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Gaurav Seth 6 years, 11 months ago
Let 1st number (numerator) be x
let 2nd number (denominator) be y
they are in ratio so theri ratio = x / y
as per sum
1 condition = x / y = 5 / 6
6x = 5y
6x - 5y = 0 ----- 1
2 condition = x-8 / y-8 = 4 / 5
5x - 4y = -32 + 40
5x - 4y = 8 ------- 2
By substitution method
x = 5y / 6 (from 1)
putting value of x in equation 2
5 ( 5y / 6 ) - 4y = 8
25y - 24y / 6 = 8
25y - 24y = 8 X 6
y = 48
putting value of y in equation 1
6x - 5(48) = 0
6x = 240
x = 240 / 6
x = 40
Value of x = 40
Value of y = 48
Gaurav Seth 6 years, 11 months ago
Let the two no be x and y
therefore x / y = 5 / 6
6 * x = 5 * y
6 * x - 5 * y = 0 * 5
30 * x - 25 * y = 0 ..................(i)
also (x - 8)/(y - 8) = 4 / 5
5 * x - 40 = 4 * y - 32
5 * x - 4 * y - 8 = 0 *6
30 * x - 24 * y - 48 = 0 .....................(ii)
subtracting (i) from (ii)
y - 48 = 0
y = 48
putting y = 48 in x/y = 5/6
x/48 = 5/6
x = 5 * 48 / 6
x = 5 * 8
x = 40
therefore the two numbers are 40 and 48
Posted by Devashish Kumar 6 years, 11 months ago
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Sia ? 6 years, 6 months ago
Given,
5(x+1) + 5(2 - x) = 53 +1
{tex}\Rightarrow{/tex} 5x.5 + 52.5-x = 125 + 1
{tex}\Rightarrow{/tex} 5x.5 + {tex}\frac { 25 } { 5 ^ { x } }{/tex} = 126
Let {tex}y =5^x{/tex}
{tex}\Rightarrow{/tex} 5y + {tex}\frac { 25 } { y }{/tex} = 126
{tex}\Rightarrow5y^2+25=126y{/tex}
{tex}\Rightarrow{/tex} 5y2 - 126y + 25 = 0
Factorise the equation,
{tex}\Rightarrow{/tex} 5y2 - 125y - y + 25 = 0
{tex}\Rightarrow{/tex} 5y(y - 25) -1 (y - 25) = 0
{tex}\Rightarrow{/tex}(y - 25) (5y - 1) = 0
{tex}\Rightarrow{/tex} y - 25 = 0 or 5y - 1 = 0
{tex}\Rightarrow{/tex} y = 25 or y = 1/5
Now,
{tex}\Rightarrow{/tex} 5x = 25 or 5x = 1/5
{tex}\Rightarrow{/tex} 5x = 52 or 5x = 5-1
{tex}\Rightarrow{/tex} x = 2 or x = -1.
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Sia ? 6 years, 6 months ago
Total number of days to visit the shop = 6
Two customers can visit the shop on two days in 6 {tex}\times{/tex} 6 = 36 ways
So total number of outcomes = 36
- Two customers can visit the shop on same day of the week in 6 ways i.e.
(M, M), (T, T), (W, W), (Th, Th), (F, F), (S, S)
Favourable number of ways = 6
{tex}\therefore{/tex}P(both will reach on same day) = {tex}\frac{6}{36}{/tex}={tex}\frac{1}{6}{/tex} - Two customers can visit the shop on consecutive days in 5 ways i.e.
(M, T), (T, W), (W, Th), (Th, F), (F, S)
Favourable number of ways = 5
P(both will reach on consecutive days) = {tex}\frac{5}{36}{/tex}. - We know, Probability of occurrence of an event + Probability of non occurrence of event = 1
i.e. P(E) + {tex}P ( \overline { E } ){/tex} = 1
P(both will reach on same day) = {tex}\frac{1}{6}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{1}{6}{/tex} + {tex}P ( \overline { E } ){/tex} = 1
{tex}\Rightarrow{/tex} {tex}P ( \overline { E } ){/tex}= 1 - {tex}\frac{1}{6}{/tex}
{tex}\Rightarrow{/tex} {tex}P ( \overline { E } ){/tex} = {tex}\frac{5}{6}{/tex}
Hence, P(both will reach on different day) = {tex}\frac{5}{6}{/tex}
Posted by Rounak Singh Rajputana 6 years, 11 months ago
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Ricky Roy 6 years, 11 months ago
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