If u and v are positive real numbers, is u>v?
1. u^3/v < 1
2. (u^1/3) /v < 1
C
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If u and v are positive real numbers, is u>v?
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clock60
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hi viv-gmat
to me very hard problem, and my try
(1)u^3<v, we can cross-multiply, provided that both u and v are +ve
u^3<u<v, on the number line from (0,1) here the answer is no u<v.but
u^3<v<u also possible case and the answer is yes
to be more clear
u^3=1/8, u=1/2, v=1/4. here 1/8<1/4 (u^3<v) but u=1/2>v=1/4 (1/2>1/4) the answer is yes, on the other hand
u^3=1/8, u=1/2,v=3/4, here 1/8<3/4 (u^3<v) but u=1/2<v=3/4 (1/2<3/4) the answer is no
so 1 st insuff
(2)again cross multiply, given that u^1/3<v
u^1/3<u on the number line (u>1)and again possible cases are
u^1/3<u<v the answer is no
u^1/3<v<u the answer is yes, it is possible also to plug numbers, but i need more practice to do this quickly
so 2 st also insuff
both
u^3<v, and u<v^3, here i raised in 3th power both parts
sum both
u^3+u<v+v^3
u^3-v^3+(u-v)<0, u^3-v^3=(u-v)*(u^2+uv+v^2)
(u-v)*(u^2+vu+v^2)+(u-v)<0
(u-v)(u^2+uv+v^2+1)<0 it is clear that u^2+uv+v^2+1 is alsways greated than 0 so it happens that (u-v) must be <0 for the product to be -ve
is u-v<0, and u<v
both are suff
to me very hard problem, and my try
(1)u^3<v, we can cross-multiply, provided that both u and v are +ve
u^3<u<v, on the number line from (0,1) here the answer is no u<v.but
u^3<v<u also possible case and the answer is yes
to be more clear
u^3=1/8, u=1/2, v=1/4. here 1/8<1/4 (u^3<v) but u=1/2>v=1/4 (1/2>1/4) the answer is yes, on the other hand
u^3=1/8, u=1/2,v=3/4, here 1/8<3/4 (u^3<v) but u=1/2<v=3/4 (1/2<3/4) the answer is no
so 1 st insuff
(2)again cross multiply, given that u^1/3<v
u^1/3<u on the number line (u>1)and again possible cases are
u^1/3<u<v the answer is no
u^1/3<v<u the answer is yes, it is possible also to plug numbers, but i need more practice to do this quickly
so 2 st also insuff
both
u^3<v, and u<v^3, here i raised in 3th power both parts
sum both
u^3+u<v+v^3
u^3-v^3+(u-v)<0, u^3-v^3=(u-v)*(u^2+uv+v^2)
(u-v)*(u^2+vu+v^2)+(u-v)<0
(u-v)(u^2+uv+v^2+1)<0 it is clear that u^2+uv+v^2+1 is alsways greated than 0 so it happens that (u-v) must be <0 for the product to be -ve
is u-v<0, and u<v
both are suff
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goalevan
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u, v > 0
u > v?
Statement 1) u^3 < v
With u of 1/2 and v of 1/4, (1/2)^3 = 1/8 < 1/4, but 1/2 > 1/4
With u of 2 and v of 100,000, 2^3 = 8 < 100,000, and 2 < 100,000
Insufficient.
Statement 2) u^(1/3) < v, or v^3 > u
Using the same logic as above, this statement is insufficient.
Combined) For me this come fairly intuitively, if u^3 is less than some quantity v, and u^(1/3) is less than the same quantity, then u is less than v. Even 0 < v < u < 1 is not possible, since u^3 and u^(1/3) would not both be less than v.
Sufficient. C
u > v?
Statement 1) u^3 < v
With u of 1/2 and v of 1/4, (1/2)^3 = 1/8 < 1/4, but 1/2 > 1/4
With u of 2 and v of 100,000, 2^3 = 8 < 100,000, and 2 < 100,000
Insufficient.
Statement 2) u^(1/3) < v, or v^3 > u
Using the same logic as above, this statement is insufficient.
Combined) For me this come fairly intuitively, if u^3 is less than some quantity v, and u^(1/3) is less than the same quantity, then u is less than v. Even 0 < v < u < 1 is not possible, since u^3 and u^(1/3) would not both be less than v.
Sufficient. C
