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Friday, August 19, 2016

Latest TCS Placement Papers Fully Solved.

Latest TCS Placement Papers Fully Solved. Sample TCS Test Paper For Freshers. TCS Off Campus Aptitude Test Paper. Most Frequently Asked Placement Test Questions In TCS Off Campus Interview.



1. The face of the square and an equilateral triangle are equal with 12 inches.Find the quantitative relation of their area.

a) 3:4 b) 2:3 c) 4:sqrt3 d) 2:sqrt3

Answer : c) 4:sqrt3

Solution :

Area of a square with side a = a2
Here, a = 12 inches then a2 = 122
Area of an equilateral triangle with side b = (sqrt3)(b2)/4.
Here, b = 12 inches then (sqrt3)(b2)/4 = (sqrt3)(122)/4
Now, the required relation = Area of the square / Area of the triangle = 122 / [(sqrt3)(122)/4]
= 4 x 122 / (sqrt3)(122) = 4/sqrt3.
Hence, the answer is 4:sqrt3.

2. The area of a triangle and circle are equal.If the radius of the circle is 6 cm and height of the triangle is 7 cm then the length of the base of the triangle is:

a) 32.32 cm b) 28.12 cm c) 19.19 cm d) 27.27 cm

Answer : a) 32.32 cm

Solution :
Radius of the circle = 6 cm.
Area of the circle = (pi)r2 = 22/7 x 6 x 6 cm2.
Height of the triangle = 7 cm
Let the base of the triangle be b cm.
Area of the triangle = (1/2) x b x h = 1/2 x b x 7 cm2.
Since the areas are equal, we can equate as
22/7 x 6 x 6 = 1/2 x b x 7
b = 22 x 6 x 6 x 2 / 7 x 7 = 32.32 cm
Hence, the answer is option a.

3. An equilateral triangle whose face is 4 cm and an isosceles triangle whose bottom is 8 cm have equal areas. What is the length (in cm) of the other side of the isosceles triangle?

a) sqrt(21) b) 17 c) sqrt(19) d) 15

Answer : c) sqrt(19).

Solution :
Side of the equilateral triangle is 4 cm.
Area = (sqrt3)(a2)/4 = sqrt(3)(42) / 4 = 4sqrt(3)...(1)
Base b of the isosceles triangle is 8 cm.
Area = (b/4)sqrt(4a2 - b2) = (8/4)sqrt(4a2 - 82)
= 2 x sqrt(4a2 - 64)....(2)
Since (1) and (2) are equal,we have
4sqrt(3) = 2sqrt(4a2 - 64)
2sqrt(3) = sqrt(4a2 - 64)
Squaring both sides, we get,
4 x 3 = 4a2 - 64
12 + 64 = 4a2
76 = 4a2
a2 = 19
a = sqrt(19).

4. If 1 can of pure milk is to be mixed with 3 cans of water to make coffee then how many six litre can of pure milk are needed to prepare hundred servings of 3 litre of coffee?

a) 12 b) 13 c) 14 d) 15

Answer : b) 13.

Solution :
Let the number of 6 litre cans of pure milk required be X.
Total quantity of pure milk required = 6 x X litres = 6X litres.
Since 1 can of pure milk is mixed with 3 cans of water then number of 6 litre cans of water required is 3X.
And, the total quantity of water required = 3X x 6 litres = 18X litres.
Note that, there are 100 servings of 3 litres.
Therefore, total quantity of coffee required for serving = 100 x 3 litres = 300 litres.
We have 6X + 18X = 300
24X = 300
X = 12.5
i.e., twelve and half cans are required.
Hence, the answer is 13.

5. On his holiday, a man watches N number of musical competitions from 5 different channels in the repeating pattern of A, B, C, D and E. If he takes up with
the channel A and ends with the channel D then N would be:

a) 35 b) 42 c) 59 d) 63.

Answer : c) 59.

Solution :
Given that, A B C D E is the channel pattern and it repeats and ends with the channel D.
The choices to watch channel D comes in the order 4th, 9th, 14th, 19th, and so on.
The choice for ending with channel D can be calculated by subtracting 4 from the number of competitions watched and the resultant value should be divisible by 5.
i.e., 5 must be a divisor of N-4.(N is the number of competitions watched)
From the options,
35 - 4 does not divisible by 5
43 - 4 does not divisible by 5.
59 - 4 is divisible by 5.
63 - 4 also does not divisible by 5.
Thus he had watched 59 musical competitions.

6. A bead seller arranged some beads in multi-layer box. The first layer of the box was square shaped with 4 rows and 6 columns. Each layer was 1 less in each
dimension of the previous layer. What will be the maximum number of beads that could have been in the fourth layer from the first?

a) 3 b) 1 c) 12 d) 24

Answer : a) 3.

Solution :
Given that, the dimension of first layer = 4 rows x 6 columns.
Maximum number of beads in the first layer = 4 x 6 = 24 beads.
Each layer dimension was 1 less compared to the previous layer,i.e., the dimension of second layer = 3 rows x 5 columns.
Maximum number of beads in the second layer = 3 x 5 = 15 beads.
Similarly, the dimension of third layer = 2 rows x 4 columns.
Therefore the maximum number of beads in this layer = 2 x 4 = 8 beads.
And, the dimension of fourth layer = 1 row x 3 columns.
So, the maximum number of beads in this layer = 1 x 3 = 3 beads.
Hence, the required answer is 3.

7. There are 4 machines namely P, Q, R and S in a factory. P and Q running together can finish an order in 10 days. If R works twice as P and S works 1/3 as much as Q then the same order of work can be finished in 6 days. Find the time taken by P alone to complete the same order.

a) 11.5 days b) 12.5 days c) 13.5 days d) 14.5 days

Answer : b) 12.5days

Solution :
Let P's 1 day work be X and Q's 1 day work be Y ...(1)
Given that, the time taken to complete the order by (P+Q) = 10 days.
Then (P+Q)'s 1 day work = 1/10 …(2)
Therefore, from (1) and (2), X+Y = 1/10 …(3)
Suppose, R works twice as P then R's 1 day work 2X.
And, S works 1/3 as much as Q then S's 1 day work be Y/3.
Now, the time taken to complete the order by (R+S) = 6 days
2X+Y/3 = 1/6
12X+2Y = 1 ….(4)
Solving the above two equations(3) and (4), we get X = 4/50.
Thus, P's 1 day work = 4/50.
Hence P alone can complete the entire order of work in 50/4 days = 12.5 days.

8. X and Y individually can complete a task in 30 days and 40 days respectively.If X and Y worked together for 12 days and B alone worked for the remaining part of task. Then how many part of the task is completed by B alone ?

a) 2/7 b) 3/10 c) 8/15 d) 1/3

Answer : b) 3/10.

Solution :
X can complete the task in 30 days.
Then X's 1 day work = 1/30
Y can complete the task in 40 days then Y's 1 day work = 1/40.
Therefore, (X+Y)'s 1 day's work = 1/30 + 1/40 = 7/120.
And (X+Y)'s 12 days work = 12 x 7/120 = 7/10.
Remaining work = 1 - 7/10 = 3/10.
Hence, B alone completes 3/10 of the given task.

9. Two pipes P and Q together can fill a tank in 18 hours.If P alone takes 1/2 of the time of thrice Q's then the time taken by Q alone to fill the tank is:

a)29 hours b)32 hours c)28 hours d)30 hours.

Answer : d)30 hours.

Solution :
Suppose Q alone takes X hours to fill the tank.
Then P takes = 1/2 of 3X = 3X/2 hours.
Now, Q's 1 hours work = 1/X and P's 1 hours work = 2/3X.
Given that, (P+Q) takes = 18 hours.
Then (P+Q)'s 1 hour's work = 1/18.
Therefore, 1/18 = 1/X + 2/3X.
5/3X = 1/18
X = 30.
Hence the answer is 30 hours.

10. A certain company retirement plan has a rule that allows an employee to retire when 100 minus years of employment with the company is 2/3 of employee's age when he retired from the company. Then what is the experience(in years) of an employee joined on his 25th age be eligible to retire under the rule?

a) 40 b) 45 c) 50 d) 65

Answer : c) 50

Solution:
Let X be the required year of experience of the employee.
And let Y be his age when he retired.
Then given condition, 100 - X = 2/3 of Y
That is, 2Y/3 + X = 100
Now, discuss with the given options,
a) If X = 40 then Y becomes 25 + 40 = 65
Then 2/3 (65) + 40 is not equal to 100
b) If X = 45 then Y becomes 25 + 45 = 70
Then 2/3 (70) + 45 is not equal to 100
c) If X = 50 then Y becomes 25 + 50 = 75
Then 2/3 (75) + 50 = 50 + 50 = 100.
Hence the required answer is 50.

11. In what year could an employee joined in 1959 on his 29th age be eligible to retire from the company, which has a retirement rule such as an employee to retire when the difference between employee's age when he hired by the company and years of employment with the company is 5 ?

a) 1983 b) 1999 c) 1985 d) 1992

Answer : a) 1983

Solution :
Given that, the age of an employee when he joined in the company is 29.
Let X years be his experience in the company.
From the given question, we have 29 - X = 5
X = 29 - 5 = 24
That is, he retired after 24 years from 1959.
Therefore, he retired in 1983.

12. In what year could an employee hired in 1995 on his 30th age be eligible to retire from his company with a retirement plan which has a rule that allows an employee to retire when the sum of employee's age and years of employment with the company is 90?

a) 2025 b) 2055 c) 2005 d) 2015

Answer : a) 2025

Solution :
Let X be the employee's age and Y be the years of employment.
Then the condition for retirement is
X + Y = 90 ....(1)
Given that his age is 30 in 1995, that means he was born in 1965, and if Z is the current year
X = Z - 1965,
and Y = Z - 1995
From (1),
Z - 1965 + Z - 1995 = 90
2Z = 3960 + 90
Z = 1980 + 45
Z = 2025
Hence, he will retire in the year 2025.

13. In a mathematical test, Ram was asked to find the average of some numbers. By using the average 51 2/3 of X,Y and Z, he found 6 values X, Y, Z, X+Y, Y+Z and Z+X. Find the average of all the seven values.

a) 87.12 b) 88.57 c) 79.92 d) 78.12

Answer : b) 88.57

Solution :
Given that, average of X, Y, Z = 51 2/3 = 155/3
i.e., X + Y + Z / 3 = 155/3
Then X + Y + Z = 155.
Average of seven values = (X+Y+Z + X + Y + Z + X+Y + Y+Z + Z+X) / 7
= [(X+Y+Z) + (X + Y + Z) + (Y + Z+X) + (X+Y + Z)]/7
= [155 + 155 + 155 + 155 ] / 7
= 620 / 7 = 88.57
Hence the required average is 88.57

14. The average of three numbers is 42. if we add one more number then the average becomes 40 and if we replace the first number by a number which is 3 more than the recently added number the new average becomes 38. Find the First number.

a) 45 b) 42 c) 37 d) 38

Answer : a) 45

Solution :
Let the first three numbers be a, b and c.
And the 4th number (recently added) be d.
Given average of a, b and c = 42.
Then, a + b + c = 42 x 3 = 126 ....(1)
If we add d, then a + b + c + d / 4 = 40
a + b + c + d = 160 .....(2)
From (2)-(1), d = 160 - 126 = 34
Now, let the new number used to replace the 1st number a be e.
Then e = d + 3 = 34 + 3 = 37
Therefore, b + c + d + e / 4 = 38
b + c + d + e = 38 x 4 = 152.
Now, a + b + c + d - (b + c + d + e) = 160 - 152 = 8
a - e = 8
a = e + 8 = 37 + 8 = 45
Hence the required number is 45.

15. The average score of three participants is 55 points. If the average score of 1st two is 50 points and that of the last two be 53 points. Then the score of 2nd participant is:

a) 41 b) 42 c) 43 d) 44

Answer : a) 41

Solution :
Let a, b and c be the score of three participants.
Then we have, a + b + c / 3 = 55
a + b + c = 55 x 3 = 165 ....(1)
And a + b = 50 x 2 = 100 ....(2)
b + c = 53 x 2 = 106 ....(3)
Adding (2) and (3), we get
a + 2b + c = 100 + 106 = 206 ....(4)
Subtracting (1) from (4) we have
b = 206 - 165 = 41.
Hence the required score is 41.

16. A survey was taken among American tourists to find whether they like the Architecture of Indian Temples or Indian Culture. The data collected are,

1) 38 tourists like Architecture of Indian Temples .
2) 20 tourists like Indian Culture.
3) 13 likes both Architecture of Indian Temples and Indian Culture.
4) 8 likes neither Architecture of Indian Temples nor Indian Culture.
Find how many tourists
1) were included in the survey ?
2) like Culture not Architecture ?
3) like Architecture not Culture ?
a) 45, 7, 25 b) 37, 7, 25 c) 53, 7, 25 d) 53, 15, 25

Answer : c) 53, 7, 25.

Solution :

Let n(a) be the number of tourist who likes Architecture of Indian Temples, n(a) = 38.
Let n(c) be the number of tourist who likes Indian Culture, n(c) = 20.
Let n(a & c) be the number of tourist who likes both Architecture and Indian Culture, n(a AND c) = 13.
Remember the formula "n(a OR b) = n(a) + n(b)- n(a AND b)"
Number of tourist who likes either Architecture or Culture = n(a OR c) = n(a) + n(c)- n(a AND c)
= 38 + 20 - 13 = 45
8 tourist doesn't like both Architecture and Culture.
Therefore, total tourist included in the survey is 45 + 8 = 53.
Number of tourist who likes only Architecture = n(a) - n(a AND c) = 38 - 13 = 25
Number of tourist who likes only Culture = n(c) - n(a AND c) = 20 - 13 = 7.
Hence the answer is 53, 7, 25.

17. A survey made among 280 college students highlighted the following facts:

1) 50 students are from village, comes by government bus and takes canteen lunch.
2) 110 students are from village and takes canteen lunch
3) 160 students are from village
4) 90 students comes by government bus and takes canteen lunch
5) 130 students comes by government bus
6) 30 students are from village, comes by government bus but do not take canteen lunch
7) 50 students are not from village, do not come by government bus and do not take canteen lunch
Find how the number of students who take canteen lunch ?

a) 220 b) 120 c) 170 d) none of these

Answer : c) 170

Solution :
Total number of students participated in survey is 280.
Let A be the set of students who are from village.
Let B be the set of students who come to college by government bus.
Let C be the set of students who takes lunch in college canteen.
Given that, 50 students are not from village, do not come by government bus and do not take canteen lunch
Then n(A OR B OR C) = 280 - 50 = 230.
We have to find n(C).
Remember the formula that, n(A OR B OR C) = n(A) + n(B) + n(C) - n(A AND B) - n(B AND C) - n(A AND C) + n(A AND B AND C)
From statement (1), we have, n(A AND B AND C) = 50
From statement (2), n(A AND C) = 110
From statement (3), n(A) = 160
From statement (4), n(B AND C) = 90
From statement (5), n(B) = 130
Note that, n(A AND C), n(A), n(B AND C) and n(B) must contain 50 students which was described in statement 1.
But statement (6) has 30 who are only comes from village and comes to college by government bus.
Therefore we have n(A AND B) = 30 + 50 = 80.
Now, from above formula,
230 = 160 + 130 + n(C) - 80 - 90 - 110 + 50
n(C) = 230 - 160 - 130 + 80 + 90 + 110 - 50 = 170.

18. In a school competition, 24 students were participated in dance, 11 participated in drama, 25 participated in vocal solo song, 7 participated in both dance and drama, 4 participated in both drama and vocal solo song, 12 participated in dance and vocal solo song and 3 students participated in all the three. If totally there were 50 students, then the number of students who participated in none of the three competition is:

a) 10 b) 20 c) 15 d) 25

Answer : a) 10

Solution :
Let n(A) be the number of students who participated in dance, n(A) = 24
Let n(B) be the number of students who participated in drama, n(B) = 11
Let n(C) be the number of students who participated in vocal solo song, n(C) = 25
n(A AND B) = number of students participated in dance and drama = 7
n(B AND C) = number of students participated in vocal solo song and drama = 4
n(A AND C) = number of students participated in dance and vocal solo song = 12
And n(A AND B AND C) = number of students participated in all the three = 3.
n(A OR B OR C) = n(A) + n(B) + n(C) - n(A AND B) - n(B AND C) - n(A AND C) + n(A AND B AND C)
n(A OR B OR C) = 24 + 11 + 25 - 7 - 4 - 12 + 3 = 63 - 23 = 40
Therefore, number of students participated in either dance or drama or vocal solo song is 40.
Total number of students = 50.
Number of students participated in none of the three = 50 - 40 = 10.
Hence the required answer is 10.

19. Let ABCD be a square. Alex wants to draw 5 circles of equal radius 'r' with their centres on BD such that the two extreme circles touch two sides AB, BC and AD, CD of the square respectively and each middle circle touches two circles on either side. Find the ratio of r to that of BD.

a) 2 : 11.828 b) 5 : 19.214 c) 1 : 10.828 d) 1 : 12.515

Answer : c) 1 : 10.828

Solution :
As per the statements in question, Alex would have drawn a diagram as shown below.
circle
Let the radius of each circle be r.
Let the side of the square ABCD be a.
Then Diagonal BD = AC = a x sqrt 2 (You will get this formula by applying Pythagoras theorem)
We have to find this diagonal length as follows:
Given, he had drawn 5 circles on diagonal BD.
After sketching 5 circles according to the question, we see that some gap between corner of square and extreme circles. Now, let us calculate that space:
Draw perpendicular lines from sides AB and BC to extreme circle. Now you get a small square of side r and of diagonal r x sqrt2.
In the below diagram, Here, Y is the centre of left extreme circle, BXYZ is a small square with side r and diagonal r sqrt2.
Similar observation can be made on the other extreme circle as well.
circle
Summing up our observations in the main diagram, we get,
circle
Hence the diagonal of big square = r sqrt2 + r + 6r + r + r sqrt2 = 8r + 2r sqrt2
Then the required ratio = r : 8r + 2r sqrt2 = r : r(8 + 2sqrt2)
We can remove r on both sides as it is common. Therefore, we get,
= 1 : 8+2 sqrt2
= 1 : 8+2(1.414) = 1:10.828
Hence the required ratio is 1 : 10.828.

20. In an exam, students were asked to find the perimeter of a square which contains three circles such that their centres on the diagonal of the square, the middle circle touches two extreme circles on either side and two extreme circles touch two sides of the square and the radius of the three are equal. And at what times the radius equals the perimeter?

a) 19.312 b) 12.312 c) 21.312 d) 17.312

Answer : a) 19.312

Solution:
Let the radius of circle be r.
Let the side of the square be a.
Then Diagonal = a(sqrt2)
By applying a similar logic as that of problem 1, we will get a diagram as follows.
circle
a(sqrt2) = diagonal = r sqrt2 + r + 2r + r + r sqrt2 = 4r + 2r sqrt2
a = (4r + 2r sqrt2) / sqrt2
a = 4r/sqrt2 + 2r sqrt2 / sqrt2
a = (2 x 2r) / sqrt2 + 2r
But we know, 2/sqrt2 = sqrt2. Therefore, above equation becomes,
a = 2r sqrt2 + 2r = 2r(sqrt2+1)
Then perimeter of the square = 4a = 4 x 2r(sqrt2+1) = 8r(sqrt2+1)
= 8r(1.414+1) = 8r(2.414) = r(19.312)
Hence 19.312 time radius equals the perimeter of the square.

21. Find the length of diagonal BD of a square ABCD when 7 circles of radius 2 cm are located in ABCD such that their centres on BD, two extreme circles touch two sides of ABCD and each middle circle touches two circles on either side.

a)32.98cm b)29.65cm c)31.12cm d)28.56cm

Answer : b)29.65cm

Solution :
Given that, ABCD is a square where AB, BC, CD and AD are sides of square and BD, AC are diagonals of the square.
Applying similar logic to that of first and second questions, we will get the below diagram corresponding to this :
circle
The length of the diagonal BD = r sqrt2 + r + 10r + r + r sqrt2
= 2 rsqrt2 + 12r
Given that r = 2 cm
Then BD = 2(2)sqrt2 + 12(2) = 4 sqrt2 + 24 = 4(1.414) + 24 = 29.65 cm.
Hence the required answer is 29.65cm

22. Let N = 80pq2pq (7digit number). If N is exactly divisible by 120 then the sum of the digits in N is equal to:

a)18 b)22 c)24 d)12

Answer : a)18

Solution :
Given N = 80pq2pq is exactly divisible by 120.
Since the last digit of the dividend is 0, the last digit of N also should be 0.
Then obviously, q = 0.
i.e., N = 80p02p0
Now we have to find p.
We know 120 = 3 x 5 x 8 and 3, 5 and 8 are co-primes.
If N is divisible by 120 then it is divisible by 3, 5 and 8.
We know that, "If a number is divisible by 3 then the sum of its digits also divisible by 3"
Then, 8 + 0 + p + 0 + 2 + p + 0 = 10 + 2p
ie., 10 + 2p must be divisible by 3
Then the possible values of p are 1,4,7,10,13 and so on
Since p is a digit, the possible values are 1, 4, 7
Now, N must be divisible by 8 also.
We know that, "If a number is divisible by 8 then its last 3 digits also divisible by 8"
Here 2p0 is a multiple of 8.
Now put all the above possible values of p then we have 2p0 = 210 or 240 or 270
From these, 240 is a multiple of 8.
Then the value of p = 4.
and N = 8040240
Hence the required sum = 8 + 0 + 4 + 0 + 2 + 4 + 0 = 18.

23. If 21pq33pq is a 8-digit number which is divisible by 12 then how many 2 digit numbers as pq are possible?

a)8 b)18 c)12 d)28

Answer : a)8

Solution :
Given 8 digit number is 21pq33pq which is exactly divisible by 12.
we know that the factors of 12 are 3 and 4 and that are co-primes.
Also, if a number is divisible by 12 then it must be divisible by 3 and 4 also.
Then 21pq33pq is divisible by 3 and 4.
We know that, "If a number is divisible by 4 then its last 2 digits are divisible by 4".
Here pq is a multiple of 4.
ie., pq = 04, 08, 12, 16 and so on.
Remember that, "If a number is divisible by 3 then the sum of its digits also divisible by 3".
Now, the sum of digits = 2 + 1 + p + q + 3 + 3 + p + q = 9 + 2(p + q)
The number 9 + 2(p + q) is a multiple of 3 and then 2(p + q) is also a multiple of 3.
Then the possible values of p + q are 3, 6, 9, 12, 15, 18,21 and so on.
The largest possible sum of p and q is 18. (suppose p,q = 9)
Note that pq is a multiple of 4.
If p + q = 3 then (p,q) = (0,3), (1,2), (2,1) and (3,0)
From these, 12 is possible for pq. ie., p = 1 and q = 2
If p + q = 6 then p,q = (0,6),(1,5),(2,4),(3,3),(4,2),(5,1),(6,0)
Then possible of pq = 24, 60.
Similarly if p + q = 9 then pq = 36,72
If p + q = 12 then pq = 84, 48
If p + q = 15 then pq = 96
If p + q = 18 then only possible of p and q is 9. ie., pq = 99 but which is not a multiple of 4.
Then the required two digit numbers are 12,24,60,36,72,84,48 and 96.
Hence the number of possible two digit number is 8.

24. If 14p0p0p4 which is a 8 digit number and is divisible by 12 then the number of possible values of p is:

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

Answer : b)5

Solution :
If 14p0p0p4 is divisible by 12 then it should also be divisible by 3 and 4.
If 14p0p0p4 is divisible by 3 then the sum of its digits 1 + 4 + p + p + p + 4 is divisible by 3
ie., 9 + 3p is divisible by 3.
This is true for p = 0,1,2,...,9 ...(1)
Now, if 14p0p0p4 is divisible by 4 then the last two digit p4 is divisible by 4.
Now the possible value of p = 0, 2, 4, 6 and 8 ...(2)
From (1) and (2), the required number of possible values of p is 5 and possible 8-digit numbers are 14000004, 14202024, 14404044, 14606064 and 14808084.
Hence the answer is 5.

25. If 53p26p3 is a 7 digit number divisible by 9 and if 757qp is divisible by 8 then the minimum value of p + q is:

a)4 b)8 c)12 c)16

Answer : a)4

Solution :
Given that, 53p26p3 is a 7 digit number divisible by 9.
We know that, if a number is divisible by 9 then the sum of digits is also divisible by 9.
Then 5 + 3 + p + 2 + 6 + p + 3 = 19 + 2p is divisible by 9
If p = 1 then 19 + 2p = 21 which is not divisible by 9.
If p = 2 then 19 + 2p = 24 which is not divisible by 9.
Proceeding like this,
we would have 19 + 2p divisible by 9 when p = 4
Given that, the 2nd number 757qp = 757q4 is divisible by 8.
Then the last 3 digits 7q4 is divisible by 8.
Then the possible values are 704, 714, 724, 734, 744, 754, 764, 774, 784, 794.
From these, 704, 744 and 784 are the multiples of 8.
Then the possible value for q are 0,4 and 8.
We have to find the minimum value of p + q
p + q = 4 + 0 = 4, or 4 + 4 = 8 or 4 + 8 = 12
Hence the required answer is 4.

26. Sunil, Subha and Sathish are working in an IT company.If Sunil, Subha and Sathish works together they can complete 216 files in 240 minutes. In 60 minutes, Sathish can complete as many files more than Subha as Subha can complete more than Sunil. In 300 minutes Sathish can complete as many files as Sunil can complete in 420 minutes. Find how many files can each of them complete in 1 hour.

a)12, 20, 22 b)12, 18, 21 c)10, 16, 28 d)15, 18, 21

Answer : d)15, 18, 21

Solution :
Let the number of files completed in 1 hour by Sunil, Subha and Sathish be X, Y and Z respectively.
Then in 1 hour they can complete (X + Y + Z) files ...(1)
Working together they can complete 216 files in 240 minutes (240 minutes = 240/60 hours = 4 hours).
Then in 1 hour they can complete = 216 / 4 = 54 files ...(2)
From (1) and (2), (X + Y + Z) = 54 ...(3)
In 60 minutes or 1 hour, Sathish can complete as many files more than Subha as Subha can complete more than Sunil.
That means Z - Y = Y - X
2Y = X + Z ...(4)
Note that, in 300 minutes or 5 hours Sathish can complete 5Z files and in 420 minutes or 7 hours Sunil can complete 7X files.
In it is given that in 300 minutes Sathish can complete as many files as Sunil can during 420 minutes
i.e., 5Z = 7X , X = 5Z/7 ...(5)
Substituting this X value in (3) we get
(5Z/7 + Y + Z) = 54
Multiplying both the left hand side and right hand sides by 7, we get,
5Z + 7Y + 7Z = 7x54 = 378
= 12Z + 7Y = 378 ...(6)
Substituting X value from(5) in (4) we get
And 2Y = 5Z/7 + Z = 12Z/7
= 14Y = 12Z
= 12Z - 14Y = 0 ...(7)
(6)-(7) => 21Y = 378
= Y = 378/21 =18
Substitute Y = 18 in (7) we get,
12Z - 14 X 18 = 0
12Z = 252
Z = 252/12 = 21
Substitute Z = 21 IN (5) we get,
X = 5(21)/7 = 15
Therefore we have found, X = 15, Y = 18 and Z = 21
Hence in 1 hour, Sunil, Subha and Sathish can complete 15, 18 and 21 files respectively.

27. Mr.P, Mr.Q and Mr.R takes a project and they can complete in 36 hours, 54 hours and 72 hours respectively. Unfortunately Mr.P met an accident and left from the project 8 hours before the completion while Mr.Q left 12 hours before the completion. Then for how many hours did Mr.R worked?

a) 24 hours b) 18 hours c) 42 hours d) 36 hours.

Answer : a) 24 hours

Solution :
Assume that the project has to be completed in X hours.
So R has worked for X hours.
Since P has left 8 hours before completion, he worked for X - 8 hours and
Q had left 12 hours before, his participation in work is for X - 12 hours.
Therefore (X - 8) + (X - 12) + X = 1(completion of project)
P can complete a project in 36 hours.
In (X - 8) hours P can complete = (X - 8) / 36
Similarly, Q can complete a project in 54 hours.
In (X - 12) hours P can complete = (X - 12) / 54
In X hours R can complete = X / 72
(X - 8) / 36 + (X - 12) / 54 + X / 72 = 1
6 (X - 8) + 4(X - 12) + 3X = 216
13X = 312
X = 24
Hence the required answer is 24 hours.

28. A project is assigned to P, Q and R. They decided to work together and make some adjustment with them such that P can leave 3 weeks before the completion and R works only for 4 weeks. If P, Q and R can complete in 24, 36 and 48 weeks respectively then the project will complete in:

a)12 weeks b)24 weeks c)13 weeks d)18 weeks

Answer : c)13 weeks

Solution:
P's 1 week work = 1/24
Q's 1 week work = 1/36
R's 1 week work = 1/48
(P + Q + R)'s 1 week work = 1/24 + 1/36 + 1/48 = 13/144
Work done by (P + Q + R) in 4 weeks = 4 x 13 / 144 = 13/36
work done by Q in 3 weeks = 3/36 = 1/12
Remaining work = 1 - (13/36 + 1/12) = 5/9
5/9 work should be done by P and Q.
(P + Q)'s 1 week work = 1/24 + 1/36 = 5/72
5/9 work is done by P and Q in(72/5 x 4/9) = 8 weeks
Hence the total time taken to complete = 4 + 3 + 5 = 15 weeks

29. In a team , the ratio of number of juniors and seniors is 7:8. If 52 juniors and 13 seniors are added, the ratio becomes 11:9.Find the number of juniors initially in the team.

a)91 b)99 c)77 d)71

Answer : a) 91

Solution :
Let X and Y be the initial number of juniors and seniors in the team respectively.
Present ratio of juniors to seniors = 7:8
i.e., X/Y = 7/8
X = 7Y/8 ...(1)
Juniors are increased by 52 numbers,i.e., number of juniors = X + 52
Seniors are increased by 13, then the number of seniors = Y + 13
Given that, X + 52 : Y + 13 = 11:9
X + 52 / Y + 13 = 11/9
9(X + 52) = 11(Y + 13)
9X + 468 = 11Y + 143
9X - 11Y = -325
Sub X = 7Y / 8,
9 x 7Y/8 - 11Y = -325
-25Y / 8 = -325
Y = 104
Then the initial number of juniors = X = 7(104)/8 = 91.
Hence the required answer is 91.

30. Three numbers are in the ratio 2:3:5. If the sum of the new numbers which are formed by increasing 15%, 10% and 20% of old numbers respectively is 9860. Then what will be 2nd number before increasing?

a)8500 b)6875 c)2550 d)none of these

Answer : c) 2550

Solution :
Let A, B and C be the numbers.
Given that they are in the ratio 2:3:5
Then the numbers are A = 2X, B = 3X and C = 5X
A is increased by 15%, then the new number is 2X + 15(2X)/100 = 230X / 100 = 23X / 10
B is increased by 10%,then the new number is 3X + 10(3X)/100 = 330X / 100 = 33X / 10
B is increased by 20%,then the new number is 5X + 20(5X)/100 = 600X / 100 = 6X
Therefore the new numbers are 23X/10, 33X/10 and 6X
Given that, 23X/10 + 33X/10 + 6X = 9860
23X + 33X + 60X = 98600
116X = 98600
X = 98600/116 = 850
Then the required number is 3X = 3(850) = 2550

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