@albiesriram said:
This is an olymipiad question which I had done before.
The answe comes out to be 568.
Not for CAT in any way
The answe comes out to be 568.
Not for CAT in any way
The cost of 5 pens, 8 erasers, 11 sharpeners is 54. 3 pens, 5 erasers , 7 sharpners is 34. 1pen+1eraser+1sharpener=?. puys, need sm help here....hw to approach such type of probs?
@pinaknagaon said:The cost of 5 pens, 8 erasers, 11 sharpeners is 54. 3 pens, 5 erasers , 7 sharpners is 34. 1pen+1eraser+1sharpener=?. puys, need sm help here....hw to approach such type of probs?
5P + 8E + 11S = 54 --- (1)
3P + 5E + 7S = 34 ---(2)
(1) - (2)
2P + 3E + 4S = 20 ---(3)
4P + 6E + 8S = 40 --- (4)
(4) - (2)
P + E + S = 6
@ScareCrow28 said:5P + 8E + 11S = 54 --- (1)3P + 5E + 7S = 34 ---(2)(1) - (2) 2P + 3E + 4S = 20 ---(3)4P + 6E + 8S = 40 --- (4)(4) - (2)P + E + S = 6
hw did u approach it???.... i mean is der any specific method to solve such probs....
@pinaknagaon said:The cost of 5 pens, 8 erasers, 11 sharpeners is 54. 3 pens, 5 erasers , 7 sharpners is 34. 1pen+1eraser+1sharpener=?. puys, need sm help here....hw to approach such type of probs?
5P + 8E + 11S = 54 --- 1
3P + 5E + 7S = 34 ---2
3P + 5E + 7S = 34 ---2
multiply (1) by 2 and (2) by 3 and subtract
P + E + S = 6
EDITED
@pinaknagaon said:The cost of 5 pens, 8 erasers, 11 sharpeners is 54. 3 pens, 5 erasers , 7 sharpners is 34. 1pen+1eraser+1sharpener=?. puys, need sm help here....hw to approach such type of probs?
5x + 8y + 11z = 54...i
3x + 5y + 7z = 34...ii
multiply i by 2 and ii by 3,subtract
10x + 16y + 22z = 108
9x + 15y + 21z = 102
x + y + z = 108 - 102 = 6 ?
@pinaknagaon said:hw did u approach it???.... i mean is der any specific method to solve such probs....
No specific approach brother! 😃 You just have to find it on your own, see @techgeek2050 approach, it's better! 
So for these type of questions, you have to see through your own approach
So for these type of questions, you have to see through your own approach@techgeek2050 said:5P + 8E + 11S = 54 --- 13P + 5E + 7S = 34 ---2multiply (1) by 8 and (2) by 5 and subtract P + E + S = 6
Oye..ye kaise aya ??
Equation to solve ni hogi directly
@
@ScareCrow28 said:No specific approach brother! You just have to find it on your own, see @techgeek2050 approach, it's better! So for these type of questions, you have to see through your own approach
so itzz like hit and trial ?? 
@ScareCrow28 said:Oye..ye kaise aya ??Equation to solve ni hogi directly
Bhai 2 aur 3 se multiply kia.. ghalti se 8 n 5 likh dia
@pinaknagaon said:@ so itzz like hit and trial ??
no need fr hit n trial
suppose u multiply 1 by x and 2 by y
5P + 8E + 11S = 54 --- 1
3P + 5E + 7S = 34 ----2
3P + 5E + 7S = 34 ----2
P + E + S = ?
the coefficeint of P on subtraction will be (5x - 3y) and E will be (8x - 5y)
we need coeff to be 1
so
5x - 3y = 1
8x - 5y = 1
=> x = 2 , y = 3
@techgeek2050 said:no need fr hit n trialsuppose u multiply 1 by x and 2 by y 5P + 8E + 11S = 54 --- 13P + 5E + 7S = 34 ----2P + E + S = ?the coefficeint of P on subtraction will be (5x - 3y) and E will be (8x - 5y)we need coeff to be 1so 5x - 3y = 18x - 5y = 1=> x = 2 , y = 3
nyc approach... hope it works in odr probs as well... nahi to hit and trial zidabaad!!!
@albiesriram
44=2*2*11
So the only case where we get the product of five integers as 44 is
1*(-1)*2*(-2)*11
Hence , upon adding the above values , we get the sum n =11
solution..
How many increasing arithmetic progression of three terms can be formed from first 20 perfect squares?
Don't have the OA.
Any approach other than trial and error?