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

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

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.

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.

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

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 %.

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.

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 %

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 threequestions 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 %

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