Latest Wipro Aptitude Question SOLUTION: Two series are 16, 21, 26.... And 17, 21, 25.....What is the sum of first hundred common numbers (a) 101100 (b) 110100 (c) 101110 (d) 110101

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(#M40007676) WIPRO Logical Reasoning question number Keep an EYE

Two series are 16, 21, 26.... And 17, 21, 25.....What is the sum of first hundred common numbers
(a) 101100
(b) 110100
(c) 101110
(d) 110101

Asked In Wipro K BASHA (12 years ago)

Unsolved Read Solution (12)

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Vignesh  (12 years ago)

Ans - a)101100
Expln:
By Expanding both series like
16 21 26 31 36 41 46 51 56 61.......
17 21 25 29 33 37 41 45 49 53 57 61...
we can see that common terms are:21 41 61 81 101....
it is in the form of AP wit d=20 we r askd to find sum of first 100.
so the solution is 100/2(2*21+(100-1)20)= 101100

Is this solution Helpfull? Yes (17) | No (2)  

IQBAL MEHDI  (11 years ago)

(a)101100 ans.
explanation:-
The series of first hundred common numbers will be:
21,41,61,81,...,100th
since,tn=a+(n-1)*d
where, a=1st term=21
d=common difference=41-21=61-41=...=20
so,
tn=21+(100-1)*20=2001
therefore,
last term(l)=2001
hence,
Sn=n/2(a+l)
S100=100/2(21+2001)=50*2022=101100 ans.

Is this solution Helpfull? Yes (10) | No   

Rakesh Kumar Kohli  (12 years ago)

Ans - a)101100
Both series are Arithmatical series with common difference 4 and 5 respectively. So the terms of the series are :
16 21 26 31 36 41 46 51 56 61.......
17 21 25 29 33 37 41 45 49 53 57 61...
The series with common numbers of both the series is an Arithmatical series with first term 21 and common difference 20.
So the series of common numbers is 21 41 61 81 101....
The sum of first 100 numbers = 100/2(2*21+(100-1)20)= 100*(21+990)
= 100*1011 = 101100

Is this solution Helpfull? Yes (7) | No (1)  

Reck  (12 years ago)

16,21,26....here diff b/w two term is 5
17,21,25....here diff b/w two term is 4
5*4=20=4*5
so,21,41,61,81,101......
it is in A.P.
so sum is=100/2(2*21+(100-1)*20)=50(42+1980)=101100

Is this solution Helpfull? Yes (6) | No (1)  

GANESH PANDIAN.S  (12 years ago)

=50(42+1980)
=101100

Is this solution Helpfull? Yes (1) | No (7)  

Manjunath  (9 years ago)

1st series is with increment of 5 and 2nd with 4 and lcm of 4 &5 is 20 so it repeats after 20 increment with previous and find last no. with n=a+(n-1)d and calculate sum with sum=n(1st term+last term )/2

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GuruSelva Punitha  (9 years ago)

for what reason we find the last number??? pls help me frnds

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avinash  (9 years ago)

can not be determined...bcoz any 1 among C and F can be third to left from D

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Chirag Ranjan Lahiri  (9 years ago)

16,21,26....here diff b/w two term is 5
17,21,25....here diff b/w two term is 4
5*4=20=4*5
so,21,41,61,81,101......
it is in A.P.
so sum is=100/2(2*21+(100-1)*20)=50(42+1980)=101100

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sravani  (9 years ago)

number will be 21,41,61,81,101.......
a=21
n=100
d=20
so,sum=n/2(2a+(n-1)d)=
100/2(2*21+99*20)=101100...

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vinay gupta  (9 years ago)

common term
16 21 26 31 36 41 46 51 56 61.......
17 21 25 29 33 37 41 45 49 53 57 61... so n=100,a=21,d=20..formula -----sum=n/2(2*a +(n-1)*d)
sum=100/2(2*21+(100-1)*20)=101100 ans

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RAKESH KUMAR  (9 years ago)

16, 21, 26, & 17, 21, 25,
Common number from the two series
16 + 5n = 17 + 4m → 5n -1 = 4 m
The first number is 21
And the second :at n = 5 and m = 6
The number = 16+25= 41

The series become 21, 41, 61 U₁₀₀= 21+99 . 20= 2001 S₁₀₀= ½.100(21+2001)= 50 .2022 = 101100

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