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Basic Numeracy/Quantitative Aptitude


Some Tips on Basic Numeracy/Quantitative Aptitude to help the students solve such questions are presented below:

Problem Solving by Substitution:

In this type of questions, the candidate is provided with substitutes for various mathematical symbols. It is followed by a question involving calculation of an expression or selecting the correct/incorrect equation. The real signs are to be put in the given equation to arrive at the right choices.

Remember:

Fundamental Operations:


Fundamental operations are mathematical expressions which include division, multiplication, addition and subtraction. In complex expressions there are other operations namely 'of' and 'brackets'.

Brackets are of different types. Some of the most commonly used brackets are:

Bracket symbol Name
() Paranthesis or common brackets
{} Braces or curly brackets
[] Brackets or square brackets or box brackets
- Vinculum


The order in which various mathematical operations are to be performed is the rule of 'BODMAS'. where

B -> Brackets : - (), {}, []
O -> Of
D -> Division : ÷
M -> Multiplication : x
A -> Addition : +
S -> Subtraction : -

According to the rule of BODMAS, first of all brackets should be removed followed by Of, Division, Multiplication, Addition and Subtraction.

Rules for Multiplication:

+ x + = + e.g. 4 x 4 = 16
+ x - = - e.g. 4 x (-4) = -16
- x + = - e.g. (-4) x 4 = -16
- x - = + e.g. (-4) x (-4) = 16

Rules for Division:

+ ÷ + = + e.g. 4 ÷ 4 = 1
+ ÷ - = - e.g. 4 ÷ (-4) = -1
- ÷ + = - eg. (-4) ÷ 4 = -1
- ÷ - = + (-4) ÷ (-4) = 1

Rules for Removal of Brackets:

In order to simplify complex expressions involving more than one brackets, the following steps are taken:

Step I: If an expression contains a vinculum then perform operations under it.
Step II: Perform operations within the innermost bracket.
Step III: Remove innermost bracket by using the following rules:

Rule 1: If a bracket is preceded by plus sign, remove it by writing the terms as they are.
Rule 2: If a bracket is preceded by minus sign, the positive signs are changed to negative and vice-versa.
Rule 3: If there is no sign between a number and a grouping symbol, then it means multiplication.
Step IV: Remove the second innermost brackets by using the rules given in step III. Continue this process till all the brackets are removed.

Arithmetical Reasoning

The following important topics are described under the heading arithmetical reasoning:

[A] Profit and losss:

A shopkeeper buys goods either directly from a manufacturer or through a wholeseller.
Cost Price: The money paid by the shopkeeper to buy goods from a manufacturer or wholeseller is called the cost price of the shopkeeper. Cost price is denoted as C.P.

Selling Price: The price at which the shopkeeper sells the goods is called the selling price of the shopkeeper. Selling price is denoted as S.P.

If (S.P.) > (C.P.), then there is a gain or profit. And gain = (S.P.) - (C.P.)

If (S.P.) < (C.P.), then there is a loss. And loss = (C.P.) - (S.P.)

Gain as well as loss is always calculated on C.P.

(i) Gain % = {Gain/C.P.x100}
(ii) Loss % {Loss/C.P.x100}

Generally a shopkeeper has to bear additional expenses such as freight charges, labour charges and maintenance charges for the goods, before they are sold. Such charges are known as overhead charges.

The real cost price = total investment = (payment made for buying goods + overhead charges)


[B] Unitary method:

The method of finding first the value of one (unit) article from the value of the given number of articles and then the value of the required number of articles is known as the unitary method. The general rules used in this method are:

(i) To get more, we multiply
To get less, we divide

If 12 mangoes cost Rs. 60, what is the cost of 29 mangoes?

Cost of 12 mangoes = Rs. 60
Cost of 1 mango = Rs. 60/12 [less mangoes, less cost]
Cost of 29 mangoes = Rs. 60/12 x 29 = Rs. 145 [more mangoes, more cost]


[C] Speed, Time and Distance:

Speed: 'Speed' is defined as distance travelled in unit time. A speed of '30 km per hour' or '30 km/hr' means that a distance of 30 km is covered in 1 hour.
Similarly a speed of '20 m per second' or 20 m/sec' means that a distance of 20 m is covered in 1 second.

Formulae:

(i) Distance = Speed x Time

(ii) Time = Distance / Speed

(iii) Speed = Distance / Time

(iv) To convert the speed in km/hr to m/s multiply the numerical value by 1000/(60 x 60) =10/36 = 5/18.

(v) To convert the speed in m/s to km/hr multiply the numerical value by (60 x 60)/1000 = 36/10 = 18/5.

(vi) Average speed = Total Distance covered / Total time taken

(vii) One kilometer per hour = 5/18 meter per second

viii) One meter per second = 18/5 kilometer per hour

(ix) If a train is to cross a telegraph post or a man standing near a railway line, it has to cover its own length with a given speed in a given time.

(x) If a train is to pass a railway bridge or a railway platform, it has to cover the length of the bridge or a platform in addition to its own length.

(xi) When a train is passing the other train completely (whether moving in the same or in opposite direction), one has to cover a distance equal to the sum of the lengths of the two trains.

(xii)If a body travels a distance at the rate of x km. p.h. and another equal distance at the rate of y km. p.h. the average speed for the whole distance is 2xy/(x + y).

(xiii)If two bodies are moving in the same direction with speeds u and v km. p. hour starting from the same point at the same time, their relative speed is given by (u - v) km. p. hour and it is (u + v) km. p. hour if they are moving in the opposite direction.

(xiv)If a man changes his speed in the ratio of u : v, then to cover the same distance, the ratio of the times shall be v : u e.g. if a man is to travel a distance of 20 kms, he takes 4 hours at 5 km. p. hour. Thus ratio of speed is 5 : 4 and that of time is 4 : 5.

(xv) If a man can row boat at the rate of u km. per hour in still water and if v km. per hour is the speed of the river, then (u + v) is the speed of the boat down stream and (u - v) is that of up stream. The speed of the boat down the river is called down rate and against is called up rate.

The u + v = down rate and u - v is up rate.

∴ u = 1/2[down rate + up rate]

and v = 1/2[down rate - up rate].

[D] Mensuration:

Rectangle:

Area of Rectangle = length x breadth = l x b
Perimeter of rectangle = 2 (length + breadth) = 2 (l + b)
Diagonal of the rectangle = √ {(lenght)2 + (breadth)2} = √ (l2 + b2)

Square:

Area of square = (side)2 = a2
Side of square = √area
Perimeter of square = 4 x sides = 4a
Diagonal of square = √2 x side = √2a
Side of square = Perimeter/4

Parallelogram:

Opposite sides of parallelogram are equal and parallel. Diagonals of a parallelogram bisect each other.
Area of parallelogram = base x height
Perimeter of parallelogram = 2(length + breadth)

Triangle:

Area of triangle = 1/2 x base x altitude

Circle:

Circumference of circle = 2Πr
Area of circle = Πr2

Cube:

A rectangular parallelopiped whose faces be squares, and length, breadth and height be equal is called a cube.
Volume of a cube = (side)3
Curved surface area of cube = 4a2
Total surface area of cube = 6a2
Side of cube =∛(Volume)
Side of cube = √(total surface area)/6

Cuboid:

A parallelopiped whose faces are rectangle is called a rectangular parallelopiped or a cuboid.

Volume of cuboid = length x breadth x height = lbh
Curved surface area = 2h (l+b)
Total surface area of cuboid = 2 (lb + bh + hl)
Length of the diagonal = √(l2 + b2 + h2)
Length of cuboid = volume/(breadth x height)
Breadth of cuboid = volume/(length x height)
Height of cuboid = volume/(length x breadth)
Area of four walls = 2 (lenght + breadth) x height

[E] Time:

We know that 24 hours make a day. We divide each day into two halves; for this reason we use two Latin terms a.m. and p.m.
a.m. indicates the times between 12 midnight and 12 noon (i.e, before noon or midday)
p.m. indicates the time between 12 noon and 12 midnight (i.e, after noon or midday)

Generally in railway stations 24 - hour clock timing are used. To escape from the confusion of this time table, we should remember that:

(a) 12 noon is expressed as 12:00;
(b) 12 midnight is expressed as 24:00 or 00:00;

Movement of Clock:

Just watch any clock and observe the angle which is formed by the hour hand and the minute hand. Measure the degree by which a minute hand moves every 5 minutes or the hour hand moves every 15 minutes.

A- Movement of hour hand 1. 360 degree in 12 hour
2. 30 degree in 1 hour
3. 15 degree in 1/2 hour
B- Movement of Minute hand 4. 360 degree in 60 minute
5. 6 degree in 1 minute


When you have to calculate the acute angle between a time say 3:20PM then the movement of hour hand should also be taken into account. The calculation is like this:

If hour hand is exactly at 3 hr and minute hand is at 20 minute then the angle between them will be 30 degree. The movement of hour hand in 20 minutes will be (30 degree in 1 hour or 60 minutes i.e. (30/60)*20 = 10 degrees). So the acute angle is 30 - 10 =20 degree.

Some questions asked on clock sequencing are as follows:

It is 0 hr 0 min at present. How many degrees will the minute hand move by the time it is 2 hr 30 min

a. 720
b. 900
c. 880
d. 450

A clock shows the time as 4:00. After hour hand has mover 150 degrees, the time would be

a. 8:00
b. 10:00
c. 7:00
d. 9:00

[F] Problem based on ages:

The present ages of the father and son are respectively 50 years and 5 years. After how many years will the age of the father become 6 times that of his som?

(a) 2
(b) 4
(c) 6
(d) 8

Let the required number of years be x.
The present age of father = 50 years.
The present age of son = 5 years.
∴ After x years
The age of father will be (x + 50) years.
and the age of son will be (x + 5) years.
Therefore, as per the condition given;
x + 50 = 6(x + 5)
x + 50 = 6x + 30
- 5x = - 20
x = 4 years
∴ the required number of years is 4 years.

Some more thoughts on Quantitative aptitude:

[1] Numbers:

Introductory: 1. A unit is a single thing; as one boy, one book, one table.
2. A number such as four or five, not attached to any particular things or units is called an abstract number.
3. A number of particular units, such as four boys or five men, is called a concrete number.
4. All numbers are written by the means of the symbols 0, 1, 2, 3, 4, 4, 6, 7, 8, 9 which are called digits.
5. 2, 4, 6, 8 are called even digits as they are divisible by 2.
6. 1, 3, 5, 7, 9 are called odd digits as they are not divisible by 2.
7. The number to be divided is called the dividend.
8. The number by which it is divided is called the divisor.
9. The number which tells how many times the divisor is contained in the dividend is called the quotient. 10. If the dividend does not contain the divisor an exact number of times, and we take away from the dividend as many times the divisor we can, what is left is called the remainder. When there is no remainder, the division is said to be exact. 11. In exact division:
Dividend ÷ Divisor = Quotient
or Quotient x Divisor = Dividend

In inexact division:
Quotient x Divisor + Remainder = Dividend
12. For the number of coins in the successive groups, each of which is formed by putting together ten of the next smaller groups, we have the following names:

a unit,
ten,
one hundred,
one thousand,
ten thousand,
one hundred thousand,
one million,
ten million,
one hundred million,
one thousand million,
ten thousand million,
one hundred thousand million,
one billion, etc.

e.g. we may express the number 9373 in words as nine thousand, three hundred and seventy three.
13. Local value of the number is the face value of the number e.g. the local value of 9 in 9373 is nine.
14. Place value of the number is the value in whose unit's place it occupies e.g. the place value of 9 in 9373 is nine thousand.

Factors and Prime Numbers

15. If two or more numbers be multiplied together, the result is called the product, and the numbers multiplied are called the factors of the product.
Thus 6, being the result of multiplying 2 by 3, is called the product of 2 and 3; 2 and 3 are called the factors of 6.
All numbers are multiples of unity, and each contains itself once. Any number can therefore be divided by itself and unity, but if it cannot be divided without remainder by any other number, it is said to be prime and is called a prime number. we may therefore define a prime number as "a number which has no factor or divisor but itself and unity".
16. Numbers which are not prime are called composite. Taking the numbers in order it will be seen that 2, 3, 5, 7, 11, 13, 17, 19, etc. are prime numbers and 4, 6, 8, 10, 12, 14, 16, 18, etc. are composite numbers. it is seen that no even number can be prime except 2.
17. A series of numbers in which each is greater by 1 than that which precedes it, such as 5, 6, 7, 8 are called consecutive numbers.
18. An even number is of the form 2n, and odd number is of the form 2n + 1 or 2n - 1 where n is a whole number or any integer.

Rules of Divisibility:

19.
(i) 2 is a factor of all numbers whose digit can be divided by 2.
(ii) 4 is a factor if the number composed of the last two digits can be divided by 4.
(iii) 8 is a factor if the number composed of the last three digits can be divided by 8.
(iV) 5 is a factor if the last digit be either 0 or 5.
(v) 3 is a factor if the sum of the digits can be divided by 3.
(vi) 9 is a factor if the sum of the digits can be divided by 9.
(vii) 6 is a factor if both 2 and 3 are factors.
(viii) 11 is a factor if, when 1st, 1rd, 5th etc. digits are added together, and also the 2nd, 4th, 6th, etc. the difference between these sums is 0 or can be divided by 11.
(ix) 12 is a factor if both 3 and 4 are factors.
(x) 25 is a factor if last two digits are both zero or 25, 50 or 75.
(xi) 125 is a factor if last three digits are either zeros or any number divisible by 125 namely 125, 250, 375, 500, 625, 750 or 875.
20. The following are the prime numbers between 1 and 1000.

Prime Numbers

235711131719232931374143475359616771
7379838997101103107109113127131137139149151157163167173
179181191193197199211223227229233239241251257263269271277281
283293307311313317331337347349353359367373379383389397401409
419421431433439443449457461463467479487491499503509521523541
547557563569571577587593599601607613617619631641643647653659
661673677683691701709719727733739743751757761769773787797809
811821823827829839853857859863877881883887907911919929937941
9479539679719779839919971009


21. If a and b represent any two numbers, the following general principles are frequently useful:

(i) (a + b) x (a - b) = a2 - b2
(ii) (a2 - b2)/(a + b) = (a - b)
(iii) (a2 - b2)/(a - b) = (a + b)
(iv) (a + b)2 = a2 + 2ab + b2
(v) (a - b)2 = a2 - 2ab + b2
(vi) (a + b)3 = a3 + 3a2b + 3ab2 + b3
(vii) (a - b)3 = a3 - 3a2b + 3ab2 - b3
(viii) (a3 + b3)/(a2 - ab + b2) = (a + b)
(ix) (a3 - b3)/(a2 + ab + b2) = (a - b)
(x) (a3 + b3 + c3 -3abc)/a2 + b2 + c2 - ab - bc - ca) = (a + b +c)
(xi) {(a + b)2 + (a - b)2}/(a2 -b2) = 2
(xii) (a + b)2 - (a - b)2 = 4ab

22. A number which is a factor of two or more numbers is said to be a common factor, or common measure, of the numbers. We exclude unity, which is a common measure of all numbers. The greatest number which will divide each of two or more numbers is called their Highest Common Factor, or Greatest Common Measure, and is denoted by the letters H.C.F. or G.C.M.
For example, it is required to find the H.C.F. of 8 and 12.
Factors of 8 are 1, 2, 4, 8 and
Factors of 12 are 1, 2, 3, 4, 6, 12
The common factors are 1, 2, 4; but the highest of these is 4, hence 4 is the H.C.F.

23. A number which can be divided without remainder by two others, or more, is a common multiple of them. Thus 12 is a common multiple of 2, 3, 4 and 6. The least multiple which can be so divided is called their Least Common Multiple. and is denoted by letters L.C.M.
(i) If A and B are two numbers, then the product of their L.C.M. and H.C.F. is equal to the product of the two numbers i.e. L>C.M. x H.C.F. = A x B.
Reasoning: Let 6 and 8 be two numbers
H.C.F. = 2; L.C.M. = 24
2 x 24 = 6 x 8
hence the above property is established.
(ii) Lowest Common Divisor (L.C.D.) = 1 ÷ (G.C.D.)

24. If a/b, c/d, e/f be the proper fractions, then their L.C.M. is given by
(L.C.M. of numberators a, c, e, ----)/(H.C.F. of denominator b, d, f, ...)
where as their H.C.F. is equal to (H.C.F. of numberators a, c, e, ----)/(L.C.M. of denominator b, d, f, ...)
e.g. H.C.F. of 1/2, 3/5, 4/7 and 5/21 is equal to
(H.C.F. of 1, 3, 4, 5)/(L.C.M. of 2, 5, 7, 21) = 1/210
L.C.M. of 1/2, 3/5, 4/7 and 5/21 is equal to
(L.C.M. of 1, 3, 4, 5)/(H.C.F. of 2, 5, 7, 21) = 60/1 = 60.

25. For all types of arithmetical simplification, the rule of BODMAS is very useful. The letters, B, O, D, M, A, S in order of preference, are explained as follows:
B stands for brackets
O stands for of (means multiplication)
D stands for division
M stands for multiplication
A stands for addition
S stands for subtraction
Caution: The above order of preference is to be strictly maintained, any carelessness shown, results in wrong conclusion.

26. Law of indices:
(i) ax x ay = ax + y (ii)ax ÷ ay = ax - y
(iii)(ax)y = axy = yx = (ay)x
(iv) If ax = bx -> a = b
(v) If ax = ay -> x = y
(vi) If ax = 1, then x is 0 for all values of a (except 0)

27. Metric Measures:
(a) Measures of weight is in GRAMS (g)
(b) Measures of length is in METRES (m)
(c) Measures of capacity is in LITRES (l)
TABLE (Metric system)

10 milli (of a unit) = centi (of a unit)
10 centi (of a unit) = 1 deci (of a unit)
10 deci (of a unit) = unit
10 units = 1 deca unit
10 deca units = 1 hecta unit
10 hecta units = kilo unit
units may be grams, metres or litres.
One quintal = 100 kilograms
One tonne = 10 quintals = 1000 kilograms
One tonne = 2204 pounds

Land Measure:
100 centiares = 1 are = 100 sq. metres
100 ares = 1 hectare = 10000 sq. metres
100 hectares = 1 sq. kilometre

Recurring Decimal:

A decimal in which a digit or a set of digits is repeated continually is called a Recurring decimal. Recurring decimals are written in a shortened form, the digits which are repeated being marked by dots placed over the first and the last of them, thus
8/3 = 2.666. ..... = 2.6
1/7 = .142857142857142857.... = .142857
21/22 = .9545454.... = .954
Such a decimal as .142857, in which all the digits recur, is called a pure circulator, and such a decimal as .954, in which some do not recur is called a mixed circulator. The digit, or set digits, which is repeated is called the Period of the decimal. In the decimal equivalent 8/3, the period is 6; in 1/7 it is 142857 and in 21/22 it is 54.

Square Roots and Approximation:

SQUARE ROOT of a number is that number which when multiplied by itself is equal to the given number. Thus 2 is the square root of 4 and 9 is the square root of 81.
SURDS are those numbers whose square roots cannot be exactly found. Thus √2, √3, √5 cannot be exactly found and hence they are known as surds. The square roots of the numbers can be found by means of prime factors as well as by general method of long division. If we are asked a question of the type:
"By what least number, the number 3888 be multiplied or divided by so that the resulting number is perfect square"
We use the method of prime factor.
Ans if the problem is put as"what least number should be subtracted from or added to 15410 to make the result a perfect square",
We use the general method of long division.
Let us find the square root of 1296 by prime factors
1296 = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3.
= 22 x 22 x 32 x 32
Square root = 2 x 2 x 3 x 3 = 36.

Ratio & Proportion:

1. Ratio:
The relation which one quantity bears to another of the same kind with respect to magnitude is called the ratio of the one to the other. A ratio may be expressed in the form of a vulgar fraction. A ratio, for example, of 3 kg. of tea to 5 kg. of tea may be expressed either in the form 3 : 5 or in the form 3/5.

A ratio is said to be in its simplest form when both terms are integers. and when these integers are prime to one another. We may multiply or divide both terms of a ratio by any, the same number, without affecting the value of the ratio. But the ratio is not changed, if we add equal numbers or subtract them from the terms, except when the terms are equal.

2. Proportion is an equality of ratios, for example 12 : 20 = 3 : 5 or 12/20 = 3/5. Each of these expressions indicates a proportion. The terms 12, 20, 3, 5 are called proportionals and are distinguished as the 1st, 2nd, 3rd and 4th proportional respectively. The first and fourth proportionals are called extremes. The second and third proportionals are called the means. It can be seen in the above proportionals that 12 x 5 = 20 x 3.
∴ When four numbers are in proportions, the rule is that
"product of the extremes = product of the means"
Now if a : b = c : d or a/b = c/d, then a, b, c, d are called proportionals and are said to be in proportion. ∴ a x d = b x c.

3. Compound Ratio:
When the antecedents (Numerators) and the consequents (denominators) of two or more ratios are multiplied to get a new antecedent and the new consequent, the new ratio is called their compound ratio.
e.g. the compound ratio of 2 : 3, 3 : 4, 5 : 6 = 2/3 x 3/4 x 5/6 = 5/12 or 5 : 12

4. Direct Ratio:
If the increase or decrease in the value of x is the increase or decrease in the value of y, then x and y are said to be in the direct ratio and is written as x : y.

5. Indirect Ratio:
If the increase or decrease in the value of x is the decrease or increase in the value of y, then x and y are said to be in indirect ration and is written as y : x.

e.g. "If the cost of 7 shirts is Rs. 350, what is the cost of 9 shirts."
Here increase in the number of shirts results in increase in the value of money.
∴ direct ratio of shirts is also the direct ratio of money
-> 7 shirts : 9 shirts : : Rs. 350 : Rs. x
∴ 7x = Rs. 350 x 9
x = (350 x 9)/7 = Rs. 450.

Unitary Method:

The method by which we first find the value of a unit is called Unitary method.
If 17/18 of an estate be worth Rs. 850. What is the value of 3/5 of the estate?
Let x be the value of the estate, so that 17/18x = Rs. 850.
x = Rs. 850 x 18/17, then
3/5x = Rs. 850 x 18/17 X 3/5 = Rs. 540.

Chain Rule:

When a series of quantities are so connected with one another, that we know how much of the first kind is equivalent to a given quantiry of second, how much of the second is equivalent to a given quantity of the third and son on, the rule by which we find how much of the last kind is equivalent to a given quantity of the first kind is called the "chain rule".

Rule: Beginning with x, place the given relations in the form of equations so that each right hand quantiry is of the same kind as the next left hand quantity and the last right hand quantity is the same kind as the first left hand one. Then the product of the numbers on the left side is equal to the product of the numbers on the right side. From the equation, x can be found out.

Six men earn as much as 8 women, 2 women earn as much as 3 boys and 4 boys earn as much as 5 girls. If a girl earns 50 p. a day, what does a man earn a day?
Let Rs. x = Earning of 1 man
6 men = 8 women
2 women = 3 boys
4 boys = 5 girls
1 girl = 1/2 rupee
x X 6 X 2 X 4 X 1 = 1 X 8 X 3 X 5 X 1/2
x = (8 X3 X X 5 X 1/2)/(6 X 2 X X 4 X 1)
= Re. 1.25 P

Partnership:

Introduction. Partnership is an association of two or more partners who put in money together in order to carry on a certain business. Partnership is of two kinds: Simple and Compound.

(1) Simple Partnership. When the capitals of partners are invested for the same time, the partnership is called simple. In such cases profit or loss is distributed in proportion to the numbers representing the capitals expressed in the same denomination.

(2) Compound Partnership. When the capitals, whether equal or unequal, of the partners are invested for different lenghts of time, the partnership is called compound. In such cases, profit or loss is distributed in proportion to number representing the products of the capital and the periods of their investments, each expressed in the same denomination.

(3) A sleeping partner only provides capital for the business whereas a working partner provides capital for the business but also takes part in the management of the business.

A and B are partners in a business. A invests Rs. 1000 for 8 months and B Rs. 1500 for 6 months. They gain Rs. 340, what is B's share?
A puts Rs. 1000 for 8 months. If he wants to have same amount of profit as before in one month, he has to increase his investment 8 times.
∴ He must invest Rs. 8000 for 1 month.
Similarly B must also invest Rs. 6 x 1500 i.e. Rs. 9000 for 1 month.
Hence, Rs. 340 must be divided in the ratio of
8000 : 9000 or 8 : 9
∴ B's share is Rs. 8/17 x 340 = Rs. 160.

If A invests Rs. 3000 for 1 year in the business, how much B should invest so that profits after 1 year may be distributed in the ration of 3 : 2.
As the profits are always divided in proportion to the investments of the partners.
∴ B's investment is Rs. 2/3 x 3000 = Rs. 2000.

Percentage:

(1) A fraction with 100 as its denominator is called a percentage and the numerator of the fraction is called rate per cent. The term per cent means for every hyndred. The abbreviation p.c. or % are used to denote percent. 100 per cent means 100/100 = 1.

(2) Remember the following points. Their direct use helps in solving objective type problems on percentages.

100% =1; 50% = 1/2; 25% = 1/4
12 1/2% 1/8; 6 1/4% = 1/16; 37 1/2% = 3/8
62 1/2% = 5/8; 87 1/2% = 7/8; 75% = 3/4
20% = 1/5; 30% = 3/10; 70% = 7/10

(3) If we are given two numbers x and y, one of the numbers can be expressed in term of the percentage of the other.
Rule. (i) 'x' as a percentage of 'y' = x/y x 100
(ii) 'y' as a percentage of 'x' = y/x x 100

What rate percent is 3 minutes 36 seconds to an hour?
3 minutes 36 seconds = (3 x 60 + 36) seconds = 216 seconds

Number of seconds in an hour = 3600 seconds
∴ Rate percent = (216 x 100)/3600 % = 6 %

A man spent Rs. 229.50 P which is 85% of what he earned. How much does he earn?
85% of earning = Rs. 229.50 P.
His earning = Rs. 229.50 x 100/85 = Rs. 270.

Allegation or Mixture:

Find the ratio in which rice at Rs. 7.20 P a kg. be mixed with rice at Rs. 5.70 P a kg. to produce a mixture worth Rs. 6.30 P a kg.

Difference of the 1st kind and the mean price = Rs. 7.20 - Rs. 6.30 = 90 P
Difference of the mean price and the 2nd kind = Rs. 6.30 - Rs. 5.70 = 60 P
Now mixture must be in the inverse ratio i.e. 60 : 90 or 2 : 3.

How much water must be added to 14 litres of milk worth Rs. 5.40 P a litre so that the value of the mixture may be Rs. 4.20 P a litre?

The mean value is Rs. 4.20 P a litre and the price of water is zero paisa (i.e. free of cost) per litre
By allegation method:
Milk/water = (Rs. 4.20 - 0)/(Rs. 5.40 - Rs. 4.20) = 420/120 = 7/2
∴ Milk and water must be mixed in the ratio 7 : 2.
Since the milk is 14 litres, so water to be mixed is 4 litres.

Average:

(1) If a person reckons his expenditure during an ordinary year, and divides the amount by 365, the quotient gives what is called his average daily expenditure.
If a man buys 3 horses for Rs. 300, Rs. 400 and Rs. 500 respectively, he buys 3 horses for Rs. 1200, at an average price of Rs. (1200 ÷ 3) i.e. Rs. 400 each.
If a train travels 200 km. in 5 hours, the speed may vary from time to time, but the average rate is 40 km per hour.
∴ the average of any number of quantities of the same kind is the result of dividing the the sum of the quantities by the number, average is also called the mean of the quantities.
In general the average of n quantities of the same kind, denoted by a, b, c, d, ..... is 1/n (a + b + c + d...); and conversely, if A be the average of n quantities, the sum of the quantities will be nA.

(2) Compound Average:If the average value of m quantities of one kind is a and the average of n quantities of the same kind is b, the sum of the (m + n) quantities in the two sets is am + bn. Therefore the average value of the whole is (am + bn)/(m + n)

A man's daily expenditure is Rs. 10 during May, Rs. 14 during June and Rs. 15 durinng July. Find the average daily expenditure for the three months.

The total expenditure = (10 x 31 + 14 x 30 + 15 x 31) rupees
= (310 + 420 + 465) = Rs. 1195
The number of days = 31 +30 + 31 = 92
∴ the average daily expenditure = Rs. 1195/92 = Rs. 13 approx.

Simple Interest and Compound Interest

Simple Interest: Suppose I borrow a sum of Rs. 100 from a bank and keep it with me for a year. At the end of one year, I return the sum of Rs. 100 to the bank with some extra money say Rs. 5. This extra payment is called interest. The sum of Rs. 100 borrowed by me is called the Principal.The sum of Rs. (100 + 5) i.e. Rs. 105 paid by me is called the amount.
thus "amount = Principal + Interest."
If A stands for amount; P stands for principal; I stands for Interest; T stands for time (in years); R stands for rate per cent per annum (per year); the following relations may be remembered.

I = P x R X T/100
P = 100 x I/R x T
R = 100 x I/P x T
T = 100 x I/P x R
P = 100A/(100 + RT)

Compound Interest: In business transaction if interest as it becomes due is not paid to the lender but is added on the principal, the money is said to bent at Compound Interest and the total sum owed after a given time is called the Amount at compound interest for that time. The difference between an amount and the original principal is called the compound interest (C.I.)
[Note: Unless there is a mention of compound interest, it should be understood as simple interest.]

Rule: (i) A = P(1 + r/100)n and (ii) C.I. = P[(1 + r/100)n -1].
where n is the number of times the interest is compounded.

At what rate per cent per annum will be a sum of money treble itself in 25 years?
Rs. 100 becomes Rs. 300 in 25 years i.e, interest earned on Rs. 100 is Rs. 200 over a period of 25 years.
rate % = 200/25 = 8%.

If a sum of money doubles itself in 7 years, it becomes 5 times in how many years?
Rs. 100 amounts to Rs. 200 in 7 years i.e. Rs. 100 is earned after every 7 years on Rs. 100.
Rs. 100 becoming Rs. 500 means, interest earned is Rs. 400 which shall take time 7 x 4 = 28 years.

A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to eight times itself?
If x be the sum it becomes 2x in four years at a certain rate say r%
2x = x(1 + r/100)4
or 2 = (1 + r/100)4
cube both sides
23 = (1 + r/100)3x4
8 = (1 + r/100)3x4
or 8x = x(1 + r/100)3x4

Hence the required number of years = 12 years.

Time and Work, Work and Wages, Pipes and Cisterrn (i) If a man can do a piece of work in 4 days, it is clear that he can do 1/4 of the work in 1 day. Again, if a man can do 1/4 of the work in 1 day, he will take 4 days to finish the work.
(ii) If A can do a work in 3 days and B in 8 days, the ratio of the work done by A and B in the same time is 8 : 3.
A's 1 day's work is 1/3 amd B's 1 day work is 1/8
(1 + r/100)3x4
Ratio of the work is 1/3 : 1/8 or 8 : 3.
(iii) If A is thrice as good a workman as B, then ratio of the work done by A and B is 3 : 1.

A can do piece of work in 6 days, and B can do it in 12 days. What time will they require to do it working together?
Part of the work done by A in one day = 1/6
Part of the work done by B in one day = 1/12
Part of the work done by A and B in one day = 1/6 + 1/12 = 3/12 = 1/4
Time required by A and B working together to finish the work = 4 days.

A and B undertake to do a piece of work for Rs. 200. A can do it in 6 days and B in 8 days. With the help of C they finish it in 3 days. How much is paid to C?
Each of three is working for three days.
A's work for 3 days = 3 x 1/6 = 1/2
∴ A gets 1/2 x Rs. 200 = Rs. 100
B's work for 3 days = 3 x 1/8 = 3/8
∴ B gets 3/8 x Rs. 200 = Rs. 75
∴ C gets the rest i.e. Rs. 200 - (100 + 75) = Rs. 25

Three pipes A, B and C can fill a cistern in 12, 15 and 20 minutes respectively. Pipes A and C work for one minute and Pipes B and C work in next minute. In how many minutes the cistern will be full if theis proceedure is continued?
Work done by pipes A and C in one minute = 1/12 + 1/20 = (5 +3)/60 = 8/60 of the cistern.
Work done by pipes B and C in the next minute = 1/15 + 1/20 = (4 + 3)/60 = 7/60 of the cistern
Part of the cistern filled in 2 minute = 8/60 + 7/60 = 15/60 = 1/4 ∴ The cistern shall be full in 8 minutes.

Stocks and Shares:

1. Stock: It is the capital of a company, the unit of which is usually Rs. 100 unless stated otherwise.
2. Cash: It is the money spent or received while buying or selling the stock.
Note: Stock is always purchased, sold or held but cash is invested or received.
3. Market Price: The rate at which the stock is sold or brought is called the Market Price or Market Value (M.V.)
Note:"4 per cent at 90": it means that Rs. 100 stock can be purchased for Rs. 90 and Rs. 4 is the annual income on Rs. 100 or on Rs. 90 cash.
4. Brokerage: It is the commission paid to a stock dealer on the sale and purchase of stocks.
Note: While buying stock, add brokerage to the market value and while selling stock, subtract brokerage from the market value.
5. Stock at Par: When the selling price or cash price of Rs. 100 stock is Rs. 100 cash, the stock is said to be at Par.
6. Stock above Par at a premium: When the selling price of Rs. 100 stock is more than Rs. 100 cash, it is said to be at a premium or above par.
7. Stock below Par: When the selling price of Rs. 100 stock is less than Rs. 100 cash, it is said to be at a discount or below par.

Summary work:

1. Stock = (Investment x 100)/M.V. Also
2. Stock = (Income x 100)/Rate
3. Investment = (Stock x M.V.)/100 Also
4. Investment = (Income x M.V.)/Rate
5. Income = (Stock x Rate)/100 Also
6. Income = (Investment x Rate)/M.V.

How much of 5% stock at 5 above par can be purchased by investing Rs. 7980?
Investment = Rs. 7980;
M.V. = 100 + Rs. 5 = Rs. 105
Stock = (Rs. 7980 x 100)/105 = 7600.

Income Tax, Rates and Insurance:

A man's net income after paying income tax of 8 paise in the rupee is Rs. 146.50. Find his total income?
Suppose his gross income= Re. 1 or 100 paise
His net income = 100 P - 8 P = 92 P
If net income is Rs. 92/100, the gross income = Re. 1
If net income is Rs. 1460.50, the gross income = Rs. 1 x 100/92 x 1460.50
= Rs. 1587.50

The income of a man increases by Rs. 3000 but the rate of income tax decreases from 10% to 7%. He pays the same amount of income tax as before. Find his income?
If the income of the man be Rs. x, then the tax in the first case @ 10% is 10%x.
In the second case, his income becomes (x + 3000) rupees and he now pays the tax at the rate of 7%.
∴ tax in this case = 7%(x + 3000).
Thus 10%x = 7%(x + 3000)
-> 10x = 7x + 21000
-> 3x = 21000
x = 7000
Thus his income is Rs. 7000.

For what sum should goods worth Rs. 1150 be insured at 8% so that in case of loss, the owner may recover the premium as well as the goods?
If the owner insures his goods for Rs. 100, then in case of loss he would recover Rs. 8 premium and Rs. (100 - 8) or Rs. 92 as the value of goods.
Now Rs. 92 must be insured for Rs. 100
Rs. 1150 must be insured for Rs. 100/92 x 1150 = Rs. 1250
Note: Hence to find the covered insurance, we must use the following rule:
[Multiply the value of the goods by 100/(100 - rate)]

Some more tips on Mensuration:

1. A land measure unit is are. One are measures an area equal to 100m2. One hectate means 100 ares and is therefore equal to 10000 m2. Whenever the area of a plane or that of a solid is given in Hectates and Ares this must be converted into m2(Very important)
2. Planes are two dimensionals and occupy surface area.
3. Perimeter:The lengths of all the sides of a polygon is called a perimeter of the polygon. The perimeter of a circle is called the circumference.
4. Rectangle:If the length and breadth of the rectangle be 'l' and 'b' and 'd' be the length of the diagonal, the P (perimeter) = 2(l + b)
A (surface area or area) = l x b
D (diagonal) = √(l2 + b2)
5. Square: If x be the side of a square, then
P = 4x or x = 1/4 P
A = x2 or x = √A
D (diagonal) = √2x or D2 = 2x2 or D2 = 2A.
6. Rhombus: If x be the side of the rhombus, all its sides are equal and diagonal bisect at right angle.
P = 4x or x = 1/4 P
A = 1/2 x Product of the diagonals = 1/2 x AC x BD.
7. Triangle: This is a polygon with minimum number of sides i.e. three
(i) Area of any triangle = 1/2 x base x height
(ii) Area of a triangle when its sides are given as a, b and c = √(s(s-a) (s-b) (s-c)) where s = 1/2 (a + b +c).
(iii) Area of an equilateral triangle = √3/4 x (side)2.
(iv) Area of an isosceles triangle whose base is 'b' and equal sides are 'a' each = 1/4 x b x √(4a2 - b2)
Note: If a regular hexagon (6 equal side polygon) is inscribed in a circle of radius x the area of the hexagon is equal to six times the area of the equilateral triangle of side x.
8. Area of four walls of a room = 2 x height (length + breadth) = 2 h (l + b)
9. Circle:It is a polygon with maximum number of sides. given the perimeter, the circle is the plane with maximum area.
10. What is Π? This is the ratio of the circumference of the circle with its diameter and remains constant. Its approximate value is 22/7 or 3.1416. Thus Π = circumference/diameter
11. Circumference of a circle = 2Πr
Area = Πr2 or r = √(Area/Π)
12. The radius R of a circle whose area is equal to the sum of the areas of series of circles with radii r1, r2, r3 etc. is equal to R = √(r12 + r22 + r32 + ....)
For example, the radius of a circle whose area is equal to the sum of the areas of two circles of radii 4 cms. and 8 cms. is = √(62 + 82) cm = √100 = 10 cm.
13. Volume of Solids: A solid occupies space and the space it occupies is known as its volume. A sphere, a cube are the examples of solids. It is measured in cubic units.
14. Cuboid: OA, OB, OC are called three coterminus edges of the cuboid and may be considered as length, breadth and height i.e. 'l', 'b', 'h'.
(i) Surface area = 2 x (lb + bh + lh)
(ii) Volume = l x b x h.
(iii) Diagonal = √(l2 + b2 + h2)
Note: If we want to find the quantity of paint required for painting the wooden or iron box or iron sheet required for making a box, formula (i) is used. In order to to find, how much space a box can occupy or how much a box can contain, formula (ii) can used. Formual (iii) is used to find the length of the longest pole which can be placed in a big hall.
15. Cube: A cube has all its edges equal say 'e' be its length. All its faces are square faces. It has 12 edges.
(i) Perimeter = 12 x edge
(ii) Area = 6 x (edge)2
(iii) Volume = (edge)3 or edge = (volume)1/3
(iv) Diagonal = √3 x edge
(v) Edge 'E' of a cube whose volume, is equal to the sum of volume of cubes of edges e1, e2, e3, .... is given by E = (e13 + e23 + e33 + ...)1/3
16. Sphere: A rubber ball is an example of a sphere. If the radius of the sphere is r, then
(i) Surface area = 4Πr2
(ii) Volume = 4/3 Πr3
(iii) Area of the solid hemisphere = 3Πr2
(iv) Radius 'R' of the sphere whose volume is equal to the sum of the volumes of spheres with radii r1, r2, r3, .... is given by R = (r13 + r23 + r33 + ...)1/3
17. Cylinder: When a rectangle is revolved about one of its side, a solid so formed is a cylinder.
(i) Curved surface of the cylinder = base circumference x height = 2Πr x h
(ii) Total surface of a closed cylinder = 2Πr (r + h)
(iii) Volume = base area x height = Πr2 x h
(iv) Height of the cylinder = Volume of the cylinder/base area of cylinder
18. Cone: When a right angle triangle is revolved about one of its side, a solid so formed is a cone. If r be the base radius, h the height and l, the slant height, then l = √(r2 + h2)
(i) Curved surface = Πrl = Πr x √(r2 + h2)
(ii) Total surface = Πr (l + r)
(iii) Volume = 1/3 Πr2h
Note: Volume of the cone = 1/3 Volume of the cylinder
∴ Volume of cone/Volume of cylinder = 1/3
For example, if a conical vessel can hold 20 litres of milk, then a cylindrical vessel can hold 3 x 20 i.e. 60 litres of milk.
19. If it is to find the canvas required for making the conical tent with given dimensions, formula (i) is used. For finding how much iron sheet is required to construct a conical vessel, formula (ii) is used. To find the capacity of the cone, formula (iii) is used.
20. Very Important Note: If the ratio of the two similar planes or solids are as x : y, then their areas are as x2 : y2 and volumes are x3 : y3. This is explained by the following examples.

(i) If the sides of a rectangel are doubled, what percent of its area is increased?
Now it is very clear that the area of the corresponding rectangle is becomes four times the original and thus the increase is 3 times or 300%.
(ii) If the cost of levelling the field in the form of a square is Rs. 25, what is the cost of levelling another square field whose side is three times the side of the first.
Since the area of the other field is 9 times the original, so the cost is (25 x 9) i.e. Rs. 225
(iii) The perimeter of a regular hexagone is 48 cm, what is the area of the hexagon.
The perimeter being 48 cm, so the side of the hexagon is (48 ÷ 6) i.e. 8 cm. Hence the area of the hexagon is 6 x √3/4 x (8)2 cm2 or 96√3 cm2.
(iv) The perimeter of a cube is 36 cm. What is its volume?
Now side of the cube = 1/12 x 36 cm = 3 cm
∴ Volume = 33 cm3 or 27 cm3.
(v) The weight of an iron ball of radius 2 cm is 25 grams. What is the weight of another similar sphere whose radius is 6 cm.
Since the radius of the other sphere is 3 times the first the volume is 33 = 27 times the first. Now weights are proportional to their volume, so the weight of the other sphere is 27 x 25 grams or 675 grams.
(vi) If the ratio of areas of two cylindrical vessels are as 25 : 49, what will be the ratio of their volumes?
Since area ratio is 25 : 49
∴ dimension ratio = √25 : √49 = 5 : 7
or volume ratio = 53 : 73 = 125 : 343

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