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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. |
Find adjoint of the matrix\left[ \begin{array}{lll}{1} & {2} & {1} \\ {3} & {2} & {2} \\ {1} & {1} & {2}\end{array}\right] |
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| 2. |
1 31. Find the Adjoint of the matrix2 |
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| 3. |
Example 1.2. Find the adjoint of the matrix A1 2321 31 2 |
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1 -52L5 -2 23. Show that the Adjoint of the matrix A 1 0 1 is A itself.4 4 3 |
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| 5. |
ustingQuestions Page 6xideQx. Why should a magnesium ribbon be cleaned beforeISAI-2010]burning in air? |
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Answer» Magnesium ribbon must be cleaned before burning in air so that the layer of magnesium oxide (which is formed due to reaction of magnesium with air) can be removed in order to get the desired chemical reaction. |
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| 6. |
PAGE NOtahin36 |
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Answer» (a+b)+c = (-2/3+ 5/6)+ -5/8 = 1/6-5/8 = 4-15/24= -11/24 similarlya+(b+c) = -2/3+(5/6-5/8)= -2/3+5/24 =-11 /24 verified both are equal |
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| 7. |
DatePage1322-5-6) it |
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Answer» 14 ÷ { 3 × 2 - ( 5 - 6 ) } + 9 = 14 ÷ { 6 - (-1) } + 9 = 14 ÷ { 6 + 1 } + 9 = 14 ÷ 7 + 9 = 2 + 9 = 11 If you find this answer helpful then like it. |
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| 8. |
PageC IS6Prove that Ac |
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| 9. |
Page No.Datel3x2_X-5x2+x-6 |
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| 10. |
6.InFig.I,OisthecentreofthecirclewithchordsAPandBPbeingproducedtoRandndI ZQPR 35, find the measure of ZAOB35° |
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Answer» ∠APB =∠RPQ = 35° Now,∠AOB and∠APB are angles subtended by an arc AB at centre and at the remaining part of the circle . ∴∠AOB = 2∠APB = 2× 35° = 70° |
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| 11. |
Find adjoint of the matrix2 34 2 |
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| 12. |
Unlessstated otherwise, use π =Find the area of the shaded region in Fig. 5.19, if PQ= 24 cm, PR = 7 cm and O is the centre ofthe circle.1. |
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Answer» Given thatPQ=24cm,PR=7cm,QR=? From Pythagoras theoremQR=√PQ^2+PR^2 =√24^2+7^2 =√576+49 =√625 =25cmhere redius of the circle,r=25/2cmthen area of shaded region=area of semicircle-area of ∆PQR=1/2πr^2-1/2PR×PQ=1/2{πr^2-PR×PQ}=1/2{22/7×25/2×25/2-7×24}=1/2{11×625/14-168}=6875/28-168/2=6875-2352/28=4523/28 cm^2 |
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| 13. |
1) The sum of the AP with 21 terms is 210 then the 11 term isa) 810c) 15d) 18 |
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Answer» Sum=(21/2)*(2a+20d)210=(21/2)*(2a+20d)20=2a+20da+10d=1011th term will be a+10dSo, 11th term is 10Option b is correct |
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| 14. |
18.In the ngure. Pisthe centre ofthe circle, prove that APZ-2(A ZP +LYNZ). |
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Answer» angle XPY = 2 angle YZX _____(1)since angle subtended by an arc at the centre is double the angle subtended by it on any other part of the circle. angle YPZ = 2 angle XZY_______(2)since angle subtended by an arc at the centre is double the angle subtended by it on any other part of the circle. eq(1)+(2)angleXPY+angle YPZ = 2angle YZX+ 2 angle XZYangle XPZ = 2 ( angleYZX+angleXZY)( taking 2 as common) |
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| 15. |
(b) In ther given figure O is the centre ofthe circle, given ARST-57e. Find (O 4 POR (i) refex POR 13] |
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Answer» Angle PSR=180⁰-57⁰=123⁰(linear pair)so angle PQR=180⁰-123⁰=57⁰(The sum of opposite angles in a cyclic quadrilateral is 180⁰) Reflex POR=2*123⁰=246⁰. |
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| 16. |
adjoint method1 2 |
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Answer» Inverse = adj A/ mod A adj A= [4 -2] [-1 7] mod A= (28-2)= 26 A inverse= 1/26[ 4 -2] [ -1 7] |
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| 17. |
L Laledww. 115 sides are in the ratio 9:7:6. Find its area.of a right triangle measures 126 m and the difference in length of itsand other side is 42 m. Find the measure of its two unknown sides and12. One side of ahypotenuse andcalculate its area.perimeter and area of a triangle whose pidon nan 19 |
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| 18. |
(C)AB#AC(D) BC ACIn the figure PQ il BC, AP- 2cm, PB 6cm, PQ 3 cm then BC (in cm)26(R) Q |
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| 19. |
Find the square roots of the following decimal numbers(i) 2.56 ii)18.49iii)68.89 (iv) 84.64 |
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Answer» The square root of 18.49 = 4.34.3x4.3=18.49 It is written 4.9 instead of 4.3 please change it thanks but it's government text I will change immediately sister |
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| 20. |
ATICSFind the square root ofthe following decimal numbers.(ii) 7.293.2.56(v) 31:36(ili) 51.84(iv) 4225 |
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| 21. |
3.Find the square root of the following decimal numbers.(1) 2.56(v) 31.36(i) 7.29(ii) 51.84(iv) 42.25 |
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| 22. |
Convert the following decimal numbers in the form:1. (i) 0.152. () 0.6(ii) 0.00026(in) 0.12(iii) 8.0025(it) 0.6315(iv) 3.0440625(iv) 0.571428 |
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Answer» 1.१-15/100 २-26/100000 ३-80025/10000 ४-30440625/100000002•१-6/10 २-12/99 ३-6315/9999 ४-571428/999999 |
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| 23. |
following dea5 decimal numbersExpress Cach of thein form2i) 0.5 |
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Answer» 0.555..... =x. (equation 1)hear periodicity is 1 so multiply on both sides by 1010x=0.555...×1010x=5.5555 (equation 2)substitute equation 1 and 2 10x=5.55 (-) x=0.555 = 9x=5 = x=5/9hence0.5555....=5/9
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| 24. |
How to find adjoint and cofactor of a matrix |
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| 25. |
1 2 3Find the adjoint of 0 5 02 4 3 |
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| 26. |
: Find the adjoint of the matrix A =C a |
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| 27. |
6. In the given figure, if PQ il RS, MXQ135 and AMvR- 40, find A135'that ABC ZFDE. |
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| 28. |
Evaluate : √3 up to four decimal place |
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Answer» 1.7320 is the correct answer. √3=1.7320 is the answer |
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| 29. |
If m times the mth term of an AP is equal to n times the nth term andm+n, show that its (m + n)th term is zero. |
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Answer» Let the first term of AP = acommon difference = dWe have to show that (m+n)th term is zero ora + (m+n-1)d = 0 mth term= a + (m-1)dnth term = a + (n-1) d Given thatm{a +(m-1)d} = n{a + (n -1)d}⇒am +m²d -md = an + n²d - nd⇒am - an +m²d - n²d -md + nd = 0⇒a(m-n) + (m²-n²)d-(m-n)d = 0⇒a(m-n) + {(m-n)(m+n)}d -(m-n)d = 0⇒ a(m-n) + {(m-n)(m+n) -(m-n)} d = 0⇒a(m-n) +(m-n)(m+n -1) d = 0⇒ (m-n){a+ (m+n-1)d} = 0⇒a + (m+n -1)d = 0/(m-n)⇒a + (m+n -1)d = 0 Proved! thanks brother |
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| 30. |
19. In the given figure PQ Il ST, <POR<110°, andRST=130°find«QRS |
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| 31. |
Inotherwulus,L,ceptsEXERCISE 8.2ABCD is a quadrilateral in which P, Q, R andmid-points of the sides AB, BC, CD a(see Fig 8.29). AC is a diagonal. Show that:,R and S arend DAi) SR Il AC and SRAC2(ii) PQ SR(ii) PORS is a parallelogram. |
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Answer» thanks |
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| 32. |
Evaluate : √5 up to four decimal place |
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Answer» 2.236 is the correct answer. |
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| 33. |
1.A rational number equivalent |
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Answer» -6/8=-3/4 this is the correct answer A rational number equivalent to -3/4 is -6/8. |
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| 34. |
Tues April 30 Series / Page 61 Equivalent measures are shown in thetable below.Number ofmetersNumber ofcentimeters5009001.30013Based on the information in the table, howmany centimeters are equivalent to 21meters?A 2,001 B 2,100 C 1,321 D 1,021 |
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Answer» the answer is 2100 centimetres 2100 is the right answer B)2100 is the right answer |
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| 35. |
8. The rational number equivalent to-24is45 |
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Answer» There is a lot of rational numbers equivalent to - 24/45but we have to write only one equivalent rational number Thus, Multiply both numerator and denominator by 2 -24×2/45×2 = - 48/90 7000/30 |
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| 36. |
le place value of 2 in the following decimal numbers:() 2.56cu form:(i) 200.037. Dinesh went from place A to place B and from(i) 21.37there to place C. A is 7.5 km from B and B is.(1) 10.2512.7 km from C. Ayub went from(iv) 9.42D and from the(iv) 2.0(v) 63.352. |
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Answer» 1)22)203)2/104)2/1005)2/1000 |
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| 37. |
If V2 = 1.4 and 3 = 1:7, find the valueof each of the following, correct to onedecimal place :(1) 13-12(ii) 3+272V3 |
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Answer» 1) 1/(0.3)= 10/33.33 2) 1/3+2.8= 1/3.1= 10/31 |
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| 38. |
1. In an AP, if d =-2, n-5 and an-0, then fnd the value of a. |
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Answer» an=a+(n-1)dPutting the values0=a+4×-20=a-8a=8 |
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| 39. |
25Fig. 15.197Fig. 15.19815.198, AB is a diameter of the circle such that LA = 35° and LQ =29, fndLPBR |
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| 40. |
λso that λα may bea unit3.(2,-4) +51)then fnd the value of" |
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| 41. |
Poteelevation clänges to3u Fnd the height of the tree.16. Evaluate 2 cos260+3 sin245-3sin230 2cos290 |
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Answer» 2*(1/2)² + 3(1/√2)² - 3*(1/√2)²*2*(0) = 1/2 + 3/2 -0 = 4/2 = 2 |
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| 42. |
2. Fnd the 12 oof a G.P wose m ia 192 and the common ratio is 224 |
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| 43. |
Fndthe largest number of 2 digits which is a perfect square. |
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Answer» wrong |
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| 44. |
2x+5y+1=0,x+y=0 by inversion method |
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| 45. |
Find the inverse of the matrix A-by adjoint method |
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Answer» Inverse A= adjoint A/ |A|adjoint = (5 -2 -4 3)mod = 7 Inverse= 1/7(5 -2 -4 3) |
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| 46. |
Solve the following equations by the method ofinversion(i) 2x-y=2, 3x +4y=3 |
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Answer» 2x-y=-2..........13x+4y=3........2multiply eq.1 by 3 and eq.2 by 26x-3y=-66x+8y=6subtract 1 from 211y=12y=12/11put value of y in eq. 1x=-5/11 |
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| 47. |
Fnd the solution set of the inequality $ x+5 \leq 2 x-3, $ when $ x $ is a whole number. |
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| 48. |
विज रे NIV VIS NN(i) 2x2+7x+3=0k-99 (i NS/ |
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Answer» what is your school name |
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| 49. |
7. Lach side of a polygon is 3.8 cm in length and its perimeter is 358.8 cm. How many sides does the polygonhave?47 |
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Answer» Let the number of sides of the polygon ben.Length of each side of the polygon = 3.8 cm∴ Perimeter of the polygon = (3.8××n) cmBut it is given that its perimeter is 22.8 cm.∴ (3.8××n) cm = 22.8 cm⇒n=(22.83.8)22.83.8=2283822838= 6 Hence, the given polygon has six sides Let the number of sides of the polygon ben.Length of each side of the polygon = 3.8 cm∴ Perimeter of the polygon = (3.8××n) cmBut it is given that its perimeter is 22.8 cm.∴ (3.8××n) cm = 22.8 cm⇒n=(22.83.8)22.83.8=2283822838= 6 Hence, the given polygon has six sides. |
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| 50. |
13. Show that 9" cannot end with 2 for any natural number. |
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Answer» Let, P(n) denotes the event that 9^n can not end with the digit 2 When, n=1, P(1): 9^1 = 9 does not end with 2. Let us assume that the statement is true for n=k;Then, P(k): 9^k does not end with 2. Now for n=k+1, P(k+1): = 9^k × 9^1 which does not end with 2 since 9^k and 9 both does not end with 2. Therefore, assuming P(n) is true for n=k we have that P(n) is true for n=k+1. Thus, by the principle of mathematical induction, P(n): 9^n does not end with 2. |
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