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Written Math For Bank Exam

Two math problems with detailed answer
Course

Statistics and Probability (MAT 3103)

45 Documents
Students shared 45 documents in this course
Academic year: 2018/2019
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Question - 1: A machine X can print one lakh books in 8 hours, machine Y can print the same number of books in 10 hours while machine Z can print them in 12 hours. All the machines are started at 9 A. while machine X is closed at 11 A. and the remaining two machines complete work. Approximately at what time will the work be completed? Solution : X needs 8 days Y needs 10 days Z needs 12 days Total work be 120 units [LCM of 8,10,12] Efficiency of X = Total work / No. of days = 120/ = 15 units X can do 15 units per day Efficiency of Y = 120/ = 12 units Y can do 12 units per day Efficiency of Z = 120/ = 10 units Z can do 10 units per day Per day , total work by X+Y+Z = 15+12+ = 37 units From 9 AM to 11 AM , they do 2 hrs together Then , X is closed. Finally Y+Z do the remaining task. In 2 days , They do = 37 × 2 = 74 units Remaining Task = 120 - 74 = 46 units Which is done by Y+Z. Combined efficiency of Y+X = 12+10 = 22 units So , Time = Remaining work / Efficiency = 46/22 = 2 hrs So , The work completed at = 11 + 2 = 1 P. Answer: 1 P.

Question - 2: Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is: Solution : Work Anatomy : 6 [ A + B + C ] = 2 [ A + B + C ] + 7 [ A + B ] => 6A + 6B + 6C = 2A + 2B + 2C + 7A + 7B => - 3A - 3B = - 4C => - 3 [ A + B ] = - 4C => [ A + B ] : C = 4 : 3 [ Efficiency ] Applying MDH Method : M₁D₁E₁H₁ / W₁ = M₂D₂E₂H₂ / W₂ => [ 4 + 3 ] × 6 = 3 × D => D = 14 days Answer : 14 days

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Written Math For Bank Exam

Course: Statistics and Probability (MAT 3103)

45 Documents
Students shared 45 documents in this course
Was this document helpful?
Question - 1: A machine X can print one lakh books in 8 hours, machine Y can print the same number of
books in 10 hours while machine Z can print them in 12 hours. All the machines are started at 9 A.M.
while machine X is closed at 11 A.M. and the remaining two machines complete work. Approximately at
what time will the work be completed ?
Solution : X needs 8 days
Y needs 10 days
Z needs 12 days
Total work be 120 units [LCM of 8,10,12]
Efficiency of X = Total work / No. of days
= 120/8
= 15 units
X can do 15 units per day
Efficiency of Y = 120/10
= 12 units
Y can do 12 units per day
Efficiency of Z = 120/12
= 10 units
Z can do 10 units per day
Per day , total work by X+Y+Z = 15+12+10
= 37 units
From 9 AM to 11 AM , they do 2 hrs together
Then , X is closed. Finally Y+Z do the remaining task.
In 2 days , They do = 37 × 2 = 74 units
Remaining Task = 120 - 74 = 46 units
Which is done by Y+Z .
Combined efficiency of Y+X = 12+10 = 22 units
So , Time = Remaining work / Efficiency
= 46/22 = 2.09 hrs
So , The work completed at = 11 + 2.09 = 1.09 P.M.
Answer: 1.09 P.M.