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Home / Aptitude / Quantitative Aptitude Shortcut Tricks For IBPS Exam – Part III

Quantitative Aptitude Shortcut Tricks For IBPS Exam – Part III

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.

Tricks-and-Formulaes-Quantitative-Aptitude

Source

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:

Squesre Roots and Power Roots

Source

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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 = p1 * x / (p1 + q1)

Quantity of P in pot 2 = p2 * x / (p2 + q2)

Quantity of P in pot n = pn * x / (pn + qn)… and so on

Similarly,

Quantity of Q in pot 1 = q1 * x / ( p1 + q1)

Quantity of Q in pot 2 = q2 * x / ( p2+ q2)

Quantity of Q in pot n = qn * x / (pn + qn)… 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 = [(p1x1 / p1 + q1) + (p2 x2 / p2+ q2) + ……….+ (pnxn / pn + qn)] / [(q1 x1 / p1 + q1) + (q2 x2 / p2+ q2) + ……….+ (qn xn / pn + qn)]

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                                                                                                                         (…….to be continued in Part IV)

About Puzzleduniya

I am a quiz, puzzle lover, who always intended to post different type of study materials, which will help the all the competitive job seekers all around the world.

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