Question

1 )the mean is 42 find the missing frequency of X and y if the total frequency is 100. class interval 0-10, 10-20 ,20-30, 30-40 ,40-50 ,50-60 ,60-70, 70-80
frequency 7, 10, X ,13, y, 10, 14, 9​



2) Mohan lends 4,800 to John for 9/2 years
and * 2,500 to Shyam for 6 years andreceives a total sum of 2,196 as interest
Find the rate per cent per annum, it being the
same in both the cases.


​
3) . The sum of two natural
numbers is 25 and their
difference is 7. Find the
numbers​



4) Find a number such that if you add six and divide by three you will get the same answer as if you subtracted four from the original number and divided by two


solve all questions pls I need explain pls ​

Answer 1

Answer:

Given that mean of data is 42

sum of frequencies (N) or ∑f=100

Class interval 0−10 10−20 20−30 30−40 40−50 50−60 60−70 70−80

Frequency 7 10 x 13 y 10 14 9

Class interval frequency (fir) mid value of (xi) fixi

i) fixi

0−10 7 5 35

10−20 10 15 150

20−30 x 25 25x

30−40 13 35 455

40−50 y 45 45y

50−60 10 55 550

60−70 14 65 910

70−80 9 75 675

______________ ___________

63+x+y 2775+25x+45y

Given N=100

⇒63+x+y=100

x+y=100−63=37

x+y=37 ,,,,,,,,,(1)

we know mean =∑fi xi /∑ fi

$42 \frac{2775 + 25 x + 45y}{100}$

4200=2775+25x+145y

⇒25x+45y=1425

5(5x+9y)=5×285

5x+9y=285 .............(2)

(1)×9=9x+9y=333

(2)=5x+9y=285

___________________

4x=48

⇒x=12

(1) ⇒x+y=37

⇒y=37−x=37+12=25

y=25.

_____________________________

_____________________________

According to the question,we have

$\frac{4800 \times r \times 9}{100 \times 2} + \frac{2500 \times 6 \times r}{100} = 2196$

$216R+150R=2196$

$366R = 2196$

$R = \frac{2196}{366} = 6persent$

_____________________________

_____________________________

Let the number be x

Thus

one no. is x

second no. is x + 7

Therefore

x+x+7=25

2x+7=25

2x=25-7

2x=18

x=18÷2

x=9

so, one number is 9

another 9+7=16

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sorry I don't know ans of 4th question.

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