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MKU QUANTITATIVE APTITUDE : Unit -I SMTJN61

MKUQA1

Contents

Chapter 1 Ratio and Proportion

1.1 Important Facts and Formulae

    1. Ration : The ratio of two quantities \(a\) and \(b\) in the same units, is the fraction \(\dfrac {a}{b}\) and we write it as \(a:b\).

    Example : The ratio \(5: 9\) represents \(\dfrac {5}{9}\) with antecedent \(=5\), consequent \(=9\).

    Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio.

    Example : \(4: 5=8: 10=12: 15\) etc Also, \(4: 6=2: 3\).

    2. Proportion : The equality of two ratios is called proportion.

    If \(a: b=c: d\), we write, \(a: b:: c: d\) and we say that \(a, b, c, d\) are in proportion. Here \(a\) and \(d\) are called extremes, while \(b\) and \(c\) are called mean terms,

    Product of means \(=\) Product of extremes. Thus,

    \[a: b:: c: d \Leftrightarrow (b \times c)=(a \times d).\]

      (i) Fourth Proportional : If \(a:b = c:d\), then \(d\) is called the fourth proportional to \(a,b,c\).

      (ii) Third Proportional : If \(a: b = b: c\), then \(c\) is called the third proportional to \(a\) and \(b\).

      (iii) Mean Proportional : Mean proportional between \(a\) and \(b\) is \(\sqrt {a b}\).

    3.

      (i) Comparison of ratios: We say that \((a: b)>(c: d) \Leftrightarrow \dfrac {a}{b}>\dfrac {c}{d}\).

      (ii) Compounded ration : The compounded ratio of the ratios \((a: b),(c: d),(e: f)\) is \((ace: bdf)\).

    4.

      (i) Duplicate ratio of \((a: b)\) is \(\left (a^2: b^2\right )\).

      (ii) Sub-duplicate ratio of \((a: b)\) is \((\sqrt {a}: \sqrt {b})\).

      (iii) Triplicate ratio of \((a ; b)\) is \(\left (a^3: b^3\right )\).

      (iv) Sub-triplicate ratio of \((a: b)\) is \(\left (a^{\dfrac {1}{3}}: b^{\dfrac {1}{3}}\right )\).

      (v) If \(\dfrac {a}{b}=\dfrac {c}{d}\), then \(\dfrac {a+b}{a-b}=\dfrac {c+d}{c-d}\). (componendo and dividendo)

    5. Variation :

      (i) We say that \(x\) is directly proportional to \(y\), if \(x=k y\) for some constant \(k\) ad we write, \(x \propto y\).

      (ii) We say that \(x\) is inversely proportional to \(y\), if \(x y=k\) for some constant \(k\) and we write, \(x \propto \dfrac {1}{y}\).

1.2 Solved Problems

    1. If \(a:b=5:9\) and \(b:c=4:7\), find \(a:b:c\).

    Solution :
    \(a:b=5:9\) and \(b:c=4:7 = \left ( 4 \times \dfrac {9}{4} \right ) : \left ( 7 \times \dfrac {9}{4} \right ) = 9 : \dfrac {63}{4} \)

    \[ \Rightarrow a: b: c=5: 9: \dfrac {63}{4}=20: 36: 62 . \]

    2. Find

      (i) the fourth proportional to 4, 9, 12,

      (ii) the third proportional to 16 and 36 ,

      (iii) the mean proportional between \(0.08\) and \(0.18\).

    Solution :

      (i) ) Let the fourth proportional to \(4,9,12\) be \(x\)

      Then, \(4: 9:: 12: x \Rightarrow 4 \times x=9 \times 12 \Rightarrow x=\dfrac {9 \times 12}{4}=27\).

      Fourth proportional to 4, 9, 12 is 27.

      (ii) Let the third proportional to 16 and 36 be \(x\). Then,

      \[16: 36:: 36: x \Rightarrow 16 \times x=36 \times 38 \Rightarrow x=\dfrac {36 \times 36}{16}=81.\]

      Third proportional to 16 and 36 is 81 .

      (iii) Mean proportional between \(0.08\) and \(0.18\)

      \[ =\sqrt {0.08 \times 0.18}=\sqrt {\dfrac {8}{100} \times \dfrac {18}{100}}=\sqrt {\dfrac {144}{100 \times 100}}=\dfrac {12}{100}=0.12 \]

    3. If \(x: y=3: 4\), find \((4 x+5 y):(5 x-2 y)\).

    Solution :

    \[\dfrac {x}{y}=\dfrac {3}{4} \Rightarrow \dfrac {4 x+5 y}{5 x-2 y}=\dfrac {4\left (\dfrac {x}{y}\right )+5}{5\left (\dfrac {x}{y}\right )-2}=\dfrac {\left (4 \times \dfrac {3}{4}+5\right )}{\left (5 \times \dfrac {3}{4}-2\right )}=\dfrac {(3+5)}{\left (\dfrac {7}{4}\right )}=\dfrac {32}{7}\]

    4. Divide Rs. 672 in the ratio \(5: 3\).

    Solution :
    Sum of ratio terms \(=(5+3)=8\).

    First part \(=\) Rs. \(\left (672 \times \dfrac {5}{8}\right )=\) Rs. \(420 ;\)

    Second part \(=\) Rs. \(\left (672 \times \dfrac {3}{8}\right )=\) Rs. 252.

    5. Divide Rs. 1162 among A, B, C in the ntio \(35: 28: 20\).

    Solution :
    Sum of ratio terms \(=(35+28+20)=83\).

    A’s share \(=\) Rs. \(\left (1162 \times \dfrac {35}{83}\right )=\) Rs. 490;

    B’s share \(=\) Rs. \(\left (1162 \times \dfrac {28}{83}\right )=\) Rs 392;

    C’s share \(=\) Rs. \(\left (1162 \times \dfrac {20}{83}\right )=\) Rs. 280.

    6. A bag contains \(50 \mathrm {p}, 25 \mathrm {p}\) and \(10 \mathrm {p}\) coins in the ratio \(5: 9: 4\), amounting to R. 206. Find the number of coins of each type.

    Solution :
    Let the number of \(50 \mathrm {p}, 25 \mathrm {p}\) and \(10 \mathrm {p}\) coins be \(5 x, 9 x\) and \(4 x\) respectively.

    Then, \(\dfrac {5 x}{2}+\dfrac {9 x}{4}+\dfrac {4 x}{10}=206\)

    \(\Rightarrow \quad 50 x+45 x+8 x=4120 \Rightarrow 103 x=4120 \Rightarrow \quad \Rightarrow \quad x=40\).

    \(\therefore \quad \) Number of \(50 \mathrm {p}\) coins \(=(5 \times 40)=200\);

    Number of \(25 \mathrm {p}\) coins \(=(9 \times 40)=360\);

    Number of \(10 \mathrm {p}\) coins \(=(4 \times 40)=160\).

    7. A mixture contains alcohol and water in the ratio \(4:3\). If 5 litres of water is added to the mixture, the ratio becomes \(4:5\). Find the quantity of alcohol in the given mixture.

    Solution :
    Let the quantity of alcohol and water be \(4x \) litres and \(3x\) litres respectively. Then

    \[ \dfrac {4x}{3x+5} = \dfrac {4}{5} \Rightarrow 20x = 4 (3x+5) \Rightarrow 4x = 20 \Rightarrow x = 2.5 \]

    Quantity of alcohol \(=(4 \times 2.5)\) litres = 10 litres.

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