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Solve the following problems: 1. If 540 oranges are divided equally among 10 pupils, how many oranges does each pupil get? 2. How many times is 14 contained in 3500? 3. Change the following mixed numb

Mathematics

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Here are the solutions to your homework questions:

1.

Step 1: Divide the total number of oranges by the number of pupils.

Oranges per pupil=540 oranges10 pupils\text{Oranges per pupil} = \frac{540 \text{ oranges}}{10 \text{ pupils}}

Step 2: Calculate the result.

Oranges per pupil=54\text{Oranges per pupil} = 54

Each pupil will have 54 oranges.

54 oranges\boxed{\text{54 oranges}}

2.

Step 1: Divide 3500 by 14 to find how many times 14 is contained in 3500.

350014\frac{3500}{14}

Step 2: Perform the division.

350014=250\frac{3500}{14} = 250

14 is contained 250 times in 3500.

250\boxed{250}

3. Changing them to improper fractions:

Part a· 2 steps

1141 \frac{1}{4}

  1. Multiply the whole number by the denominator and add the numerator.

    Keep the same denominator.

    114=(1×4)+141 \frac{1}{4} = \frac{(1 \times 4) + 1}{4}

  2. Simplify the numerator.

    4+14=54\frac{4 + 1}{4} = \frac{5}{4} 54\boxed{\frac{5}{4}}

Part b· 3 steps

2362 \frac{3}{6}

  1. Multiply the whole number by the denominator and add the numerator.

    Keep the same denominator.

    236=(2×6)+362 \frac{3}{6} = \frac{(2 \times 6) + 3}{6}

  2. Simplify the numerator.

    12+36=156\frac{12 + 3}{6} = \frac{15}{6}

  3. Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 3.

    15÷36÷3=52\frac{15 \div 3}{6 \div 3} = \frac{5}{2} 52\boxed{\frac{5}{2}}

Part c· 2 steps

3233 \frac{2}{3}

  1. Multiply the whole number by the denominator and add the numerator.

    Keep the same denominator.

    323=(3×3)+233 \frac{2}{3} = \frac{(3 \times 3) + 2}{3}

  2. Simplify the numerator.

    9+23=113\frac{9 + 2}{3} = \frac{11}{3} 113\boxed{\frac{11}{3}}

Part d· 2 steps

6346 \frac{3}{4}

  1. Multiply the whole number by the denominator and add the numerator.

    Keep the same denominator.

    634=(6×4)+346 \frac{3}{4} = \frac{(6 \times 4) + 3}{4}

  2. Simplify the numerator.

    24+34=274\frac{24 + 3}{4} = \frac{27}{4} 274\boxed{\frac{27}{4}}

Part e· 11 steps

101410 \frac{1}{4}

  1. Multiply the whole number by the denominator and add the numerator.

    Keep the same denominator.

    1014=(10×4)+1410 \frac{1}{4} = \frac{(10 \times 4) + 1}{4}

  2. Simplify the numerator.

    40+14=414\frac{40 + 1}{4} = \frac{41}{4} 414\boxed{\frac{41}{4}}

    4.
    This is an inverse proportion problem: more time means fewer pupils are needed.

  3. Set up the inverse proportion relationship

    P1T1=P2T2P_1 T_1 = P_2 T_2.

    27 pupils×3 hours=P2×6 hours27 \text{ pupils} \times 3 \text{ hours} = P_2 \times 6 \text{ hours}

  4. Calculate the total "pupil-hours" of work.

    81 pupil-hours=P2×6 hours81 \text{ pupil-hours} = P_2 \times 6 \text{ hours}

  5. Solve for P2P_2.

    P2=81 pupil-hours6 hoursP_2 = \frac{81 \text{ pupil-hours}}{6 \text{ hours}} P2=13.5P_2 = 13.5

    Since you can't have half a pupil, and you need enough pupils to eat the food, you would need 14 pupils. (If the question implies minimum whole pupils, it's 14. If it implies exact rate, it's 13.5. Assuming practical application, round up).

    14 pupils\boxed{\text{14 pupils}}

    5.
    This is an inverse proportion problem: fewer days mean more men are needed.

  6. Set up the inverse proportion relationship

    M1D1=M2D2M_1 D_1 = M_2 D_2.

    6 men×10 days=M2×3 days6 \text{ men} \times 10 \text{ days} = M_2 \times 3 \text{ days}

  7. Calculate the total "man-days" of work.

    60 man-days=M2×3 days60 \text{ man-days} = M_2 \times 3 \text{ days}

  8. Solve for M2M_2.

    M2=60 man-days3 daysM_2 = \frac{60 \text{ man-days}}{3 \text{ days}} M2=20M_2 = 20

    20 men can build the shed in 3 days.

    20 men\boxed{\text{20 men}}

    6.
    This is a direct proportion problem: if the distance doubles, the time taken will also double at the same rate.

  9. Identify the given information.

    D1=50 km,T1=15 hoursD_1 = 50 \text{ km}, T_1 = 15 \text{ hours} D2=100 km,T2=?D_2 = 100 \text{ km}, T_2 = ?

  10. Notice that D2D_2 is twice D1D_1.

    D2=2×D1D_2 = 2 \times D_1

  11. Since the rate is the same, T2T_2 will also be twice T1T_1.

    T2=2×T1T_2 = 2 \times T_1 T2=2×15 hoursT_2 = 2 \times 15 \text{ hours} T2=30 hoursT_2 = 30 \text{ hours}

    It will take 30 hours to travel 100 kilometres.

    30 hours\boxed{\text{30 hours}}
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