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Placement Papers with Solutions

(#M40165002) ORACLE QUESTION Percentage Keep an EYE

The length, breadth, and height of a cube are incresed by 10%, 10%, and 20% respectively. The change in the volume of the resultant cuboid is(nearly)_________
a) 40% b) 55% c) 45% d) 50%

Asked In Oracle ravi kumar (7 years ago)

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(#M40165001) ORACLE QUESTION Probablity Keep an EYE

Three dice are thrown together. The probability the three dice show the same number is .............
a) 5/36 b) 35/36 c) 1/6 d) 1/36

Asked In Oracle ravi kumar (7 years ago)

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(#M40165000) ORACLE QUESTION Mixture Keep an EYE

A 20% salt solution of 100 liters is mixed with a quantity of 10% solution. the resulting solution was 12% salt. How much was 10% solution added?
a) 175 lit b) 200 lit c) 400 lit d) 500 lit

Asked In Oracle ravi kumar (7 years ago)

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(#M40164999) ORACLE QUESTION Probablity Keep an EYE

In a class, there are 10 boy children and 13 girl children. What is the probability of selecting two children, as a first girl and then a boy in succession without replacement?
a) 1/130 b) 13/10 c) 120/130 d) 130/506

Asked In Oracle ravi kumar (7 years ago)

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(#M40164998) ORACLE QUESTION Data structure Keep an EYE

Which of the following operations cannot be implemented in O(1) for a unsorted singly linked list with head and tail pointers?
a) deletion of the last node
b) Insertion at the end
c) Deletion of the front node
d) Insertion at front

Asked In Oracle ravi kumar (7 years ago)

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(#M40164997) MATHS QUESTION Keep an EYE

FIND THE GREATEST FOUR DIGIT NUMBER WHICH WHEN DIVIDED BY 6,12,16 LEAVES THE REMAINDER 2 IN EVERY CASE

Asked In Maths AMIT KUMAR MANDAL (7 years ago)

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(#M40164996) ELITMUS QUESTION Permutation and combination Keep an EYE

How many different sums can be formed with the following coins:
5 rupee, 1 rupee, 50 paisa, 25 paisa , 10 paisa,3 paisa, 2 paisa,1 paisa?

Asked In Elitmus Abhay katiyar (8 years ago)

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(#M40164995) MATHS QUESTION Keep an EYE

The one rupee coin is placed on plain paper. How many coins of the same size can be placed round it
so that each one touches the centre and adjacent coins?

Asked In Maths Satish Kumar Sahu (8 years ago)

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(#M40164994) BANK EXAM QUESTION TRICK Keep an EYE

displaystyle{text{Number of pairs * Sum of each pair} = (frac{n}{2})(n+1) = frac{n(n+1)}{2}}
which is the formula above.

Wait — what about an odd number of items?
Ah, I’m glad you brought it up. What if we are adding up the numbers 1 to 9? We don’t have an even number of items to pair up. Many explanations will just give the explanation above and leave it at that. I won’t.

Let’s add the numbers 1 to 9, but instead of starting from 1, let’s count from 0 instead:

0 1 2 3 4
9 8 7 6 5
By counting from 0, we get an “extra item” (10 in total) so we can have an even number of rows. However, our formula will look a bit different.

Notice that each column has a sum of n (not n+1, like before), since 0 and 9 are grouped. And instead of having exactly n items in 2 rows (for n/2 pairs total), we have n + 1 items in 2 rows (for (n + 1)/2 pairs total). If you plug these numbers in you get:

displaystyle{text{Number of pairs * Sum of each pair} = (frac{n + 1}{2})(n) = frac{n(n+1)}{2}}

which is the same formula as before. It always bugged me that the same formula worked for both odd and even numbers – won’t you get a fraction? Yep, you get the same formula, but for different reasons.

Technique 2: Use Two Rows
The above method works, but you handle odd and even numbers differently. Isn’t there a better way? Yes.

Instead of looping the numbers around, let’s write them in two rows:

1 2 3 4 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Notice that we have 10 pairs, and each pair adds up to 10+1.

The total of all the numbers above is

displaystyle{text{Total = pairs * size of each pair} = n(n + 1)}

But we only want the sum of one row, not both. So we divide the formula above by 2 and get:

displaystyle{frac{n(n + 1)}{2}}

Now this is cool (as cool as rows of numbers can be). It works for an odd or even number of items the same!

Asked In Bank Exam KICK (8 years ago)

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(#M40164993) SURYA INFORMATICS SO QUESTION Technical Questions Keep an EYE

1) The default visibility mode while inheriting is?
a) Public
b) Protected
c) Private
d) All of the above

2) Constructor is
a) A class automatically called whenever a new object of this class is created
b) A class automatically called whenever a new object of this class is destroyed
c) A function automatically called whenever a new object of this class is created
d) A function automatically called whenever a new object of this class is destroyed

3) Declaring pointer more than one can cause
a) Trap
b) Abort a program
c) Error
d) None of the above

Asked In Surya Informatics Solutions private Limi Priya (8 years ago)

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