jnu doubts

[aastha]

1.how many of the 4 digit numbers can be formed if no digit is used more than once , which are divisible by 5 ?
a. 1008
b. 952
c. 896
d. none of the above

2.the graph of the 2 eqns y=ax^2+bx+c and y=Ax^2+Bx+C such that a and A have different signs and that the quantities b^2-4ac and B^2-4AC are both negative,
a. intersect at two points
b.intersect at one point
c.do not intersect
d.none of the above

3.equation x^2 +0.5mx +1 =0 has two distince real solutions , if
a.m =3
b. m=4
c. m=5
d. none of the above

[kratika]

for ques1- option b (952)

for ques3- im confused between option c and d because to have real and distinct roots we need to have b^2- 4ac>0 which in this case gives the result that m>4....so this equation will have real and distinct roots for all values of m greater than 4

[faizy]

hi kratika........ans will be c ...........as m should b greater than 4 and option c is 5. we reject option d because it is saying none of d above but we  have m =5........hence ans is option c.......

[aastha]

@kratika

could you please elaborate as to how did u get option b in q1 ? thanks in advance

[faizy]

kratika..can u please elaborate d ans of ques 1.how u reached 2 d conclusion?

[kratika]

sure....since we need the 4 digit number to be a multiple of 5 means the number should end with either 0 or 5 also we do not have repetitions so

case1- number ending with 0...in this case we can place remaining 9 digits in three places as 9*8*7=504

case2- numbers ending with 5....in this case we can place only 8 digits in thousands place(i.e, excluding 0 and 5) so we have 8*8*7=448

hence total cases is 448+504= 952

[faizy]

thankz a lot kratika..............

[honey]

solution 4 d second question in list plzzzzzz