find th sum

Q1. number of integers between 200 and 400=199
number of integers divisible by 7=>199/7=28

set of integers: 203, 210......399
sum of the integer set: 28*(203+399)/2=301*28=8428
hence, A


Q2. sum of integers from 1 to 100 divisble by 2
=2+4+6+8+.....+100
=2(1+2+...+50)
=2*50*51/2=2550........a

sum of integers from 1 to 100 divisble by 5
=5+10+15+...+100
=5(1+2+2...+20)
=5*20*21/2=5*10*21=1050.....b

sum of integers divisibe by 2 or 5
=2550+1050
=3600
hence, E


Q3. 2 digit numbers which leave a remainder 1 when divided by 4
of the form 4k+1, k must take values 3, 4, .....,24 as we only need 2 digit numbers

set of numbers: 13+17+21+....+97

Tn=97=13+(n-1)4, solving for n we get n=22

mean of the set: (13+97)/2=110/2=55

sum of the set: 55*22=1210
hence, C

scoobydooby wrote:Q1. number of integers between 200 and 400=199
number of integers divisible by 7=>199/7=28

set of integers: 203, 210......399
sum of the integer set: 28*(203+399)/2=301*28=8428
hence, A


Q2. sum of integers from 1 to 100 divisble by 2
=2+4+6+8+.....+100
=2(1+2+...+50)
=2*50*51/2=2550........a

sum of integers from 1 to 100 divisble by 5
=5+10+15+...+100
=5(1+2+2...+20)
=5*20*21/2=5*10*21=1050.....b

sum of integers divisibe by 2 or 5
=2550+1050
=3600
hence, E


Q3. 2 digit numbers which leave a remainder 1 when divided by 4
of the form 4k+1, k must take values 3, 4, .....,24 as we only need 2 digit numbers

set of numbers: 13+17+21+....+97

Tn=97=13+(n-1)4, solving for n we get n=22

mean of the set: (13+97)/2=110/2=55

sum of the set: 55*22=1210
hence, C

Acc to me Ans of First Ques is 8729

scoobydooby wrote:Q2. sum of integers from 1 to 100 divisble by 2
=2+4+6+8+.....+100
=2(1+2+...+50)
=2*50*51/2=2550........a

sum of integers from 1 to 100 divisble by 5
=5+10+15+...+100
=5(1+2+2...+20)
=5*20*21/2=5*10*21=1050.....b

sum of integers divisibe by 2 or 5
=2550+1050
=3600
hence, E

Except that you've double counted 10, 20, 30, ... 100, which are divisible by both 2 AND 5.

So, we have to subtract the multiples of both 2 and 5.

3600 - (10 + 20 + 30 + ... + 100)

= 3600 - 10(55) = 3600 - 550 = 3050... choose (B).

[The shortcut I used to calculate that series, which could have been used on all of these questions (and which Scooby did use), is merely an application of the average forumla.

Sum of terms = (# of terms)(average)

For a set of consecutive numbers, the average is simply:

(1st # + Last #)/2

or in this case:

(10 + 100)/2 = 110/2 = 55]

hawlia wrote:

scoobydooby wrote:Q1. number of integers between 200 and 400=199
number of integers divisible by 7=>199/7=28

set of integers: 203, 210......399
sum of the integer set: 28*(203+399)/2=301*28=8428
hence, A


Q2. sum of integers from 1 to 100 divisble by 2
=2+4+6+8+.....+100
=2(1+2+...+50)
=2*50*51/2=2550........a

sum of integers from 1 to 100 divisble by 5
=5+10+15+...+100
=5(1+2+2...+20)
=5*20*21/2=5*10*21=1050.....b

sum of integers divisibe by 2 or 5
=2550+1050
=3600
hence, E


Q3. 2 digit numbers which leave a remainder 1 when divided by 4
of the form 4k+1, k must take values 3, 4, .....,24 as we only need 2 digit numbers

set of numbers: 13+17+21+....+97

Tn=97=13+(n-1)4, solving for n we get n=22

mean of the set: (13+97)/2=110/2=55

sum of the set: 55*22=1210
hence, C

Acc to me Ans of First Ques is 8729

Different way to work the first is as follows.

1/2*28*(7(29+57))
= 8428

OA:
8729
3050 Thanks Stuart
1210