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Friday, August 19, 2016

Most Frequently Asked Wipro Placement Papers Fully Solved.

Most Frequently Asked Wipro Placement Papers Fully Solved. Latest Wipro Placement Papers Download. Wipro Placement Papers PDF To Download. Wipro Sample Placement Papers. 


Question 1

In an A.P., if the sum of first X terms is Y and the sum of first Y terms is X then what will be sum of (X+Y) terms?

a) 2(X+Y) b) (X+Y) c) -(X+Y) d) -(2X+Y)

Answer : c) -(X+Y).

Solution :

We know that, sum of first n-terms of an A.P. with first term a and common difference d is S(n) = n/2 (2a+(n-1)d).
Given that, sum of first X terms = Y
i.e., S(X) = X/2 (2a + (X-1)d) = Y
X(2a + (X-1)d) = 2Y....(1)
And, sum of first Y terms = X
i.e., S(Y) = Y/2 (2a + (Y-1)d) = X
Y(2a + (Y-1)d) = 2X ....(2)
Subtracting, (1) and (2), we get,
2X - 2Y = Y(2a + (Y-1)d) - X(2a + (X-1)d)
2X - 2Y = Y2a + Y(Y-1)d - X2a - X(X-1)d
2(X - Y) = (Y - X)(2a) + (dY2) - Yd - (dX2) + Xd
-2(Y - X) = (Y - X) (2a) - (Y-X)d - d(X2 - Y2)
-2(Y - X) = (Y - X) (2a - d) - d(X2 - Y2)
-2(Y - X) = (Y - X) (2a - d) - d(X-Y)(X+Y)
-2(Y - X) = (Y - X) (2a - d) + d(Y-X)(X+Y)
-2 = (2a-d) + d(X+Y)
-2 = 2a + d(X+Y-1)....(3)
We have to find the sum of X+Y terms.
i.e., S(X+Y) = [(X+Y) / 2] x (2a + (X+Y-1))d
Sub. eqn (3) in above eqn, we have,
S(X+Y) = [(X + Y) / 2] x (-2) = -(X + Y)
Hence, the required answer is -(X + Y).

Question 2

In an A.P., if the sum of first X terms is Z(X2), the sum of first Y terms is Z(Y2) and X is not equal to Y, then which of the following equals the sum of first Z terms of the A.P.?

a) Z3 b) (X + Y)Z2 c) (X - Y)Z2 d) Z2

Answer : a) Z2.

Solution :

Given that, sum of first X terms = Z(X2)
i.e., S(X) = X/2 (2a + (X-1)d) = Z(X2)
(2a + (X-1)d) = 2XZ
2a + Xd - d = 2ZX....(1)
And, sum of first Y terms = Z(Y2)
i.e., S(Y) = Y/2 (2a + (Y-1)d) = Z(Y2)
(2a + (Y-1)d) = 2ZY
2a + Yd - d = 2ZY....(2)
Subtracting (1) and (2), we get,
Xd - Yd = 2ZX - 2ZY
(X - Y)d = 2(X - Y)Z
d = 2Z....(3)
Sub. d value in eqn (1), we have
(2a + (X - 1)2Z) = 2XZ
2a + 2XZ - 2Z = 2XZ
2a = 2Z
a = Z...(4)
We have to find the sum of first Z terms, i.e., S(Z) = Z/2 (2a + (Z-1)d).
Substitute a and d values in above eqn,
S(Z) = Z/2 (2Z + (Z-1) 2Z) = Z(Z + (Z-1)Z) = Z2 + Z3 - Z2 = Z3.

Question 3

If S1, S2 and S3 are sums of first n terms of three A.P's with common difference 1, 2, and 3 respectively and the first term of each A.P. is 1, then which of the following relation is true?

(i) S3 - S1 = n(n+1)
(ii) 2(S1 + S2) = n(3n-1)
(iii) S1 + S3 = 2n2

a) (i)&(ii) only b) (ii)&(iii) only c) (iii) only d)(i),(ii)&(iii).

Answer : c) (iii) only

Solution :

Given that, the first term of each A.P. is 1 and the common difference are 1, 2 and 3.
Since S1 is the sum of first n terms of the first A.P., S1 = n/2 (2a + (n-1)d) = n/2 (2x1 + (n-1)1)
S1 = n/2 (2 + (n-1)1)
S1 = n/2 (n+1)....(1)
Since S2 is the sum of first n terms of the second A.P., S2 = n/2 (2a + (n-1)d) = n/2 (2x1 + (n-1)2)
S2 = n/2 (2 + (n-1)2) = n/2 (2n)
S2 = n2....(2)
Since S3 is the sum of first n terms of the third A.P., S3 = n/2 (2a + (n-1)d) = n/2 (2x1 + (n-1)3)
S3 = n/2 (3n-1)....(3)
Now, we have to find the given relations S3 - S1, 2(S1 + S2) and S1 + S3.
S3 - S1 = n/2 (3n-1) - n/2 (n+1)
= n/2 [3n - 1 - n - 1]
= n/2 (2n-2) = n(n-1)
Therefore, (i) is not true.
S1 + S2 = n/2 (n+1)+ n2 = [n (n+1) + 2n2]/2 = n(3n+1)/2
2(S1 + S2) = n (3n + 1)
Therefore, (ii) is not true.
S1 + S3 = n/2 (3n-1) + n/2 (n+1) = n/2 (3n - 1 + n + 1) = 2n2
Therefore, (iii) is true.
Hence, the answer is option c.

Question 4

If 85 + 17 = 51, 76 + 19 = 42, 91 + 13 = 73 then 120 + 15 = ?

a) 68 b) 84 c) 96 d) none of these

Answer : b) 84.

Solution :

The ten's place of answers in the given question are obtained as follows:
85 + 17 = 85 / 17 = 5,
76 + 19 = 76 / 19 = 4,
91 + 13 = 91 / 13 = 7,
So, 120 + 15 = 120 / 15 = 8
Observing the given question, the unit digit of the answers are 1, 2 and 3 respectively.
So, the unit digit of required answer is 4.
Hence, the required answer is 84.

Question 5

If 9 + 7 + 8 = 2, 8 + 5 + 7 = 6, 7 + 4 + 6 = 2 then 5 + 2 + 5 = ?

a) 4 b) 6 c) 5 d) 7

Answer : c) 5.

Solution :

The general pattern of the given question is a + b + c = d where, a x c = bd. That is,
9 x 8 = 72 which gives 2 as the answer.
8 x 7 = 56 which gives 6 as the answer.
7 x 6 = 42 which gives 2 as the answer.
Then, 5 x 5 = 25 which gives 5 as the answer.

Question 6

If 51 + 23 = 37, 62 + 34 = 32 and 59 + 49 = 27 then 84 + 56 = ?

a) 28 b) 38 c) 39 d) 29.

Answer : a) 28

Solution :

General form is add the given two numbers and divide the results by 2, 3, 4 and 5 respectively, to get the answer.
That is,
51 + 23 = 74 = 74 / 2 = 37
62 + 34 = 96 = 96 / 3 = 32
59 + 49 = 108 = 108 / 4 = 27
And, 84 + 56 = 140 = 140 / 5 = 28.
Hence, the required answer is 28.
Point to note:
If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of r-th kind,such that (p1 + p2 + ... pr) = n, then number of permutations of these n objects is = n!/ (p1!).(p2)!.....(pr!)

Question 7

If the letters of the word SUPPOSE are placed in a row then what is the probability that the vowels always come together?

a) 3/7 b) 4/7 c) 1/7 d) 6/7

Answer : c) 1/7.

Solution :

There are 7 letters in the word SUPPOSE which can be arranged in 7! ways.
Let, S = The letters of the word SUPPOSE be arranged.
Then, n(S) = 7!
Let, E = The letters of the word SUPPOSE be arranged in such a way that the vowels always come together.
From the above word, U, O and E are the vowels. Considering these vowels as one letter, we can arrange the letters in a row in 5! ways.
Also, three vowels can themselves be arranged in 3! ways.
Therefore, the total number of arrangements in which three vowels come together are 5! x 3!.
And, n(E) = 5! x 3!.
Hence, the required probability = p(S) = n(E)/n(S) = 5! x 3! / 7! = 3! / 6 x 7 = 1/7.

Question 8

If the letters of the word STATION are arranged randomly, then the probability that two 'T's do not come together is:

a) 5/7 b) 4/21 c) 2/3 d) 3/5

Answer : a) 5/7.

Solution :

There are 7 letters in the word STATION.
out of the letters in the word STATION two letters T are alike.
Therefore, number of permutations = 7!/2! = 7!/2.
Let, S = The letters of the word STATION be arranged.
Then, n(S) = 7!/2 ....(1)
Let, E = the letters of the word STATION be arranged in such a way that two 'T's do not come together.
Now, we have to find the number of words in which two 'T' are together.
Considering the two T's in the word STATION as one letter, we can arrange 6 letters in 6! ways.
Number of words in which two 'T's are never together = total number of words - number of words in which two 'T' are together
= (7!/2) - 6! = (7! - 2 x 6!)/2 = 6!(7 - 2)/2 = 6! x 5/2.
Therefore, n(E) = 6! x 5/2 ....(2)
Required probability = p(S) = n(E)/n(S) = (6! x 5/2)/(7!/2) = 6! x 5 / 7! = 5/7.

Question 9

Find the probability that in a random arrangement of letters of the word 'IMMEDIATE', A is always next to D.

a) 2/9 b) 3/7 c) 1/3 d) 1/9

Answer : d) 1/9

Solution :

There are 9 letters in the word IMMEDIATE.
Out of the letters in the word IMMEDIATE, two letters I are alike, two letters M are alike and two letters E are alike.
Therefore, number of random arrangement of the letters in the word IMMEDIATE = 9!/(2! x 2! x 2!) = 9!/8.
Let, S = The letters of the word IMMEDIATE be arranged randomly.
Then, n(S) = 9!/8 ....(1)
Let, E = The letters of the word IMMEDIATE be arranged in such a way that A is always next to D.
If A next to D, they can be considered as one and the 8 letters can be arranged in 8!/(2! x 2! x 2!) = 8!/8.
Therefore, n(E) = 8!/8
Now, the required probability = p(S) = n(E)/n(S) = (8!/8)/(9!/8) = 8!/9! = 1/9.
Hence, the required answer is 1/9.
Point to note:
If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of r-th kind,such that (p1 + p2 + ... pr) = n, then number of permutations of these n objects is = n!/ (p1!).(p2)!.....(pr!)

Question 10

If the letters of the word SUPPOSE are placed in a row then what is the probability that the vowels always come together?

a) 3/7 b) 4/7 c) 1/7 d) 6/7

Answer : c) 1/7.

Solution :

There are 7 letters in the word SUPPOSE which can be arranged in 7! ways.
Let, S = The letters of the word SUPPOSE be arranged.
Then, n(S) = 7!
Let, E = The letters of the word SUPPOSE be arranged in such a way that the vowels always come together.
From the above word, U, O and E are the vowels. Considering these vowels as one letter, we can arrange the letters in a row in 5! ways.
Also, three vowels can themselves be arranged in 3! ways.
Therefore, the total number of arrangements in which three vowels come together are 5! x 3!.
And, n(E) = 5! x 3!.
Hence, the required probability = p(S) = n(E)/n(S) = 5! x 3! / 7! = 3! / 6 x 7 = 1/7.

Question 11

If the letters of the word STATION are arranged randomly, then the probability that two 'T's do not come together is:

a) 5/7 b) 4/21 c) 2/3 d) 3/5

Answer : a) 5/7.

Solution :

There are 7 letters in the word STATION.
out of the letters in the word STATION two letters T are alike.
Therefore, number of permutations = 7!/2! = 7!/2.
Let, S = The letters of the word STATION be arranged.
Then, n(S) = 7!/2 ....(1)
Let, E = the letters of the word STATION be arranged in such a way that two 'T's do not come together.
Now, we have to find the number of words in which two 'T' are together.
Considering the two T's in the word STATION as one letter, we can arrange 6 letters in 6! ways.
Number of words in which two 'T's are never together = total number of words - number of words in which two 'T' are together
= (7!/2) - 6! = (7! - 2 x 6!)/2 = 6!(7 - 2)/2 = 6! x 5/2.
Therefore, n(E) = 6! x 5/2 ....(2)
Required probability = p(S) = n(E)/n(S) = (6! x 5/2)/(7!/2) = 6! x 5 / 7! = 5/7.

Question 12

Find the probability that in a random arrangement of letters of the word 'IMMEDIATE', A is always next to D.

a) 2/9 b) 3/7 c) 1/3 d) 1/9

Answer : d) 1/9

Solution :

There are 9 letters in the word IMMEDIATE.
Out of the letters in the word IMMEDIATE, two letters I are alike, two letters M are alike and two letters E are alike.
Therefore, number of random arrangement of the letters in the word IMMEDIATE = 9!/(2! x 2! x 2!) = 9!/8.
Let, S = The letters of the word IMMEDIATE be arranged randomly.
Then, n(S) = 9!/8 ....(1)
Let, E = The letters of the word IMMEDIATE be arranged in such a way that A is always next to D.
If A next to D, they can be considered as one and the 8 letters can be arranged in 8!/(2! x 2! x 2!) = 8!/8.
Therefore, n(E) = 8!/8
Now, the required probability = p(S) = n(E)/n(S) = (8!/8)/(9!/8) = 8!/9! = 1/9.
Hence, the required answer is 1/9.

Question 13

Find the sum of the first 50 common terms of 12,16,20,... and 18,24,30,....

a) 15900 b) 12700 c) 19990 d) 18400

Answer : a) 15900

Solution :

Given series are 12,16,20,... and 18,24,30,....
The terms in 12,16,20,... are multiples of 4
and the terms in 18,24,30,.... are also multiples of 4.
Therefore the common terms are even multiples of 6.
That are, 4x6, 6x6, 8x6,... (multiples of both 4 and 6).
Therefore, we have to find the sum of the terms in the series 24,36,48,...
This is an A.P with a = 24 and d = 12.
Then sum of the first n-terms of A.P = (n/2)[2a + (n-1)d]
Here, n = 50 and required sum = (50/2)[2(24) + 49(12)] = 25[636]= 15900.
Hence the answer is 15900.

Question 14

10, 16, 22,... 10,13,16,... find the sum of the 100 common terms of the given two series.

a) 30100 b) 29900 c) 30700 d) 31100

Answer : c) 30700

Solution :

Given series are 10, 16, 22,... and 10,13,16,...
Just expanding the given series,
10,16,22,28,34,40,... and 10,13, 16, 19, 22, 25, 28, 31,34,37,40,....
Every term in the first lies in the 2nd series.
Every term in 1st is a common term of both series.
Therefore we have to find 10+16+22+28+...(100 terms)
This is an A.P with a = 10, d = 6 and n = 100.
Then required sum = (n/2)[2a+(n-1)d]
(100/2)[2(10)+(99)6] = 50{20+594} = 30700.
Hence the answer is 30700.

Question 15

Find the sum of first 25 common terms of 7, 11, 15,... and 9, 13, 17, 21

a) 1410 b) 1375 c) 1655 d) cannot be determined

Answer : d) cannot be determined

Solution :

Given two series are 7, 11, 15,... and 9, 13, 17, 21...
The common difference of two series is 4.
Just expanding the two series 7, 11, 15,19,23,27,31,35... and 9,13,17,21,25,29,33,37,...
So there is no common term between given two series.
So, we cannot find the sum of the common terms.
Hence the answer is option d.

Question 16

10,15,20,... and 11,15,19,... find the sum of the 100 common terms of given two series.

a) 100100 b) 100500 c) 100300 d) 100900

Answer : b) 100500

Solution :

Given two series are 10,15,20,... and 11,15,19,...
Expanding the given series,we get
10,15,20,25,30,35,40... and 11,15,19,23,27,31,35,39,43,47,51,55,...
Therefore common terms are 15,35,55,...
This is an A.P with a = 15 and d = 20
Then required sum = (n/2)[2a+(n-1)d]
(100/2)[2(15)+(99)20] = 50[30+1980] = 100500.
Hence the answer is 100500.
Note :
To form 3232, you would need 2 3s & 2 2s. 4322 requires a 4, a 3, & 2 2s, but you would not count the numbers again that you had already counted from making 3232.

Question 17

The minimum number of digits required to form every number from 1 to 10,000 is:

a) 60 b) 9 c) 41 d) 10

Answer : c)41

Solution :

Unit place of a number from 1 to 10,000 can be formed using 0 to 9 digits. i.e., 10 digits.
10's place of a number from 1 to 10,000 can be formed using 0 to 9 digits. i.e., 10 digits.
100's place of a number from 1 to 10,000 can be formed using 0 to 9 digits. i.e., 10 digits.
1000's place of a number from 1 to 10,000 can be formed using 0 to 9 digits. i.e., 10 digits.
10,000's place of a number from 1 to 10,000 can be formed using only one digit, 1. i.e., 1 digit.
Therefore, total number of numbers required is 10 + 10 + 10 + 10 + 1 = 41.

Question 18

The minimum number of digits required to form every number which is greater than 900 and less than 9,000 is:

a) 39 b) 72 c) 49 d) 81

Answer : a) 39

Solution :

We have to find the minimum number of numbers required to form every number from 900 to 9,000.
Unit place of a number from 900 to 9,000 can be formed using 0 to 9 digits. i.e., 10 digits.
10's place of a number from 900 to 9,000 can be formed using 0 to 9 digits. i.e., 10 digits.
100's place of a number from 900 to 9,000 can be formed using 0 to 9 digits. i.e., 10 digits.
1000's place of a number from 900 to 9,000 can be formed using 1 to 9 digits. i.e., 9 digits.
Therefore, total number of numbers required is 10 + 10 + 10 + 9 = 39.

Question 19

Find the difference of the minimum number of numbers required to form every number from 1 to 1234 and 1 to 4321.

a) 9 b) 7 c) 4 d) 3

Answer : d) 3

Solution :

1st part :
Unit place of a number from 1 to 1234 can be formed using 0 to 9 digits. i.e., 10 digits.
10's place of a number from 1 to 1234 can be formed using 0 to 9 digits. i.e., 10 digits.
100's place of a number from 1 to 1234 can be formed using 0 to 9 digits. i.e., 10 digits.
1000's place of a number from 1 to 1234 can be formed using only one digit, 1. i.e., 1 digit.
Therefore, total number of numbers required is 10 + 10 + 10 + 1 = 31.
2nd part :
Unit place of a number from 1 to 4321 can be formed using 0 to 9 digits. i.e., 10 digits.
10's place of a number from 1 to 4321 can be formed using 0 to 9 digits. i.e., 10 digits.
100's place of a number from 1 to 4321 can be formed using 0 to 9 digits. i.e., 10 digits.
1000's place of a number from 1 to 4321 can be formed using 1 to 4 digits 1. i.e., 4 digits.
Therefore, total number of numbers required is 10 + 10 + 10 + 4 = 34.
Hence the required difference = 34 - 31 = 3.
Alternately,
The number of numbers needed form 1's,10's and 100's place are same for both 1 to 1234 and 1 to 4321.
And 1000's place can be filled by only 1 in 1 to 1234 and 1 to 4 in 1 to 4321. So 4 digits required.
Therefore, difference 4 - 1 = 3.

Question 20

The minimum number of numbers required to form a number from 9 to 9000 which are multiples of 5 is:

a) 31 b) 41 c) 42 d) 32

Answer : a) 31

Solution :

The multiples of 5 must end with 0 or 5.
Therefore, units place of a number from 9 to 9000 can be filled by only two digits 0 and 5. i.e., 2 digits.
10's place of a number from 9 to 9000 can be formed by 0 to 9 digits. i.e., 10 digits.
100's place of a number from 9 to 9000 can be formed by 0 to 9 digits. i.e., 10 digits.
1000's place of a number from 9 to 9000 can be formed by by 1 to 9 digits 1. i.e., 9 digits.
Therefore, total number of numbers required is 2 + 10 + 10 + 9 = 31.

Question 21

Madhuri went to a shopping mall and she want to go to the down floor through escalator. The length of the stairway in the escalator is 46 steps. If Madhuri takes X steps, then she would reach down floor in 60 seconds and if she takes Y steps, then she would reach in 36 seconds. Find the values of X and Y.

a) 13,17 b) 15,21 c) 19,30 d) 26,34

Answer : d)26,34

Solution :

The stairway have totally 46 steps.
She covers 46 - X steps in 60 seconds.
Then speed = 46 - X / 60 steps/sec (speed = distance/time)
And she covers 46 - Y steps in 36 seconds.
Here, speed = 46 - Y / 36
We have to find the values of X and Y.
Assume that speed in either case is same.
Then we would have,
46 - X / 60 = 46 - Y / 36
46 - X / 5 = 46 - Y / 3
3(46 - X) = 5(46 - Y)
230 - 138 = -3X + 5Y
92 = -3X + 5Y .....(1)
Now, substitute the given options in eqn(1), to find the answer.
If X = 13 and Y = 17 then -3X + 5Y = -3(13) + 5(17) = 46
If X = 15 and Y = 21 then -3X + 5Y = -3(15) + 5(21) = -45 + 105 = 60
If X = 19 and Y = 30 then -3X + 5Y = -3(19) + 5(30) = -57 + 150 = 93
If X = 26 and Y = 34 then -3X + 5Y = -3(26) + 5(34) = -78 + 170 = 92.
Hence, the required answer is option d.

Question 22

In airport, Vikash has to catch the flight so he chooses the escalator to reach the top. The escalator is moving at the constant speed, if he takes 25 steps he would reach the top in 9 seconds. If he takes 7 steps he would reach in 15 seconds. Find the total number of steps in the escalator.

a) 52 b) 42 c) 68 d) 62

Answer : a) 52

Solution :

Let the escalator have n steps
Now, Vikash covers n - 25 steps in 9 seconds and n - 7 steps in 15 seconds.
Given that the speed in either case is same.
Then we would have (n - 25)/9 = (n - 7)/15
(n - 25)/3 = (n - 7)/5
5n - 3n = 125 - 21 = 104.
n = 104/2
n = 52 steps
Hence the required answer is 52.

Question 23

Avinash went to a shopping mall in Bangalore. He reaches top in 13 seconds taking 16 steps. If he takes 36 steps he would reach in 8 seconds. Find the speed of the escalator in steps per second.

a) 3 b) 4 c) 5/7 d) 2/3

Answer : b) 4.

Solution :

Let the number of steps in escalator be n.
Given that, he covers n - 16 steps in 13 seconds and n - 36 steps in 8 seconds.
Then the required speed is either (n - 16)/13 or (n - 36)/8.
Now, we have (n - 16)/13 = (n - 36)/8
8(n - 16) = 13(n - 36)
13n - 8n = 468 - 128
5n = 340
n = 68 steps
Therefore speed, n - 16/13 steps per second = 68 - 16 / 13 = 4
Hence, the answer is 4 steps per second.

Question 24

In a NSS camp, a work can be completed by a group of students in a certain number of days. After 20 days, 2/5 of the students left from the camp and it was found that the remaining student take time as long as before. What will be the time taken to complete the entire work?

a) 50 b) 58 c) 59 d) 52

Answer : a) 50

Solution :

Initially, there were X numbers of students to complete the work in Y days.
we have to find the value of Y.
After 20 days, X students would complete the work in (Y - 20) days.
Then, the work completed by X students in (Y - 20) days = X(Y - 20)
If 2/5th of student left, then the remaining students = X - 2X/5
Therefore, (X - 2X/5) students work for Y days.
Now, the work completed by (X- 2X/5) students in Y days = (X - 2X/5)Y
Then we would have X(Y - 20) = (X - 2X/5)Y
5X(Y - 20) = (5X - 2X)Y
5XY - 100X = 3XY.
2XY - 100X = 0
(Y - 50)= 0
Y = 50.
Hence the required number of days is 50.

Question 25

A team of 200 wagers undertakes building work of a bridge. The total time allocated to build the entire bridge is 20 days. After 10 days since start, 200 more wagers join the team and together the team completes the bridge in required time. If the original team do not get those 200 extra wagers, how many days they would be behind schedule to complete building the bridge.

a) 10 days b) 20 days c) 15 days d) 1 days

Answer : a) 10 days

Solution :

First, 200 wagers worked for 10 days.
Amount of total work completed in 10 days = no of wagers x no of days = 200 x 10
= 2000 wager-days ...(1)
(here the unit of measurement is wager-days. Unit of work can be expressed in the form of ”persons x time”)
After adding 200 more wagers, we have 400 wagers in total for the remaining 10 days.
During the second 10 days (i.e from 11 day till the end of 20th day), amount of total work completed will be = 400x10
= 4000 wager-days ...(2)
For the entire duration of 20 days, total work completed = 2000 + 4000 = 6000 wager-days
6000 wager- days amount of work is required to complete the bridge
If 200 wagers are involved, the number of days they would require to complete the bridge = total wager days / 200 = 6000/200 = 30 days
We know that the addition of extra 200 wagers helped the team to complete in 20 days.
This means, without those 200 additional wagers, the original team would have taken, 30-20 = 10 days more to complete the work.

Question 26

A team P of 20 engineers can complete a task in 32 days. Another team Q of 16 engineers can complete the same task in 30 days. Then the ratio of working capacity of 1 member of P to that of a member of Q is:

a) 3:2 b) 4:3 c) 2:5 d) 3:5

Answer : b) 4:3

Solution :

Given, 20 engineers of P can complete the task in 32 days.
This means, to complete the entire task in a single day, number of engineers required would be (20 x 32) engineers = 640 engineers.
640 engineers can complete the work in 1 day.
Therefore, 1 engineer when working alone will take 640 days to complete the work.
In other words, 1 engineer when working alone can finish 1/640 work in 1 day.
Similarly, 1 engineer from Q team alone can complete 1/ 480 (1/30 x 16) work in 1 day
So, required ratio = 1/640 : 1/480 = 64:48 = 4:3.

Question 27

The price of Machine D equals the sum of the prices of machine A, B and C whose price are in the ratio 2:3:4 respectively. If weights of A, B, C and D varies as square of its individual price and difference of weight of D and A, B and C together is 9880 kg. Then what is the weight of D?

a) 15390 kg b) 14790 kg c) 15800 kg d) none of these

Answer : a) 15390 kg

Solution :

Note that,"We say that x varies as y, if x = ky for some constant k"
Given ratio of prices of A, B and C = 2:3:4
Let their price be 2X, 3X and 4X respectively.
Then D's price = 2X + 3X + 4X = 9X
Weight varies as square of price.
Then A's weight = k 4X2
B's weight = k 9X2
C's weight = k 16X2
Sum of their weight = k 29X2
And D's weight = k 81X2
Therefore, k81X2 - k29X2 = 9880 kg
k52X2 = 9880
KX2 = 190
Hence, D's weight = 81kX^2 = 81 x 190 = 15390 kg.

Question 28

If 6:4:3 is the ratio of volumes of three boxes which are filled by mixtures of two variety rice. The mixture consists type 1 and type 2 rice in the ratio 5:3, 4:3 and 3:1 respectively. The contents of all three boxes are transferred into a new box. The ratio of type 1 and type 2 rice in the new box is:

a) 52:19 b) 43:21 c) 51:30 d) 58:33

Answer : d) 58:33

Solution :

Three boxes contains 6X, 4X and 3X kg of mixtures respectively.
1st box contains mixture in the ratio 5:3.
Type 1 in 1st mixture = 6X x 5/8 = 15X/4 kg.
And type 2 in 1st mixture = 6X - 15X/4 = 9X/4 kg
2nd box contains mixture in the ratio 4:3.
Type 1 in 2nd mixture = 4X x 4/7 = 16X/7 kg.
And type 2 in 2nd mixture = 4x - 16X/7 = 12X/7 kg
3rd box contains mixture in the ratio 3:1.
Type 1 in 3rd mixture = 3X x 3/4 = 9X/4 kg.
And type 2 in 3rd mixture = 3X - 9X/4 = 3X/4 kg
Totally, type 1 in final mixture = 15X/4 + 16X/7 + 9X/4 = 232X/28 = 58X/7
Totally, type 2 in final mixture = 9X/4 + 12X/7 + 3X/4 = 132X/28 = 33X/7
Required ratio = 58X/7 : 33X/7 = 58:33

Question 29

If 4:6 is the ratio of number of girls and boys in a computer coaching class.If 50% of girls and 40% of boys are degree holders then the percentage of the candidates who are non-degree holders is:

a) 12% b) 38% c) 56% d) 49%

Answer : c) 56%

Solution :

Given ratio of girls and boys = 4:6 = 2:3
Let boys = 3X and girls = 2X.
Number of girls who are non-degree holders = (100 - 50)% of 2X = 50% of 2X
Number of boys who are non-degree holders = (100 - 40)% of 2X = 60% of 3X.
Total non-degree holders = 50% of 2X + 60% of 3X.
= X + 9X/5 = 14X/5
Required percentage = [(14X/5) / 5X] x 100 = 14X/5 x 1/5X x 100 = 14 X 4 = 56%.

Question 30

The time showed by an analog clock at a moment is 11 am then 1234567890 hours later it will show the time as:

a)11am b)11pm c)5am d)4pm

Answer : c)5am

Solution :
We know a day has 24 hours.
So 1234567890 hours can be expressed as,
1234567890 = 24(51440328) + 18 (i.e., divide the given hour(1234567890) by 24)
1234567890 hours = 51440328 days + 18 hours
Now the clock shows 11am, after 1234567890 hours it will show 11 am + 18 hours
11am + 12 hours + 6 hours
11pm + 6 hours
5 am
Hence the required time is 5 am

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