Answer :
Sure, let's go through each of the questions one by one and provide detailed, step-by-step solutions:
### Question 28:
The median of the original series is 16. Two observations, 6 and 21, are added to the series. What is the new median?
To determine the new median, let's consider the effects of adding two observations on the position of the median:
- If the original series had [tex]\( n \)[/tex] values, the median was at the [tex]\( \left(\frac{n+1}{2}\right) \)[/tex]-th position.
- Now the series has [tex]\( n + 2 \)[/tex] values.
- The median position of the new series is at the [tex]\( \left(\frac{n+2+1}{2}\right) = \left(\frac{n+3}{2}\right) \)[/tex]-th position.
Since adding two values only slightly adjusts the series, and given that the original median 16 is centrally located, the new median is likely around the previous median value 16. Thus, the answer is:
c) 16
### Question 29:
Which one of the following is not a median?
a) [tex]\(5^{\text {th }}\)[/tex] decile
b) [tex]\(5^{\text {th }}\)[/tex] percentile
c) [tex]\(2^{\text {nd }}\)[/tex] quartile
d) None
- The [tex]\(5^{\text {th }}\)[/tex] decile is the same thing as 50%, or the median.
- The [tex]\(2^{\text {nd }}\)[/tex] quartile is also 50%, or the median.
- The [tex]\(5^{\text {th }}\)[/tex] percentile, however, is only 5%, which is not the median.
Thus, the answer is:
b) [tex]\(5^{\text {th }}\)[/tex] percentile
### Question 30:
If mean [tex]\(= [3\)[/tex] median - mode], then the value of [tex]\( y \)[/tex] is:
Given the relationship, we can set up:
Mean [tex]\( = 3 \times \text{Median} - \text{Mode} \)[/tex]
This equation defines [tex]\( y \)[/tex] as being equal to the constant relationship difference observed:
[tex]\[ y = 2 \][/tex]
Thus, the answer is:
a) 2
### Question 31:
After [tex]\(10\%\)[/tex] devaluation of Nepali currency, [tex]$2000 can be exchanged for Rs 231,000. Find the Nepali rupees equivalent to $[/tex]1 before devaluation.
Given the devaluation rate:
- After 10% devaluation, we have [tex]\( 0.9X \)[/tex] Rs per [tex]$ - Given 2000 \times \( 0.9X \) Rs = 231000 Rs - Thus, \( 0.9X \) Rs = 115.5 Rs per $[/tex]
- So, [tex]\( X = \frac{115.5}{0.9} \)[/tex]
So,
d) Rs. 128.3
### Question 32:
How many paise per rupee per year will Rs 510 amount to Rs 595 in 8 months on simple interest?
First, let's calculate the total interest earned:
Total interest = Rs 595 - Rs 510 = Rs 85.
Using the Simple Interest formula:
[tex]\[ I = P \times \frac{R \times T}{100} \][/tex]
[tex]\[ 85 = 510 \times \frac{R \times \frac{8}{12}}{100} \][/tex]
[tex]\[ 85 = 510 \times \frac{2R}{30} \][/tex]
[tex]\[ 85 = 34R \][/tex]
[tex]\[ R = \frac{85}{34} = 2.5 \][/tex]
Here, R is the annual interest rate in percent. Rs is calculated based on paise per rupee per year:
100 paise is 1 Rs, thus 2.5Rs = 25 paise.
Hence, [tex]\(\text{R}\)[/tex] can be thus converted:
[tex]\[ 2.5 \times 10 \text{ paise} = 25 \text{ paise per year} \][/tex]
Thus,
a) 25 paise per year per rupee
### Question 28:
The median of the original series is 16. Two observations, 6 and 21, are added to the series. What is the new median?
To determine the new median, let's consider the effects of adding two observations on the position of the median:
- If the original series had [tex]\( n \)[/tex] values, the median was at the [tex]\( \left(\frac{n+1}{2}\right) \)[/tex]-th position.
- Now the series has [tex]\( n + 2 \)[/tex] values.
- The median position of the new series is at the [tex]\( \left(\frac{n+2+1}{2}\right) = \left(\frac{n+3}{2}\right) \)[/tex]-th position.
Since adding two values only slightly adjusts the series, and given that the original median 16 is centrally located, the new median is likely around the previous median value 16. Thus, the answer is:
c) 16
### Question 29:
Which one of the following is not a median?
a) [tex]\(5^{\text {th }}\)[/tex] decile
b) [tex]\(5^{\text {th }}\)[/tex] percentile
c) [tex]\(2^{\text {nd }}\)[/tex] quartile
d) None
- The [tex]\(5^{\text {th }}\)[/tex] decile is the same thing as 50%, or the median.
- The [tex]\(2^{\text {nd }}\)[/tex] quartile is also 50%, or the median.
- The [tex]\(5^{\text {th }}\)[/tex] percentile, however, is only 5%, which is not the median.
Thus, the answer is:
b) [tex]\(5^{\text {th }}\)[/tex] percentile
### Question 30:
If mean [tex]\(= [3\)[/tex] median - mode], then the value of [tex]\( y \)[/tex] is:
Given the relationship, we can set up:
Mean [tex]\( = 3 \times \text{Median} - \text{Mode} \)[/tex]
This equation defines [tex]\( y \)[/tex] as being equal to the constant relationship difference observed:
[tex]\[ y = 2 \][/tex]
Thus, the answer is:
a) 2
### Question 31:
After [tex]\(10\%\)[/tex] devaluation of Nepali currency, [tex]$2000 can be exchanged for Rs 231,000. Find the Nepali rupees equivalent to $[/tex]1 before devaluation.
Given the devaluation rate:
- After 10% devaluation, we have [tex]\( 0.9X \)[/tex] Rs per [tex]$ - Given 2000 \times \( 0.9X \) Rs = 231000 Rs - Thus, \( 0.9X \) Rs = 115.5 Rs per $[/tex]
- So, [tex]\( X = \frac{115.5}{0.9} \)[/tex]
So,
d) Rs. 128.3
### Question 32:
How many paise per rupee per year will Rs 510 amount to Rs 595 in 8 months on simple interest?
First, let's calculate the total interest earned:
Total interest = Rs 595 - Rs 510 = Rs 85.
Using the Simple Interest formula:
[tex]\[ I = P \times \frac{R \times T}{100} \][/tex]
[tex]\[ 85 = 510 \times \frac{R \times \frac{8}{12}}{100} \][/tex]
[tex]\[ 85 = 510 \times \frac{2R}{30} \][/tex]
[tex]\[ 85 = 34R \][/tex]
[tex]\[ R = \frac{85}{34} = 2.5 \][/tex]
Here, R is the annual interest rate in percent. Rs is calculated based on paise per rupee per year:
100 paise is 1 Rs, thus 2.5Rs = 25 paise.
Hence, [tex]\(\text{R}\)[/tex] can be thus converted:
[tex]\[ 2.5 \times 10 \text{ paise} = 25 \text{ paise per year} \][/tex]
Thus,
a) 25 paise per year per rupee
