We read your photo as:

The price of a book is thrice that of an exercise book. If a student spent #9600 on books and at 8000 on exercise books he has 86 items altogether. a) find the price of an exercise book. b) What is the difference between the price?

Mathematics

ScanSolve AI Answer

Answers

2 parts
a

560043 Naira\frac{5600}{43} \text{ Naira}

b

1120043 Naira\frac{11200}{43} \text{ Naira}

Step-by-step solution

4 steps
  1. Define variables and set up equations.

    Let PbP_b be the price of one book and PeP_e be the price of one exercise book.
    Let NbN_b be the number of books and NeN_e be the number of exercise books.

    From the problem statement, we have the following relationships:

    1. The price of a book is thrice that of an exercise book: Pb=3PeP_b = 3 P_e
    2. A student spent N9600 on books: NbPb=9600N_b P_b = 9600
    3. A student spent N8000 on exercise books: NePe=8000N_e P_e = 8000
    4. The total number of items (books and exercise books) is 86: Nb+Ne=86N_b + N_e = 86
  2. Express the number of books (NbN_b) and exercise books (NeN_e) in terms of PeP_e.

    Substitute Pb=3PeP_b = 3 P_e into the equation for the cost of books:

    Nb(3Pe)=9600N_b (3 P_e) = 9600 3NbPe=96003 N_b P_e = 9600 NbPe=96003N_b P_e = \frac{9600}{3} NbPe=3200  ⟹  Nb=3200PeN_b P_e = 3200 \implies N_b = \frac{3200}{P_e}

    From the equation for the cost of exercise books:

    NePe=8000  ⟹  Ne=8000PeN_e P_e = 8000 \implies N_e = \frac{8000}{P_e}
  3. Solve for the price of an exercise book (PeP_e).

    Substitute the expressions for NbN_b and NeN_e into the total number of items equation (Nb+Ne=86N_b + N_e = 86):

    3200Pe+8000Pe=86\frac{3200}{P_e} + \frac{8000}{P_e} = 86

    Combine the terms on the left side:

    3200+8000Pe=86\frac{3200 + 8000}{P_e} = 86 11200Pe=86\frac{11200}{P_e} = 86

    Now, solve for PeP_e:

    Pe=1120086P_e = \frac{11200}{86}

    Simplify the fraction by dividing the numerator and denominator by 2:

    Pe=560043P_e = \frac{5600}{43}
  4. Calculate the price of a book (PbP_b).

    Using the relationship Pb=3PeP_b = 3 P_e:

    Pb=3×560043P_b = 3 \times \frac{5600}{43} Pb=1680043P_b = \frac{16800}{43}

    a) Find the price of an exercise book.
    The price of an exercise book is PeP_e.

    Pe=560043 NairaP_e = \frac{5600}{43} \text{ Naira}

    As a decimal, Pe≈130.23P_e \approx 130.23 Naira.
    The price of an exercise book is 560043 Naira\boxed{\frac{5600}{43} \text{ Naira}}.

    b) What is the difference between the price?
    The difference between the price of a book and an exercise book is Pb−PeP_b - P_e.

    Difference=Pb−Pe\text{Difference} = P_b - P_e Difference=1680043−560043\text{Difference} = \frac{16800}{43} - \frac{5600}{43} Difference=16800−560043\text{Difference} = \frac{16800 - 5600}{43} Difference=1120043 Naira\text{Difference} = \frac{11200}{43} \text{ Naira}

    As a decimal, the difference ≈260.47\approx 260.47 Naira.
    The difference between the prices is 1120043 Naira\boxed{\frac{11200}{43} \text{ Naira}}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question