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41. The charges per hour of internet surfing is increased by 25% then find the percentage decrease in the time period of surfing user (a net savy) who can afford only 10% increase in expenditure:
100 × 100 = 10000
X × 125 = 110 [25% increase in rate, user can afford only 10% increase]
X = = 88%
Thus, decrease in time = 12%
Solution:
Time × Rate = total charges 100 × 100 = 10000
X × 125 = 110 [25% increase in rate, user can afford only 10% increase]
X = = 88%
Thus, decrease in time = 12%
42. Ram Lal is a renowned packager of fruits in Varanasi. He packs 70 mangoes or 56 guavas every day working 7 hours a day. His wife also helps him. She packs 30 mangoes or 24 guavas working 6 hours per day. Ram Lal has to pack 3300 mangoes and 2400 guavas with help of his wife. They works alternately, each day 10 hours. His wife started packaging firs day and works every alternate days. Similarly, Ram Lal started his work second day and and worked alternatively till the completion of the work. In how many days the work will finished?
Guavas Packing rate of Ram Lal = = 8 per hour
Mango Packing rate of Ram Lal's Wife = = 5 per hour
Guavas Packing rate of Ram Lal's Wife = = 4 per hour
On comparing,
8 Guavas = 10 Mangoes
1 Guavas = Mangoes
2400 Guavas = = 3000 Mangoes
This means that they need to pack (6300 + 3000) Mangoes
Combined Mango Packing rate of Both = 10 + 5 = 15 per hour
We will take 2 days = 1 time unit.
Working 10 hours a day they will pack 150 Mangoes in 1 time unit
Total time taken to pack 6300 Mangoes = = 42 time unit
So, total time taken = 2 × 42 = 84 days
Solution:
Mango Packing rate of Ram Lal = = 10 per hourGuavas Packing rate of Ram Lal = = 8 per hour
Mango Packing rate of Ram Lal's Wife = = 5 per hour
Guavas Packing rate of Ram Lal's Wife = = 4 per hour
On comparing,
8 Guavas = 10 Mangoes
1 Guavas = Mangoes
2400 Guavas = = 3000 Mangoes
This means that they need to pack (6300 + 3000) Mangoes
Combined Mango Packing rate of Both = 10 + 5 = 15 per hour
We will take 2 days = 1 time unit.
Working 10 hours a day they will pack 150 Mangoes in 1 time unit
Total time taken to pack 6300 Mangoes = = 42 time unit
So, total time taken = 2 × 42 = 84 days
43. A group of men decided to do a job in 4 days. But since 20 men dropped out every day, the job completed at the end of the 7th day. How many men were there at the beginning?
According to the question,
4X = X + (X - 20) + (X - 40) + (X - 60) + (X - 80) + (X - 100) + (X - 120)
⇒ 4X = 7X - 420
⇒ 3X = 420
⇒ X =
⇒ X = 140 men
Solution:
Let X be the initial number of men then,According to the question,
4X = X + (X - 20) + (X - 40) + (X - 60) + (X - 80) + (X - 100) + (X - 120)
⇒ 4X = 7X - 420
⇒ 3X = 420
⇒ X =
⇒ X = 140 men
44. Two persons having different productivity of labour, working together can reap a field in 2 days. If one-third of the field was reaped by the first man and rest by the other one working alternatively took 4 days. How long did it take for the faster person to reap the whole field working alone?
First Person completes work = = 33.33% [In 2 days]
Rest work will be completed by Second man = = 66.66% [In 2 days]
So, efficiency of second person is greater.
Efficiency of second person = = 33.33% per day
Then, Second person will complete whole work in,
= = 3 days.
Solution:
Total efficiency of two persons = 50% [As they complete work in 2 days. ]First Person completes work = = 33.33% [In 2 days]
Rest work will be completed by Second man = = 66.66% [In 2 days]
So, efficiency of second person is greater.
Efficiency of second person = = 33.33% per day
Then, Second person will complete whole work in,
= = 3 days.
45. A group of workers was put on a job. From second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in 55% of the time. How many workers were there in the group?
Now,Using work equivalence method,
X + (X - 1) + (X - 2) + . . . . . + 1 = X × 55% of X
[series is in AP. Sum of AP = {No. of terms (first term + last term)/2}]
X = 10 workers.
Solution:
Let initially X number of workers be there.Now,Using work equivalence method,
X + (X - 1) + (X - 2) + . . . . . + 1 = X × 55% of X
[series is in AP. Sum of AP = {No. of terms (first term + last term)/2}]
X = 10 workers.
46. A single reservoir supplies the petrol to the whole city, while the reservoir is fed by a single pipeline filling the reservoir with the stream of uniform volume. When the reservoir is full and if 40, 000 litres of petrol is used daily, the supply fails in 90 days. If 32, 000 litres of petrol used daily, it fails in 60 days. How much petrol can be used daily without the supply ever failing?
90X + L = 40000 × 90 ----------- (1)
60X + L = 32000 × 60 ----------- (2)
Solving the equation,
X = 56000 litres
Thus, 56000 litres per day can be used without the failure of supply.
Solution:
Let X litres be the per day filling and L litres be the capacity of the reservoir, then90X + L = 40000 × 90 ----------- (1)
60X + L = 32000 × 60 ----------- (2)
Solving the equation,
X = 56000 litres
Thus, 56000 litres per day can be used without the failure of supply.
47. Working together B and C take 50% more number of days than A, B and C together take and A and B working together, take more number of days than A, B and C take together. If A, B and C all have worked together till the completion of the work and B has received Rs. 120 out of total earnings of Rs. 450, then in how many days did A, B and C together complete the whole work?
= 5x : 4x : 6x
Number of days required by A and B = ------ (1)
Number of days required by A, B and C = ------ (2)
Number of days required by A, B and C
=
=
= 4 days
Solution:
Ratio of efficiencies of A, B and C,= 5x : 4x : 6x
Number of days required by A and B = ------ (1)
Number of days required by A, B and C = ------ (2)
Number of days required by A, B and C
=
=
= 4 days
48. Two men and women are entrusted with a task. The second man needs three hours more to cope up with the job than the second man and the woman would need working together. The first man, working alone, would need as much time as second man and the woman working together. The first man working alone, would spend eight hours less than the double period of the time second man would spend working alone. How much time would the two men and the women need to complete the task if they all asked together?
tb - ta = 3 . . . . . . . . . ..(1)
Also, B alone be take = (ta + 3) h
According to the question,
2tb - ta = 8
2 × (ta + 3) - ta = 8 [Using equation (1)]
ta = 2 hours.
Now B and woman together take 2 hours and A also take 2 hours, so time required will be half when all 3 work together. So in 1 hour work would be completed.
Solution:
Difference in times required by the first man (A) and second man (B) = 3 hours. Also, if ta and tb are the respective times, thentb - ta = 3 . . . . . . . . . ..(1)
Also, B alone be take = (ta + 3) h
According to the question,
2tb - ta = 8
2 × (ta + 3) - ta = 8 [Using equation (1)]
ta = 2 hours.
Now B and woman together take 2 hours and A also take 2 hours, so time required will be half when all 3 work together. So in 1 hour work would be completed.
49. At a Tech Pvt Ltd. there are some engineering students employed as trainee engineers, belong to two eminent institutions of India. One group belongs to IIT and another to NIT. Each student of IIT works for 10 hours a day till 60 days and each student of NIT works for 8 hours a day till 80 days on the two same project. The ratio of students of IIT and that of NIT is 4:5 respectively. Students of which institution is slower in work and by how much?
Using work equivalence method,
4 × 10 × 60 × E1 = 5 × 8 × 80 × E2
As, 3E1 = 4E
% less efficient of NITians = = 25%
Thus, each engineer from NIT is 25% less efficient than that of IIT.
Solution:
Let E1 and E2 are the working efficiency of each student of IIT and NIT per hour respectively.Using work equivalence method,
4 × 10 × 60 × E1 = 5 × 8 × 80 × E2
As, 3E1 = 4E
% less efficient of NITians = = 25%
Thus, each engineer from NIT is 25% less efficient than that of IIT.
50. 42 women can do a piece of work in 18 days, How many women would be required do the same work in 21 days.
Now, using Work Equivalence Method:
42 × 18 = X × 21
X = 36.
Number of women required = 36
Solution:
Let X be the number of women required to finish the work in 21 days.Now, using Work Equivalence Method:
42 × 18 = X × 21
X = 36.
Number of women required = 36