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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.
| 42451. |
If the distance between the point (4,p) and (1,0) is 5 then the value of P is. |
| Answer» Therefore we can write the formula as follows:By squaring on both sides we get:Therefore, the\xa0value of p is +4 or -4 | |
| 42452. |
anybody PLZZ tell me how to check which questions are wrong in test |
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Answer» I have Sumita Arora ka book chahie computer |
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| 42453. |
Case Study Questions Class 10th Exam 2020-21 solutions with explaination |
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Answer» I have Yes. I also need it. |
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| 42454. |
Find the roots of the equation x2+3x-(m+2)(m+5)=0 where m is constant |
| Answer» x2 +3x - (m+2)(m+5) = 0⇒ x2 + (m+5)x - (m+2)x - (m+2)(m+5) = 0⇒ x (x + m+5) - (m+2)(x+m+5) = 0⇒ (x+m+5)(x-m-2) = 0⇒ x = - m-5. and m+2Roots are (-m-5) and (m+2) | |
| 42455. |
1.2ka3 |
| Answer» 3.6 | |
| 42456. |
Question no-1 of excercise-6.5 |
| Answer» 1. Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.(i) 7 cm, 24 cm, 25 cm(ii) 3 cm, 8 cm, 6 cm(iii) 50 cm, 80 cm, 100 cm(iv) 13 cm, 12 cm, 5 cmSolution:(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.Squaring the lengths of the sides of the, we will get 49, 576, and 625.49 + 576 = 625(7)2\xa0+ (24)2\xa0= (25)2Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.Length of Hypotenuse = 25 cm(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.Squaring the lengths of these sides, we will get 9, 64, and 36.Clearly, 9 + 36 ≠ 64Or, 32\xa0+ 62\xa0≠ 82Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.Hence, the given triangle does not satisfies Pythagoras theorem.(iii)\xa0Given, sides of triangle’s are 50 cm, 80 cm, and 100 cm.Squaring the lengths of these sides, we will get 2500, 6400, and 10000.However, 2500 + 6400 ≠ 10000Or, 502\xa0+ 802\xa0≠ 1002As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.Therefore, the given triangle does not satisfies Pythagoras theorem.Hence, it is not a right triangle.(iv) Given, sides are 13 cm, 12 cm, and 5 cm.Squaring the lengths of these sides, we will get 169, 144, and 25.Thus, 144 +25 = 169Or, 122\xa0+ 52\xa0= 132The sides of the given triangle are satisfying Pythagoras theorem.Therefore, it is a right triangle.Hence, length of the hypotenuse of this triangle is 13 cm.\xa0 | |
| 42457. |
Pair of linear equations exercise 3.2 |
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Answer» pura exercise hi????? Chal nikal furst mai |
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| 42458. |
Riya choudhary In which chapter board may ask more questions |
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Answer» Algebra in which chapter Thanks ಇನ್ ರಿಯ ಚೌಧರಿ ಚಾಪ್ಟರ್ ಬೋರ್ಡ್ ಮೇ ಆಸ್ಕ್ ಮೋರ್ ಕ್ಯೂಸ್ಷನ್ಸ್???? Unit-wise wieghtage to be followed in the CBSE Class 10 Maths Exam 2021 is as follows:\tUnitMarksI. NUMBER SYSTEMS06II. ALGEBRA20III. COORDINATE GEOMETRY06IV. GEOMETRY15V. TRIGONOMETRY12VI. MENSURATION10VII. STATISTICS & PROBABILITY11Total80\t |
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| 42459. |
X+1/X 3. 6x |
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| 42460. |
(7x)^2+(8x)+9 |
| Answer» | |
| 42461. |
Will graph questions come in board exams??? Or is it deleted?! |
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Answer» yes (hai graphs hai co-ordinate ka bhi aur quadratic equation ka bhi bs co ordinate me area of triangle na puchega) HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.The CBSE syllabus has been rationalized keeping intact the learning outcomes so that the core concepts of students can be retained.Deleted syllabus of CBSE Class 10 Mathematics\xa0\xa0D I think it is there as it is teachen and done to us.. |
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| 42462. |
How many solutions can we get by solving these equations ?3x+2y=5,2x-3y=7 |
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Answer» You will get an unique solution The two equations are3x\xa0+ 2y\xa0= 5 …… (1)2x\xa0− 3y\xa0= 7 …… (2)Here,Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. That is, there is only one solution. |
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| 42463. |
How to solve case study based question |
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Answer» Be practice By practice |
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| 42464. |
For what value of \'a\'Quadratic equation 30q- 6x+1= 0 has no equal roots? |
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| 42465. |
How can we get perfection in trigonometry because its like puzzle? |
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Answer» You can get perfection in trigonometry by only practising more and more Do more and more practice of Trigonometry\'s questions and in case of having any doubt, ask your teachers.I wish you get what you want. |
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| 42466. |
What is frustum ? |
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Answer» In geometry, a frustum is the portion of a solid that lies between one or two parallel planes cutting it. A right frustum is a parallel truncation of a right pyramid or right cone. In computer graphics, the viewing frustum is the three-dimensional region which is visible on the screen. The portion of a cone or pyramid which remains after its upper part has been cut off by a plane parallel to its base or which is intercepted between two such planes, known as frustum. |
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| 42467. |
Displacement and rotation of geometrical figure |
| Answer» Displacement is defined as the moving something from its position. Rotation is defined as the action of rotating around an axis or center. A geometrical figure can be displaced or rotated in either order to produce the same end result.\xa0There are some general rules for the rotation of objects using the most common degree measures (90\xa0degrees, 180\xa0degrees, and 270\xa0degrees). The general rule for rotation of an object 90\xa0degrees\xa0is (x, y) --------> (-y, x). | |
| 42468. |
If HCF of (26,91)is 13 then LCM of(26,91) |
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Answer» The product of LCM and HCF of two numbers is\xa0equal\xa0to the product of two numbers. \xa0LCM = product of the numbers\xa0/\xa0HCF LCM = 26×91 /13LCM = 2366 / 13LCM = 182
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| 42469. |
Which digit would occupy the units place of 6 to the power of 100 |
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| 42470. |
1/(x+4)-1/(x-7)=11/30 |
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Answer» 1 and 2 x = 1, 2 |
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| 42471. |
Solve : 3(n-1)=21 |
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Answer» 3(n-1) =21n-1=21/3n-1= 7n= 7+1=8 8 3(n-1)=21\xa03n - 3 = 213n = 21 + 33n = 24n = 24/3 = 8 Ans=8 3n-3=21 3n= 21+33n= 24n= 24/3n= 8 |
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| 42472. |
Anyone pls send the sample paper maths class 10 2020-21 standard paper -3 pls |
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Answer» You can download it from this app also of any subject I am having exam tmrw .So that only why |
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| 42473. |
How to solve case based questions tricks |
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Answer» Read carefully and understand the situation and answer it carefully I think he is right Case study based questions are very easy .You just have to read the given information very carefully and what is asking in the question.I hope this will help you☺ |
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| 42474. |
Case study question |
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Answer» Which one Case study answer |
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| 42475. |
4x^2+5x-2√3=0 then,find roots |
| Answer» Question galat hai recheck it | |
| 42476. |
If the equation x2-bx+1=0does not posses real roots ।then |
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| 42477. |
From where I get case study questions on this app?? |
| Answer» For case study you can refer byjus or toppr or extra marks online(I don\'t promote any particular classes just for your help) | |
| 42478. |
Subbha rao started |
| Answer» an=7000 | |
| 42479. |
To prove:- volume of frustum of cone is πh/3(R²+r²+Rr) |
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Answer» Frustum se related koi questions nahi aane wale he board me Bro, volume of frustum of cone paper me nahi ae ga |
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| 42480. |
. If sin? and sec? (0 < ? < 90) are the roots of the equation √3 x2+kx+3 =0, then find thevalue of k |
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| 42481. |
find the zeroes of x square minus root 5 |
| Answer» There might be some error could you please share the photograph of question so that I could help you | |
| 42482. |
Cos²a + sin²a + 1/tan²a = ? |
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Answer» Cosec²a Cot²a Sin 2 A |
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| 42483. |
Sin theta-cos theta +1/sin theta + cos theta -1=1/sec theta -tan theta |
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| 42484. |
In myCBSEguide from where can we get case study based questions if we want to practice them |
| Answer» U can go to the cbse sample paper basic or standard ( 2020 -21 ) in this app only | |
| 42485. |
Find HCF of 306 and 657. And also find their LCM using their HCF |
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Answer» Find HCF and LCM of 306 and 657 AnswerUse Euclid s division algorithm lemma to find HCF a= bq+ r where 0<= r < b 657 and 306657= 306 * 2+ 45 306=45*6+945=36*1+936=9*4+0 Therefore HCF = 9 Product of prime = Product of LCM and HCF 306*657= LCM* HCF306*657= LCM *9 306*657/9=LCM34*957= LCM22,338 = LCM. Therefore LCM= 22 ,338 and HCF=9 The given numbers are 306 and 657.The factor form of given numbers areIt is clear that 3 and 3 are common factors. So, the HCF of both numbers isIf a and b are two numbers thenTherefore the LCM of both numbers is 22338. |
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| 42486. |
Find two consecutive positive integer sum of whose is 365 |
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Answer» The two numbers are x=13 and x=14 Hyee Hello |
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| 42487. |
Maths project on class 10 working model |
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Answer» Can you please tell me where I found the project . We too have the same which states topic Yes Cbse project for class tenth mathematics is it |
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| 42488. |
Q5. In ∆ABC and ∆DEF, AB/DE = BC/EF, then they will be similar when * |
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Answer» Angle ABC= Angle DEF OR AB/DE=BC/EF=AC/DF Two triangles are similar when there corresponding angles are equal and their corresponding sides are proportional.So, angle B=angleD |
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| 42489. |
What is full Form of BODMAS Formula |
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Answer» Bracket Order of Division Multiplication Addition Subtraction B- bracket,O- order of ,D-division ,M-multiplication,A-addition,S-substration a n s w e rAs per BODMAS rule, we have to calculate the expressions given in the brackets first. The full form of BODMAS is Brackets, Orders, Division, Multiplication, Addition and Subtraction BODMAS is an acronym and it stands for Bracket, Order, Division, Multiplication, Addition, and Subtraction. Bracket Of Division Multiplication Addition and Subtraction |
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| 42490. |
If lateral surface area of cylinder is94.2cm2 and height is 5cm then find radius of its base |
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Answer» Ok Given,\xa0h=5cm⇒Let radius\xa0=r(i) Curved Surface Area\xa0⇒94.2cm2=2πrh⇒94.2\xa0cm2=2×3.14×r×5cm⇒r=3cm(ii) Volume of cylinder\xa0⇒πr2h=3.14×3×3×5=141.2cm3? |
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| 42491. |
Find the value of tan15o.tan25o.tan30otan65o.tan85o |
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Answer» 1 1 |
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| 42492. |
નીચે આપેલા પ્રશ્નો ના માગ્યા મુજબ ઉતાર લાખો |
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Answer» Which language is this? ???? |
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| 42493. |
If the alpha and bita are the zeroes of polynomial f(x)=ax^2-px-q=0. Evaluate the value alpha-bita |
| Answer» -q/a | |
| 42494. |
Explain the figure |
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| 42495. |
NCERT se kaun sa exercise nahi ayega |
| Answer» | |
| 42496. |
Quadratic equation NCERT me Jain as exercise bhi ayega board me |
| Answer» | |
| 42497. |
If one zero of the quadratic polynomial x²-5x-6 is 6, then find the other zero. |
| Answer» p(x) =\xa0x²-5x-6-b/a = 5c/a = -6p(6) = 0Let the other zero be "a"sum of zeros = 6 + a = -b/a\xa06 + a = 5a = 5 - 6 = -1The other zero is -1 | |
| 42498. |
NCERT math se kaun kaun sa exercise nahi ayega |
| Answer» Sab exercise aayega | |
| 42499. |
भाई जिसके पास सब्सक्रिप्शन हो गया है लॉक वाला जितना भी स्टडि मटेरियल है मुझ में सेंड कर दो प्लीज |
| Answer» | |
| 42500. |
V=375 cm³,l=20 cm,b=7.5 cm h=? |
| Answer» V=375 cm³,l=20 cm,b=7.5 cm h=?V = lbh20 (7.5) h = 375150 h = 375h = 375/150 = 2.5 cm\xa0 | |