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Quantitative Ability
– POINTS TO REMEMBER
Eg: x3+3x2+2x+6=0 has no positive roots
ü
For a cubic
equation ax3+bx2+cx+d=o
ü
For a
bi-quadratic equation ax4+bx3+cx2+dx+e =
0
a1x+b1y=c1 a2x+b2y=c2 Then, ü
If a1/a2
= b1/b2 = c1/c2, then we have
infinite solutions for these equations. ü
If a1/a2
= b1/b2 <> c1/c2, then
we have no solution. ü
If a1/a2
<> b1/b2, then we have a unique solution.
else, |a| + |b| >= |a + b|
ü
If for two
numbers x + y=k (a constant), then their PRODUCT is MAXIMUM if x=y (=k/2).
The maximum product is then (k2)/4. ü
If for two
numbers x*y=k (a constant), then their SUM is MINIMUM if
ü
a>AM>GM>HM>b
(where AM, GM ,HM stand for arithmetic, geometric , harmonic means
respectively) ü
(GM)^2 = AM
* HM
ü
(a + b + c)
* (1/a + 1/b + 1/c)>=9
ü
2<= (1 +
1/n)^n <=3
If a=b=c, then the case of equality holds good.
Note:
2 < e < 3
ü
The sum of
first n natural numbers = n(n+1)/2 ü
The sum of
squares of first n natural numbers is n(n+1)(2n+1)/6 ü
The sum of
cubes of first n natural numbers is (n(n+1)/2)2/4 ü
The sum of
first n even numbers= n (n+1) ü
The sum of
first n odd numbers= n2
ü
the total
number of factors is (x+1)(y+1)(z+1) .... ü
the total
number of relatively prime numbers less than the number is ü
the sum of
relatively prime numbers less than the number is ü
the sum of
factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x * y *...) ü
Total no.
of prime numbers between 1 and 50 is 15 ü
Total no.
of prime numbers between 51 and 100 is 10 ü
Total no.
of prime numbers between 101 and 200 is 21 ü
The number
of squares in n*m board is given by m*(m+1)*(3n-m+1)/6 ü
The number
of rectangles in n*m board is given by n+1C2 * m+1C2
ü
210 =
45 = 322 = 1024 ü
38 =
94 = 812 = 6561 ü
7 * 11 * 13
= 1001 ü
11 * 13 *
17 = 2431 ü
13 * 17 *
19 = 4199 ü
19 * 21 *
23 = 9177 ü
19 * 23 *
29 = 12673
This is given by (sum of digits) * (no. of digits-1)! * 1111…1
(i.e. based on the no. of digits) Eg) Find the sum of all 3-digit nos. formed using the digits
2, 3, 5, 7 & 8. Sum = (2+3+5+7+8) * (5-1)! * 11111 (since 5 digits are there) = 25 * 24 * 11111 =6666600
As per Fermat’s Last Theorem, the above equation will not
have any solution whenever n>=3.
145 = 1! + 4! + 5!
ü
Where a no.
is of the form a^n – b^n, then, ·
The no. is
always divisible by a - b ·
Further,
the no. is divisible by a + b when n is even and not divisible by ü
Where a no.
is of the form a^n + b^n, then, ·
The no. is
usually not divisible by a - b ·
However,
the no. is divisible by a + b when n is odd and not divisible by
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
The traditional formula for computing CI is CI = P*(1+R/100)^N – P Using
Pascal’s Triangle, Number of Years (N)
-------------------
1
1
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? Step 1: 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. Step 2: 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
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
ü
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 ü
The
difference between 3 years’ simple interest and compound interest is
given by (P * R2 *(300+R))/1003 ü
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. ü
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. ü
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 ü
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)
ü
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 ü
When an
unbiased coin is tossed even no. (2n) of times, then,
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! = 1/6 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! =
1/2
ü
Calendar
repeats after every 400 years. ü
Leap year-
it is always divisible by 4, but century years are not leap years unless
they are divisible by 400. ü
Century has
5 odd days and leap century has 6 odd days. ü
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. ü
January 1,
1901 was a Tuesday. 51.
ü
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) ü
For any
regular polygon, the sum of interior angles =(n-2)*180 degrees So measure of one angle is (n-2)/n *180 ü
If any
parallelogram can be inscribed in a circle, it must be a rectangle. ü
If a
trapezium can be inscribed in a circle it must be an isosceles trapezium
(i.e. oblique sides equal).
ü
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. ü
For a
cyclic quadrilateral, area = root((s-a) * (s-b) * (s-c) * (s-d)), where Further, for a cyclic quadrilateral, the measure of an
external angle is equal to the measure of the interior opposite angle. ü
Area of a
Rhombus = Product of Diagonals/2
[(a + e)/2, (b + f)/2] = [(c + g)/2, (d + h)/2]
ü
1/2*base*altitude ü
1/2*a*b*sin
C (or) 1/2*b*c*sin A (or) 1/2*c*a*sin B ü
root(s*(s-a)*(s-b)*(s-c))
where s=(a+b+c)/2 ü
a*b*c/(4*R)
where R is the circumradius of the triangle ü
r*s ,where
r is the inradius of the triangle
ü
a=b*cos C +
c*cos B ü
b=c*cos A +
a*cos C ü
c=a*cos B +
b*cos A ü
a/sin
A=b/sin B=c/sin C=2R, where R is the circumradius ü
cos C =
(a^2 + b^2 - c^2)/2ab ü
sin 2A = 2
sin A * cos A ü
cos 2A =
cos^2 (A) - sin^2 (A)
In a triangle ABC, if AD is the median to side BC, then AB2 + AC2 = 2(AD2 + BD2)
or 2(AD2 + DC2) ü
In an
isosceles triangle, the perpendicular from the vertex to the base or the
angular bisector from vertex to base bisects the base. ü
In any
triangle the angular bisector of an angle bisects the base in the ratio of
the other two sides.
WD2 + WB2 = WC2 + WA2
ü
Distance
between a point (x1, y1) and a line represented by
the equation ü
Distance
between 2 points (x1, y1) and (x2, y2)
is given by
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