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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. |
What is the deleted portion for term 1 exams |
| Answer» ? | |
| 2. |
Lcm is 1600 , hcf is 50 it is possible to have |
| Answer» Lcm is 2^6 5^2 = 1600Hcf is 5×5× 2 = 50 | |
| 3. |
Find the smallest number which when divided by 28 and 32 leave remainders 8 and 12 respectively |
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Answer» Who are u ?utkarsh rana Have u got your?? Sorry suhani your whatsapp no. 5 and 4 |
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| 4. |
If Sin tetha + 2Cos tetha = 1 , prove 2Sin tetha - Cos tetha = 2 |
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Answer» Cos theta is 90 degree Hlo Hii Hii I am her brother aapke ghar khaa hai Hii I am boy aap konsi class me ho |
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| 5. |
Find the coordinates of the points of trisection of the line segment joining (4,-1)and (-2,-3) |
| Answer» | |
| 6. |
Exercise 4.3 full |
| Answer» Hlo | |
| 7. |
Case study ? ch 1 real number |
| Answer» Go to see more days | |
| 8. |
ha seervi jazz pakhi pehli pitch bare Paoli pehli |
| Answer» | |
| 9. |
(1+tan theta+sec theta)(1+cot theta-cosec theta)=? |
| Answer» | |
| 10. |
Hcfof 9and 25 is ? What is lcm? |
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Answer» HCF =1 and LCM =225 2 |
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| 11. |
If sinA+sin^2A=1, then cos^A+cos^4A= |
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Answer» ?????????????????????????????????????????????????????????????????????????????????????????????????????????????? Hii |
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| 12. |
Two poles of hight 8m and 13m are standing in 12m apart the distance their tops is |
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| 13. |
(CosecA-sinA)(secA-cosA)=1/tanA-cosA |
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Answer» Hence Proved Check the ncert always Refer to NCERT |
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| 14. |
If the product of the zeroes of the polynomial ax²_ |
| Answer» Please write the question full-fledged manner. | |
| 15. |
If the LCM and hcf of two numbers is 1260 and their LCM is 900 more than their hcf |
| Answer» Let the HCF be xLCM=HCF+900LCM=x+900 --(1)And\xa0LCM+HCF=1260LCM+x=1260 --(2) From (1) and (2)(x+900)+x=12602x+900=12602x=1260−9002x=360x=2360\u200bx=180HCF\xa0=180And, from (1),LCM=HCF+900LCM=180+900LCM=1080By formula, product of the numbers is equal to the product of their HCF and LCM.Product of numbers\xa0=HCF×LCM =180×1080Product\xa0=194400 | |
| 16. |
How to identify polynomials |
| Answer» Polynomials //Linear polynomial – of degree one.Quadratic Polynomial- of degree two.Cubic Polynomial – of degree three.Degree of polynomial is never negative nor fractional also variable should not be in denominator or in root | |
| 17. |
Find the point on the x-axis which is equidistant from (2,-5)and (-2,9) |
| Answer» 2,-2 | |
| 18. |
lcm for 24, 60,150 |
| Answer» 600 | |
| 19. |
Chapter 8 trigonometry Hindi medium all formulas |
| Answer» https://www.examtricksadda.com/2017/02/trigonometry-formula-short-tricks-with-example-hindi.html?m=1//Visit through this link | |
| 20. |
All formulas of chapter 4 |
| Answer» | |
| 21. |
If cot A=15/8 find sinA, cosA, cosecA |
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Answer» cotA=b/pGiven,b/p=15/8By applying Pythagoras theorem we get,P^2+b^2=h^28^2+15^2=h^254+225=h^2h=√64+225therefore,H=17SinA=p/h =8/17CosA=b/h=15/17 and,CosecA=h/b=17/15Hope you will like this answer Given that:- cot A=15/8And we know that:- cot A=b/pSo that, base is 15 & perpendicular is 8 and by the Pythagoras theorem we find 17 as a hypotenuse.Hence,•sin A=p/h=8/17•cos A=b/h=15/17•cosec A=h/p=17/8Hope you understand this very simple way. Cot A=b/pSin A=p/hCosA=b/hCosecA=h/pGiven Cot A=15/8b=15 p=8We find h from Pythagoras theorm h²=b²+p²h=17So,SinA=8/17,Cos=15/17 and Cosec=17/8 |
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| 22. |
Probability of 52 mondays in a leap year |
| Answer» 1/7 | |
| 23. |
Find tan p – cot R |
| Answer» Can you please write the full question | |
| 24. |
What is linr |
| Answer» What is line segnment | |
| 25. |
2×2² |
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Answer» 6 2×2^22×(2×2)2×48Answer : 8 8 8 8 |
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| 26. |
Why( 33×33+33-33÷0 ) shows the error in calculator......... Pls check and just inform me |
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Answer» 0 Any number that is divided by 0 the answer is not defined.. |
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| 27. |
Exercise 3.3 2 |
| Answer» x = -2y = 5m = -1 | |
| 28. |
Solve this equation by cross multiplication method . X/a + y/ b = a+b , x/a2 + y/ b2 =2 |
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| 29. |
What is the triangle |
| Answer» Triangle is a closed figure, having 3 edges, 3 vertices and 3 sides. | |
| 30. |
Can you find the HCF of 1.2 and 0.12 by using euclid division algorithm? Justify your answer |
| Answer» The HCF of 1.2 and 0.12 is\xa00.12We will convert the given decimals into like decimals in which the number of digits after the decimal point are the same. So, now our numbers are 120 and 12. Now, we will find the HCF of 120 and 12. Clearly, the Highest Common Factor (HCF) of 120 and 12 is 12. | |
| 31. |
tan² A – 5 sec² A + 1 is equal to(a) 6(6) -5(c) 1(d) -4 |
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Answer» -4 -4/cos2 Answer d ,(-4) |
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| 32. |
Find the centre of the circle passing through (5 ,-8),(2, -9)and (2, 1) |
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| 33. |
Solve x and y hence find a for which y= ax-4 2/x + 2/3y = 1/6 3/x + 2/y =0 |
| Answer» | |
| 34. |
Ex6.6 12 |
| Answer» Wo optional hai nahi aayega exam me | |
| 35. |
prove that the diagonals of a trapezium intersect each other in the same ratio |
| Answer» | |
| 36. |
PROVETHAT: COSA÷1+SINA+1+SINA÷COSA=2SECA |
| Answer» | |
| 37. |
Solve the quadratic equation by factorisation : 1) 1/ x - 1/x-2 =3 |
| Answer» Nhaush | |
| 38. |
4x-3y=86x-y=29/3By substituting method |
| Answer» 4x-3y=8 ...eq(1)6x-y=29/3 ....e(q2)From eq 1 4x-3y=8X=8+3y/4Putting the value of x in eq 2 we get;6 | |
| 39. |
The hcf of 4095 and 378 |
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Answer» 63 63 The HCF of 4095 and 378 is 63. |
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| 40. |
The hcf of 4095 and 378 is |
| Answer» HCF of 4095 and 378 is 63? | |
| 41. |
Two quadrants are inscribed under square . Find the area of shaded portion. |
| Answer» Area of square - area of semi circle | |
| 42. |
Ch 12 example 6 by another method |
| Answer» Pls give me and quickly | |
| 43. |
1/4,1/3 |
| Answer» | |
| 44. |
Find the centre of a circle passing through 5, - 8 2,-9 2,1 |
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| 45. |
If product of the zeroes of the polynomial kx sqaure- 11x+30 is 2, then the value of k is |
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Answer» 15 K 2 |
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| 46. |
show that any integer is if the form 3q 3q+1 |
| Answer» Let any integer be nBy Euclid geography lemmaP=n.q+r Let remainder be 0,1,2,3When remainder be 0N=3q +0=3qWhen r=1N=3q+1 | |
| 47. |
If tanA =4/3 then find (SinA)(cosA)/(cosec A)(secA) |
| Answer» 1 | |
| 48. |
In∆ODC SIMILAR ∆OBA,ANGLE BOC=125° and |
| Answer» | |
| 49. |
The zeroes of the quadratic polynonial x^2+99x+127 |
| Answer» | |
| 50. |
The zeros of the polynomial p(x)x^2-7x-30 are 3,10 |
| Answer» | |