Maths shortcuts for Competitive Exams

7/29/2019 Maths shortcuts for Competitive Exams

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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 performs the above again.

If the result is 0 or is divisible by 7 the original number is alsodivisible by 7.

Example 1 ) 2599*2= 18.25-18 = 7 which is divisible by 7 so 259 is also divisible by 7.

Example 2 ) 27933*2= 6279-6= 273

now 3*2=627-6= 21 which is divisible by 7 so 2793 is also divisible by 7 .

Now find out if following are divisible by 7

1) 28412) 38733) 13934) 2877

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

1) Subtract the number from 50 getting result A.2) Square A getting result X.3) Subtract A from 25 getting result Y4) Answer is xy

EXAMPLE 1 : 4450-44=6

For More Visit http://www.yoursyllabus.blogspot.com


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Sq of 6 =3625-6 = 19So answer 1936

EXAMPLE 2 : 47

50-47=3Sq 0f 3 = 0925-3= 22So 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 STEP2.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 carryoverfrom Step 4.

EXAMPLE :

SQ (431)

STEP 1. Last digit = last digit of SQ(1) =1STEP 2. Second Last Digit = 2*3*1 + any carryover from STEP1.= 6STEP 3. Third Last Digit 2*4*1+ Sq(3) + any carryover from STEP2.= 2*4*1 +9= 17. so 7 and 1 carryoverSTEP 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 carryoverfrom 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 aredifferent, then you need to carry out first two steps only , thussaving time. You may save up to 30 seconds on eachcalculations and if there are 4 such questions you save 2minutes which may really affect UR Percentile score.


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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 thussaving 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 differencein UR PERCENTILE score.

The unique set of pythagoras triplets with the Hypotenuse less than 100 or one ofthe 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 willagain 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 derivedtriplet (derived by multiplying a constant to a unique triplet), all the threenumbers cannot be odd.

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

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

3 4 55 12 137 24 259 40 4111 60 61

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


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

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

Example 2: First side = 11so 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 checkfor other derived triplets.

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

4 3 5

8 15 1712 35 3716 63 6520 99 101

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

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

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

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

PROFIT AND LOSS : In every exam there are from one to threequestions on profit and loss, stating that the cost was firstincreased by certain % and then decreased by certain %. Hownice it would be if there was an easy way to calculate the finalchange in % of the cost with just one formula. It would really help

you in saving time and improving UR Percentile. Here is theformula for the same :

Suppose the price is first increase by X% and then decreasedby Y% , the final change % in the price is given by the followingformula

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


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

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

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

EXAMPLE 2 : The price of T.V set is increased by 25 % of costprice 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.Vsold.

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

So if 25 % = 5,000then 100 % = 20,000.C.P = 20,000S.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 JUST10 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 12days. In how many days Blake and Chandana can finish thework together ?

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

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


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TIME AND WORK :

1. If A can finish work in X time and B can finish work in Y timethen 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 finishwork 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 timethen 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 Ctogether 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 howmuch % should the consumption be reduced so that theexpense remain the same.

TYPE 2 : Price of a commodity is decreased by 60 %. By howmuch % can the consumption be increased so that the expenseremain the same.

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

Now do the following questions for UR Practice:

Q1. Price of a commodity is increased by 10 %. By how much %should the consumption be reduced so that the expense remainthe same.

Q2. Price of a commodity is decreased by 10 %. By how much %

can the consumption be increased so that the expense remainthe same.



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Q4. Price of a commodity is decreased by 15 %. By how much %can the consumption be increased so that the expense remainthe same.

Q5. Price of a commodity is increased by 25 %. By how much %

should the consumption be reduced so that the expense remainthe same.


Q7. Price of a commodity is increased by 50%. By how much %should the consumption be reduced so that the expense remainthe same.



Q10. Price of a commodity is decreased by 20 %. By how much% can the consumption be increased so that the expenseremain the same.

To find the number of factors of a given number, express the number as a productof powers of prime numbers.

In this case, 48 can be written as 16 * 3 = (24 * 3)

Now, increment the power of each of the prime numbers by 1 and multiply the

result.

In this case it will be (4 + 1)*(1 + 1) = 5 * 2 = 10 (the power of 2 is 4 and the powerof 3 is 1)

Therefore, there will 10 factors including 1 and 48. Excluding, these twonumbers, you will have 10 2 = 8 factors.


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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 first n even numbers= n (n+1)

The sum of first n odd numbers= n^2

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To find the squares of numbers near numbers of which squares are known

To find 41^2 , Add 40+41 to 1600 =1681

To find 59^2 , Subtract 60^2-(60+59) =3481

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If an equation (i:e f(x)=0 ) contains all positive co-efficient of any powers of x , ithas no positive roots then.eg: x^4+3x^2+2x+6=0 has no positive roots .

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For an equation f(x)=0 , the maximum number of positive roots it can have is thenumber of sign changes in f(x) ; and the maximum number of negative roots itcan have is the number of sign changes in f(-x) .Hence the remaining are the minimum number of imaginary roots of theequation(Since we also know that the index of the maximum power of x is thenumber of roots of an equation.)

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For a cubic equation ax^3+bx^2+cx+d=o

sum of the roots = b/asum of the product of the roots taken two at a time = c/aproduct of the roots = -d/a++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

For a biquadratic equation ax^4+bx^3+cx^2+dx+e = 0

sum of the roots = b/asum of the product of the roots taken three at a time = c/asum of the product of the roots taken two at a time = -d/aproduct of the roots = e/a+++++++++++++++++++++++++++++++++++++++++++++++++++++++++

If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM ifx=y(=k/2). The maximum product is then (k^2)/4

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If for two numbers x*y=k(=constant), then their SUM is MINIMUM ifx=y(=root(k)). The minimum sum is then 2*root(k) .

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|x| + |y| >= |x+y| (|| stands for absolute value or modulus )(Useful in solving some inequations)

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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 eachother

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For any regular polygon , the sum of the exterior angles is equal to 360 degreeshence measure of any external angle is equal to 360/n. ( where n is the number ofsides)

For any regular polygon , the sum of interior angles =(n-2)180 degrees


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So measure of one angle in

Square =90

Pentagon =108

Hexagon =120

Heptagon =128.5

Octagon =135

Nonagon =140

Decagon = 144

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++If any parallelogram can be inscribed in a circle , it must be a rectangle.

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If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i:eoblique sides equal).


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For an isosceles trapezium , sum of a pair of opposite sides is equal in length tothe sum of the other pair of opposite sides .(i:e AB+CD = AD+BC , taken in order).

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Area of a regular hexagon : root(3)*3/2*(side)*(side)

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For any 2 numbers a>b

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

(GM)^2 = AM * HM

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For three positive numbers a, b ,c

(a+b+c) * (1/a+1/b+1/c)>=9

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For any positive integer n

2


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a^2+b^2+c^2 >= ab+bc+caIf a=b=c , then the equality holds in the above.

a^4+b^4+c^4+d^4 >=4abcd

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(n!)^2 > n^n (! for factorial)

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If a+b+c+d=constant , then the product a^p * b^q * c^r * d^s will be maximumif a/p = b/q = c/r = d/s .

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Consider the two equations

a1x+b1y=c1a2x+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 for these equations.(means not equal to )If a1/a2 b1/b2 , then we have a unique solutions for these equations..+++++++++++++++++++++++++++++++++++++++++++++++++++++++

+++ For any quadrilateral whose diagonals intersect at right angles , the area ofthe quadrilateral is0.5*d1*d2, where d1,d2 are the lenghts of the diagonals.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++Problems on clocks can be tackled as assuming two runners going round a circle ,


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one 12 times as fast as the other . That is ,the minute hand describes 6 degrees /minutethe hour hand describes 1/2 degrees /minute .

Thus the minute hand describes 5(1/2) degrees more than the hour hand per

minute .

The hour and the minute hand meet each other after every 65(5/11) minutes afterbeing together at midnight.(This can be derived from the above) .

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If n is even , n(n+1)(n+2) is divisible by 24

If n is any integer , n^2 + 4 is not divisible by 4

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++Given the coordinates (a,b) (c,d) (e,f) (g,h) of a parallelogram , the coordinates ofthe meeting point of the diagonals can be found out by solving for[(a+e)/2,(b+f)/2] =[ (c+g)/2 , (d+h)/2]

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Area of a triangle

1/2*base*altitude = 1/2*a*b*sinC = 1/2*b*c*sinA = 1/2*c*a*sinB = 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 isthe inradius of the triangle .

In any trianglea=b*CosC + c*CosBb=c*CosA + a*CosC


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c=a*CosB + b*CosA

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If a1/b1 = a2/b2 = a3/b3 = .. , then each ratio is equal to(k1*a1+ k2*a2+k3*a3+..) / (k1*b1+ k2*b2+k3*b3+..) , which isalso equal to(a1+a2+a3+./b1+b2+b3+.)

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(7)In any trianglea/SinA = b/SinB =c/SinC=2R , where R is the circumradius

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .+ a^(n-1) ) Very useful forfinding multiples .For example (17-14=3 will be a multiple of 17^3 14^3)

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e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ..to infinity2 < e < 3

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log(1+x) = x (x^2)/2 + (x^3)/3 (x^4)/4 to infinity [ Note the alternatingsign . .Also note that the ogarithm is with respect to base e ]

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+++

In a GP the product of any two terms equidistant from a term is always constant .

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For a cyclic quadrilateral , area = root( (s-a) * (s-b) * (s-c) * (s-d) ) , wheres=(a+b+c+d)/2

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+++

For a cyclic quadrilateral , the measure of an external angle is equal to themeasure of the internal opposite angle.

(m+n)! is divisible by m! * n! .

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If a quadrilateral circumscribes a circle , the sum of a pair of opposite sides isequal to the sum of the other pair .

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The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term andcommon ratio of the GP .++++++++++++++++++++++++++++++++++++++++++++++++++++++++++The equation whose roots are the reciprocal of the roots of the equationax^2+bx+c is cx^2+bx+a

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++The coordinates of the centroid of a triangle with vertices (a,b) (c,d) (e,f)is((a+c+e)/3 , (b+d+f)/3) .

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+++The ratio of the radii of the circumcircle and incircle of an equilateral triangle is2:1 .

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Area of a parallelogram = base * height

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APPOLLONIUS THEOREM:

In a triangle , if AD be the median to the side BC , thenAB^2 + AC^2 = 2(AD^2 + BD^2) or 2(AD^2 + DC^2) .

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for similar cones , ratio of radii = ratio of their bases.

The HCF and LCM of two nos. are equal when they are equal .


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Volume of a pyramid = 1/3 * base area * height

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In an isosceles triangle , the perpendicular from the vertex to the base or theangular bisector from vertex to base bisects the base.

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In any triangle the angular bisector of an angle bisects the base in the ratio of theother two sides.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++The quadrilateral formed by joining the angular bisectors of anotherquadrilateral isalways a rectangle.

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Roots of x^2+x+1=0 are 1,w,w^2 where 1+w+w^2=0 and w^3=1

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|a|+|b| = |a+b| if a*b>=0else |a|+|b| >= |a+b|

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2


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+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++WINE and WATER formula:

If Q be the volume of a vessel

q qty of a mixture of water and wine be removed each time from a mixturen be the number of times this operation be doneand A be the final qty of wine in the mixture

then ,A/Q = (1-q/Q)^n

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Area of a hexagon = root(3) * 3 * (side)^2

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(1+x)^n ~ (1+nx) if x


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Area of a trapezium = 1/2 * (sum of parallel sids) * height = median * heightwhere median is the line joining the midpoints of the oblique sides.

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when a three digit number is reversed and the difference of these two numbers istaken , the middle number is always 9 and the sum of the other two numbers isalways 9 .

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ANy function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y) .

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Let W be any point inside a rectangle ABCD .ThenWD^2 + WB^2 = WC^2 + WA^2

Let a be the side of an equilateral triangle . then if three circles be drawn insidethis triangle touching each other then eachs radius = a/(2*(root(3)+1))

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Let x be certain base in which the representation of a number is abcd , then thedecimal value of this number is a*x^3 + b*x^2 + c*x + d+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

when you multiply each side of the inequality by-1, you have to reverse the

direction of the inequality.

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To find the squares of numbers from 50 to 59

For 5X^2 , use the formulae

(5X)^2 = 5^2 +X / X^2

Eg ; (55^2) = 25+5 /25

=3025

(56)^2 = 25+6/36

=3136

(59)^2 = 25+9/81

=3481

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many of u must b aware of this formula, but the ppl who dont know it must buseful for them.a+b+(ab/100)

this is used for succesive discounts types of sums.like 1999 population increses by 10% and then in 2000 by 5%so the population in 2000 now is 10+5+(50/100)=+15.5% more that was in 1999

and if there is a decrease then it will be preceeded by a -ve sign and likeiwse


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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)/2.4.6Funda : if a number n is represented asa^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 thenumber 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 ofprimes; 10 = 2*5

divide 203 with 2 and 5 individually203/2 = 101

101/2 = 5050/2 = 2525/2 = 1212/2 = 66/2 = 33/2 = 1thus power of 2 in 203! is, 101 + 50 + 25 + 12 + 6 + 3 + 1 = 198


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divide 203 with 5203/5 = 4040/5 = 88/5 = 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 valueof 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 squaresshud 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*11When 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 2squares?

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 waysFunda is that, all four ways of expressing can be used to findout distinct a,bvalues,for example take 9*81now since 9*81 = (a-b)(a+b) by solving the system a-b = 9 and a+b = 81 we canhave 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 withdenominator as 3?


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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 andadding 1 we will have a perfect square

10) When 987 and 643 are divided by same number n the reminder isalso 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 + rand 643 = bn + r and thus987 643 is divisible by r and987 643 = 344 = 86 * 4 = 43 * 8 and thus the prime is 43hence r is 43

11) when a number is divided by 11,7,4 the reminders are 5,6,3respectively. what would be the reminders when the same number isdivided 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) + 5that is 302 + LCM(11,7,4) and thus the rems when the same number is divided by4,7,11 respectively are,

302 mod 4 = 275 mod 7 = 510 mod 11 = 10

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


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13) if a+b+c = 0 then a^3 + b^3 + c^3 = 3abc

EXAMPLE: 40^3-17^3-23^3 is divisible bysince 40-23-17 = 0, 40^3-17^3-23^3 = 3*40*23*17 and thus, the number isdivisible by 3,5,8,17,23 etc.

14) There is a seller of cigarette and match boxes who sits in thenarrow lanes of cochin. He prices the cigarettes at 85 p, but foundthat there are no takers. So he reduced the price of cigarette andmanaged to sell all the cigarettes, realising Rs. 77.28 in all. What is thenumber of cigarettes?

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

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

Quote:

i have given this question to make the funda clear

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

16) In any number system 121 is a perfect square,SOLN: let the base be nthen 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:


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(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 = 234.3434343434and 10 x = 2.3434343434thus, 990 x = 232and 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 bycertain number 9 are 6, 8, 1 respectively. What would be the reminder when youdivide 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 2and thus, the remainder left when 1073 * 1079 * 1087 is divided by 119 is2*8*16 mod 119 and that is 256 mod 119 and that is (238 + 18 ) mod 119 and that

is 18


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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


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To find out if a number is divisible by seven:

Take the last digit, double it, and subtract it from the rest of thenumber.If the answer is more than a 2 digit number perform the aboveagain.If the result is 0 or is divisible by 7 the original number is alsodivisible by 7.

Example 1 ) 2599*2= 18.25-18 = 7 which is divisible by 7 so 259 is also divisible by 7.

Example 2 ) 27933*2= 6279-6= 273

now 3*2=627-6= 21 which is divisible by 7 so 2793 is also divisible by 7 .

Now find out if following are divisible by 7

1) 28412) 38733) 13934) 2877

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

1) Subtract the number from 50 getting result A.2) Square A getting result X.3) Subtract A from 25 getting result Y4) Answer is xy

EXAMPLE 1 : 4450-44=6Sq of 6 =36


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25-6 = 19So answer 1936

EXAMPLE 2 : 4750-47=3

Sq 0f 3 = 0925-3= 22So 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 STEP2.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 carryoverfrom Step 4.

EXAMPLE :

SQ (431)

STEP 1. Last digit = last digit of SQ(1) =1STEP 2. Second Last Digit = 2*3*1 + any carryover from STEP1.= 6STEP 3. Third Last Digit 2*4*1+ Sq(3) + any carryover from STEP2.= 2*4*1 +9= 17. so 7 and 1 carryoverSTEP 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 carryoverfrom 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


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different, then you need to carry out first two steps only , thussaving time. You may save up to 30 seconds on eachcalculations and if there are 4 such questions you save 2minutes which may really affect UR Percentile score.

PYTHAGORAS THEROEM :

In any given exam there are about 2 to 3 questions based on pythagoras theorem.Wouldnt it be nice that you remember some of the pythagoras triplets thussaving up to 30 seconds in each question. This saved time may be used to attemptother questions. Remember one more right question may make a lot of differencein UR PERCENTILE score.

The unique set of pythagoras triplets with the Hypotenuse less than 100 or one ofthe 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 willagain 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 derivedtriplet (derived by multiplying a constant to a unique triplet), all the threenumbers cannot be odd.

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


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

3 4 55 12 13

7 24 259 40 4111 60 61

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

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

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

Example 2: First side = 11so 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 and15, which has been obtained from multiplying 3 to the triplet of 3,4,5. You maycheck for other derived triplets.

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

4 3 58 15 17

12 35 3716 63 6520 99 101

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

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

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

Other side = 17-2=15

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


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PROFIT AND LOSS : In every exam there are from one to threequestions on profit and loss, stating that the cost was firstincreased by certain % and then decreased by certain %. Hownice it would be if there was an easy way to calculate the finalchange in % of the cost with just one formula. It would really help

you in saving time and improving UR Percentile. Here is theformula for the same :

Suppose the price is first increase by X% and then decreasedby Y% , the final change % in the price is given by the followingformula

Final Difference % = X- Y XY/100.

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

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

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

EXAMPLE 2 : The price of T.V set is increased by 25 % of costprice 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.Vsold.

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

So if 25 % = 5,000then 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 %.


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First Increased by 20 % and then decrease by 15 %.

TIPS TO IMPROVE UR PERCENTILE :

HOW ABOUT SOLVING THE FOLLOWING QUESTION IN JUST10 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 12days. In how many days Blake and Chandana can finish thework together ?

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

NOW CAREFULLY READ THE FOLLOWING TO SOLVE THETIME 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 timethen 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 finishwork 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 timethen 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 Ctogether 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

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


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TYPE 2 : Price of a commodity is decreased by 60 %. By howmuch % can the consumption be increased so that the expenseremain the same.

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

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Maths shortcuts for Competitive Exams