Quiz – 1

f(-3) = a(-3)4 - b(-3)2 + (-3) + 5 = 81a - 9b + 2 = 2 So 81a - 9b = 0
f(3) = a(3)4 - b(3)2 + (3) + 5 = 81a - 9b + 8
Substituting the value of 81a - 9b = 0 in the above we get f(3) = 8

f(1) = 0
f(2) = f(1+1) = f(1)+f(1)+4(9×1×1 – 1) = 0+0+4×8 = 32
f(4) = f(2+2) = f(2)+f(2)+4(9×2×2 – 1) = 32+32+4×35 = 204
f(8) = f(4+4) = f(4)+f(4)+4(9×4×4 – 1) = 204+204+4×143 = 980
f(16) = f(8+8) = f(8)+f(8)+4(9×8×8 – 1) = 980+980+4×575 = 4260
f(17) = f(1+16) = f(16)+f(1)+4(9×16×1 –1) = 4260+0+ 4×143 = 4832

Given that 10a + 25b = 1000 ⇒ 2a + 5b = 200
One general solution for the above equation is when a = 0, then b = 40.
To get other solutions, aa is increased by 5 and b is decreased by 2. So next solution is (5, 38)
Now b is reduced upto 0.
So total solutions = ((40−0)/2)+1 = 21

Product of 5 terms equal to 1127. As all the five terms are integers, the given product should be a product of 5 numbers. Now factorize 1127.
1127 = 72 × 23 = 7 × 7 × 23
But given that all the a, b, c, d, e are distinct. And we are getting only 3 terms with 7 repeats.
Now the logic is, integers mean positive and negative, 7 and - 7 possible and 1, - 1 also possible. As a,b, c, d, e are in ascending order, the factors should be in decreasing order. So (23, 7, 1, -1, -7)
Now,
76 - a = 23
a = 76 - 23 = 53
76 - b = 7
b = 76 - 7 = 69
76 - c = 1
c = 76 - 1 = 75
76 - d = -1
d = 76 + 1 = 77
76 - e = -7
e = 76 + 7 = 83
a + b + c +d= 53 + 69 + 75 + 77
= 274

Given that
30L + 3Q = 1167
30L + 6Q = 1284
Solving we get Q = 39, L = 35
So cost of first 10 kgs of apples = 35 × 10 = 350

Step 1: (3 x 2) - 1 = 5 ( 2^{2} + 1)
Step 2: (5 x 2) - 1 = 9 (2^{3} + 1)
Step 3: (9 x 2) - 1 = 17 (2^{4} + 1)
Step 4: (17 x 2) - 1 = 33 (2^{5} + 1)
So After 30 steps we have 2^{31} + 1

F(x+y)=f(x)+f(y)+7xy+2
so
f(1+1)=f(1)+f(1)+7*1*1+2
=4+4+7+2
=17
f(2)=17
f(2+2)=f(2)+f(2)+7*2*2+2
=17+17+28+2
=64
f(4)=64
f(4+1)=f(4)+f(1)+7*4*1+2
=64+4+28+2
=98
f(5)=98
now
f(2)+f(5)= 17+98
=115

Let the initial total fruits based on the ratio be 27 k(7k+11k+9k0

The final total fruits are 8 m(2m+3m+3m)

We know 27 k - 8 m = 21, now we have to try trial and error, as 8m is always even, 27 k should be odd, so try k = 1,3,5,7 and so on.

The values match when k=7 and m = 21

So, the initial fruits are 189 and the final fruits are 168

We have to form equations.
311 + E = x - - - - -(1)
212 + A + C = x - - - - -(2)
319+ F = x - - - - -(3)
212 + B + D = x - - - - -(4)
317 + A = x - - - - -(5)
214 + E + B = x - - - - -(6)
214 + C + F = x - - - - -(7)
309 + D = x - - - - -(8)
196 + F + D = x - - - - -(9)
327 + C = x - - - - -(10)
Here 7 variables but 10 equations. So can be solved easily.
From (2) and (10)
212 + A + C = 327 + C
⇒ A = 115
From (5), x = 317 + 115 = 432
From (1), E = 432 - 311 = 121
From (3), F = 432 - 319 = 113
From (8), D = 432 - 309 = 123
From (4), B = 432 - 123 - 212 = 97
From (10), C = 432 - 327 = 105
So (A*B)+(C*D) - (E*F)
= (115*97) + (105*123) - (121*113) = 10,397

Let us assume rightmost two squares are a, b
Then Sum of all the squares = 18 x 4 + a + b .......... (1)
Also Sum of the squares before 7 = 18
Sum of the squares between 7, Y = 18
and sum of the squares between Y , 8 = 18
So Sum of the 14 squares = 18 + 7 + 18 + Y + 18 + 8 + a + b (2)
Equating 1 and 2 we get Y = 3