[akhilmisra]

Originally Posted by priyalli

Find the possible value of p:q given that
21p^2+pq =15????
3pq - q^2



pls give elaborate solution

21p^2 +pq =45pq -15q^2
21p^2+15q^2 =44pq

21(p/q) + 15(q/p) = 44

let p/q=x
21x +15/x=44
21x^2 -44x + 15=0
solving the quadratic eqation we get
x=3/7 or 5/3
p/q = 3/7 or 5/3

pls confirm....
cheers...

[$U!\!$H|!\!E]

Originally Posted by prateek7563

oops i got excited and forgot about sum as 18 and 27

see. first check for numbers having sum of digits = 9

0 + 9 = 9000 and 3330
3+6 = 3222 and 2223, 6111 and 1116

Now for 18:

0 + 18 = 6660
3+ 15 = 3555 & 5553
6+12 = 6444 and 4446
9 + 9 = 3339 and 9333

now for 27 =

8883, 3888, 7776, 6777, 9990,9666,6669.


hence equal to 20. check for a + b =9 or 18 or 27 where a or b is divisible by 3

Answer shoub be 21:
NB: Divisibility by 9:
Sum of the digits of the number must be divisible by 9.
Set 1:{1,1,1,6} - 2 ways eg: 1116, 6111
Set 2:{2,2,2,3} - 2 ways eg: 2223,3222
Set 3:{3,3,3,(0,9)} - 2+1 = 3 ways eg: 3339,9333,3330
Set 4:{4,4,4,6} - 2 ways eg: 4446,6444
Set 5:{5,5,5,3} - 2 ways eg: 5553,3555
Set 6:{6,6,6,(0,9)} - 2+1 = 3 ways eg: 6669,9666,6660
Set 7:{7,7,7,6} - 2 ways eg: 7776,6777
Set 8:{8,8,8,3} - 2 ways eg: 8883,3888
Set 9:{9,9,9,(0,9)} - 1+1 = 2 ways eg: 9999,9990
Set 10:{9,0,0,0} - 1 ways eg: 9000
So (2+2+3+2+2+3+2+2+2+1) = 21

[shilpakokkanti]

thank you ..

[priyalli]

A student took 6 papers in an examination where the maximum marks were the same for each paper .In all papers together ,the candidate obtained 58% of the total marks .His marks in these papers were in the ratio of 12:13:14:15:16:17.Then the number of papers in which he got more than 55% is ?

ans :4 pls elaborate the steps

[protoplasm]

Source: TestFunda

In the given diagram, AB and DC are arcs of concentric circles with centre O. The perimeter of the figure ABCD is 22 cm. What is the area of the figure ABCD?


Question :
i don't understand this step in the given solution

Perimeter of ABCD = length of arc AB + 6 + length of arc DC + 6 = 22 cm

∴ Length of arc AB + length of arc DC = 22 – 12 = 10 cm …(i)

∵ Sector OAB and sector ODC are similar,

// Similarity holds for triangles ...but these are sectors ...when two sectors can be told similar ? whats the rule here ?



∴ From equation (i),

Length of arc AB = 2 cm and length of arc DC = 8 cm

// Can we deduce this really ? we could claim AB = 2k and DC=8k ...k is any integer . we cant directly tell AB=2 and DC=8 ......how can we say this ?

Now, area of figure ABCD = area of sector ODC – area of sector OAB



= 32 – 2

= 30 cm2


Thanks

[prakharc]

Originally Posted by priyalli

A student took 6 papers in an examination where the maximum marks were the same for each paper .In all papers together ,the candidate obtained 58% of the total marks .His marks in these papers were in the ratio of 12:13:14:15:16:17.Then the number of papers in which he got more than 55% is ?

ans :4 pls elaborate the steps

total marks= (12+13+14+15+16+17)x = 87x

So 87x/6n= 0.58 => x= 6*0.58n/87

ax/n >=0.55 => a>= 0.55n/x = (0.55*87)/(6*0.58 ) =13.75

a>=13.75

Hence 4 papers (a=14,15,16,17)

[vineet.nitd]

Originally Posted by priyalli

A student took 6 papers in an examination where the maximum marks were the same for each paper .In all papers together ,the candidate obtained 58% of the total marks .His marks in these papers were in the ratio of 12:13:14:15:16:17.Then the number of papers in which he got more than 55% is ?

ans :4 pls elaborate the steps

12x+13x+14x+15x+16x+17x=87x

Total marks= 100*87x/58= 150x=> max marke in each paper = 25x

13/25 = .52, while 14/25=.56

Hence, in 4 papers he obtained more than 55%

[shilpakokkanti]

Hi Guyz,
I have one more question in time and distance

towns P and Q are 265 km apart.Car A started from P at 60 kmph at 8.00 am towards Q. Car B started from Q at 40km/hr at 10.00 am towards P.At 11.00 am B stopped for half an hour.A stopped for 8 minutes at town S,which is at a distance of 210 km from P.Findthe time at which B crossed A.


I am getting the answer as 11.52.50 am.

But the answer is 11.54.30 am..


Plz plz help me out..

[prakharc]

Originally Posted by protoplasm

Question :
i don't understand this step in the given solution

Perimeter of ABCD = length of arc AB + 6 + length of arc DC + 6 = 22 cm

∴ Length of arc AB + length of arc DC = 22 – 12 = 10 cm …(i)

∵ Sector OAB and sector ODC are similar,

// Similarity holds for triangles ...but these are sectors ...when two sectors can be told similar ? whats the rule here ?

l= r*angle => angle= l/r ...since angle in both case are equal we can simply put l1/r1= l2/r2

Originally Posted by protoplasm

Length of arc AB = 2 cm and length of arc DC = 8 cm

// Can we deduce this really ? we could claim AB = 2k and DC=8k ...k is any integer . we cant directly tell AB=2 and DC=8 ......how can we say this ?

Thanks

Perimeter =22

Let's say it's 2k,8k ..then 10k+12=22 => k=1..hence 2,8

[vineet.nitd]

Originally Posted by protoplasm

Source: TestFunda

In the given diagram, AB and DC are arcs of concentric circles with centre O. The perimeter of the figure ABCD is 22 cm. What is the area of the figure ABCD?


Question :
i don't understand this step in the given solution

Perimeter of ABCD = length of arc AB + 6 + length of arc DC + 6 = 22 cm

∴ Length of arc AB + length of arc DC = 22 – 12 = 10 cm …(i)

∵ Sector OAB and sector ODC are similar,

// Similarity holds for triangles ...but these are sectors ...when two sectors can be told similar ? whats the rule here ?



∴ From equation (i),

Length of arc AB = 2 cm and length of arc DC = 8 cm

// Can we deduce this really ? we could claim AB = 2k and DC=8k ...k is any integer . we cant directly tell AB=2 and DC=8 ......how can we say this ?

Now, area of figure ABCD = area of sector ODC – area of sector OAB



= 32 – 2

= 30 cm2


Thanks

The arc lenth is given by R* Angle subtended by the arc at the center.
Since for both the arcs the angle subtended is the same, hence ratio of arc lengths will be ratio of radii only.

Also, the sum of arc AB and DC is 10 hence a ratio of 2:8 mean their respective lengths are 2 am and 8 cm only.