If the price of 23 toys is 276, then what will the price of 12 toys ?

Answer: Option C

Explanation:

As it is given that food is there for 54 days for 2000 students,
But after 15 days, remaining food was sufficient for 2000 students for 39 days only ( as 54-15 = 39 )
Let x number of students came in hostel after 15 days,
=> Total number of students after 15 days = 2000 + x
=> Remaining food was enough for (2000 + x) students for 20 days
So here is indirect proportion, as more people and less food
=> 2000 : ( 2000 + x ) :: 20 : 39
=> 2000 * 39 = ( 2000 + x ) * 20
=> 3900 = 2000 + x
=> x = 3900 - 2000
=> x = 1900

Answer: Option C

Explanation:

Please always remember, that what exactly we need to calculate before applying chain formula. As in this question we need to find how many mats, RIGHT !
So we will compare everything with "number of mats"
As,

More men, More number of mats (Direct Proportion)
More days, More number of mats (Direct Proportion)

So, it can be solved as,

\begin{aligned}
\begin{bmatrix}Weavers & 4 & 8 \\Days & 4 & 8 \end{bmatrix} :: 4 : x \\

=> x*4*4 = 8*8*4 \\
=> x = 16
\end{aligned}

Answer: Option B

Explanation:

Less shadow, Less Height (Direct Proportion)
So, let height of building be x metre
then,
\begin{aligned}
40.25 : 17.5 :: 28.75 : x \\

=> x = \frac{17.5*28.75}{40.25} \\
=> x = 12.5
\end{aligned}

So height of building should be 12.5 metre

Answer: Option C

Explanation:

Although we can first calculate the cost of one toy and then can multiply it with 40 to get the result.
but we will solve it with chain rule so that we can prepare ourself for other complex problems which we can solve easily by applying this rule.

Let the required cost be Rs. x
More toys means More Cost. (Direct Proportion)

\begin{aligned}
15:40 :: 240:x \\
=> x = \frac{240*40}{15} \\
=> x = 640
\end{aligned}

So 40 toys will cost Rs 640

Answer: Option B

Explanation:

Less Cows, More days (Indirect Proportion)
Less bags, Less days (Direct Proportion)

\begin{aligned}
\begin{bmatrix}Cows & 1 & 40 \\Bags & 40 & 1 \end{bmatrix} :: 40 : x \\
=> x * 40 * 1= 40 * 1 * 40 \\
=> x = 40
\end{aligned}