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 Page 1


Question 1 
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. 
Question 2 
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 
Page 2


Question 1 
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. 
Question 2 
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 
Question 3 
In a college during the start of the 1 year course, the number of boys and girls are in the ratio 5 : 
4. After few months some of the students left the college in the ratio 3 : 2. At the end of the year 
the number of boys and girls who have completed the course is 6 and 16 respectively.Find the 
initial number of boys in the class. 
a)50 b)60 c)58 d)72 
Answer : b)60 
Solution : 
Given that boys and girls are in the ratio 5:4 
Let 5X be the number of boys and 4X be the number of girls. 
Ratio of boys and girls who left the course is 3:2. 
Let 3Y and 2Y be the number of boys and girls who left from the course. 
Course has been successfully completed by 6 boys and 16 girls. 
i.e., 5X - 3Y = 6 ....(1) and 
4X - 2Y = 16 ....(2) 
Multiplying eqn 1 by 2 we get, 
10X - 6Y = 12 ....(3) 
Multiplying eqn 2 by 3 we get, 
12X - 6Y = 48 ....(4) 
Subtracting eqn 4 from eqn 3 we get, 
X = 12 and Y = 16 
Initial number of boys is 5X = 60. 
Hence the required answer is 60. 
Question 1 
Page 3


Question 1 
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. 
Question 2 
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 
Question 3 
In a college during the start of the 1 year course, the number of boys and girls are in the ratio 5 : 
4. After few months some of the students left the college in the ratio 3 : 2. At the end of the year 
the number of boys and girls who have completed the course is 6 and 16 respectively.Find the 
initial number of boys in the class. 
a)50 b)60 c)58 d)72 
Answer : b)60 
Solution : 
Given that boys and girls are in the ratio 5:4 
Let 5X be the number of boys and 4X be the number of girls. 
Ratio of boys and girls who left the course is 3:2. 
Let 3Y and 2Y be the number of boys and girls who left from the course. 
Course has been successfully completed by 6 boys and 16 girls. 
i.e., 5X - 3Y = 6 ....(1) and 
4X - 2Y = 16 ....(2) 
Multiplying eqn 1 by 2 we get, 
10X - 6Y = 12 ....(3) 
Multiplying eqn 2 by 3 we get, 
12X - 6Y = 48 ....(4) 
Subtracting eqn 4 from eqn 3 we get, 
X = 12 and Y = 16 
Initial number of boys is 5X = 60. 
Hence the required answer is 60. 
Question 1 
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 question 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) 
Page 4


Question 1 
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. 
Question 2 
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 
Question 3 
In a college during the start of the 1 year course, the number of boys and girls are in the ratio 5 : 
4. After few months some of the students left the college in the ratio 3 : 2. At the end of the year 
the number of boys and girls who have completed the course is 6 and 16 respectively.Find the 
initial number of boys in the class. 
a)50 b)60 c)58 d)72 
Answer : b)60 
Solution : 
Given that boys and girls are in the ratio 5:4 
Let 5X be the number of boys and 4X be the number of girls. 
Ratio of boys and girls who left the course is 3:2. 
Let 3Y and 2Y be the number of boys and girls who left from the course. 
Course has been successfully completed by 6 boys and 16 girls. 
i.e., 5X - 3Y = 6 ....(1) and 
4X - 2Y = 16 ....(2) 
Multiplying eqn 1 by 2 we get, 
10X - 6Y = 12 ....(3) 
Multiplying eqn 2 by 3 we get, 
12X - 6Y = 48 ....(4) 
Subtracting eqn 4 from eqn 3 we get, 
X = 12 and Y = 16 
Initial number of boys is 5X = 60. 
Hence the required answer is 60. 
Question 1 
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 question 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. 
Question 2 
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 
Page 5


Question 1 
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. 
Question 2 
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 
Question 3 
In a college during the start of the 1 year course, the number of boys and girls are in the ratio 5 : 
4. After few months some of the students left the college in the ratio 3 : 2. At the end of the year 
the number of boys and girls who have completed the course is 6 and 16 respectively.Find the 
initial number of boys in the class. 
a)50 b)60 c)58 d)72 
Answer : b)60 
Solution : 
Given that boys and girls are in the ratio 5:4 
Let 5X be the number of boys and 4X be the number of girls. 
Ratio of boys and girls who left the course is 3:2. 
Let 3Y and 2Y be the number of boys and girls who left from the course. 
Course has been successfully completed by 6 boys and 16 girls. 
i.e., 5X - 3Y = 6 ....(1) and 
4X - 2Y = 16 ....(2) 
Multiplying eqn 1 by 2 we get, 
10X - 6Y = 12 ....(3) 
Multiplying eqn 2 by 3 we get, 
12X - 6Y = 48 ....(4) 
Subtracting eqn 4 from eqn 3 we get, 
X = 12 and Y = 16 
Initial number of boys is 5X = 60. 
Hence the required answer is 60. 
Question 1 
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 question 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. 
Question 2 
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. 
Question 3 
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  
 
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