Given, Principal = Rs.40,000 and Amount = Rs.44,100

Time = 2 years

A = P(1 + R/100)ⁿ

⇒ 44100/40000 = (1 + R/100)²

⇒ 21/20 = 1 + R/100

⇒ R = 5%

2/5 = 0.40  and 4/9 = 0.44

Fraction between them by options - 

3/7 = 0.42
2/3 = 0.67
4/5 = 0.80
1/2 = 0.50

so 3/7 is the answer.

As we know, area of a rhombus = (d₁×d₂)/2 = (8×10)/2 = 40 m²

Hence, option A is the correct answer.

The largest five-digit number is 99999.

Now, divide 99999 by 97 and obtain remainder

Here, remainder is 89.

Therefore, the largest 5 digit number which is divisible by 97, is 99910 (99999 – 89).

4 m+6 w=10 days

40 m+60 w=1 days-----1

And,

3 m+7 w=12 days

36 m+84 w=1 days -----2

Now, From 1 and 2 we get,

40 m+60 w=36 m+84 w

4 m=24 w

m=6 w

m:w = 6:1

from 1,

(40×6)w + 60w = 300w will complete the work in 1 day 

No. of women to complete the work in 20 days-

300 = 20 × w

So, required women = 15 women

Let original average expenditure was 'a'.

Total expenditure = 40a

Now, According to question,

40a = (a - 3) × 45

40a = 45a - 135

5a = 135

a = Rs 27

So, Initial expenditure would be 27 × 40 = Rs 1080.

Formula used:

Ratio of time time taken by P and Q = 17: 51 = 1:3

Given that the length of train P is 3/4th of train Q

P = 3/4 Q

So, ratio of lenght of train P and train Q = 3: 4

Required, Ratio of speed of train P and Train Q = 

Therefore, the ratio of speed P and Q = 9: 4

Let the numbers be a,  b and c  then a : b = 2 : 3

and  b : c  = 4 : 5

So a : b : c = 8 : 12 : 15

Total 8 + 12 + 15 = 35 units = 140

Therefore, 1 unit = 140/35 = 4

Hence the second number b  = 12 × 4 = 48