how ????
@bs0409 said:Correct...................A four digit number numbered from 0000 to 9999 is said to be lucky, if the sum of the first two digits is equal to the sum of the last two digits. Find how many such numbers are possible ?
670
2(1^2 + 2^2 + .... + 9^2) + 10^2 = 670
2(1^2 + 2^2 + .... + 9^2) + 10^2 = 670
@bs0409 said:If x , y and z are the lengths of sides of a triangle and ,k=(x^2+y^2+z^2)/(xy+yz+xz), which of the following best describes ka) 0
x+y > z ==> x> y-z ==> x^2 > (y-z)^2 ___ (1)
y+z > x ==> y> x-z ==> y^2 > (x-z)^2 ___ (2)
x+z > y ==> z> y-x ==> z^2 > (y-x)^2 ___ (3)
add 1 , 2 and 3 we get
(x^2 + y^2 + z^2)/(xy+yz+zx)
.
also..
(x^2 + y^2 + z^2)/3 >= cuberoot (x^2 * y^2 * Z^2)___ (5)
and
(xy+yz+zx)/3 >= cuberoot (x^2 * y^2 * Z^2)___ (6)
dividing 5 by 6 we get
(x^2 + y^2 + z^2)/(xy+yz+zx)>= 1 ___ (b)
.
therefore from (a) and (b) ..
1
1
check the ans .. it should be option D
@adityaknsit said:can u explain ur approach a bit?
Sum of first two digits = 0 ....... 1 Case
Sum of first two digits = 1 ....... 2 Case
.
.
Sum of first two digits = 10 ...... 9 Case
Sum of first two digits = 11 ...... 8 Case
.
.
Sum of first two digits = 18 ...... 1 Case
So Total will be = 2(1^2 + 2^2 + .... + 9^2) + 10^2 = 670
Sum of first two digits = 1 ....... 2 Case
.
.
Sum of first two digits = 10 ...... 9 Case
Sum of first two digits = 11 ...... 8 Case
.
.
Sum of first two digits = 18 ...... 1 Case
So Total will be = 2(1^2 + 2^2 + .... + 9^2) + 10^2 = 670
@shattereddream said:Sum of first two digits = 0 ....... 1 CaseSum of first two digits = 1 ....... 2 Case..Sum of first two digits = 10 ...... 9 CaseSum of first two digits = 11 ...... 8 Case..Sum of first two digits = 18 ...... 1 CaseSo Total will be = 2(1^2 + 2^2 + .... + 9^2) + 10^2 = 670
Thanks I get it now....a silly ignorance from my part...i assumed that the last two digits had to be the same as the first two with only their order being changed...how silly of me........
There were 60 questions in an exam. If 1 mark was awarded for every correct answer and 1/3rd mark was deducted for every wrong answer, how many different net scores were possible in the exam?
@ravi.theja said:There were 60 questions in an exam. If 1 mark was awarded for every correct answer and 1/3rd mark was deducted for every wrong answer, how many different net scores were possible in the exam?
student attempts all questions ?
@gs4890 said:student attempts all questions ?
post ur solution for both the cases bhai 😃 i.e., if its compulsory attempting all and nt compulsory
@ravi.theja said:There were 60 questions in an exam. If 1 mark was awarded for every correct answer and 1/3rd mark was deducted for every wrong answer, how many different net scores were possible in the exam?
2^60
now the series is -20, -19 2/3, -19 1/3, 19.....0....59,59 1/3 , 59 2/3, 60.
the total elements in this series are (60-(-20)/(1/3)+1 = 241
but 3 options are not possible viz. 59 1/3 , 59 2/3 and 58 1/3 for obvious reasons..
thus answer is 241-3=238..
the total elements in this series are (60-(-20)/(1/3)+1 = 241
but 3 options are not possible viz. 59 1/3 , 59 2/3 and 58 1/3 for obvious reasons..
thus answer is 241-3=238..
@ravi.theja said:There were 60 questions in an exam. If 1 mark was awarded for every correct answer and 1/3rd mark was deducted for every wrong answer, how many different net scores were possible in the exam?
241??
60,59 1/3,58 2/3.............-20
60-(-20)/1/3+1=241 @ravi.theja
@ishu1991 said:now the series is -20, -19 2/3, -19 1/3, 19.....0....59,59 1/3 , 59 2/3, 60.the total elements in this series are (60-(-20)/(1/3)+1 = 241but 3 options are not possible viz. 59 1/3 , 59 2/3 and 58 1/3 for obvious reasons..thus answer is 241-3=238..
y these above three cases are not possible?
@sos2god said:how to solve this question??? not getting it...please help
consider this in this way.
for every element of A we have 3 possibilities. how?
either present in B asa in C.
present in B, not present in C.
not present in B.
so total cases = 3^5
stil not clear , revert.
for every element of A we have 3 possibilities. how?
either present in B asa in C.
present in B, not present in C.
not present in B.
so total cases = 3^5
stil not clear , revert.
@mailtoankit
hw can 59 1/3 posssible there be a case where u attempt all 60 question 59 crct 1 wrng or all 60 crct thts it
hw can 59 1/3 posssible there be a case where u attempt all 60 question 59 crct 1 wrng or all 60 crct thts it
@ravi.theja said:There were 60 questions in an exam. If 1 mark was awarded for every correct answer and 1/3rd mark was deducted for every wrong answer, how many different net scores were possible in the exam?
1 marks for each right answer. So max marks can be 60.
1/3 marks deducted for negative answer Min marks can be -20.
1/3 marks deducted for negative answer Min marks can be -20.
No of scores possible in b/w will be n.
So 60=-20+(n-1)1/3 =>n=241.
But some of scores are not possible: 592/3,591/3 and 581/3.
So different net scores possible is 241-3 =>238//
So different net scores possible is 241-3 =>238//
Instead of calculating every time the marks not possible.just subtract sigma(n-1) from total values possible to get final answer. If +1 for right answer and -1/n for negative answer is given in question
@mailtoankit
to get 59(2/3) we should have 60 correct and 1 wrong....(total 60 questions ...if u get 60 correct how can u get one wrong ...)
similary for 59(1/3) we should have 60 correct and 2 wrong....(that means there shud be a total of 62 questions...not possible)
also for 58(1/3) we shud have 59 correct and two wrong.....(that means there shud be a total of 61 questions...not possible)
hope ths will help
to get 59(2/3) we should have 60 correct and 1 wrong....(total 60 questions ...if u get 60 correct how can u get one wrong ...)
similary for 59(1/3) we should have 60 correct and 2 wrong....(that means there shud be a total of 62 questions...not possible)
also for 58(1/3) we shud have 59 correct and two wrong.....(that means there shud be a total of 61 questions...not possible)
hope ths will help
@ishu1991 said:@mailtoankitto get 59(2/3) we should have 60 correct and 1 wrong....(total 60 questions ...if u get 60 correct how can u get one wrong ...)similary for 59(1/3) we should have 60 correct and 2 wrong....(that means there shud be a total of 62 questions...not possible)also for 58(1/3) we shud have 59 correct and two wrong.....(that means there shud be a total of 61 questions...not possible)hope ths will help
yeah got it...
