Maths Shortcuts for Competitive Exams in INDIA

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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 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 For Education and Jobs in India -http://naukari-times.blogspot.com

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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.
















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 pythagora pythagorass 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 pythagora pythagorass 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. 345 5 12 13 7 24 25
















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 . 435 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 :
















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 40-25-(40*25/100)= /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 25-40-(25*45/100)= /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 ?
















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 % 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 remain the same. Q2. Price of a commodity is decreased by 10 %. By how much % can the consumption be increased so that the expense remain the same.


















Q4. Price of a commodity is decreased by 15 %. By how much % can the consumption be increased so that the expense remain the same. Q5. Price of a commodity is increased by 25 %. By how much % should the consumption be reduced so that the expense remain the same. Q6. Price of a commodity is decreased by 25 %. By how much % can the consumption be increased so that the expense remain the same. Q7. Price of a commodity is increased by 50%. By how much % should the consumption be reduced so that the expense remain the same. Q8. Price of a commodity is decreased by 50 %. By how much % can the consumption be increased so that the expense remain the same. Q9. Price of a commodity is increased by 20 %. By how much % should the consumption be reduced so that the expense remain the same. Q10. Price of a commodity is decreased by 20 %. By how much % can the consumption be increased so that the expense remain the same.
















+++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++

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

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
















If an equation (i:e f(x)=0 ) contains all positive co-efficient co-efficient of any powers of x , it has no positive roots then. eg: x^4+3x^2+2x+6=0 has no positive 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 number of negative roots it can have is the number of sign changes in f(-x) . Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.) +++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++

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
















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

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
















So measure of one angle in

Square

=90

Pentagon

=108

Hexagon

=120

Heptagon

=128.5

Octagon

=135

Nonagon

=140

Decagon

= 144


















+++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++ 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) . +++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++ Area of a regular hexagon : root(3)*3/2*(side)*(side) +++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++ For any 2 numbers a>b a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa respectively)
















+++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++ a^2+b^2+c^2 >= ab+bc+ca If a=b=c , then the equality holds in the above. a^4+b^4+c^4+d^4 >=4abcd +++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++ (n!)^2 > n^n (! for factorial) +++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++

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 .
















one 12 times as fast as the other . That is , the minute hand describes 6 degrees /minute the 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 after being together at midnight. (This can be derived from the above) . +++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++ 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
















c=a*CosB + b*CosA +++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++ If a1/b1 = a2/b2 = a3/b3 = ………….. , then each ratio is equal to (k1*a1+ k2*a2+k3*a3+…………..) k2*a2+k3*a3+…………..) / (k1*b1+ k2*b2+k3*b3+…………..) k2*b2+k3*b3+…………..) , which is also equal to (a1+a2+a3+…………./b1+b2+b3+……….) +++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++ (7)In any triangle a/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 for finding multiples .For example (17-14=3 will be a multiple of 17^3 – 14^3)
















For a cyclic quadrilateral , area = root( (s-a) * (s-b) * (s-c) * (s-d) ) , where s=(a+b+c+d)/2 +++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++ For a cyclic quadrilateral , the measure of an external angle is equal to the measure of the internal opposite angle. (m+n)! is divisible by m! * n! . +++++++++++++++++++++++++++++++++++++++++++++++++++++++ ++++++++ If a quadrilateral circumscribes circumscribes a circle , the sum of a pair of opposite sides is equal to the sum of the other pair . +++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++


















Area of a parallelogram = base * height +++++++++++++++++++++++++++++++++++++++++++++++++++++++ +++ APPOLLONIUS THEOREM: In a triangle , if AD be the median to the side BC , then AB^2 + AC^2 = 2(AD^2 + BD^2) or 2(AD^2 + DC^2) .
















+++++++++++++++++++++++++++++++++++++++++++++++++++++ Volume of a pyramid = 1/3 * base area * height +++++++++++++++++++++++++++++++++++++++++++++++++++++ In an isosceles triangle , the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.
















+++++++++++++++++++++++++++++++++++++++++++++++++++++++ ++++++++ 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 mixture n be the number of times this operation be done and A be the final qty of wine in the mixture then , A/Q = (1-q/Q)^n +++++++++++++++++++++++++++++++++++++++++++++++++++++++
















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
















To find the squares of numbers from 50 to 59

For 5X^2 , use the formulae

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


















Numbers
















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
















ANS : sum = 10.33 + 10.66 + 11.33 11 .33 + 11.66 + 12.33 + 12.66 + 13.33 + 13.66……. = 21 + 23 + …… = 300
















13) if a+b+c = 0 then a^3 + b^3 + c^3 = 3abc EXAMPLE: 40^3-17^3-23^3 is divisible by EXAMPLE: 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.
















(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





































































































Maths Shortcuts for Competitive Exams in INDIA

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