Mathematical Aptitude (Ratio and Proportion,Average,Percentage)

Mathematical Aptitude

Ratio And Proportion-

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 Ratio : In ratio multiplication or division of each and every term of a ratio by the same

non- zero number does not affect the ratio.

Proportion-

The idea of proportions is that two ratios 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.

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If there is given three quantities like a, b, c of same

kind then then we can say the 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 ).

Examples-

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

Solution-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: 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

Solution-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.

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

Solution-.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 .

Average

An average or an arithmetic mean of given data is the sum of the given observations divided by the number of observations.

Averages of a group is defined as the ratio of sum of all the items in the group to the number of items in the group.

Average = (Sum of all items in the group)/ Number of items in the group .

Some Important Concepts:

1.Average =total of data/No.of data

2.If the value of each item is increased by the same value a, then the average of the group or items will also increase by a.

3.If the value of each item is decreased by the same value a, then the average of the group of items will also decrease by a.

4.If the value of each item is multiplied by the same value a, then the average of the group or items will also get multiplied by a.

5.If the value of each item is multiplied by the same value a, then the average of the group or items will also get divided by a.

6.If we know only the average of the two groups individually, we cannot find out the average of the combined group of items.

7.Average of n natural no’s=(n+1)/2

8.Average of even No’=(n+1)

9.Average of odd No’= n

10.General Formula=(1st number +Last number)/2

Eg.

.The average age of 8 persons in a committee is increased by 2 years when two men aged 35 years and 45 years are substituted by two women.Find the average age of these two women.

(a)48

(b)36

(d)29

(e)none of these

Solution- Total increase = 8 * 2 = 16 years

So, total age of two women = 35 + 45 +16= 96

Average age of two women=96/2=48 years

Important Formulas Related to Average of numbers

1. Average of first n natural number=(n+1)/2

2. Average of first n even number= (n+1)

3. Average of first n odd number= n

4. Average of consecutive number= (First number+Last number)/2

5. Average of 1 to n odd numbers= (Last odd number+1)/2

6. Average of 1 to n even numbers= (Last even number+2)/2

7. Average of squares of first n natural numbers=[(n+1)(2n+1)]/6

8. Average of the cubes of first n natural number=[n(n+1)^2]/4

9. Average of n multiples of any number=[Number*(n+1)]/2

Average Concepts

If the average of n_1 observations is a_1; the average of n_2 observations is a_2 and so on, then

Average of all the observations=(n_1* a_1+n_2 *a_2+……)/(n_1+n_2+……..)

If the average of m observations is a and the average of n observations taken out of is b, then Average of rest of the observations=(ma-nb)/(m-n)

Example 1 : A man bought 20 cows in RS. 200000. If the average cost of 12 cows is Rs. 12500, then what will be the average cost of remaining cows?

Here m = 20 , n = 12 , a = 10000 , b = 12500

average cost of remaining cows ( 20-8) cows = (20*10000 – 12*12500)/(20-8) =Rs 6250

If the average of n students in a class is a, where average of passed students is x and average of failed students is y, then

Number of students passed=[Total Students (Total average-Average of failed students)]/(Average of passed students-Average of failed students)

=[n(a-y)]/(x-y)

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Example 2: In a class, there are 75 students whose average marks in the annual examination is 35. If the average marks of passed students is 55 and average marks of failed students is 30, then find out the number of students who failed.

Here , n = 75 , a = 35 , x = 55 , y = 30

Number of students who passed = 75(35- 30)/(55- 30) = 15

Number of students who failed = 75- 15 = 60

If the average of total components in a group is a, where average of n components (1st part) is b and average of remaining components (2nd part) is c, then Number of remaining components (2nd part)=[n(a-b)]/(c-a)

Example 3 : The average salary of the entire staff in an office is Rs. 200 per day. The average salary of officers is Rs. 550 and that of non-officers is Rs. 120. If the number of officers is 16, then find the numbers of non-officers in the office.

Here n= 16 , a = 200 , b = 550 , c = 120

Number of non – officer = 16(200- 550)/(120- 200) = 70

Average Speed

Average speed is defined as total distance travelled divided by total time taken.

Average speed=Total distance travelled/ Total time taken

Case 1

If a person covers a certain distance at a speed of A km/h and again covers the same distance at a speed of B km/h, then the average speed during the whole journey

will be

2AB/A+B

Case II

If a person covers three equal distances at the speed of A km/h, B km/h and C km/h respectively, then the average speed during the whole Journey will be

3ABC/(AB+BC+CA)

Case III

If distance P is covered with speed x, distance Q is covered with speed y and distance R is covered with speed z, then for the whole journey,

Average speed=(P+Q+R+…..)/(P/x+Q/y+R/z+…)

Example 4 : A person covers 20 km distance with a speed of 5 km/h, then he covers the next 15 km with a speed of 3 km/h and the last 10 km is covered by him with a speed of 2 km/h. Find out his average speed for the whole journey.

Average speed = ( 20 +15 +10)/(20/5+15/3+10/2) = 3(3/14)

Percentage:

The term percent means ‘for every hundred’.

“A percent as a fraction whose denominator is 100 and the numerator of the fraction is called the rate per cent.” Per cent is denoted by the sign ‘%’.

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

Likewise if it is say 30% then it simply means I want to know the 30/100th or 3/10th part value of something.parveeenaa.com

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%.

Q-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.

If we see anywhere % of something. Just convert the no. into its 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 we will just multiply it by 0.47

no. of Boys will be 0.47*1000 = 470.

Concept to Calculate Percent

If we have to find y% of x, then

y% of x=(x*y)/100

Conversion of Percent into Fraction

Expression per cent (x%) into fraction.

Required fraction=x/100 Conversion of fraction into Percentage

Expressing a fraction (x/y) in percent.

Required percentage=(x/y)*100)% Expressing One Quantity as a Percent with Respect to Other

To express a quantity as a percent with respect to other quantities the following formula is used.

(The quantity to be expressed in percent)/ (2nd quantity (in respect of which the percent has to be obtained))X100%

Important Concept and Tricks

1. If x% of A is equal to y% of B, then

z% of A=(yz/x)% of B

2. When a number x is increased or decreased by y%, then the new number will be

(100+y)*x/100

3. When the value of an object is first changed (increased ) by a% and then changed (increased ) by b%,then

Net effect=[a+b +ab/100]%

4. Suppose in an examination, x% of total number of students failed in subject A and y% of total number of students failed in subject B and z% failed

in both the subjects. Then,

(i) Percentage of students who passed in both the subjects=[100-(x+y-z)]%

(ii) Percentage of students who failed in either subject=(x+y-z)%

5. If due to r% decrease in the price of an item, a person can buy A kg more in Rs.x, then

Actual price of that item= Rs (rx)/((100-r)A) Per kg

Example :If due to a 10% decrease in the price of sugar ,Ram can buy 5 kg more sugar in Rs 100 , then find the actual Price of sugar ?

solution : Here r = 10 % ,x = 100 and A = 5 kg

Actual price of sugar = 10*100/((100-10 )*5) = Rs. 2(2/9)

6.If the population of a town is P and it increases at the rate of R% per annum, then

(i) Population after n yr= P(1+r/100)^n

(ii) Population, n yr ago=P/(1+r/100)^n

7. If the present population of a city is P and there is a increment of R1%, R2% ,R3% in first, second and third year respectively, then

Population of city after 3 yr=P(1+R1/100)(1+R2/100)(1+R3/100)

Example : Population of a city in 20004 was 1000000. If in 2005 there is an increment of 15 % , in 2006 there is a decrease of 35 % and in 2007 there is an increment of 45 %, then find the population of the city at the end of the year 2007.

solution : Required population = P (1 + R1/100)(1 – R2/100)(1 + R3/100)

= P (1 + 15/100)(1 – 35/100)(1 + 45/100)

= 1083875