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Dear Students, Kindly find the binder of Grade 10th Maths. Regards, Team askIITians

Dear Students,
 
Kindly find the binder of Grade 10th Maths.
 
Regards,
Team askIITians

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Grade:Upto college level

10 Answers

Jatin Soni
36 Points
2 years ago
Corrrect answer is option C
Jatin 
chandra 21
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Vikas TU
14149 Points
2 years ago
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 ,
 
Krutarth Thakkar
13 Points
2 years ago
It's Option c
KRUTARTH THAKKAR
GRADE 10 
BINDER MATH
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Jatin Soni
36 Points
2 years ago
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
 
 
Jatin 
chandra 21
Prakhar Gupta
36 Points
2 years ago
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
Nigel Martin Jose
14 Points
2 years ago
Correct answer is option C.
 
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DHRUVI SACHIN SHAH
17 Points
2 years ago
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
DHARA SACHIN SHAH
15 Points
2 years ago
a is the 6th number, so a and b option arent valid since the the first house numbers will be negative which is not possible.….that means, a has to be atleast 2 + 2*5 = 12.....now option c means that the first house number will be either 4 or 6 or 8 …. which is possible(since numbers are positive)...hence we check these as to get a sum of 170.. we find that when a is 18, first term is 8.... now suppose there are n terms...so 8+....+{8 + 2*(n-1)} = 170.... this means {2(8) + 2(n-1)}*n / 2 = 170 ....which means that (n+7)*n=170=10*17 → n=10 by subtituting value of n we get first term is 8 and hence 6th number is 18 → a=18 
there fore correct answer is option (C)
 
Utkal
30 Points
2 years ago
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.....       ....                                                     
Zainab Raza
15 Points
2 years ago
 
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 
 

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