Quant Boosters - Vikas Saini - Set 1



  • Rohit + Dhawan = 2 x 280 = 560.
    Virat + Rahane = 2 x 200 = 400.
    Total runs = 560 + 400 = 960.
    Average = 960 / 4 = 240.

    Short approach :-
    Take average of any two
    let we take Rohit & Dhawan,which is nothing but 280.(base)
    now average of virat & rahane 200.
    which is -80 (from base)
    Average = 280 - (80/2)
    = 280 - 40
    = 240.

    We can solve it by take 200 too..
    base = 200.
    average of Rohit & Dhawan is +80(from base).
    average = 200 + (80/2)
    = 240.
    Here divided by 2, because that is average of two person..



  • Q11) The average weight of a class of 29 students is 40 kg. If the weight of teacher be included,the average rises by 500gm. What is the weight of the teacher ?



  • previous weight of class = 29 x 40 = 1160 kg.
    new weight (including teacher) = 30 x 40.5 = 1215.
    Teacher's weight = 1215 - 1160 = 55 kg.

    Short approach :-
    base = 40.
    including teacher average increased by 0.5 kg.
    Teacher's weight = 40 + 30 x 0.5
    = 40 + 15
    = 55 kg.



  • Q12) By selling 18 oranges profit is equal to SP of 3 oranges. Find profit percentage.



  • Sp of 1 orange = x.
    cp of 1 orange = y.
    18x – 18y = 3x
    = > 18 y = 15 x.
    Profit percentage = (3x/18y)x100
    = (3x/15x)x100
    = 20%.

    Short approach :-
    Profit percentage = [profit (quantity) / (Total quantity – profit)] x 100
    = [(3)/(18-3)] x 100
    = [3/15] x 100
    = 20%



  • Q13) By selling 18 chocolates, vendor loss selling price of 2 chocolates. Find loss percentage.



  • sp of 1 chocolate = x
    cp of 1 chocolate = y
    18 y – 18 x = 2x.
    18 y = 20 x.
    Loss percentage = (2x/18 y) x 100
    = (2x/20x)x100
    = 10%.

    Short approach :-
    Loss percentage = [Loss (quantity) / (Total quantity + loss) ] x 100
    = [ 2 / (18+2) ] x 100
    = 10%.



  • Q14) On selling an article for rs 576 a trader losses 4%. In order to gain 25/6 % , he must sell article for
    a. 655 rs
    b. 676 rs
    c. 625 rs
    d. 600 rs.



  • Cost price = [576 / (100 – 4)] x 100 = 600 rs.
    Selling price = 600 x (1 + 25/(6 x 100) = 625.

    Short approach :-
    Direct formula to find sp in such problems
    = previous selling price x 100 x (1+ gain percent)/(1 – loss percent)
    = 576 x 100 x (1+25/600)/(1 – 4/100)
    = 625 rs.



  • Q15) A goldsmith has 361 rings of gold. He sells some of them at a loss of 4% and rest at a profit of 15%. Overall profit is 8%. Find no of rings sold at profit of 15%.



  • Suppose he sold x no of rings at profit of 15%.
    (361-x)x(1-4/100)+x (1+15/100) = 361 x (1+8/100)
    361 x 0.96 – 0.96x + 1.15x = 361x1.08
    0.19 x = 361(1.08-0.96)
    X = 361 x (0.12) / (0.19) = 228.

    Short approach :-
    By allegation

    -4 -- 15

    -- 8 --

    7 -- 12

    No of rings = [(12)/(12+7)] x 361 = 228.



  • Q16) On selling 17 balls at rs 720, there is loss equal to the cost price of 5 balls.The cost price of a ball is
    a. 45 rs
    b. 50rs
    c. 55rs
    d .60 rs.



  • Short approach :-
    Cp of ball = selling price of all balls / (total balls – loss(no of balls)
    = 720 / (17-5 )
    = 60 rs.



  • Q17) A, B and C start simultaneously from X to Y. A reaches Y, turns back and meet B at a distance of 11 km from Y. B reached Y, turns back and meet C at a distance of 9 km from Y. If the ratio of the speeds of A and C is 3:2, what is the distance between X and Y?



  • Suppose distance between X and Y is D km.
    A = D+11, B = D-11.
    B = D+9, C = D-9.
    A/C = 3:2.
    (AB) / (BC) = 3/2.
    (D+11) (D+9) / (D-11) (D-9) = 3 / 2.
    D^2 + 20D + 99 / D^2 – 20D + 99 = 3 / 2
    2(D^2 + 20D + 99) = 3(D^2 – 20 D + 99)
    D^2 – 100 D + 99 = 0.
    D = 99.



  • Q18) A shopkeeper uses a weight of of 460 gm instead of 500 gm and sells the article at the cost price. What is profit percentage ?
    a. 40%
    b. 23%
    c. 8 +16/23 %
    d. 20%



  • Short approach :
    Profit percentage = [ (500 – 460)/460] x 100
    = (40/460) x 100



  • Q19) A 30% solution of alcohol is mixed with a 50% solution of alcohol to form a 10 litre solution of 45% alcohol. How much of 30% solution was used.
    (a) 2 ltr
    (b) 2.5 ltr
    (c) 2.7 ltr
    (d) 3.2 ltr
    (e) 3.7 ltr



  • By allegation
    30 --- 50
    -- 45 --
    (50-45 ) : (45-30)
    5 : 15 = 1 : 3.
    Ratio = 1 : 3
    30% solution was added = 10 x 1 / (1 + 3) = 2.5 ltr.



  • Q20) The percentage volumes of milk in three solutions A,B and C form a geometric progression in that order. If we mix the first, second and third solutions in the ratio 2:3:4, by volume we obtain solution containing 32% milk. If we mix them in the ratio 3:2:1, by volume, we obtain a solution containing 22% milk. What is the percentage of milk in A?
    a) 6%
    b) 12%
    c) 18%
    d) 24%


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