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Math Shortcuts for Competitive Examinations – Part 3

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

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  %

 

 

Some good books on Math shortcuts useful for competitive exams are given below: