Wipro online test questions
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In an election contested by two parties, Party D secured 12% of the total votes more than Party R. If party R got 132,000 votes, by how many votes did it lose the election?
A46000
B33000
C16000
D36000
Answer: Option D
Explanation:Let the percentage of the total votes secured by Party D be x%
Then the percentage of total votes secured by Party R = (x – 12)%
As there are only two parties contesting in the election, the sum total of the votes secured by the
two parties should total up to 100%
i.e., x + x – 12 = 100
2x – 12 = 100
or 2x = 112 or x = 56%.
If Party D got 56% of the votes, then Party got (56 – 12) = 44% of the total votes.
44% of the total votes = 132,000 i.e. 44/100 * T = 132,000
T = (132000 * 100)/44 = 300000 votes.
The margin by which Party R lost the election = 12% of the total votes
= 12% of 300000 = 36000.
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332 / 362
The G.C.D. of 1.08, 0.36 and 0.9 is:
A0.03
B0.9
C0.18
D0.108
Answer: Option C
Explanation:Given numbers are 1.08 , 0.36 and 0.90
H.C.F of 108, 36 and 90 is 18 ( G.C.D is nothing but H.C.F)
So H.C.F of given numbers = 0.18
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Structure can be used
ATo hold different datatypes
BHave pointers to structure
CTo assign to one another
DAll the above
Answer: Option D
Explanation:Here is no explanation for this answer
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334 / 362
In which of the following cases dynamic arrays are not preferred?
AIf the array holds less number of elements
BIf the size of the array is unknown
CIf the size of the array changes after few iterations
DIf the memory reallocation takes more time i.e. expensive
Answer: Option A
Explanation:Dynamic arrays are preferred when the size of the array is unknown during memory allocation or the size changes after few iterations or the memory reallocation is expensive. If array holds less number of elements, the physical size is reduced and reduction takes more time. In that case, we can use normal arrays instead of dynamic arrays.
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335 / 362
<b>I.</b> public void store(int row_index, int col_index, Object val)
{
if (row_index < 0 && row_index > N)
{
System.out.println("column index out of bounds");
return;
}
if (col_index < 0 && col+index > N)
{
System.out.println("row index out of bounds");
return;
}
sparse_array[row_index].store(col_index, val);
}
<b>II.</b> public void store(int row_index, int col_index, Object val)
{
if (row_index < 0 || row_index > N)
{
System.out.println("row index out of bounds");
return;
}
if (col_index < 0 || col+index > N)
{
System.out.println("column index out of bounds");
return;
}
sparse_array[row_index].store(col_index, val);
}
<b>III.</b> public void store(int row_index, int col_index, Object val)
{
if (row_index < 0 || row_index > N)
{
System.out.println("column index out of bounds");
return;
}
if (col_index < 0 || col+index > N)
{
System.out.println("row index out of bounds");
return;
}
sparse_array[row_index].store(col_index, val);
}
<b>IV.</b> public void store(int row_index, int col_index, Object val)
{
if (row_index < 0 && row_index > N)
{
System.out.println("row index out of bounds");
return;
}
if (col_index < 0 && col+index > N)
{
System.out.println("column index out of bounds");
return;
}
sparse_array[row_index].store(col_index, val);
}
AI
BII
CIII
DIV
Answer: Option B
Explanation:Each row in a sparse matrix acts as a sparse array, hence this row with the specified col_index is the array and the specified position where the element is stored.
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336 / 362
What is the time complexity of the code that uses self balancing BST for determining the number of inversions in an array?
AO(n log n)
BO(n2)
CO(n)
DO(log n)
Answer: Option A
Explanation:When a self balancing BST like an AVL tree is used to calculate the number of inversions in an array then the time complexity is O(n log n) as AVL insert takes O(log n) time.
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Which of the following is the predefined function for array reversal in javascript?
Arev()
Breverse()
Carr_reverse()
Darray_reverse()
Answer: Option B
Explanation:The predefined function for reversing an array is reverse() in javascript. It does not requires any argument.
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#include <bits/stdc++.h>
using namespace std;
int min(int x, int y)
{ return (x < y)? x: y; }
int func(int arr[], int n)
{
int *jump = new int[n];
int i, j;
if (n == 0 || arr[0] == 0)
return INT_MAX;
jump[0] = 0;
for (i = 1; i < n; i++)
{
jump[i] = INT_MAX;
for (j = 0; j < i; j++)
{
if (i <= j + arr[j] && jumps[j] != INT_MAX)
{
jump[i] = min(jump[i], jump[j] + 1);
break;
}
}
}
return jump[n-1];
}
int main()
{
int arr[] = {1, 3, 6, 1, 9,7};
int size = sizeof(arr)/sizeof(int);
cout<< func(arr,size);
return 0;
}
Aerror
B1
C2
D3
Answer: Option D
Explanation:The given code finds the minimum number of steps required to reach the end of the array by using dynamic programming. So the output will be 3.
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Reference bit is used for
AImplementing LRU page replacement algorithm
BImplementing NRU algorithm
CTo check the page table entry in the cache memory
DNone of Above
Answer: Option B
Explanation:Here is no explanation for this answer
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340 / 362
What profit percentage is obtained by selling an article at certain price if by selling at 2/3 of that price there is a loss of 16%?
A24%
B26%
C20%
D22%
Answer: Option B
Explanation:Let a certain selling price = x
so, 2/3 of that price = 2x/3
in 16% loss, selling price 84 then buying price 100
selling price 1 then buying price 100/84
selling price 2x/3 then buying price = 100*2x/252 = 0.79365x
therefore, profit = x - 0.79365x = 0.20635x
required(%) = 0.20635x*100/0.79365x=26
Therefore, profit is 26%
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