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Kindly find the binder of Grade 10th Maths.

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Corrrect answer is option C

Jatin 

chandra 21

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Dear student 

kc+ k(k-1) = k(c+k-1) = 170 = 2:5:17
k is a factor of 170 , and then the constraint k is greater than 6 
Limit it to the fllowing numbers
10, 17 , 34 , 85, 170 
17*16 = 272 which is greater than 170 
We must have k = 10 
we get c =8 
c+10 = 18 
Ans is option C ,

 

It's Option c

KRUTARTH THAKKAR

GRADE 10 

BINDER MATH

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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
 
CSum 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
 
Option C

 

 

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
 
CSum 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
 
Option C
By Prakhar Gupta 
Chandra 21

Correct answer is option C.

 

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Dhruvi Shah

Chandra 20

Answer is option ‘C’

Reason is – 

If the number of houses are k then let the number on each house be n,n+2,n+4.......n+2(k-1) 

Sum = 170 = k/2[n+n+2(k-1)] = k/2 (2n+2k-2) = k(n+k-1) 

Let k(n+k-1) be equation 1

Now k is a factor of 170

k is greater than equal to 6 (since there are at least 6 houses in the row)

now k can be either 10 or 17 or 34 or 85 or 170 (since k is factor of 170)

k can not be 17 anything more than that since by substituting it in eq 1 we get the number more than 170

So, k=10

Sub k = 10 in eq 1,

n = 8

Number on house 6 = n+2(k-1) = 8+2(5) = 8+10 = 18 

So option c is correct

 

Answer is option c 14

By utkal Chandra 22.          Take 6th house as n, then 1st house  as n-10,n-8,n-6,n-4,n-2,n,n+2.....       ....                                                     

 

The sum of all houses is 170 so Sn=170 

let the no.of houses be n and x be the no. of first house 

since numbers are increasing consecutively by even no.s ,d will be 2

Sn=n/2[2a+(n-1)d]

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

over here n is the factor of 170  

170=5x2x17

we know there are at least 6 houses so the possibility of n are 10,34,85,17,170

0+2+4+6+8.......2(n-1)=n(n-1)

10(10-1)=90

Hence n is equal to 10 
sub n=10 in 170=n(x+n-1)

x=8

a=8+5x2=18

so option c is correct