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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.
| 8401. |
the ratio of the length of a rod and its shadow is equal find the angle of elevation |
| Answer» 45° | |
| 8402. |
Find HCF of 96 and 404? |
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Answer» 4 Simmi fagna your answer is wrong check your answer HCF OF( 96 AND 404) =4 404=2×2×101=2²×101¹ 96=2×2×2×2×2×3=2²×3¹ |
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| 8403. |
Show that equation 2xsquare -7x+6=0 has 2 as a root. |
| Answer» Alpha =2 and bita =3/2 | |
| 8404. |
If cosec thita -cot thita=1/3, than find cosec thita+cot thita. |
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Answer» I have used the following identity:a²-b²=(a+b)(a-b)1+cot²∆=cosec²∆ Consider ∆ to be theta.Cosec²∆-cot²∆=1(cosec∆+cot∆)(cosec∆-cot∆)=1(Cosec∆+cot∆)1/3=1cosec∆+cot∆=3 |
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| 8405. |
In the given fig CB Parallel QR and CA parallel. If AQ=12cm,AR=20cm,PB=CQ=15cm, calculate PC and BR. |
| Answer» Bro questions sahi sa likho kya CB. QR CA kya hai parallelogram hai ya aur kuch hai | |
| 8406. |
Table of 9 |
| Answer» 9×1=99×2=189×3=279×4=369×5=45 9×6=549×7=639×8=729×9=819×10=90 | |
| 8407. |
If sin thita -cos thita =0,find thita |
| Answer» 45° | |
| 8408. |
A tangent to circle intersects it in. Point |
| Answer» No point interested | |
| 8409. |
show that there is no value of n fir which (2n×3n)ends in 5. |
| Answer» It is terminationg | |
| 8410. |
A well whose diameter is 7m has been dug 22.5 m deep |
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Answer» Write full question here. What to find here? What to find here??? |
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| 8411. |
798789×987789987 |
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Answer» 798789×98778987=789035775925743 ? 789,035,775,925,743 |
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| 8412. |
Perimeter of a square EFHI where E(-2,0) F(3,0) H(3,5) I(-2,5) |
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Answer» using distance formula to find the sides of squareEF=√{3-(-2)}^2+(0-0)^2EF=√(5)^2+0EF=√25EF=5we know that sides of a square are equal then th distance of sides EF=5 FG=5 GH=5 HE=5PERIMETER= sum of all sides5+5+5+5=20ans Sin A = 3/4 =perpendicular /base P=3 H=4 We can find tha base by pythagoras theorem in triangle ABC BA²=AC²-BC² BA²=4²-3² BA²=16-9 BA²=7 BA=√7 SecA = hypotenuse /base SecA=4/√7 First we find distance between EF, FH, HI and IE respectively after than add the distance all |
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| 8413. |
(a2-b3) |
| Answer» (a2-b3)=(a²‐b²)ka formula | |
| 8414. |
a(x-a) +b(x-b) /(a-b) (a+b)=2 |
| Answer» | |
| 8415. |
Prove that root5+root3 is irrational |
| Answer» Let root5+root3be a rational number which can be written in the form of p upon q and q is not equal to 0root5+root3=p\\qroot3=p\\q-root5root3=p-root5\\q So that p-root5 is rational number Where root3 is irrational numberSo our contradict assumptions is wrong Thus\'root 5+root 3 is irrational numberProved | |
| 8416. |
Proofs in mathematics |
| Answer» A mathematical proof is an inferential argument for a mathematical statement, showing that the stated assumptions logically guarantee the conclusion. | |
| 8417. |
If sinA=3/4 then find the value of secA |
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Answer» Sec=4 by root 7 Sin A = 3/4 =perpendicular /base P=3 H=4 We can find tha base by pythagoras theorem in triangle ABC BA²=AC²-BC² BA²=4²-3² BA²=16-9 BA²=7 BA=√7 SecA = hypotenuse /base SecA=4/√7 Diya soni you copied my answer Sin A = 3/4 =perpendicular /base P=3 H=4 We can find tha base by pythagoras theorem in triangle ABC BA²=AC²-BC² BA²=4²-3² BA²=16-9 BA²=7 BA=√7 SecA = hypotenuse /base SecA=4/√7 Sin A = 3/4 =perpendicular /baseP=3H=4We can find tha base by pythagoras theorem in triangle ABCBA²=AC²-BC²BA²=4²-3²BA²=16-9BA²=7BA=√7SecA = hypotenuse /base SecA=4/√7 |
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| 8418. |
a*a+b*b |
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Answer» if the question is a^2+b^2= (a^2+b^2)-2ab A square + b square??what do u want to ask?? What is this ? |
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| 8419. |
Tan1 tan2 tan3 ...tan89 |
| Answer» Ans is 1 | |
| 8420. |
Express 17/25 in decimal form without actual division |
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Answer» Answer is 0.68 17×2×2/5×5×2×2=6810×10=10068/100=0.68 ans 17×2×2/5×5×2×2=68/10×10=68/100=0.68 ANS |
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| 8421. |
Find the area of shaded region pq=24cm pr =7cm and o is the centre of circle |
| Answer» | |
| 8422. |
Find the area of the segment AYB ,if radius of the circle is 28cm and angle AOB is 90° |
| Answer» Area of segment is 224 cm^2 | |
| 8423. |
Prove cosA-sinA+1/cosA+sinA-1=cosecA+cotA |
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Answer» LHS=cosA+sinA−1cosA−sinA+1\u200bdividing Nr and Dr by\xa0sinA\xa0we get,=sinAcosA\u200b+sinAsinA\u200b−sinA1\u200bsinAcosA\u200b−sinAsinA\u200b+sinA1\u200b\u200b=cotA+1−cosecAcotA−1+cosecA\u200b=cotA+1−cosecAcotA+cosecA−(cosec2A−cot2A)\u200b=cotA+1−cosecA(cotA+cosecA)(1−cosecA+cotA)\u200b=cotA+cosecA=RHS Multiplying the numerator and denominator with sinA :SinA [CosA – SinA +1]/SinA [CosA + SinA –1]SinACosA–Sin^2A+SinA/SinA [CosA +SinA –1]SinACosA+SinA –[1–Cos^2A]/SinA[CosA+SinA–1]SinACosA+SinA–[(1–CosA)(1+CosA)] /SinA[CosA+SinA–1]SinA[CosA+1]–[(1–CosA)(1+CosA)] /SinA [CosA + SinA–1](CosA+1)(SinA–1+CosA)/SinA (CosA + SinA–1)(CosA+1)(SinA+CosA—1)/SinA (CosA+SinA–1)(CosA+SinA–1) will get cancelledSo , CosA+1/SinA is leftDividing (CosA) and (1) individually by (SinA), we get CotA + CosecA Which is equal to R.H.SHence Proved |
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| 8424. |
Case study of coordinator geometry |
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Answer» Where is question Send me the questions |
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| 8425. |
Is ncert book enough for math standard board 2021? |
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Answer» No There\'s no need to refer 100s of books just make ur concepts clear Yes but make your concept clear Try rs aggrawal or rd Sharma book No |
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| 8426. |
A number x is choosen from the number -3,-2,-1,0,1,2. Find the probability that 1*1 |
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Answer» Probably is 1 Because total no. of favourable outcomes/total no. Of outcomes 5/6 The probability is |
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| 8427. |
Find the largest number that divides 31 and 99 leaving remainder 5 and 8 respectively |
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Answer» Answer is 2?? Now, we use Euclid\'s division algorithm/lemma 31-5 and 99-5, we get 26 and 94 First we leave remainder 5 8 and 5 |
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| 8428. |
Find a,b,c if it is given that the numbers a,7,b,23,c are in ap |
| Answer» (a+d)=7 and (a+3d)=23 by subtracting equations we get d=8. putting this value in a+d=7 we get a=-1. b=15 and c=31 | |
| 8429. |
Mam please buy a bio |
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Answer» ?????? ?????? Sir I can ask doubt |
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| 8430. |
Is ncert maths sufficient for board 2021?? |
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Answer» If you have taken basic then practice NCERT 100 % and any other side book you wish . But you have taken standard the you should practice NCERT full as well as side book to , like, examplar. Ya ncert prefer Yes If you have taken basic then practice NCERT 100 % and any other side book you wish . But you have taken standard the you should practice NCERT full as well as side book to , like, examplar. |
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| 8431. |
Find the 21st term of the A.P: -4,-3,-1,... |
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Answer» The 21st term of the A.P:16 16 answer is tn=16 The terms are not AP. Please recheck your question..... It is not an A.P. Please verify the question. |
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| 8432. |
Solve the following equation 2 X square + X - 4 is equal to zero |
| Answer» Here we have to use quadratic formula.» D=(b)^2-4ac.» D=(1)^2-4(2)(-4).» D= 1+32.» D= 33.Now using quadratic formula:» x=(- b±√D)/2a.» x=(-1±√33)/4.» x=(-1+33)/4,(-1-33)/4.» x=32/4 & -34/4.» x=8 and -17/2.?????????? | |
| 8433. |
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts. |
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Answer» Easy ? Easy question ? |
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| 8434. |
Suggest some ways improve marks in mathematics |
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Answer» Do Practice... Practice practice practice.... Daily practice help you to get mark.u should be through in ncert book. Refer some guide.daily atleast spend an hour to maths subject it\'s enough to get good mark. 1. Practise practise and practice2. Ncert should be like God3. Sample papers..anyone in perfect 3 he sittings |
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| 8435. |
if √3sinθ-cosθ=0, and 0 |
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Answer» Correct answer is 60° We would get tan thitta is 1/√3. Then, the angle would be 30°. 60° |
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| 8436. |
If (x^2+y^2)(a^2+b^2)=(ax+by)^2. Prove that x/a+y/b. |
| Answer» Check the question | |
| 8437. |
x^2 + 7x - 1 |
| Answer» b^2-4ac 7×7-4×1×1=45Formulae is ( -b plus or minus root of (b^2-4ac))/2×aSubstitute value in the formulaeThe answer will be (-7+_root of 45)/2 | |
| 8438. |
6x^2 - x + 12 solve the quadratic equations |
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Answer» Ya yogeshwaran bro your answer is correct It does not contain real roots ie apply b^2-4ac it will less than 0 |
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| 8439. |
Which are the theorems in mathematics comes in 2021 board |
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Answer» bpt BPT,converse of BPT and Pythagoras Manve karna ki Chamak Divya Chamak Ke a Saransh Bpt theorem |
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| 8440. |
Find the value of m so that (3,2) is a solution of the equation y=mx-5. |
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Answer» 2= m(3)-5......... 7=m(3).......... m=7/3 Thanks✨✨ 2=m(3) -57=3mm=7/3 |
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| 8441. |
Write an equation of the line parallel to the given line 3x+4y=14. |
| Answer» 6x+8y=59. | |
| 8442. |
Case stady |
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Answer» Is there any figure Write the complete question |
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| 8443. |
simplify: (1-sin²θ) -cos²θ |
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Answer» 0 (1_SIN^2)_Cos^2Cos^2-cos^2=0 0 Cos²@ - cos²@ = 0 |
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| 8444. |
In an AP. a=8, an=62, sn=210 find n and d |
| Answer» an=a+(n-1)d62=8+(n_1)d54=nd-d (1)Sn=n/2 (2a+(n-1)d)210=n/2 (2×8+(n-1)d)420=n (16+nd-d )420=16n+n (nd-d )420=16n+n (54) (from 1)420=16n+54n420=70n420/70=nHence n=6 Now put the value of n in equation 1,we get54=nd-d 54=6d-d54/5=d | |
| 8445. |
If 2sin^2theta-cos^2theta=2,then find the value if theta. |
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Answer» Thita = 90° 2sin^2theta-cos^2theta=2Theta=90°. |
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| 8446. |
If the perimeter of semicircular protractor is 36cm,find its diameter. |
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Answer» 11.45 is wrong answer ? 2r+πr=36......... r(2+π)=36......... r{2+22/7}=36........ r(36/7) =36 ...... r=7 ...... Diameter = 2(7) =14cm 11.45 |
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| 8447. |
Taking theta=30°,verify cos3theta=4cos(cube)theta-3cos theta |
| Answer» | |
| 8448. |
What is meant by Geometric Mean and Arithmetic Mean m |
| Answer» | |
| 8449. |
Divide: (x3-3x2+5x-3) by (x2-2) |
| Answer» x^2-2)x^3-3x^2+5x-3(x-3 (-)x^3-2x ―――――――― -3x^2+7x-3 (-)-3x^2+6 ―――――――― 7x-9 Since, degree(r) is less than degree(q) It will not be further divided.For complete division subtract 7x-9 from p(x)....... | |
| 8450. |
Give me Chapter 13 formules |
| Answer» | |