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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.
| 28101. |
Prove that CosA-sinA+1/cosA+sinA-1=cosecA-cosA |
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| 28102. |
What is x?? |
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Answer» X is an alphabet????????????????? Apko kya lgta h |
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| 28103. |
If a and B are the zeroes of quadratic polynomial f(x)=6x square +x-2, find tha values of a/B+ B/a? |
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Answer» Factorise the eqn first then let one root \'a\' and other \'b\' now put the values of a & b in the eqn a/b + b/a , solve it and you will get your answer equivalenat to -25/12 -25/12 -25/12 -25/12 |
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| 28104. |
Sums for Practice |
| Answer» ???? | |
| 28105. |
25-53 |
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Answer» -28...common sense -28..... what a silly question yu are asking ? |
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| 28106. |
Sample paper buy nhi ho rahe hai |
| Answer» Iss app ke..?! | |
| 28107. |
Sample paper buy nhii ho rahe |
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| 28108. |
What is the relation between alfa and beta |
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Answer» Alfa + Beta = -b/a.Alfa × Beta = c/a. alpha: aa , beta : bbaa + bb : -b/aaa × bb : c/a |
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| 28109. |
Use Euclid \'s algorithm to find hcf of 1190 and 1445. Express the HCF in the form 1190m+1445n. |
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Answer» 1145=1190*1+2551190=255*4+170255=170*1+85179=85*2+0With this solve 1190m+1445n First find hcf of 1190 and 1445 by division and also write the Euclid\'s algorithm. with the algorithm find m and n |
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| 28110. |
Show that root2 + root3 is irrational... |
| Answer» Let √2 + √3 = (a/b) is a rational no.On squaring both sides , we get2 + 3 + 2√6 = (a2/b2)So,5 + 2√6 = (a2/b2) a rational no.So, 2√6 = (a2/b2) – 5Since, 2√6 is an irrational no. and (a2/b2) – 5 is a rational no.So, my contradiction is wrong.So, (√2 + √3) is an irrational no. | |
| 28111. |
Op is equal to diameter of circle .Then prove that ABP is an equilateral triangle |
| Answer» Construction : Join A to Bwe have,OP = diameter{tex}\\Rightarrow {/tex}\xa0OQ + QP = diameter{tex}\\Rightarrow {/tex}\xa0Radius + QP = diameter{tex}\\Rightarrow {/tex}\xa0OQ - PQ = radiusThus OP is the hypotenuse of right angled\xa0{tex}\\triangle {/tex}AOPSo, In\xa0{tex}\\angle A O P , \\sin \\theta = \\frac { A O } { O P } = \\frac { 1 } { 2 }{/tex}\xa0{tex}\\theta = 30 ^ { \\circ }{/tex}Hence,\xa0{tex}\\angle A P B = 60 ^ { \\circ }{/tex}Now, In {tex}\\triangle A O P{/tex}AP = ABSo,\xa0{tex}\\angle P A B = \\angle P B A = 60 ^ { \\circ }{/tex}{tex}\\therefore \\triangle A P B{/tex} is an equilateral triangle. | |
| 28112. |
Change 10 min into hour |
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Answer» 1/6 hour 1/6hour 0.16 hours |
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| 28113. |
Our Real Indian?? super hero came back ...Vikram Abinadan .. from pak. |
| Answer» Bharat humko jaan se pyara hai ...... | |
| 28114. |
Find the common difference if 17th term of an A.P exceeds its 10th term by 14 |
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Answer» A/Q,a+16d-a-9d=14. or, 7d=14 or,d=2 Let a be the first term and d the common difference.It is given a_{17} - a_{10} = 7⇒ (a + 16d) - (a + 9d) = 7⇒ 7d = 7 ⇒ d = 1 d=2 |
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| 28115. |
Find the largest number which divided 615 and 963 leaving remainder 6 in each case |
| Answer» 87 | |
| 28116. |
Ex 9.1 ex 6 cbse ncert book |
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| 28117. |
An Easy solution of q no \'6\' of ex 12.3 chapter area related to circles |
| Answer» 12.1 q 4 | |
| 28118. |
When is \'n\' ask in AP and when Tn |
| Answer» complete ur question | |
| 28119. |
factorise x^2_x_12 |
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Answer» x2 - x - 12= x2 - 4x + 3x - 12= x(x - 4) + 3(x - 4)= (x -4) (x +3) Surface area |
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| 28120. |
What is the nature of roots of quadratic equation X square - 2x + 4 =0 |
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Answer» No real root as D is less than 0 Non- Real roots Thank you No real root |
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| 28121. |
Using Euclid lemma find hcf of 56 96 and 104 |
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| 28122. |
If 7times the 7th term of an AP is equal to 11 times the 11th term .find 18 th term |
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Answer» 7(a+(7-1)d)=11(a+(11-1)d)7a+42d=11a+110d-4a=68d-a=17dWe have to find 18th termSo, a18=a+(18-1)d=a+17dAs we got value of 17d=-a,so put in this equation=a+(-a)=a-a=0 0 0 |
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| 28123. |
Find modal daily expenditure |
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| 28124. |
What is blade suit |
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| 28125. |
If secA + tan A = P then find value of cosec A |
| Answer» Refer sample paper of 2019 | |
| 28126. |
Evaluate : cos 30+sin45-1/3tan60+cos90. |
| Answer» Pls give me answer. ... | |
| 28127. |
NTSE result declared !!!! How many of you are qualified for stage 2 |
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| 28128. |
Solve for x - 1/a+b+x= 1/a + 1/b + 1/x |
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| 28129. |
7+9+11+12+15+15........ Upto 70 terms find the sum of the series |
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Answer» A=7 , d=2 a = 7 =d=2 ...n =70= Sn= n/2(2a +( n-1)d) = S70 = 70÷2(2×7 + ( 70-1)2) = 35(14+138 ) = 35× 152 =2320 5320 |
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| 28130. |
Find the quadratic polynomial who\'s Alpha equal to under root 2 and beta 1/ 3 |
| Answer» Firstly find alpha+beta than (alpha)(beta ) put the values in formula (x^2 — (sum of zeros ) + (product of zeros ) , (x^2 — (alpha +beta ) + (alpha )(beta ) | |
| 28131. |
Do we mean adding up the series if we are asked to combine two series ? |
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| 28132. |
A man goes to 12m due south and then 35m due west. Jow far is he from the starting point? |
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Answer» Use Pythagoras 37m |
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| 28133. |
In triangle opq, right angke at p,op=7cm and oq-pq=1 cm determine the value of sin q and cos q |
| Answer» SinQ=7/25 and cosQ=24/25 | |
| 28134. |
The mth term of an ap is n and the nth term is m.find the nth term |
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Answer» 0 it will be mnth term Bo mnth term hoga nth term given to h |
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| 28135. |
x /a cosA + y/b sinA = 1 x/a sinA - y/ b cosA = -1 Prove that :- x^2/a^ + y^2/b^2 =2 |
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| 28136. |
Which term of the AP: 21,18,15,----------- is -81 also,is any term 0 koi solution bata do plz |
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Answer» An= -81 -81=21+(n-1)-3n=35 -81 ko last term lekr solve kr |
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| 28137. |
PT and PS are 2 tangents are drawn to a circle and radius r OP=2r show angle ots=ost=30 |
| Answer» Given,In {tex}\\triangle{/tex}OTS,{tex}\u200b\u200b\u200bOT =OS {/tex}{tex}\\Rightarrow \\quad \\angle O T S = \\angle O S T{/tex}\xa0...(i)In right\xa0{tex}\\triangle OTP,{/tex}\xa0{tex}\\frac { \\mathrm { OT } } { \\mathrm { OP } } = \\sin \\angle \\mathrm { TPO }{/tex}{tex}\\Rightarrow \\frac { r } { 2 r } = \\sin \\angle \\mathrm { TPO }{/tex}{tex}\\sin \\angle \\mathrm { TPO } = \\frac { 1 } { 2 } \\Rightarrow \\angle \\mathrm { TPO } = 30 ^ { \\circ }{/tex}Similarly\xa0{tex}\\angle \\mathrm { OPS } = 30 ^ { \\circ }{/tex}{tex}\\Rightarrow \\angle T P S = 30 ^ { \\circ } + 30 ^ { \\circ } = 60 ^ { \\circ }{/tex}Also\xa0{tex}\\angle \\mathrm { TPS } + \\angle \\mathrm { SOT } = 180 ^ { \\circ }{/tex}{tex}\\Rightarrow \\quad \\angle \\mathrm { SOT } = 120 ^ { \\circ }{/tex}In {tex}\\triangle{/tex}SOT,{tex}\\angle \\mathrm { SOT } + \\angle \\mathrm { OTS } + \\angle \\mathrm { OST } = 180 ^ { \\circ }{/tex}{tex}\\Rightarrow 120 ^ { \\circ } + 2 \\angle \\mathrm { OTS } = 180 ^ { \\circ }{/tex}{tex}\\Rightarrow \\angle \\mathrm { OTS } = 30 ^ { \\circ }{/tex}\xa0...(ii)From (i) and (ii)\xa0{tex}\\angle \\mathrm { OTS } = \\angle \\mathrm { OST } = 30 ^ { \\circ }{/tex} | |
| 28138. |
Find the k if equation have no solution (3k+1)x+3y=2,(k square +1)x+(k-2)y=5 |
| Answer» Given linear equation is(3k + 1)x + 3 y - 2 = 0 .......... (i)(k2 + 1)x + (k - 2)y - 5 = 0 ............ (ii)Compare with a1x + b1y + c = 0 and a2x + b2y and c2 = 0a1 = 3k + 1 , b1 = 3 , c1 = -2and a2 = k2+ 1 , b2 = k - 2, c2 = -5The given system of equations will have no solution, if{tex} \\frac { a_1 } { a_2 } = \\frac { b_1 } { b_2} \\neq \\frac { c_1 } { c_2 }{/tex}{tex} \\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \\frac { 3 } { k - 2 } \\neq \\frac { - 2 } { - 5 }{/tex}{tex}\\Rightarrow \\quad \\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \\frac { 3 } { k - 2 } \\text { and } \\frac { 3 } { k - 2 } \\neq \\frac { 2 } { 5 }{/tex}Now,\xa0{tex}\\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \\frac { 3 } { k - 2 }{/tex}{tex}\\Rightarrow{/tex}\xa0(3k + 1)(k - 2)=3(k2 + 1){tex}\\Rightarrow{/tex}\xa03k2 - 5k - 2 =3k2 + 3{tex}\\Rightarrow{/tex}\xa0-5k - 2 =3{tex}\\Rightarrow{/tex}\xa0-5k = 5{tex}\\Rightarrow{/tex}\xa0k = -1Clearly,\xa0{tex}\\frac { 3 } { k - 2 } \\neq \\frac { 2 } { 5 }{/tex}\xa0for k = -1.Hence, the given system of equations will have no solution for k = -1. | |
| 28139. |
Find the distance of the point (-4,-7) from y-axis. |
| Answer» Apply distance formula and take x=0 | |
| 28140. |
Sec +tan=p find the value of cosec |
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Answer» Soneb prefer sample paper of 2019 Miss your answer is wrong (p²+1)/(p²-1) |
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| 28141. |
Tan1 Tan10 Tan 20 Tan 70 Tan 80 Tan 89 |
| Answer» 1 is answet ...If more problem then click thanks on my answer | |
| 28142. |
Ye heart hacker ko kya ho raha hai |
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| 28143. |
if tan+cot =5 find the value of tan²+cot² |
| Answer» tan = 5- cotcot=5-tanab khud karle value put kerke | |
| 28144. |
In triangle DEF,if DN is the median,then show that DE^2+DF^2=2(DN^2EN^2) |
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| 28145. |
The diagonal of a rectangular field is 60 m more than the shortest side . If the |
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| 28146. |
In a class test the sum of shefali\'s marks in maths and english is 30 |
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Answer» Arihant Ncert solution me diya hai .. 3.4 |
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| 28147. |
the sum of four observations is( p)2+31p if mean is 8 find the value of p |
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| 28148. |
Therorem 6.2 |
| Answer» Book me hai | |
| 28149. |
2(ax-by)+(a+4b)=02(bx+ay)+(b-4a)=0Solve above eqation by cross multiplucation method |
| Answer» x=-1/2 and y=2 | |
| 28150. |
In trapizium ABCD AB parallelCD DC equals 2AB EFparallelAB where Eand F lie on BC and AD |
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