Numbers - Aptitude Questions and Answers

Numbers - Aptitude Questions and Answers

Monday, April 7, 2014 /

Questions One ::

Option A
Option B
Option C
Option D

If you click on the "Hide" button, I will disappear.

Wrong Answer

Questions Two ::

Option A
Option B
Option C
Option D

If you click on the "Hide" button, I will disappear.

Wrong Answer

Questions Three::

Option A
Option B
Option C
Option D

If you click on the "Hide" button, I will disappear.

Wrong Answer

Questions Four::

Option A
Option B
Option C
Option D

If you click on the "Hide" button, I will disappear.

Wrong Answer

Questions Five::

Option A
Option B
Option C
Option D

If you click on the "Hide" button, I will disappear.

Wrong Answer

Questions Six ::

Option A
Option B
Option C
Option D

If you click on the "Hide" button, I will disappear.

Wrong Answer

Questions Seven ::

Option A
Option B
Option C
Option D

If you click on the "Hide" button, I will disappear.

Wrong Answer

Questions Eight ::

Option A
Option B
Option C
Option D

If you click on the "Hide" button, I will disappear.

Wrong Answer

Questions One ::

Option A
Option B
Option C
Option D

If you click on the "Hide" button, I will disappear.

Wrong Answer

Questions Ten ::

Option A
Option B
Option C
Option D

If you click on the "Hide" button, I will disappear.

Wrong Answer

Compound Interest - Aptitude Concepts (Important Facts And Formulae)

Compound Interest - Aptitude Concepts (Important Facts And Formulae)

Monday, March 17, 2014 /

IMPORTANT FACTS AND FORMULAE

Compound Interest: Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, let say it yearly or half-yearly or may be quarterly to settle the previous account.

In such cases, amount after the first unit of time becomes the principal for second unit, amount after second unit becomes the principal for the third unit and so on.

After a specified period of time, the difference between the amount and the money borrowed is called the Compound Interest (abbreviated as C.I.) for that particular period.

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

I. When interest is compound Annually:
   

Amount = P(1+R/100)^n

II. When interest is compounded Half-yearly:

     Amount = P[1+(R/2)/100]^2n

III. When interest is compounded Quarterly:

      Amount = P[ 1+(R/4)/100]^4n

IV. When interest is compounded AnnuaI1y but time is in fraction, say 3(2/5) years.

     Amount = P(1+R/100)^3 x (1+(2R/5)/100)

V. When Rates are different for different years, say Rl%, R2%, R3% for 1st, 2nd and 3rd year respectively.
     Then,

Amount = P(1+R1/100)(1+R2/100)(1+R3/100)

VI. Present worth of Rs.x due n years hence is given by :

Present Worth = x/(1+(R/100))^n

Volume And Surface Area - Aptitude Concepts (Important Facts And Formulae)

Volume And Surface Area - Aptitude Concepts (Important Facts And Formulae)

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IMPORTANT FORMULAE

I. CUBOID

Let length = 1, breadth = b and height = h units. Then,

1. Volume = (1 x b x h) cubic units.

2. Surface area= 2(lb + bh + lh) sq.units.

3. Diagonal.=l^2 +b^2 +h^2 units

II. CUBE

Let each edge of a cube be of length a. Then,

1. Volume = a3 cubic units.

2. Surface area = 6a2 sq. units.

3. Diagonal = 3 a units.

III. CYLINDER

Let radius of base = r and Height (or length) = h. Then,

1. Volume = (pi r^2h) cubic units.

2. Curved surface area = (2pi rh). units.

3. Total surface area =2pi*r (h+r) sq. units

IV. CONE

Let radius of base = r and Height = h. Then,

1. Slant height, l = h^2+r^2

2. Volume = (1/3) pi*r2h cubic units.

3. Curved surface area = (pi*rl) sq. units.

4. Total surface area = (pi*rl + pi*r2 ) sq. units.

V. SPHERE

Let the radius of the sphere be r. Then,

1. Volume = (4/3)pi*r3 cubic units.

2. Surface area = (4pi*r2) sq. units.


VI. HEMISPHERE

Let the radius of a hemisphere be r. Then,

1. Volume = (2/3)pi*r3 cubic units.

2. Curved surface area = (2pi*r2) sq. units.

3. Total surface area = (3pi*r2) units.

    Remember: 1 litre = 1000 cm3.

Problems On Train - Aptitude Concepts (Important Facts And Formulae)

Problems On Train - Aptitude Concepts (Important Facts And Formulae)

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IMPORTANT FACTS AND FORMULAE

1. a km/hr= (a* 5/18) m/s.

2. a m / s = (a*18/5) km/hr.

3. Time taken by a train of length 1 metres to pass a pole, a standing man, a signal post is equal to the time taken by the train to cover 1 metres.

4. Time taken by a train of length 1 metres to pass a stationary(static) object of length b metres is equal to the time taken by the train to cover (1 + b) metres.

5. Let two trains or say two bodies are moving in the same direction at u m/s and v m/s,
    where u > v,
    then their relatives speed =

(u - v) m / s.

6. Let two trains or say two bodies are moving in opposite directions at u m/s and v m/s,
   then their relative speed is =

(u + v) m/s.

7. If the two trains of length a metres and b metres are moving in the opposite directions at u m / s and v m/s, then in this case time taken by the trains to cross each other =

(a + b)/(u+v) sec.

8. If the two trains of length a metres and b metres are moving in the same direction at u m / s and v m / s, then in this case time taken by the faster train to cross the slower train =

(a+b)/(u-v) sec.

9. If the two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively,then

(A's speed) : (B’s speed) = (b1/2: a1/2).

Mainly Used Formulae

Permutation And Combination - Aptitude Concepts (Important Facts And Formulae)

Permutation And Combination - Aptitude Concepts (Important Facts And Formulae)

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IMPORTANT FACTS AND FORMULAE

Factorial Notation: Let n be a positive integer. Then, factorial n is denoted by n! and is defined as:

n! = n(n-1)(n-2)........3.2.1.

Examples: (i) 5! = (5x 4 x 3 x 2 x 1) = 120; (ii) 4! = (4x3x2x1) = 24 etc.

We define, 0! = 1.


Permutations: The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Ex. 1. All permutations (or arrangements) made with the letters a, b, c by taking two at a time are:

(ab, ba, ac, bc, cb).

Ex. 2. All permutations made with the letters a,b,c, taking all at a time are:

(abc, acb, bca, cab, cba).

Number of Permutations: Number of all permutations of n things, taken r at a time, given by:

nPr = n(n-1)(n-2).....(n-r+1) = n!/(n-r)!

Examples: (i) 6p2 = (6x5) = 30. (ii) 7p3 = (7x6x5) = 210.

Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects, is called a combination.

Ex. 1. Let we want to select two boys out of three boys A, B, C. Then, the possible selections are AB, BC and CA.
Note :: AB and BA represent the same selection.

Ex. 2. All combinations formed by a, b, c, taking two at a time are ab, bc, ca.

Ex. 3. The only combination that can be formed of three letters say a, b, c taken all at a time is ::  abc.

Ex. 4. Various groups of two out of four persons say A, B, C, D are: AB, AC, AD, BC, BD, CD.

Ex. 5. Note :: ab and ba are two different permutations but they represent same combination.

Number of Combinations: The number of all the combination of n things, taken r at a time is:

nCr = n! / (r!)(n-r)! = n(n-1)(n-2).....to r factors / r!

Note that: ncr = 1 and nc0 = 1.

An Important Result: ncr = nc(n-r).

Example: (i) 11c4 = (11x10x9x8)/(4x3x2x1) = 330.
              (ii) 16c13 = 16c(16-13) = 16x15x14/3! = 16x15x14/3x2x1 = 560.

Logarithms - Aptitude Concepts (Important Facts And Formulae)

Logarithms - Aptitude Concepts (Important Facts And Formulae)

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IMPORTANT FACTS AND FORMULAE

I. Logarithm: If a is a positive real number, other than 1 and am = X,
   then we write: m = loga x and we say that the value of log x to the base a is m.

Example:
(i) 103 = 1000 => log10 1000 = 3
(ii) 2-3 = 1/8 => log2 1/8 = - 3
(iii) 34 = 81 => log3 81=4
(iiii) (.1)2 = .01 => log(.l) .01 = 2.


II. Properties of Logarithms:

1. loga(xy) = loga x + loga y

2. loga (x/y) = loga x - loga y

3.logx x=1

4. loga 1 = 0

5.loga(xp)=p(logax) 1

6. logax =1/logx a

7. logax = logb x/logb a=log x/log a.

Remember: When base is not mentioned, it is taken as 10.


III. Common Logarithms:

Logarithms to the base 10 are known as common logarithms.


IV. The logarithm of a number contains two parts, namely characteristic and mantissa.

Characteristic: The integral part of the logarithm of a number is called its characteristic.

Case I: When the number is greater than 1.
             The characteristic is one less than the number of digits in the left of the decimal point in the given number.

Case II: When the number is less than 1.
               The characteristic is one more than the number of zeros between the decimal point and the first significant digit of the number and it is negative.

Instead of - 1, - 2, etc. we write, 1 (one bar), 2 (two bar), etc.

Area - Aptitude Concepts (Important Facts And Formulae)

Area - Aptitude Concepts (Important Facts And Formulae)

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FUNDEMENTAL CONCEPTS

I.RESULTS ON TRIANGLES:

1.Sum of the all angles of a triangle is 180 degrees.

2.Sum of any two of sides in a triangle is greater than the third side.

3.Pythagoras theorem:

  In a right angle triangle,

(Hypotenuse)^2 = (base)^2 + (Height)^2

c^2 = b^2 + a^2

4.The line joining the midpoint of a side of a triangle to the opposite vertex is called the MEDIAN

5.The point where three medians of a triangle meet is called CENTROID of trianlge . Centroid divides each of the medians in the ratio of 2:1.

6.In an isosceles triangle, the altitude from the vertex bi-sects the base.

7.The median of a triangle divides it into two triangles of the same area.

8.Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle.


II.RESULTS ON QUADRILATERALS:

1. The diagonals of a parallelogram bisects each other .

2. Each diagonal of a parallelogram divides it into two triangles of the same area

3. The diagonals of a rectangle are equal and bisect each other.

4. The diagonals of a square are equal and bisect each other at right angles.

5. The diagonals of a rhombus are unequal and bisect each other at right angles.

6. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.

7. Of all the parallelograms of a given sides , the parallelogram which is a rectangle has the greatest area.

IMPORTANT FORMULAE

I. 1.Area of a rectangle=(length*breadth)
    Therefore length = (area/breadth) and breadth=(area/length)

  2. Perimeter of a rectangle = 2*(length+breadth)

II.  Area of a square = (side)^2 =1/2(diagonal)^2

III. Area of four walls of a room = 2*(length + breadth)*(height)

IV. 1.Area of the triangle=1/2(base*height)
     
       2. Area of a triangle = (s*(s-a)(s-b)(s-c))^(1/2),
          where a,b,c are the sides of a triangle and s=½(a+b+c)

       3.Area of the equilateral triangle =((3^1/2)/4)*(side)^2

       4.Radius of incircle of an equilateral triangle of side a=a/2(3^1/2)

       5.Radius of circumcircle of an equilateral triangle of side a=a/(3^1/2)

       6.Radius of incircle of a triangle of area del and semiperimeter S=del/S

V. 1.Area of the parellogram = (base *height)

     2.Area of the rhombus=1/2(product of the diagonals)

     3.Area of the trapezium=1/2(size of parallel sides)*distance between them


VI. 1.Area of a circle = pi*r^2,where r is the radius

      2. Circumference of a circle = 2ΠR.

      3. Length of an arc = 2ΠRθ/(360) where θ is the central angle

      4. Area of a sector = (1/2) (arc x R) = pi*R^2*θ/360.


VII. 1. Area of a semi-circle = (pi)*R^2.

        2. Circumference of a semi-circle = (pi)*R.

Time And Work - Aptitude Concepts (Important Facts And Formule)

Time And Work - Aptitude Concepts (Important Facts And Formule)

Thursday, March 13, 2014 /

IMPORTANT FACTS AND FORMULAE

1. If a person say A can do a piece of work in n days, then

A's 1 day's work = (1/n).

2. On the other hand If A’s 1 day's work = (1/n),then

A can finish the work in n days.

3. A is twice as good a workman as B, then:

    The Ratio of work done by A and B = 2 : 1.
    The Ratio of times taken by A and B to finish a work = 1 : 2

SOLVED EXAMPLES

Ex. 1. Worker A takes 8 hours to do a particular job. Worker B takes 10 hours to do the same Job. Then how long should it take both A and B, working together but independently, to do the same job?

Sol.  A’s 1 hour's work = 1/8

        B's 1 hour's work = 1/10

       (A + B)'s 1 hour's work = (1/8) +(1/10)=9/40

       Both A and B will finish the work in 40/9 days.

Ex. 2. A and B together can complete a piece of work in 4 days. If A alone can complete the same work in 12 days, in how many days can B alone complete that work? 

Sol. (A + B)'s 1 day's work = (1/4). A's 1 day's work = (1/12).
 
       B's 1 day's work =((1/4)-(1/12))=(1/6)

       Hence, B alone can complete the work in 6 days.

Ex 3 .45 men can complete a work in 16 days. Six days after they started working, 30 more men joined them. How many days will they now take to complete the remaining work?

Sol.  (45 x 16) men can complete the work in 1 day.
 
         1 man's 1 day's work = 1/720
 
        45 men's 6 days' work =(1/16*6)=3/8

        Remaining work =(1-3/8)=5/8
 
        75 men's 1 day's work = 75/720=5/48

Now, 5/48 work is done by them in 1 day.

5/8 work is done by them in (48/5 x 5/8) = 6 days.

Ex. 4. A and B working separately can do a piece of work in 9 and 12 days respectively, If they work for a day alternately, A beginning, in how many days, the work will be completed?

Sol. (A + B)'s 2 days' work =(1/9+1/12)=7/36

       Work done in 5 pairs of days =(5*7/36)=35/36

       Remaining work =(1-35/36)=1/36

       On 11th day, it is A’s turn. 1/9 work is done by him in 1 day.

       1/36 work is done by him in(9*1/36)=1/4 day

       Total time taken = (10 + 1/4) days = 10 1/4days.

Ex. 5. A can do a piece of work in 80 days. He works at it for 10 days B alone finishes the remaining work in 42 days. In how much time will A and B working together, finish the work?

Sol. Work done by A in 10 days =(1/80*10)=1/8

       Remaining work = (1- 1/8) =7/ 8

       Now,7/ 8 work is done by B in 42 days.

      Whole work will be done by B in (42 x 8/7) = 48 days.

       A’s 1 day's work = 1/80 and B's 1 day's work = 1/48

      (A+B)'s 1 day's work = (1/80+1/48)=8/240=1/30

      Hence, both will finish the work in 30 days.

Partnership - Aptitude Concepts (Important Facts And Formulae)

Partnership - Aptitude Concepts (Important Facts And Formulae)

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IMPORTANT FACTS AND FORMULAE 

1. Partnership :: When two or more than two persons start or run a business jointly, at that time they are called partners and the deal between/among them is known as partnership.


2. Ratio of Division of Gains:

(i) When the investments of all the partners are for the same time, the loss or gain is distributed a among the partners in the ratio of their investments.

Let A and B invest Rs. x and Rs. y respectively for a year in a particular business, then at the end of the year ::

(A’s share of profit) : (B's share of profit) = x : y.

(ii) When the investments are for different time periods, then the equivalent capitals are calculated for a unit of time by taking (capital x number of units of time).
Now, gain or loss is divided into the ratio of these capitals.

Let A invests Rs. x for m months and B invests Rs. y for n months, then

(A’s share of profit) : (B's share of profit) = xm : yn.

3. Working and Sleeping Partners :: A partner who mainly manages all the business is known as a working partner and on the other hand the one who simply invests the money is known as sleeping partner.

SOLVED EXAMPLES

Ex. 1. Shubham, Himanshu and Arpit started a business by investing Rs. 1,20,000, Rs. 1,35,000 and ,Rs.1,50,000 respectively. Find out the share of each, out of an annual profit of Rs.56,700.

Sol. Ratio of shares of A, Band C = Ratio of their investments

= 120000 : 135000 : 150000 = 8 : 9 : 10.

Shubham’s share = Rs. (56700 x (8/27))= Rs. 16800.
Himanshu's share = Rs. ( 56700 x (9/27)) = Rs. 18900.
Arpit's share = Rs. ( 56700 x (10/27))=Rs. 21000.

Ex. 2. Shubham, Himanshu and Arpit start a business each investing Rs. 20,000. After 5 months Shubham withdrew Rs.6000 Himanshu withdrew Rs. 4000 and Arpit invests Rs. 6000 more. At the end of year, a total profit of Rs. 69,900 was recorded. Find the share of each.

Sol. Ratio of the capitals of Shubham, Himanshu and Arpit

= 20000 x 5 + 15000 x 7 : 20000 x 5 + 16000 x 7 : 20000 x 5 + 26000 x 7

= 205000:212000 : 282000 = 205 : 212 : 282.

Shubham’s share = Rs. 69900 x (205/699) = Rs. 20500 I

Himanshu's share = Rs. 69900 x (212/699) = Rs. 21200;

Arpit's share = Rs. 69900 x (282/699) = Rs. 28200.

Ex. 3. Shubham, Himanshu and Arpit enter into partnership. Shubham invests 3 times as much as Himanshu and Himanshu invests two-third of what Arpit invests. At the end of the year, the profit of Rs. 6600 was recorded. What is the share of B ?

Sol. Let Arpit's capital = Rs. x. Then, Himanshu's capital = Rs. (2/3)x
Shubham’s capital = Rs. (3 x (2/3).x) = Rs. 2x.

Ratio of their capitals = 2x : (2/3)x :x = 6 : 2 : 3.
Hence, Himanshu's share = Rs. ( 6600 x (2/11))= Rs. 1200.

Ratio And Proportion - Aptitude Concepts (Important Facts And Formulae)

Ratio And Proportion - Aptitude Concepts (Important Facts And Formulae)

Wednesday, March 12, 2014 /

IMPORTANT FACTS AND FORMULAE

I. RATIO :: The ratio of the two quantities say a and b in the same units, is the fraction a/b and we write it or we denote it as- 

a:b.

In the ratio a:b, 

                        a - antecedent. (first term)

                        b - consequent. (second term)

Ex. The ratio 5: 7 represents 5/7 with antecedent = 5, consequent = 7.

Rule :: The multiplication or division of each term(both antecedent and consequent) of a ratio by the same non-zero number does not affect the ratio.

Ex. 4: 6 = 8: 12 = 12: 24 etc. Also, 4: 6 = 2: 3.

2. PROPORTION :: It is define as the equality of two ratios is called proportion.

If a: b = c: d, we can write it as, a: b:: c : d and we say that a, b, c, d are in proportion. Here a and d are called extremes terms, while b and c are called mean terms.

a and d = extremes terms.

b and c = mean terms.

Product of means = Product of extremes. Thus, 

a: b:: c : d <=> (b x c) = (a x d).

3. (i) Fourth Proportional :: If a : b = c: d, then here d is called as the fourth proportional to a, b, c.

    (ii) Third Proportional :: If a: b = b: c, then here c is called as the third proportional to a and b.

    (iii) Mean Proportional :: Mean proportional between a and b is the square root of ab

4. (i) COMPARISON OF RATIOS ::

 (a: b) > (c: d) <=> (a/b)>(c /d).

    (ii) COMPOUNDED RATIO :: The compounded ratio of the ratios -

(a: b), (c: d), (e : f) is (ace: bdf)

5. (i) Duplicate ratio of (a : b) is (a2 : b2).

    (ii) Sub-duplicate ratio of (a : b) is (√a : √b).

    (iii)Triplicate ratio of (a : b) is (a3 : b3).

    (iv) Sub-triplicate ratio of (a : b) is (a ⅓ : b ⅓ ).

    (v) If (a/b)=(c/d), then ((a+b)/(a-b))=((c+d)/(c-d)) (Componendo and dividendo)

6. VARIATION ::

(i) x is directly proportional to y, if x = ky for some constant L and we write it as, 

x ∞ y.

(ii) x is inversely proportional to y, if xy = k for some constant L and we write, 

x∞(1/y)

Profit And Loss - Aptitude Concepts (Important Facts And Formulae)

Profit And Loss - Aptitude Concepts (Important Facts And Formulae)

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IMPORTANT FACTS

Cost Price :: This is the price at which a particular item is purchased, (Abbreviated as C.P.)

Selling Price :: This is the price at which a particular item is sold.

Profit Or Gain :: If selling price is greater than the cost price at that time the selling price is said to have profit or gain.

Loss :: If Selling Price is less than the Cost Price at that time the seller is said to incurred a loss.

FORMULAE

1. GAIN=(SP)-(CP).

2. LOSS=(CP)-(SP).

3. LOSS OR GAIN IS ALWAYS RECKONED ON CP

4. GAIN %={GAIN*100}/CP.

5. LOSS%={LOSS*100}/CP.

6. SP={(100+GAIN%) /100}*CP.

7. SP={(100-LOSS%)/100}*CP.

8. {100/(100+GAIN%)} *SP

9. CP=100/(100-LOSS%)}*SP

10. IF THE ARTICLE IS SOLD AT A GAIN OF SAY 30%, THEN SP =130% OF CP

11. IF A PARTICULAR ARTICLE IS SOLD AT A LOSS OF SAY 30%. THEN SP=70% OF CP.

12. WHEN A PERSON SELLS TWO ITEMS,ONE AT A GAIN OF X% AND THE OTHER AT A LOSS OF X%.THEN IN THAT CASE THE SELLER ALWAYS INCURES A LOSS GIVEN:

{LOSS%=(COMMON LOSS AND GAIN ) 2}/10.=(X/10) 2

13. IF A TRADER PROFESSES TO SELL HIS GOODS AT CP BUT USES FALSE WEIGHTS,THEN

GAIN=[ERROR/(TRUE VALUE)-(ERROR)*100]%

Percentage - Aptitude Concepts (Important Facts And Formulae)

Percentage - Aptitude Concepts (Important Facts And Formulae)

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IMPORTANT FACTS AND FORMULAE

1. Concept of Percentage : By a certain percent ,we mean that many hundredths. Thus the x percent means x hundredths and can be written as x%.

To express x% as a fraction : We have , x% = x/100.

Thus, 

        20% =20/100 =1/5; 

        48% =48/100 =12/25, etc.

To express a/b as a percent : We have, a/b =((a/b)*100)%.

Thus, 

        ¼ =[(1/4)*100] = 25%; 

        0.6 =6/10 =3/5 =[(3/5)*100]% =60%.

2. If the price of a particular commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is 

[R/(100+R))*100]%.

If the price of a particular commodity decreases by R%, then the increase in consumption so as to decrease the expenditure is 

[(R/(100-R)*100]%.

3. Results on Population : Let the population of the city be P now, and suppose it increases at the rate of R% per annum.  Then :

I. Population after n years = P [1+(R/100)]^n.

II. Population n years ago = P /[1+(R/100)]^n.

4. Results on Depreciation : Let the present value of a particular machine be P. Let it depreciates at the rate R% per annum. Then,

I. Value of the machine after n years = P[1-(R/100)]n.

II. Value of the machine n years ago = P/[1-(R/100)]n.

5. If A is R% more than B, then B is less than A by

[(R/(100+R))*100]%.

If A is R% less than B , then B is more than A by

[(R/(100-R))*100]%.

Surds And Indices - Aptitude Concepts

Surds And Indices - Aptitude Concepts

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Iq Mantra provides you all the concepts about the surds and indices. The main formulae that are use to solve the problems on surds and indices are given below with the help of these concepts you will be able to solve the problems based on surds and indices. An example is also given in this you can easily understand it if you have learn all important facts and formulae for solving the surds and indices problems. 

IMPORTANT FACTS AND FORMULAE

Example : 

Square Roots And Cube Roots - Basic Concepts

Square Roots And Cube Roots - Basic Concepts

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IMPORTANT FACTS AND FORMULAE

Square Root: If a2 = b, we say that the square root of b is a and we write, √b = a.

                       Thus, √4 = 2, √9 = 3, √196 = 14.

Cube Root: The cube root of a given number a is the number whose cube is a. We denote the cube root of a by 3√a.

Thus, 3√8 = 3√2 x 2 x 2 = 2, 3√343 = 3√7 x 7 x 7 = 7 etc.

Note: 1.√ab = √a * √b 

           2. √(a/b) = √a / √b = (√a / √b) * (√b / √b) = √ab / b

SIMPLIFICATION - IMPORTANT CONCEPTS

SIMPLIFICATION - IMPORTANT CONCEPTS

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IMPORTANT CONCEPTS

I. ’BODMAS’ Rule: It is the rule that depicts the correct sequence in which the operations are to be executed, so as to find out the value of a given expression.

Here, ‘B’ stands for ’bracket’ ,

          ’O’for ‘of’ , 

          ‘D’ for’ division’ 

          ‘M’ for ‘multiplication’, 

          ‘A’ for ‘addition’ and 

          ‘S’ for ‘subtraction’.

Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order(), {} and [].

After removing the brackets, we must use the following operations strictly in the order:

(1)of (2)division (3) multiplication (4)addition (5)subtraction.

II. Modulus of a real number : Modulus of a real number x can be defined as

Thus, |5|=5 and |-5|=-(-5)=5.

III. Virnaculum (or bar): Whenever an expression or question contains Virnaculum, then before applying the ‘BODMAS’ rule, we first simplify the given expression under the Virnaculum.

H.C.F. AND L.C.M. OF NUMBERS - Basic Concepts

H.C.F. AND L.C.M. OF NUMBERS - Basic Concepts

Tuesday, March 11, 2014 /

IMPORTANT FACTS AND FORMULAE

I. Factors and Multiples :- If a number A(say) divides another number B(say) exactly, in that case we say that A is a factor of B and in this case, B is called a multiple of A.

II. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.): The H.C.F.(Highest Common Factor) of two or more than two numbers is the greatest number that divides each of them exactly.

There are mainly two methods for finding the H.C.F. of a given set of numbers and these are as follows:

1. Factorization Method : Express the each one of given numbers as the product of prime factors.The product of least powers of common prime factors gives the H.C.F.

2. Division Method: Suppose we have to find the H.C.F. of two given numbers. 

1. Divide the larger number by the smaller one. 
2. Now, divide the divisor by the remainder. 
3. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. 
4. The last divisor is the required H.C.F.

Finding the H.C.F. of more than two numbers : In the case where we have to find the H.C.F. of three numbers, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given numbers and Similarly, the H.C.F. of more than three numbers may be obtained.

III. Least Common Multiple (L.C.M.) : The least number which is exactly divisible by each one of the given numbers is called their L.C.M(Least Common Multiple).

There are mainly two methods for finding the L.C.M. of a given set of numbers and these are as follows:

1. Factorization Method of Finding L.C.M.: 

1. Resolve the each one of given numbers into a product of the prime factors. 
2. Then, the L.C.M. is the product of highest powers of all factors.

2. Common Division Method or Short-cut Method for Finding L.C.M.: 

1. Arrange given numbers in a row in any order. 
2. Divide by a number which divides exactly at least two of the given numbers & carry forward the numbers which are not divisible. 
3. Repeat above process till no two of the numbers are divisible by the same number except 1. 
4. The product of divisors and the undivided numbers is required L.C.M. of the given numbers.

IV. Product of two numbers =Product of their H.C.F. and L.C.M.

V. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.

VI. H.C.F. and L.C.M. of Fractions:

1. H C F= H.C.F. of Numerators / L.C.M. of Denominators
2. L C M = L.C.M of Numerators / H.C.F. of Denominators 

Numbers - Important Facts And Formulae

Numbers - Important Facts And Formulae

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IMPORTANT FACTS AND FORMULAE 

I. Numeral : In Hindu Arabic system, there are ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and these symbols called digits to represent any number.

A group of digits that denoting a number is called a numeral.

We can represent a number, say 689745132 as shown below :

We read it as :- 'Sixty-eight crores, ninety-seven lacs, forty-five thousand, one hundred and thirty-two'.

II. Local Value or Place Value of a Digit in a Numeral : 

In the above numeral : 

Place value of 2 is (2 x 1) = 2; Place value of 3 is (3 x 10) = 30; 

Place value of 1 is (1 x 100) = 100 and so on. 

Place value of 6 is 6 x 108 = 600000000 

III.Face Value : It is the value of the digit itself in a numeral at whatever place it may be. In the above numeral, 

face value of 2 is 2; 

face value of 3 is 3 and so on

IV.TYPES OF NUMBERS :

1.Natural Numbers : Counting numbers 1, 2, 3, 4, 5, 6, 7..... are called natural numbers. 

2.Whole Numbers : All the counting numbers together with the zero form the set of whole numbers. Thus, 

 (i) 0 is only the whole number which is not a natural number(zero is not a natural number). 

 (ii) Every natural number is a whole number(except zero). 

3.Integers : All the natural numbers, 0 and negatives of counting numbers i.e., {…, - 3 , - 2 , - 1 , 0, 1, 2, 3,…..} together form the set of the integers. 

(i) Positive Integers : {1, 2, 3, 4, 5 …..} is the set of all positive integers. 

(ii) Negative Integers : {- 1, - 2, - 3, -4…..} is the set of all negative integers. 

(iii) Non-Positive and Non-Negative Integers : 0 is neither positive nor negative. So, {0, 1, 2, 3, 4….} represents the set of non-negative integers, while {0, - 1 , - 2 , - 3 , -4 …..} represents the set of non-positive integers. 

4. Even Numbers : A number which is divisible by 2 is called an even number, e.g., 2, 4, 6, 8, 10, 12 etc. 

5. Odd Numbers : A number which is not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7, 9, 11, etc. 

6. Prime Numbers : A number that is greater than 1 is called a prime number and if it has exactly two factors, namely 1 and the number itself. 

Prime numbers up to 100 are : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. 

7.Composite Numbers : Numbers that are greater than 1 which are not prime, are known as the composite numbers, e.g., 4, 6, 8, 9, 10, 12. 

Note : (i) 1 is neither prime nor composite. 

(ii) 2 is the only even number which is prime. 

(iii) There are 25 prime numbers between 1 and 100. 

8. Co-primes : Two numbers say a and b are said to be co-primes, if their H.C.F. is 1. e.g., (2, 3), (4, 5), (7, 9), (8, 11), etc. are co-primes, 

V. TESTS OF DIVISIBILITY 

1. Divisibility By 2 :- A number is divisible by 2, if its unit's digit is any of :- 0, 2, 4, 6, 8. 

Ex. 849372 is divisible by 2, while 659315 is not. 

2. Divisibility By 3 :- A number is divisible by 3, if the sum of its digits is divisible by 3. 

Ex.5924823 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2+3) = 33, which 

is divisible by 3. 

But, 8643292 is not divisible by 3, since sum of its digits =(8 + 6 + 4 + 3 + 2 + 9+2) = 34, 

which is not divisible by 3. 

3. Divisibility By 4 :- A number is divisible by 4, if the number formed by the last two digits is divisible by 4. 

Ex. 892644 is divisible by 4, since the number formed by the last two digits is 44, which is divisible by 4. 

But, 749281 is not divisible by 4, since the number formed by the last two digits is 81, which is not divisible by 4. 

4. Divisibility By 5 :- A number is divisible by 5, if its unit's digit is either 0 or 5. Thus, 

20820 and 50345 are divisible by 5, while 30934 and 40946 are not. 

5. Divisibility By 6 :- A number is divisible by 6, if it is divisible by both 2 and 3. Ex. The number 35256 is clearly divisible by 2. 

Sum of its digits = (3 + 5 + 2 + 5 + 6) = 21, which is divisible by 3. Thus, 35256 is divisible by 2 as well as 3. Hence, 35256 is divisible by 6. 

6. Divisibility By 8 :- A number is divisible by 8, if the number formed by the last three digits of the given number is divisible by 8. 

Ex. 953360 is divisible by 8, since the number formed by last three digits is 360, which is divisible by 8. 

But, 529418 is not divisible by 8, since the number formed by last three digits is 418, which is not divisible by 8. 

7. Divisibility By 9 :- A number is divisible by 9, if the sum of its digits is divisible by 9. 

Ex. 60732 is divisible by 9, since sum of digits * (6 + 0 + 7 + 3 + 2) = 18, which is divisible by 9. 

But, 68956 is not divisible by 9, since sum of digits = (6 + 8 + 9 + 5 + 6) = 34, which is not divisible by 9. 

8. Divisibility By 10 :- A number is divisible by 10, if it ends with 0.

Ex. 96410, 10480 are divisible by 10, while 96375 is not. 

9. Divisibility By 11 :- A number is divisible by 11, if the difference of 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. 

Ex. The number 4832718 is divisible by 11, since : 

(sum of digits at odd places) - (sum of digits at even places) - (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11. 

10. Divisibility By 12 :- A number is divisible by 12, if it is divisible by both 4 and 3. 

Ex. Consider the number 34632. 

(i) The number formed by last two digits is 32, which is divisible by 4, 

(ii) Sum of digits = (3 + 4 + 6 + 3 + 2) = 18, which is divisible by 3. Thus, 34632 is 

divisible by 4 as well as 3. Hence, 34632 is divisible by 12. 

11. Divisibility By 14 :- A number is divisible by 14, if it is divisible by 2 as well as 7. 

12. Divisibility By 15 :- A number is divisible by 15, if it is divisible by both 3 and 5. 

13. Divisibility By 16 :- A number is divisible by 16, if the number formed by the last 4 digits is divisible by 16. 

Ex.7957536 is divisible by 16, since the number formed by the last four digits is 7536, 

which is divisible by 16. 

14. Divisibility By 24 :- A given number is divisible by 24, if it is divisible by both3 and 8. 

15. Divisibility By 40 :- A given number is divisible by 40, if it is divisible by both 5 and 8. 

16. Divisibility By 80 :- A given number is divisible by 80, if it is divisible by both 5 and 16. 

Note : If a number is divisible by p as well as q, where p and q are co-primes, then the given number is divisible by pq. 

If p arid q are not co-primes, then the given number need not be divisible by pq, even when it is divisible by both p and q. 

Ex. 36 is divisible by both 4 and 6, but not divisible by (4x6) = 24, since 4 and 6 are not co-primes.