In a bank, principal increases continuously at the rate of r% pe

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 Multiple Choice QuestionsLong Answer Type

161.

For the differential equation xy dy over dx space equals space left parenthesis straight x plus 2 right parenthesis thin space left parenthesis straight y plus 2 right parenthesis comma find the solution curve passing through the point (1, -1).

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162. Find the equation of a curve passing through the point (0, – 2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
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163. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, – 3). Find the equation of the curve given that it passes through ( 2, 1).
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164. Find the equation of the curve passing through the point open parentheses 0 comma space straight pi over 4 close parentheses whose  differential equation is sin x cos y dx + cos x sin y dy = 0.
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165. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y' = ex sin x.
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166.

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

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167. In a bank, principal increases continuously at the rate of 5% per year. In how many years Rs. 1000 double itself ?
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168. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 double itself in 10 years. 
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169. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs. 100 double itself in 10 years.  (e0·5 = 1·648).


Let P denote the principal at any time t.
From given condition,
            dP over dt space equals space left parenthesis 0.05 right parenthesis thin space straight P
Separating the variables and integrating,
                       integral 1 over straight P dP space equals space 0.05 space integral 1. space dt
therefore            log space straight P space equals space left parenthesis 0.05 right parenthesis space straight t space plus space log space straight c space space space space space space rightwards double arrow space space space space space space log space straight P over straight c space equals space left parenthesis 0.05 right parenthesis space straight t
therefore                straight P space equals space straight c space straight e to the power of 0.05 space straight t end exponent                              ...(1)
Now              straight P space equals space 1000 space space space space space space space space space space space when space space space straight t space equals space 0
therefore space space space space space space 1000 space equals space straight c space straight e to the power of 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight c space equals space 1000
therefore space space space space from space left parenthesis 1 right parenthesis comma space space space straight P space equals space 1000 space straight e to the power of 0.05 space straight t end exponent
When                 straight t space equals space 10 space years comma space then
                          straight P space equals space 1000 space cross times space straight e to the power of 0.05. space 10 end exponent
                               equals space 1000 space cross times straight e to the power of 0.5 end exponent space equals space 1000 space cross times space 1.648 space equals space 1648
therefore space space Rs. space 1000 will become Rs.  1648 after 10 years.
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170.

In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?

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