A pack of 240 ml, mixture of wine and water contains in the ratio 5: 7. How much wine must be mixed in the pack, so that the ratio of wine and water in the pack becomes 5:4?

- A) 75 ml B) 100 ml C) 115 ml D) 125 ml

Wine = 100 ml, water = 140 ml

(100 + x) /140 = 5/4 => 100 + x = 175

x = 175 - 100 = 75 ml

One sample of milk contain 22.22% of water by volume, while the other sample of milk contain 20% of water by volume. A glass is filled with 5 parts of first sample and 7 parts of second sample. Find out the percentage volume of the water in the mixture.

Hello Samyak ,

22.22% = 2/9 , 20% = 1/5

using allegations

2/9 1/5

k

5 7

=> (2/9 - k)/(k -1/5) = 7/5

k = 113/540

Hence , 20.9 %

A local grocer mixed 3 qualities of coffee X , Y , Z at Rs. 56 per kg , Rs, 64 per kg and Rs. 80 per kg in the ratio 1:2:4. To 4 kg of this mixture , he added p kg of X and 4p Kg of Z .The final mixture so formed is sold for Rs. 87.60 per kg at 20% profit. Find p.

What are the options ?

Answer is 4/11 ?

Yes , it's the correct answer. Can you please share the solution?

Does the Copy NMAT also has a deadline like the copy CAT's within which we have to attempt the exam?

A shopkeeper bought two kinds of wheat. The per kg rate of dearer wheat is 25% more than that of the cheaper wheat. In what ratio should the cheaper and the dearer wheat be mixed by the shopkeeper so that if he sold them at the per kg rate of dearer wheat , he has a gain of 10% per kg?

let the cheaper rate be 100 than dearer rate becomes 125(25% more than cheaper its given)

now we have to sell on the dearer rate( sp=250) and the profit should be 10%

so the cp=125*100/110 =1250/11

now apply alligation

100 125

1250/11

125/11 150/11

5 : 6 (eq ratio)

A tank contains 20 litres of pure milk . Mr B takes out 4 litres of milk from the tank and replaces it with water . Then again he takes out 4 litres of mixture of milk and water and replaces it with water . He repeates this process four times . Then , the final amount of water left in the tank is ?

4 lt taken out from 20 so 1/5 part remaining milk =4/5 ,

this process is done for four times

milk remained=20*(4/5)^4

water= 20-milk

solve and get answer

Three solutions of Bhang have respective content of 1% , 2% , & 3% concentrated bhang . If Ravi wants to make a 1.5% solution , the ratio of solution 1 to solution 2 to solution 3 must be maintained at ?

what are the options ?

A)5:4:1 B)4:2:1 C)5:2:1 D)Has a no. of solutions

let the ratio be x:y:z

(1*x+2*y+3*z)/(x+y+z)=1.5

solving this we will get this equation

y+3z=x (option c satisfies this equation)

but,

by putting random values of y and z and we will get different x

so there will be no. of solutions

ratio of water and syrup=3:5

total quantity =8

new ratio should be 1:1

for equal quantity ratio =4:4

part of water replaced will be same as the quantity of syrup taken out

part of syrup taken out=5-4/5=1/5

part of water replaced =1/5 (option c)

Thank you 🙂

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

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)