WRITOPEDIA: SHORTCUTS

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Friday, 21 November 2014

QUANT SHORTCUTS 3

COMMON ADMISSION TEST – MBA ENTRANCE

Quantitative Ability – POINTS TO REMEMBER

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.

ü 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

ü 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,

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.

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

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
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

ü a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)

ü (GM)^2 = AM * HM

For three positive numbers a, b, c

ü (a + b + c) * (1/a + 1/b + 1/c)>=9

For any positive integer n

ü 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 = 2YZ + any carryover from STEP 1
3	Third last digit 2XZ+ Sq(Y) + any carryover from STEP 2
4	Fourth last digit is 2XY + 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 = 231 + any carryover from STEP 1=6+0=6
3	Third last digit 241+ Sq(3) + any carryover from STEP 2 = 8+9+0 = 17 i.e. 7 with carry over of 1
4	Fourth last digit is 243 + 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.

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

If a number ‘N’ is represented as a^x * b^y * c^z… where {a, b, c, …} are prime numbers, then

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
N * (1-1/a) * (1-1/b) * (1-1/c)....

ü the sum of relatively prime numbers less than the number is
N/2 * N * (1-1/a) * (1-1/b) * (1-1/c)....

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

If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational.

Certain nos. to be remembered

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

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!

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
a + b when n is odd

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
a + b when n is even

The relationship between base 10 and base ‘e’ in log is given by
log₁₀N = 0.434 log_eN

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

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

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.

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)

In case ‘n’ faced die is thrown k times, then, probability of getting atleast one more than the previous throw = ⁿC₅/n⁵

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,
P (no. of heads=no. of tails) = 1-(²ⁿC_n/2²ⁿ)

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!

= 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

CALENDAR

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.

52.

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).

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)

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
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.

ü Area of a Rhombus = Product of Diagonals/2

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]

Area of a triangle

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

In any 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)

The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1

Appollonius Theorem

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.

The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle.

Let W be any point inside a rectangle ABCD, then,

WD² + WB² = WC² + WA²

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))

ü 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)

ü Distance between 2 points (x₁, y₁) and (x₂, y₂) is given by
Sq((x₁-x₂)²+ (y₁-y₂)²)

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.

QUANT SHORTCUTS 2

Numbers

1) 2^2n-1 is always divisible by 3
2^2n-1 = (3-1)^2n -1
= 3M +1 -1
= 3M, thus divisible by 3

2) What is the sum of the divisors of 2^5.3^7.5^3.7^2?
ANS : (2^6-1)(3^8-1)(5^4-1)(7^3-1)/2.4.6
Funda : if a number 'n' is represented as
a^x * b^y * c^z ....
where, {a,b,c,.. } are prime numbers then

Quote:

(a) the total number of factors is (x+1)(y+1)(z+1) ....
(b) the total number of relatively prime numbers less than the number is n * (1-1/a) * (1-1/b) * (1-1/c)....
(c) the sum of relatively prime numbers less than the number is n/2 * n * (1-1/a) * (1-1/b) * (1-1/c)....
(d) the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x*y*...)

3) what is the highest power of 10 in 203!ANS : express 10 as product of primes; 10 = 2*5

divide 203 with 2 and 5 individually
203/2 = 101
101/2 = 50
50/2 = 25
25/2 = 12
12/2 = 6
6/2 = 3
3/2 = 1
thus power of 2 in 203! is, 101 + 50 + 25 + 12 + 6 + 3 + 1 = 198

divide 203 with 5
203/5 = 40
40/5 = 8
8/5 = 1

thus power of 5 in 203! is, 49

so the power of 10 in 203! factorial is 49

4) x + y + z = 7 and xy + yz + zx = 10, then what is the maximum value of x? ( CAT 2002 has similar question )
ANS: 49-20 = 29, now if one of the y,z is zero, then the sum of other 2 squares shud be equal to 29, which means, x can take a max value at 5

5) In how many ways can 2310 be expressed as a product of 3 factors?
ANS: 2310 = 2*3*5*7*11
When a number can be expressed as a product of n distinct primes,
then it can be expressed as a product of 3 numbers in (3^(n+1) + 1)/2 ways

6) In how many ways, 729 can be expressed as a difference of 2 squares?
ANS: 729 = a^2 - b^2
= (a-b)(a+b),
since 729 = 3^5,
total ways of getting 729 are, 1*729, 3*243, 9*81, 27*27.
So 4 ways
Funda is that, all four ways of expressing can be used to findout distinct a,b values,
for example take 9*81
now since 9*81 = (a-b)(a+b) by solving the system a-b = 9 and a+b = 81 we can have 45,36 as soln.

7) How many times the digit 0 will appear from 1 to 10000
ANS: In 2 digit numbers : 9,
In 3 digit numbers : 18 + 162 = 180,
In 4 digit numbers : 2187 + 486 + 27 = 2700,
total = 9 + 180 + 2700 + 4 = 2893

8 ) What is the sum of all irreducible factors between 10 and 20 with denominator as 3?
ANS :
sum = 10.33 + 10.66 + 11.33 + 11.66 + 12.33 + 12.66 + 13.33 + 13.66…….
= 21 + 23 + ……
= 300

9) if n = 1+x where x is the product of 4 consecutive number then n is,
1) an odd number,
2) is a perfect square
SOLN : (1) is clearly evident
(2) let the 4 numbers be n-2,n-1,n and n+1 then by multing the whole thing and adding 1 we will have a perfect square

10) When 987 and 643 are divided by same number 'n' the reminder is also same, what is that number if the number is a odd prime number?
ANS : since both leave the same reminder, let the reminder be 'r',
then, 987 = an + r
and 643 = bn + r and thus
987 - 643 is divisible by 'r' and
987 - 643 = 344 = 86 * 4 = 43 * 8 and thus the prime is 43
hence 'r' is 43

11) when a number is divided by 11,7,4 the reminders are 5,6,3 respectively. what would be the reminders when the same number is divided by 4,7,11 respectively?
ANS : whenever such problem is given,
we need to write the numbers in top row and rems in the bottom row like this

11 7 4
| \ \
5 6 3

( coudnt express here properly

)
now the number is of the form, LCM ( 11,7,4 ) + 11*(3*7 + 6) + 5
that is 302 + LCM(11,7,4) and thus the rems when the same number is divided by 4,7,11 respectively are,

302 mod 4 = 2
75 mod 7 = 5
10 mod 11 = 10

12) a^n - b^n is always divisible by a-b

13) if a+b+c = 0 then a^3 + b^3 + c^3 = 3abc
EXAMPLE: 40^3-17^3-23^3 is divisble by
since 40-23-17 = 0, 40^3-17^3-23^3 = 3*40*23*17 and thus, the number is divisible by 3,5,8,17,23 etc.

14) There is a seller of cigerette and match boxes who sits in the narrow lanes of cochin. He prices the cigerattes at 85 p, but found that there are no takers. So he reduced the price of cigarette and managed to sell all the cigerattes, realising Rs. 77.28 in all. What is the number of cigerattes?

a) 49
b) 81
c) 84
d) 92

ANS : (d)
since 77.28 = 92 * 84, and since price of cigarette is less than 85, we have (d) as answer

Quote:

i have given this question to make the funda clear

15) What does 100 stand for if 5 X 6 = 33
ANS : 81
SOLN : this is a number system question,
30 in decimal system is 33 in some base 'n', by solving we will have n as 9
and thus, 100 will be 9^2 = 81

16) In any number system 121 is a perfect square,
SOLN: let the base be 'n'
then 121 can be written as n^2 + 2*n + 1 = (n+1)^2
hence proved

17) Most of you ppl know these, anyways, just in case

Quote:

(a) sum of first 'n' natural numbers - n*(n+1)/2
(b) sum of the squares of first 'n' natural numbers - n*(n+1)*(2n+1)/6
(c) sum of the cubes of first 'n' natural numbers - n^2*(n+1)^2/4
(d) total number of primes between 1 and 100 - 25

18 ) See Attachment to know how to find LCM, GCF of Fractions

Quote:

CAT 2002 has 2 questions on the above simple concept

19) Converting Recurring Decimals to Fractions

let the number x be 0.23434343434........

thus 1000 x = 234.3434343434......
and 10 x = 2.3434343434.........
thus, 990 x = 232
and hence, x = 232/990

20) Reminder Funda

(a) (a + b + c) % n = (a%n + b%n + c%n) %n
EXAMPLE: The reminders when 3 numbers 1221, 1331, 1441 are divided by certain number 9 are 6, 8, 1 respectively. What would be the reminder when you divide 3993 with

9? ( never seen such question though

)
the reminder would be (6 + 8 + 1) % 9 = 6

(b) (a*b*c) % n = (a%n * b%n * c%n) %n
EXAMPLE: What is the remainder left when 1073 * 1079 * 1087 is divided by 119 ? ( seen this kinda questions alot

)
1073 % 119 = ?
since 1190 is divisible by 119, 1073 mod 119 is 2
and thus, "the remainder left when 1073 * 1079 * 1087 is divided by 119 " is 2*8*16 mod 119 and that is 256 mod 119 and that is (238 + 18 ) mod 119 and that is 18

Glossary : % stands for reminder operation

find the number of zeroes in 1^1* 2^2* 3^3* 4^4.............. 98^98* 99^99* 100^100

the expresion can be rewritten as (100!)^100 / 0!* 1!* 2!* 3!....99!

Now the numerator has 2400 zeros

the formular for finding number of zeros in n! is

[n/5]+[n/5^2]...[n/5^r]
where r is such that 5^r<=n<5^(r+1)

and [..] is the grestest integer function

for the numerator find the number of zeros using the above formulae..

for 0!...4! number of zeros ..0
5!...9!.number os zeros ..1
9!...14!... 2
15!..19!..................3
20!..24!..................4!
now at 25! the series makes a jump to 6
25!...29!.................6
30!...34!.................7
this goes on and again makes a jump at 50!
and then at 75!

so the number of zeros is...

5(1+2....19) + 25+ 50+ 75

the last 3 terms 25 50 and 75 are because of the jumps..

this gives numerator has 1100 zeros

now total number of zeros in expression is no of zeros in denominator - no of zeros in numerator
2400 - 1100

the Answer 1300

SOME MORE SHORTCUTS:

To find out if a number is divisible by seven:
Take the last digit, double it, and subtract it from the rest of the
number.
If the answer is more than a 2 digit number perform the above
again.
If the result is 0 or is divisible by 7 the original number is also
divisible by 7.

Example 1 ) 259
9*2= 18.
25-18 = 7 which is divisible by 7 so 259 is also divisible by 7.

Example 2 ) 2793
3*2= 6
279-6= 273

now 3*2=6
27-6= 21 which is divisible by 7 so 2793 is also divisible by 7 .

Now find out if following are divisible by 7

1) 2841
2) 3873
3) 1393
4) 2877

TO FIND SQUARE OF A NUMBER BETWEEN 40 to 50
Sq (44) .

1) Subtract the number from 50 getting result A.
2) Square A getting result X.
3) Subtract A from 25 getting result Y
4) Answer is xy

EXAMPLE 1 : 44
50-44=6
Sq of 6 =36
25-6 = 19
So answer 1936

EXAMPLE 2 : 47
50-47=3
Sq 0f 3 = 09
25-3= 22
So answer = 2209

NOW TRY To Find Sq of 48 ,26 and 49

TO FIND SQUARE OF A 3 DIGIT NUMBER :

LET THE NUMBER BE XYZ

SQ (XYZ) is calculated like this

STEP 1. Last digit = last digit of SQ(Z)
STEP 2. Second Last Digit = 2*Y*Z + any carryover from STEP 1.
STEP 3. Third Last Digit 2*X*Z+ Sq(Y) + any carryover from STEP
2.
STEP 4. Fourth last digit is 2*X*Y + any carryover from STEP 3.
STEP 5 . In the beginning of result will be Sq(X) + any carryover
from Step 4.

EXAMPLE :

SQ (431)

STEP 1. Last digit = last digit of SQ(1) =1
STEP 2. Second Last Digit = 2*3*1 + any carryover from STEP
1.= 6
STEP 3. Third Last Digit 2*4*1+ Sq(3) + any carryover from STEP
2.= 2*4*1 +9= 17. so 7 and 1 carryover
STEP 4. Fourth last digit is 2*4*3 + any carryover (which is 1) . =
24+1=25. So 5 and carry over 2.
STEP 5 . In the beginning of result will be Sq(4) + any carryover
from Step 4. So 16+2 =18.

So the result will be 185761.

If the option provided to you are such that the last two digits are
different, then you need to carry out first two steps only , thus
saving time. You may save up to 30 seconds on each
calculations and if there are 4 such questions you save 2
minutes which may really affect UR Percentile score.

PYTHAGORAS THEROEM :

In any given exam there are about 2 to 3 questions based on pythagoras theorem. Wouldn't it be nice that you remember some of the pythagoras triplets thus saving up to 30 seconds in each question. This saved time may be used to attempt other questions. Remember one more right question may make a lot of difference in UR PERCENTILE score.

The unique set of pythagoras triplets with the Hypotenuse less than 100 or one of the side less than 20 are as follows :

(3,4,5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), (12, 35, 37), (9, 40, 41), (28, 45, 53), (11, 60, 61), (33, 56, 65), (16, 63, 65), (48, 55, 73), (36, 77, 85), (13, 84, 85), (39, 80, 89), and (65, 72, 97).
(15,112,113),   (17,144,145),   (19,180,181),   (20,99,101)

If you multiply the digits of the above mentioned sets by any constant you will again get a pythagoras triplet .

Example : Take the set (3,4,5).
Multiply it by 2 you get (6,8,10) which is also a pythagoras triplet.
Multiply it by 3 you get ( 9,12,15) which is also a pythagoras triplet.
Multiply it by 4 you get (12,16,20) which is also a pythagoras triplet.
You may multiply by any constant you will get a pythagoras triplet

Take another example (5,12,13)
Multiply it by 5,6 and 7 and check if you get a pythagoras triplet.

TIPS FOR SMART GUESSING :

You will notice that in any case, whether it is a unique triplet or it is a derived triplet (derived by multiplying a constant to a unique triplet), all the three numbers cannot be odd.

In case of unique triplet , the hypotenuse is always odd and one of the remaining side is odd the other one is even.

Below are the first few unique triplets with first number as Odd.

3    4    5
5    12   13
7    24   25
9    40   41
11   60   61

You will notice following trend for unique triplets with first side as odd.

Hypotenuse = (Sq(first side) +1) / 2
Other side = Hypotenuse -1

Example : First side = 3 ,
so hypotenuse = (3*3+1)/2= 5 and other side = 5-1=4

Example 2: First side = 11
so hypotenuse = (9*9+1)/2= 41 and other side = 41-1=40

Please note that the above is not true for a derived triplet for example 9,12 and 15, which has been obtained from multiplying 3 to the triplet of 3,4,5. You may check for other derived triplets.

Below are the first few unique triplets with first number as Even .

4    3    5
8    15   17
12   35   37
16   63   65
20   99   101

You will notice following trend for unique triplets with first side as Even.

Hypotenuse = Sq( first side/ 2)+1
Other side = Hypotenuse-2

Example 1. First side =8
So hypotenuse = sq(8/2) +1= 17
Other side = 17-2=15

Example 2. First side = 16
So hypotenuse = Sq(16/2) +1 =65
Other side = 65-2= 63

PROFIT AND LOSS : In every exam there are from one to three
questions on profit and loss, stating that the cost was first
increased by certain % and then decreased by certain %. How
nice it would be if there was an easy way to calculate the final
change in % of the cost with just one formula. It would really help
you in saving time and improving UR Percentile. Here is the
formula for the same :

Suppose the price is first increase by X% and then decreased
by Y% , the final change % in the price is given by the following
formula

Final Difference % = X- Y - XY/100.

EXAMPLE 1. : The price of T.V set is increased by 40 % of the
cost price and then decreased by 25% of the new price . On
selling, the profit for the dealer was Rs.1,000 . At what price was
the T.V sold.

From the above mentioned formula you get :
Final difference % = 40-25-(40*25/100)= 5 %.

So if 5 % = 1,000
then 100 % = 20,000.
C.P = 20,000
S.P = 20,000+ 1000= 21,000.

EXAMPLE 2 : The price of T.V set is increased by 25 % of cost
price and then decreased by 40% of the new price . On selling,
the loss for the dealer was Rs.5,000 . At what price was the T.V
sold.

From the above mentioned formula you get :
Final difference % = 25-40-(25*45/100)= -25 %.

So if 25 % = 5,000
then 100 % = 20,000.
C.P = 20,000
S.P = 20,000 - 5,000= 15,000.

Now find out the difference in % of a product which was :
First increased by 20 % and then decreased by 10 %.
First Increased by 25 % and then decrease by 20 %.
First Increased by 20 % and then decrease by 25 %.
First Increased by 10 % and then decrease by 10 %.
First Increased by 20 % and then decrease by 15 %.

TIPS TO IMPROVE UR PERCENTILE :

HOW ABOUT SOLVING THE FOLLOWING QUESTION IN JUST
10 SECONDS

Ajay can finish work in 21 days and Blake in 42 days. If Ajay,
Blake and Chandana work together they finish the work in 12
days. In how many days Blake and Chandana can finish the
work together ?

(21*12 )/(24-12) = (21*12)/9= 7*4= 28 days.

NOW CAREFULLY READ THE FOLLOWING TO SOLVE THE
TIME AND WORK PROBLEMS IN FEW SECONDS.

TIME AND WORK :

1.   If A can finish work in X time and B can finish work in Y time
then both together can finish work in (X*Y)/ (X+Y) time.

2.   If A can finish work in X time and A and B together can finish
work in S time then B can finish work in (XS)/(X-S) time.

3.   If A can finish work in X time and B in Y time and C in Z time
then they all working together will finish the work in
(XYZ)/ (XY +YZ +XZ) time

4.    If A can finish work in X time and B in Y time and A,B and C
together in S time then :
C can finish work alone in (XYS)/ (XY-SX-SY)
B+C can finish in (SX)/(X-S)
and A+ C can finish in (SY)/(Y-S)

Here is another shortcut to improve URPERCENTILE.

TYPE 1 : Price of a commodity is increased by 60 %. By how
much % should the consumption be reduced so that the
expense remain the same.

TYPE 2 : Price of a commodity is decreased by 60 %. By how
much % can the consumption be increased so that the expense
remain the same.

Solution :
TYPE1 : (100* 60 ) / (100+60) = 37.5 %
TYPE 2 : (100* 60 ) / (100-60) = 150 %

Friday, 21 November 2014

QUANT SHORTCUTS 3

QUANT SHORTCUTS 2

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