# 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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