If both side of coin is having Head. So, probability of obtaining a tail is zero.

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Applying successive %age method: -6-10+(60/100) = -15.4% -15.4 -15 +(15*15.4)/100 = 28.09%

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Let the roots be y and y+4. Sum of roots=a and product of root=b. putting these values in option d satisfy the equation.

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When we read the question carefully, it is a problem similar to the one where we are asked to find the least number divisible by given numbers and leaving a common remainder. In such problems we find the LCM.

2 16, 18, 20, 25

2 8, 9, 10, 25

2 4, 9, 5, 25

2 2, 9, 5, 25

3 2, 3, 5, 25

5 2, 3, 1, 5

LCM = 2*2*2*2*2*3*3*5*5 = 3600

Remainder = 4

So, the required number is 3600 + 4 = 3604

Jagdish takes 10 days to complete the work, so we can say that he completes 110th of the work in a day.

Similarly, Narender completes 112th and Sumit completes 115th of the work in a day.

The time taken to complete the work can be calculated in two ways. First method is by adding their individual one day’s work and then taking the reciprocal. Second is the short method (D1∗D2D1+D2), where D1 is the number of days taken by first man and D2 is the number of days taken by second man.

Let us find the combined work of these men taking two at a time.

Jagdish and Narender: D1 = 10 days; D2 = 12 days.

They both will take (10∗1210+12) 12022 or 6011 days.

Sumit and Narender: D1 = 15 days; D2 = 12 days.

They both will take (15∗1215+12) 15027 or 509 days.

Jagdish and Sumit: D1 = 10 days; D2 = 15 days.

They both will take (10∗1510+15) 15025 or 6 days.

Jagdish and Sumit will take 6 days to complete the work while working together.