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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 80601. |
show that a-=B anu -a.ху18) Ifx+iy= (a+ib)”, show that * + =4185= 1-i, show that (5a-7b) = 0.a+3i19) If 2+ib-3i)(2-3i)latib20) If xtiy= Vc+id'a²+b2prove that (x2+y2)2 =ber.c2+d?1+i21) If (a+ib) =, then prove that (a +b) =1;tik22) Show thatT-id})=is rea |
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Answer» 20) if a+3i/2+ib=1-i, show that (5a-7b)=0;; a+3i= (1-i)(2+ib) a+3i=2+b+1(b-2); equating real and imaginary parts; a=2+b 3=b-2; b=5; a=7; 5a-7b=35-35=0 |
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| 80602. |
21. The sides of a quadrilateral ABCD taken in order are 6 cm, 8 cm, 12 cmnd 14 cm respectively and the angle between the first two sides is aright angle. Find its area. (Given, V6 2.45.) |
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Answer» page 1 |
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| 80603. |
the bisectors of ZB and Z C intersecte ABC, with AB = AC, the bisectorsShow that:isosceles triangle ABCher at O. Join A to O. Show that .other areOB=0CABC AD IST(ii) AO bisects ZAAD is the perpendicular bisector of BC730). Show that A ABC is an isosceleser Fje 7.30). Showde in which AB=AC. |
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Answer» 2 question answer:AD = AD (common)angle ADB = angle ADC (90 degree given)angle BAD = angle CAD (AD bisects angle A)So, Triangle BDA congruent to Triangle ADC by ASA congruence rule.AB = AC (CPCT)Triangle ABC is an isosceles Triangle |
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| 80604. |
Show that everyptivale meShow that any positive odd integer e of the f |
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Answer» Letabe any positive integer andb= 2.Applying Euclid’s algorithm, we have:a= 2q+r, for some integerq≥ 0, and 0 ≤r< 2a= 2qor 2q+ 1Ifa= 2q, thenais an even integer.Now, a positive integer can either be even or odd. Thus, any positive odd integer is of the form 2q+ 1. |
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| 80605. |
\begin{aligned} \overline{c} &=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k} \\|\vec{c}|=& \sqrt{\left(\frac{1}{\sqrt{3}}\right)^{3}+\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(-\frac{1}{\sqrt{3}}\right)^{2}}=\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=\sqrt{\frac{3}{3}}=\end{aligned} |
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Answer» ratio to percentage |
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| 80606. |
f Athen show that | 2A1 4JA4 2, then show that | 2A4|A |
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| 80607. |
( a + b ) ^ { 3 } + ( b + c ) ^ { 3 } + ( c + a ) ^ { 3 } - 3 ( a + b ) ( b + c ) ( c + a ) = 2 ( a ^ { 3 } + b ^ { 3 } + c ^ { 3 } - 3 a b c ) |
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Answer» thank you |
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| 80608. |
62(2+3x)In the given figureperpendicular to lirbetween rays OPProve that ZRCONIn the given figure lines AB and CD intersectat O. If ZAOC + Z BOE = 70° and Z BOD AR= 40°, find Z BOE and reflex 2 COE.It is given thatDraw a figure4 LINES ANObserve thenects the other leo distinct pointnsversal. It is ao distinct poirints 'P'andlines m andMRIn the given figure lines XY and MN intersectat O. If ZPOY = 90° and a: b = 2:3, find c.X |
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| 80609. |
Find the area of the following figures:(b) 8 cnm5 cm4 cm5 cm10 cm2 cm15 cm C |
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Answer» a) Area= 1/2(AB)(BC) + (CD)(DE)=1/2(8)(15) + (17)(3)=60 + 51 = 111cm^2 b)Area=(1/2)(8+4)(5) +(1/2)(2+4)(5)=30+15=45cm^2 c)Area of circle=π(r)^2=3.14(10)^2=314cm^2 |
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| 80610. |
(0) Quadrilateral ABCDAB-4.5 cmBC 5.5 cmCD 4 cmAD 6 cm |
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| 80611. |
How many millimetres does a kilometre have ?2. 56 bottles are equally filled by 14 litres of soft drink. How many millilitres of soft drink does ending the Namhave ?L i 2 Linarams?stimes they may saythe Numbers:the situation and regmoting to the more |
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Answer» 1 kilometer has 10,00,000 millimeters. the right answer is 10,00,000mm |
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| 80612. |
. In the given figure, a circle of diameter 21 cm is given. Inside this circle two circles with diameters-andof the diameter of the big circle have been drawn, as shown in the given figure. Find the area of theshaded regionFig, 16.27 |
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| 80613. |
Determine LP+ LQ+LR+ LS+ LT.24(a) 4 right angle(c) 2 right angle(b) 3 right angle(d) 6 right angle |
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Answer» Bylinearpairaxiomwehave,∠DEP=180-∠5∠EDP=180-∠4In△PDE,sumofallangleswillbe180.⇒∠P+180-∠5+180-∠4=180⇒∠P=∠5+∠4-180Similarly∠Q=∠5+∠1-180∠R=∠1+∠2-180∠S=∠2+∠3-180∠T=∠3+∠4-180Addingabove5equations,weget,∠P+∠Q+∠R+∠S+∠T=2∠1+∠2+∠3+∠4+∠5-900InpentagonABCDE,∠1+∠2+∠3+∠4+∠5=540⇒∠P+∠Q+∠R+∠S+∠T=2540-900=1080-900⇒∠P+∠Q+∠R+∠S+∠T=180= 2 right angleAnswer |
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| 80614. |
Solved ExamplesFactorise 3xy +6x23xy = 3 xxxyG24 =2 x 3 x x x x x y x y: 3xy + 62-3 xxxy 2 x 3 x x x x x yx yThe common factors are 3, x and y.. By taking out common factors, we have3 xxxyx (1 +2 xxx y)1.: 3xy+ 6x2- 3xy(1 + 2x) |
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| 80615. |
\left| \begin{array} { c c c } { x - 2 } & { 2 x - 3 } & { 3 x - 4 } \\ { x - 4 } & { 2 x - 9 } & { 3 x - 16 } \\ { x - 8 } & { 2 x - 27 } & { 3 x - 64 } \end{array} \right| = 0 |
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Answer» subtract first row from second row and third row|x-2 2x-3 3x-4||-2 -6 -12 | =0|-6 -24 -60|(x-2)(360-288)-(2x-3)(120-72)+(3x-4)(48-36)=072x-144-96x+144+36x-48=012x-48=012x=48x=4 thnq very much sir |
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| 80616. |
Show that only one out of n, n+2, n+4 is divisible by 3 |
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Answer» We applied Euclid Division algorithm on n and 3.a = bq +r on putting a = n and b = 3n = 3q +r , 0<r<3i.e n = 3q -------- (1), n = 3q +1 --------- (2), n = 3q +2 -----------(3) n = 3q is divisible by 3 or n +2 = 3q +1+2 = 3q +3 also divisible by 3or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3 Hence n, n+2 , n+4 are divisible by 3. |
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| 80617. |
45.His office is in the first[A] level[C] stage[B] grounddj floor |
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Answer» Answer:D) FloorHis office is in the first floor |
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| 80618. |
34. Solve the following differential equationsdjcos(x+y) +sin(x +y) |
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Answer» thanks |
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| 80619. |
EXERCISE 4.1Construct the following quadrilaterals.1. Quadrilateral ABCD.AB 4.5 cmBC-5.5 cmCD 4 cmAD = 6 cmAC=7 cm |
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| 80620. |
\left. \begin{array} { l } { PL = 4 cm } \\ { LA = 6.5 cm } \\ { \angle P = 90 ^ { \circ } } \\ { \angle A = 110 ^ { \circ } } \\ { \angle N = 85 ^ { \circ } } \end{array} \right. |
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| 80621. |
EXERCISE 4.1lowing quadrilateralsa)QundrilateralL Constructthe folJU# 3.5 cmUM 4 cmMP 5 cmPJ4.5 cmPU 6.5 cmo Qundrilateral ABCDAB 45cmBC 5.5 cmCD-4cmAD-6 cmAC 7 cm(iv) Rhombus BESTParalelogram MOREBE 4.5 om0Rz6cmRE-4.5 cmEO: 7.5 cmET- 6 cm |
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| 80622. |
3. In the given figure, if LA =<B = 90, OB = 4.5 cm, OA =6 cm and AP = 4 cm, then find QB. |
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| 80623. |
\left. \begin{array} { l } { \operatorname { log } _ { a } ( 1 + 2 + 3 ) = \operatorname { log } _ { a } 1 + \operatorname { log } _ { a } 2 + \operatorname { log } _ { a } 3 } \\ { \operatorname { log } \frac { 35 } { 78 } = \operatorname { log } 7 + \operatorname { log } 5 - \operatorname { log } 2 - \operatorname { log } 3 - \operatorname { log } 13 } \\ { 7 \operatorname { log } \frac { 16 } { 15 } + 5 \operatorname { log } \frac { 25 } { 24 } + 3 \operatorname { log } \frac { 81 } { 80 } = \operatorname { log } 2 } \end{array} \right. |
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| 80624. |
-log(12, 10) %2B log(81, 10)/4 %2B log(9, 10) %2B 2*log(6, 10) |
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Answer» As log(mn)= logm- lognand logx^n= nlogx |
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| 80625. |
\left. \begin{array} { l } { \operatorname { log } ( x + 3 ) + \operatorname { log } ( x - 3 ) = \operatorname { log } 16 } \\ { \operatorname { log } ( 3 x + 2 ) - \operatorname { log } ( 3 x - 2 ) = \operatorname { log } 5 } \\ { 2 \operatorname { log } _ { 10 } x = 1 + \operatorname { log } _ { 10 } ( x + ( \frac { 11 } { 10 } ) ) } \end{array} \right. |
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Answer» plz solve this |
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| 80626. |
If the sum of first m terms of an A.P is n and the sum of first n terms is m, then show thatthe sum of its first (m+ n) terms is -(m + n).24. |
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| 80627. |
If the sum of first m terms of an A P is same as the sum of its first n terms (m+n), show that the sum ofits first (m + n) terms is zero.9. |
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| 80628. |
If the sum of first m terms of an A.P. is the same as that of its first n terms,show that the sum of its first (m+n) terms is zero. |
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| 80629. |
() the total atfiounlt savcu Dj50. If the sum of first m terms of an A.P. is n and sum of first n terms of the same A.P. ism.Show that sum of first (m + n) terms of it is(m+ n). |
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Answer» Let a be the first term and d be c.d. of the A P .ThenSm=nn= m/2{2a+ (m-1)d} 2n= 2am+ m( m-1)d. ........(1)andSn= mm= n/2{2a+(n-1)d}2m= 2an+ n(n-1)d. ...........(2)Subtracting eq.(2)- (1), we get2a(m-1)+{m(m-1)- n(n-1)}d = 2n-2m2a(m-n) +{(m^2-n^2)-(m-n)}d = -2(m-n)2a +(m+n-1) d = -2. [On dividing both sides by ( m-n)]………(3) Now,Sm+n=m+n/2{2a +(m+n-1)d}Sm+n=m+n/2(-2) ………[using (3)]Sm+n=-(m+n) |
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| 80630. |
Ifthe surm of first 9 terms ofan А.Р. is 81 and that of first 1 7 terms is 289, find the sum of first n terms.. |
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| 80631. |
Quadrilateral ABCDAB 4.5 cmBC= 5.5 cmCD 4 cmAD 6cmAC 7 cm |
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| 80632. |
EXERCISE 4.1T. Construct the following quadrilaterals.(i)Quadrilateral ABCD.AB 4.5 emBC-5,5 cmCD 4 emAD-6cmAC 7 cm |
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| 80633. |
(ii) In rectangle ABCD, AB6cm, and AD 4cm, then what is the area of APAB(Fig. 12.52)?4 cmLAvbFig. 12.52 |
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| 80634. |
TWO MARK QUESTIONSI. In the figure ABCD is a rectanglein which CD- 6cm. AD 8cm. Find the area of parallelogram CDEFD 6em C8 cm |
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| 80635. |
Write the following decimals in the place value table.(a) 19.4 (b) 0.3 (c) 10.6 (d) 205.9Write each of the following as decimals:(a) Seven-tenths(c) Fourteen point six.(b) Two tens and nine(d) One hundred andt |
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| 80636. |
\log _{10} 15\left(1+\log _{15} 30\right)+\frac{1}{2} \log _{10} 16\left(1+\log _{4} 7\right)-\log _{10} 6\left(\log _{6} 3+1+\log _{6} 7\right) |
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| 80637. |
log(4, 10) %2B log(2, 10) %2B log(3, 10)=log(x, 10) |
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| 80638. |
(5,12,13) is Pythagorean triplet.? If yes then state the reason. |
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Answer» 5²+12²=25+144=169=13² Hence, (5,12,13) is a Pythagorean triplet. |
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| 80639. |
find the sum of9. If the sum of first 7 terms of an APis 49 and that of 17 terms is 289,first n terms. |
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| 80640. |
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum offirst n terms |
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| 80641. |
(c)Subtract\begin{array}{l}{4 p^{2} q-3 p q+5 p q^{2}-8 p+7 q-10 \text { from }} \\ {18-3 p-11 q+5 p q-2 p q^{2}+5 p^{2} q}\end{array} |
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| 80642. |
Ifthe sum of first 9 terms of an A.P. is 81 and that of first 17 terms is 289, find the sum of first n terms. |
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| 80643. |
(b) 112(d) 130(a) 103(c) 12076. 4, 200, 369, 513, 634, ?(a) 788( 734(b) 715(d) 755nh and answer the |
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Answer» 734 is the right answer. |
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| 80644. |
nce A, B, C by suitable numerals.Replace A, B, C by1.5A+87CB32. 4 CB6+ 369 A8173 |
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| 80645. |
name the place value of five in each of the followinga 512 |
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Answer» Place value of 5 in 512 ishundredth |
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| 80646. |
Write the place value of 7 in following numbersa. 372,156,932b. 712,156,200 c. 152,373,129 d. 4,037,152 |
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Answer» Place value of 7 is:a) 70000000b) 700000000c) 70000d) 7000 PLEASE HIT THE LIKE BUTTON tq |
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| 80647. |
in which PO-25m PR-17m15Fig. 10.13s. In Fig 10.14 4B-scm, BC-6cmCDFig. 10.14DC |
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Answer» Kindly post one question per post to experience the instant solution feature of scholr at its best. |
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| 80648. |
ite the following decimals in the place value table.0.29 (b) 2.08 (c) 19.60 (d) 148.32 (e) 200.812 |
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| 80649. |
\begin{array}{l}{\int \frac{k^{\sqrt{x}}}{\sqrt{x}} d x=} \\ {\text { (A) } k^{\sqrt{x}} \log e^{k}+e \quad \text { (B) } 2 k^{\sqrt{x}} \log e^{k}+c} \\ {\text { (C) } 2 k^{\sqrt{x}} \log 10^{k}+c}\end{array} |
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| 80650. |
\begin{array}{l}{\int \frac{k^{\sqrt{x}}}{\sqrt{x}} d x=} \\ {\text { (A) } k^{\sqrt{x}} \log e^{k}+e} \\ {\text { (C) } 2 k^{\sqrt{x}} \log 10^{k}+c \quad \text { (D) } 2 k^{\sqrt{x}} / \log e^{k}+c}\end{array} |
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Answer» answer nhi aa raha h |
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