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Saturday, September 5, 2015

UPSC Maths Challenge : A Problem A Day! (Day 25) - 5 Number Based Quantitative Aptitude Questions

All the questions solved here are relevant to Quantitative Aptitude questions such as those asked in examinations conducted by UPSC. Read on to find out the complete solution to this problem.





Today's Problem(05/09/2015)


Questions Credit : http://careerbless.com



1.
{(481 + 426)2 - 4 x 481 x 426} = ?
A. 3025
B. 4200
C. 3060
D. 3210

2.
(64 - 12)2 + 4 x 64 x 12 = ?
A. 5246
B. 4406
C. 5126
D. 5776

3. 
121 x 5^4= ?
A. 68225
B. 75625
C. 72325
D. 71225

4.
If (2^32 + 1) is completely divisible by a whole number, which of the following numbers is completely divisible by this number?

A. (2^96 + 1)
B. (7 x 2^23 )
C. (2^16 - 1)
D. (2^16 + 1)

5.
What is the unit digit in (6324)^1797× (615)^316× (341)^476?
A. 1
B. 2
C. 4
D. 0


Solutions

1.The numbers are in the format (a + b)^2 - 4ab = a^2 + b^2 -2ab = (a - b)^2
Therefore,{(481 + 426)2 - 4 x 481 x 426} = (481 - 426)^2 = (55)^2 = 3025

2.The numbers are in the format (a -b)^2 + 4ab = a^2 + b^2 + 2ab = (a + b)^2
Therefore,(64 - 12)2 + 4 x 64 x 12 = (64 + 12)^2 = 5776
D.5776

B.75625

4. (2^96 + 1) = (2)^32*3 + (1)^3 = (2^32 + 1)(2^64 - 2^32 + 1),which is divisible by (2^32 + 1).
A.(2^96 + 1)

5. Just keep an eye on the last digits of each number. If you multiply odd numbers of 4,you will get 4 as the one's digit whereas multiplying even numbers of 4 will give you 6 as one's digit.
In the first place,we have  (6324)^1797,now 1797 is an odd number,so the one's digit you will get is 4.
No matter how many times you multiply 5,you will get only 5 as the one's digit.
Again,no matter how many times you multiply 1,you will get 1 as the one's digit.
So,now you multiply 4x5x1 = 20,where the unit digit (or one's digit) is 0.
Therefore,the correct answer is 0.
D.0
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2 comments:

  1. For the 4th, there is an error in the answers. It gave me trouble.

    In the Fifth, it's much simpler.
    615 is a multiple of 5. So 615^316 is a multiple of 5^316. The same deduction gives that (6324)^1797 is a multiple of 2^1797 with is a multiple of 2^316.
    Hence, the big product is a multiple of 10^316, and there is 316 zeros!

    ReplyDelete
    Replies
    1. There was error in the question itself.I missed the exponents.I have corrected it now.

      And yes,this solution is definitely cooler! :)

      Delete

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