If a, b, and c are postitive numbers, is (a^2)(b + c) > (c^2)(b + a) ?
(1) a > b > c
(2) b = 10 and a + c = 1
(1) a > b > c
(2) b = 10 and a + c = 1
I think its A.
wat is OA?
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Good one
Source: Beat The GMAT — Data Sufficiency |
yeah.
we have an equation with a,b and c +ve.
a^2b + a^2c > c^2a + c^2b
statement 1) - a>b>c
a>c (both being positive)
so a^2b > c^2b and a^2c > c^2a ,hence suff.
2) b=10, a+c = 1
here we know the value of b is 10 but we cant figure out which is greater of a and c.
so this statement is insuff.
hence A.
we have an equation with a,b and c +ve.
a^2b + a^2c > c^2a + c^2b
statement 1) - a>b>c
a>c (both being positive)
so a^2b > c^2b and a^2c > c^2a ,hence suff.
2) b=10, a+c = 1
here we know the value of b is 10 but we cant figure out which is greater of a and c.
so this statement is insuff.
hence A.
a^2)(b + c) > (c^2)(b + a)
= (a-c)(ab+bc+ca)
if a>b>c then (a-c) >0 and (a-c)(ab+bc+ca) >0
(A) suff
b=10 and a+c=1
can't derive the value of (a-c)(ab+bc+ca) insuff
(A)
= (a-c)(ab+bc+ca)
if a>b>c then (a-c) >0 and (a-c)(ab+bc+ca) >0
(A) suff
b=10 and a+c=1
can't derive the value of (a-c)(ab+bc+ca) insuff
(A)
Drill baby drill !
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If a, b, and c are postitive numbers, is (a^2)(b + c) > (c^2)(b + a) ?
(1) a > b > c
(2) b = 10 and a + c = 1
Question Rephrase:
Is (a^2)(b + c) > (c^2)(b + a)?
Or Is a^2*b+a^2*c>c^2*b+c^2*a
Or Is a^2*b-c^2*b>c^2*a-a^2*c
Or Is b(a^2-c^2)>ca(c-a)
Or Is b(a+c)(a-c)>ca(c-a)
Or Is b(a+c)<ac? or is ab+bc<ac?
From A:a>b>c Hence ab+ac >ac Thus Sufficient.
From B:b= 10;a+c=1;
Now A.M>=G.M
Or a+c/2>Root ac
or 1/2 > Root ac
or 1/4 > ac
Hence 10*1 >1/4 ;Thus b(a+c)>ac And sufficient
Hence D
(1) a > b > c
(2) b = 10 and a + c = 1
Question Rephrase:
Is (a^2)(b + c) > (c^2)(b + a)?
Or Is a^2*b+a^2*c>c^2*b+c^2*a
Or Is a^2*b-c^2*b>c^2*a-a^2*c
Or Is b(a^2-c^2)>ca(c-a)
Or Is b(a+c)(a-c)>ca(c-a)
Or Is b(a+c)<ac? or is ab+bc<ac?
From A:a>b>c Hence ab+ac >ac Thus Sufficient.
From B:b= 10;a+c=1;
Now A.M>=G.M
Or a+c/2>Root ac
or 1/2 > Root ac
or 1/4 > ac
Hence 10*1 >1/4 ;Thus b(a+c)>ac And sufficient
Hence D
