81279 + Interview Questions in Mathematics QA in Mathematics

1.	18 Explain the structure and functioning of human eye. How are we able to see nearby aswell as distant objects?
Answer» The structures and functions of the eyes are complex. Each eye constantly adjusts the amount of light it lets in, focuses on objects near and far, and produces continuous images that are instantly transmitted to the brain. Theorbitis the bony cavity that contains the eyeball, muscles, nerves, and blood vessels, as well as the structures that produce and drain tears. Each orbit is a pear-shaped structure that is formed by several bones. The outer covering of the eyeball consists of a relatively tough, white layer called thesclera(or white of the eye). Near the front of the eye, in the area protected by the eyelids, the sclera is covered by a thin, transparent membrane (conjunctiva), which runs to the edge of the cornea. The conjunctiva also covers the moist back surface of the eyelids and eyeballs. Light enters the eye through thecornea, the clear, curved layer in front of the iris and pupil. The cornea serves as a protective covering for the front of the eye and also helps focus light on the retina at the back of the eye. After passing through the cornea, light travels through thepupil(the black dot in the middle of the eye). Theiris—the circular, colored area of the eye that surrounds the pupil—controls the amount of light that enters the eye. The iris allows more light into the eye (enlarging or dilating the pupil) when the environment is dark and allows less light into the eye (shrinking or constricting the pupil) when the environment is bright. Thus, the pupil dilates and constricts like the aperture of a camera lens as the amount of light in the immediate surroundings changes. The size of the pupil is controlled by the action of the pupillary sphincter muscle and dilator muscle. Behind the iris sits thelens. By changing its shape, the lens focuses light onto the retina. Through the action of small muscles (called the ciliary muscles), the lens becomes thicker to focus on nearby objects and thinner to focus on distant objects. Theretinacontains the cells that sense light (photoreceptors) and the blood vessels that nourish them. The most sensitive part of the retina is a small area called themacula, which has millions of tightly packed photoreceptors (the type called cones). The high density of cones in the macula makes the visual image detailed, just as a high-resolution digital camera has more megapixels.

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18 Explain the structure and functioning of human eye. How are we able to see nearby aswell as distant objects?

Answer»

The structures and functions of the eyes are complex. Each eye constantly adjusts the amount of light it lets in, focuses on objects near and far, and produces continuous images that are instantly transmitted to the brain.

Theorbitis the bony cavity that contains the eyeball, muscles, nerves, and blood vessels, as well as the structures that produce and drain tears. Each orbit is a pear-shaped structure that is formed by several bones.

The outer covering of the eyeball consists of a relatively tough, white layer called thesclera(or white of the eye).

Near the front of the eye, in the area protected by the eyelids, the sclera is covered by a thin, transparent membrane (conjunctiva), which runs to the edge of the cornea. The conjunctiva also covers the moist back surface of the eyelids and eyeballs.

Light enters the eye through thecornea, the clear, curved layer in front of the iris and pupil. The cornea serves as a protective covering for the front of the eye and also helps focus light on the retina at the back of the eye.

After passing through the cornea, light travels through thepupil(the black dot in the middle of the eye).

Theiris—the circular, colored area of the eye that surrounds the pupil—controls the amount of light that enters the eye. The iris allows more light into the eye (enlarging or dilating the pupil) when the environment is dark and allows less light into the eye (shrinking or constricting the pupil) when the environment is bright. Thus, the pupil dilates and constricts like the aperture of a camera lens as the amount of light in the immediate surroundings changes. The size of the pupil is controlled by the action of the pupillary sphincter muscle and dilator muscle.

Behind the iris sits thelens. By changing its shape, the lens focuses light onto the retina. Through the action of small muscles (called the ciliary muscles), the lens becomes thicker to focus on nearby objects and thinner to focus on distant objects.

Theretinacontains the cells that sense light (photoreceptors) and the blood vessels that nourish them. The most sensitive part of the retina is a small area called themacula, which has millions of tightly packed photoreceptors (the type called cones). The high density of cones in the macula makes the visual image detailed, just as a high-resolution digital camera has more megapixels.

2.	Math:-tVisit a nearby vegetable shop and observe how they weigh vegetables using difterentunits of mass and write that in shots. S
Answer» They use grams and kilograms to weigh vegetables

3.	A contractor has the target of completing work indays. He employed 20 persons who completed (1/4) ofthe work in 10 days and left. The number of persons hehas to employ to finish the remaining part as per target is:
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6.	A family of 4 members consumes 6kg of sugar in a month what will be the monthly consumption of sugar if the number of family members becomes 6 ?
Answer» 4members --- 6kg/month1member ---- (6/4) kg/month6 members ---- 6*(6/4) = 9 kg/month

7.	।।।।।।।।।।५।१५।५।।[3 3 5]प्रश्न 10. यदि A =\|2 3 4, तो सिद्ध कीजिए कि AJ = IA5 2 3](2014)[3 3 5][1 0 0]
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11.	Q10 Area of a sector of a circle of radius 36cm is 54x cm^2. Find the length of the corespondingarc of the sector.
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16.	pIC rthe sector2 Find the area of a quadrant of a circle whose circumference is 22 cm
Answer» solution ma ta batava hA

17.	समीकरण से » »% 2 के मान ज्ञात कीजिए--ककe 0 b5 Gie
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20.	है न 2 लि के कक गे जा ५ सतत है कि एए पं कि vam~ < b 0 / dWM_,‘VL« ‘U oहि।। >> नजर 1!v . वि o i = ,/’—HA <
Answer» Shivam = 1.5Mohan Mohan = 2/3 Shivam so Mohan's salary 33% lower than Shivam.

22.	यदि किसी बेलन का वक्रपृष्ठ 110 सेमी तथा ऊँचाई 5 सेमी है, तो उस बेलन कीत्रैज्या ज्ञात कीजिए। 2b5 i . ! % न
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24.	Solve the equation $ \frac{4}{x}-3=\frac{5}{2 x+3} ; x \neq 0,-\frac{3}{2}, $ for $ x $
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4.	-2/3 X 3/5 + 5/2 - 3/5 X 1/6
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13.	-1/10 %2B (3/5)*((-2)/3) %2B 5/2
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14.	Poove thatthe 756 å insational
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15.	37.Find Å¿ sinxcos2xdx
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25.	x ^ { 2 } - \frac { 5 } { 2 } x + 7 \text { from } \frac { 1 } { 3 } x ^ { 2 } + \frac { 1 } { 2 } x + 2
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26.	By AA similarity criterion which is the reguleuQ. 6, In figure, ΔΑΒΕ AACD, show that AAD15AABC
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28.	\frac { 3 ( x - 2 ) } { 5 } \leq \frac { 5 ( 2 - x ) } { 3 }
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29.	\frac { 5 } { 2 x } + \frac { 2 } { 3 y } = 7 , \frac { 3 } { x } + \frac { 2 } { y } = 12 ( x \neq 0 , y \neq 0 )
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31.	Solve the following equation:\frac{x+3}{2}+3 x=\frac{5}{2}(x-3)
Answer» (x + 3)/2 + 3x = 5(x-3)/2 x/2 + 3/2 + 3x = 5x/2 -15/2x/2 + 3x -5x/2 = -15/2 - 3/2-4x/2 + 3x = -18/2 -2x + 3x = -9x = -9

33.	6) Name two elements having independent existence
Answer» The atoms of only a few elements called noble gases (Such as helium, neon, argon and kryton, etc.) are chemically unreactive and exist in the free state or independently.

34.	QuelTheI findsumtheof nAP andAP = 5n²-3terms ofloth term.
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39.	\begin{array}{l}{\text { If } \lambda=\lim _{n \rightarrow \infty} \sqrt{n} \text { then the value of }} \\ {\operatorname{Sec}^{-1}(\lambda)+\tan ^{-1}(\lambda)=}\end{array}
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40.	\frac 5 2 x %2B \frac 2 3 y = 7 , \frac 3 x %2B \frac 2 y = 12 , x \neq 0 , y \neq 0
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41.	(AB)-1 = B-A-1.[cos x - sin x 0]5. आव्यूह A = \| sin x cos x 0\| 0 0 1 \|\|का व्युत्क्रम आव्यूहज्ञात कीजिए।
Answer» what is your question

Explore topic-wise InterviewSolutions in Current Affairs.

18 Explain the structure and functioning of human eye. How are we able to see nearby aswell as distant objects?

Math:-tVisit a nearby vegetable shop and observe how they weigh vegetables using difterentunits of mass and write that in shots. S

A contractor has the target of completing work indays. He employed 20 persons who completed (1/4) ofthe work in 10 days and left. The number of persons hehas to employ to finish the remaining part as per target is:

-2/3 X 3/5 + 5/2 - 3/5 X 1/6

(5+2√3) (5-2√3)

A family of 4 members consumes 6kg of sugar in a month what will be the monthly consumption of sugar if the number of family members becomes 6 ?

।।।।।।।।।।५।१५।५।।[3 3 5]प्रश्न 10. यदि A =|2 3 4, तो सिद्ध कीजिए कि AJ = IA5 2 3](2014)[3 3 5][1 0 0]

(5**6*((3/5)**2)**3)/(3**3)**2

-5*x %2B 3 %2B 2*(3*5 %2B 5^2) - 3

2 rthe aree of trapezium is 2.8m and parallel sides are 0.8m and 60cm respectively, find thealtitude.

Q10 Area of a sector of a circle of radius 36cm is 54x cm^2. Find the length of the corespondingarc of the sector.

-1/10 %2B (3/5)*(-2/3) %2B 5/2

-1/10 %2B (3/5)*((-2)/3) %2B 5/2

Poove thatthe 756 å insational

37.Find Å¿ sinxcos2xdx

pIC rthe sector2 Find the area of a quadrant of a circle whose circumference is 22 cm

समीकरण से » »% 2 के मान ज्ञात कीजिए--ककe 0 b5 Gie

(a^x)^4*(a^(3*x %2B 5))^2=a^(8*x %2B 12)

ि (BB bt bt 5८ o x m;a;;unue.e 9B BB b ha % 0=94 40+ xe न हिs

है न 2 लि के कक गे जा ५ सतत है कि एए पं कि vam~ &lt; b 0 / dWM_,‘VL« ‘U oहि।। &gt;&gt; नजर 1!v . वि o i = ,/’—HA &lt;

he peanok beadta of Ha Îłeelanghe .x-2

यदि किसी बेलन का वक्रपृष्ठ 110 सेमी तथा ऊँचाई 5 सेमी है, तो उस बेलन कीत्रैज्या ज्ञात कीजिए। 2b5 i . ! % न

If A 100&amp;B5a) A &gt; Bthenb) A &lt; Bc) A B

Solve the equation $ \frac{4}{x}-3=\frac{5}{2 x+3} ; x \neq 0,-\frac{3}{2}, $ for $ x $

x ^ { 2 } - \frac { 5 } { 2 } x + 7 \text { from } \frac { 1 } { 3 } x ^ { 2 } + \frac { 1 } { 2 } x + 2

By AA similarity criterion which is the reguleuQ. 6, In figure, ΔΑΒΕ AACD, show that AAD15AABC

o 1 (के 0w= (909500~ 100+ |) (9 998 +QUEL+]) (1) I‘

\frac { 3 ( x - 2 ) } { 5 } \leq \frac { 5 ( 2 - x ) } { 3 }

\frac { 5 } { 2 x } + \frac { 2 } { 3 y } = 7 , \frac { 3 } { x } + \frac { 2 } { y } = 12 ( x \neq 0 , y \neq 0 )

\frac { 5 } { 2 x } + \frac { 2 } { 3 y } = 7 , \frac { 3 } { x } + \frac { 2 } { y } = 12 , x \neq 0 , y \neq 0

Solve the following equation:\frac{x+3}{2}+3 x=\frac{5}{2}(x-3)

x : \frac 4 x - 3 = \frac 5 2 x %2B 3

6) Name two elements having independent existence

QuelTheI findsumtheof nAP andAP = 5n²-3terms ofloth term.

5. ABCD is a trapezium in which AB I CD and AD-BC, Show that :(iii) aABCAADHoard Term II 2012, Set 01] [NCERT)

2*x^2 %2B 5*x - 12=0

2*x^2 - 5*x/6 %2B 1/12

12*x^2 %2B 17*x - 5

\begin{array}{l}{\text { If } \lambda=\lim _{n \rightarrow \infty} \sqrt{n} \text { then the value of }} \\ {\operatorname{Sec}^{-1}(\lambda)+\tan ^{-1}(\lambda)=}\end{array}

\frac 5 2 x %2B \frac 2 3 y = 7 , \frac 3 x %2B \frac 2 y = 12 , x \neq 0 , y \neq 0

(AB)-1 = B-A-1.[cos x - sin x 0]5. आव्यूह A = | sin x cos x 0| 0 0 1 ||का व्युत्क्रम आव्यूहज्ञात कीजिए।

\cos 3 x + \cos x - \cos 2 x = 0

cosx1-sinx=tan(Ď+x10.Pove thatOr

\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x

tan | Ď +tan + )Prove that(1- sinx)

Prove that 1-sinx = tan(4+2)10.[41(1

2Xwase 22 KuixaQueland the cubic Polynomial with crum, somof product of two zerols taken at a timeand product of zeroes as 5, 6 and 20respectively.

I with the sum, sum of the product of its zeroes taken twwo. fits zero estakenttime, and the product of its zeroes as 2, -7, -14 respectivel

tan O + tan 20 + tan 0. taअथवा/ एन

(4V3-5V5) + (1V3-6v5) = a3 + b5 -

(5**6*((3/5)2)3)/(33)2

-5x %2B 3 %2B 2(3*5 %2B 5^2) - 3

(a^x)^4(a^(3x %2B 5))^2=a^(8*x %2B 12)

है न 2 लि के कक गे जा ५ सतत है कि एए पं कि vam~ < b 0 / dWM_,‘VL« ‘U oहि।। >> नजर 1!v . वि o i = ,/’—HA <

If A 100&B5a) A > Bthenb) A < Bc) A B

2x^2 %2B 5x - 12=0

2x^2 - 5x/6 %2B 1/12

12x^2 %2B 17x - 5

cosx1-sinx=tan(Ď+x10.Pove thatOr

tan | Ď +tan + )Prove that(1- sinx)

47.	2Xwase 22 KuixaQueland the cubic Polynomial with crum, somof product of two zerols taken at a timeand product of zeroes as 5, 6 and 20respectively.
Answer» polynomial=x^3-x^2(sum of zeros)+x(sum of product of two)-product of threex^3-5x^2+x(-6)-(-2)x^3-5x^2-6x+2

50.	(4V3-5V5) + (1V3-6v5) = a3 + b5 -
Answer» Please hit the like button if this helped you