In the previous part (**Part II**), I have already discussed the shortcut methods, tips and tricks to solve problems from Problems on Ages, Average, and Permutation and Combination. **This section (Part III)** includes another three very important section from quantitative aptitudes including *“**Problems on H.C.F and L.C.M**”, “**Square Root and Cube Root Problems**”, and “**Alligation and Mixture**”*.

Shortcut methods and tricks to solve questions from ‘**Problems on ****H.C.F and L.C.M** are given below:

There are two important methods to solve the L.C.M (Least Common Factor) – division and factorization.

**Factorization:**

If x and y are two different numbers (or same numbers) and x divided y, then we say x is a **factor** of y, and y is called **multiple** of x.

Products of x and y = H.C.F (xy) * L.C.M (xy)

H.C.F of given fractions = H.C.F of numerator / L.C.M of denominator

L.C.M of given fraction = L.C.M of numerator / H.C.F of denominator

If a is the H.C.F of two positive integer p and q, then there exist unique integer c and d such that a = pc + qd

Largest number which divides a,b,c to leave same remainder = h.c.f. of b-a, c-b, c-a.

Largest number which divides a,b,c to leave same remainder R = h.c.f. of a-R, b-R, c-R.

Largest number which divides a,b,c to leave same remainder x,y,z = h.c.f. of a-x, b-y, c-z.

Least number which when divided by p,q,r and leaves a remainder R in each case = ( L.C.M. of p,q,r) + R

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Shortcut methods and tricks to solve questions from ‘**Square Root and Cube Root Problems**’ are given below:

Firstly you have to memorize the below power roots and square roots for your **quantitative aptitude test:**

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Shortcut methods and tricks to solve questions from ‘**Alligation or Mixture**’ are given below:

**Rules of Alligation:**

Alligation rule is used to solve the problems related to any mixture and it’s ingredients.

Quantity of cheaper ingredient = yc,

Cost price (C.P.) of cheaper ingredient = xc,

Quantity of dearer or costlier ingredient = yd,

Cost price of costlier or dearer ingredient = xd.

Consider, mean price of mixture as pm and quantity of mixture as qm.

We know, qm = yc + yd

Then we get,

[(yc * xc) + (yd * xd) = qm * pm = (yc + yd) * pm

→ yc ( pm – xc) = yd (xd – xc)

→ yc / yd = (xd – xc) / ( pm – xc)

Finally we get the traditional and important relation for allegation as:

**(Quantity of cheaper ingredient / Quantity of dearer ingredient) = (C.P. of dearer ingredient – Mean Price) / (Mean Price – C.P. of cheaper ingredient)**

The representation is shown in detail below:

C.P.M.1 = Cost price of material1 in a mixture

C.P.M.2 = Cost price of material2 in a mixture

C.P.M.3 = Cost price of mixture

So, **(C.P.M.1 : C.P.M.2) = (C.P.M.1 = C.P.M.3) – (C.P.M.2 – C.P.M.3)**

or, ( **Quantity of cheaper ingredient) : ( Quantity of dearer ingredient) = (xd – pm) : (pm – xc)**

__Another important consideration:__

Suppose a container contains p units of liquid mixture. Now q unit of that mixture is taken out of that container and replaced by water.

If the operation is done n times, then the quality of pure liquid mixture = [_{p (1 – q/p)}n] units.

**Rules of Mixture:**

In a mixture of p liters, the ratio of soluble chemical solution and water is x : y. If this ratio is to be m : n further, then the quantity of water to be further added is:

In original mixture

Quantity of soluble chemical solution = p * x/(x + y) liters

Quantity of water = p * y/(x + y) liters

Let quantity of water to be added further be w litres.

Therefor in new mixture:

Quantity of soluble chemical solution = p * x/(x + y) liters → Equation(1)

Quantity of water = [p * y/(x + y) ] + w liters → Equation (2)

→ m / m = Equation (1) / Equation (2)

Quantity of water to be added further, w = p * [(xn – ym) / m (x + y)]

__Quantity of ingredient to be added to increase the content of ingredient in the mixture to ____y%__

If P liters of a mixture contains x% ingredient in it. Find the quantity of ingredient to be added to increase the content of ingredient in the mixture to y%.

Let the quantity of ingredient to be added = Q liters

Quantity of ingredient in the given mixture = x% of P = x/100 * P

Percentage of ingredient in the final mixture = Quantity of ingredient in final mixture / Total quantity of final mixture.

Quantity of ingredient in final mixture = [x/100 * P] + Q = [ P*x + 100 * Q] / 100

Total quantity of final mixture = P + Q

→ y/100 = [[ P*x + 100 * Q] / 100]/[P + Q]

→ y[P + Q] = [P*x + 100 * Q]

The quantity of ingredient to added Q = P * [(y – x) / (100 – y)]

__If n different pots of equal size are filled with the mixture of P and Q__

If n different pots of equal size are filled with the mixture of P and Q in the ratio p1 : q1, p2 : q2, ……, pn : qn and content of all these pots are mixed in one large pots, then

Let x liters be the volume of each pots,

Quantity of P in pot 1 = p_{1} * x / (p_{1} + q_{1})

Quantity of P in pot 2 = p_{2} * x / (p_{2} + q_{2})

Quantity of P in pot n = p_{n} * x / (p_{n} + q_{n})… and so on

Similarly,

Quantity of Q in pot 1 = q_{1} * x / ( p_{1} + q_{1})

Quantity of Q in pot 2 = q_{2} * x / ( p_{2}+ q_{2})

Quantity of Q in pot n = q_{n} * x / (p_{n} + q_{n})… and so on.

Therefore, when content of all these pot are mixed in one large pot, then

Quantity of P / Quantity of Q = Sum of quantities of P in different pots / Sum of quantities of Q in different pots

**Quantity of P / Quantity of Q = [(p _{1}x_{1} / p_{1} + q_{1}) + (p_{2 }x_{2} / p_{2}+ q_{2}) + ……….+ (p_{n}x_{n} / p_{n} + q_{n})] / [(_{q1 }x_{1} / p_{1} + q_{1}) + (_{q2 }x_{2} / p_{2}+ q_{2}) + ……….+ (_{qn }x_{n} / p_{n} + q_{n})]**

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