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Ancylostomiasis is a disease caused by Ancylostoma duodenale or Necator americanus, both hookworms belonging to the nematode phylum (roundworms). Ancylostomiasis caused by these worms is also called hookworm disease.

Since the parasites nourish themselves on human blood the infection causesanemia, hypoproteinemia and the patient often seems pale.

The main prophylactic measures against ascariasis are: efficient washing of vegetables and other foods; basic sanitary conditions and appropriate destination of feces; hygiene education for people; combat against insects that can carry the eggs of the parasite, like flies and cockroaches.

Ascaris is a monoxenous parasite, its life cycle is dependent only on one host and so it does not have intermediate host.

The intermediate host of the schistosome is a gastropod mollusc, a snail of the Planorbidae family and Biomphalaria genus. The snail vector of schistosomiasis lives in freshwater, as in lagoons and creeks.

We Know that,

r = 0.529 x \(\frac{n^2}{z}\)pm

Where,

n = Orbit Number 

(Principle Quantum Number)

z = Atomic Number

∴ r = 0.529 x \(\frac{4^2}{1}\) 

= 0.529 x 4 x 4

= 8.464 pm

The title of “After Twenty Years” is ironic because two old friends are reuniting after a long time, but they are now on opposite sides of the law. When Bob sees Jimmy, he does not recognize him. They were young men when they last saw each other.

Adult hookworms within the human intestine release eggs that are eliminated with the human feces. Under adequate conditions of moisture and temperature the eggs mature in the soil and generate larvae. The larvae differentiate into thread-like infective larvae that can penetrate the human skin, generally through the feet. The larvae them gain the human circulation and reach the lungs from where they go to the airway and the pharynx. When the larvae are swallowed they enter the small intestine and develop into adult worms and the cycle restarts.

The correct option is (c) 1:-1

as K.E : Total energy = KZe²/2r  :  - KZe²/2r  

                                   = 1:-1

Both Ancylostoma duodenale and Necator americanus have mouthparts with hooks or “teeth” that help the fixation of the parasite in the human intestine wall and facilitate the tissue injury necessary to drain blood from the host. The structures are evolutionary adaptations for the parasitic way of life of these animals.

Hookworms are monoxenous, i.e., their life cycle depends only on one host.

0.4x+0.3y=1.7...(1),
0.7x−0.2y=0.8....(2).

Multiply both the equations by 10, we get
4x+3y=17...(1)
7x−2y=8...(2)

Multiply (1) by 2 and (2) by 3,
8x+6y=34
21x−6y=24
Adding both the equations
29x=58
x=2

From (1):4×2+3y=17
⇒8+3y=17
⇒3y=17−8=9
y=3

Answer : x=2,y=3

33/427 D Cabin 

Ahmedabad 

15 July 20XX

Dear 

Varun

You will be glad to learn that I have secured 80th rank in the CBSE-PMT competition. I have got admission in a prestigious institution – B J Medical College,Ahmedabad. I want to share a few happy moments of my life in the company of my old Mends at a dinner in the Hotel Kanishka at 9.00 p.m. on 23 July, 20XX. Please join the celebrations and merry-making.

Yours Rahul 

sincerely 

(i) \((\frac{-3}{17})+(\frac{-12}5)=(-\frac{12}5)+(\frac{-3}{17})\) By Commutative law

(ii) \(-9+(\frac{-21}8)=(\frac{-21}8+(-9))\) (By Commutative law)

(iii) \((\frac{-8}{13}+\frac37)+(-\frac{13}4)=(\frac{-8}{13})+[\frac37+(\frac{-13}4)]\) (By associative law)

(iv) \(-12+[\frac7{12}+(\frac{-9}{11})]=(-12+\frac7{12})+(\frac{-9}{11})\) (By associaive law)

97 years out of every 400 are leap years, 
In 100 years there are 24 leap years and 76 normal years.
400 years =4×24=96
400 year has 96 leap year + 1 year (i.e, 400 is leap year)
So, 29th February will come 97 times in 400 years.

Accommodation Required 

REQUIRED a posh Bungalow for rent in the area of Pratap Nagar, Delhi to be used as company guest house. Bungalow will be rented by EVL company. Bungalow owner may contact Ashish, General Manager, EVL Company, Surat Nagar, Delhi. Contact number 0xxxxxxxxxx.

x = 300 + 0.15 (2500 - 1100)

 = 300 + 0.15 x 1400

 = 300 + 210

 = 510

y = 400+ 0.10 (2500 - 1300)

 = 400 + 0.10 x 1200

 = 400 + 120

 = 520

∴ x - y = 510 - 520 = -10

For each of the 3 girls to be seated between 2 boys, the minimum number of boys needed is 4. The arrangement in which this happens is:

BGBGBGB

Since there are 5 boys, there is one boy left over, who can be placed next to any of the 4 boys already placed. So there are 4 possible arrangements if each child is identified only by their gender. They are:

BBGBGBGB

BGBBGBGB

BGBGBBGB

BGBGBGBB

If each child is distinguishable (not identified only by their gender) then the number of possible arrangements is

(4) x (5!) x (3!)

= 4 x 120 x 6

= 2,880.

x = (1 + 2x) e^y/x 

Taking log both side, we get

log x = log(1 + 2x) + y/x log e

\(\Rightarrow\) log x = log(1 + 2x) + y/x (∵ log e = 1)

\(\Rightarrow\) y = x log x - x log (1 + 2x) .....(1)

Differentiating (1) w.r.t x, we get

y' = log x + x/x - log(1 + 2x) - 2x/(1 + 2x)

= log x - log(1 + 2x) + \(\frac{1 + 2x - 2x}{1 + 2x}\)

\(\Rightarrow\) y' = log x - log(1 + 2x) + \(\frac{1}{1 + 2x}\) ......(2)

Differentiating (2) w.r.t x we get

\(y'' = \frac{1}{x} - \frac{2}{1 + 2x} - \frac{2}{(1 + 2x)^2}\)

\(=\frac{(1 + 2x)^2 - 2(1 + 2x)x - 2x}{x(1 + 2x)^2}\) \(=\frac{4x^2 + 4x + 1 - 4x^2 - 2x - 2x}{x(1 + 2x)^2}\)

\(=\frac{1}{x(1 + 2x)^2}\)

Now xy' - y = x log \(\frac{x}{1 + 2x} + \frac{x}{1 + 2x} - x log \frac{x}{1 + 2x}\) = \(\frac{x}{1 + 2x}\)

∴ (xy' - y)² = \(\frac{x^2}{(1 + 2x)^2}\) = x³ × \(\frac{1}{x(1 + 2x)^2}\) = x³y''

Hence, x³y'' = (xy' - y)² Hence proved.

The relation shows the relationship between INPUT and OUTPUT. Whereas, a function is a relation which derives one OUTPUT for each given INPUT.

Correct option:

(A) 0 

(A) cos^-14/21

(C) a₁b₁ + a₂b₂ + a₃b₃ 

Let the number of children be x.

It is given that Rs. 250 is divided equally amongst x childrens. 

∴ Money received by each child = Rs. \(\frac{250}{x}\). 

If there were 25 more children then money received by each child = Rs. \(\frac{250}{x+25}\) 

From the given information,

\(\frac{250}{x}\) - \(\frac{250}{x+25}\) = \(\frac{50}{100}\)

⇒ \(\frac{250(x+25)−250x}{ x(x+25)}\) = \(\frac{1}{2}\) 

⇒ 2 × 6250 = x² + 25x 

⇒ x² + 25x − 12500 = 0 

⇒ x² + 125x − 100x − 12500 = 0 

⇒ x(x + 125) − 100(x + 125) = 0 

⇒ (x − 100)(x + 125) = 0 

⇒ x = 100 or x = −125. 

∵ x ≠ −125 

(Because number of children never be negative.) 

∴ x = 100. 

Hence,

The number of children is 100.

Correct option:

(C) -1

(A) l₁l₂ + m₁m₂ + n₁n₂ = 0 

Correct option:

(C) (0,0,1) 

Correct option:

(B) n × m 

Correct option:

(B)  tan^-1 3/4

(c) 3CuSO₄ + 2PH₃ → Cu₃ P₂ + 3H₂ SO₄

    3HgCl₂ + 2PH₃ → Hg₃ P₂ + 6HCl

(d) SH–bond is weaker than, O–H bond. Hence H₂ S will furnish more H⁺ ions

(c) Air is liquified by making use of the joule-Thompson effect (cooling by expansion of the gas) Water vapour and CO₂ are removed by solidification. The remaining major constituents of liquid air i.e., liquid oxygen and liquid nitrogen are separated by means of fractional distillation (b.p. of O₂ = –183°C : b. P. of N₂ = – 195.8°C)