Multiplying Two Numbers Close to the Same Power of 10 Using Speed Mathematics
Example $1$

Calculate $98×93$

Solution

Select the closest base (a power of $10$). In this case we can select $100$ as a base. Subtract base from these numbers. That is $98-100=-2$ and $93-100=-7.$ So write it as

$98\qquad -2\\93\qquad -7$

Left side of the answer will be the diagonal sum including their signs. That is $98-7=91.$ So $91$ will be the left side of the product. (You can take the other diagonal sum also, that is $93-2=91.$ Always these diagonal sums will be same).

To find the right side, just multiply the differences including their signs. That is $-2×-7=14$

So we got that left side of the product is $91$ and right side is $14.$ So answer is $9114$

Example $2$

Calculate $96×112$

Solution

Use the same method. Let's select $100$ as the base (close to both the numbers). Subtract base from these numbers. That is $96-100=-4$ and $112-100=12.$ So write it as

$96\qquad -4\\112\qquad 12$

Take the diagonal sum to get the left side of the product. That is $96+12=108$ (Or $112-4=108$).

For the right side, find the product of the differences. That is $-4×12=-48.$ Since it is -ve, we need to make it as +ve. For this borrow $1$ from our left side $108.$ This borrowed $1$ becomes $100$ (because base is $100$) and our right side becomes $100+(-48)=52.$

Since we borrowed one from left side, left side becomes $107$

So answer is $10752$

Example $3$

Calculate $103×115$

Solution

Use the same method. Let's select $100$ as the base (close to both the numbers). Subtract base from these numbers. That is $103-100=3$ and $115-100=15.$ So write it as

$103\qquad 3\\115\qquad 15$

Take the diagonal sum to get the left side of the product. That is $103+15=118$ (Or $115+3=118$).

For the right side, find the product of the differences. That is $3×15=45$

So answer is $11845$

Example $4$

Calculate $122×89$

Solution

Use the same method. Let's select $100$ as the base (close to both the numbers). Subtract base from these numbers. That is $122-100=22$ and $89-100=-11.$ So write it as

$122\qquad 22\\89\qquad -11$

Take the diagonal sum to get the left side of the product. That is $122+(-11)=111$ (Or $89+22=111$).

For the right side, find the product of the differences. That is $22×(-11)=-242.$ Since it is -ve, we need to make it as +ve. For this borrow $3$ from our left side $111.$ This borrowed $3$ becomes $300$ (because base is $100$ and $3×100=300$) and our right side becomes $300+(-242)=58$

Since we borrowed $3$ from left side, left side becomes $108$

So answer is $10858$

Example $5$

Calculate $1024×989$

Use the same method. Let's select $1000$ as the base (close to both the numbers). Subtract base from these numbers. That is $1024-1000=24$ and $989-1000=-11.$ So write it as

$1024\qquad 24\\989\qquad -11$

Take the diagonal sum to get the left side of the product. That is $1024+(-11)=1013$ (Or $989+24=1013$).

For the right side, find the product of the differences. That is $24×(-11)=-264.$ Since it is -ve, we need to make it as +ve. For this borrow $1$ from our left side $1013.$ This borrowed $1$ becomes $1000$ (because base is $1000$) and our right side becomes $1000+(-264)=736$

Since we borrowed $1$ from left side, left side becomes $1012$

So our answer is $1012736$

Example $6$

Calculate $997×986$

Use the same method. Let's select $1000$ as the base (close to both the numbers). Subtract base from these numbers. That is $997-1000=-3$ and $986-1000=-14.$ So write it as

$997\qquad-3\\986\qquad -14$

Take the diagonal sum to get the left side of the product. That is $997+(-14)=983$ (Or $986-3=983$).

For the right side, find the product of the differences. That is $(-3)×(-14)=42.$ But here $42$ has only $2$ digits whereas our base $1000$ has three zeros. So write $42$ as $042$

So our answer is $983042$

Comments(46) Sign in (optional)
showing 1-10 of 46 comments,   sorted newest to the oldest
mani
2016-02-01 09:04:18
i can't understand the example 2
107 how comes.
(0) (0) Reply
sam
2016-02-02 20:38:49
Initially we got 108 as LHS.

Later, 1 was borrowed from 108 for making RHS positive.
hence LHS became 107.
(0) (0) Reply
vijay
2015-06-25 16:07:16
405 x 397
(0) (0) Reply
B RAJEN TOPNO
2015-08-01 13:43:43
405 x 397

This can be done using base 100 or 1000. But it will hardly help in solving speedily.

Using base 100

405 :305
397 :297

Left side answer = 702 (405+297 or 397+305)
Right side answer = 90585

Final answer
Right side answer = 85 ( 2 left side of the number 90585)
Left side answer = 1607 (702+ 905)

1607 85

A better method is to apply the formula (x + a)(x –b) = x2 + x(a-b) – ab

405= (400+5)
397= (400-3)

405 × 397 = (400+5)×(400-3)=1600+400(5-3)-15=160000+800-15=160785
(0) (0) Reply
swethashri
2014-09-27 19:20:59
how to know whether to borrow 3 or 1
(0) (0) Reply
rajkumar
2014-12-29 15:21:57
if you multiply  24*(-11) the answer will be -264. to make it as a positive,we need to borrow 3,since it makes 300. then 300-264=36,the answer will be positive.if we borrow 1,it makes 100,then 100-264 will be again negative number
(0) (0) Reply
Jayasri
2016-01-09 07:42:12
Consider Example 5.

In this sum,why we have borrowed 1 rather than 3?

If we have borrowed 3 then it may become 300-264=36.
36 is also +ve number know.Then why we have borrowed 1 ?
(0) (0) Reply
sam
2016-01-11 21:13:27
In example 5, base is 1000

So, borrowed 1 becomes 1000
No need to borrow 3 as it becomes 3000
(0) (0) Reply
ketan
2014-07-02 08:28:23
how we caculate the figure of 557*613??
(0) (0) Reply
B RAJEN TOPNO
2015-08-01 13:07:16
Taking base as 1000 ( Power of 10)

557 : -443 (1000-443)
613 : -387(1000-387)

Left side of the answer will be the diagonal sum including their signs : 170(557-387 or 613-443)

Right side of the answer will be the product of the differences : -443 × -387 = 171441 (First three digits will add up with the left side of the answer)

So the final answer is 341(170+171) 441 (Left side answer Right side answer) ie 341 441

Although this procedure is applicable but it is rather lengthy for this example.
I feel it is more appropriate for nos. which are nearer to 10,100,1000, etc.
(0) (0) Reply
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