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(49^2 - 35^2)/14 =

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(49^2 - 35^2)/14 =  [#permalink]

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New post Updated on: 10 Dec 2017, 19:52
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\(\frac{(49^2 - 35^2)}{14} =\)

A. 74
B. 76
C. 78
D. 79
E. 84

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Originally posted by adkikani on 10 Dec 2017, 17:51.
Last edited by Bunuel on 10 Dec 2017, 19:52, edited 1 time in total.
Renamed the topic, edited the question
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(49^2 - 35^2)/14 =  [#permalink]

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New post 10 Dec 2017, 18:03
1
adkikani wrote:
\((49^2 - 35^2)\) /14 =

A. 74
B. 76
C. 78
D. 79
E. 84



Ofcourse it is \(a^2-b^2=(a-b)(a+b)\)..
Here (49-35)(49+35)/14=14*84/14=84..

Do your calculations or observer the properties of number involved, of the calculations are difficult, and answer in a sec, ...

Both 49 and 35 have a 7 in them..
So we will get 7*7 in the product, one 7 from addition and other from subtraction..
One 7 gets cancelled out with 7 in 14..
Our ANSWER should be a MULTIPLE of 7..
Only E left
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Re: (49^2 - 35^2)/14 =  [#permalink]

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New post 10 Dec 2017, 19:56
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1
adkikani wrote:
\(\frac{(49^2 - 35^2)}{14} =\)

A. 74
B. 76
C. 78
D. 79
E. 84


(49² - 35²)/14 = (49 + 35)(49 - 35)/14
= (84)(14)/14
= 84
= E

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(49^2 - 35^2)/14 =  [#permalink]

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New post 10 Dec 2017, 21:33
adkikani wrote:
\(\frac{(49^2 - 35^2)}{14} =\)

A. 74
B. 76
C. 78
D. 79
E. 84

\(\frac{(49^2 - 35^2)}{14}=?\)

Start writing what you know, breaking numbers into pieces: \(49^2 = (49)(49)\), for example. And there are factors of 7 everywhere.

Factor everything, with cancellation in mind if you can, to reduce errors.

\(\frac{(49)(49) - (35)(35)}{(2*7)}=\)

\(\frac{(7*7*7*7) - (7*5*7*5)}{(2*7)}=\)

Notice two 7s in numerator's two terms, hence

\(\frac{49(7*7 - 5*5)}{(2*7)}=\)

\(\frac{49(49 - 25)}{(2*7)}=\frac{49*(24)}{2*7}=\)

\(\frac{(7*7)(2*12)}{2*7}= (7*12) = 84\)

Answer E
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Re: (49^2 - 35^2)/14 =  [#permalink]

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New post 13 Dec 2017, 19:58
adkikani wrote:
\(\frac{(49^2 - 35^2)}{14} =\)

A. 74
B. 76
C. 78
D. 79
E. 84


Responding to a pm:
Quote:
Can this be also solved using unit digits?


I am not sure what you mean. How do you intend to use unit's digit concept here?

You are probably confusing it with using units digit concept for remainders when dividing by 2, 5 or 10. Check it out here:

https://www.veritasprep.com/blog/2015/1 ... questions/
https://www.veritasprep.com/blog/2015/1 ... ns-part-2/
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Re: (49^2 - 35^2)/14 =  [#permalink]

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Re: (49^2 - 35^2)/14 =   [#permalink] 09 Feb 2019, 13:22
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