Quiz 1

as x^3 is divisible by 16, hence x^3 is even and so x must be even,
therefore we can write x^3/16=(2*2*2*something)/16 or
(8*something)/16 or (something)/2 which is equal to zero
as x^3 is divisible by 16. Therefore x is atleast a multiple of 4.
Hence it is clear that x^2 is divisible by 8.
From (b)
x is divisible by 18, we can only guess x is an even number,
nothing else. x^2 may or may not be divisible by 8.

So answer is (a) alone is sufficient

Explanation :

since pq=24 and p > q, so 24=24*1 or 12*2 or 8*3 or 6*4
statement 1) 24*1 and 12*2 satisfying statement 1. but we are getting two values for p. hence we arent getting unique value of p. so statemt 1 cant solve.
statement 2) 8*3 satisfying the conditions of statement 2.. so p=8

Let's analyze the given statements:
Statement 1: 9x > 10x Statement 2: (2)x + 3 is positive.
To determine if x is a negative number, we need to evaluate both statements:
Statement 1: 9x > 10x
If we subtract 9x from both sides, we get: 0 > x
This inequality tells us that x must be a negative number in order for the statement to be true. Therefore, Statement 1 implies that x is a negative number.
Statement 2: (2)x + 3 is positive.
In order for (2)x + 3 to be positive, the expression (2)x + 3 must be greater than zero. This means that x could be positive, negative, or zero, as long as the expression is positive. Therefore, Statement 2 does not provide conclusive information about whether x is a negative number.
In conclusion, based on the given information, Statement 1 implies that x is a negative number, while Statement 2 does not provide enough information to determine the sign of x.

It requires both option 1 and 2,
because
whule number are 0,1,2,3,..... and so on

->so if we know x is a multiple of 42 then it might be
possible that y be any fraction which gives (24*y) a fraction
value so then c will not be a whole number.

->also if we take only y is a multiple of 48 then also
same condition can arise.

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1.m/2 is not an even integer --> m2≠evenm2≠even could occur when mm is odd as well as when mm is even (10 and 5 for example) --> m2=102=5≠evenm2=102=5≠even and m2=52=2.5≠evenm2=52=2.5≠even. Not sufficient.

(2) m-3 is an even integer --> m−odd=evenm−odd=even --> m=even+odd=oddm=even+odd=odd. Sufficient.

(1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient.

(2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient.

(1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient.

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