@sujamait said:Let N be the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors. Find the sum of the digits of N?OPTIONS1) 18 2) 9 3) 27 4) 15
180 should be the number
Sum should be 9??
@sujamait said:Let the roots of x^3 + 3x^2 + 4x – 11 = 0 be a, b, c and the roots of x^3 + rx^2 + sx + t = 0 are a + b, b + c, c + a the find the value of t?OPTIONS1) 0 2) –11 3) 23 4) 98 thak gaye
x^3 + 3x^2 + 4x – 11 = 0 has a, b, c so a + b + c = -3, ab + ac + bc = 4 and abc = -(-11) = 11.
Now t will be
= -[(a+b)(a+c)(b+c)] which after some shuffling expands to
= -[(a+b+c)(ab + ac + bc) - abc] which is
= -[(-3) *4 - 11]
= 23
(Nice one
)regards
scrabbler
@sujamait said:Let N be the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors. Find the sum of the digits of N?OPTIONS1) 18 2) 9 3) 27 4) 15
N=180=2^2*3^2*5.....total factors=18
Total odd factors =6
total even factors=18-6=12
sum =9?
@sujamait said:Let N be the smallest positive integer with six positive odd integer divisors and twelve positive even integer divisors. Find the sum of the digits of N?OPTIONS1) 18 2) 9 3) 27 4) 15
9
@sujamait said:Let the roots of x^3 + 3x^2 + 4x €“ 11 = 0 be a, b, c and the roots of x^3 + rx^2 + sx + t = 0 are a + b, b + c, c + a the find the value of t?OPTIONS1) 0 2) €“11 3) 23 4) 98 thak gaye
23
@sujamait said:Let the roots of x^3 + 3x^2 + 4x – 11 = 0 be a, b, c and the roots of x^3 + rx^2 + sx + t = 0 are a + b, b + c, c + a the find the value of t?OPTIONS1) 0 2) –11 3) 23 4) 98 thak gaye
very tricky question and the answer to this can be done verbally ..(I got it in around 45 secs without using pen and paper hence sharing it)
Calculation steps: put x = -3 in original equation and multiply it with -1
= -(-27+27 -12 -11) =23.
Logic
since x^3 + 3x^2 + 4x – 11 = 0 has a,b,c as roots so we can write it as (x-a)(x-b)(x-c). Also, notice that sum of roots is -3
(a+b) = (a+b+c) - c = (-3) - c
(b+c) = (a+b+c) - a = (-3) - a
(c+a) = (a+b+c) - b = (-3) - b
hence t = - (a+b)(b+c)(c+a) = -(-3 -a)(-3 -b)(-3 -c) [which is equivalent of putting -3 in the original equation and multiplying it with -1]
ATDH.
A train meets with an accident 150 km from the originating station. It completetd the remaining journey at half of the usual speed and reached an hour late at the destination. Had the accident took palce 30 km later, it would have been only half an hour late. Find the distance between the two stations.
@veertamizhan said:A train meets with an accident 150 km from the originating station. It completetd the remaining journey at half of the usual speed and reached an hour late at the destination. Had the accident took palce 30 km later, it would have been only half an hour late. Find the distance between the two stations.
Let total distance = D kms.
Forming equations using Case I and II.
1. --> t + 1 = 150/s + (D - 150)/(s/2)
2. --> t + 1/2 = 180/s + (D - 180)/(s/2)
Subtracting the two we have,
=> 1/2 = -30/s + 60/s
=> 30/s = 1/2
Thus, s = 60 km/hr.
Substituting in 1 and putting D/s at t, we have :
=> t + 1 = 150/60 + 2*t - 300/60
=> t = 6 - 2.5 = 3.5 hours.
Thus, Distance = 3.5*60 = 210 kms. :D
@veertamizhan said:"What should be the length of a race be so that two driver with speeds with speeds of 22 m/s and 25 m/s reach the end point simultaneously even though the slower one had a head start of 6 minutes.
Head Start of 6 mint => 22*6*60 = 132*60 = 7920 m.
Now, Relative Speed = 25 - 22 = 3 m/sec.
Thus, 7920 metres can be covered in 7920/3 = 2640 secs.
Thus, Total Length = 25*2640 = 66000 = 66 km. :)
P.S. Head Start of 6 "minutes" is too much. 😛 :P
@Estallar12 said:Let total distance = D kms.Forming equations using Case I and II.1. --> t + 1 = 150/s + (D - 150)/(s/2)2. --> t + 1/2 = 180/s + (D - 180)/(s/2)Subtracting the two we have,=> 1/2 = -30/s + 60/s=> 30/s = 1/2Thus, s = 60 km/hr.Substituting in 1 and putting D/s at t, we have :=> t + 1 = 150/60 + 2*t - 300/60=> t = 6 - 2.5 = 3.5 hours.Thus, Distance = 3.5*60 = 210 kms.
cross checking because the answer was none of the above option.
@veertamizhan said:cross checking because the answer was none of the above option.
Where and What are the Options..??
@veertamizhan said:@Estallar12 kitab main. options were 190, 220, 300 and none of the above.
Yeah so 210 will be None of the Above. :)
@veertamizhan said:A train meets with an accident 150 km from the originating station. It completetd the remaining journey at half of the usual speed and reached an hour late at the destination. Had the accident took palce 30 km later, it would have been only half an hour late. Find the distance between the two stations.
No need of any calculation.
The answer is given in statement.
Statement says it would have taken 30 mins less if accident had taken 30 km later, that means the distance he needs to travel to reach station is 60 km, hence he is taking 60 mins..
So total distance is 150+60=210km
how many six digit numbers contain exactly 4 different digits ?
first to choose 4 places where i can fill distinct numbers i used c(6,4), then 10*9*8*7*4*4
4*4 in last two because i will choose from the 4 digits i have already written.so my answer was c(6,4)*10*9*8*7*4*4. but m not getting answer.can anyone tell what is wrong with my approach?
@nole said:how many six digit numbers contain exactly 4 different digits ?first to choose 4 places where i can fill distinct numbers i used c(6,4), then 10*9*8*7*4*44*4 in last two because i will choose from the 4 digits i have already written.so my answer was c(6,4)*10*9*8*7*4*4. but m not getting answer.can anyone tell what is wrong with my approach?
You didn't remove those cases when first digit can be 0.
Then C(6, 4)*10*9*8*7*4*4 is also not correct. Lets take a look at the following two cases
i) In C(6, 4), positions 1, 2, 3, 4 are choosen and numbers are 9, 8, 7, 6. Rest two digits are 6 and 6. So number will be 987666
ii) In C(6, 4), positions 1, 2, 3, 5 are choosen and numbers are 9, 8, 7, 6. Rest two digits are 6 and 6. So number will be 987666
So, we got the same number using 2 different ways (means it should be counted just once)
My take:-
There are two cases:-
i) One digit is used 3 times and rest used once
Choose the 4 digits in C(10, 4) ways, then the one which appears thrice in 4 ways. Now they can be permuted in 6!/3! ways
So, C(10, 4)*4*6!/3! such numbers
ii) One digit is used twice and another digit also used twice and rest two used once
Choose the 4 digits in C(10, 4) ways, then the two digits which appears twice in C(4, 2) ways. Now they can be permuted in 6!/(2!2!) ways
So, C(10, 4)*C(4, 2)*6!/(2!2!) such numbers
So, C(10, 4)*C(4, 2)*6!/(2!2!) such numbers
Now, we have considered 0 at the first position also. So to remove those cases multiply by (9/10) as all of the digits will appear equal number of times at the first place.
So, total such numbers = (9/10)*C(10, 4){4*120 + 6*180} = 294840
Someone give sums to solve, mmm? ^_^
@ankita14 said:Someone give sums to solve, mmm? ^_^
PQRS is a parallelogram, /_PQR=30. Find the minimum possible ratio of (PR)^2 to the area of PQRS
.A. 2 - 3^(1/2)
B. 4 - 2*3^(1/2)
C. 3 - (3)^(1/2)
D. 4 - (3)^(1/2)
B. 4 - 2*3^(1/2)
C. 3 - (3)^(1/2)
D. 4 - (3)^(1/2)
@Brooklyn said:PQRS is a parallelogram, /_PQR=30. Find the minimum possible ratio of (PR)^2 to the area of PQRS.A. 2 - 3^(1/2)B. 4 - 2*3^(1/2)C. 3 - (3)^(1/2)D. 4 - (3)^(1/2)
A^2+b^2-2abcos30/absin30
A^2+b^2/2>=Ab (AM>GM) i. e a rhombus
option 2.
@veertamizhan said:A train meets with an accident 150 km from the originating station. It completetd the remaining journey at half of the usual speed and reached an hour late at the destination. Had the accident took palce 30 km later, it would have been only half an hour late. Find the distance between the two stations.
d/v=t
(150)/v + (d-150)*2/v =t+1
180/v + (d-180)*2/v =t+1/2
subtracting
30/v -60/v =-1/2
v=60
putting
5/2 +(60t-150)/30 = t+1
t=7/2
so d=210