High School

1. The product of two decimal numbers is 131.58. If one of them is 2.15, find the other number.

2. The price of one meter of cloth is ₹62.85. Find the price of 23 meters of cloth.

3. A pile of books is 54.4 cm high. If each book is 3.2 cm thick, how many books are there in the pile?

4. A shirt requires 2.7 meters of cloth. How many such shirts can be made from a piece of cloth that is 40.5 meters long?

5. Shubham saves 0.2 part of his salary every month. If his monthly salary is ₹55,000, how much will he save in one year?

6. If 1 meter is equal to 39.37 inches, how many inches will be equal to 16 meters?

7. A school purchased 15 steel chairs for ₹1706.25. Find the cost of one steel chair.

8. How much would a dozen oranges cost if the price of one orange is ₹6.45?

9. In a school, 0.4 part of the total number of students is boys, and the number of girls is 360. Find the number of boys in the school.

10. The total weight of 73 bags of cement is 4106.25 kg. Find the weight of 100 such bags of cement.

11. A bus takes 4.5 hours to cover the distance of 180 km between Delhi and Roorkee. How much time will it take to cover a distance of 280 km?

12. The thickness of 12 sheets of paper is 2.16 mm. Find the thickness of 1 sheet of paper.

Answer :

Let's solve each question one by one:

  1. Finding the other number when the product of two decimals is given:
    We know that the product of two numbers is 131.58, and one of the numbers is 2.15.

    Let the other number be [tex]x[/tex]. So, the equation is:

    [tex]2.15 \times x = 131.58[/tex]

    To find [tex]x[/tex], divide both sides by 2.15:

    [tex]x = \frac{131.58}{2.15} = 61.2[/tex]

  2. Finding the price of multiple metres of cloth:
    The price of one metre of cloth is ₹62.85.

    For 23 metres, the total cost would be:

    [tex]62.85 \times 23 = ₹1445.55[/tex]

  3. Finding the number of books in a pile based on thickness:
    A pile of books is 54.4 cm high, with each book 3.2 cm thick.

    The number of books [tex]n[/tex] is:

    [tex]n = \frac{54.4}{3.2} = 17[/tex]

  4. Calculating how many shirts can be made from a given length of cloth:
    One shirt requires 2.7 m of cloth, and the available cloth is 40.5 m.

    The number of shirts [tex]n[/tex] is:

    [tex]n = \frac{40.5}{2.7} = 15[/tex]

  5. Calculating savings in a year from monthly savings:
    Shubham saves 0.2 part of his monthly salary, which is ₹55,000.

    Monthly savings: [tex]0.2 \times 55000 = ₹11,000[/tex]

    Yearly savings: [tex]11,000 \times 12 = ₹132,000[/tex]

  6. Converting meters to inches:
    Given 1 m = 39.37 inches. For 16 m:

    [tex]16 \times 39.37 = 629.92 \text{ inches}[/tex]

  7. Finding the cost of one steel chair:
    The total cost for 15 chairs is ₹1706.25. So, the cost of one chair is:

    [tex]\frac{1706.25}{15} = ₹113.75[/tex]

  8. Calculating the cost of a dozen oranges:
    The price of one orange is ₹6.45.

    A dozen (12) oranges will cost:

    [tex]12 \times 6.45 = ₹77.40[/tex]

  9. Finding the number of boys in a school:
    Given 0.4 part of students are boys and the number of girls is 360.

    If the total number of students is [tex]T[/tex], then:

    [tex]0.6T = 360[/tex]
    [tex]T = \frac{360}{0.6} = 600[/tex]
    Number of boys = [tex]0.4 \times 600 = 240[/tex]

  10. Calculating the weight of 100 bags of cement:
    Total weight of 73 bags is 4106.25 kg.

    Weight of one bag [tex]W[/tex] is:

    [tex]W = \frac{4106.25}{73} \approx 56.25 \text{ kg}[/tex]

    Weight of 100 bags:

    [tex]100 \times 56.25 = 5625 \text{ kg}[/tex]

  11. Calculating time to cover a different distance at the same speed:
    A bus covers 180 km in 4.5 hours, so its speed is [tex]\frac{180}{4.5} = 40 \text{ km/h}[/tex].

    Time to cover 280 km:

    [tex]\text{Time} = \frac{280}{40} = 7 \text{ hours}[/tex]

  12. Finding the thickness of one sheet of paper:
    The thickness of 12 sheets is 2.16 mm.

    Thickness of one sheet [tex]t[/tex] is:

    [tex]t = \frac{2.16}{12} = 0.18 \text{ mm}[/tex]

These calculations provide clear solutions to each problem.