All cash purchases are entered in the _____________book
Select one:
a. Trial balance
b. Journal proper
c. Purchase
d. Cash
Answers
All cash purchases are entered in the _____________book.
Answer: Cash Book.
All cash purchases are entered in the Cash book.
When the goods are purchased and the payment is done immediately by cash or bank, it is called as cash purchases.
In the cash book, all the cash as well as bank transactions are recorded. It is a book including the original entries. A cash book has two sides, Receipt and Payment side.
A simple cash book is prepared so as to maintain a record of all cash transactions and to find out the cash balance at the end of an accouting period.
Answer:
To find : The distance between the school and his house
Solution :
Speed of a student = 2 ½ km/h = 5/2km/h
★ A student reaches his school 6 minutes late.
Consider the time be x
60min = 1 hour
Time = (x + 6) min = (x + 6/60) = (x + 1/10)h
As we know that
★ Speed = distance/time
Consider the distance be y
\implies \sf \dfrac{5}{2} = \dfrac{y}{ x + \dfrac{1}{10}}⟹25=x+101y
\implies \sf \dfrac{5}{2} = \dfrac{y}{ \dfrac{10x + 1}{10} }⟹25=1010x+1y
\implies \sf \dfrac{5}{2} = y \times \dfrac{10}{10x +1}⟹25=y×10x+110
\implies \sf \dfrac{5}{2} = \dfrac{10y}{10x +1}⟹25=10x+110y
\implies \sf 5(10x +1) = 20y⟹5(10x+1)=20y
\implies \sf 50x +5=20y⟹50x+5=20y
\implies \sf 50x - 20y = - 5 \: \: \: \bf(Equation \: 1)⟹50x−20y=−5(Equation1)
★ Next day starting at the same time, he increases his speed by 1 km/hour and reaches 6 minutes early.
\implies \sf \dfrac{5}{2} + 1 = \dfrac{y}{ x - \dfrac{1}{10}}⟹25+1=x−101y
\implies \sf \dfrac{5 + 2}{2} = \dfrac{y}{\dfrac{10x - 1}{10}}⟹25+2=1010x−1y
\implies \sf \dfrac{7}{2} = \dfrac{10y}{10x - 1}⟹27=10x−110y
\implies \sf 7(10x - 1) = 20y⟹7(10x−1)=20y
\implies \sf 70x - 20y = 7 \: \: \: \bf(Equation \: 2)⟹70x−20y=7(Equation2)
Subtract both the equations
→ 50x - 20y - (70x - 20y) = -5 - 7
→ 50x - 20y - 70x + 20y = - 12
→ - 20x = - 12
→ x = 12/20 = 3/5
Put the value of x in eqⁿ (1)
→ 50x - 20y = - 5
→ 50 × 3/5 - 20y = - 5
→ 30 - 20y = - 5
→ 30 + 5 = 20y
→ 35 = 20y
→ y = 35/20 = 7/4
→ y = 1 3/4 km
•°• The distance between school and house is 1 3/4 km.