1. Question: The current birth rate per thousand is `32` whereas corresponding death rate is `11` per thousand. The net growth rate in terms of population increase in percent is given by.

    A
    `0.021%`

    B
    `0.0021%`

    C
    `21%`

    D
    `2.1%`

    Note: Net growth on `1000 = (32 - 11)` `= 21` Net growth on`100 = (21/1000 xx 100)` `= 2.1%`.
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  2. Question: A reduction of `21%` in the price of what enables a person to buy `10.5kg`more for Rs.`100`. What is the reduced price per kg ?

    A
    `Rs.2`

    B
    `Rs. 2.22`

    C
    `Rs. 2.30`

    D
    `Rs.2.50`

    Note: Let original price = Rs. x∕kg. Reduced price `= (79/100 x)∕kg` `(100)/(79x)/(100) - 100/x` `= 10.5 => (10000)/(79x) - 100/x` `= 10.5` `10000 - 7900` `= 10.5 xx 79x` `or x = 2100/(10.5 xx 79)` :. Reduced price`Rs. (79/100 xx 2100/(10.5 xx 79)∕kg` `= Rs. 2 per kg.`
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  3. Question: A man's basic pay for a `40` hour week is Rs.`20`. overtime is paid for at `25%` above the basic rate. In a center week, he worked overtime and his total wage was Rs. `25.` He therefore, worked for a total of ;

    A
    `45 hours`

    B
    `47 hours`

    C
    `48 hours`

    D
    `50 hours`

    Note: Basic rate = Rs.`20/40` per hour. overtime rate = Rs. `(125/100 xx 20/40)` per hour `= Rs. 5/8` per hr. Let overtime hours = x Then,`5/8 x = 5 => x = (5 xx 8)/5` `= 8 hours` So, he worked for (40 + 8) hours `= 48 hours`
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  4. Question: Salaries of A,B and C are in the ratio `1 : 2 : 3`. Salary of B and C together is Rs. `6000`. By what percent is salary of C more than that of A ?

    A
    `300`

    B
    `600`

    C
    `100`

    D
    `200`

    Note: Let A = x, B = 2x and C = 3x. Then, `2x + 3x = 6000 => x = 1200.` `:. A = 1200 and C = 3600` Required excess `= (2400/1200 xx 100)%` `= 200%`
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  5. Question: The price of a table is Rs. `400` more than that of a chair. If `6` tables and `6` chairs together cost Rs.`4800` by what percent is the price of the chairs less than that of the table ?

    A
    `200`

    B
    `400`

    C
    `100`

    D
    None of these

    Note: `T = C + 400. Now 6T + 6C` `= 4800` `:. 6 (C + 400) + 6C` `= 4800 or C = 200.` Thus, `C = 200 & T =(200 + 400)` `= 600`. Required percerntage `= (400/600 xx 100)%` `= 66 2/3%`
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  6. Question: If the side of a square is increased by `25%`then its area is increased by :

    A
    `25%`

    B
    `55%`

    C
    `40.5%`

    D
    `56.25%`

    Note: Let side` = 10 cm.` Then, area `= 100 cm^2` New side`= 125% of 10 = 12.5 cm.` Area = (12.5)^2 `= 156.25.` :. Incresae percent`= 56.25%`
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  7. Question: The length and breadth of a square are in increased by `40%` and `30%` respectively. The area of the resulting rectangle exceeds the area of the square by :

    A
    `42%`

    B
    `62%`

    C
    `82%`

    D
    None

    Note: Let `∕ = 10m & b = 10m.` Then, area`= 100 m^2`. length`= 140% of 10 = 14 m.` breadth`= 130% of 10 = 13 m.` New area`= (14 xx 13) m^2` `= 182 m^2` Increases in area`= 82%`.
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  8. Question: The length of a rectangle is increased by `20%` and the width is decreased by `20%`. The area decreases by :

    A
    `0.8%`

    B
    `1.2%`

    C
    `4%`

    D
    `8%`

    Note: `Let∕`= 10 m & b = 10 m. Then, area`= 100 m^2` New length`= 120% of 10 = 12 m.` New breadth`= 80% of 10 = 8 m`. New area`= (12 xx 8) m^2` `= 96 m^2` Decrease`= 4%`.
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  9. Question: The length of a rectangle is increased by `60%` and the width is decreased by `20%`. The area decreases by :

    A
    `37 1/2%`

    B
    `60%`

    C
    `75%`

    D
    None

    Note: Let length = l and breadth = b Let the required decrease in breadth be x%. Then,`160/100 l xx ((100 - x))/100 xx b` `= lb => 160 (100 - x)` `= 100 xx 100` `or 100 - x` `= 10000/160` `= 125/2 => x` `= (100 - 125/2)` `= 37 1/2.`
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  10. Question: If the radius of a circle is decreased by `50%`, its area is reduced by :

    A
    `25%`

    B
    `50%`

    C
    `75%`

    D
    None

    Note: Let original radius be R. Then, area` = π R^2` New radius `= 50% of R = 50/100 R = R/2.` New area`= π xx (R/2)^2` `= (πR^2)/4` decrease`= (πR^2 - (π R^2)/4)` `= (3 π R^2)/4` :. Decrease `% = ((3 πR^2)/(4) xx 1/(π R^2) xx 100)%` `= 75%`.
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