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Sia ? 6 years, 5 months ago
The largest number which divides 650 and 1170 = HCF (650,1170)
By Euclid’s division algorithm,
1170 = 650 {tex}\times{/tex} 1 + 520
650 = 520 {tex}\times{/tex} 1 + 130
520 = 130 {tex}\times{/tex} 4 + 0
So, HCF = 130
So 130 is the largest number which divides 650 and 1170
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Sia ? 6 years, 5 months ago
Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
a2 = (3q)2 or (3q + 1)2 or (3q + 2)2
a2=9q2 or 9q2 +6q + 1 or 9q2 + 12q + 4
=3 × (3q2) or 3(3q2+2q) +1 or 3(3q2+4q)+1
=3k1 or 3k2+1 or 3k3+1
Where k1 , k2 , and k3 are some positive integers
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Prateek Chawla 7 years, 8 months ago
So , on middle term splitting
K²-200k+100k-20000=0
K(k-200)+100(k-200)=0
(K+100)(k-200)=0
K=-100,200
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Sia ? 6 years, 6 months ago
| 100-150 | 24 | 125 | -200 | -4 | -96 |
| 150-200 | 40 | 175 | -150 | -3 | -120 |
| 200-250 | 33 | 225 | -100 | -2 | -66 |
| 250-300 | 28 | 275 | -50 | -1 | -28 |
| 300-350 | 30 | 325 | 0 | 0 | 0 |
| 350-400 | 22 | 375 | 50 | 1 | 22 |
| 400-450 | 16 | 425 | 100 | 2 | 32 |
| 450-500 | 7 | 475 | 150 | 3 | 21 |
| {tex} N = \sum f _ { i } = 200{/tex} | {tex} \sum f _ { i } u _ { i } = - 235{/tex} |
Let the assumed mean be A=325.
N = 200, A =325, h=50, and {tex} \Sigma f _ { i } u _ { i }{/tex}=-235
{tex} mean=\overline { x} = A + h \frac { 1 } { N } \Sigma f _ { i } u _ { i }{/tex}
{tex} \Rightarrow \quad \overline { x } = 325 + 50 \times \left\{ \frac { - 235 } { 200 } \right\}{/tex}
{tex} \Rightarrow \quad \overline { x } = 325 - \frac { 235 } { 4 } = 325 - 58.75 = 266.25{/tex}
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Sia ? 6 years, 6 months ago
Let the present age of A = x years and B = y years
According to question,
x + y = 48 ....(i)
x = 5 [ y - (x - y)]
x = 5 [2y - x]
x = 10y - 5x
3x = 5y
3(48 - y) = 5y
{tex}\Rightarrow{/tex} y = 18 years
and x = 48 - 18 years
x = 30 years
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