Business Mathematics and Quantitative Techniques

Let X’s 1 day work = $\frac{1}{50}$
(Since X does $\frac35$ work in 30 days, his 1 day work $=\frac{3}{5}\times\frac{1}{30}=\frac{1}{50}$)

Remaining work after 30 days: $1-\frac35=\frac25$

Let Y alone take $x$ days to complete the work.
So Y’s 1 day work $=\frac{1}{x}$.

X and Y work together for 5 days, then X alone works for 5 days.

[ 5\left(\frac{1}{50}+\frac{1}{x}\right)+5\left(\frac{1}{50}\right)=\frac25 ]

[ 5\left(\frac{1}{50}+\frac{1}{x}\right)+\frac{1}{10}=\frac25 ]

[ \frac{1}{10}+\frac{5}{x}+\frac{1}{10}=\frac25 ]

[ \frac{1}{5}+\frac{5}{x}=\frac25 ]

[ \frac{5}{x}=\frac25-\frac15=\frac15 ]

[ x=25 ]

Therefore, Y alone can complete the work in 25 days.

Let the principal be Rs $P$.

Simple interest is half the sum, so:

$SI = \frac{P}{2}$

Let the annual rate of interest be $R\%$ and time be $T$ years.

Given:

$T = 2R$

Simple interest formula:

$SI = \frac{P \times T \times R}{100}$

Substituting $SI = \frac{P}{2}$:

$\frac{P}{2} = \frac{P \times T \times R}{100}$

$\frac{1}{2} = \frac{T R}{100}$

$TR = 50$

Now put $T = 2R$:

$2R \times R = 50$

$2R^2 = 50$

$R^2 = 25$

$R = 5$

So,

$T = 2R = 2 \times 5 = 10$

Answer:
Rate of interest = 5% per annum
Time = 10 years

Total percentage spent on insurance and food:

$24\% + 34\% = 58\%$

Total spent on all items:

$24\% + 34\% + 19\% + 17\% = 94\%$

Remaining percentage:

$100\% - 94\% = 6\%$

Given, remaining amount = Rs 540.

So,

$6\% = \text{Rs } 540$

$1\% = \frac{540}{6} = \text{Rs } 90$

Total amount:

$100\% = 90 \times 100 = \text{Rs } 9,000$

Amount spent on food and insurance policy together:

$58\% \text{ of Rs } 9,000 = \frac{58}{100} \times 9000 = \text{Rs } 5,220$

Therefore, Rohan spends Rs 5,220 on food and insurance policy together.

Let the marked price of the watch be Rs (x) and its cost price be Rs (y).

Case 1: 15% discount is allowed and there is a profit of Rs 210.
[ 0.85x-y=210 ] [ y=0.85x-210 \quad \text{......(i)} ]

Case 2: 30% discount is allowed and there is a loss of Rs 500.
[ y-0.70x=500 ] [ y=0.70x+500 \quad \text{......(ii)} ]

From (i) and (ii): [ 0.85x-210=0.70x+500 ] [ 0.15x=710 ] [ x=\frac{710}{0.15}=\frac{7100}{15}=4733.33 ]

Therefore, the marked price of the watch is Rs 4,733.33 (or Rs (4,733\frac{1}{3})).