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8 8 votes

In a locality, the houses are numbered in the following way:

The house-numbers on one side of a road are consecutive odd integers starting from $301$, while the house-numbers on the other side of the road are consecutive even numbers starting from $302$. The total number of houses is the same on both sides of the road.

If the difference of the sum of the house-numbers between the two sides of the road is $27$, then the number of houses on each side of the road is

  1. $27$
  2. $52$
  3. $54$
  4. $26$

3 Answers

Best answer
8 8 votes
let's say both the side has $n$ houses

$S_1= 301, 303,... , (300+2n-1)$

$S_2= 302,304,..., (300+2n)$

$\sum S_2 -\sum S_1= 27$

$(302 + 304+ 306+ \dots+n )-(301 + 303+ 305 + .. +n )  =27$

$(302-301) + (304-303)+ \dots+ (300+2n-300-2n+1)=27$

$\underbrace{1+ 1+1+1+1+\dots +1}_{\text{n times}}$ = $27$

$n.1 = 27$

now from here we can see that $n$ must be $27$
moved by
3 3 votes

Let try option by option. 

Option A: 27

The houses starting with 301 will go on till 353 that is {301 , 303 , 305 , 307 , 309 ...... , 353} . The summation of the odd numbered houses will be 8829.

The houses starting with 302 will go on till 354 that is {302 , 304 , 306 , 308 , 310 ...... , 354} . The summation of the even numbered houses will be 8856.

The difference will be 8856-8829 that is 27. Therefore, the Correct Answer is Option A) 27 for this question.

 

PS: No need to check other options.

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2 2 votes

When an Arithmetic Progression (AP) series is multiplied/divided by a constant or a constant is added/subtracted to/from each term of the series, the resulting series remains an AP

Thus, from the given series, if we subtract 300 from each term, 
We get 1,3,5,7... on one side of the road and 2,4,6,8... on other side of the road.
Number of houses on both sides is equal.
Thus, let number of houses be n.

sum of first n odd natural numbers  = n²
sum of first n even numbers  = n(n+1)

Difference of the sum is given 27.
Thus, n(n+1) -  n² = 27
Solving , we get n = 27. 

 

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