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Sia ? 6 years, 6 months ago
It is given that the mean of 100 observations is 50 . The value of the largest observation is 100.
Mean = {tex}\frac { \Sigma f x } { \Sigma f }{/tex}
{tex}\Rightarrow{/tex} 50 = {tex}\frac { \Sigma f x } { 100 }{/tex}
{tex}\Rightarrow \quad \Sigma f x {/tex} = 5000
It was later found that it is 110 not 100.
Correct, {tex}\Sigma f x ^ { \prime }{/tex} = 5000 + 110 - 100 = 5010
Therefore, Correct Mean = {tex}\frac { \Sigma f x' } { \Sigma f }{/tex} {tex}= \frac { 5010 } { 100 } {/tex}= 50.1
It is given that the median of 100 observations is 52.
Median will remain same i.e. = 52
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Sia ? 6 years, 6 months ago
Let speed of the train be x km/hr and speed of the car be y km/hr
Total distance = 600 Km
In first case, he travelled 320 km by train and rest i.e. 600 -320 =280 Km by Car.
In 2nd case, he travelled 200 Km by train and rest i.e. 400 Km by Car.
Using formula ; {tex}time=\frac{distance}{speed}{/tex} & according to question,
{tex}\frac{320}{x}+\frac{280}{y}=\frac{26}{3}{/tex} ..(i) (first case, here time = 8hr 40 minutes= 26/3 hrs)
{tex}\frac{200}{x}+\frac{400}{y}=\frac{55}{6}{/tex} ..(ii) (2nd case, here time = 30 minutes i.e. 1/2 hrs greater than that of 1st case)
multiplying eq. (i) by 5 and eq. (ii) by 8, we get
{tex}\frac{1600}{x}+\frac{1400}{y}=\frac{130}{3}{/tex} ..(iii)
{tex}\frac{1600}{x}+\frac{3200}{y}=\frac{220}{3}{/tex} ..(iv)
subtracting (iii) by (iv), we get
{tex}\frac{-1800}{y}=\frac{-90}{3}{/tex}
y = {tex}\frac{1800 \times 3}{90}{/tex}
y = 60 km/hr
Putting y = 60 in eq. (i), we get
{tex}\frac{320}{x}+\frac{280}{60}=\frac{26}{3}{/tex}
{tex}\frac{320}{x}=\frac{26}{3}-\frac{14}{3}{/tex}
{tex}\frac{320}{x}=\frac{12}{3}{/tex}
x = {tex}\frac{320 \times 3}{12}{/tex}
x = 80 km/hr
Hence, speed of the train = 80 Km/hr
speed of the car = 60 Km/hr
Posted by T. Yashwanth? 6 years, 10 months ago
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Avika Chaudhary 6 years, 10 months ago
Anjali Kumari? 6 years, 10 months ago
Yogita Ingle 6 years, 10 months ago
A(2,3) ------------------P(4,m)--------------B(6,-3)
Applying section formula ,
Let λ is the ratio by which P divides the line joining A and B.
x = (λx₂ + x₁)/(λ + 1)
here , x = 4 , x₁ = 2 , x₂ = 6
Now, 4 = (6λ + 2)/(λ + 1)
4(λ + 1) = 6λ + 2
4λ + 4 = 6λ + 2
2 = 2λ ⇒ λ = 1
Hence, p divides the line AB into 1 : 1 ratio.
Now, for y - co - ordinate
y = (λy₂ + y₁)/(λ + 1)
m = (1 × -3 + 3)/(1 + 1) = 0
Hence , m = 0
Anjali Kumari? 6 years, 10 months ago
Posted by Siddharth Xtry 6 years, 10 months ago
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Posted by Lionel Messi⚽️ 6 years, 10 months ago
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Yogita Ingle 6 years, 10 months ago
Given AP is:
121, 117, 113, ................
First term a = 121
Common difference d = 117-121 = -4
Let nth term is the first negative term in AP
Now nth tern in AP = a + (n-1)*d
= 121 + (n - 1)*(-4)
= 121 - 4n + 4
= 125 - 4n
Now, we have to find the suitable value of n so that the value of (125 - 4n) is negative.
⇒ 125 - 4n < 0
⇒ 125 < 4n
⇒ 4n > 125
⇒ n > 125/4
=> n > 31.25
Since the term can not be in decimal form (term is always taken a positive integer)
So, n = 32
Now, by taking value of n = 32, we get
nth term = 125 - 4× 32 = 125 - 128 = -3
This is the first negative term in the series.
So, n = 32
Hence, the 32nd term is the first negative term in AP.
Posted by Tushaar Sharma 6 years, 10 months ago
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Yogita Ingle 6 years, 10 months ago
Let x1 = -1 , x2 = 4 , y1 = 7 and y2 = -3 , m1 = 2 and m2 = 3
Using Section Formula to find coordinates of point which divides join of (–1, 7) and (4, –3) in the ratio 2:3, we get
x = (m1x2 + m2x1)/(m1 + m2) = (8-3)/ 5 = 5/5 = 1
y = (m1y2 + m2y1)/(m1 + m2) = (-6 + 21)/ 5 = 15/5 = 3
Therefore, the coordinates of point are (1, 3) which divides join of (–1, 7) and (4, –3) in the ratio 2:3.
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