Bharat me harit kranti ka janak kise kaha jata hai

Explanation:

Correct option is A)

When there is an interchange of the sentence from comparative to positive, without changing its meaning, a negative word 'no' or 'not' is added. Here the sentence is in the comparative degree. As 'Harry' is taller than 'brother', the brother is shorter than 'Harry.' In other words, the brother is 'not as tall as' Harry. Option A is correct because the sentence is of the positive degree of comparison using 'not as tall as', without changing the meaning of the sentence. Option B is incorrect because the adjective 'very' is not a comparative and the sentence is not of the positive degree. Option C is incorrect because it is opposite in meaning to that of the given sentence, where the 'brother' is taller than 'Harry'. Option D is grammatically incorrect.

Answer:

The degree of comparison that is used in the sentence "Ram is taller than Mohan" is the comparative degree of comparison.

Explanation:

• The three degrees of comparison that can be used are positive, comparative and superlative.
• The positive degree of comparison establishes the existence of quality is mentioned. This sentence can be rewritten to represent the positive degree as "Ram is as tall as Mohan".
• The structure "as + adjective +as" denotes the positive structure and the adjective is used in its root word form.
• The comparative degree is used to compare two things and establish an order of superiority. The given sentence is in this form.
• The comparative degree uses the adjective with the suffix "-er."
• The superlative degree establishes something as superior to everything else. This sentence can be rewritten to represent the superlative degree as "Ram is the tallest".
• The superlative degree uses the adjective along with the suffix "-est".

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Answer:

60.05 gm /mol

Explanation:

Molar mass ={2×12.011(C)+4×1.00794(H)+2×15.999(O)}.g.mol

−1

=60.05.g.mol

The molar mass of acetic acid is 60.05⋅g⋅mol

[tex]\star[/tex] Concept

Molecular Mass is the sum of individual atoms in the molecule .Start the question by writing chemical formula of Acetic acid and the add the molar mass of individual atoms.

Complete step by step answer :

[tex]\sf\implies[/tex] Chemical formula of Acetic acid [tex]\rightarrow[/tex] (C [tex]H_{3}[/tex] COOH)

[tex]\rightarrow[/tex] Atomic mass of C = 12

[tex]\rightarrow[/tex] Atomic mass of O = 16

[tex]\rightarrow[/tex] Atomic mass of H = 1

Molecular mass :

• Since , there are 2 carbon atoms

[tex]\rightarrow[/tex] 2 × atomic mass of C

• Since, there are 2 atoms of oxygen

[tex]\rightarrow[/tex] 2 × atomic mass of O

• Since, there are 4 atoms of hydrogen

[tex]\rightarrow[/tex] 4 × atomic mass of one hydrogen atom

[tex]\sf\implies[/tex] 2× 12 + 2 × 16 + 4 × 1

[tex]\sf\implies[/tex] 24 + 32 + 4

[tex]\sf\implies[/tex] 60

Since , Molecular mass is donated by u

[tex]\therefore[/tex] Molecular mass of Acetic acid is 60 u.

Step-by-step explanation:

answer is given above

I hope this help you

Given,

The student's ages are in A.P.

The common difference  = 3 months.

The younger student's age = 12.

The total age = 375yrs.

To Find,

The count of students in the group.

Solution,

Given that,

The ages of students are in A.P.

The difference between their ages = 3months.

3/12 = 1/4yrs

The total of their age = 375yrs.

Younger one's age =12yrs.

In A.P, the sum of the terms

S = n/2 (2a +(n-1)d)

a =12.

S = n/2 (2*12+ (n-1)1/4)

S = 375

n/2(2*12+(n-1)1/4) = 375

n/2(24+(n-1)1/4) = 375

n(24+(n-1)1/4) = 750

n(24*4+(n-1)) = 750 * 4

n(96+(n-1)) = 3000

96n + [tex]n^{2}[/tex] -n = 3000

95n + [tex]n^{2}[/tex] = 3000

[tex]n^{2}[/tex] + 95n -3000 = 0

Let's factorise it.

[tex]n^{2}[/tex] + 120n -25n -3000 = 0

n(n+120) -25(n +120) = 0

So,

n = -120 or 25

Hence, 25is the total number of students in the group.

Answer:

Astha borrowed approx. [tex]\text{Rs}30021[/tex].

Step-by-step explanation:

Let the sum Astha borrowed be [tex]$x$[/tex].

Then, using [tex]S.I=\frac{p*r*t}{100}[/tex], where [tex]p,r,t[/tex] are principle, rate  & time respectively.

[tex]$$\begin{aligned}&\left(\frac{x \times 7 \times 2}{100}\right)+\left(\frac{x \times 6 \times 4}{100}\right)+\left(\frac{x \times 9 \times 1}{100}\right)=14110 \\&\left(\frac{14 x}{100}+\frac{24 x}{100}+\frac{9 x}{100}\right)=14110 \\\frac{47 x}{100}=14110 \\& x=\frac{14110\times 100}{47}=30021 .27\end{aligned}$$[/tex]

Hence, sum borrowed[tex]$=$ Rs 30021.27[/tex] or approximately Rs. [tex]30021[/tex].

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Answer:

Total amount of money Astha borrowed= Rs 17000

Step-by-step explanation:

In the above question:

Given:

1. R(rate)= 7%

   T(time)= 2 years

2. R(rate)= 6%

    T(time)= 4 years

3.    R(rate)= 9%

    T(time)= 5 years

To find:

Total amount borrowed?

Solution:

Total interest= Rs 14110

We know, the formula for Simple interest as:

S.I= P×R×T/100

P= principle amount(here borrowed amount)

R= Rate of interest

T= Time

Let the total amount borrowed be 'x'

Then, total amount borrowed at the end of 11years will be;

S.I= P1×R1×T1/100+P2×R2×T2/100+P3×R3×T3/100

14110= x×7×2/100+x×6×4/100+x×9×5/100

14110= 14x/100+24x/100+45x/100(common denominator)(x is also common)

14110= 14x+24x+45x/100          

14110= 83x/100

  83x= 14110×100

      x= 1411000/83

      x= 17000

The amount of money she borrowed was Rs17000

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