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If x is equal to the sum of the integers from 30 to 50, inclusive, and
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15 Sep 2021, 01:14
15 Sep 2021, 01:14
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If x is equal to the sum of the integers from 30 to 50, inclusive, and y is the number of even integers from 30 to 50, inclusive,what is the value of x + y ?
(A) 810
(B) 811
(C) 830
(D) 850
(E) 851
(A) 810
(B) 811
(C) 830
(D) 850
(E) 851
Re: If x is equal to the sum of the integers from 30 to 50, inclusive, and
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15 Sep 2021, 07:51
15 Sep 2021, 07:51
1
Integers from \(30\) to \(50\)
Sum \(x = \frac{n}{2}(a+l)\) , where \(n =\) no of terms, \(a =\) first term & \(l =\) last term
= \(\frac{21}{2}(30+50)\)
\(x = 840\)
\(y =\) no of even integers between \(30\) and \(50 = 11\)
\(x + y = 840 + 11 = 851\)
Answer E
Sum \(x = \frac{n}{2}(a+l)\) , where \(n =\) no of terms, \(a =\) first term & \(l =\) last term
= \(\frac{21}{2}(30+50)\)
\(x = 840\)
\(y =\) no of even integers between \(30\) and \(50 = 11\)
\(x + y = 840 + 11 = 851\)
Answer E
If x is equal to the sum of the integers from 30 to 50, inclusive, and
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16 Sep 2021, 00:00
16 Sep 2021, 00:00
1
for x:
Numbers between (30,50) are 50-30+1=21 so median is 30+10=40 as avg=median=sum/21
=>sum=avg*21=40*21=840 [as numbers are equally spaced hence average=median]
for y:
Even Numbers between (30,50) are (50-30)/2 +1 = 11
=>y=11
so x+y=840+11=851 , Answer (E)
Numbers between (30,50) are 50-30+1=21 so median is 30+10=40 as avg=median=sum/21
=>sum=avg*21=40*21=840 [as numbers are equally spaced hence average=median]
for y:
Even Numbers between (30,50) are (50-30)/2 +1 = 11
=>y=11
so x+y=840+11=851 , Answer (E)
