Given : 

A number consists of three digits of which the middle one is zero and their sum is 4. If the number formed by interchanging the first and last digits is greater than the number itself by 198. 

Calculations :

Let the number be 100x + 10y + z 
Then the number of reversed digit will be 100z + 10y + x 

According to the question 

100z + x - (100x + z) = 198        (y = 0)

99z - 99x = 198 

z - x = 2

∴ The difference of the 1st and last digit is 2.

Let the smaller number is x and larger number is y. 

Given that sum of both numbers is 80. 

∴ x + y = 80. ... (1) 

Given that the larger number exceeds four times the smaller one by 5. 

∴ y = 4x +5. ... (2) 

By putting y = 4x + 5 in equation (1), we get x + (4x +5) = 80. 

⇒ 5x = 80 – 5 = 75 

⇒ x = \(\frac{75}{5}\) = 15. 

Now, putting x = 15, in equation (2), we get 

y = 4 × 15 + 5 = 60 + 5 = 65. 

Hence, the numbers are 15 and 65.

. L.H.S = (secA + tanA) (1 – sinA) 

= (\(\frac{1}{cosA}+\frac{sinA}{cosA}\))(1 – sinA) (∵ secA = \(\frac{1}{cos A}\) & tanA = \(\frac{sinA}{cosA}\)) 

= ( \(\frac{1+sin^2A}{cosA}\) ) (1 – sinA) 

= (\(\frac{1 − sin^2A }{cosA}\)) [∵ (a + b)(a − b) = a² − b² ] 

= \(\frac{cos^2A}{cosA}\) = cosA = R.H.S. (∵ sin²A + cos²A = 1 ⇒ 1 − sin²A = cos²A) 

Hence proved

We have to prove \(\frac{1}{secθ − tanθ}\) – \(\frac{1}{cosθ}\) = \(\frac{1}{cosθ}\) – \(\frac{1}{secθ + tanθ}\) .

⇒ \(\frac{1}{secθ − tanθ}\) + \(\frac{1}{secθ + tanθ}\) = \(\frac{1}{cosθ}\) + \(\frac{1}{cosθ}\)

 ⇒ \(\frac{1}{secθ − tanθ}\)  + \(\frac{1}{secθ + tanθ}\) = \(\frac{2}{cosθ}\).

To proving \(\frac{1}{secθ − tanθ}\) - \(\frac{1}{cosθ}\) = \(\frac{1}{cosθ}\) - \(\frac{1}{secθ + tanθ}\) 

we have required to prove \(\frac{1}{secθ + tanθ}\) + \(\frac{1}{secθ + tanθ}\) = \(\frac{2}{cosθ}\).

Now, \(\frac{1}{secθ − tanθ}\) + \(\frac{1}{secθ + tanθ}\) = \(\frac{(sec\theta + tan \theta)(sec\theta -tan\theta)}{(sec\theta-tan\theta)(sec\theta +tan \theta}\) = \(\frac{2sec\theta}{sec^2 \theta - \tan^2\theta}\)

(∵ (a + b)(a – b) =a² − b² )

= 2 secθ = \(\frac{2}{cos \theta}\). (∵ sec²θ − tan²θ = 1 & secθ =\(\frac{1}{cosθ}\))

Hence, \(\frac{1}{sec\theta - tan \theta} + \frac{1}{sec\theta + tan \theta}\) = \(\frac{2}{cos\theta}\) = \(\frac{1}{cos\theta} + \frac{1}{cos\theta}\)

⇒ \(\frac{1}{sec\theta-tan \theta} - \frac{1}{cos \theta}\) = \(\frac{1}{cos \theta}\) - \(\frac{1}{sec\theta +tan \theta}\).

Hence proved

x=sint
Differentiating w.r.t. t, we get,

dx/dt = cos t

y = cos t

Differentiating w.r.t. t, we get,

dy/dt = -sin t

Thus, dy/dx = - sin t / cos t

dy/dx = -tan t

(i) four

(ii) four

(iii) four, collinear

(iv) two

(v) four

(vi) two

(vii) 360°

(viii) opposite

(ix) four

(x) less than

(xi) the interior

(xii) interiors

(xiii) vertices

The equation of circle whose radius is r and centre is (2, 3) is

\((x -2)^2 + (y -3)^2 = r^2\)

It is passing through points (2, - 1).

\(\therefore (2 - 2)^2 + (- 1 - 3)^2 = r^2\)

⇒ \(r^2 = (-4)^2 = 16\)

\(\therefore r = 4\)

\(\therefore \) Equation of circle is

\((x -2)^2 + (y -3)^2 = 4^2\)

⇒ \((x -2)^2 +(y - 3)^2 = 16\)

or \(x^2 + y^2 - 4x - 6y + 4 + 9= 16\)

⇒ \(x^2 + y^2 - 4x -6y = 3\).

The equation of circle whose radius is r and centre is (2, 3) is

\((x -2)^2 + (y -3)^2 = r^2\)

It is passing through points (2, - 1).

\(\therefore (2 - 2)^2 + (- 1 - 3)^2 = r^2\)

⇒ \(r^2 = (-4)^2 = 16\)

\(\therefore r = 4\)

\(\therefore \) Equation of circle is

\((x -2)^2 + (y -3)^2 = 4^2\)

⇒ \((x -2)^2 +(y - 3)^2 = 16\)

or \(x^2 + y^2 - 4x - 6y + 4 + 9= 16\)

⇒ \(x^2 + y^2 - 4x -6y = 3\).

The diagonals intersect each other. The non-parallel sides in the trapezium are unequal except in isosceles trapezium. The line that joins the mid-points of the non-parallel sides is always parallel to the bases or parallel sides which is equal to half of the sum of parallel sides.

cube root of 4=4^(1/3)=2^(2/3)

cube root of 2=2^(1/3)

a=2^(2/3) +2^(1/3) +1

3/a + 3/a² + 1/a³= (3a²+3a+1) /a³

A second-hand shop is a shop which sells used goods.

Environmental Pollution Caused by Waste

• The burning of fossil fuels like oil, gas or coal.
• The exhaust fumes from your vehicles.
• The mismanagement of landfill waste caused by garbage pollution.
• The harmful fragrance or off-gasing from plastic production, paints and so on.
• Nuclear accidents or radiation spills.

Surface water is the water that is available on land in the form of rivers, ocean, seas, lakes and ponds. Groundwater is the underground water that seeps into the soil and is located in large aquifers under the ground. This water can be accessed by digging wells and using motors.

3√2×4√3=12√6 is the answer

Economic growth means an increase in real national income / national output. Economic development means an improvement in the quality of life and living standards, e.g. measures of literacy, life-expectancy and health care. Ceteris paribus, we would expect economic growth to enable more economic development.

Sustainability focuses on meeting the needs of the present without compromising the ability of future generations to meet their needs. The concept of sustainability is composed of three pillars: economic, environmental, and social—also known informally as profits, planet, and people. Increasingly, companies are making public commitments to sustainability through actions like reducing waste, investing in renewable energy, and supporting organizations that work toward a more sustainable future.

A sustainable society is one that ensures the health and vitality of human. life and culture and of nature's capital, for present and future generations. Such. a society acts to slop the activities that serve to destroy human life and culture and.

Let the other number be x.

According to the question,

x * (-6) = 18

So,

\(x = {18 \over -6}\)

x = -3

Real Number

One of the fact you must know about real number is “It is named as real number because it is not imaginary.” It is a part of imaginary number.

Formally real number is defined as the number which can be expressed on number line. Any number you are thinking beyond is real number except zero and neglecting Imaginary number.

There is a unique real number corresponding to every point on the number line. Also, corresponding to each real number, there is a unique point on the number line.

There are some types of real numbers:

1. Natural Number
2. Whole Number
3. Integers
4. Rational Number
5. Irrational Number
6. Fraction

Formulas Related to Real Number:

1. \(\sqrt{ab}\:=\: \sqrt{a}\sqrt{b}\)
2. \(\sqrt{\frac{a}{b}} \:=\:\frac{\sqrt{a}}{\sqrt{b}}\)
3. \((\sqrt{a}\:+\:\sqrt{b})(\sqrt{a}\:-\:\sqrt{b})\) = a - b
4. (a + \(\sqrt{b}\))(a - \(\sqrt{b}\)) = a² – b
5. (\(\sqrt{a}\:+\:\sqrt{b})\)² = a + 2\(\sqrt{ab}\) + b
6. To rationalise the denominator of \(\frac{1}{\sqrt{a}\:+\:b}\) multiply it by \(\frac{\sqrt{a}\:-\:b}{\sqrt{a}\:-\:b}\), where a and b are integers.
7. a^p . b^q = a^p+q
8. (a^p)^q = a^pq
9. \(\frac{a^{p}}{a^{q}}\) = a^p-q
10. a^pb^p = (ab)^p

Euclid’s division lemma:

[A lemma is a proven statement used for proving another statement].

If there are two positive integers a and b, there exist whole numbers q and r satisfying
a = bq + r   [0 ≤ r < b].

Euclid’s division algorithm:

[An algorithm is a series of well-defined steps which gives a procedure for solving a type of problem].

This algorithm provided the way to calculate HCF.It is based on on Euclid’s division lemma.

According to this algorithm, the HCF of any two positive integers a and b, with a > b, is obtained as follows:

Step 1: Apply the division lemma to find q and r where a = bq + r, 0 ≤ r < b.

Step 2: If r = 0, the HCF is b. If r ≠ 0, apply Euclid’s lemma to b and r.

Step 3: Continue the process until the remainder is zero. The divisor at this stage will be HCF (a, b). Also, HCF(a, b) = HCF(b, r).

Fundamental Theorem of Arithmetic:

It is also known as “unique factorization theorem” or “unique prime factorization theorem”.

Every composite number can be expressed (factorised) as a product of prime numbers, and this factorisation is unique, apart from the order in which the prime factors occur.

Polynomial

To understand about polynomials Let us first break the word poly+nomial. Where “poly” means “many” and “nomial” means “terms”. As the meaning itself suggests that it must be the mathematical expression which contains many terms. Exactly it is what is being said. It contains variables, coefficients, constants, and follows addition, subtraction and multiplication and also it contains non-negative exponents. There we listed out polynomial examples.

It is also a broader part of algebra which has its own implications in solving mathematical expressions in equations.

Terms used in polynomial:

1. Variable: The term ‘x’ generally used in algebra is variable which takes its value according to different mathematical expression. It is also called
2. Coefficient: The numerical which is attached to variables is known as coefficient.
3. Constant: The numerical term in an equation which has no variable attached.

Polynomial Examples: In expression 2x+3, x is variable  and 2 is coefficient and 3 is constant term.

Notation of polynomial:

Polynomial is denoted as function of variable as it is symbolized as P(x).

Different kinds of polynomial:

There are several kinds of polynomial based on number of terms.

Monomial: The polynomial expression which contain single term.

Binomial: The polynomial expression which contain two terms.

Trinomial: The polynomial expression which contain two terms.

Polynomial Examples:
4x²y is a monomial.
2xy³ + 4y is a binomial.
4xy + 2x² + 3 is a trinomial.

Degree of polynomial

The degree of polynomial with single variable is the highest power among all the monomials.

In terms of degree of polynomial polynomial.

1. Quadratic Polynomial: A polynomial of degree 2 is called quadratic polynomial.
2. Cubic polynomial: A polynomial of degree 3 is called cubic polynomial.

Zeroes of polynomial:

A real number ‘a’ is a zero of a polynomial p(x) if p(a) = 0. In this case, a is also called a root of the equation p(x) = 0.

Every linear polynomial in one variable has a unique zero, a non-zero constant polynomial has no zero, and every real number is a zero of the zero polynomial.

Remainder Theorem:

If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by the linear polynomial (x – a), then the remainder is p(a).
expressed as p(x) = g(x).q(x) + r(x) where, r(x) = 0 or [degree r(x)] < [degree g(x)].

Factor Theorem:

(x – a) is a factor of the polynomial p(x), if p(a) = 0. Also, if (x – a) is a factor of p(x), then p(a) = 0.

Formulae to Learn:

• (x + y + z)²= x² + y² + z² + 2xy + 2yz + 2zx
• (x + y)³ = x³ + y³ + 3xy(x + y)
• (x – y)³ = x³ – y³– 3xy(x – y)
• x³ + y³ + z³ – 3xyz = (x + y + z)(x²+ y²+ z²– xy – yz – zx

Polynomial Examples:

Find the remainder when x⁴ + x³ – 2x² + x + 1 is divided by x – 1.

Solution:

Here, p(x) = x⁴ + x³ – 2x² + x + 1, and the zero of x – 1 is 1.
So, p(1) = (1)⁴ + (1)³ – 2(1)² + 1 + 1
= 2
So, by the Remainder Theorem, 2 is the remainder when x4 + x3 – 2x2 + x + 1 is divided by x – 1.

Verify whether 2 and 0 are zeroes of the polynomial x² – 2x.

Solution:

Let p(x) = x² –  2x
Then p(2) = 2² – 4 = 4 – 4 = 0
and p(0) = 0 – 0 = 0
Hence, 2 and 0 are both zeroes of the polynomial x² – 2x.

The following observations took place:

(i) zero of a polynomial need not be 0.

(ii) 0 may be a zero of a polynomial.

(iii) Every linear polynomial has one and only one zero.

(iv) A polynomial can have more than one zero.

LCM is the multiplication of one common factor of the numbers and the other different factors of the numbers.

Write the LCM = 120 into factored form, that is

120 = 2 × 2 × 2 × 3 × 5

= 4(2 × 3 × 5)

⇒ 4 is the common factor of the numbers.

So, the HCF of three numbers is a multiple of 4.

Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, ...

Since 35 is not the multiple of 4, therefore 35 cannot be their HCF.

(a) 39=(3×10)+(9×1)=(3×10¹)+(9×10⁰)
(b) 52=(5×10)+(2×1)=(5×10¹)+(2×10⁰)
(c) 106=(1×100)+(0×10)+(6×1)=(1×10²)+(0×10¹)+(6×10⁰)

2 is the smallest prime number

(b) cosec^-1x

Answer: India

Inventory facilitates bulk purchase of raw materials at discount rates.

Type I error is taking a wrong decision to reject the null hypothesis, when it is actually true.

a) (B)

b) 1. Enhances the entry glucose into the cell.

2. Converts glucose to glycogen in liver and muscles.

c) Sufficient quantity of glucose i not reaching the cell. Energy production decreases. Excess amount of glucose is eliminated through urine.

If the number of non-zero allocation are equal to (m + n – 1) then, the solution is called non-degenerate.

Assignable cause of variation.

Deficiency of thyroxine (Hypothyroidism) leads to cretinism in infants and myxoedema in adults. Excess production of thyroxine (Hyperthyroidism) leads to a condition, known as the Graves disease.

A. Medulla 

B. Aldosterone 

C. Cortisol 

D. Norepinephrine

The normal level of glucose is 70-110 mg/100 ml blood. The level of glucose in blood is maintained by the combined action of insulin and glucagon of the Islets of Langerhans tissues of the pancreas. Insulin, released from the beta cells of Islets of Langerhans, helps to reduce blood sugar by accelerating the process of cellular uptake of glucose and Conversion of glucose in to glycogen. When blood glucose level falls, glucagon, released from the alpha cells of Islets of Langerhans, converts glycogen to glucose and synthesizes glucose from amino acids.

Let x and y the number of shirts and points 

The objective function / L.P.P. is:- 

Maximize z= 100 x + 170 y 

Subject to constraints: 

Stiching: x + 2y ≤ 20 

Cutting: 2x + 3y ≤ 36 and x,y ≥ 0.

The hormone, parathormone prevents the deposition of calcium in bones resulting its weakening.

Melatonin, the secretion of the pineal gland helps to . maintain rhythm of our daily activities. 

Therefore pineal gland is called as the biological clock.

• Decreased production of insulin
• Malfunctioning of insuline
• Destruction of Beta Cells
• Inactive insulin

Cortisol of adrenal gland. It cannot be given to diabetic patients as it increases the level of glucose in blood.

While fasting glucagon converts glycogen or amino acids into glucose. When taking heavy food insulin enhances cellular uptakes of glucose and converts glucose into glycogen.

Insulin is helpful to reduce the excess glucose in the blood and to maintain its normal level

From I:

Let the age of A, B and C be a, b, and c respectively

a + b + c = 15

c > b > a and a > 3

If a = 4, b = 5 and c = 6, the sum of the age of A, B and C is 15. No other case is possible,

So, the ratio of age of A and B will be 4 : 5

Hence statement I alone is sufficient.

From II:

Let the age of A, B and C be a, b, and c respectively

b – c = 2 and a + c = 12

b – c + a + c = 2 + 12

⇒ b + a = 14

But we cannot determine the ratio of age of A and B.

Hence Statement II alone is not sufficient.

From III:

Let the age of A, B and D be a, b, and d respectively

a – d = d - c

⇒ a + c = 2d

⇒ a + c = 2 × 30

⇒ a + c = 30

But we cannot determine the ratio of age of A and B.

Hence statement III alone is not sufficient.

From I:

Let the rate of interest be R%.

A × R × 4/100 = 2A –A

⇒ R = 25%

Hence statement I is alone sufficient

From II:

Let the rate of interest be R%.

Simple interest earned by mamta = 5000Rt/100

Simple interest earned by sajan = 3000RT/100

5000Rt/3000RT = 4/7

⇒ 5t/3T = 4/7

⇒ But we cannot determine the rate of interest.

Hence, statement II alone is not sufficient.

From III:

Let the rate of interest be R%.

S[R/100 + 1]² = 4S

⇒ [R/100 + 1] = 2

R = 100%

Hence, statement III is alone sufficient.

Lets number of employees who plays cricket is F₁,

Number of employees who play pool is F₂,

Number of employees who play both cricket and pool is G and

Number of employees who doesn’t play any game is H.

Total employees = F₁ + F₂ – G + H = 360      ----(1)

F₁ = 288

From I:

Employees who plays only chess = F₁ – G = 216

G = 288-216 = 72

But, we cannot find the value of H.

So. I alone is not sufficient.

From II:

Number of employees who play either cricket or pool or both = (F₁ + F₂ - G)

Number of employees who does not play any game = H = (F₁ + F₂ - G) – 216

⇒ (F₁ + F₂ – G) = H + 216      ----(2)

From equation (1) and (2)

⇒ H + 216 + H = 360

⇒ 2H = 144

⇒ H = 72

Hence, total employes who doesn’t play any game is 72.

So, II alone is sufficient.

Kelly’s method and simple geometric mean of price relatives method.

‘Interpolation’ is the technique of estimating the value of the dependent variable (Y) for any intermediate value of the independent variables (X). Extrapolation is the technique of estimating the value of dependent variable (Y) for any value of the independent variable (X) which is outside the range of the given series.

Controlling the quality of the product during the manufacturing process itself is called process control. Controlling the quality of the finished products/manufactured products is called product control.

Index number for 2012 = 180