TCS Ninja Questions on Mixtures and Allegations Part 1

Explanation: -

Let the amount of 90% solution of concentrated acid added be A.

Let the amount of 75% solution of concentrated acid added be B.

Since total volume must be 30 L,

A + B = 30                                                              .......equation 1

Amount of acid in A L of 90% concentrated acid = A x

= 0.9 A

Amount of acid in B L of 75% concentrated acid = A x

= 0.75 B

Amount of acid in 30 L of 90% concentrated acid = 30 x

= 23.4

Thus 0.9 A + 0.75 B = 23.4                                          ....equation 2

By multiplying equation 1 by 0.9 we get

0.9 A + 0.9 B = 27                                                         .....equation 3

Subtracting equation 2 from 3

We get 0.15 B = 3.6

B = = 24

From equation 1

A + B = 30

A = 30 - B

= 30 - 24

= 6

a+b+c=48*3=144
a+b+c+d=47*4=188
b+c+d+e=48*4=192
d=188-144=44
e=3+44=47
b+c+d=192-47=145
a=188-145=43

Video solution:

let the no of i10 cars be x and i20 be y
now the rejected i10 cars are 4x/100 and i20 cars are 8y/100

and it is given that 4x/100 +8y/100 = 7(x+y)/100
so we get y = 3x
so the ratio is 3:1

total distance travelled by the car=40000km
so total distance travelled by 4 wheels=4*40000=160000
as all tyres(4+1) are equally used
so average distance travelled by the each tyre=160000/5=32000
Video solution:

Let the actual amount of A be 4x and actual amount of B be 3x since A and B are in the ratio 4/3

amount of first basic element in the new alloy T (5/8)*4x + (1/3)*3x = (7x)/2

amount of second basic element in the new alloy T (3/8)*4x + (2/3)*3x= (7x)/2

so ratio of first basic element to second basic element: [(7x)/2 ] / [ 7x/2] = 1/1 = 1:1 ( Answer A)

Video solution:

In questions of replacement where we initially have some quantity of pure element (like petrol), and we keep on replacing a fixed portion of this pure liquid every time with some other element (like water) thus making it a mixture, we apply the following formula for finding the quantity of pure element after ‘n’ replacements :
P x [1 – (R / P)]n, where
P is the initial quantity of pure element
R is the quantity replaced every time
n is the number of replacements

729(1-n/729)^6

Video solution:

Percentage of milk in first mixture

= [5/6] x 100 = 250/3 %

Percentage of milk in second mixture

= [7/9] x 100 = 700/9

Using Alligation method,



So, Required Ratio = [20/9] : [10/3] = 2:3

Video solution:

let the mixture contain \'x\' lit of milk and already given 16lit of water
as water don\'t cost (x+16) lit of mixture = x lit of pure milk
(x+16)90=(x*1.08*100) /*since given in paise*/
x=80lit

Video solution:

it is 33.33%
given 25 liters of sol has 20% sugar .
qty of sugar =(20%/100%)*25 =5 lit of sugar
(assume sugar qty in liter for easy calculation ).
therefore 25 lit sol contain 20 lit of water and 5 liter of sugar .
now the sol boils losing 10 liter of water .
therefore the new 15 liter sol contains 10 liter water and 5 liter sugar . there new % of sugar is 5/15 * 100 =33.33%

Video solution:

Consider 90 litre of solution1 and 60 litre of solution2

90 litre of solution1 contains 90*1/5=18 litre of milk and (90-18)=72 litre of water
60 litre of solution2 contains 60*1/3=20 litre of milk and (60-20)=40 litre of water

Mixture will have 18+20=38 litre of milk 72+40=112 litre of water
i.e., Quantity of milk : Quantity of water = 38 : 112 = 19 : 56

Take 75 litre of solution1. Quantity of milk = 75*1/5=15 litre. Profit earned in selling solution is 20%
assume cost price of 1 litre of milk = Rs.1
Cost price of 75 litre of solution1 = Rs.15.
Selling price of 75 litre of solution1 = 15*120/100 = Rs.18

Take 75 litre of mixture. Quantity of milk = 75*19/75=19 litre.
Cost price is Rs.19
Selling price will be same and is Rs.18
so there will be a loss
Loss = Rs.1 in selling 75 litre of the mixture
Percentage loss = 1*100/19 = 5.25% loss