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24 Feb 2020, 02:14
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56%
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wrong
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If a, b, c, d, and e are positive numbers and \(\sqrt{a} + \sqrt{b} = \sqrt{c} + \sqrt{d} + \sqrt{e}\), is a < c?
(1) c = d
(2) \(\sqrt{b}+\sqrt{d}<\sqrt{e}\)
Are You Up For the Challenge: 700 Level Questions
24 Feb 2020, 04:35
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a,b,c,d,e are all positive
Q. is a < c?
(1) We can only deduce from c=d that √a+√b=2√c+√e. We cannot determine whether a < c
NOT SUFFICIENT
(2) √a+√b +√d = √c+√d+√e +√d
If √b+√d < √e, then √a > √c + 2√d --> 2√d is positive, thus √a > {√c + positive}.
We can determine that it must be true that √a > √c and therefore, a > c
SUFFICIENT
FINAL ANSWER IS (B)
Q. is a < c?
(1) We can only deduce from c=d that √a+√b=2√c+√e. We cannot determine whether a < c
NOT SUFFICIENT
(2) √a+√b +√d = √c+√d+√e +√d
If √b+√d < √e, then √a > √c + 2√d --> 2√d is positive, thus √a > {√c + positive}.
We can determine that it must be true that √a > √c and therefore, a > c
SUFFICIENT
FINAL ANSWER IS (B)
24 Feb 2020, 09:15
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given
√a+√b=√c+√d+√e determine whether a<c
#1
c=d
we can have
√9+√4= √4+√4+√1
3+2= 2+2+1
no
and
√4+√9= √4+√4+√1
we get yes
insufficient
#2
√b+√d=√e
in that case √e>√d
so
for any value of b,d ,e
to get √a+√b=√c+√d+√e
a has to be greater than c is not >a
IMO B; sufficient
If a, b, c, d, and e are positive numbers and a‾√+b‾√=c√+d‾‾√+e√a+b=c+d+e, is a < c?
(1) c = d
(2) b‾√+d‾‾√<e√
√a+√b=√c+√d+√e determine whether a<c
#1
c=d
we can have
√9+√4= √4+√4+√1
3+2= 2+2+1
no
and
√4+√9= √4+√4+√1
we get yes
insufficient
#2
√b+√d=√e
in that case √e>√d
so
for any value of b,d ,e
to get √a+√b=√c+√d+√e
a has to be greater than c is not >a
IMO B; sufficient
If a, b, c, d, and e are positive numbers and a‾√+b‾√=c√+d‾‾√+e√a+b=c+d+e, is a < c?
(1) c = d
(2) b‾√+d‾‾√<e√
