In this article, you will learn very important formulas from the Topic Interest, Mensuration and Modern maths. These topics are very important for the Placement and for the Government Competitive Exams like SSC, BANKING, RAILWAYS, etc. These formulas will help you in solving problems, very quickly.
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Important Formulas
1. INTEREST – Interest is nothing but the amount of money earned by a lender or any financial institution by the borrower. It is one of the most important and easy topics. One can solve questions from this topic by using direct formulas.
a. Formula to find the Simple Interest :
SI = (P × R × T)/100
where P is the principal, R is the Rate of interest and T is the time period.
b. Formula to find Compound Interest :
A = P + CI = P[1 + (R / 100 )]n
where P is the principal amount, R is the rate , n means the time period.
Compound Interest = Amount – Principal
c. When the Interest is Compounded Annually :
Amount = P( 1 + (R / 100))n
where P is the principal amount, R is the rate , n means the time period.
d. When the Interest is Compounded Half-Yearly :
Amount =P [1 + (R / 200)]n×2
where P is the principal amount, R is the rate , n means the time period.
e. When the Interest is Compounded Quarterly :
Amount = P[1+ (R / 400)]n×4
where P is the principal amount, R is the rate , n means the time period.
2. Mensuration – In this topic, we will learn about the Volume, TSA, and CSA of 3D figures. To solve questions from this topic requires imagination as well which helps one to make figures in mind or make a raw figure.
a. Cube –
Volume- a3
Total surface area = 6 × a × a = 6a2 sq unit.
Curved surface area = 4a2 sq unit.
Diagonal of the cube = √3 × a unit
Perimeter=12 × a unit.
where a is the side of cube.
b. Cuboid –
Volume = l × b × h cubic unit.
Total surface area = 2 × (l × b + b × h + h × l). sq unit.
Curved surface area = 2 × (l + b) × h sq unit.
Perimeter = 4 × (l + b + h). unit
where ‘l’ is the length, ‘b’ is the breadth, and ‘h’ is the height of a cuboid.
c. Cylinder –
Volume = (π × r × r × h) cubic unit
Curved Surface Area = 2 × π × r × h. sq unit.
Total Surface Area = 2 × π × r × (r + h). sq unit
where ‘r’ is the radius of cylinder, π is 3.14 and ‘h’ is height of a cylinder.
d. Cone –
Volume- (π × r × r × h)/3
Curved Surface Area -π × r × l.
Total Surface Area- π × r × ( r + l).
where ‘r’ is the radius of cone, π is 3.14, ‘h’ is the height of cone, and ‘l’ is the slant height of the cone
e. Sphere –
Volume = 4/3 × π × r3 cubic unit.
Curved Surface Area = 3 × π × r2 sq unit
Total Surface Area- 4 × π × r2 sq unit
where r is the radius of sphere , π is 3.14
f. Hemi-Sphere –
Volume = 2/3 × π × r3 cubic unit.
Curved Surface Area = 2 × π × r2 sq unit.
Total Surface Area = 3 × π × r2 sq unit.
where r is the radius of Hemi-sphere , π is 3.14
g. Frustum –
Volume of frustum of cone = π × h × (R2 + R × r + r2)/3 cubic unit.
Total Surface Area of Frustum of Cone = π × L× (R + r) + π × R2 + π × r2 sq unit.
Curved Surface Area of Frustum of Cone = π × L× (R + r) sq unit.
R is the radius of the bottom base, and r is the radius of the top base, L is the slant height, h is the height of the cone , π is 3.14
h. Hollow-Sphere –
Volume- 4 × (π × r1× r1 × r1 )/3 – 4 × (π × r2 × r2 × r2 )/3
Curved Surface Area – 4 × π × r1 × r1 – 4 × π × r2 × r2
Total Surface Area- 4 × π × r1× r1 – 4 × π × r2× r2
where r1 is the outer radius and r2 is the inner radius of Hollow – sphere , π is 3.14.
3. Permutation and combination: This topic comes under the category of modern maths.
a. Multiplication theorem (And) :
If there are ‘m’ ways to do a particular task and ‘n’ ways to do another task, then the total number of ways of doing both the task is “m × n”. The formula
“m × n” ways is valid if and only if the decisions are independent of each other.
b. Addition theorem (OR) :
If there are ‘m’ ways to do a particular task and ‘n’ ways to do another task, then one can perform either of the two tasks (not both) in (m + n) ways.
c. Permutation : nPr = [n!/(n–r)!]
d. Permutation of n objects containing some repeated objects :
Let there be some objects (n) of which ‘p’ number of objects are of one kind, ‘q’ number of objects are of a second kind, ‘r’ number of objects are of the third kind and the rest are different, the number of permutation of the n things taken all together will be equal to = {n!/ (p! × q! × r!) }
e. Permutation of ‘n’ objects when an object can be placed repeatedly :
It is known that the number of permutation of n things taken ‘r’ at a time is the same as the number of ways in which r places can be filled up with
n things. The first place can be filled up in n ways as any of the n things can be placed there. After filling the first place, the second place can also be filled
up in n ways as things can be repeated. Proceeding in this way, one observe that the power of n is the same as the number of places filled up at this stage.
Therefore, the number of ways of filling up r places = nr
f. GEOMETRICAL PROPERTIES :
A. In a plane if there are n points of which no three are collinear, then
- The number of straight lines formed by joining them = nC2
- The number of triangles formed by joining them = nC3
- The number of polygons of k sides formed by joining them = nCk
B. In a plane if there are n points out of which m points are collinear, then :
- The number of straight lines formed by joining them = nC2 – mC2 + 1.
- The number of triangles formed by joining them = nC3 – mC3.
- The number of polygons of k sides formed by joining them = nCk – mCk
g. De-arrangement :
De-arrangement of objects means that none of the objects occupies its original place. It implies that if ‘n’ distinct items are arranged in a row, then
the number of ways they can be rearranged such that none of them occupies its original position is given by,
= n! [1/0! – 1/1! + 1/2! – 1/3! + 1/4!…….(-1)n/n! ]
4. Probability: This topic comes under the category of modern maths.
a. Probability of occurring of event A = (Favourable outcomes)/(total number of outcomes).
b. Number of outcomes in case of coins –
When 1 coin is tossed – H or T – 2 outcomes
When 2 coins are tossed – (HH), (HT), (TH), (TT) – 4 outcomes.
When n coins are tossed – total outcomes = 2n where n is the number of coins.
c. Number of outcomes in case of Dice.
When 1 dice is thrown – 1, 2, 3, 4, 5, 6 = 6 outcomes
When 2 dice are thrown – 36 outcomes.
(1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1)
(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2)
(1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3)
(1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4)
(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)
(1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6)
When n dice are thrown then number of outcome = 6n where n is the number of dice.
d. Cards :
Total cards = 52.
Colours = 2 Black and 2 Red 26 each.
Suits = 4 (Club, Diamond, Spade, Heart) 13 each.
Face cards = (Jack, queen, king) = 3 × 4 = 12
(Numbered card) = (2 to 10) = 9 × 4 = 36 card.
Ace cards = 4 (each colour)
e. Odds in favour and against of an event:
Odds in favor of an event = (favourable outcomes)/(unfavourable outcomes)
Odds in against of an event = (unfavourable outcomes)/(favourable outcomes)
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