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A commercial bank buys and sells 100 Japanese Yen in Kenya shillings at the rates shown. A tourist ended his tour in Kenya and was left with Ksh 20 000 which he converted to Japanese yen through the commercial bank. How many Japanese yen, to the nearest whole number, did he get?

Mathematics

ScanSolve AI Answer

Answer

D

Step-by-step solution

11 steps

Here are the solutions to your questions:

:
The tourist has Ksh 20,000 and wants to convert it to Japanese Yen. This means the bank will be selling Japanese Yen to the tourist.
The bank's selling rate is Ksh 83.96 for 100 Japanese Yen.

  1. Determine the amount of Japanese Yen per Kenya Shilling.

    JPY per Ksh=100 JPY83.96 Ksh\text{JPY per Ksh} = \frac{100 \text{ JPY}}{83.96 \text{ Ksh}}
  2. Calculate the total Japanese Yen received for Ksh 20,000.

    Total JPY=20000 Ksh×100 JPY83.96 Ksh\text{Total JPY} = 20000 \text{ Ksh} \times \frac{100 \text{ JPY}}{83.96 \text{ Ksh}} Total JPY=200000083.96\text{Total JPY} = \frac{2000000}{83.96} Total JPY≈23821.093\text{Total JPY} \approx 23821.093
  3. Round the total Japanese Yen to the nearest whole number.

    Total JPY≈23821\text{Total JPY} \approx 23821

    The correct option is D.
    The tourist received 23,821 Japanese Yen.

    :
    We need to find the missing mean mark for Section II using the weighted mean formula.
    The overall mean mark is given by:

    Overall Mean=∑(Number of questions×Mean mark)∑(Number of questions)\text{Overall Mean} = \frac{\sum (\text{Number of questions} \times \text{Mean mark})}{\sum (\text{Number of questions})}
  4. Calculate the total number of questions.

    Total questions=43+48+45=136\text{Total questions} = 43 + 48 + 45 = 136
  5. Set up the equation for the overall mean.

    Let xx be the mean mark for Section II.

    64.22=(43×65.18)+(48×x)+(45×62.37)13664.22 = \frac{(43 \times 65.18) + (48 \times x) + (45 \times 62.37)}{136}
  6. Calculate the sum of products for the known sections.

    43×65.18=2802.7443 \times 65.18 = 2802.74 45×62.37=2806.6545 \times 62.37 = 2806.65
  7. Substitute these values back into the equation and solve for xx.

    64.22=2802.74+48x+2806.6513664.22 = \frac{2802.74 + 48x + 2806.65}{136} 64.22×136=5609.39+48x64.22 \times 136 = 5609.39 + 48x 8733.92=5609.39+48x8733.92 = 5609.39 + 48x 48x=8733.92−5609.3948x = 8733.92 - 5609.39 48x=3124.5348x = 3124.53 x=3124.5348x = \frac{3124.53}{48} x≈65.094375x \approx 65.094375
  8. Round the result to 2 decimal places.

    x≈65.09x \approx 65.09

    The correct option is B.
    The missing mean mark for section II is 65.09\boxed{\text{65.09}}.

    :
    We need to find the three consecutive days with the highest total number of eggs sold by reading the bar graph.

  9. Read the number of eggs sold for each day from the graph.

    • Monday: 45 eggs
    • Tuesday: 75 eggs
    • Wednesday: 15 eggs
    • Thursday: 60 eggs
    • Friday: 30 eggs
    • Saturday: 70 eggs
    • Sunday: 40 eggs
  10. Calculate the sum of eggs for each set of three consecutive days.

    • Monday, Tuesday, Wednesday: 45+75+15=13545 + 75 + 15 = 135 eggs
    • Tuesday, Wednesday, Thursday: 75+15+60=15075 + 15 + 60 = 150 eggs
    • Wednesday, Thursday, Friday: 15+60+30=10515 + 60 + 30 = 105 eggs
    • Thursday, Friday, Saturday: 60+30+70=16060 + 30 + 70 = 160 eggs
    • Friday, Saturday, Sunday: 30+70+40=14030 + 70 + 40 = 140 eggs
  11. Identify the set of days with the highest total.

    The highest total is 160 eggs, which occurred on Thursday, Friday, and Saturday.
    The correct option is D.
    The three consecutive days with the highest total number of eggs sold were Thursday, Friday, Saturday.

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