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Given:

A, B, and C entered into a business and the ratio of their investments was 5 ∶ 3 ∶ 4.

Calculations:

Let the ratio of initial investments be 5x, 3x, and 4x.

Ratio of their investments = 5x × 12 ∶ {3x × 6 + (3x + 1000) × 6} ∶ {4x × 8 + (4x + 2000) × 4}

⇒ 30x ∶ (18x + 3000) ∶ (24x + 4000)

According to the question,

\(\Rightarrow \frac{30x}{18x\ + \ 3000}\ =\ \frac{45}{42}\)

\(\Rightarrow \frac{30x}{18x\ + \ 3000}\ =\ \frac{15}{14}\)

⇒ 28x = 18x + 3000

⇒ 10x = 3000

⇒ x = 300

The B’s investment = 3x = 3 × 300 = Rs. 900

∴ The B's investment is Rs. 900.

Given:

A invested = Rs. 70,000 for 10 years

B invested = Rs. 80,000 for 8 years

C invested = Rs. 96,000 for 9 years

Formula:

Ratio of Profit = ratios of product of Amount invested and time

Calculation:

Required ratio = 70000 × 10 ∶ 80000 × 8 ∶ 96000 × 9

⇒ 700 ∶ 640 ∶ 864

⇒ 175 ∶ 160 ∶ 216

Given:

Ratio of investment of Kamal and Dinesh = 5 : 7

Time after which Manohar invested = 4 months

Formula Used:

Profit = Investment × Time

Calculations:

Let the profits of Kamal, Dinesh and Manohar at the ned of year be x, y and z respectively.

Ratio of investment of Kamal and Dinesh = 5 : 7

Capital of Manohohar = Capital of Dinesh

⇒ Ratio of investment of Kamal, Dinesh and Manohar = 5 : 7 : 7

Now, Profit = Investment × Time

⇒ x : y : z = 5 × 12 : 7 × 12 : 7 × (12 – 4)

⇒ x : y : z = 60 : 84 : 56

⇒  x : y : z = 15 : 21 : 14

∴ The ratio of their profits at the end of the year is 15 : 21 : 14.

Given:

A invested = Rs. 500000 for 12 months

B invested = Rs. 800000 for 8 months

C invested = Rs. 80000 for 6 months

Formula:

Ratio of Profit = ratios of product of Amount invested and time

Calculation:

Required ratio = 50 × 12 ∶ 80 × 8 ∶ 8 × 6

⇒ 75 ∶ 80 ∶ 6

∴ The ratio of their profits after 1 years is 75 ∶ 80 ∶ 6.

Given number is :

\(\frac{2.34 \times172.23}{4.08}\)

n = \(\frac{403.0182}{4.08}\)

= 98.779

Approx value = n_a = 99

Error = n_a - n

= 99 - 98.779

= 0.221

Present error = \(\frac{n_a-n}{n}\times100\)%

= \(\frac{0.221}{98.779}\times100\)%

= 0.22%

Teaching career is appropriate for those people who are really interested in it one can choose one’s career in elementary schools, middle schools, high schools and schools and collages according to one’s educational qualification. Teachers enjoy lots of respect among their students because the teacher contributes a lot to the society. It also provide inner satisfaction.

Milkman X

Water = 3L

Milk = 10 L

Mixture = 10 + 3 = 13L

Milk percentage = 10/13 × 100 = 76.92%

Milkman Y

Water percentage = 18%

Milk percentage = 82%

Milkman Z

Water content = 6/31

Milk content = 1 - (6/31) = 25/31

Milk percentage = 25/31 × 100 = 80.64%

In decreasing order of the quality of the milk is  Y, Z, X.

i. It should be based on the inter related sub-topics of physical education.
ii. It should cater to the requirements of different people.
iii. It should be capable of facing the challenges of the present and the future.
iv. It should be capable of providing comprehensive and deep knowledge about physical education.
v. It should motivate an individual to engage in a lifelong healthy and active lifestyle.
vi. It should be able to develop social and emotional skill among people.

(i) Medical Examination : It is very important for the success of programmes related to adapted physical education. Otherwise, it will be difficult to find out what kind of disability, the student is suffering from. Therefore, it is imperative to conduct medical examination of the students.

(ii) Programmes according to the interest of the students : Programmes should be made keeping in mind the interest, capacity and previous experience of the students. The teachers should also have deep knowledge about. It, then only they can make any successful programme.

(iii) Equipment should be appropriate : Students should be provided with equipment as per disability concerned. For example, students suffering from visual impairment should be given a ball with bell so that they many catch the ball as it rolls because of the sound. Thus such students can make out. The direction and distance of the ball.

(iv) Proper Environment : The play area also should be limited because of the limited speed capacity of the children. For example, speech impaired children should be given rest in between the games. The play area should be limited to top smaller area.

(v) Modification of Rules : Rules and regulations of the games and sports should be modified depending on the specific needs of students. In order to learn new skill they may be given extra time, extra effort, extra rest and 2 marks in place of 1 mark.

Thus, they might be given the opportunity for all round development.

B. the isotopes are chemically the same

(4x² +1+4x )- (9x²  +4+12x)=0

4x² +1+4x - 9x² - 4-12x=0

-(-5x² -8x - 3)=0

5x

5x² +5x+3x+3=0

5x(x+1)+3x(x+1)=0

(x+1)(5x+3x)=0

x=-1, x=0

Solution: 
Let us assume, to the contrary that 5 + 3√2 is rational. 
So, we can find coprime integers a and b(b ≠ 0) 
such that 5 + 3√2 = a/b 
=> 3√2 = a/b - 5

=> √2 = (a - 5b)/3b 
Since a and b are integers,  (a - 5b)/3b is rational. 
So,  √2 is rational. 
But this contradicts the fact that √2 is irrational. 
Hence, 5 + 3√2  is irrational.

For pipe, r = 1cm 

Length of water flowing in 1 sec, h = 0.7m = 7cm 

Cylindrical Tank, R = 40 cm, rise in water level = H

Volume of water flowing in 1 sec = πr²h = π x 1 x 1 x 70

= 70π 

Volume of water flowing in 60 sec = 70π x 60 

Volume of water flowing in 30 minutes = 70π x 60 x 30

Volume of water in Tank = πr²H = π x 40 x 40 x H 

Volume of water in Tank = Volume of water flowing in 30 minutes 

π x 40 x 40 x H = 70π x 60 x 30 

H = 78.75cm

Factor Theorem Procedure

Step 1 : If f(-c)=0, ( x+ c) is a factor of the polynomial f(x).

Step 2 : If p(d/c)= 0, (cx-d) is a factor of the polynomial f(x).

Step 3 : If p(-d/c)= 0, (cx+d) is a factor of the polynomial f(x).

Step 4 : If p(c)=0 and p(d) =0, then (x-c) and (x-d) is a factor of the polynomial.

If p(d/c)= 0, (cx-d) is a factor of the polynomial f(x).

https://www.sarthaks.com/?qa=blob&qa_blobid=5057042858246142522

Z₁ = (2, -1) & Z₂ = (6, 3)

\(\therefore\) Z₁ - Z₂ = (2, -1) - (6, 3)

 = (2 - 6, -1 -3)

 = (-4, -4)

z₁ & z₂ are two non zero distinct complex numbers and |z₁| = |z₂|

Now, \(\frac{z_1+z_2}{z_1-z_2}=\frac{(z_1+z_2)}{(z_1-z_2)}\times\frac{\overline{z_1-z_2}}{\overline{z_1-z_2}}\)

\(=\frac{(z_1+z_2)(\bar{z_1}-\bar{z_2})}{|z_1-z_2|}\) (\(\because z\bar z = |z|^2\))

\(=\frac{z_1\bar{z_1}-z_1\bar{z_2}+z_2\bar{z_1}-z_2\bar z_2}{|z_1-\bar z_2|^2}\) 

\(=\frac{|z_1|^2-|z_2|^2-z_1\bar z_2+z_2\bar z_1}{|z_1-z_2|^2}\)

\(=\frac{z_2\bar z_1-z_1\bar z_2}{|z_1-z_2|^2}\)....(1)

Let z₁ = a + ib & z₂ = c + id

Then \(\bar z_1\) = a - ib & \(\bar z_2\) = c - id

\(\therefore\) \(z_1\bar z_2\) = (a + ib)(c - id) = ac + bd + i(bc - ad)

& \(z_2\bar z_1 = \) (c + id) (c - id) = ac + bd + i(ad - bc)

\(\therefore\) \(z_2\bar z_1-z_1\bar z_2 = \) (ac + bd + i(ad - bc)) - (ac + bd) - i(ad - bc)

 = 2i(ad - bc)

\(\therefore\) From (1)

\(\frac{z_1+z_2}{z_1-z_2}=\frac{2i(ad-bc)}{|z_1-z_2|^2}\) which is a purely imaginary number.

From the first equation, |z₁|=|z₂|

From the second equation, arg(z₁)+arg(z₂)=0, the complex numbers are conjugate of each other.

So |z₁|²=|z₂|²=z₁z₂=Re²(z)+Im(z)². Option (B) is correct!!

We have |z₁ + z₂| = √3

and |z₁| = |z₂| = 1

∵ |z₁ + z₂|² + |z₁ - z₂|² = 2(|z₁|² + |z₂|²)

⇒ |z₁ - z₂|² = 2(|z₁|² + |z₂|²) - |z₁+z₂|²

= 2(1+1) - 3  (∵ |z₁| = |z₂| = 1 and |z₁ + z₂| = √3)

= 4 - 3 = 1

(D) Increased ECF volume

(d) the linear momentum is conserved

Limitations of Dimensional analysis

1. This method gives no information about the dimensionless constants in the formula like 1, 2, ….. π, e, etc. 

2. This method cannot decide whether the given quantity is a vector or a scalar. 

3. This method is not suitable to derive relations involving trigonometric,exponential and logarithmic functions.

4. It cannot be applied to an equation involving more than three physical quantities. 

5. It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation.

For example, 

using dimensional analysis, s = ut + \(\frac{1}{3}\)at² is dimensionally correct whereas the correct relation is s = ut + \(\frac{1}{2}\)at²