Let the three real numbers be a, b, c. c is taken as 73 instead of 37 
So, 73ab = 37ab + 720 
36 ab = 720 
ab = 720/36 = 20 
We know, AM ≥ GM 
(a²+b²) / 2 ≥ ab 
(a 2 + b 2 ) ≥ 2ab 
(a 2 + b 2 ) ≥ 40 
(Or) 
ab = 20 
(a,b) = (√20, √20) 
(a 2 + b 2 ) = 20 + 20 = 40

As the triangle progresses infinitely and the side length decreases, it follows an infinite GP series 
As the sides decrease by half, their areas decrease by 1/4 
We know, Area of an Equilateral Triangle = √3/4 × a 2 
Area of T1 = √3/4 × 24 × 24 = 144 √3 sq cms 
Sum of an Infinite GP = a / 1−r where a = 144 √3 , r = 1/4 
Sum of areas ( T1, T2, T3,) = 144√3 / {1−(1/4)} = 4×144√3 / 3 
Therefore, Sum of areas= 192 √3 sq cms. 

Its given that B needs 25% more time than A 
So, if Time taken by A = 4x days, Time taken by B = 5x days 
A does half the work, So A works alone for 2x days --- (1) 
A and B work for 4 days --- (2) 
B works alone for 5x/20= x/4 days --- (3) 
Total working days = 13 
(1) + (2) + (3) = 13 
2x + 4 + x/4 = 13 we get, x = 4 
So, B alone can do the work in 5 × 4 = 20 days 
Hence, the answer is 20 days

Let the wholesaler bought walnuts and peanuts at Rs 3x and Rs x respectively 
He sold 8kgs of Peanuts to the shopkeeper at 10% Profit 
Cost price of the Peanuts bought by the shopkeeper = Rs. 1.1x per kg 
Similarly, he sold 16 kgs of Walnuts to the shopkeeper at 20% Profit 
Cost price of the Walnuts bought by the shopkeeper = 1.2 × 3x = Rs. 3.6x per Kg 
He lost 5 kgs of Walnuts and 3 Kgs of Peanuts in transit 
Remaining = 16 - 5 = 11 kgs of Walnuts & 8-3 = 5 kgs of Peanuts 
He mixes them together and sells them at Rs. 166 per kg, making an overall Profit of 25% 
Selling Price = 5/4 × Cost Price 
Overall Cost Price = Rs.(3.6 × 16x + 1.1 × 8x) 
Overall Selling Price = Rs. 16 × 166 
Overall Selling Price = 5/4 × Overall Cost Price 
(3.6 × 16x + 1.1 × 8x) × 5/4 = 16 × 166 
( ( 57.6 × 8.8) x ) × 5/4 = 16 × 166 
x = Rs. 32 
Cost price per Kg of Walnuts bought by the Wholesaler = Rs. 3x = 3 × 32 = >Rs. 96

Let Tea A and B be sold at Rs.a and Rs.b respectively 
When A and B are mixed in the ratio 3 : 2 and sold at a profit of 10% at Rs.40 per Kg 
3a+2b / 5 = 40/1.1------(1) 
When A and B are mixed in the ratio 2 : 3 and sold at a profit of 5% at Rs.40 per Kg 
2a+3b/5 = 40/1.05 ------(2) 
Solving equations (1) and (2), we get 
1.1 × (3a+2b/5) = 1.05 × (2a+3b/5) 
11/10 × (3a + 2b) = 21/20 × (2a + 3b) 
66a + 44b = 42a + 63b 
24a = 19b 
a/b = 19/24 
So, the cost prices per Kg of A and B are in the ratio 19 : 24 

Let there be 'n' tests and the Overall Average score be k 
Average (n) = k => Total marks = nk 
So, when we ignore the first 10 questions, 
Average (n-10) = k+1 
Similarly, if we ignore the last 10 questions, 
Average (first (n-10)) = k-1 
It is given that when the first 10 tests are not considered, the overall average increases by 1
(Each question carries 20 marks) 
Total marks – 20×10 = (k+1) (n-10) 
kn – 200 = (k+1) (n-10) ---(1) 
Similarly, If the last 10 tests are not considered, the overall average decreases by 1 (Each
question carries 30 marks) 
kn - 300 = (k-1) (n-10) ---(2) 
Solving (1) and (2), we get 
kn - 200 = (k+1) (n-10) 
(-) kn - 300 = (k-1) (n-10) 
------------------------------------- 
100 = 2(n-10) 
50 = n-10 
n = 60 questions