Hello sir , kindly share the solutions to ques 28 and 29 -
Hi Nikita,
PFA solution to Q28 below:
Applying simple alligation, we write 5% in left as this is what she consumes at present. We further write 100% on right because she will now consume pure glucose (which means 100% glucose) and 10% in the middle as this is what she desires to consume.
29th
30% of 200 ml means 60ml of from first , applying alligation
Sir in ques 28 how we got 540 and 30?
because by allegation ratio of 5% solution to pure solution is 90:5
so when 5% solution is 540 =(90*6) , then pure solution should be 30 = (5*6)
Hello sir , kindly share the solution of this question -
By alligation rule
[6% ] [ 50%]
10%
[ 55 - x ] [ x ]
(55 - x )/ x = ( 10- 6)/ ( 50 - 10)
x = 5 .
100 liters of milk solution with concentration of 40% is heated so as to evaporate the water. After the heating process is over, concentration of milk solution increases to 50%. What is the final quantity of the milk solution after heating?
50 liters
60 liters
M:W=2:3
after evaporating ratio becomes 1:1 ,since milk will not evaporate
M:W=2:2
earlier 2 x+3x=5x=100 now 2+2= 4x=80lt
i am getting 80 lt
in case if the answer is 60 lt and milk in it is 50% which is 30lt , how can it happen that milk gets evaporated because the initial solution had 40 lt of milk
solution for questions 14&16
14.
A= 7:5 and B= 1:1
First we have to make these quantities equal
7+5=12 and 1+1=2 so we have to multiply 6 in B
So now A= 7:5 AND B= 6:6
Now we have to take 2 of A and 3 of B so multiply it
A= 14:10 and B = 18:18
Now mix them so W:M In C = (14+18):(10+18)
= 8:7
M:W= 7:8 (answer)
solution please
Hello Sir , kindly share the solution to this problem -
Since, the Selling Price is not increased, to keep the profit percentage same the cost price of total mixture needs to be brought back to the original level by mixing the impurity
CP ___________New CP
100___________120
to bring back to the original level
120 ___________ 100 , % change = (20/120 ) x 100 = 16.66% = 1/6
So, impurity mixed is 1/6th of 12kg = 2kg .
Alternative Approach:
Use alligation rule
[120] [0]
[100]
[100] [20]
5 : 1
Hence impurity mixed is 1/6 of 12 = 2kg.
A mixture of milk and water contain 30%milk. What % of mixyure should be replaced with milk such tjat resultant mixture contain milk and water in the ratio 11:14.
Hello Nancy!
Let the total volume of mixture be 100ml
and the volume of the replaced mixture be 10x ml
[(30- 3x )+ 10x]/(70- 7x) = 11/14
on solving we get, x = 2
Hence replaced volume : 2 x 10 = 20 ml, which is 20% of the mixture.
Alternative Approach:
% of milk in original mixture : 30%
pure milk contains 100% milk
% of milk in Final mixture : 11/25 x 100 = 44%
[ 30%] [100%]
44%
[1- x] [ x]
56 14i.e
(1- x )/x = ( 100- 44)/(44-30)
x = 1/5 i.e 20%
We can assume anything x , 2x , 3x …10x,…
Since ratio of milk and water is 3 : 7 so it is easier to divide 10x in this ratio.
two glasses A and B contain milk and water mixed in the ratio 5:3 and 2:3. The ratio in which the liquid of these two glasses should be combined to form a new mixture containing half milk and half water is :
ratio of milk to total - 5/8 and 2/5
let's mix them in k:1
so ((5/8) *k + (2/5)*1)/(K+1) = 1/2
(25k + 16)/40 = (1/2)(k+1)
k =4/5
hence they should be mixed in the 4:5 ratio
a family uses a mixture of a blend of 2 coffees costing Rs30 and Rs 60 respectively. If the family uses only the expensive variety, it will have to spend Rs 450 more. The annual consumption of coffee of the family is 20 Kgs. In what ratio are the two blends being used ?
cost difference per kg = 30 (ie 30Rs will be extra paid when family shifts from cheaper to expensive coffee)
total difference 450, hence total extra kgs = 450/30 = 15 kg (ie 15 kg of cheaper coffee is replaced by 15 kg of expensive coffee)
so in the current blend there is 15 kg of cheaper coffee and 5 kg of expensive coffee.
Hence the ratio is 3:1
ritik and batik share a piece of land. the ratio of the area of ritik's portion to the area of batik's portion is 3:2. they each grow ghiya and tori on their piece of land. the entire piece of land is covered by ghiya and tori in the ratio 7:3 (by area). on ritik's portion of land, the ratio of area under ghiya to area under tori is 4:1. what is the ratio of ghiya to tori on batik's portion?
piece of land
Ritik - 3 (Ghiya - (4/5)*3 = 2.4
Batik -2 (Ghiya -(k*2) = 2k
Total -5 (Ghiya - (7/10)*5) = 3.5
so 2.4+2k = 3.5
k = 1.1/2 = 11/20
hence ratio of Ghiya to Tori = 11:9
a parantha shop in murthal gets an order of paranthas and 5 litres of lassi. the shop only has 3 litres of lassi at present. it decided to add water and curd to get 5 litres of lassi, which has 30% of curd by volume. if the original quantity of lassi had 40% curd by volume, how much water, in litres, did the shop owner add?
original lassi - 3 liters (40% curd)--> 1.2 litres curd (1200 ml), 1.8 liters water
5 liters lassi (30% curd) --> 1.5 liters (curd) and 3.5 liters water
Hence added quantity is 0.3 liters curd and 1.7 liters water
Bebe Sehgal is a musician who consumes a mixture of a concentrated drink 'Glucozade'
and Flavoured water 'Flavele' before a live show. She drank a 540 ml solution consisting
of 5% Glucozade and 95% Flavele but she feels most energetic when she consumes Glucozade and Flavele in the ratio 1:9 What quantity of pure Glucozade must she
consume to feel her best for the show?
A 30 ml
B. 60 ml
C. 120 ml
D. 240 ml
1:9 ration is 10% concentration
so using 5% and 100% , we have to make 10%
5 100
10
90 : 5
540 : 30
hence 30 ml of 100% concentration is required
hello sir,
kindly help me out to solve this question.
A chemist sells 3 variants of a medicine of concentrations 30%, 50% , 80%. The original cost of 100% medicine is Rs 80/unit. If he wants to make a mixture of these 3 variants and sell this mixture at Rs 60/unit , what cant be the respective ratios to enable him to earn a profit of 25%? (Assume that Cp of the medicine equals only the cost of medicine contained in it)
From a vessel full of pure petrol, 6 litres of petrol was drawn out and replaced by kerosene. Again 6 litres of mixture was drawn out from the vessel and replaced by kerosene. the ratio of petrol to kersene left in vessel was 25:24. Then volume of vessel was?
A. 14
B. 15
C. 18
D.21
(1-6/x)^2 = 25/49 (Ratio of petrol to total is 25/49)
from above x= 21