Cube
Example :
6859, 5832, 4913, 4096, 3375, ?
Answer - 2744

2. Difference Series

Example :
1348, 1338, 1318, 1288, 1248, ?
Answer - 1198

3. Ratio Series

Example :
336, 168, 84, 42, 21, ?
Answer - 10.5

4. Mixed Series

Example :
222, 441, 1321, 2639, 7915, ?
Answer - 15827

5. Geometric Series

Example 1. 5, 35, 245, 1715, ?
Ans. 12005
Examples 2. 43923,3993, 363, 33, ?
Ans. 3

Other Type

To find the odd number from the number series. In this type of series the above rules are also followed.

Some Examples ;

 2, 3, 7, 22, 89, 440, 2677, 18740
Solution : ×1+1, ×2+1, ×3+1, ×4+1, ×5+1 ........
 So, 440 is replaced by 446
5, 6, 14, 40, 89, 170, 291
Solution : +12, +32, +52, +72, +92
 So, 14 is replaced by 15.
445, 221, 109, 46, 25, 114, 4
Solution : -3÷2, -3÷2........
So,46 is replaced by 53.
12, 26, 56, 116, 244, 498, 1008
Solution : ×2+2, ×2+4 ×2+6,  ........
So, 116 is replaced by 118
8, 27, 64, 125, 217, 343
Solution : 23, 33, 43, 53,.....
So, 217 is replaced by 216