Simplification
 

  1. Question: In a certain office,`1/3` of the workers are women,`1/2`of the women are married and `1/3`of the married women have children.If`3/4`of the men are married and `3/4`of the married and `2/3` of the married men have children,what part of workers are without children ?

    A
    `5/18`

    B
    `4/9`

    C
    `11/18`

    D
    `17/36`

    Note: Let the total number of workers be x. Then,`w =x/3`and `M = (2x)/3` Women having children `= 1/3 of 1/2 of 1/3 x = x/18` Men having children `= 2/3 of 3/4 of (2x)/3 = x/3` Workers having children `= (x/18 + x/3) = (7x)/18` Workers having no children `= (x - (7x)/18) = (11x)/18` `= 11/18` of all workers.
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  2. Question: A train starts full of passengers and takes `280` more .At the seceond station,it drops one half of the new total anf takes `12`more .On arriving at the third station ,it is found to have `248`passengers .The number of passsengers in the begining was :

    A
    `156`

    B
    `288`

    C
    `564`

    D
    `608`

    Note: Let the number of passengers in the begining be x, After 1st station ,passengers `= (x - x/3) + 280` `= ((2x)/3 + 280)` After 2nd starion ,passengers `= ((2x)/3 +280) - 1/2 ((2x)/3 + 280) + 12` `:. 1/2((2x)/3 + 280) + 12= 248` `or (2x)/3 + 280` `= 2 xx 236` `or (2x)/3 = 192` `:. x = (192 xx 3/2) = 288`.
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  3. Question: One test tube contains some acid and another test tube contains an equal quantity o f water. To prepare a soluation ,`20`grams of the acid is poured into the second test tube .Then two thirds of the so formed solition is poured from the second tube into the first.If the fluid in the first test tube is four times that in the second ,what quantity of water was taken intially,

    A
    `40 grams`

    B
    `60 grams`

    C
    `80 grams`

    D
    `100 grams`

    Note: Suppose each tube contains x gms initially. `4[1/3(x + 20)]` `= x + 2/3 (x + 20)` or `2/3 (x + 20) = x` or` x/3 = 40/3` `:. x = 40`
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  4. Question: Four a group of boys and girls `15` girls leave.There are then left `2` boys for each girl.After this, `45` boys leave. There are then `5`girls for each boy.The number of girls in the begining was:

    A
    `40`

    B
    `43`

    C
    `29`

    D
    `50`

    Note: Let at present there be x boys. Then,the number of girls at present `= 5x`. Before the boys have left: Number of boys`= x + 45` and number of girls`= 5x` :.`x + 45 ` `= 2 (5x) ⇒ 9x` `= 45 ⇒ x = 5`. Hence,the number of girls in the beginig `= 5x + 15` `= 25 + 15` `= 40`
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  5. Question: A woker may claim `15` p for each km which he travels by taxi and `5`p for each km which he drives his own car.If in one week he claimed Rs `5` for travlling `80`km how many kms did he travel by taxi ?

    A
    `10`

    B
    `20`

    C
    `30`

    D
    `40`

    Note: Let the distance covered by taxi be x km Then, distance covered by car `= (80 - x)` km. `15/100 (x) + 5/100 (80 - x) = 5` or `(3x)/20 + (80 - x)/20 = 5` or `3x + 80 -x = 100` or `x = 10 km`
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  6. Question: In an examination , a student scores `4`marks for every correct answar and loses `1`mark for every wrong answar.If he attempts all `75` question and secures `125`marks, the number of question he attempts correctly,is :

    A
    `35`

    B
    `40`

    C
    `42`

    D
    `46`

    Note: Let the number of correct answars be x, Number of incorrect answars `= (75 - x)`. `4x - (75 - x) = 125` `or 5x = 200` `or x = 40`.
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  7. Question: Each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys in a class of `60`students .If the total contribution thus collected is Rs `1600`,how many boys are there in the class ?

    A
    `30`

    B
    `25`

    C
    `50`

    D
    Data inadequate

    Note: Let the number of boys `= x` Then,number of girls`= (60 - x)`. :.`x(60 - x) + (60 - x) x = 1600` or `60x - x^2 + 60x - x^2 = 1600` or `2x^2 - 120x + 1600 = 0` or `x^2 - 60x + 800 = 0` or `(x - 40)(x - 20)=0` :.`x = 40` or `20` So,we are not definite .Hence data is inadequate.
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  8. Question: The value of `36`coins of `10`p and `20`p is Rs `6.60`.The number of `20` p coins is :

    A
    `16`

    B
    `20`

    C
    `28`

    D
    `30`

    Note: Let the number of `20`paise coins be x. Then,number of `10` paise coins `= (36 - x). `10 (36 - x) + 20x = 660` `or 10x = 300` `or x = 30`.
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  9. Question: Madhuri had `85`currency notes in all, some of which are of Rs `100` denomination and the remaining of Rs `50`denomination.The total amount of all these currency notes was Rs `5000`.How much amount (in Rs) did she have in the denomination of Rs `50`.

    A
    `3500`

    B
    `70`

    C
    `1500`

    D
    None of these

    Note: Let the number of `50`rupee notes `= x` Then, the number of `100`-rupee notes `= (85 - x)`. `50x + 100 (85 - x) = 5000` `or x + 2 (85 - x)= 100` `or x = 70`
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  10. Question: The sum of three fraction is `2 11/24`.When the largest fraction is divided by the smallest, the fraction thus obtained is `7/6` which is `1/3` more than the middle one. The fractions are :

    A
    `3/5,4/7,2/3`

    B
    `7/8,5/6,3/4`

    C
    `7/9,2/3,3/5`

    D
    None of these

    Note: Let largest fraction be x and smallest be y. Then` x/y = 7/6` `or y = 6/7 x` Let the middle one be z Then,` x + 6/7 x + z = 59/24` `:. z = (59/24 - (13x)/7)` `:. 59/24 - (13x)/7 + 1/3 = 7/6` `or (13x)/7 = 59/24 + 1/3 - 7/6` `= (59 + 8 - 28)/24` `= (13x)/7 = 39/24` `or x = (39/24 xx 7/13) = 7/8` `:. x = 7/8, y = 6/7 xx 7/8= 3/4 ` `& z = 59/24 - 13/7 xx 7/8` `= 20/24` `= 5/6` :. Fraction are `7/8,5/6,3/4`
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