Let quantity of alcohol and water be 4x litres and 3x litres respectively

20x = 4(3x+5)

8x = 20

x = 2.5

Quantity of alcohol = (4 x 2.5) litres = 10 litres.

Let speed of boat in still water be x kmph. Then,

Speed downstream = (x + 3) kmph,

Speed upstream = (x - 3)kmph.

`(x - 3) * 2 = (x + 3) * 3/2`

4x - 12 = 3x + 9

x = 21kmph.

Number of terms in series = 10.
(We can get it easily by pointing number of zeros in power of terms.
In 1st term number of zero is 1, 2nd term 2, and 3rd term 3 and so on.)

10¹⁰/7, Written as,
(7+3)^(4*2+2)/7
The remainder will depend on,
3²/7
So, remainder will be 2.
10¹⁰⁰/7, Can be written as,
3¹⁰⁰/7.
Or, 3^25*4/7
Or, 3⁴/7.
Remainder = 1
10¹⁰⁰⁰/7, Remainder = 2
10¹⁰⁰⁰⁰/7, Remainder = 1

So, we get alternate 2 and 1 as remainder, five times each.
So, required remainder is given by

(2+1+2+1+2+1+2+1+2+1)/7
= 15/7

Remainder: 1

Let edge = `a` cm 

So increase by `50%` = `a+a/2 = 3a/2`

Total surface Area of original cube = `6a^2`

TSA of new cube = `6(3a/2)^2`

`6 xx 9a^2/4 = 13.5a^2`

Increase in area = `13.5a^2 - 6a^2 = 7.5a^2`

 `7.5a^2` Increase % = `(7.5a^2)/(6a^2) xx 100 = 125%`

A runs 25% as fast as B.
That is, if B runs 100m in a given time, n A will run 125m in same time
In or words, if A runs 5m in a given time, n B will run 4m in same time.

Therefore, if length of a race is 5m, n A can give B a start of 1m so that y finish race in a dead heat.

Start : length of race :: 1 : 5

In this question, we know that start is 7m.

Hence, length of race will be 7 * 5 = 35m.

The only way to roll higher on one die is if magicians rolls between 2 and 5, inclusive, with two dice. Where he to roll a six with two dice than re is no way he could eclipse that number by rolling one die.

Below is probability of rolling a certain number with two dice.

?2? ? 1/36

?3? ? 2/36

?4? ? 3/36

?5?-  4/36

Now only numbers a magician can roll with one die and win is between 3 and 6, inclusive. The chances of rolling any are always 1/6.

Next, we have to combine probability distribution relating to two dice with that relating to one.

The chances of a magician rolling any given number with one die are 1/6. So let?s start with lowest number he can roll: a ?3.? To win with this roll, he will have to roll a ?2? with two dice, odds of which are 1/36. So .

So chances of him rolling a ?3? with one die and winning are 1/216.

Let?s repeat this logic for next roll, ?4.? Chances of rolling are 1/6. Only way a ?4? wins is if he rolls a ?2? or a ?3? with two dice.

Odds of rolling a ?2? with two dice  + odds of rolling a ?3? with two dice = . Combine this with odds of rolling a ?4? ( which is 1/6): .

If he rolls a ?5? with one die, he can win if he rolls a ?4? with two dice, probability is 3/36. He can also win if he rolls a ?2? or a ?3? with two dice, number outcomes we just found: 3/36. So we add

Next, if he rolls a ?6? on one die, he can beat ?2? through ?5? with two dice. Number of ways to roll a ?5? = 4/36. Combining this with odds of rolling a ?6? on one die with odds or rolling ?2?, ?3?, ?4?, or ?5? with two dice we get: 10/216.

You may be wondering why I left denominator as 216. Well, this allows us to add up all instances he can possibly win:

Let,
Principal = Rs. 100.
Amount = Rs. 200.
Rate = r%
Time = 4 years.
Now,
`A = P*[1+ (r/100)]^n`;
`200 = 100*[1+(r/100)]^4`;
`2 = [1+(r/100)]^4`; ........... (i)
If sum become 8 times in time n years,
n,
`8 = (1+(r/100))^n`;
`2^3 = (1+(r/100))^n`; ........(ii)
Using eqn (i) in (ii), we get;
`([1+(r/100)]^4)^3 = (1+(r/100))^n`;
`[1+(r/100)]^12 = (1+(r/100))^n`;
Thus, n = 12 years.

As win leads to multiplying amount by 1.5 and loss leads to multiplying amount by 0.5, we will multiply initial amount by 1.5 thrice and by 0.5 thrice (in any order).

The overall resultant will remain same.

So final amount with person will be (in all cases):

=64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs 27

Hence final result is: 64-27 =37

h(x) = x ² - 4 x + 9 = x ² - 4 x + 4 - 4 + 9 = (x - 2) ² + 5

(x - 2) ² >= 0

(x - 2) ² + 5 >= 5

Hence minimum value h(x) can have is 5 and maximum can go upto + infinity. So range is [5 , + infinity)

Area = 5.5 � 3.75 sq. metre.
Cost for 1 sq. metre. = Rs. 800
Hence total cost = 5.5 � 3.75 � 800 = 5.5 � 3000 = Rs. 16500