A simple one :

An equilateral triangle has sides of length 43. A point Q is situated inside the triangle so that the
perpendicular distances from two of the sides of the triangle are 1 and 2. What is the perpendicular
distance to the third side?

getting 3 !!!

Last years XAT question

There are 545 people numbered from 1 to 545 are standing around a circle. That 1,2,3,....545,1
Now every alternate people is eliminated. Like 2,4,..
1.Which is the number of last man standing ?
2. Say for 2nd round it was every 300th people getting eliminated. Now one proficient math teacher does some quick calculation and find that last no standing would be 437 if there were 542 people. What would be the no of last man standing for 545 people ?

Options for second one:-
3
197
249
437
543

Panditjii..answer mat batana yaar (you know it)

No bhai, It's like 12345 can be be anywhere but they must be in the sequence where 5 is after 4 is after 3 is after 2 is after 1 and 6 should not be in a sequence..
So, 1,2,3,4,5,9,8,7,6 is wrong and 1,9,8,2,3,6,4,5,7 is correct

In a set of ten speakers what is probability of A speaks always before B and B speaks always before C.

I have seen this solved using 10!/3!.

Why should not we use the same as here.

Please explain my mistake

A simple one :

An equilateral triangle has sides of length 43. A point Q is situated inside the triangle so that the
perpendicular distances from two of the sides of the triangle are 1 and 2. What is the perpendicular
distance to the third side?

area of triangle sqrt 3 /4 16x3 = 12sqrt 3

ab 3 hisso mein baant lo triangle ko

1/2x 43x 1+ 1/2x 43x 2 +1/2x 43xa = 123
solving a = 3

A simple one :

An equilateral triangle has sides of length 43. A point Q is situated inside the triangle so that the
perpendicular distances from two of the sides of the triangle are 1 and 2. What is the perpendicular
distance to the third side?

3?

Equating area of Equi Trianlge = Sum of area of 3 smaller triangles

pandit ji................... where r u????????????///
answer to bata ke sojao................. for ur 916..................

fungus Says

Good Catch Bro. One lil correction 2^7 hoga... 1152 ka factor. :)

ya 2^7.. i tot i myt be wrong.............thanku fungus

A simple one :

An equilateral triangle has sides of length 43. A point Q is situated inside the triangle so that the perpendicular distances from two of the sides of the triangle are 1 and 2. What is the perpendicular distance to the third side?

let the perpendicular distance to the third side is h unit
equating the areas of the three small triangles with the big one,
1/2*4sqrt(3)* = sqrt(3)/4*4sqrt(3)*4sqrt(3)
solving, h = 3 unit

1152 = 2^8 * 3^2

so all terms from 4!^4 will be divisible by 1152..

so remainder = 1+8+216 = 225.. ??

Good Catch Bro. One lil correction 2^7 hoga... 1152 ka factor.

A simple one :

An equilateral triangle has sides of length 43. A point Q is situated inside the triangle so that the
perpendicular distances from two of the sides of the triangle are 1 and 2. What is the perpendicular
distance to the third side?

yar,

12345_ _ _ _=18ways
_12345_ _ _=18ways
_ _12345_ _=18ways
_ _ _12345_=18ways
_ _ _ _12345=24ways

total=96

missing something :splat::splat:

If i understood it correctly

12345_ _ _ _

6 can be at 3 places,
3*3! => 18
_ 12345_ _ _

_ _ _ 12345_ => 3*3!
_ _ _ _ 12345 => 4!

18*4 + 24 => 96 numbers

No Groans please

No bhai, It's like 12345 can be be anywhere but they must be in the sequence where 5 is after 4 is after 3 is after 2 is after 1 and 6 should not be in a sequence..
So, 1,2,3,4,5,9,8,7,6 is wrong and 1,9,8,2,3,6,4,5,7 is correct

hi5blast Says

Pagalguy me naya thread kaise banate hai??muje to aata hi nahi hai ye

wah!! go to forums and waha pe post a new thread!!

Bingo!! 2520 it is

We have to find number of numbers hjaving 1, 2, 3,4 ,5 in order and then subtract numbers having 1, 2, 3, 4, 5,6 in order.

So, 9!/5! - 9!/6! = 2520

Me too

XXXXXXXXX (9 places)

7,8,9 can go in 9*8*7 places.
And, '6' cannot be in the rightmost place. So, we are left with 5 places for 6.
The position of 1 to 5 will be fixed after this.

Hence, the solution = 9*8*7*5 = 2520

the number 916238457 is an example of a nine-digit number which contains each of the digits 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. How many such numbers are there?

Off for some vocab

9c5*4!
forgot to minus the cases when 123456 are in ascending order ...

9c5*4!-9C6*3!

The number 916238457 is an example of a nine-digit number which contains each of the digits 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. How many such numbers are there?

If i understood it correctly

12345_ _ _ _

6 can be at 3 places,
3*3! => 18
_ 12345_ _ _

_ _ _ 12345_ => 3*3!
_ _ _ _ 12345 => 4!

18*4 + 24 => 96 numbers

No Groans please

yar,

12345_ _ _ _=18ways
_12345_ _ _=18ways
_ _12345_ _=18ways
_ _ _12345_=18ways
_ _ _ _12345=24ways

total=96

missing something :splat::splat:

1st one is not possible i suppose as 1 to 6 will be in order...

The number 916238457 is an example of a nine-digit number which contains each of the digits 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. How many such numbers are there?
no. of ways of selecting 5 places in 9 places = 9C5

in this in the number of cases in which 6 is right to 5 has to be subtracted...
if 5 in 5th place 4 case
if 5 in 6th place 5C4*3 cases
if 5 in 7th 6C4*2
if 5 in 8th 7C4...

9C5= 126
126- (4+ 15 + 30 + 35) = 42... i think sum mistake myt hav crept in please let me know..
42*3! (rest of d numbers 2 b arranged) = 252

yup 225 is the ans.

The number 916238457 is an example of a nine-digit number which contains each of the digits 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. How many such numbers are there?

off for some vocab

yar,

12345_ _ _ _=18ways
_12345_ _ _=18ways
_ _12345_ _=18ways
_ _ _12345_=18ways
_ _ _ _12345=24ways

total=96

missing something :splat::splat:

Pagalguy me naya thread kaise banate hai??muje to aata hi nahi hai ye

123465
123645
126345
162345
612345
these will be the 5 cases
so number of numbers are= 5*(9!/6!)
hope not missing anything...
what is OA?

Bingo!! 2520 it is

We have to find number of numbers hjaving 1, 2, 3,4 ,5 in order and then subtract numbers having 1, 2, 3, 4, 5,6 in order.

So, 9!/5! - 9!/6! = 2520

Q.Tap A fills water in a tank in 8 hrs, tap B fills the same tank with milk in 10 hrs. A man who wanted to fill the tank with water and milk ,starts tap A
in a tank containing milk up to 5% of tank capacity initially.After 4 hrs ,he starts tap B till the tank gets filled completely. In what proportion should he mix this solution with the other one containing water and milk in the ratio 1:3?
a)1:1
b)1:2
c)2:1
d)1:3 (correct option)

take capacity of 80ltrs

efficiency of tap A(water)=10ltrs/hr
efficiency of tap B(milk)=8ltrs/hr

5% of 80=4(milk containing tank)

In 1st 4hrs tap A threw 40ltrs of water,36 short to fill the tank completely
In next 2hrs tank will get completely filled

now,total milk in tank=4+8+8=20
water=60
ration=water:milk=3:1

after this didnt understand what the question is??

The number 916238457 is an example of a nine-digit number which contains each of the digits 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. How many such numbers are there?

off for some vocab

123465
123645
126345
162345
612345
these will be the 5 cases
so number of numbers are= 5*(9!/6!)
hope not missing anything...
what is OA?

from arun sharma:-
what is the remainder when (1!)^3+(2!)^3+(3!)^3+..............+(1152!)^3 is divided by 1152?

1152 = 2*24^2
from 4!^3 onwards all the numbers will be divisible by 1152
Checking for 1!^2+2!^3+3!^3 => 1+8+216 = 225/1152 = 225 remainder?

from arun sharma:-
what is the remainder when (1!)^3+(2!)^3+(3!)^3+..............+(1152!)^3 is divided by 1152?

1152 = 2^8 * 3^2

so all terms from 4!^4 will be divisible by 1152..

so remainder = 1+8+216 = 225.. ??

The number 916238457 is an example of a nine-digit number which contains each of the digits 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. How many such numbers are there?

off for some vocab

.. bad post!

from arun sharma:-
what is the remainder when (1!)^3+(2!)^3+(3!)^3+..............+(1152!)^3 is divided by 1152?

---snipped-----

The number 916238457 is an example of a nine-digit number which contains each of the digits 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. How many such numbers are there?

off for some vocab

Not sure if it is right..
9c5*4c1*3! = 3024 numbers?

Sirf 1-6 ke liye karna hai :/
Wrong approach ! :/

There is a circle of radius 1cm. Each member of a sequence of regular polygon S1(n) where n = 4, 5, 6 ---, where n is the number of sides of the polygon, is circumscribing the circle; and each member of the sequence of regular polygons S2(n), n = 4, 5, 6, --- where n is the number of sides of the polygon, is inscribed in the circle. Let L1(n) and L2(n) denote the perimeters of the corresponding polygons of S1(n) and S2(n).
Then {L1(13) + 2p} / L2(17) is

(1) greater than p/4 and less than 1
(2) greater than 1 and less than 2
(3) greater than 2
(4) less than p/4