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Chapter 18: Problem 745

A student obtain \(75 \%, 80 \%\) and \(85 \%\) in three subjects. If the marks of another subject are added then his average cannot be less then (a) \(60 \%\) (b) \(65 \%\) (c) \(80 \%\) (d) \(90 \%\)

Short Answer

Expert verified
The lowest possible average the student can obtain is \(75\%\) which is achieved when the student scores a minimum of \(60\%\) in the fourth subject. Therefore, the answer is (a) \(60 \%\) in the fourth subject.

Step by step solution

01

Calculate the total percentage of the first three subjects.

To do this, simply add the percentages together: \(75\% + 80\% + 85\% = 240\%\).
02

Consider the first option (60%).

If the student scores a minimum of 60% in the fourth subject, the total percentage would be \(240\% + 60\% = 300\%\). Now, to find the average percentage, divide the total percentage by 4 (number of subjects): \(300\% ÷ 4 = 75\%\).
03

Consider the second option (65%).

If the student scores a minimum of 65% in the fourth subject, the total percentage would be \(240\% + 65\% = 305\%\). Now, to find the average percentage, divide the total percentage by 4 (number of subjects): \(305\% ÷ 4 = 76.25\%\).
04

Consider the third option (80%).

If the student scores a minimum of 80% in the fourth subject, the total percentage would be \(240\% + 80\% = 320\%\). Now, to find the average percentage, divide the total percentage by 4 (number of subjects): \(320\% ÷ 4 = 80\%\).
05

Consider the fourth option (90%).

If the student scores a minimum of 90% in the fourth subject, the total percentage would be \(240\% + 90\% = 330\%\). Now, to find the average percentage, divide the total percentage by 4 (number of subjects): \(330\% ÷ 4 = 82.5\%\).
06

Compare the calculated averages.

It was found that if the student scores a minimum of (a) 60%, the average would be 75%; (b) 65%, the average would be 76.25%; (c) 80%, the average would be 80%; and (d) 90%, the average would be 82.5%. Based on these calculations, the lowest possible average is 75%, which is achieved when the student scores a minimum of 60% in the fourth subject. Answer: (a) \(60 \%\).

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Most popular questions from this chapter

A five digit number is chosen at random. The probability that all digits are distinct and digits at odd places are odd and digits at even places are even is (a) \((1 / 60)\) (b) \((2 / 75)\) (c) \((1 / 50)\) (d) \((1 / 75)\)

Using \(1,2,3,4,5,6\) four digit numbers without repetition of any digit are formed. If one number is taken from these what is the probability that the selected number is divisible by \(4 ?\) (a) \((96 / 6 !)\) (b) \(\left(96 /{ }^{6} \mathrm{P}_{4}\right)\) (c) \(\left(84 /{ }^{6} \mathrm{P}_{4}\right)\) (d) None

For observations \(x_{1}, x_{2}, \ldots \ldots . . x_{n}\). If \(\sum_{i=1}(x i+1)^{2}=9 n\) and \({ }^{n} \sum_{i=1}(x i-1)^{2}=5 n\) then standard deviation of the data is (a) \(\sqrt{3}\) (b) \(\sqrt{5}\) (d) \(\sqrt{2}\) (d) \(\sqrt{10}\)

Variance of \(1,3,5,7 \ldots \ldots \ldots(4 n+1)\) is (a) \([\\{2 n(2 n-1)\\} / 3]\) (c) \((1 / n) \sqrt\left[\left(n^{2}-1\right) / 3\right] 100\) (d) \([\\{4 n(n+1)\\} / 3]\)

The median of first \(\mathrm{n}+3\) natural number is (a) \([(n+4) / 2]\) (b) \([(n-4) / 2]\) (c) \([(n-1) / 2]\) (d) \([(n+3) / 4]\)
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