Introduction
Interest is the amount paid on the money paid by the borrower to the lender for the use of money lent for the given fixed period of time. The amount borrowed is called the principle, time for which it is borrowed is called as the period and amount paid as a interest per year is called as the rate of interest.
Interest is of two types:
1. Simple interest
2. Compound interest
If the interest reckoned is uniform then it is known as the simple interest and if the interest is computed at the end of certain fixed period such that the amount at the end of the period become the principal for the next intervals, then the total interest calculated over all the intervals is called compound interest. The compound interest is calculated either yearly, half yearly or quarterly. Thus the compound interest at the end of certain specified period is equal to the difference between the amount at the end of the period and the original principal.
C.I. = Amount - Principal
Conversion Period
The fixed period over which the interest is calculated is known as the conversion period. After conversion period the interest is added to the principal and the amount is taken as the principal for the next period. If it is done after every one year then it is called compounded yearly, if it is done after every six month, then it is called half yearly and if it is done after every three month then it is called quarterly. If no conversion period in the question then we normally take it as yearly. 
Simple Interest
Simple interest is the interest calculated on the original principal for a given fixed time in which the accumulated interest from previous period is not added up to the previous principal. Simple interest is normally used for a single period.
Simple Interest \[=\frac{P\times R\times T}{100}\]
Where,
P = Principal (original amount borrowed or loaned)
T = Time periods
R = The rate of interest charged over the principal
Amount
When interest is added to the principal, the value is called amount.
It is denoted by A.
Mathematically,
Amount = Principle + Interest, or A = P + I
Conversion Relations
The various parameters we can be calculated from the relation for the simple interest.
Formula of Time: \[T=\frac{I\times 100}{P\times R}\]
Formula of Principal: \[P=\frac{I\times 100}{T\times R}\]
Formula of Rate: \[R=\frac{I\times 100}{P\times T}\] 
Compound Interest
If the interest is calculated at the end of certain fixed period and principal for the next is the amount after adding the interest of the previous period to the principal.
Formula for Calculating Compound Interest
\[A=P{{\left( 1+\frac{r}{100} \right)}^{n}}\]
\[C.I.=P\left[ {{\left( 1+\frac{r}{100} \right)}^{n}}-1 \right]\]
Thus, \[\mathbf{C}\mathbf{.I}\mathbf{.=A-P}\]
Where,
P = Principal Amount
r = Interest Rate
n = Number of times the amount is compounded
A = Amount after time t
C.I. = Compound interest.
Arwin lent Rs. 10,000 on first January 2000 to his friend Robert at 5% simple interest which amounts to Rs. 18000 after certain period of time. Find the date on which he will get the amount.
(a) \[\text{3}{{\text{1}}^{\text{st}}}\] December 2016
(b) \[\text{3}{{\text{1}}^{\text{st}}}\] December 2015
(c) \[\text{3}{{\text{1}}^{\text{st}}}\] December 2014
(d) \[\text{3}{{\text{1}}^{\text{st}}}\] December 2013
(e) None of these
Answer: (a)
Codi borrows Rs. 10000 for 120 days at 10% simple interest per year from his friend Lorentz for admission of his son in the school. Find the interest he has to pay to his friend after 120 days.
(a) Rs. 382.20
(b) Rs. 255.65
(c) Rs. 328.76
(d) Rs. 192.72
(e) None of thee
Answer: (c)
Johnson has Rs. 4000 he wants to invest it in two types of bond. The first bond pays him 7% and second pays an interest 9% per annum. He uses the first bond for 12 years and other for 6 years such that interest on first type is double to that more... 
Introduction
This chapter deals with the concept of finding the surface area and volume of the regular figures. By regular figures we mean to say that the figures whose parameters are known to us. This chapter also deals with the application of geometry to our real life problems like finding the area of the land or finding the volume of the buckets or many other such items.
Previously we have learnt to find the perimeter and area of the rectilinear figures, but now onwards we will learn to find the surface area of some of the two dimensional figures or the volume of some of the three dimensional figures. In this chapter we will basically discuss about the figures such as cube, cuboid, cylinder, cone, sphere, and frustums. 
Surface Area and Volume of Cuboids
A cuboid is a closed rectangular solid which has six rectangular faces.
Length of the cuboid = AB = CD = EF = GH. (Equal length)
Width of the cuboid = AD = BC = EH = FG. (Equal width)
Height of the cuboid = AE = BF = CG = DH.
Diagonal of the cuboid \[~=\text{A}{{\text{B}}^{\text{2}}}+\text{B}{{\text{C}}^{\text{2}}}+\text{B}{{\text{F}}^{\text{2}}}\]
Diagonal of a Cuboid \[=\sqrt{(Lengt{{h}^{2}}+Breadt{{h}^{2}}+Heigh{{t}^{2}})}\]
\[=\sqrt{{{l}^{2}}+{{b}^{2}}+{{h}^{2}}}\]
Lateral surface area of cuboid \[=2h(l+b)\]
Total surface area of cuboid \[=2(lb+bh+hl)\]
Volume of the cuboid \[=l\times \text{b}\times \text{h}\] 
Surface Area and Volume of Cube
Cube
It is a cuboid of equal sides.
Diagonal of a Cube
Diagonal of a cube =\[\sqrt{3}\](side)
In the above given figure alongside of the cube,
side = AB = CD = EF = 6H = AD = BC = EH = FG = AE = BF = CG = DH.
Diagonal of the cube \[=\sqrt{3}\times 1\]
Lateral surface area of the cube \[=4\times {{l}^{2}}\]
Total surface area of the cube \[=6\times {{l}^{2}}\]
Volume of the cube \[={{l}^{3}}\] 
Surface Area and Volume of Cylinder
Surface Area of Cylinder
Total surface area of cylinder = curve surface area + 2 (area of face)
Total surface area \[=2\pi r(r+h)\]
Curve surface area \[=2\pi rh\]
Volume of Cylinder
Volume of cylinder \[=\pi {{r}^{2}}h\]
Where r is the radius of cylinder and h is the height of cylinder.
Surface Area of Cone
A cone is a solid with a circular base. It has a curved surface which tapers (i.e. decreases in size) towards the vertex along the top.
The height of the cone is the perpendicular distance from the base to the vertex.
Now, \[\frac{Area\,of\,\sec tor\,ABC}{Area\,of\,circle\,with\,centre\,at\,C}=\] \[\frac{Arc\,length\,AB\,of\sec tor\,ABC}{Circumference\,of\,circle\,with\,centre\,at\,C}\]
Area of sector ABC \[=\frac{r}{l}\times \pi {{l}^{2}}=\pi rl\].
The curved surface area of a cone = The area of sector ABC.
\[\therefore \] The curved surface area of a cone\[=\pi rl\]
Total surface area of the cone = Area of the base + Curved surface area
\[\therefore \] \[TSA=\pi {{r}^{2}}+\pi rl\]
or, \[TSA=\pi r(r+l)\].
The volume of the cone is given by,
\[V=\frac{1}{3}\pi {{r}^{2}}h\] 
Surface Area of Sphere
Any round figure like football, cricket ball, etc. are known as sphere.
Sphere is a three dimensional figure having certain radius and volume.
The surface area of the sphere \[=4\pi {{r}^{2}}\]
Volume of the sphere \[=\frac{4}{3}\pi {{r}^{3}}\] 
Surface Area of Hemisphere
It is obtained by cutting the plane through the centre of a sphere into two equal parts and each part is known as the hemisphere.
The curved surface area of the hemisphere \[=2\pi {{r}^{2}}\]
The total surface area of the hemisphere \[=3\pi {{r}^{2}}\]
Volume of the hemisphere \[=\frac{2}{3}\pi {{r}^{3}}\] 
Surface Area of Frustum
It is the figure obtained by cutting the cone by a plane parallel to the base of the cone, then the portion between the plane and the base is known as the frustum of the cone.
The curved surface area of frustum\[=\pi l(R+r)\]
Where, \[{{l}^{2}}={{h}^{2}}+{{(R-r)}^{2}}\].
Total surface area of the frustum \[\pi =({{R}^{2}}+{{r}^{2}}+l(R+r)\]
Volume of the frustum \[=\frac{1}{3}\pi h({{R}^{2}}+{{r}^{2}}+Rr)\]
The guest room of Mary's flat is in the form of rectangle. If the width of the room is three fourth of its length and its area is given as 600\[{{\text{m}}^{\text{2}}}\]. The difference between the length and width of the room is:
(a) \[5\sqrt{2}\,m\]
(b) \[7\sqrt{2}\,m\]
(c) \[9\sqrt{2}\,m\]
(d) \[3\sqrt{2}\,m\]
(e) None of these
Answer: (a)
Explanation:
Let the length of the room \[=x\].
Then the width of the room \[=\frac{3}{4}x\]
Area of the room \[=\frac{3}{4}{{x}^{2}}\]
But the area of the room \[~=\text{6}00\,{{\text{m}}^{\text{2}}}\]
Therefore, \[x=\text{6}00{{\text{m}}^{\text{2}}}\]
\[x=\sqrt{800}=20\sqrt{2}\]
The ratio between the length and breath of a field is 10 : 6. The area of the field is 3840 m2. Find the difference between the length and width of the field.
(a) 32m
(b) 24m
(c) 20m
(d) 18m
(e) None of these
Answer: (a)
Explanation:
Let the length of the field be \[10x\] and breath be \[6x\].
Then the area of the field \[=10x\times 6x=60{{x}^{2}}\]
Therefore, \[\text{6}0{{x}^{2}}=\text{384}0\] \[x=8\,m\]
Therefore, length of the field = 80 m and breath of the field = 48 m
Difference \[=\text{8}0-\text{48}=\text{32 m}\]
A room of the hall is such that the ratio of the height of the room to its semi perimeter is 6 : 10 and the cost of paper wall of the room to cover the wall of the room is Rs. 1700, when the width more... 
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