41. The relationship between base 10 and base ‘e’ in log is given by log10N = 0.434 logeN
42. WINE and WATER formula
Let Q - volume of a vessel, q - qty of a mixture of water and wine be removed each time from a mixture, n - number of times this operation is done and A - final qty of wine in the mixture, then,
A/Q = (1-q / Q)^n
43. Pascal’s Triangle for computing Compound Interest (CI)
The traditional formula for computing CI is
CI = P*(1+R/100)^N – P
Using Pascal’s Triangle,
Number of Years (N)
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
… 1 .... .... ... ... ..1
Eg: P = 1000, R=10 %, and N=3 years. What is CI & Amount?
Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331
The coefficients - 1,3,3,1 are lifted from the Pascal's triangle above.
CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331 (leaving out first term in step 1)
If N =2, we would have had,
Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs.1210
CI = 2 * 100 + 1* 10 = Rs.210
44. Suppose the price of a product is first increased by X% and then decreased by Y% , then, the final change % in the price is given by:
Final Difference% = X - Y - XY/100
Eg) The price of a T.V set is increased by 40 % of the cost price and then is decreased by 25% of the new price. On selling, the profit made by the dealer was Rs.1000. At what price was the T.V sold?
Applying the formula,
Final difference% = 40 – 25 - (40*25/100) = 5 %.
So if 5 % = 1,000
Then, 100 % = 20,000.
Hence, C.P = 20,000
& S.P = 20,000+ 1000= 21,000
45. Where the cost price of 2 articles is same and the mark up % is same, then, marked price and NOT cost price should be assumed as 100.
-> Where ‘P’ represents principal and ‘R’ represents the rate of interest, then, the difference between 2 years’ simple interest and compound interest is given by P * (R/100)2
-> The difference between 3 years’ simple interest and compound interest is given by (P * R2 *(300+R))/1003
-> If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time.
-> If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time.
-> If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time
-> If A can finish a work in X time and B in Y time and A, B & C together in S time then
· C can finish that work alone in (XYS)/ (XY-SX-SY)
· B+C can finish in (SX)/(X-S); and
· A+C can finish in (SY)/(Y-S)
48. In case ‘n’ faced die is thrown k times, then, probability of getting atleast one more than the previous throw = nC5/n5
-> When an unbiased coin is tossed odd no. (n) of times, then, the no. of heads can never be equal to the no. of tails i.e. P (no. of heads=no. of tails) = 0
-> When an unbiased coin is tossed even no. (2n) of times, then, P (no. of heads=no. of tails) = 1-(2nCn/22n)
50. Where there are ‘n’ items and ‘m’ out of such items should follow a pattern, then, the probability is given by 1/m!
Eg) 1. Suppose there are 10 girls dancing one after the other. What is the probability of A dancing before B dancing before C?
Here n=10, m=3 (i.e. A, B, C)
Hence, P (A>B>C) = 1/3!
Eg)2. Consider the word ‘METHODS’. What is the probability that the letter ‘M’ comes before ‘S’ when all the letters of the given word are used for forming words, with or without meaning?
P (M>S) = 1/2!
-> Calendar repeats after every 400 years.
-> Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400.
-> Century has 5 odd days and leap century has 6 odd days.
-> In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day.
-> January 1, 1901 was a Tuesday.
-> For any regular polygon, the sum of the exterior angles is equal to 360 degrees, hence measure of any external angle is equal to 360/n (where n is the number of sides)
-> For any regular polygon, the sum of interior angles =(n-2)*180 degrees
So measure of one angle is (n-2)/n *180
-> If any parallelogram can be inscribed in a circle, it must be a rectangle.
-> If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i.e. oblique sides equal).
53. For an isosceles trapezium, sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides (i.e. AB+CD = AD+BC, taken in order)
-> For any quadrilateral whose diagonals intersect at right angles, the area of the quadrilateral is
0.5*d1*d2, where d1, d2 are the length of the diagonals.
-> For a cyclic quadrilateral, area = root((s-a) * (s-b) * (s-c) * (s-d)), where s=(a + b + c + d)/2
Further, for a cyclic quadrilateral, the measure of an external angle is equal to the measure of the interior opposite angle.
-> Area of a Rhombus = Product of Diagonals/2
55. Given the coordinates (a, b); (c, d); (e, f); (g, h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for
[(a + e)/2, (b + f)/2] = [(c + g)/2, (d + h)/2]
56. Area of a triangle
-> 1/2*a*b*sin C (or) 1/2*b*c*sin A (or) 1/2*c*a*sin B
-> root(s*(s-a)*(s-b)*(s-c)) where s=(a+b+c)/2
-> a*b*c/(4*R) where R is the circumradius of the triangle
-> r*s ,where r is the inradius of the triangle
57. In any triangle
-> a=b*cos C + c*cos B
-> b=c*cos A + a*cos C
-> c=a*cos B + b*cos A
-> a/sin A=b/sin B=c/sin C=2R, where R is the circumradius
-> cos C = (a^2 + b^2 - c^2)/2ab
-> sin 2A = 2 sin A * cos A
-> cos 2A = cos^2 (A) - sin^2 (A)
58. The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1
59. Appollonius Theorem
In a triangle ABC, if AD is the median to side BC, then
AB2 + AC2 = 2(AD2 + BD2) or 2(AD2 + DC2)
-> In an isosceles triangle, the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.
-> In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides.
61. The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle.
62. Let W be any point inside a rectangle ABCD, then,
WD2 + WB2 = WC2 + WA2
63. Let a be the side of an equilateral triangle, then, if three circles are drawn inside this triangle such that they touch each other, then each circle’s radius is given by a/(2*(root(3)+1))
-> Distance between a point (x1, y1) and a line represented by the equation ax + by + c=0 is given by ax1+by1+c/Sq(a2+b2)
-> Distance between 2 points (x1, y1) and (x2, y2) is given by Sq((x1-x2)2+ (y1-y2)2)
65. Where a rectangle is inscribed in an isosceles right angled triangle, then, the length of the rectangle is twice its breadth and the ratio of area of rectangle to area of triangle is 1:2.