Question

Q 1: A shopkeeper allows his customers 10% off on the marked price of goods ans still gets a profit of 25%. What is the actual cost to him on an article
marked rs. 250
Q 2: If a shopkeeper marks the price of goods 50% more than their C.P and allows a discount of 40%. What is his gain or loss percent?
Q 3: A trader allows two successive discounts of 10% and 5%. If the M.P of an article is rs.300, find his net S.P.
Q 4: Which offer as better , two successive discounts of 10% and 8% or a single discount of 18%.
Q 5 : A scooter dealer allows a discount of 16% on the marked price.However . he still makes a profit of 20% on the C.P . Find the profit percent he would have made, had he sold the scooter at the M.P.

please do help me frnds its urgent and is to br submitted by tomorrow itself. and plzz show the procedure

Answer 1

1 ans)discount=marked price and discount %
250x10/100=25
selling price=250-25=225
cost price=100/(100+profit%)xsp
=100/(100+25)x225
=180 is the cost

Answer 2

Answer:

MP= Marked Price

SP= Selling Price

CP=Cost Price

p= profit

l= loss

p%=profit %

d= discount

I). Marked Price = Rs 250

Discount= 10%

$=250 \times \frac {10}{100}= 250 \times 0.1 = 25$

Margin=250-25=225

Margin =25%

Cost price

$\begin{array} { c } { = \frac { 100 } { 100 + \text { profit } } \times \text { selling price } } \\\\ { = \frac { 100 } { 100 + 25 } \times 225 = 0.8 \times 225 = R s .180 } \end{array}$

Actual cost of article is Rs.180

Ii). Assume original cost price= x

Marked price= 40% of x +x

$\begin{array} { l } { = \frac { 50 \mathrm { x } } { 100 } + \mathrm { x } = \frac { 50 x + 100 x } { 100 } } \\\\ { = \frac { 150 x } { 100 } = \frac { 3 x } { 2 } } \end{array}$

Selling price

$\begin{array} { c } { = \frac { 3 \mathrm { x } } { 2 } - 40 \% \text { of } \frac { 3 \mathrm { x } } { 2 } } \\\\ { = \frac { 3 \mathrm { x } } { 2 } - \frac { 40 } { 100 } \times \frac { 3 \mathrm { x } } { 2 } = \frac { 3 \mathrm { x } } { 2 } - \frac { 2 } { 5 } \times \frac { 3 \mathrm { x } } { 2 } }\\ \\ { = \frac { 3 \mathrm { x } } { 2 } - \frac { 3 \mathrm { x } } { 5 } = \frac { 15 \mathrm { x } - 6 \mathrm { x } } { 10 } = \frac { 9 \mathrm { x } } { 10 } } \end{array}$

Loss=CP-SP

C.P is taken as x,

$\begin{array} { l } { \text { Loss } = \mathrm { x } - \frac { 9 \mathrm { x } } { 10 } = \frac { \mathrm { x } } { 10 } } \\\\ { \text { LOSS } \% = \frac { \text { Loss } } { \mathrm { CP } } \times 100 } \\\\ { = \frac { \mathrm { x } } { 10 \mathrm { x } } \times 100 } \\\\ { \text { Loss } = 10 \% } \end{array}$

III). Given that

Discount, d=10%

$d= \frac {10}{100} \times 300=30$

Selling Price, SP=300-30

SP=270

Discount d=5%

$\begin{array} { l } { \mathrm { d } = \frac { 5 } { 100 } \times 270 } \\\\ { \mathrm { d } = 13.5 } \end{array}$

Selling Price, SP=270-13.5

SP=Rs 256.5.

IV). Given that

Discount, d1= 10%

Discount d2=8%

Formula for two successive discounts is

$\begin{array} { c } { = \mathrm { d } 1 \times \mathrm { d } 2 - \frac { \mathrm { d } 1 \times \mathrm { d } 2 } { 100 } } \\\\ { = 10 + 8 - \frac { 80 } { 100 } = 18 - \frac { 4 } { 5 } } \\\\ { = \frac { 90 - 4 } { 5 } = \frac { 86 } { 5 } = 17.2 \% } \end{array}$

Single offer of 18% is better.

v). Given that  

Selling Price,

$\begin{aligned} \mathrm { SP } & = \mathrm { MP } \left( \frac { 100 - 16 } { 100 } \right) \\\\ & = \frac { 84 } { 100 } \times \mathrm { MP } \\\\ \mathrm { SP } & = 0.84 \mathrm { MP } \ldots . ( 1 ) \end{aligned}$

$\begin{array} { c } { \mathrm { SP } = \mathrm { CP } \left( \frac { 100 + 20 } { 100 } \right) } \\\\ { \mathrm { SP } = \frac { 120 } { 100 } \times \mathrm { CP } } \\\\ { \mathrm { SP } = 1.2 \mathrm { CP } \ldots \ldots ( 2 ) } \end{array}$

Substitute equation (1) in (2)

$\begin{array} { c } { 0.84 \mathrm { MP } = 1.2 \mathrm { CP } } \\\\ { \mathrm { MP } = \frac { 120 } { 84 } \mathrm { CP } } \end{array}$

Profit if he sold at

$\begin{array} { c } { \mathrm { MP } = \frac { \frac { 120 } { 84 } \mathrm { CP } - \mathrm { CP } } { \mathrm { CP } \times 100 } } \\\\ { = \frac { 1.428 \mathrm { CP } - \mathrm { CP } } { \mathrm { CP } \times 100 } = \frac { 0.428 \mathrm { CP } } { \mathrm { CP } \times 100 } = \frac { 0.428 } { 100 } } \\\\ { \mathrm { MP } = 42.8 \% } \end{array}$