IMO C

Tin A contains wheat and rice in ratio 2:3. Let's suppose concrete values are 2x and 3x for wheat and rice respectively.
Similarly, Tin B has ratios 3:7. Therefore, concrete values are 3y and 7y.

Since we need to have 5kg wheat and 11kg rice:
2x+3y = 5
3x+7y = 11
Solving for x,y we get x=2/5 and y=7/5

Mixture to be taken from Tin A = 2x+3x = 5x = 2kgs
Mixture to be taken from Tin B = 3y+7y = 10y = 14kgs

thanks jasz .... but how do i use the alligation method...on the above problem ??

Hello,

For allegation , I guess you can use the following method.

Let X be the quatity of mixture taken from tin A.
Then automatically 16 - X will be the quantity taken from Tin B. ( As 5:11 has a total of 16 parts )

Now we know that we need 5kg of wheat and 11 kg of rice.

So in the first tin A --> Wheat is 2/5. So we shall take 2/5X of mixture from tin A
In Tin B, wheat is 3/10. So lets take 3/10 ( 16 - X )

Totally , we get
2/5X = 3(16-X)/10 = 5.

Solve, we get the answer of X as 2.
16-X will be 14

Hope this helps.

Regards and All the best.

the equations are
(2/5)A + (3/10)B = 5
(3/5)A + (7/10)B = 11
solving gives
(1/10)A = 2/10. thus C

weighted average question.

2x/5+3y/10=5
3x/5+7y/10=11

solve for x, y
X = 2kg and y = 14 kg.

Check: Weight of rice and wheat should be same as weight taken from mixture x and y

5 + 11 = x + y =16

You can also solve it using a simpler method of choosing values and plugging them in.
You know that you need 5 kg of wheat and 11 kg of rice.

we know that,
mixture A is 40% [2/(2+3)] wheat.
mixture B is 30% [3/(3+7)] wheat.

test out the options to see which produces 5 kg of wheat.

1) (3kg * .4) + (13kg * .3) = 5.1
2) (8kg * .4) + (8kg * .3) = 5.6
3) (2kg * .4) + (14kg * .3) = 5
4) (6kg * .4) + (10kg * .3) = 5.4


without even having to address the rice situation, you've eliminated all but option 3.