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Simplify the following ratios and solve for the missing value: a) 30:54 b) 20:40 c) 45:125 d) A:B = 2:3, if A=30 e) Share sh. 190 in the ratio 2:3.

Mathematics

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RATIOS Section

Part a· 2 steps

  1. Write the ratio as a fraction.

    3054\frac{30}{54}
  2. Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 6.

    30÷654÷6=59\frac{30 \div 6}{54 \div 6} = \frac{5}{9}

    The ratio is 5:9\boxed{\text{5:9}}.

Part b· 2 steps

  1. Write the ratio as a fraction.

    2040\frac{20}{40}
  2. Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 20.

    20÷2040÷20=12\frac{20 \div 20}{40 \div 20} = \frac{1}{2}

    The ratio is 1:2\boxed{\text{1:2}}.

Part c· 2 steps

  1. Write the ratio as a fraction.

    45125\frac{45}{125}
  2. Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 5.

    45÷5125÷5=925\frac{45 \div 5}{125 \div 5} = \frac{9}{25}

    The ratio is 9:25\boxed{\text{9:25}}.

Part d· 4 steps

  1. Set up the proportion.

    AB=23\frac{A}{B} = \frac{2}{3}
  2. Substitute the given value A=30A = 30.

    30B=23\frac{30}{B} = \frac{2}{3}
  3. Cross-multiply to solve for B.

    30×3=2×B30 \times 3 = 2 \times B 90=2B90 = 2B
  4. Divide by 2.

    B=902B = \frac{90}{2} B=45B = 45

    The value of B is 45\boxed{45}.

Part e· 3 steps

  1. Find the total number of parts in the ratio.

    2+3=5 parts2 + 3 = 5 \text{ parts}
  2. Find the value of one part.

    sh. 1905 parts=sh. 38 per part\frac{\text{sh. } 190}{5 \text{ parts}} = \text{sh. } 38 \text{ per part}
  3. Calculate each share.

    First share: 2×sh. 38=sh. 762 \times \text{sh. } 38 = \text{sh. } 76
    Second share: 3×sh. 38=sh. 1143 \times \text{sh. } 38 = \text{sh. } 114
    The shares are sh. 76 and sh. 114\boxed{\text{sh. } 76 \text{ and sh. } 114}.

Part f· 3 steps

  1. Find the total number of parts in the ratio.

    2+3+5=10 parts2 + 3 + 5 = 10 \text{ parts}
  2. Find the value of one part.

    1870 kg10 parts=187 kg per part\frac{1870 \text{ kg}}{10 \text{ parts}} = 187 \text{ kg per part}
  3. Calculate each share.

    First share: 2×187 kg=374 kg2 \times 187 \text{ kg} = 374 \text{ kg}
    Second share: 3×187 kg=561 kg3 \times 187 \text{ kg} = 561 \text{ kg}
    Third share: 5×187 kg=935 kg5 \times 187 \text{ kg} = 935 \text{ kg}
    The shares are 374 kg, 561 kg, and 935 kg\boxed{374 \text{ kg, } 561 \text{ kg, and } 935 \text{ kg}}.

Part g· 4 steps

  1. Set up the proportion.

    LS=35\frac{L}{S} = \frac{3}{5}
  2. Substitute the given value L=60L = 60.

    60S=35\frac{60}{S} = \frac{3}{5}
  3. Cross-multiply to solve for S.

    60×5=3×S60 \times 5 = 3 \times S 300=3S300 = 3S
  4. Divide by 3.

    S=3003S = \frac{300}{3} S=100S = 100

    The value of S is 100\boxed{100}.

Part h· 3 steps

  1. Set up the proportion.

    KH=79\frac{K}{H} = \frac{7}{9}
  2. Substitute the given value H=54H = 54.

    K54=79\frac{K}{54} = \frac{7}{9}
  3. Multiply both sides by 54 to solve for K.

    K=79×54K = \frac{7}{9} \times 54 K=7×6K = 7 \times 6 K=42K = 42

    The value of K is 42\boxed{42}.

Part h(i)· 21 steps

  1. Set up the proportion.

    NP=13\frac{N}{P} = \frac{1}{3}
  2. Substitute the given value P=42P = 42.

    (Assuming the question meant to find N, as P is given).

    N42=13\frac{N}{42} = \frac{1}{3}
  3. Multiply both sides by 42 to solve for N.

    N=13×42N = \frac{1}{3} \times 42 N=14N = 14

    The value of N is 14\boxed{14}.

    j)
    The question asks to find P, but P is not present in the given equation xy=213\frac{x}{y} = \frac{2}{13}. This question is incomplete or contains a typo. If it meant to find the ratio xy\frac{x}{y}, then it is already given.

    k)
    The question asks to find n, but the value of m is not given. This question is incomplete or contains a typo.

    l)

  4. Write the equation.

    117Q=534\frac{1}{17} Q = \frac{5}{34}
  5. Multiply both sides by 17 to solve for Q.

    Q=534×17Q = \frac{5}{34} \times 17 Q=5×1734Q = \frac{5 \times 17}{34} Q=5×172×17Q = \frac{5 \times 17}{2 \times 17}
  6. Simplify the fraction.

    Q=52Q = \frac{5}{2} Q=2.5Q = 2.5

    The value of Q is 2.5\boxed{2.5}.

    m)

  7. Write the ratio as a fraction.

    2575\frac{25}{75}
  8. Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 25.

    25÷2575÷25=13\frac{25 \div 25}{75 \div 25} = \frac{1}{3}

    The simplified ratio is 1:3\boxed{\text{1:3}}.

    n)

  9. Write the ratio as a fraction.

    836\frac{8}{36}
  10. Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4.

    8÷436÷4=29\frac{8 \div 4}{36 \div 4} = \frac{2}{9}

    The simplified ratio is 2:9\boxed{\text{2:9}}.

    o)

  11. Set up the proportion.

    AB=13\frac{A}{B} = \frac{1}{3}
  12. Substitute the given value B=240B = 240.

    A240=13\frac{A}{240} = \frac{1}{3}
  13. Multiply both sides by 240 to solve for A.

    A=13×240A = \frac{1}{3} \times 240 A=80A = 80

    The value of A is 80\boxed{80}.

    p)

  14. Set up the proportion.

    AB=97\frac{A}{B} = \frac{9}{7}
  15. Substitute the given value B=175B = 175.

    A175=97\frac{A}{175} = \frac{9}{7}
  16. Multiply both sides by 175 to solve for A.

    A=97×175A = \frac{9}{7} \times 175 A=9×25A = 9 \times 25 A=225A = 225

    The value of A is 225\boxed{225}.

    q)

  17. Set up the proportion.

    PT=911\frac{P}{T} = \frac{9}{11}
  18. Substitute the given value P=198P = 198.

    198T=911\frac{198}{T} = \frac{9}{11}
  19. Cross-multiply to solve for T.

    198×11=9×T198 \times 11 = 9 \times T 2178=9T2178 = 9T
  20. Divide by 9.

    T=21789T = \frac{2178}{9} T=242T = 242

    The value of T is 242\boxed{242}.

    r)

  21. Write

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