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The house on one side of a road are numbered using consecutive positiv 21 Jun 2019, 18:58
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31% (02:44) correct 69% (02:56) wrong based on 127 sessions

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The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth, then

A. 2 ≤ a ≤ 6
B. 8 ≤ a ≤ 12
C. 14 ≤ a ≤ 18
D. 20 ≤ a ≤ 24
E. 26 ≤ a ≤ 30
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C
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The house on one side of a road are numbered using consecutive positiv 21 Jun 2019, 20:03
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The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth, then

A. 2 ≤ a ≤ 6
B. 8 ≤ a ≤ 12
C. 14 ≤ a ≤ 18
D. 20 ≤ a ≤ 24
E. 26 ≤ a ≤ 30
Show more

1. Use choices


a is the 6th, so A and B are not possible as the first will be negative then. a has to be atleast 2+2*5=12.
Now C means the first is 4,6 or 8...check if any if these we can get 170 as the sum...
You will get the answer when a is 18 and first term is 8..let there be n terms, so 8+....+(8+2*(n-1)=170
\(\frac{8+8+2(n-1)}{2}*n=170....(n-7)*n=170=10*17\), yes..
First term is 8 and a is 18

C

2. Algebraic


Sum of first n positive even numbers is 2+4+6..2n=2(1+2+3+...n)=n(n+1)
Clearly n(n+1) cannot be 170, so we can subtract initial numbers one by one, that is adding the numbers one by one to 170..
First, 170+2=172=2*86=4*43..not in form of n(n+1).
Next, 170+2+4=176=2*88=8*11..not in form of n(n+1)
Next, 170+2+4+6=182=2*91=2*7*13=13*14.... in form of n(n+1) so first term is 8 and 6th term is 8+2*(6-1)=8+10=18

C
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The house on one side of a road are numbered using consecutive positiv 21 Jun 2019, 20:51
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chetan2u
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The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth, then

A. 2 ≤ a ≤ 6
B. 8 ≤ a ≤ 12
C. 14 ≤ a ≤ 18
D. 20 ≤ a ≤ 24
E. 26 ≤ a ≤ 30

1. Use choices


a is the 6th, so A and B are not possible as the first will be negative then. a has to be atleast 2+2*5=12.
Now C means the first is 4,6 or 8...check if any if these we can get 170 as the sum...
You will get the answer when a is 18 and first term is 8..let there be n terms, so 8+....+(8+2*(n-1)=170
\(\frac{8+8+2(n-1)}{2}*n=170....(n-7)*n=170=10*17\), yes..
First term is 8 and a is 18

C

2. Algebraic


Sum of first n positive even numbers is 2+4+6..2n=2(1+2+3+...n)=n(n+1)
Clearly n(n+1) cannot be 170, so we can subtract initial numbers one by one, that is adding the numbers one by one to 170..
First, 170+2=172=2*86=4*43..not in form of n(n+1).
Next, 170+2+4=176=2*88=8*11..not in form of n(n+1)
Next, 170+2+4+6=182=2*91=2*7*13=13*14.... in form of n(n+1) so first term is 8 and 6th term is 8+2*(6-1)=8+10=18

C
Show more

Hello Chetan sir.

Whats wrong with my approach.

I assumed 'a' as the last term .So we can have
a
a-2
a-4
a-6
a-8
a-10

so 6a-30 = 170

6a = 200

a= 33.33
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The house on one side of a road are numbered using consecutive positiv 21 Jun 2019, 21:11
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'a' is not necessarily the last house, as there are at least 6 houses.
Also, 'a' must be positive even integer, but in your case a=33.33. Hence your assumption that 6th house is the last house is wrong.

prabsahi
chetan2u
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The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth, then

A. 2 ≤ a ≤ 6
B. 8 ≤ a ≤ 12
C. 14 ≤ a ≤ 18
D. 20 ≤ a ≤ 24
E. 26 ≤ a ≤ 30

1. Use choices


a is the 6th, so A and B are not possible as the first will be negative then. a has to be atleast 2+2*5=12.
Now C means the first is 4,6 or 8...check if any if these we can get 170 as the sum...
You will get the answer when a is 18 and first term is 8..let there be n terms, so 8+....+(8+2*(n-1)=170
\(\frac{8+8+2(n-1)}{2}*n=170....(n-7)*n=170=10*17\), yes..
First term is 8 and a is 18

C

2. Algebraic


Sum of first n positive even numbers is 2+4+6..2n=2(1+2+3+...n)=n(n+1)
Clearly n(n+1) cannot be 170, so we can subtract initial numbers one by one, that is adding the numbers one by one to 170..
First, 170+2=172=2*86=4*43..not in form of n(n+1).
Next, 170+2+4=176=2*88=8*11..not in form of n(n+1)
Next, 170+2+4+6=182=2*91=2*7*13=13*14.... in form of n(n+1) so first term is 8 and 6th term is 8+2*(6-1)=8+10=18

C

Hello Chetan sir.

Whats wrong with my approach.

I assumed 'a' as the last term .So we can have
a
a-2
a-4
a-6
a-8
a-10

so 6a-30 = 170

6a = 200

a= 33.33
Show more
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The house on one side of a road are numbered using consecutive positiv 21 Jun 2019, 21:21
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nick1816
'a' is not necessarily the last house, as there are at least 6 houses.
Also, 'a' must be positive even integer, but in your case a=33.33. Hence your assumption that 6th house is the last house is wrong.

The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth, then


C
Show more

Hello Chetan sir.

Whats wrong with my approach.

I assumed 'a' as the last term .So we can have
a
a-2
a-4
a-6
a-8
a-10

so 6a-30 = 170

6a = 200

a= 33.33[/quote][/quote]


Thanks Nick for replying back.

I agree 'a' is not a number here .The fact that its asking for a range of values .

So,I am looking for starting point in the range for options by considering only 6 houses first.( 'AT LEAST 6 houses.')
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The house on one side of a road are numbered using consecutive positiv 21 Jun 2019, 21:55
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IMO approach is not wrong. This approach can be time consuming if you don't get your answer in first 5-6 values of 'n'. You are not sure that for how many values of 'n' you gotta check to reach to the answer.

You can modify your approach a bit though to make it efficient.

Let first house number is 'x'.

n/2[2x+(n-1)*2]=170
n[x+n-1]=170
n is a factor of 170 and is greater than 6.
Possible values of n are 10,17,35 85 and 170.

x is +ve only when n=10











prabsahi
nick1816
'a' is not necessarily the last house, as there are at least 6 houses.
Also, 'a' must be positive even integer, but in your case a=33.33. Hence your assumption that 6th house is the last house is wrong.

The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth, then


C

Hello Chetan sir.

Whats wrong with my approach.

I assumed 'a' as the last term .So we can have
a
a-2
a-4
a-6
a-8
a-10

so 6a-30 = 170

6a = 200

a= 33.33
Show more
[/quote]


Thanks Nick for replying back.

I agree 'a' is not a number here .The fact that its asking for a range of values .

So,I am looking for starting point in the range for options by considering only 6 houses first.( 'AT LEAST 6 houses.')[/quote]
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The house on one side of a road are numbered using consecutive positiv 22 Jun 2019, 00:06
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[quote="nick1816"]IMO approach is not wrong. This approach can be time consuming if you don't get your answer in first 5-6 values of 'n'. You are not sure that for how many values of 'n' you gotta check to reach to the answer.

You can modify your approach a bit though to make it efficient.

Let first house number is 'x'.

n/2[2x+(n-1)*2]=170
n[x+n-1]=170
n is a factor of 170 and is greater than 6.
Possible values of n are 10,17,35 85 and 170.

x is +ve only when n=10


Thanks Nick !!I guess I know where I went wrong now and the fact is same.Taking n as 6 and assuming corresponding numbers is wrong
since both are variable here.

I tried the below approach and I still don't know where am I going wrong .

We can imagine first term as x and so last term will be a=x+(n-1)d =

Now,

n/2( 2x + (n-1)2) = 170

n( 2x + 2n -2 ) = 170

n(x+n -1) = 170

Given that n has to be greater than 6 and n*integer = 170 so n can be 5,10,15.. (5 is ruled out) .For 6,7,8,9 ruled out since 170 has factors(5,2,17)

Now replacing n as 10

10(x+9) = 170

x= 8

Therefore a(last term )= 8 + 9*2= 26
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The house on one side of a road are numbered using consecutive positiv 22 Jun 2019, 01:31
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chetan2u
nick1816
The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth, then

A. 2 ≤ a ≤ 6
B. 8 ≤ a ≤ 12
C. 14 ≤ a ≤ 18
D. 20 ≤ a ≤ 24
E. 26 ≤ a ≤ 30

1. Use choices


a is the 6th, so A and B are not possible as the first will be negative then. a has to be atleast 2+2*5=12.
Now C means the first is 4,6 or 8...check if any if these we can get 170 as the sum...
You will get the answer when a is 18 and first term is 8..let there be n terms, so 8+....+(8+2*(n-1)=170
\(\frac{8+8+2(n-1)}{2}*n=170....(n-7)*n=170=10*17\), yes..
First term is 8 and a is 18

C

2. Algebraic


Sum of first n positive even numbers is 2+4+6..2n=2(1+2+3+...n)=n(n+1)
Clearly n(n+1) cannot be 170, so we can subtract initial numbers one by one, that is adding the numbers one by one to 170..
First, 170+2=172=2*86=4*43..not in form of n(n+1).
Next, 170+2+4=176=2*88=8*11..not in form of n(n+1)
Next, 170+2+4+6=182=2*91=2*7*13=13*14.... in form of n(n+1) so first term is 8 and 6th term is 8+2*(6-1)=8+10=18

C
Show more

If C is the answer, then 14 and 16 can also be a right ? But if we take 14 or 16 as a we won't get sum as 170. Can you please help me understand how "C" is the answer here ?
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The house on one side of a road are numbered using consecutive positiv 22 Jun 2019, 02:27
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Muralidhar
chetan2u
nick1816
The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth, then

A. 2 ≤ a ≤ 6
B. 8 ≤ a ≤ 12
C. 14 ≤ a ≤ 18
D. 20 ≤ a ≤ 24
E. 26 ≤ a ≤ 30

1. Use choices


a is the 6th, so A and B are not possible as the first will be negative then. a has to be atleast 2+2*5=12.
Now C means the first is 4,6 or 8...check if any if these we can get 170 as the sum...
You will get the answer when a is 18 and first term is 8..let there be n terms, so 8+....+(8+2*(n-1)=170
\(\frac{8+8+2(n-1)}{2}*n=170....(n-7)*n=170=10*17\), yes..
First term is 8 and a is 18

C

2. Algebraic


Sum of first n positive even numbers is 2+4+6..2n=2(1+2+3+...n)=n(n+1)
Clearly n(n+1) cannot be 170, so we can subtract initial numbers one by one, that is adding the numbers one by one to 170..
First, 170+2=172=2*86=4*43..not in form of n(n+1).
Next, 170+2+4=176=2*88=8*11..not in form of n(n+1)
Next, 170+2+4+6=182=2*91=2*7*13=13*14.... in form of n(n+1) so first term is 8 and 6th term is 8+2*(6-1)=8+10=18

C

If C is the answer, then 14 and 16 can also be a right ? But if we take 14 or 16 as a we won't get sum as 170. Can you please help me understand how "C" is the answer here ?
Show more

No, we have to answer whether a lies in that range, and 16 is in the range given in option C.

Something like if I were to ask you a<0, then what must be true..
a<2 will be true because it gives me that range a<0, but not a<-2.
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The house on one side of a road are numbered using consecutive positiv 22 Jun 2019, 02:32
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prabsahi
chetan2u
nick1816
The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth, then

A. 2 ≤ a ≤ 6
B. 8 ≤ a ≤ 12
C. 14 ≤ a ≤ 18
D. 20 ≤ a ≤ 24
E. 26 ≤ a ≤ 30

1. Use choices


a is the 6th, so A and B are not possible as the first will be negative then. a has to be atleast 2+2*5=12.
Now C means the first is 4,6 or 8...check if any if these we can get 170 as the sum...
You will get the answer when a is 18 and first term is 8..let there be n terms, so 8+....+(8+2*(n-1)=170
\(\frac{8+8+2(n-1)}{2}*n=170....(n-7)*n=170=10*17\), yes..
First term is 8 and a is 18

C

2. Algebraic


Sum of first n positive even numbers is 2+4+6..2n=2(1+2+3+...n)=n(n+1)
Clearly n(n+1) cannot be 170, so we can subtract initial numbers one by one, that is adding the numbers one by one to 170..
First, 170+2=172=2*86=4*43..not in form of n(n+1).
Next, 170+2+4=176=2*88=8*11..not in form of n(n+1)
Next, 170+2+4+6=182=2*91=2*7*13=13*14.... in form of n(n+1) so first term is 8 and 6th term is 8+2*(6-1)=8+10=18

C

Hello Chetan sir.

Whats wrong with my approach.

I assumed 'a' as the last term .So we can have
a
a-2
a-4
a-6
a-8
a-10

so 6a-30 = 170

6a = 200

a= 33.33
Show more


What you have got is the MAX possible value of a, be it integer or non-integer.
But we are looking for INTEGER value.
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The house on one side of a road are numbered using consecutive positiv 28 Jun 2019, 01:44
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nick1816
The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth, then

A. 2 ≤ a ≤ 6
B. 8 ≤ a ≤ 12
C. 14 ≤ a ≤ 18
D. 20 ≤ a ≤ 24
E. 26 ≤ a ≤ 30
Show more


I solved following way
given that houses are numbered in consecutive even no
so series ; first be house x then x+2 , x+4 ... so on
for 10 houses we will have ; 10a+90=170 and a= 8
so first house is 8=x
in that case 6th house will be x+10 ; 8+10 ; 18
IMO C :)
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chetan2u
nick1816
The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170. If there are at least 6 houses in that row and a is the number of the sixth, then

A. 2 ≤ a ≤ 6
B. 8 ≤ a ≤ 12
C. 14 ≤ a ≤ 18
D. 20 ≤ a ≤ 24
E. 26 ≤ a ≤ 30

1. Use choices


a is the 6th, so A and B are not possible as the first will be negative then. a has to be atleast 2+2*5=12.
Now C means the first is 4,6 or 8...check if any if these we can get 170 as the sum...
You will get the answer when a is 18 and first term is 8..let there be n terms, so 8+....+(8+2*(n-1)=170
\(\frac{8+8+2(n-1)}{2}*n=170....(n-7)*n=170=10*17\), yes..
First term is 8 and a is 18

C

2. Algebraic


Sum of first n positive even numbers is 2+4+6..2n=2(1+2+3+...n)=n(n+1)
Clearly n(n+1) cannot be 170, so we can subtract initial numbers one by one, that is adding the numbers one by one to 170..
First, 170+2=172=2*86=4*43..not in form of n(n+1).
Next, 170+2+4=176=2*88=8*11..not in form of n(n+1)
Next, 170+2+4+6=182=2*91=2*7*13=13*14.... in form of n(n+1) so first term is 8 and 6th term is 8+2*(6-1)=8+10=18

C
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The house on one side of a road are numbered using consecutive positiv 06 Sep 2019, 04:15
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i tried this way

Given sum=170
Sn=n/2(a1+an);--(1)
Prime factorizing 170=17*5*2
possible combinations in equation (1) can be sn=17*10
apply n=10-->s10=10/2(a1+a10);
s10=5(a1+a10)=170;
now possible value for(a1+a10) is 34 only. Sum of the first and last term has to be 34 to satisfy the above equation.
possible a1=8, n=10, a10=26;
a6=18
option c) is the answer
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The house on one side of a road are numbered using consecutive positiv 15 Jan 2026, 02:46
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The house on one side of a road are numbered using consecutive positive even numbers. The sum of the numbers of all the houses in that row is 170.

If there are at least 6 houses in that row and a is the number of the sixth, then

Number the number of houses in the row be n. n>=6

Let the starting house no is s.

The sum of the numbers of all the houses in that row = s + s + 2 + .. + s + 2(n-1) = ns + n(n-1) = 170.

n & s both are even since n(n-1) is even and 170 is even

n = 6; 6s + 30 = 170; s = (170-30)/6 ; Not an integer
n = 8; 8s + 56 = 170; s = 14.25; Not an integer
n = 10; 10s + 90 = 170; s = 8; Integer; a = 8 + (6-1)*2 = 18
n = 12; 12s + 132 = 170; s is NOT an integer
n = 14; 14s + 182 = 170; s can not be negative;
Other value of n are not feasible since s can not be negative.

Since a = 18

IMO C
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Tuck School Moderator
853 posts