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by m&m » Mon May 11, 2009 5:50 pm
sum is avg * num terms

(a) 100 to 179
S = 139*80

(b) 200 to 259
S=230*60

(c) 300 to 339
S=319*40

(d) 400 to 429
S=414*30

(e) 500 to 519
S=509*20


note that
80 = 2*40 --> a< c
60 = 2*30 --> d< b
40 = 2*20 --> e< c

so compare b and c only --> c>b

answer is C
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by avenus » Mon May 11, 2009 11:45 pm
m&m wrote:sum is avg * num terms

(a) 100 to 179
S = 139*80

(b) 200 to 259
S=230*60

(c) 300 to 339
S=319*40

(d) 400 to 429
S=414*30

(e) 500 to 519
S=509*20


note that
80 = 2*40 --> a< c
60 = 2*30 --> d< b
40 = 2*20 --> e< c

so compare b and c only --> c>b

answer is C
c < b

answer is B
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Posts: 59
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by Pranay » Tue May 12, 2009 1:49 am
avenus wrote:
m&m wrote:sum is avg * num terms

(a) 100 to 179
S = 139*80

(b) 200 to 259
S=230*60

(c) 300 to 339
S=319*40

(d) 400 to 429
S=414*30

(e) 500 to 519
S=509*20


note that
80 = 2*40 --> a< c
60 = 2*30 --> d< b
40 = 2*20 --> e< c

so compare b and c only --> c>b

answer is C
c < b

answer is B
I agree with Avenues :)

Another way of solving the problem ..

Evaluate options,

A -> 100 + 101 + ... + 179 => 100*80 + 1 + 2 + 3 ........ 79
B -> 200*60 + 1 + 2 + 3 + ......... + 59
c -> 300*40 + 1 + 2 + 3 + .... 39
D -> 400*30 + 1 + 2 + 3 + ......... 29
E -> 500*20 + 1 + 2 + 3 + ..... 19.

By looking at the options, straight away you can eliminate, A and E.

Now among B,C and D,

CLearly, B has the highest numbers.

Thus B is the answer.

Hope it helps.
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