Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

In figure 5.38, points X, Y, Z are the midpointsAB, side BC and side AC of \Delta ABCAB = 5 cm, AC = 9 cm andBC = 11 cm. Find the length of XY, YZ, XZ.

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AB = 5CM , AC = 9CM , BC = 11CMIF points X,Y,Z are the midpoints of side AB, side BC and side AC of ΔABC.then,AX = BX = 5/2 = 2.5CM BY = CY = 11/2 = 5.5 CMAZ = CZ = 9/2 = 4.5 CMFrom the property, we can say that XY is parallel to AC. Similarly, YZ is parallel to AB and XZ is parallel to BC.then ,BY = XZ = 5.5CMYZ = XB = 2.5CMXY = CZ = 4.5CM

Hence , (XY, YZ, XZ.) = ( 4.5cm , 2.5cm , 5.5cm)

2.

4.Shikha larns $ 1500 for working in 16 days.How much will she larm if she works for30 days?

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She earns 7500 in 16 daysso in one day she will earn 7500/16nowin 30days

= 7500*16/30750*16/3

4000

3.

outtohe eigit at the ones place of the product 25 x 16 x 7 without workingthe problem.

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lets take unit digit of each number so they are 5,6,7 so multiplication of it 5*6*7=210 so 0 will be at unit place in 25^2*16^3*7^2

4.

Ram cam finish a work in 16 days and the same work finished by shyam in 8 days if how many both of them will finish the work while working together

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5.

72°+36°+v°=180°

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72 +36+v =180 v =72

72°+36°+v°=180°108°+v°. =180° v°=180°-108° v°=72°

6.

Each side of a square is 9 cm. Find its:(i) perimeter(i) area.

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7.

The length of the sides of a triangle are 9 cm, 12 cm and 15 cm. Find the length ofaltitude corresponding to the shortest side.

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8.

(v)(4x + 1)² + (2x+3)4r? +12x+96136

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using a_square+2ab+b_square

9.

( v ) x ^ { 2 } - 4 x - 192 \quad ( v i ) x ^ { 4 } - 5 x ^ { 2 } + 4 \quad ( v i 1 ) x ^ { 4 } - 13 x y ^ { 2 } + 36 y

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10.

find the V8-36

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√8.356/√2.635 = 1.8796787...

this question is solve logarthoms mathod

y = √8.356/√2.635

logy = log(√8356/2635) = 1/2*[log(8356)-log(2635)] = 1/2*(3.92-3.42) = 1/2*(0.5) = 0.25so, y = 10^0.25 = 1.7782

approximate Value.. is this, you can follow the steps on your own

11.

(i) dy+ (3x +cot x) dx = 0

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12.

A metallio sphere of diameter 1 cm is to be melted and recasted into a hollowthickness 1 mm Find the internal and extemal radi of the sphere

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DIAMETER OF METALLIC SPHERE = OUTER DIAMETER OF HOLLOW SPHERE = 1 cm THEREFORE OUTER RADII = 0.5 cm = 5mm

THICKNESS = 1mm

THEREFORE INNER RADII = OUTER RADII - THICKNESS

= 5 mm - 1mm

= 4mm

13.

1. In a college of 300 students, every student reads5 newspapers and every newspaper is read by.60 students. The number of newspapers is(a) atleast 30(b) atmost 20(c) exactly 25(d) of these

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Total no. of students= 300.Let total no. of news papers = n.Given every student reads 5 news papers and every newspaper is read by 60 students.Total no. of newspaper reading by every student = 300*5 = 1500 .Total no. of newspaper reading which each paper will get = 60*n .which implies that, 60*n = 300 *5 => 60 *n =1500=> n = 25

Total no. of newspapers is 25

(c) is correct option

Ans =total no.of newspapers is 25 correct ans is option (c)

14.

(9.2 (जाए 8 + 0०56८8)% न (८०59 के 5600)% न एक धाए6 ने ८०9)

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15.

b) In a school of 300 students, every student reads 5 newspapers and every newspaper isread by 60 students. Find the number of news papers.

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Given newspaper read by 1 student = 5

so total newspaper read by 300 students = 300X 5 = 1500

Single newspaper is read by 60 studentso total different newspaper read = 1500/60= 25

16.

log x- 1 72[1 + (log x)2]t a2

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17.

1. Suppose ABCD is rectangle. Using RHS theorem,provetriangles ABC and ADC are congruent.? Sunn

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tanks sir

18.

” dy .= “"fnnddx--—'+(sinx) oL

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y= x^sinx + (sinx)^cosx​let x^sinx = u and (sinx)^cosx= vdy/dx = du/dx + dv/dxu= x^sinx+++Taking log on both sideslog u = log(x^sinx) = sinx log xDifferntiating both sides1/u du/dx= cosx logx + sinx/x (product rule)du/dx= u {cosxlogx+sinx/x}du/dx= x^sinx(cosxlogx+sinx/x)

Now, v = (sinx)^cosxTaking log on both sideslog v = log {(sinx)^cosx} = cosxlog sinxDifferentiating1/v. dv/dx = (-sinx)log sinx+ cos x (cos x/ sin x)dv/dx = v { cos x cot x - sinx log sinx}​dv/dx = (sinx)^cos x (cos x cot x - sinx log sinx)=> dy/dx = {x^sinx(cosxlogx+sinx/x)} + {​(sinx)^cosx (cosx cot x -sinxlogsinx)}Therefore, dy/dx = x^sinx (cosxlog x + sinx/x) + sinx^cosx (cosx cotx - sinx log sinx)

19.

In what time will rs 5600 amount to RS 6720 at 8% p.a.

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20.

Suppose you are given a circle. Give aconstruction to find its centre.

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21.

Suppose you are given a circle. Give a construction to find its

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thanks

22.

Calculate dy/dxif y=x⁴+sinx

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23.

if y = f(sinX) then dy/dx =

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y = f(sinx) dy/dx = f'(sinx)*cosx

24.

If y =sinx^3dy/dx will be:

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y=sin^3xdy/dx=3sin^2x [d (sinx)/dx]dy/dx=3sin^2xcosx

25.

dy3xtan xIf yx2 sinx +thenwill be

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26.

1. If y logx, finddydx

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27.

Suppose you are given a circle. Give a construction to find its centre.

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28.

2. Suppose you are given a circle. Give a construction to find its centre

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Ans :- Steps: Draw a line across thecircleto make a "chord" Construct the perpendicular bisector of that chord to make a diameter of thecircle. Construct the perpendicular bisector of that diameter togetthecenterof thecircle.

29.

3500 (3༯202

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(3x+5y)(3x+2y)=9x^2+6xy+15xy+10y^29x^2+21xy+10y^2

30.

1. Find the length of the hypotenuse of a right triangle, the other two sides of which measure9 cm and 12 cm.

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31.

EXERCISE 15D1. Find the length of the hypotenuse of a right triangle, the other two sides of which measure9 cm and 12 cm.

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length of the hypotenuse is 15

32.

? % of 5600-28% of 3500=1988

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May i know why it has been marked as spam?

33.

f (2y22xy, find dydx

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34.

If y = ex" then finddydx

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y = (e^x)^xTaking log on both sides, log y = x*log(e^x)Differentiating both sides with respect to x,(1/y)(dy/dx) = {x/(e^x)}*(e^x) + log(e^x)=> dy/dx = xy + y*log(e^x)

35.

Find dy/dx if y = Sin x

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dy/dx (sinx)= cosxthanks

cos X is the correct answer

y=sinxdy/dx=d/dx(sinx) =cosx

dy/dx=cosx is the right answer

Y=sinced(y)/dx=d(sinx)/dx=cosx Answer

dy/dx= cos x is the answer

dy/de(since)=coaxanswer.

36.

y= logx/1+sinx find dy/dx

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Please hit the like button if this helped you

37.

dc1,2x+3y = sin x

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38.

а82-a²82 taz-a (s² + a2)

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2a³/(s²+a²)² is the correct answer of the given question

39.

dyFind in the following:dlx1, 2x + 3y sin x4, xy + y2 tan x + y

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Given - 2x+3y = sinx

Differentiating both sides with respect to x , we get

d/dx ( 2x+3y) = d/dx (sinx)

2.d/dx(x) + x.d/dx(2) + 3.d/dx(y) + y.d/dx(3) = cosx . d/dx(x) [ using product rule ]

2(1) + x(0) + 3.(dy/dx) + y(0) = cosx (1)

2 + 3(dy/dx) = cos x

dy/dx = ( cosx - 2) / 3

thank you

40.

Given HCF(396, 82) = 2, find LCM(396, 82).

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If you find this solution helpful, Please give it a 👍

41.

किg.14. 82 : R Gbk S ?1) 63 2) 12 3) 821) 32

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Let 82:72::72:x=>82x=72×72=>x=63

Option 1 is correct.

I cannot understand

Let the missing number here(?) be x.

When 4 numbers are in continued proportion,a:b::c:d=>ad=bc

42.

\cos \theta \left[ \begin{array}{cc}{\cos \theta} & {\sin \theta} \\ {-\sin \theta} & {\cos \theta}\end{array}\right]+\sin \theta \left[ \begin{array}{cc}{\sin \theta} & {-c_{0}} \\ {\cos \theta} & {\text { si }}\end{array}\right.

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43.

$ 9 x+3 y+12=0 $$ 18 x+6 y+24=0 $

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44.

2. Suppose you are given a circle. Give a construction to find its ce

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45.

2.IP.

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46.

Write the missing numbers.3 +0 = 37 + 6=+1+ 14 =0+ 18 =18+40

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ans

3+0=3

1+14=15

7+6=6+7

0+18=18+0

3+0=3

1+14=15

7+6=6+7

0+18=18+0

3+0=31+14=157+6=6+70+18=18+0

3+0=31+14=157+6=6+70+18=18+0

47.

Fill in the blanks in the following table, given that a is the first term, d the commondifference and a, the nth term of the AP:(I120)1810018(iv) 18.92.53.6105(v)3.5

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48.

K. If the cost price of 9 pens is equal to the selling price of 11pens, the loss percentage is(A) 18%(B)187%10(0)18 %(D) of these-

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18 2/11

18×2÷11 is the correct answer

18×2÷11 is right answer of that question

49.

8. Find the zeroes of the quadratic polynomial and verify the relationship between the zeroesand the coefficients 6x-3-7x.t 3v-6 2 x _ 3v : 1 2 İs consistent, if so, solve them

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50.

9 = {p€Ip-017. Rhombus

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Given:d1=10; d2=6

Area of Rhombus= (d1 ×d2)/2=(10×6)/2=60/2=30 sq units

area of rhombus is half the product of diagnols=1/2×(d1 × d2)=1/2×10×6=60/2 =30 sq units

it is given, lenght of diagonals is 10 and 6 area=half of product of diagonals = 30

1/2 ×d1×d2 =1/2×6×10=30