Question

Q.1) Solve the following problem and
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Problem No-01
X Co. Ltd was incorporated on 01/07/2019 to take over the business of Mr. A and B with effect from 01/04/2019.
From the following information and figure related to the year ending 31 March 2020. Find out profit before and after
Incorporation.
a) The sales of the year ending 31/03/2020 were 2.500,000. divided into 1,00,000 for the period upto 01/07/2019 and
4,00,000 for the remaining period.
b) Gross Profit for the year was 1,25,000.
Particular
Particulars
c) Exenses -
2.000
24.000
Salary
1.250
9,600
Reat and Takes
4,000
Commision on Sales
Carrtege Outward
Bank Changes
Elad dep's
Preliminary Exp
Deprication
Oflice Expenses
Lectricity
750
2.000
1.500
Insurance
2.800
3.500
Director fees
2.400
Auditor fees
Intrest on Loan
1.200
Advertisement
1.000
d) Net Profit of the year 67,000
Ascertain the profit prior to and after incorporation.​

Answer 1

♣GIVEN :

Length of the floor of the room (l) = 4.84 m = 484 cm

Breadth of the floor of the room (b) = 3.1 m = 310 cm

Length of each tile (l) = 22 cm

Breadth of each tile (b) = 10 cm

Rate of paving (r) = ₹ 6.50/tile

♣TO FIND :

Cost of the tiles (c)

♣SOLUTION :

❶ First we need to find the area of the floor:-

❐ USING FORMULAE :

$\huge \boxed{ \sf{A = \bigg(l \: \times b \bigg) unit}}$

A=(l×b)unit

✯ WHERE :-

A = Area

l = length

b = breadth

★══════════════════════★

➲ Putting all values :-

➨$\:\bf{A = \bigg( 484 \times 310 \bigg)} \: cm^{2}$➨A=(484×310)cm

2

➠$\: \underline{\boxed{\bf{A = 150040} \: {cm}^{2} }}$➠

A=150040cm

2

★══════════════════════★

❷ Now we have to find the area of each tile:-

❐ USING FORMULAE :

$\huge \boxed{ \sf{A = \bigg(l \: \times b \bigg) unit}}$

A=(l×b)unit

✯ WHERE :-

A = Area

l = length

b = breadth

★══════════════════════★

➲ Putting all values :-

➨$\:\bf{A = \bigg( 22\times 10 \bigg)} \: cm^{2}$➨A=(22×10)cm

2

➠$\: \underline{\boxed{\bf{A = 220} \: {cm}^{2} }}$➠

A=220cm

2

★══════════════════════★

❸ Now let us find the number of tiles required for paving the floor:-

❐ USING FORMULAE :

$\huge \boxed{ \sf{ Tiles \: Required = \bigg( \dfrac{ Area \: of \: the \: floor }{Area \: of \: each \: tile} \bigg)}}$

★══════════════════════★

➲ Putting all values :-

➨ $\:\bf{Tiles \: required= \dfrac{15004 \cancel0}{22 \cancel0} }$➨Tilesrequired=

22

0

15004

0

➨$\:\bf{Tiles \: required= \dfrac{ \cancel{15004} ^{682}}{ \cancel{22}_{1} }}$➨Tilesrequired= 22 0

➠$\: \underline{\boxed{\bf{Tiles \: required= 682} }}$➠

Tilesrequired=682

★══════════════════════★

❹ Finally we find the cost of the tiles:-

We have :-

❍ Rate per tile = ₹ 6.50

❐ USING FORMULAE :

$\huge \boxed{ \sf{ Cost \: of \: tiles= ₹\bigg( Rate × \: tiles \: required \bigg)} }$

Costoftiles=₹(Rate×tilesrequired)

★══════════════════════★

➲ Putting all values :-

➨$\:\bf{ Cost \: of \: tiles \: = ₹\bigg(6.50 \times 682 \bigg)}$➨Costoftiles=₹(6.50×682)

➠ $\: \underline{ \boxed{\bf{ Cost \: of \: tiles \: = ₹4433 }}}$➠

Costoftiles=₹4433

♣ANSWER :

Our required answer is ₹4433