Let the number of coins of each type be x, y and z. Hence x+10y+100z=255.
Hence max value of z=2.
When z=2, y can take 6 values 0,1,2,3,4 and 5. Hence for z=2 => 6 ways.
For z=1 => 16 ways and
z=0 => 26 ways.
Hence total no. of ways = 6+16+26=48

R+W = 120 3R - W = 188 So, 4R = 308 and R equals 77

Let the speed of the elevator be $$a+k\sqrt{p}$$.
Where a and k are constant, and p is the number of people in the lift.
Speed when 9 people are there is $$a+k\sqrt{9}=100/10=10 m/s$$.
=> $$a+3k=10$$
Similarly, when 4 people are there in the lift, the speed is $$a+k\sqrt{4}= a+2k=700/35=20m/s$$
=> k=-10 and a=40 m/s
The lift will stop moving when it speed is zero. In that case
$$a+k\sqrt{p}=0$$
$$40=10\sqrt{p}$$
$$a+k\sqrt{p}=0$$
=> p=16

Since F(1) = 4, sum of the coefficients = 4 => $$a_0 + a_1 + a_2 + ... + a_n = 4$$
As $$a_0, a_1 . . $$ are non-negative integers, if $$a_n$$ when n>=4 is non-zero, then F(F(1)) would be greater than $$4^4$$ = 256. Thus, the highest non-zero coefficient $$a_n$$ possible is $$a_3$$. Also, if $$a_3$$ >=2 then F(F(1)) >= 64*2 >= 128. Hence, $$a_3$$ <= 1.

Thus there are two cases: $$a_3$$=0 or $$a_3$$=1

Consider the first case:
We have $$a_0 + a_1 + a_2 = 4$$ and $$a_0 + 4a_1 + 16 a_2 = 82$$. There is no soln for this where $$a_0, a_1, a_2$$ are non-negative.

Hence, consider the second case: $$a_3=1$$

We have $$a_0 + a_1 + a_2 +1 = 4$$ and $$a_0 + 4a_1 + 16 a_2+64 = 82$$. => $$a_0 + a_1 + a_2 = 3$$ and $$a_0 + 4a_1 + 16 a_2 = 18$$

=> $$3a_1 + 15a_2=15$$ => The only possible soln for this is $$a_1$$=0 and $$a_2$$=1 => $$a_0$$=2

Hence the function is F(x) = $$ 2 + x^2 + x^3$$. Hence F(5) = 152

4 = 0(mod 4) 7 = 3 (mod 4) 14 = 2(mod 4) and 21 = 1(mod 4)

Any amount greater than 21 paisa, can be paid using these three coins. For example, if the amount leaves a remainder of 0 when divided by 4, we can pay using 4 paisa coins only. If it leaves a remainder of 1 when divided by 4, we can pay three 7 paisa coins and the rest by 4 paisa coins. Similarly if it leaves a remainder of 2 when divided by 4, we can pay two 7 paisa coins and the rest by 4 paisa coins. Similarly if it leaves a remainder of 3 when divided by 4, we can pay one 7 paisa coin and the rest by 4 paisa coins.

Of the numbers less than 21 paisa, the maximum we can't pay using these coins is 17 paisa coins.

Hence, answer is 17 paisa.

Let m, b and p be the cost of mango, banana and pineapple. Let a be the amount with Sunil. Hence 2m+5b+3p=a+7 and 2m+2b+3p=a-2 => b=3. m+3p=a-17, 2m+3p=a-8 => m=9. Hence, 3p=a-26. As there is no other equation given, the answer cannot be determined

Let the cost of pencil, eraser, and sharpener be a, b, and c respectively.
So, 4a+2b+c=80 ---(1) and a+2b+c=35 ---(2)
Subtracting both we get, 3a=45 => a=15
Now, we want the value of 2a+2b+c. Adding a to (2), we will get 2a+2b+c=50.

Let the number of coins of each type be x, y and z. Hence x+10y+100z=255.
Hence max value of z=2.
When z=2, y can take 6 values 0,1,2,3,4 and 5. Hence for z=2 => 6 ways.
For z=1 => 16 ways and
z=0 => 26 ways.
Hence total no. of ways = 6+16+26=48

Let the number be x y. Hence $$|x-y| = 4$$ and 10(10x+y)=14(10y+x+x+y)= 14(11y+2x) = 154y+28x => 100x+10y=154y+28x = 72x = 144y => x=2y.
Hence y=4 => x=8. Hence the number is 84

Let the number be a b.
Hence the number is stored as b a.
Hence diff = 10a+b-(10b+a)=9a-9b=72.
As a and b are integers <10, a,b can be 8,0 or 9,1. As the number stored is greater than 10 => 10b + a > 10, a b=91

3x + 4y + z = a --> (1)

2x + 6y + 4z = b --> (2)

x - y - 2z = c --> (3)

Eliminating z from (1) and (2) , (1)x4 - (2) : 12x + 16y + 4z - (2x + 6y + 4z) = 4a - b

10x + 10 y = 4a - b
=> x + y = (4a - b)/10 - (4)

Eliminating z from (1) and (3), (1)x2 + 3 = 6x + 8y + 2z + (x - y - 2z) = 2a + c
=> x + y = (2a + c)/7 - (5)

Equating equations (4) and (5), we get,

(4a - b)/10 = (2a + c)/7
28a - 7b = 20a + 10c
Thus, 8a = 7b + 10c

Let the digits of X be a b c. Hence Y is c b a. Y - X =100c+10b+a-100a-10b-c = 99c-99a=198 => c-a=2. Also, b=a+c. The 3 digit numbers satisfying these conditions are 143, 264 and 385. As only 264 is even, X=264

The guaranteed return to be obtained by partially investing in scheme 2 and 3 is as follows: If we invest x in 2) and 1-x in 3) then guaranteed return is for that value of x for which return is independent of market
=> rise return = fall return
=> 20x-10(1-x) = -5x+15(1-x)
=> 50x = 25 => x=0.5
Guaranteed return is 20*.5 -10*.5 = 5%
Hence guaranteed return from 2 and 3 is 5% which is less than 10% of FD. Hence maximum return is 10%

Let the three numbers be x-7, x and x+7. Hence 5(x+7)+14=8(x-7) => 5x+35+14=8x-56 => 3x = 35+14+56 = > x=35. Avg of three numbers = x=35

The equation of the line that is parallel to this line is 3x+4y+c=0
Distance between the lines = |6-c|/$$ \sqrt{3^2+4^2}$$
Since the line has to be closer to the origin, the y intercept will be smaller in magnitude than -6/4
So 1 = (6-c)/$$ \sqrt{3^2+4^2}$$
5 = 6 - c or c = 1 (y intercept = -1/4)

Lets take an expression with small numbers and derive the formula. Let the expression be 4x+5y.
Here, all the numbers of 5k form can be expressed.
4 is of the form 5k+4, hence all numbers of the form 5k+4 from 4 can be expressed in the form of 4x+5y.
8 is of the form 5k+3, hence all numbers of the form 5k+3 from 8 can be expressed in the form of 4x+5y.
12 is of the form 5k+2, hence all numbers of the form 5k+2 from 12 can be expressed in the form of 4x+5y.
16 is of the form 5k+1, hence all the number of the form 5k+1 from 16 can be expressed in the form of 4x+5y.
Hence 11 is the highest number that cannot be expressed in the form of 4x+5y.
11 can be written as 4*5 - 4 - 5 => For 2 co-prime numbers a and b, the largest number that cannot be expressed in the form of ax+by such that x and y are integers is ab-a-b.
In our case, we need the number from which all the integers can be expressed => answer = 43*37 - 43 - 37 + 1 = 1512

Alternatively,
According to Chicken McNugget Theorem/Postage Stamp theorem, For any two relatively prime positive integers $$a$$ & $$b$$,
the greatest integer that cannot be written in the form $$a\times\ x\ +\ b\times\ y$$ for non-negative integers $$x$$ and $$y$$
is given by $$ab - a - b$$.
Hence, 43*37-43-37=(40+3)(40-3)-80=1600-9-80=1511
As per the theorem all numbers after 1511 can be expressed as 43x+37y
Hence, the right answer is 1512.

Let cholesterol levels be represented by C = Ax + By where A is the component of weight and B is the component of cigarettes.
1200 = 100A + 100B
1000 = 100A + 50B
So, B = 4 and A = 8
? = 100A + 33B
= 800 + 132 = 932

3X + 2Y = 50
=> 2Y = 50-3X
Y = (50-3X)/2
X can take all the even values from 2 to 16.
But it has been given that X<Y.
When X = 10, Y = 10.
Therefore, X has to be less than 10.  The value of X can be 2, 4, 6, or 8.
The only solutions for the equation are (2,22); (4,19);(6,16) and (8,13)

For the above set of three equations to have a unique solution, the determinant of the coefficients of the equations should not be 0.

D = det(a, 3, 4; 4, 2, -1; 7, -a, 5)

D $$\neq$$ 0

D = $$a*(10\ -\ a)\ -\ 3*(20\ +\ 7)\ +\ 4*(-4a\ -\ 14)\ =\ 10a\ -\ a^2\ -\ 81\ -\ 16a\ -\ 56\ =\ -a^2\ -\ 6a\ -\ 137$$

The roots of $$a^2\ +\ 6a\ +\ 137\ =\ 0$$ are complex. So, ‘a’ can take all real values

Let the number be a b.
Hence the number is stored as b a.
Hence diff = 10a+b-(10b+a)=9a-9b=72.
As a and b are integers <10, a,b can be 8,0 or 9,1. As the number stored is greater than 10 => 10b + a > 10, a b=91

Let cost of 1 apple, 1 banana, 1 orange be a, b and c respectively.
7a+12b+17c=219 => difference of coefficients is 5
3a+5b+7c=91 => difference of coefficients is 2
=> multiply the first equation with 2 and the second equation with 5 and then subract each other.
14a+24b+34c-15a-25b-35c = - 17
=> a+b+c = 17

6 fruits of the same kind can be only pineapples. So, the minimum number of fruits to be pick to be certain of having 6 pineapples is 3 (apples) + 5 (oranges) + 6 (pineapples) = 14.