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Bunuel
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Hey, i think that your reasoning is correct, just, it's written 250 exluding so you have to remove this multiple from the five's multiple

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Multiples of 5 upto 250 -> 50
Multiples of 7 upto 250 -> 35
Multiples of 35 upto 250 -> 7

50+35-2*7=71

we remove the multiples of 35 twice because they show up in both 5 and 7 mutiples
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so what do you think the answer is?
Matthyrou
Hey, i think that your reasoning is correct, just, it's written 250 exluding so you have to remove this multiple from the five's multiple

ribbons
Multiples of 5 upto 250 -> 50
Multiples of 7 upto 250 -> 35
Multiples of 35 upto 250 -> 7

50+35-2*7=71

we remove the multiples of 35 twice because they show up in both 5 and 7 mutiples
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Yes correct. answer has to be B 70 it says excluding so we cant count 250
Ananth2604
so what do you think the answer is?
Matthyrou
Hey, i think that your reasoning is correct, just, it's written 250 exluding so you have to remove this multiple from the five's multiple

ribbons
Multiples of 5 upto 250 -> 50
Multiples of 7 upto 250 -> 35
Multiples of 35 upto 250 -> 7

50+35-2*7=71

we remove the multiples of 35 twice because they show up in both 5 and 7 mutiples
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The condition is between 1 and 250 => 1 < x < 250, this excludes 1 and 250

No. of Multiples of 5

245 = 5 + (n-1)*5
n = 49

No. of Multiples of 7

245 = 7 + (n-1)*7
n = 35

No. of Multiples of both 5 and 7

245 = 35 + (n-1)*35
n = 7

Multiples of 5 and 7 and both => 49 + 35 - 7 (as both is counted twice) => 77

Multiples of 5 and 7 but not both => 77 - 7 (subtracting both) => 70

Answer B.
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Multiples of 5 upto 250 -> 49
Multiples of 7 upto 250 -> 35
Multiples of 35 upto 250 -> 7

49+35-2*7=70

Answer: B.
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