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Statement Sums involving Linear Equations

Quadratic Equations - Concepts

Identifying Quadratic Equations

Factors of Quadratic Equation

Factorization by Middle Term Splitting

Factorization by Grouping of Terms

Statement Sums involving Linear Equations

Factorization by Completing the Square

Quadratic Formula

Nature of Roots of Quadratic Polynomial

Equations Reducible to Quadratic Form

Statement sum Involing Quadratic Equations

Solving Statements Involving Age

Statement sums involving Coins

Solving Statements For A Two Digit Number

Class - SBI PO Prelims Subjects

Concept Explanation

Statement Sums involving Linear Equations

Statement Sums involving Linear Equation: In these sums a certain situation is given we have to analyse the statement and then suppose a certain quantity. and will then try to identify a relation with the quantity we have supposed.

The procedure to translate a word problem in the form of an equation is known as the formulation of the problem. Thus, the process of solving a word problem consists of two parts, namely, formulation and solution.

The following steps should be followed to solve a word problem:

Step I Read the problem carefully and note what is given and what is required.

Step II Denote the unknown quantity by some letters, say x, y, z, etc.

Step III Translate the statements of the problem into mathematical statements.

Step IV Using the condition (s) given in the problem, form the equation

Step V Solve the equation for the unknown.

Step VI Check whether the solution satisfies the equation.

Example: The numerator of a fraction is 4 less than the denominatr. If 1 is added to both its numerator and denominator, it becomes 1/2. FInd the fraction.

Solution Let the denominator of the fraction be x. Then,

Numerator of the fraction = x - 4

$large therefore$ Fraction $large =frac{x-4}{x}$

If 1 is added to both its numerator and denominator, the fraction becomes $large frac{1}{2}$ .

$large therefore$ $large frac{x-4+1}{x+1}=frac{1}{2}$

$large Rightarrow frac{x-3}{x+1}=frac{1}{2}$

$large Rightarrow 2(x-3)=x+1$ [ Using cross - multiplication]

$large Rightarrow 2x-6=x+1$

$large Rightarrow 2x-x=6+1$

$large Rightarrow x=7$

Putting x = 7 in (i), we get

Fraction $large =frac{7-4}{7}=frac{3}{7}.$ Hence, the given fraction is $large frac{3}{7}.$

Example : How much pure alcohol be added to 400 ml of a 15% solution to make its strength 32%?

Solution Let x ml pure alcohol be added to 400 ml of a 15% solution to make its strength 32%. Here, 15% solution means that there is 15 ml pure alcohol in a solution of 100 ml.

Now, Quantity of alcohol in 100 ml solution = 15 ml

$large therefore$ Quantity of alcohol in one ml solution $large =frac{15}{100};ml$

Quantity of alcohol in 400 ml solution $large =frac{15}{100}times 400;ml=60; ml$

Total quantity of the solution = (400 + x) ml

Total quantity of alcohol in (400 + x) ml solution = (60 + x) ml

$large therefore$ Quantity of alcohol in one ml $large =frac{60+x}{400+x};ml$

Quantity of alcohol in 100 ml $large large =frac{60+x}{400+x}times 100;ml$

$large Rightarrow$ Strength of the solution $large =left ( frac{60+x}{400+x} right )times$ 100%

But, the strength of the solution is given as 32%.

$large therefore ;;frac{60+x}{400+x}times 100=32$

$large Rightarrow ;;100(60+x)=32(400+x)$ [ Multiplying both sides by (400 + x)]

$large Rightarrow ;;6000+100x=12800+32x$

$large Rightarrow ;;100x-32x=12800-6000$

$large Rightarrow ;;68x=6800$

$large Rightarrow ;;frac{68x}{68}=frac{6800}{68}$

$large Rightarrow ;;x=100$

Thus, 100 ml alcohol must be added to make 32% strength of the solution.

Sample Questions

(More Questions for each concept available in Login)

Question : 1

Two numbers are in the ratio 5:6. If 8 is subtracted from each, the ratio becomes 4:5. Find the numbers.
A. x = 9	B. x = 5	C. x = 1	D. x = 8
Right Option : D
View Explanation

Question : 2

Two number are in the ratio 3 : 5. If 9 is subtracted from each, the new numbers are in the ratio 12 : 23. The smaller number is ____________
A. 27	B. 33	C. 49	D. 55
Right Option : B
View Explanation

Question : 3

A train, 150 m long, passes through a tunnel at a rate of 72 Km/ hr in 20 seconds. What is the length of the tunnel?
A. 200 m	B. 225 m	C. 230 m	D. 250 m
Right Option : D
View Explanation