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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. |
-95, at a: d=?41 a: b = 3:5, c: b(a) 12:36(c) 15:36. e36= 3:2, c: d=5: 6(b) 12:15(d) 11:36 |
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Answer» a)12:36 is correct answer 15:36 is the right answer option a is the right answer a is right answer for your question option A is 12:36 is correct answer 12:36 is the correct answer. |
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| 2. |
number of students in the class.S. In aâłABC,KC = 3 2 B+ 2 (ZA +2 B). Find the three aralgebraitetiong 55 and 3x-y 3. Determi |
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Answer» C = 2(A+B) => A+B=C/2 now its triangle so A + B + C = 180 so C + C/2 = 180 3C/2 = 180 so C=120 so A+B=60 now C=3B so B=40 so A=20 |
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| 3. |
Example 4 After givingfor Rs. 560. Determidiscount of 20% on mark up price, a trouser was soldne mark up price of trouser |
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| 4. |
.95. How many rupees are needed to buy 8 suchThe cost of 1 toy istoys ? |
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Answer» Cost of one toy = Rs 95 Then,Cost of 8 toys = 8*95= Rs 760 Therefore,Rs 760 needed to buy 8 such toys Cost of 1 rupee= 95 rupeesso for 8 toysmoney needed= 95*8= 760 1 toy = 95 rupees8 toys = x X = 95x8 -------- 1x = 95 x 8 = 760 cp=RS95RS need to buy 8 such toy=95×8=Rs 760 is the answer |
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| 5. |
(viii) y-318で攷司!Find the co-ordinates of a point on y-axis which are at a distance of5V2 from the point P (3,-2, 5)] |
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| 6. |
Two mobiles cost to 16000 rupees. How mush money will be required to buy 13 such mobiles? |
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Answer» cost of two mobiles is ₹16,000 Cost of one mobile is ₹ 16000/2= ₹8000 Total money requried for 13 mobiles is = ₹ 8000 × 13 = ₹ 104,000 |
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| 7. |
ich term of the AP 8, 14, 20, 26,... will be 72 moré thaf its 'tist c50.Determiterm is 20.ne the AP whose 5th term is 19 and the difference of the 8th term from the 13thNCERT Exemplor |
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| 8. |
(c) (6x + 5) (x 1)Very Short Answer Type Question4. Determine which of the followin(iii) x3+3x2 3x + 1Use factor theorem to determi(i) p(x)=x4-2x2 + 1, g(x) = :5. |
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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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| 9. |
COSCOSCOSCOS15 15 15 15. LHS 16c0s?4 s 4 |
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| 10. |
cos π/15.cos2π/15.cos4π/15.cos7π/15=1/16 |
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| 11. |
D las fadius 1.75 m, find the capacity of the tank ines have their heights in the ratio 1:3 and the radi of their bases in the ratio 3 1, findrato of their volumes.the length of the cloth used in conical pandal of haiol |
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Answer» Ratio of Volume of cones = [(1/3)*π*r*r*h]/ [(1/3)π*R*R*H]= (r*r*h)/(R*R*H)= (3*3*1)/(1*1*3)= 3/1. |
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| 12. |
4 cm3 cm3 cm-3 cm3 cm4 cm |
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| 13. |
DABCD is a square with side 10 cm. Find the area of the shadedA 3 cm3 cm B32 cmD 3 cm3 cm C10 cm |
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Answer» find the area of a square of side 335 cm |
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| 14. |
Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smallertriangle is 12 cm, then find corresponding side of the bigger triangle5. |
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| 15. |
-1000 %2B 20010=square |
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Answer» 19010 is correct answer of following question 20010 - 1000 = 19010 19010 is answer of your sum. 20010 - 1000 = 19010 20010-1000=19010 is the right answer of the followingso, accept me best. 20010-1000=19010 is the answer 19010 is correct answer 19010 is the right answer. |
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| 16. |
11. Find the value of (cosec2 θ-1) tan2 θ.[CBSE 2015)] |
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Answer» cosec^2-1=cot^2AHence cot^2A*tan^2A=1 as cotA=1/tanA |
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| 17. |
11. For what value of a the point (a, 1), (1, -1) and (11, 4) are collinear? [CBSE 20171 |
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| 18. |
Q.2 Solve the following question.(any 5)1) If 600 rupeesbuy 15 bunches of feed, how many will 1280 rupees buy? |
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Answer» 600 rupees buy 15 bunches of feed so 1280 rupees will buy 1280*15/600 = 32 bunches of feed. If you find this answer helpful then like it. |
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| 19. |
Question 5fh term of an arithmetic progression are 10 and 96 respectively.Find the first term and the common difference and hence find the sum of the first15 terms. |
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Answer» let first term be a and common Difference be dSo, According to questiona(2) = 10a(45) = 96 a + d = 10 ......(1)a + 44d = 96 .....(2)eq(2) - eq (1) 43d = 86d = 2So, from (1) a = 8Sum of First 15 term ,S(n) = n(2a+(n-1)d)/2S(15) = 15/2 ( 16 + 14×2)S(15) = 15/2 (16+ 28) = 15/2 × 44 = 330 |
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| 20. |
5. I20, 25, 15 andr are in proportion, find the value of x.(1) 14 (2) 18 (3) 21 (4) 24her urill renlace the question |
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Answer» 20*x= 25*15x=25*15/20x=18.75 |
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| 21. |
35. Which number should come in place of question mark (?)9 142 318 25 -2?-3(A) 18(B) 13(C) 30(D)-3015) |
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Answer» 7×3×2=429×2×1=185×-3×-2=30 |
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| 22. |
चŕĽ) sin Acos C +cosAsinC= = |
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Answer» sinAcosC + cosAsinC = sin(A+C) If you find this answer helpful then like it. |
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| 23. |
tIf a/b=7/3\frac{5 a+3 b}{5 a-3 b}-\frac{7 a+9 b}{7 a-9 b} |
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Answer» thnks |
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| 24. |
QuestionS9Question id: 43320 (2 marks)Rakesh lent out Rs 8000 for 5 years at 15% per annum and borrowed Rs 6000 for 3 years at 12% per annum. How much did he gain orlose?(a) Rs 1630(b) Rs 3840(c) Rs 2640(d) Rs 2120 |
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| 25. |
1. यदि a: b = 7:9और b: c = 5:7,तो a: C क्या होगा।4a) 5:9(b) 3: 5(c) 7:21(d) 7: 15 |
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Answer» yes 5:9 will be the right one bro (A) is the correct answer A) is the right answer 35/63 or 5/9 is the right answer. A is the right answer |
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| 26. |
If a/b=7/3 then find the values of the following ratios.\frac{5 a+3 b}{5 a-3 b} \quad\left(\text { ii) } \frac{2 a^{2}+3 b^{2}}{2 a^{2}-3 b^{2}} \quad\left(\text { iii) } \frac{a^{3}-b^{3}}{b^{3}} \quad(\text { iv }) \quad \frac{7 a+9 b}{7 a-9 b}\right.\right. |
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| 27. |
IfA: B = 3:4, B: C = 5:7 and C: D= 8:9 then A: Dis 29equal to ?A1 A: B = 3': 4, B : C = 5:7 in C:D=8:9 **A: D f ara ETI?(2) 3:7 (6) 7:3 (0) 21:10 (8) 10:21Harsh is 40 years old and Ritu is 60 years old. How many |
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Answer» If A : B = 3 : 4, B : C = 5 : 7 and C : D = 8 : 9 then A : D is equal to[A]10 : 21[B]21 : 10[C]3 : 7[D]7 : 3 10 : 21 Hence option [A] is the right answer. A/D = A/B × B/C × C/D = 3/4 × 5/7 ×8/9 = 1/1 × 5/7 × 2/3 = 5×2/7×3 = 10/21 = 10:21Hence option D) 10:21 is the correct answer . 10:21 is the right answer. plz like my answer |
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| 28. |
The ratio of perimeter of two similar triangle is 3:7. Find the ratio of the area(A) 7:3(B) 3:7(C) 9:49(D) 49:9 |
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Answer» Answer: C)9:49Explanation |
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| 29. |
Q2. Solve the following and show your process: (in register)a. 2+(-3) +4d.5+9+ (-1)-(-2)b. (-5)-(-1)+2e. 9-(-1)-2 +4c.(-1)+7-3f.7+(-3)+(-7)-(-3) |
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Answer» b. (-5) - (-1) +2= -5 + 1 + 2= -5 + 3= -2 c .(-1)+7-3= -4+7= 3 |
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| 30. |
3. Write each of the following ratios in the simplest form(1) 3 m 5 cm : 35 cm270 mL (vi) 4 kg: 2 kg 500 g(1) 2 6.30: R 16.80(ii) 3 weeks : 30 days10 min ful 11 35 m |
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| 31. |
7.19VALUES OF TRIGONOMETRİC FUNCTIONS AT SUMS OR DFFERENCE OF ANGLESEXERCISE 7.1LEVEL-1A-4 and cos B-, where 0 < A, B. find the values of the following:13(ii) cos(A+B)(iii)sin (A-B)2. (a) lfsin A·2 and sin B-4 , where x </: π and0c B <) sin (A B)(iv) cos (A-B), find the followin(ii) cos (A+ B(1) sin (4 +B)t) sin A-1, where A and B both le n second quadrant,find the valueof sin (A t B)INCERT. Ifcors A。一2, and os 8-5, where π«A<3x and 2<b<2n, find the following:4. Iftan A-3, cos B-41, where π < A<3xand 0 < B <프,find tan (A + B).s lf ein A-1, or, B213 , where 2 < A < π and 3: < B < 2, find tan (A-B).6. lisin A-1, dos B--', where , <Acπ and 0 < B < π ' find the following :(ii) cos (A +(i) tan (A+B)(ii)tan (A-B)7. Evaluate the following(ii) cos 47 cos 13 - sin 47 sin 13(iv) cos 80 cos 20+ sin 80° sin 20(i) sin78° cos 18"-cos 78° sin 18"(iii) sin 36° cos 9° + cos 36" sin 9。and cot B 2, where A lies in the second quadrant and B in the thirdquadrant, find the values of the following(iii)(i) sin (A +B) (ii) cos (A + B)9. Prove that: cos 12 + cos 12-sin 12tan A -tan Btan (A+ B)10.Prove that: tan 4 + tan Bsin (A -B)11. Prove that:cos 11° + sin 11°cos 11°-sin 11cos 8。.. sin 80= tan 37。cos 9° +sin 9° = tan 54°cos 90-sin 9。(ii) so in so 3712. Prove that:(iii) sin/3x-5las/프+5.cos(3x-5|sinl2.5|-113. Prove that: tan 69+ tan |
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| 32. |
11. (b + c) cos A + (c + a) cos B + (a + b) cos C = a + b + c.c - a cos Bb- acos Csin Bsin C12. (a) sbcosA cosBab-ccOSA cos C |
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| 33. |
JA 5A2If cos A =-then the value of sinsiisvalue4ㅡㅡㅡ1S2(a) 1/32(c) 11/32(b) 11/8(d) 11/16 |
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Answer» the formula for sin(A/2)sin(5A/2) is 1/2*(cos(A/2-5A/2)-cos(A/2+5A/2)) so sin(A/2)sin(5A/2) = 1/2*(Cos(-2A)-cos(3A)) given cosA = 3/4 , so cos2A = 2cos²A-1 = 2*(3/4)²-1 = 1/8and cos3A = 4cos³A -3cosA = 4*(3/4)³-3*(3/4) = 27/16-9/4= -9/16 putting these values.. in the equation we get = 1/2*(1/8-(-9/16)) = 1/2*(11/16) = 11/32. option C |
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| 34. |
11. If a cos θ + b sin θ--c, then prove that:asin θ _ b cos θ+/a2 + b2-c2 |
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| 35. |
37. यदि tan a = cos11° + sin11°cos 11° - sin 11°१०, तो a का |मान होगा-(A) 17°(C)51°(B) 34°(D) 56° |
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Answer» . answer is b option correctly |
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| 36. |
Sc11./a) The principal value of-1cosb) Write tanthe simplest form. |
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| 37. |
u JU VJ 11sin (A + 3B) + sin (3A + B)_sin 2A + sin 2B(a) 2 cos (A + B) (b) 2 sin (A - B)(c) 2 sin (A + B) (d) 2 cos (A - B) |
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Answer» your answer is 2 cos( A+B) |
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| 38. |
\left. \begin{array} { l } { \operatorname { sin } A = \frac { 1 } { 11 } } \\ { \operatorname { cos } B = \frac { 12 } { 13 } } \end{array} \right. |
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| 39. |
al # 2 and an-an _ 1 + 4 for n 22. |
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Answer» a1 = 2 a2 = a1+4 = 2+4 = 6a3 = a2+4 = 6+4 = 10a4 = a3+4 = 10+4 = 14a5 = a4+4 = 14+4 = 18 |
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| 40. |
the previsous math factorrization question |
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Answer» if the ques is the 4ab one what i did is correct onl y please check the solution properly or else post the solution image in the same thread thnq hello? please feel free to ask query in the same thread i have seen the que you posted the printed one shows 26 and you are posting 2b according to the posted image it is 26 if that us 2b then you can cancel 2b in numerator and denominator |
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| 41. |
19. From theowing, calculate (a) Debt Equity Ratio (b) Total Assets to Debt Ratiocalc(c) Proprietary Ratio.Equity Share CapitalPreference Share CapitalGeneral ReserveAccumulated ProfitsDebenturesSundry CreditorsOutstanding ExpensesPreliminary Expenses to be written-offRs. 75,000Rs. 25,000Rs. 50,000Rs. 30,000Rs. 75.000Rs. 40,000Rs. 10,000Rs. 5,000(Ans: Debt Equity Ratio 3:7: Total Asets to Debt Ratio 4:1: Proprietary Ratio7:12) |
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| 42. |
hrol the tesm independent al22+ |
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| 43. |
3. Write the first threetermns of the following sequences whoxe n" terms are given bye first three terms of the following sequences whose nth terms are given bya n(-1) n (n+2)4(i)%"," (3-2)(ii) C ,m(_ 1 )" 3.+2(iii) Zn-e |
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| 44. |
s ‘._(i) Tt x o yz, y oc ab? G z o b/, T, O - दर शी¥ बिक |
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| 45. |
कण ey L qued य0 यणण g‘Wzrwa-/—'fir,mmm Tt Y .Ty 175 |
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| 46. |
proved that https://boardmodelpaper.com/nios-10th-model-paper-national-open-school-10th-sample-paper/amp/NIOS 10th Model Paper 2020 NIOS 10th Question Paper 2020 |
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| 47. |
board question paper 2018 class 9th math |
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Answer» Please visit our blog for question papers and study materials.https://blog.scholr.com/ |
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| 48. |
send a math question paper in chapter vise of class 10cbse |
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Answer» We do not provide such facilities for now! k This app do not provide papers then which app can u suggest You may visit the internet. For sample papers you may visit www.mycbseguide.com or www.studiestoday.com.If you still not find the right paper,type sample papers for class a on Google and click on the website |
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| 49. |
EXERCISE 10.2I. Is it possible to have a triangle with the following sides?(a) 6 cm, 8 cm and 9 cmcm and 1 cm(b) 5 cm, 12 cm and 18 cm(d) 3 cm, 4 cm and 5 cm |
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| 50. |
EXERCISE 10.21. Is it possible to have a triangle with the following sides?(a) 6 cm, 8 cm and 9 cm(c) 1 cm, 1 cm and 1 cm(b) 5 cm, 12 cm and 18(d) 3 cm, 4 cm and 5 crhe odion of the triangle. Which one of |
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