(4) (a) (iv), (b) (i), (c) (ii), (d) (iii)

Electron deficient hydride → Less than 8e (B₂H₆) 

Electron precise hydride → having 8e^- without I.p. (GeH₄) 

Electron rich hydride → having 8e^- with 1.p. (HF)

(2) Enzymes are polysaccharides.

Which is incorrect statement regarding enzymes 

(1) Like chemical catalysts enzymes reduce the activation energy of bio process ⇒ This is correct statement.

(2) Enzymes are polysaccharides ⇒ This is incorrect statement because enzymes are protein in nature 

(3) Enzymes are very specific for a particular reaction and substrate ⇒ This is correct statement. 

(4) Enzymes are biocatalyst ⇒This is correct statement.

(2) Enantiomers are superimposable mirror images of each other.

Enantiomers are non-superimposable mirror images of each other.

1. As the bubbles move upward, the pressure on them decreases. This causes increase in volume. So as the bubbles move upward, their size increases.

2. Ink rises through the tube

3. When the bottle is placed in hot water, air inside the bottle becomes hot. This causes expansion of air. This pushes ink in the tube. So ink rises through the tube.

4. Ink comes down. Because as air becomes cool, Its volume decreases.

5. When temperature is increased volume of gas increases. Similarly, when temperature is decreased, The volume decreases.

Correct option is B) examples of which

Correct option is B) IV,I,II,V,III

Correct option is A) is being spring-cleaned

Correct option is B) has been used

Lakshmi was not satisfied with her husband Gangaram because he was ugly, poor and uncaring. Heera Lal, the driver took advantage of this situation and tried to entice Lakshmi. He gave her a blouse, a sari and a packet of delicious halwa. When Gangaram came to know about these gifts, he was furious. He bluntly abused her and ran to beat her. Lakshmi left the haveli in fear and humiliation.

लक्ष्मी अपने पति से सन्तुष्ट नहीं थी क्योंकि वह निर्धन, कुरूप तथा उसका ध्यान नहीं रखता। ड्राईवर हीरा लाल ने इस परिस्थिति का लाभ उठाया और लक्ष्मी को फांसने का प्रयास किया। उसने उसको ब्लाउज, सारी और लजीज हलवे का एक पैकेट दिया। जब गंगाराम को इन उपहारों के बारे में पता चला तो वह गुस्से से आग बबूला हो गया। उसने उसे बुरी तरह गालियाँ दी और उसे पीटने के लिये भागा। लक्ष्मी भय और अपमान से हवेली छोड़कर चली गई।

Correct option is B) 4,2,3,5,1

Correct option is D) were spoiled

Correct option is B) II,IV,I,III,V

Correct option is E) 3,6,1,5,4,2

The anxiety of Geeta’s parents was about Ajay’s family. The women of upper class family in Udaipur remained in purdah. They moved about with their face covered. But Geeta was a modern girl and had studied with boys. They believed that her education · would enable her to adjust herself to the new traditional family.But they thought that Ajay’s more orthodox relatives would not accept an educated modern girl.

गीता के माता-पिता की चिन्ता अजय के परिवार के विषय में थी। उदयपुर में उच्च जाति की महिलायें परदे में रहती थीं। वे अपना चेहरा ढककर ही इधर-उधर आती जाती थीं। परन्तु गीता एक आधुनिक लड़की थी और लड़कों के साथ पढ़ चुकी थी। उन्हें यकीन था कि उसकी शिक्षा उसे इस योग्य बना देगी कि वह परम्पराओं से बंधे परिवार के साथ सामंजस्य स्थापित कर लेगी। परन्तु वे सोचते थे कि अजय के कट्टर विचारधारा के रिश्तेदार एक सुशिक्षित लड़की को स्वीकार नहीं करेंगे।

Correct option is A) bitten

Correct option is E) IV, II, V, I, III

Seeing the splendour of the rooms, Geeta feels her heart filled with unusual joy. When she sees the portrait of her husband’s ancestors in their court-dress, she realizes that the people of the haveli are gentle, kind and brave. The emotions of joy, affection and pride fill her heart. She feels proud of being a bride of an honourable and royal family.

कमरे की शान शौकत देखकर गीता अपने हृदय को असाधारण आनन्द से भरा हुआ महसूस करती है। जब वह अपने पति के पूर्वजों की तसवीरें दरबारी पोशाक में देखती है तो वह समझ जाती है कि हवेली के लोग भले, दयालु और बहादुर हैं। खुशी, प्यार और गौरव की भावनायें उसके हृदय को भर देती हैं। एक सम्मानित और राजसी परिवार की वधू होने का उसे गर्व अनुभव होता है।

Correct option is A) address

Correct option is C) invented

Here,

δ = 22˚, 

H = 0.35 G, 

R = ? 

As, 

H = R cosδ 

⇒ R = \(\frac{H}{cosδ}\) 

= \(\frac{0.35}{cos22°}\)

= \(\frac{0.35}{0.9272}\)

= 0.38 G

Correct option Statement-1 is False, Statement-2 is True

Correct option(a)

Explanation:

Heating element of an electric heater must be made with a material having high melting point so that it couldn’t melt easily and high specific resistance.

(B) Heater output will decrease 

Answer is (2) 8/9

 ν₁ = (m₁ - m₂)/(m₁ + m₂)u1 = [(m - 2m)/(m + 2m)]u = -u/3

E_i = (1/2) mu², E_f = (1/2) m(-u/3)² = 1/9 x (1/2) mu².

(E_i – E_f)/E_i

= {1 – (1/9)}/1 = 8/9. 

ΔG = ΔH – TΔS 

At equilibriu, ΔG = 0 so that ΔH = TΔS or

T = ΔH/ΔS = -1000 mol^-1/-33.3JK^-1 mol^-1 = 300.03 K

(i) For spontaneity from left to right. G should be –ve for the given reaction. This will be so if T < 300.3 K 

(ii) For reverse reaction to occur, G should be +ve forward reaction. This will be so if T > 300.3 K.

T_1/2 = 28 years = 28 x 365 x 24 x 600 = 8.83 x 10⁸ seconds 

90g \(Sr_{38}^{90}\) contains 6.023 x 10²³ atoms 

λ = \(\frac{0.693}{T_{1/2}}\)

Decay constant λ = 7.85 x 10¹⁰ x 1.004 x 1.004 × 10¹⁰ 

Rate of disintegration R= λN 

R = 7.88 x 10¹⁰ Bq

R = 20Ω, L = 1.5 H, C = 35 x 10^–6F, 

V = 220V, v = 50Hz, Z = ?, I = ?

Z = \(\sqrt{R^2+(X_c-X_L)^2}\)

X_L = wL = 2πvL = 471 Ω

\(X_C = \frac{1}{\omega c}\) = \(\frac{1}{2\pi \theta c}\) = 90.99Ω

Calculation of Z = 380.53 Ω

I = \(\frac{v}{x}\) = 0.578 A

Correct answer is: (c) 3² x5² x17

3825=3² x5² x17

Factors:

We are aware that a natural number can be expressed as the product of other natural numbers. As
20 = 2 x 10 = 4 x 5
Thus, 1, 2, 4, 5, 10, 20 are the factors of 20. Of these 2 and 5 are prime factors of 20.

Similarly, there are factors of algebraic expressions as,
5ab = 5 x a x b where 5, a and b are prime factors of 5ab.

Factorisation:

When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.

Example: 12xy is already in factors but we write it as 3 x 2 x 2 x X x Y. thus factors are 2, 3, x, y.

Factorisation by Common Factors:

As the term common itself suggest that it must be some similar pattern.

A systematic way of factorising an expression is the common factor method. It consists of three steps:

• Write each term of the expression as a product of irreducible factors
• Look for and separate the common factors and
• Combine the remaining factors in each term in accordance with the distributive law.

As example: We have to factorise 3x + 9.
first let’s factorise 3x = 3 x X and 9 = 3 x 3
similar pattern (say common) among these factors of both are 3
thus we can write 3x + 9 = 3 x X + 3 x 3 = 3(x + 3)

Factorisation by regrouping terms:

Rearranging the expression allows us to form groups leading to factorisation. This is regrouping.

Let us write (2xy + 2y) in the factor form:
2xy + 2y = (2 × x × y) + (2 × y)
= (2 × y × x) + (2 × y × 1)
= (2y × x) + (2y × 1) = 2y (x + 1)

Factorisation using identities:

• (a + b)² = a² + 2ab + b²
• (a – b)² = a² – 2ab + b²
• (a + b) (a – b) = a² – b²

Factors of the product form

• (x + a).(x + b) = x² + (a + b) x + ab

Division:

We already know that division is the inverse of multiplication for numerals. It is also applicable to the division of algebraic expressions.

Division of a monomial by another monomial:

We use the simple method for division. The division of numerical is same as we divide but for the monomials, we simply subtract the power of denominator from the numerator.

Example: \(\frac{21x^{4}}{7x}\) = 3x²

Division of Algebraic Expressions Continued (Polynomial ÷ Polynomial):

By taking the common factor we divide it.
Just an example: (7x² + 14x) ÷ (x + 2)
(7x² + 14x) = (7 × x × x) + (2 × 7 × x) = 7x(x + 2)
thus, (7x² + 14x) ÷ (x + 2)
= \(\frac{7x(x + 2)}{x + 2}\) = 7x

Examples related to factorisation:

1. Factorise 12a²b + 15ab².
   Solution:
   As, 12a²b = 2 × 2 × 3 × a × a × b
   and, 15ab² = 3 × 5 × a × b × b
   by seeing the two terms we find that these two terms have 3, a and b as common factors.
   12a²b + 15ab² = (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b)
   = 3 × a × b × [(2 × 2 × a) + (5 × b)]
   = 3ab × (4a + 5b)
   = 3ab (4a + 5b)
2. Factorise 6xy – 4y + 6 – 9x.
   Solution:
   6xy – 4y = 2y (3x – 2)
   –9x + 6 = –3 (3x) + 3 (2)
   Taking them together,
   6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6
   = 2y (3x – 2) – 3 (3x – 2)
   = (3x – 2) (2y – 3)
   thus factors of (6xy – 4y + 6 – 9 x) are (3x – 2) and (2y – 3).

Introduction to Integral formulas:

The list of integral calculus formula is here with all the rules which are needed to solve integration. The formula sheet of integration include basic integral formulas, integration by parts and partial fraction, area as a sum and properties of definite integral. At first take a look at indefinite integration.

Indefinite integral formulas:

• Integration is the inverses of differentiation.
• Two indefinite integrals having same derivative lead to the same family of curves, this makes them equivalent.
• Addition rule of integration: ∫ [ f(x) + g(x) ]dx = ∫f(x) dx + ∫g(x) dx
• The multiplication rule for any real number k, ∫k f(x) dx = k ∫f(x) dx

1. \(\large \int  \; dx = x + C\)
2. \(\large \int a \; dx = ax + C\)
3. \(\large \int x^{n} \; dx = \frac{x^{n+1}}{n+1} + C; \; n \neq -1\)
4. \(\large \int x^{n} \; dx = \frac{x^{n+1}}{n+1} + C; \; n \neq -1\)

Integral formulas of trigonometric functions:

∫ sin x dx = -cos x + c

∫ cos x dx = sin x + c

∫ tan x dx = log|sec x| + c

∫ cot x dx = log|sin x| + c

∫ sec x dx = log|sec x + tan x| + c

∫ cosec x dx = log|cosec x – cot x| + c

∫ sec²x dx = tan x + c

∫ cosec²x dx = -cot x + c

∫ sec x tan x dx = sec x + c

∫ cosec x cot x dx = - cosec x + c

Integral formulas for inverse trigonometric functions:

 \(\large \int \frac{1}{\sqrt{1-x^{2}}} \; dx = \sin^{-1} x + C\)

\(\large \int \frac{1}{\sqrt{1+x^{2}}} \; dx = \tan^{-1} x + C\)

\(\large \int \frac{1}{|x|\sqrt{x^{2}-1}} \; dx = \sec^{-1} x + C\)

\(\large \int \frac{dx}{\sqrt{x^{2}+ {a}^2}} \; dx = \frac{1}{a}\;{tan}^{-1} \frac{x}{a} + C\)

\(\large \int \frac{dx}{\sqrt{a^{2}- {x}^2}} \; dx = {sin}^{-1} \frac{x}{a} + C\)

∫ sin^-1x dx = x sin^-1x + \(\sqrt{1-{x}^2}\) + c

∫ cos^-1x dx = x cos^-1x - \(\sqrt{1-{x}^2}\) + c

∫ tan^-1x dx = x tan^-1x – ln\(\sqrt{1 + {x}^2}\) + c

∫ cot^-1x dx = x cot^-1x +ln\(\sqrt{1 + {x}^2}\) + c

∫ sec^-1x dx = x sec^-1x – ln|x + \(\sqrt{|{x}^2-1|}\)| + c

∫ cosec^-1x dx = x cosec^-1x + ln|x + \(\sqrt{|{x}^2-1|}\)| + c

Integral formulas for exponential trigonometric functions:

\(\large \int \sin^{n} (x) dx = \frac{-1}{n} \sin^{n-1} (x) \cos (x) +\frac{n-1}{n} \int \sin^{n-2} (x) dx\)

\(\large \int \cos^{n} (x) dx = \frac{1}{n} \cos^{n-1} (x) \sin (x) +\frac{n-1}{n} \int \cos^{n-2} (x) dx\)

\(\large \int \tan^{n} (x) dx = \frac{1}{n-1} \tan^{n-1} (x) +\int \tan^{n-2} (x) dx\)

\(\large \int \sec^{n} (x) dx = \frac{1}{n-1} \sec^{n-2} (x) \tan (x) + \frac{n-2}{n-1} \int \sec^{n-2} (x) dx\)

\(\large \int \csc^{n} (x) dx = \frac{-1}{n-1} \csc^{n-2} (x) \cot (x) + \frac{n-2}{n-1} \int \csc^{n-2} (x) dx\)

∫ sin²x dx = \(\frac{1}{2}\)x - \(\frac{1}{4}\)sin 2x + c

∫ cos²x dx = \(\frac{1}{2}\)x + \(\frac{1}{4}\)sin 2x + c

∫ tan²x dx = tan x – x + c

∫ cot²x dx = -cot x – x + c

Integral formulas for logarithmic functions:

∫ a^x dx = \(\frac{{a}^x}{log\:a}\) + c

∫ \(\frac{1}{x}\) dx = log|x| + c

∫ \(\frac{dx}{{x}^2 –{a}^2}\) =\(\frac{1}{2a}\) log|\(\frac{x\:-\:a}{x\:+\:a}\)| + c

∫\(\frac{dx}{{a}^2 –{x}^2}\) = \(\frac{1}{2a}\) log|\(\frac{a\:+\:x}{a\:-\:x}\)| + c

∫\(\frac{dx}{\sqrt{{x}^2 –{a}^2}}\) = log |x + \(\sqrt{{x}^2 – {a}^2}\)| + c

∫\(\frac{dx}{\sqrt{{x}^2 +{a}^2}}\) = log |x + \(\sqrt{{x}^2 + {a}^2}\)| + c

Integral formulas for partial fractions:

\(\frac{px\:+\:q}{(x-a)(x-b)}\) = \(\frac{A}{x\:-\:a} + \frac{B}{x\:-\:b}\) , a ≠ b

\(\frac{px\:+\:q}{(x-a)^{2}}\) = \(\frac{A}{x\:-\:a} + \frac{B}{(x-a)^{2}}\)

\(\frac{p{x}^2+qx+r}{(x-a)(x-b)(x-c)}\) = \(\frac{A}{x\:-\:a} + \frac{B}{x\:-\:b} + \frac{C}{x\:-\:c}\)

\(\frac{p{x}^2+qx+r}{(x-a)^{2}(x-b)}\) = \(\frac{A}{x\:-\:a} + \frac{B}{(x-a)^{2}} + \frac{C}{x-b}\)

\(\frac{p{x}^2+qx+r}{(x-a) ({x}^2 + bx +c)}\) = \(\frac{A}{x-a} + {Bx+C}{{x}^2 + bx +c}\)

Where x² +bx +c can’t be factorized further.

Integral Formulas by parts:

$$\large \int u\;v\;dx=u\int v\;dx-\int\left(\frac{du}{dx}\int v\;dx\right)dx$$

In this formula there is simple rule to know first and 2^nd function i.e. u and v:

I - Inverse Function

L - Logarithmic Function

A – Algebraic Function

T- Trigonometric Function

E- Exponential Function

∫ e^x [f(x) + f’(x)] dx = ∫e^x f(x) dx + c

Some Special integral formulas:

∫ \(\sqrt{{x}^2 – {a}^2}\) dx = \(\frac{x}{2}\;\sqrt{{x}^2 – {a}^2}\) - \(\frac{{a}^2}{2}\)log|x + \(\sqrt{{x}^2 – {a}^2}\)| + c

∫\(\sqrt{{x}^2 + {a}^2}\) dx = \(\frac{x}{2}\;\sqrt{{x}^2 + {a}^2}\) + \(\frac{{a}^2}{2}\)log|x + \(\sqrt{{x}^2 + {a}^2}\)| + c

∫\(\sqrt{{a}^2 - {x}^2}\) dx = \(\frac{x}{2}\;\sqrt{{x}^2 – {a}^2}\) + \(\frac{{a}^2}{2}\) sin^-1\(\frac{x}{a}\) + c

Definite integral formulas:

Actual Rule:

$$\large \int_{a}^{\infty}f(x)dx=\lim_{b\rightarrow \infty}\left [ \int_{a}^{b}f(x)dx\right ]$$

Or we can write:

$$\large \int_{a}^{b}f(x)dx=F(b)-F(a)$$

Here a is lower limit and b, ∞ are upper limits.

Some rational formula for definite integration:

\(\large \int_{a}^{\infty }\frac{dx}{x^{2}+a^{2}}=\frac{\pi }{2a}\)

\(\large \int_{a}^{\infty }\frac{x^{m}dx}{x^{n}+a^{n}}=\frac{\pi a^{m-n+1}}{n\sin \left ( \frac{(m+1)\pi }{n} \right )},0< m+1< n\)\)

\(\large \int_{a}^{\infty }\frac{x^{p-1}dx}{1+x}=\frac{\pi }{\sin (p\pi )},0< p< 1\)

\(\large \int_{a}^{\infty }\frac{x^{m}dx}{1+2x\cos \beta +x^{2}}=\frac{\pi \sin (m\beta )}{\sin (m\pi )\sin \beta }\)

\(\large \int_{a}^{\infty }\frac{dx}{\sqrt{a^{2}-x^{2}}}=\frac{\pi }{2}\)

\(\large \int_{a}^{\infty }\sqrt{a^{2}-x^{2}}dx=\frac{\pi a^{2}}{4}\)

Definite integration formula for trigonometric functions:

\(\large \int_{0}^{\pi }\sin(mx)\sin (nx)dx=\left\{\begin{matrix} 0 & if\;m\neq n\\ \frac{\pi }{2} & if\;m=n \end{matrix}\right.\;m,n\;positive\;integers\)

\(\large \int_{0}^{\pi }\cos (mx)\cos (nx)dx=\left\{\begin{matrix} 0 & if\;m\neq n\\ \frac{\pi }{2} & if\;m=n \end{matrix}\right.\;m,n\;positive\;integers\)

\(\large \int_{0}^{\pi }\sin (mx)\cos (nx)dx=\left\{\begin{matrix} 0 & if\;m+n\;even\\ \frac{2m}{m^{2}-n^{2}} & if\;m+n\;odd \end{matrix}\right.\;m,n\;integers\)

Algebra Formula:

Do you ever think that if algebra were not there then how the mathematics would be? It would be like a human without a heart. Some of you think that it must be quiet interesting as we dont had to learn the algebra formula (jokes apart). For them, we have made it easy learning the formulas. In mathematics, algebra formula has a great use every time. we have made a list and PDF of algebra formula to ensure you better with the algebraic section.

Each formula which you need in solving Algebraic equation is listed here. Algebra formulas list for class 8, 9, 10, 11 & 12 will also help you in all competitive examinations like SSC, Railway, CBSE board etc. You must know that basic algebraic identities is in CBSE class 8 syllabus. Take a look at the algebra formula list below:

• Square algebra formula

(a + b)² = a² + 2ab + b²

(a – b)² = a² – 2ab + b²

a² – b² = (a – b)(a + b)

a² + b² = (a – b)² + 2ab = (a + b)² – 2ab

(a + b + c)² = a² + b² + c² + 2ab + 2ac + 2bc

(a – b – c)² = a² + b² + c² – 2ab – 2ac + 2bc

(a + b + c + …)² = a² + b² + c² + … + 2(ab + ac + bc + …..)

• Cube based formula:

(a + b)³ = a³ + b³ + 3ab(a + b) ; (a + b)³ = a³ + 3a²b + 3ab² + b³

(a – b)³ = a³ – 3a²b + 3ab² – b³

a³ – b³ = (a – b)(a² + ab + b²)

a³ + b³ = (a + b)(a² – ab + b²)

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a – b)³ = a³ – 3a²b + 3ab² – b³

(a + b + c)

• 4^th power algebra formula:

(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴)

(a – b)⁴ = a⁴ – 4a³b + 6a²b² – 4ab³ + b⁴)

a⁴ – b⁴ = (a – b)(a + b)(a² + b²)

• 5^th power formula:

a⁵ – b⁵ = (a – b)(a⁴ + a³b + a²b² + ab³ + b⁴)

• nth power algebra formula

If n is a natural number, aⁿ – bⁿ = (a – b)(a^n-1 + a^n-2b+.…+ b^n-2a + b^n-1)

If n is even (n = 2k), aⁿ + bⁿ = (a + b)(a^n-1 – a^n-2b +….+ b^n-2a – b^n-1)

If n is odd (n = 2k + 1), aⁿ + bⁿ = (a + b)(a^n-1 – a^n-2b +….- b^n-2a + b^n-1)

(a + b + c + …)² = a² + b² + c² + … + 2(ab + ac + bc + …..)

• Laws of Exponents (algebra formula)

(p^m)(pⁿ) = p^m+n

(pq)^m = p^mq^m

(p^m)ⁿ = p^mn

• Fractional exponents:

$$a^{0}={1}$$

$$\frac{a^{m}}{a^{n}} = a^{m-n}{(if m>n)}$$

$$={1 (if m=n)}$$

$$=\frac{1}{a^{n-m}} {(if m<n)}$$

$$p^{m}= \frac{1}{a^{-m}}$$

$$a^{-m}=\frac{1}{a^{m}}$$

Those who want to solve questions then they can download algebra formula in pdf format. You can download it and learn it easily whenever & wherever you want.

Download Algebra formula PDF:

Algebra formula PDF Chart is available here to download. From the link provided below you can download Algebraic formula, equations pdf. The light pdf here includes all the formula from class 6 to class 12th. Take a look at it.

Algebra Formula PDF Download

• Solved examples on algebra formula:

Q. Sove the question 15²-5².

Solution:

As, a² – b² = (a – b)(a + b)

So,
15²-5² = (15-5)(15+5)
= (10)(20)
=200

Q. Sove the question 5² x 5⁴.

By using exponential equation (p^m)(pⁿ) = p^m+n
so, 5²x5⁴ = 5⁶ = 78125

Q.Find the square of 4x+y.

As, (a + b)² = a² + 2ab + b²
(4x+y)²= (4x)² + 2(4xy) + (y)²
= 16x²+ 8xy + y²

Q.Express 16x²+24xy +9y² as a perfect square.

Solution:

16x²+20xy +16y²
=(4x)²+ 2.4x.3y + (3y)²
= (4x + 3y)²

A number system is a collection of various symbols which are called digits.

The Never Never Nest is a comic one-act play about a young couple who make full use of the buy-now-pay-later system. Jack and Jill were a young married couple who had a small baby. One day Aunt Jane visited them and was surprised to find that even though Jack’s salary was not high, they lived in a beautiful house with all comforts. She began to wonder whether, as a wedding gift she had given them 2000 pounds instead of 20 pounds. Otherwise how did Jack and Jill buy all these things? Then Jane understood that though they had everything, nothing really belonged to them. They bought everything on instalments.

Only a steering wheel of the car, a wheel and two cylinders had been paid for. The total amount to be paid towards instalments was more than their earnings. Aunt Jane was shocked at the way Jack and Jill ran their family. Before she left, she gave ten pounds to Jill and told them to make at least one article completely theirs, using that money. While Jack went with aunt Jane to the bus stop, Jill sent the money to Dr. Martin. Jack came back and said that he wanted to pay .two months instalments on the car using the aunt’s gift. But Jill said that by paying this money to Dr. Martin, their baby would become completely theirs.

Correct option is A) used to

Summary

No. of words given in the original passage: 280 No. of words to be written in the summary: 280/3 = 93 ± 5

Rough Draft

procrastination is a Latin word. Hundreds of studies reviewed by Steel, a psychologist states that to procrasinate is “to voluntarily delay an intended course of action despite expecting to be worse-off for the delay”. Another expert Dr. Joseph R. Ferrari, distinguishes between “chronic” or “real” procrastinators into three types: stimulation types avoiders and decisional procrastinators. Chronic procrastinators self-esteem.

Tice and Baumeister found that procrastinators get lower grades and have higher levels of stress. Chu and Choi, say that “Passive procrastinators” are more stressed and so less efficient. “Active procrastinators prefer to work under pressure” and “if something unexpectedly comes up, they will knowingly switch gears and engage in new tasks they perceive as more urgent.

Fair Draft

Procrastination

Procrastination is a Latin word. Hundreds of studies reviewed by Steel, a psychologist states that to procrastinate is “to voluntarily delay an intended course of action despite expecting to be worse-off for the delay”. Another expert, Dr. Joseph R. Ferrari, distinguishes between “chronic” or “real” procrastinators. Ferrari categorizes procrastinators into three types: stimulation types avoiders and decisional procrastinators. Chronic procrastinators have low self-esteem.

Tice and Baumeister found that procrastinators get lower grades and are stressed. Chu and Choi, say that “Passive procrastinators” are more stressed and less efficient. “Active procrastinators” work under pressure and if something more urgent comes up they change their focus.

No. of words in the summary: 104

Note-making
Title: Procrastination

1. Intro A proctn
Origin – Latin
Meaning – Put fwd till tomorrow – Idea A postponement or delay Steel (psychologist) – voluntary delay – despite expecting to be worse-off

2. Categorⁿ
a. Dr. Joseph R. Ferrari
Stimln types – thrill 4m beating a deadline
Avoiders – avoid things // make others think badly A them
Decisional – postpone making a decsn (until enough information)
Real procs – way A life (might need therapy)
Chronic procs:
Low self esteem
Focus on past

b. Tice and Baumeister
Procs got lower grades
Higher levels A stress and illness

c. Chu and Choi
Passive procs – stressed, less efficient
Active procs – prefer to work under pressure, knowingly engage in new tasks

Conclusion: The word procrastination comes from two Latin terms meaning to ‘put forward until tomorrow’.
Abbreviations: Intro – Introduction; A – of; Proctn – procrastination; Fwd – forward; Stimln – stimulation; Categorⁿ – categorization; 4m – from; // – that; decsn – decision; procs – procrastinators

Calculation:

\(\frac{n^2 + 7n + 12}{n^2 - n - 12}\)

⇒ \(\frac{{{n^2} + 3n + 4n + 12}}{{{n^2}\;- 4n + 3n + 12}}\)

⇒ \(\frac{{n\left( {n + 3} \right) + 4\left( {n + 3} \right)}}{{n\left( {n - 4} \right) + 3\left( {n - 4} \right)}}\)

⇒ \(\frac{{\left( {n + 3} \right)\left( {n + 4} \right)}}{{\left( {n - 4} \right)\left( {n + 3} \right)}}\)

⇒ \(\frac{{\left( {n + 4} \right)}}{{\left( {n - 4} \right)}}\)

⇒ \(\frac{{\left( {n - 4 + 4 + 4} \right)}}{{\left( {n - 4} \right)}}\)

⇒ \(1 + \;\frac{8}{{\left( {n - 4} \right)}}\)

Which will be maximum when (n - 4 = 8)

So, n = 12

∴ The largest positive integer is 12.

In the story Tomorrow by Joseph Conrad, Captain Hagberd plans what he will do when his son, Harry who has been away at the sea for 16 years comes back. He hopes to marry him to the sweet girl who lives in the neighbourhood. After his wife passes away he redoubles his efforts to find his son. He tells himself every day that his son will return tomorrow.

He lives in the little seaport of Colebrook and hopes that his son will return to the place where he stayed once. He builds two small cottages. He prepares one for his son's return and lets out the other to his tenants, an old blind man and his daughter Bessie. Initially the local people make fun of him but they accept his ways and his unusual attire in the course of time. They pity him.

When the young man comes home unexpectedly many secrets about the characters in the story are revealed.

The story tells about the lives of three characters, the father Captain Hagberd, the wild son who returns home from America and the girl who stays next door, who dreams of being released from her hellish existence.

His life was quite happening as he told Bessel that I’ve been everything you can think of almost but a tailor or soldier. I’ve been a boundary rider; I’ve sheared sheep, and humped my swag, and harpooned a whale. I’ve rigged ships, and prospected for gold, and skinned dead bullocks, —and turned my back on more money than the old man would have scraped in his whole life.

5

यह स्पष्ट है कि पूर्ण वर्ग के लिए एक 5 और चाहिए। 

अत:दी गई संख्या को 5 से गुणा करने पर प्राप्त गुणनफल पूर्ण वर्ग संख्या होगी। 

∴ 180 x 5 = 900 एक पूर्ण वर्ग संख्या है।

Given:

Aayush and Lohit together can finish in 17 days.

Lohit efficiency = 2 × Aayush efficiency

Calculation:

Lohit efficiency = 2

Aayush efficiency = 1

Lohit + Aayush Efficiency

⇒ 2 + 1 = 3

Aayush and Lohit together can finish in 17 days

Total work

⇒ 17 × 3

⇒ 51

Aayush can finish alone 

⇒ 51/1

⇒ 51 days

∴ Aayush can finish alone in 51 days.

As we know that, 
The dimension of Force is MLT-2
We have to derive dimension of Pressure
Also Pressure = Force / Area  
The dimension of Area = L²
The dimension of Force now = F
so,  P = FL^-2