TS Police SI Mains 2018 Arithmetic and Reasoning

Simple Interest for Rs. 15000 at 3% per annum for 3 years = $$\frac{15000\times3\times3}{100}$$ = Rs.1350

$$\therefore\ $$Gain of Q in 3 years = Rs.1350

Hence, the correct answer is Option C

Given, Principal amount = Rs.6000

Time = 4 years

Total amount = Rs.8500

Let the rate of interest = R

$$\Rightarrow$$ 8500 - 6000 = $$\frac{6000\times4\times R}{100}$$

$$\Rightarrow$$ 2500 = 240R

$$\Rightarrow$$ R = $$\frac{125}{12}$$%

Let the time required for Rs.1200 to become Rs.1825 at $$\frac{125}{12}$$% is T

$$\Rightarrow$$  1825 - 1200 = $$\frac{1200\times T\times\frac{125}{12}}{100}$$

$$\Rightarrow$$  625 = $$\frac{12\times T\times125}{12}$$

$$\Rightarrow$$  T = 5 years

Hence, the correct answer is Option A

Let the rate of interest be R and $$1+\frac{R}{100}=m$$

Hence, 108000 = P*$$m^3$$ and 192000 = P*$$m^9$$  

Dividing the second by first, we get $$m^6=\frac{16}{9}$$
Hence, $$m^3=\frac{4}{3}$$ , substituting in the first equation, we get P = 81,000

Let the principal be P, rate of interest be R, $$1+\frac{R}{100}=m$$ 
We are given that 2P = P*$$m^8$$ which gives us $$m^8=2$$

Now, let N be the number of years when sum of money becomes 4 times itself, hence:
$$4=m^N$$
$$2^2=m^N$$
$$\left(m^8\right)^2=m^N$$
Hence, N = 16 

Let the rate of interest be 2R% p.a. and let $$1+\frac{R}{100}=m$$

Hence, by the given data:
$$551250=5,00,000\cdot m^2$$
$$m^2=1.1025$$
$$m=1.05$$
$$R=5$$
Hence, rate of interest = 2R = 10% p.a. 

C = 2.25A = 0.9B 
Hence, A = 4C/9, B = 10C/9 

Total interest earned in one year = 0.08A + 0.1B + 0.12C = 0.0355C + 0.1111C + 0.12C = 0.2666C 
48,000 = 0.2666C 
C = 1,80,000 

Interest received from first person = 3000*125*t = 3,75,000*t rupees
Interest received from second person = 1000*105*t = 1,05,000*t rupees 

3,75,000t*t + 1,05,000*t = 16,80,000
4,80,000*t = 16,80,000
t = 3.5

Let the total work be = 360 units 
Hence, A can complete 360/40 = 9 units/day and B can complete 360/45 = 8 units/day 

It is given that A,B,C together finish the work in 20 days i.e. all 3 complete 18 units/day 
Hence, C alone does 18-9-8 = 1 unit/day 

Hence, share of C = (1/18)*54,000 = 3000 rupees 

Given, A takes $$7\frac{1}{2}$$ days to finish $$\frac{5}{8}^{th}$$ of the work

$$\Rightarrow$$ Number of days required to complete the total work = $$\frac{7.5}{\frac{5}{8}}$$ = $$\frac{7.5\times8}{5}$$ = 12 days

Hence, total work is done in 12 days

Let the total work be 120 units and let the per day capacities of A,B,C be a, b, c units

Hence,
a + b = 6 ..... (A)
b + c = 5 ..... (B)

Hence, a + 2b + c = 11 ..... (C) & a - c = 1 ..... (D)
Also, given that:
8a + 11b + 22c = 120 ..... (E) 

Multiplying (A) by 6 and (B) by 5, adding both, we get 6a + 11b + 5c = 61 ..... (F) 
Subtracting (F) from (E), we get: 2a + 17c = 59 ..... (G) 

Multiplying (D) by 2, subtracting from (G), we get
19c = 57 
c = 3 

Hence, C alone takes 120/3 = 40 days to complete work