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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. |
5. Eind the area of a quadrilateral one of whose diagonals is 40 cm and the lengths of theperpendiculars drawn from the opposite vertices on the diagonal are 16 cm and 12 cm |
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| 2. |
3!ARIS W& उस त्रिज्यखंड का क्षेत्रफल जिसका कोण /;* है, निम्नलिखित है: |
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| 3. |
Find a vector in the direction of the vector a=6i-2j+3k, whosemagnitude is 4 units. |
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Answer» nice work |
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| 4. |
rgo.gmi29. Solve the quadratic equationb+0r. , "t:1 ris. [CBSE March 2012] |
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| 5. |
8 Ifcos0+sin 6 =1,prove that cos 変sin 6 = F1. |
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| 6. |
B)gThe tangent to the circle x2 + y2 - 5 at the point (1, -2) also touches the circle(A) (2, 1) +6y + 20=(B)( 3,0)8.a(C)F1-1)(D) (3,-1) |
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Answer» For the circle x²+ y² = 5,the equation of tangent at (1,-2) is x-2y=5 Solving for x and y from the equation of tangent and the second circle,we get x=3 and y=-1so the point is (3,-1) |
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| 7. |
The mean of the following distribution is 50. If the sumof frequencies 120, find the value of f1 and f2. |
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| 8. |
Prove that the points A(a, O), B(O, b) andCl, 1)ar,f1are collinear, if 1/A+1/B=1 |
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| 9. |
1. Evaluate 2aluate 27+27 +2..thirty two tines22..thirty two times. |
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| 10. |
12.The greatest number which willdivide 2112 and 2792 leaving4 as the remainder in each caseIJNV 1999(i) 68 (ii) 58 (iii) 78 (iv)1881S |
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Answer» In order to find the greatest number which will divide 2112 and 2792 leaving the remainder 4 in each case follow the below steps:- Let us assume the requirednumber is x. In order to make 2112 and 2792 completely divisible by x,deduct remainder 4 fromboth 2112 and 2792. 2112 - 4 = 2108 2792 - 4 = 2788 Now we've to find the HCF of 2108 and 2788. 2108 = 4 x 17 x 312788 = 4 x 17 x 41 HCF is 4 x 17 = 68. Therefore,the greatest number which will divide 2112 and 2792 leaving the remainder 4 in each case is 68. |
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| 11. |
Find the least number which when divided by 15, 8, 7, 6 leaving a remainder 3 |
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Answer» Find lcm of 15,8,7,6 which is 840Since it leaves a remainder 3,So the no is 843 LCM of numbers = 840840 - 3 = 837Required number = 837 |
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| 12. |
8. Find the largest number which divides 2053 and 967 leaving 5 and 7remainder, respectively |
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| 13. |
8. Find the largest number which divides 2053 and 967 leaving 5 and 7remainder, respectively. |
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Answer» tnq u😃 |
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| 14. |
26, Find the value of x if8-(68 |
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Answer» Since the base is same we can equate the powersSo,x-1=3-xx+x=3+12x=4x=2.............................................. |
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| 15. |
Find area of largesttriangle that can beinscribed in a semicircleof radius r unit. |
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| 16. |
(a) < 1550(b) 1020(c) < 1430(ii) When the profit % on selling a toy for 400 is 11%, the cost price ofthe toy is(a) 440(b) 380(c) 360 |
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Answer» Let the cost price of toy be ' x ' Find the profit :-------------------- profit = 11 whole 1 / 9% = 100/ 9 = 11.11 % profit = 11.11 % of cost price profit = 11.11 % of x = 0.11x (approx. ) Find the cost price :--------------------------- since , the selling price of toy = ₹ 400 cost price + profit = selling price x + 0.11x = 400 => x ( 1 + 0.11 ) = 400 1.11 x = 400 => 111 x ÷ 100 = 400 111 x = 40000=> x = ₹ 360.36 therefore, cost price of the toy = ₹ 360.36 Answer : C.P = ₹ 360.36 --------------------------------------------------------- option D is the correct answer option B) is the correct answer option D is the correct answer option (c) is the right answer |
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| 17. |
given that HCP of (2539 240)=11 g Lem of(253 & 440)=253XRthen find the value |
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Answer» 253= 11×23440= 11×2×2××2×5LCM= 11×2^3×5×23253×R= 11×8×5×23R= 11×8×5×23/253 = 40 |
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| 18. |
16. The HCF of 72 and 120 is 12. Find their L.C.M.(A) 120(B) 240(C) 360(D) 720 |
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Answer» We know LCM×HCF =Product of numbers So, 12 × LCM = 72 × 120 LCM = 6 × 120 = 720 D) is correct option |
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| 19. |
12 Perform the following divisio |
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| 20. |
Carry out the following divisio |
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| 21. |
The greatest number among 30 4530 and 620 is |
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Answer» Taking the tenth roots, you need to compare 3^5, 4^4, 5^3, 6^2 since if a>b, a^10>b^10. These are 243, 256,125, 36. Thus the greatest is 4^40 |
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| 22. |
SECTION A. CLASS RESPONSE1. Fill in the blanks:1. The successor of 9999 is2. The predecessor of 250305 is3. The greatest number among 25286, -catest number among 25286, 25245, 25270 and 25210 is4. The standard form of 2 x 10000 + 3 x 1000 + 3 xlis5. The face value of 3 in 53201 is -6. The place value of 7 in 67891 is _ |
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Answer» 1. 100002. 2503043. 252864. 230015. 36. 7000 1 100002 2503043 252865 30006 7000 1)100002)2503043)252864)230015)36)7000 Answer:-1. 100002. 2503043. 252864. 230015. 30006. 7000 1.100002. 2503043. 252864. 230015. 36. 7000 1. 100002. 2503043. 252864. 230015. 36. 7000 10000, 250304, 252 86, 23001 , 3000, 7000 is the correct answer of the given question a. 10000b. 250304c. 25286d. 23001e. 3f. 7000 |
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| 23. |
Theorem 8.9 The line segment joining the mid-points of two sides of a triangis parallel to the third side |
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| 24. |
Reeti has thirty two marbles which is eightmore than twice the number of marblesthat Savita has. Find the number ofmarbles Savita has. |
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| 25. |
(4) Sum of the present ages of Manish and Savitais 31. Manish's age 3 years ago was 4 timesthe age of Savita. Find their present ages. |
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Answer» savita age7& monisha24 AGE'S ARE...... OF SAVITA... 7AND. MONISH'S |
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| 26. |
Sum of the present ages of Manish and Savita is31. Manish's age 3 years ago was 4 times the ageof Savita. Find their present ages. (3 marks) |
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Answer» thankyou so much... |
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| 27. |
....cqudtion(4) Sum of the present ages of Manish and Savita is 31. Manish's age 3 years agowas 4 times the age of Savita. Find their present ages. |
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Answer» let the present age of Manish and savita be x and y respectively. as per 1st condition x+y=31--------(1)as per 2nd condition (x-3) = 4(y-3)x - 3 = 4y - 12x - 4y = -12 + 3x - 4y = -9-------(2)subtract eqn (2) from (1) x + y = 31- x - 4y = -9 - + + ——————— 5y = 405y = 40 y = 40÷5y = 8substitute the value of y =8 in eqn (1)x + y = 31x + 8 = 31x = 31 - 8x = 23 |
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| 28. |
) Savita bought a calculator for 325. She sold it to Geeta at 12% profit. Find theselling price of the calculator. |
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Answer» CP = ₹325 Profit % = 12% Profit = 12% of ₹325 = 12/100 × 325 = ₹ 39 So, SP = Profit + CP = 325 + 39 = ₹364 Thank you 😊 |
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| 29. |
15.The greatest number that divides 38 and 68 leaving 8 as remainder incase is(a) 10(c) 60(b) 15(d) 30 |
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Answer» 30 is the greatest number that divides 38 and 68 leaving 8 as remainder. |
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| 30. |
A ARl o — ”/’T’/L{ नी दा है ही ताg bty (= 1 |
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| 31. |
sV *vrgi\'m ¥ सम्पूर्ण पृष्ठ का सूत्र है कर 2(अ) 2ath () arl (स) |
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Answer» volume of cone=πrl, where L=√(l²+r²) संपूर्ण पृष्ठ का क्षेत्रफल होता हैπr(r+l) |
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| 32. |
uu leunRefer to the given figure. O is thethe following:given figure. O is the centre of the circle. Fill in the blanks inis a quadrant of thecircle.(ii) AB is a of the circle.(iii) OLBM is a __ of thecircle.(iv) OM is a of the circle.- PQ is a of the circle.В |
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Answer» AB is a diameter of the circle pq is a hafe of the circle AB is a diameter of a circleOLBM is a sector of circleOM is radius of circlePQ is a chord of circle |
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| 33. |
The HCF of two numbers is 12 and also their difference is 12. The numbers2 |
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Answer» 48 and 60 is the right answer the numbers are 48 and 60 48 and 60 are right answer hai the answer is 48 and 60 |
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| 34. |
7. On dividing 55390 by 299, the remainder is 75. Find the quotient using the divisioalgorithm1100111Dible bu071 |
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Answer» 55390 = 299x + 75299x = 55315x= 185 ans 55390 = 299x + 75299x = 55315x= 185 ans 185 is the answer of the following 55390=299X+75X=185is the right answer... 55390=299x+75299x=55390-75299x=55315x= 55315÷299x=185 |
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| 35. |
30. Prove that the line segment joining mid points of two sides of a triangle is paralel to thethird side and equal to half of it. |
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| 36. |
(d) 180The sum of all interior angles of a hexagon is(a) 6 right <s (b) 8 right Ls12.(c) 9 right Zs(d) 12 right Zss. The sum of all interior angles of a regular nol |
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| 37. |
Find the mid points of the line joining the point(4,7) \&(2,1) |
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| 38. |
equation of the above data and then represent it graphically30. Prove that the line segment joining mid points of two sides of a triangle is parallel to thethird side and equal to half of it. |
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| 39. |
12 -lS(iv)ir rational numbers equivalent to: |
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Answer» -2/7=-2*2/7*2=-4/14=-2*3/7*3=-6/21=-2*4/7*4=-8/28=-2*5/7*5=-10/35 |
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| 40. |
III. LS An A.P.comsists of SO terms of which 3d term is 12 and Last term2sis 106. Find the 29h serm |
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| 41. |
Consider the figure below in which 1,1 12-mea 13y,mcb = 31y + 4, mer = 30x + 40, and mis = 130x-160.t1What are the values ofa,LD,Lr, and LS? |
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| 42. |
3. Iwo parallel lines I and m are cut by a transversal t. If the interior angles of the same side oft be (2x - 8) and (3x - 7)º, find the measure of each of these angles. |
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| 43. |
Savita borrowed4000 from her friend at the rate of 12% per annum for25 years. Find the interest and amount paid by her. $2002 |
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Answer» thanks |
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| 44. |
Harpreet borrowed ₹20000 from her friendat 12% per annum simple interest. She lends to Alam at the same rate but compounded annually. Find her gain after 2 years. |
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| 45. |
329Quadrilateralss. In the adjoining figure, M is the midpoint of Aside BC of a parallelogram ABCD such that <BAM - <DAM. Prove that AD =2CD |
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| 46. |
In the given figure, D, E and FA.are respectively the mid-pointsof the sides AB, BC and CA ofAABC. Prove that BDFE is aparallelogramDQuadrilaterals |
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| 47. |
eorem 10.7: Chord equidistant from the centre of a circle arle of equal length |
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Answer» 1 2 3 |
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| 48. |
900+9 |
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Answer» 900+9=909 is right answer 900+9=909 is correct answer of following question 900+9=909 is the best answer 909 is correct answer. |
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| 49. |
900, find |
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Answer» let the two no's be X and X+1X+X+1=1552x=154X=77hence.nos are 77 and 78 |
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| 50. |
The sides of a triangle are in the ratio 5: 12: 13 and its perimeter is 300 m, its area is? |
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