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quant - imp points 2 remembr
If an equation (i.e. f(x) = 0) contains all positive co-efficients of any powers of x, it has no positive roots.
Eg: x3+3x2+2x+6=0 has no positive roots
For an equation, if all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then it has no negative roots.
For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x)
Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i, another has to be 2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real root could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.
For a cubic equation ax3+bx2+cx+d=o
· Sum of the roots = - b/a
· Sum of the product of the roots taken two at a time = c/a
· Product of the roots = -d/a
For a bi-quadratic equation ax4+bx3+cx2+dx+e = 0
· Sum of the roots = - b/a
· Sum of the product of the roots taken three at a time = c/a
· Sum of the product of the roots taken two at a time = -d/a
· Product of the roots = e/a
If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients, then the equation has no real roots in each case (except for x=0 in the second case)
Consider the two equations
a1x+b1y=c1
a2x+b2y=c2
Then,
If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.
If a1/a2 = b1/b2 <> c1/c2, then we have no solution.
If a1/a2 <> b1/b2, then we have a unique solution.
Roots of x2 + x + 1=0 are 1, w, w2 where 1 + w + w2=0 and w3=1
|a| + |b| = |a + b| if a*b>=0
else, |a| + |b| >= |a + b|
The equation ax2+bx+c=0 will have max. value when a<0 and min. value when a>0. The max. or min. value is given by (4ac-b2)/4a and will occur at x = -b/2a
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.
If for two numbers x*y=k (a constant), then their SUM is MINIMUM if
x=y (=root(k)). The minimum sum is then 2*root (k).
Product of any two numbers = Product of their HCF and LCM. Hence product of two numbers = LCM of the numbers if they are prime to each other.
For any 2 numbers a, b where a>b
a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)
(GM)^2 = AM * HM
For three positive numbers a, b, c
(a + b + c) * (1/a + 1/b + 1/c)>=9
For any positive integer n
2<= (1 + 1/n)^n <=3
a2 + b2 + c2 >= ab + bc + ca
If a=b=c, then the case of equality holds good.
a4 + b4 + c4 + d4 >= 4abcd (Equality arises when a=b=c=d=1)
(n!)2 > nn
If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s
If n is even, n(n+1)(n+2) is divisible by 24
x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 - 14^3)
e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity
Note: 2 < e < 3
log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]
(m + n)! is divisible by m! * n!
When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9.
Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)
To Find Square of a 3-Digit Number
Let the number be XYZ
Step No. Operation to be Performed
1 Last digit = Last digit of Sq(Z)
2 Second last digit = 2*Y*Z + any carryover from STEP 1
3 Third last digit 2*X*Z+ Sq(Y) + any carryover from STEP 2
4 Fourth last digit is 2*X*Y + any carryover from STEP 3
5 Beginning of result will be Sq(X) + any carryover from Step 4
Eg) Let us find the square of 431
Step No. Operation to be Performed
1 Last digit = Last digit of Sq(1) = 1
2 Second last digit = 2*3*1 + any carryover from STEP 1=6+0=6
3 Third last digit 2*4*1+ Sq(3) + any carryover from STEP 2 = 8+9+0 = 17 i.e. 7 with carry over of 1
4 Fourth last digit is 2*4*3 + any carryover from STEP 3 = 24+1 = 25 i.e. 5 with carry over of 2
5 Beginning of result will be Sq(4) + any carryover from Step 4 = 16+2 = 18
THUS SQ(431) = 185761
If the answer choices provided are such that the last two digits are different, then, we need to carry out only the first two steps only.
The sum of first n natural numbers = n(n+1)/2
The sum of squares of first n natural numbers is n(n+1)(2n+1)/6
The sum of cubes of first n natural numbers is (n(n+1)/2)2/4
The sum of first n even numbers= n (n+1)
The sum of first n odd numbers= n2
If a number ‘N’ is represented as a^x * b^y * c^z… where {a, b, c, …} are prime numbers, then
the total number of factors is (x+1)(y+1)(z+1) ....
the total number of relatively prime numbers less than the number is
N * (1-1/a) * (1-1/b) * (1-1/c)....
the sum of relatively prime numbers less than the number is
N/2 * N * (1-1/a) * (1-1/b) * (1-1/c)....
the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x * y *...)
Total no. of prime numbers between 1 and 50 is 15
Total no. of prime numbers between 51 and 100 is 10
Total no. of prime numbers between 101 and 200 is 21
The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6
The number of rectangles in n*m board is given by n+1C2 * m+1C2
If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational.
Certain nos. to be remembered
210 = 45 = 322 = 1024
38 = 94 = 812 = 6561
7 * 11 * 13 = 1001
11 * 13 * 17 = 2431
13 * 17 * 19 = 4199
19 * 21 * 23 = 9177
19 * 23 * 29 = 12673
Where the digits of a no. are added and the resultant figure is 1 or 4 or 7 or 9, then, the no. could be a perfect square.
If a no. ‘N’ has got k factors and a^l is one of the factors such that l>=k/2, then, a is the only prime factor for that no.
To find out the sum of 3-digit nos. formed with a set of given digits
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
Consider the equation x^n + y^n = z^n
As per Fermat’s Last Theorem, the above equation will not have any solution whenever n>=3.
Further as per Fermat, where ‘p’ is a prime no. and ‘N’ is co-prime to p, then,
N^(p-1) – 1 is always divisible by p.
145 is the 3-digit no. expressed as sum of factorials of the individual digits i.e.
145 = 1! + 4! + 5!
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
a + b when n is odd
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
a + b when n is even
The relationship between base 10 and base ‘e’ in log is given by
log10N = 0.434 logeN
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
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)
-------------------
1 1
2 1 2 1
3 1 3 3 1
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
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
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)
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