eg: 87 * 94 94 is *6* less than 100 87 is *13* less than 100. so 1st two digits are *87 - 6 = 79 * last two digits are 13 * 6 = 78 therefore answer is 7978. hope it is useful for us in the upcoming CAT. Cheers, ...

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can anyone give me a link which has online quant papers for the cat and is a free website. pagalguy had some but i cant seem to find it anymore.

if u do could u please email me the link instead of posting it.

email is salil_pawah@yahoo.co.in

thanks again.

Take a 2-digit number beginning with 5.

Square the first digit.

Add this number to the second number to find the first part of the answer.

Square the second digit: this is the last part of the answer.

Example:

If the number is 58, multiply 5 5 = 25 (square the first digit).

25 + 8 = 33 (25 plus second digit).

The first part of the answer is 33 3 3 _ _

8 8 = 64 (square second digit).

The last part of the answer is 64 _ _ 6 4

So 58 58 = 3364.

See the pattern?

For 53 53, multiply 5 5 = 25 (square the first digit).

25 + 3 = 28 (25 plus second digit).

The first part of the answer is 28 2 8 _ _

3 3 = 9 (square second digit).

The last part of the answer is 09 _ _ 0 9

So 53 53 = 2809.

cheers,

Prashanth

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Take a 2-digit number beginning with 1.

Square the second digit

(keep the carry) _ _ X

Multiply the second digit by 2 and

add the carry (keep the carry) _ X _

The first digit is one

(plus the carry) X _ _

Example:

If the number is 16, square the second digit:

6 6 = 36 _ _ 6

Multiply the second digit by 2 and

add the carry: 2 6 + 3 = 15 _ 5 _

The first digit is one plus the carry:

1 + 1 = 2 2 _ _

So 16 16 = 256.

See the pattern?

For 19 19, square the second digit:

9 9 = 81 _ _ 1

Multiply the second digit by 2 and

add the carry: 2 9 + 8 = 26 _ 6 _

The first digit is one plus the carry:

1 + 2 = 3 3 _ _

So 19 19 = 361.

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I came across these formulae in some Book.Anyways I dont find it of much help.They end up confusing you most of the times.After going round and round u end up doing the same S*** multiplication.

Multiplication of 2 or 3 digit nos. is ok.It helps in case u r in deep S*** when it comes to time.

Also read Anils thread on Trachtenberg System of Speed Arithmetic

Anyways Thanks and keep up the good work

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one more to go**MULTIPLICATION FROM 13 TO 19: **

The reason why the rule is different for multiplication by 11 and by 12 is obviously because the right digits are different. The right digit, we could call the Parent Index Number (PIN). Thus in 11, the PIN is 1 and in 12 it is 2. (In 13, it is 3; in 14, it is 4 etc.) When the PIN is 1, we are simply taking each figure of the multiplicand (we could call this figure the Parent Figure PF in short ) as such and adding the right neighbour. When the PIN is 2, we are doubling the PF and then adding the right neighbour.

Obviously, if the PIN is 3 (as in 3), we would treble the PF and then add the right neighbour. If the PIN is 4 (as in 14), we would quadruple (i.e. multiply by 4) the PF and then add the right neighbour. If PIN is 9 (as in 19), we would multiply the PF by 9 and then add the right neighbour, What is the advantage of the method? We need know the tables only upto 9 and still multiply by a simple process of addition.

(1) 39942 X 13 = ? (2) 43285 X 14 = ?

2331 21132

039942 X 13 043285 X 14

519246 605990

(3) 58265 X 15 = ? (4) 36987 X 16 = ?

34132 25654

058265 X 15 036987 X 16

873975 591792

(5) 69873 X 17 = ? (6) 96325 X 18 = ?

57652 85224

069873 X 17 096325 X 18

1187841 1733850

(7) 74125 X 19 = ?

73124

074125 X 19

1408375

2. Multiplication of two 2 digit numbers

Consider the conventional multiplication of two 2 digit numbers 12 and 23

shown below:

12

X 23

36

24

Ans. 276

It is obvious from the above that

(1) the right digit 6 of the answer is the product obtained by the "vertical"

multiplication of the right digit of the multiplicand and of the multiplier.

(2) the left digit 2 of the answer is the product obtained by the "vertical"

multiplication of the left digit of the multiplicand and of the multiplier

(3) the middle digit 7 of the answer is the sum of 3 and 4. The 3 is the product

of the left digit of the multiplicand and the right digit of the multiplier; the 4

is the product of the right digit of the multiplicand and the left digit of the

multiplier. This means that, to obtain the middle digit, one has to multiply

"across" and add the two products (in our example 1 X 3 + 2 X 2)

The working in our above example can therefore be depicted as

1 2

2 3

1 X 2 / 1 X 3 + 2 X 2 / 2 X 3

and can be summarised as

1 2

2 3

2 / 3 + 4 / 6 = 276

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hey guyz,i am loading a few more formulae.

moderators i dont know whether this is redundant stuff .in case if it is i am sorry.

1. Special method for multiplication by numbers from 11 to 19. **Multiplication by 11**

Rule: 1. Prefix a zero to the multiplicand

2. Write down the answer one figure at a time, from right to left as in

any multiplication.

The figures of the answer are obtained by adding to each successive

digit of the multiplicand. its right neighbour. Remember the right

neighbour is the right, (i.e., the correct) neighbour to be added.

(1) 123 X 11 = ?

Step1: Prefix a zero to the multiplicand so that it reads 0123.

Step2: To the right digit 3, add its right neighbour.

There is no neighbour on the right; so add 0. 0123 X 11

3 + 0 =3. 3

3

To the next digit 2, add its right neighbour 3. 0123 X 11

2 + 3 = 5. 53

To the next digit 1, add the right neighbour 2. 0123 X 11

1 + 2=3. 353

To 0, add the right neighbour 1. 0123 X 11

0 + 1 = 1. 1353

Therefore, 123 X 11 = 1353 (which you can easily verify by a conventional multiplication).

cheers,

prashanth

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eg: 87 * 94

94 is **6** less than 100

87 is **13** less than 100.

so 1st two digits are **87 - 6 = 79 **

last two digits are 13 * 6 = 78

therefore answer is 7978.

hope it is useful for us in the upcoming CAT.

Cheers,

Prashanth

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