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1) Solving the inequality gives us \(x \in (-1;0) u (0;1)\).

Option 1 and Option 2 include both the solutions and non-solutions, making them insufficient.
Combination of (1) and (2) give us the exact interval, where the inequality holds.

C it is.
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Math Revolution Daily quiz (Day 29)

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Math Revolution Daily quiz (Day 30)

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Math Revolution Daily quiz (Day 31)

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Math Revolution Daily quiz (Day 32)

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Math Revolution Daily quiz (Day 33)

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Math Revolution Daily quiz (Day 34)

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Math Revolution Daily quiz (Day 35)

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Math Revolution Daily quiz (Day 36)

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Math Revolution Daily quiz (Day 37)

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Math Revolution Daily quiz (Day 38)

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Math Revolution Daily quiz (Day 39)

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Math Revolution Daily quiz (Day 37)




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My approach got me the answers as "either statement alone" i.e. "D"
From stat 1) n/8 = ax+3, implying n = 8ax + 24, implying n/6 = 3/2 (ax) + 4, 3/2 (ax) shall always give 1 as remainder therefore total remainder = 5...Suff.
From stat 2) (n-1)/9 = ax+8, implies n-1 = 9ax + 72, implies (n-1)/6 = 3/2(ax) + 12, 3/2 (ax) shall always give 1 as remainder whereas 12 is divisible terefore remiander = 1...suff

Therefore "D"

where did I go wrong:(
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Math Revolution Daily quiz (Day 40)

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saurabhsavant

Hello,

In fact, "n/8 = ax+3,(n-1)/9 = ax+8" is incorrect. It is supposed to be shown as "n=8p+3 and n-1=9q+8".
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saurabhsavant

Hello,

In fact, "n/8 = ax+3,(n-1)/9 = ax+8" is incorrect. It is supposed to be shown as "n=8p+3 and n-1=9q+8".

Aaarrgghhh...what a stupid mistake, thanx a ton for making it clear.
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Math Revolution Daily quiz (Day 37)




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My approach got me the answers as "either statement alone" i.e. "D"
From stat 1) n/8 = ax+3, implying n = 8ax + 24, implying n/6 = 3/2 (ax) + 4, 3/2 (ax) shall always give 1 as remainder therefore total remainder = 5...Suff.
From stat 2) (n-1)/9 = ax+8, implies n-1 = 9ax + 72, implies (n-1)/6 = 3/2(ax) + 12, 3/2 (ax) shall always give 1 as remainder whereas 12 is divisible terefore remiander = 1...suff

Therefore "D"

where did I go wrong:(

Hi saurabhsavant,

Condition 1)
You shouldn't write n/8 = ax + 3 and so n = 8ax + 24 = 8(ax + 3).
It doesn't mean that its remainder is 3 when n is divided by 8, but it means n is multiple of 8.
It should be expressed by n = 8a + 3


Condition 2)
You shouldn't write (n-1)/9 = ax + 8. It means n - 1 = 9 * ( ax + 8 ) and the remainder of n - 1 is 0 when n - 1 is divided by 9
It should be expressed by n - 1 = 9b + 8

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