Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 29)
29.
107 x 107 + 93 x 93 = ?
Answer: Option
Explanation:
| 107 x 107 + 93 x 93 | = (107)2 + (93)2 |
| = (100 + 7)2 + (100 - 7)2 | |
| = 2 x [(100)2 + 72] [Ref: (a + b)2 + (a - b)2 = 2(a2 + b2)] | |
| = 20098 |
Discussion:
20 comments Page 2 of 2.
Sia said:
7 years ago
@Jonavin, @Hari.
We can't use (a+b) (a-b) because there is (a^2+b^2).
Instead of (a^2-b^2).
We can't use (a+b) (a-b) because there is (a^2+b^2).
Instead of (a^2-b^2).
Gayu said:
1 decade ago
107*100+107*7 = 11449.
93*90+93*3 = 8649.
11449+8649 = 20098.
93*90+93*3 = 8649.
11449+8649 = 20098.
Jonavin said:
8 years ago
Why we can't take a+b a-b ?
S.abi said:
9 years ago
107 + 7 = 114.
7^2 = 49.
so 107^2 = 11449.
93^2 = 8649.
93-7 = 86.
7^2= 49.
Then add 111449 + 8649 = 20098.
7^2 = 49.
so 107^2 = 11449.
93^2 = 8649.
93-7 = 86.
7^2= 49.
Then add 111449 + 8649 = 20098.
Hari said:
9 years ago
why can't we take it as (a + b) (a - b) = a^2 + b^2.
(107 + 93) (107 - 93).
Please explain.
(107 + 93) (107 - 93).
Please explain.
Rakshana said:
9 years ago
@Manoj.
Is this applicable for the method of addition and multiplication?
Is this applicable for the method of addition and multiplication?
M S Husain said:
10 years ago
Note--Digital sum is only eliminate method not sure give to correct answer it's sure cancel all wrong answer.
Ajaz 110 and 125.
= (110)^2.
Break 110 as [1!10] = a^2 2*a*b b^2.
I take b as 10 that is 2 digit so we take only two digit if more than 2 carry forward.
1*1 2*1*10 100 == 12100.
Ajaz 110 and 125.
= (110)^2.
Break 110 as [1!10] = a^2 2*a*b b^2.
I take b as 10 that is 2 digit so we take only two digit if more than 2 carry forward.
1*1 2*1*10 100 == 12100.
Ajax said:
1 decade ago
@ Mr. Milind.
Your first method is not working on 110 - 125. Please give us another method.
Your first method is not working on 110 - 125. Please give us another method.
Manoj said:
1 decade ago
We can solve this by using digital sum method:
=(1+0+7)*(1+0+7)+(1+2)*(1+2)
=(8*8)+(3*3)
=64+9
=(6+4)+9
=10+9
=19
=1+9
=10
=1+0
=1
Then check the answers in the same way. If we get 1 for option A it is the answer. For looking it is difficulty but if you practice once its going to simple believe me friends.
=(1+0+7)*(1+0+7)+(1+2)*(1+2)
=(8*8)+(3*3)
=64+9
=(6+4)+9
=10+9
=19
=1+9
=10
=1+0
=1
Then check the answers in the same way. If we get 1 for option A it is the answer. For looking it is difficulty but if you practice once its going to simple believe me friends.
Milind Hanchate said:
1 decade ago
Simplest way.
(107)^2+(97)^2.
To find 107 square.
Take 1 as it is.
After that add 07+07= 14.
Next step is take square of 7 is 49.
Write as 11449.
Note: This method is applicable for number in between 100 to 125.
For number 75 to 100 is,
Subtract 93 from 100.
We get 7. Take square of 7 is 49.
Next step is subtract 7 from 93 we get 86.
Write it as 8649.
Then perform addition.
i.e. 11449+8649 = 20098.
(107)^2+(97)^2.
To find 107 square.
Take 1 as it is.
After that add 07+07= 14.
Next step is take square of 7 is 49.
Write as 11449.
Note: This method is applicable for number in between 100 to 125.
For number 75 to 100 is,
Subtract 93 from 100.
We get 7. Take square of 7 is 49.
Next step is subtract 7 from 93 we get 86.
Write it as 8649.
Then perform addition.
i.e. 11449+8649 = 20098.
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