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Evaluate (36 - 8 x 4 - 15 / 3)/(-3 x 3 - 8(-6 + 2))

Mathematics

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Answers

2 parts
a

14.29%14.29\%

b

31.25%31.25\%

Step-by-step solution

6 steps

Here are the solutions to the problems:

1. Evaluate 36−8×4−15÷3−3×3−8(−6+2)\frac{36 - 8 \times 4 - 15 \div 3}{-3 \times 3 - 8(-6 + 2)}

Step 1: Evaluate the numerator using the order of operations (BODMAS/PEMDAS).

36−8×4−15÷3=36−32−5=4−5=−136 - 8 \times 4 - 15 \div 3 = 36 - 32 - 5 = 4 - 5 = -1

Step 2: Evaluate the denominator using the order of operations.

−3×3−8(−6+2)=−3×3−8(−4)=−9−(−32)=−9+32=23-3 \times 3 - 8(-6 + 2) = -3 \times 3 - 8(-4) = -9 - (-32) = -9 + 32 = 23

Step 3: Divide the numerator by the denominator.

−123\frac{-1}{23}

The evaluated expression is −123\boxed{-\frac{1}{23}}.

2. Simplify a+b−2aa + b - 2a

Step 1: Combine the like terms involving aa.

a−2a+b=−a+ba - 2a + b = -a + b

The simplified expression is b−a\boxed{b - a}.

3. Find the greatest number which divides 181 and 236 and leaves a remainder of 5 in each case.

Step 1: Subtract the remainder from each number to find the numbers that are perfectly divisible.

181−5=176181 - 5 = 176 236−5=231236 - 5 = 231

Step 2: Find the Greatest Common Divisor (GCD) of 176 and 231.
Prime factorization of 176=24×11176 = 2^4 \times 11
Prime factorization of 231=3×7×11231 = 3 \times 7 \times 11

Step 3: Identify the common prime factors and multiply them.
The only common prime factor is 11.
The GCD is 11.
The greatest number is 11\boxed{11}.

4. A rectangle measures 20cm by 15cm. If each dimension is increased by 2.5cm, by what percentage is:

  • Original length (L1L_1) = 20 cm
  • Original width (W1W_1) = 15 cm
  • New length (L2L_2) = 20+2.5=22.520 + 2.5 = 22.5 cm
  • New width (W2W_2) = 15+2.5=17.515 + 2.5 = 17.5 cm

Part a· 3 steps

the perimeter of the rectangle increased

  1. Calculate the original perimeter.

    P1=2(L1+W1)=2(20+15)=2(35)=70 cmP_1 = 2(L_1 + W_1) = 2(20 + 15) = 2(35) = 70 \text{ cm}
  2. Calculate the new perimeter.

    P2=2(L2+W2)=2(22.5+17.5)=2(40)=80 cmP_2 = 2(L_2 + W_2) = 2(22.5 + 17.5) = 2(40) = 80 \text{ cm}
  3. Calculate the percentage increase in perimeter.

    Percentage Increase=P2−P1P1×100%=80−7070×100%=1070×100%=17×100%≈14.29%\text{Percentage Increase} = \frac{P_2 - P_1}{P_1} \times 100\% = \frac{80 - 70}{70} \times 100\% = \frac{10}{70} \times 100\% = \frac{1}{7} \times 100\% \approx 14.29\%

    The perimeter is increased by approximately 14.29%\boxed{14.29\%}.

Part b

31.25%31.25\%

the area of the rectangle increased

  1. Calculate the original area.

    A1=L1×W1=20 cm×15 cm=300 cm2A_1 = L_1 \times W_1 = 20 \text{ cm} \times 15 \text{ cm} = 300 \text{ cm}^2
  2. Calculate the new area.

    A2=L2×W2=22.5 cm×17.5 cm=393.75 cm2A_2 = L_2 \times W_2 = 22.5 \text{ cm} \times 17.5 \text{ cm} = 393.75 \text{ cm}^2
  3. Calculate the percentage increase in area.

    Percentage Increase=A2−A1A1×100%=393.75−300300×100%=93.75300×100%=0.3125×100%=31.25%\text{Percentage Increase} = \frac{A_2 - A_1}{A_1} \times 100\% = \frac{393.75 - 300}{300} \times 100\% = \frac{93.75}{300} \times 100\% = 0.3125 \times 100\% = 31.25\%

    The area is increased by 31.25%\boxed{31.25\%}.

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