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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. |
\operatorname{tg} 3 x=\sin 45^{\circ} \cos 45^{\circ}+Sin 3o then the Value of x |
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Answer» Tan3x = sin45°cos45° + sin30° = (1/√2)×(1/√2) + 1/2 = 1/2 + 1/2 = 1tan 45° = 1so, 3x = 45⇒x = 15° |
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| 2. |
2 x 30() 2 of 30 books 2x30 books3boolk2 x 45 CDs(c) of 30 books x books(d) _ of45 CDs-X 45 CDsente CEXERCISEI. Write the fraction representing the shaded portion:5-1 |
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Answer» 2/3 of 30 books=2/3*30=202/3 of 45 CDs=2/3*45=30in figure first one is 6/72nd one is 7/10 |
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| 3. |
4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to eaangle of 60° |
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| 4. |
1. A triangle and a parallelogram are on the same baseparallels, if the area of parallelogram is 32cm' find the area ofgram are on the same base and between the same |
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Answer» If a triangle and a parallelogram are on the same base and between the same parallels then prove that the area of the triangle is equal to half the area of the parallelogram. Area of triangle = 1/2(32) = 16cm² |
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| 5. |
A.I. In figure 5.22, âĄABCD is a parallelo-gram, P and Q are midpoints of side ABand DC respectively, then prove âĄAPCQis a parallelogramFig. 5.22 |
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| 6. |
कै (sin90° + cos 60° + cos 45°) X (sin30° — cos 0° + cos 45°)i |
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Answer» (sin90-cos45+cos60)*(cos0+sin45+sin30)= (1- 1/(root2)+1/2)*(1 +1/(root2)+1/2)= [3/2-1/(root2)]*[3/2+ 1/(root2)]= (3/2)^2 - [1/(root2)^2]= 9/4 - 1/2= (9-2)/4= 7/2 |
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| 7. |
One pair of adjacent sides of a parallele gram is in the ratio 3:4. If oneits angle is a right angle and the diagonal is 10 cm. Find thelengths of the sides of the parallelogram.i1) perimeter of the parallelogram |
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Answer» As adjacent sides are in the ratio 3:4 . Let AD=3x and AB=4x . Applying pythagoras theorem , in △ABD AB^2+AD^2=BD^216x^2+9x^2=100 25x^2=100x^2=4x=2 cm. So ,AB=4(2)=8 cm .AD=3(2)=6 cm .CD=AB=8 cm . (Opposite sides of parallelogram)BC=AD=6 cm . (Opposite sides of parallelogram) Perimeter of parallelogrm=2 (AB+AD)=2 (8+6)=2×14=28 cm . hit like if you find it useful |
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| 8. |
576+v196V576 196 |
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Answer» (24+14)/(24-14)38/1019/5 |
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| 9. |
a, B are roots of the equation 2 (x2 - x) + x + 5 = 0. If 2, and 2, are the two values of for which(22222 Mroots a, s are conected by the relationroots a, ß are connected by the relation+- A, on the value os= 4, then the value of2lisB |
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Answer» First we add then take LCM |
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| 10. |
Qo Pxove dtot 4squase ofhégmh e infegee 10 9६1०Bm,Sm+1 6c Sm-+Y न 9६ 4orme intege |
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Answer» consider that m=2.square of 2=4, (2×2=4).5m=5 (2)=10 4 not equals to 10 5m+1=5 (2)+1=11 4 not equals to 11 5m+4=5 (2)+4=1414 is also not equals to 4. If we take 5 as "m" then the square of 5=25,which satisfies the condition 5mhence proved. only 5 is the number which can satisfies the condition 5m. There is no other number which can satisfies the conditions 5m,5m+1,5m+4. is it a correct ans. |
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| 11. |
sind = पक == sm'('e 15" बब्र मान |
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Answer» sin theta= √3/2= theeta= 60°sin(60-15}= sin45= 1/√2please like the solution 👍 ✔️ |
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| 12. |
hind the ratio of:.di) sm tolokm |
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Answer» 5/10000m=1/2000m so 1:2000 this is right 5m/10km=5m/10000m=1/2000 m 1:2000 is the right answer |
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| 13. |
Find S.P. when (i) C.Р. = Rs 850, gain = 8%, |
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| 14. |
Use the diagram shown to find x.28850 |
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Answer» by angle sum property of a triangle we know that sym of interior angles of a triangle is 180°.therefore in this triangle 28°+85°+x=180° from this we can find the value that is 113°+x=180°that is,x=180°-113°that is equal to,x=67° 28 + 85 + x = 180; x = 180/113 =67 |
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| 15. |
हे Vtanx पद का मान है850 2 ९0०5 2. |
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| 16. |
iuunt to734 in 3years?18. At what rate per cenl per an768 in 2 years 6 months, what will2850 amount to in 3unts to850 amosame rate per cent per annum?Vea |
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| 17. |
P =850, R = 6%, T = ?, S.1.-7 178.50, A = ? |
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Answer» OK HIT THE LIKE BUTTON ✔️👍 |
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| 18. |
E Parallelogram GRAM with GR= AM-5 cm. RA= MG6.2 cm and LR-850 |
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Answer» Steps for construction1. Draw line GR = 5 cm.2. Using protractor construct angle of 85° at R.3. Now construct RA of length 6.2cm through R at angle 85.4. Since it is a parallelogram angle A = 180 - R. A= 95.5. Make angle 95 from A taking RA as base.6. Now construct AM of length 5cm through A.7. Complete the parallelogram. |
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| 19. |
Parallelogram HEARHE = 5 cmEA = 6 cm<R = 850 |
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Answer» 1 |
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| 20. |
)Parallelogram GRAM with GR = AM-5 cm, RA = MG-6.2 cm and <R-850 . |
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| 21. |
निम्नलिखित का मान ज्ञात करो।()(164) (i) (81)(vi)(125)(243) (256)*निम्नलिखित का मान ज्ञात करो।(24313 |
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Answer» ( V64)^2/3 = (8)^2/3×2 = 8 ^1/3 = 2 ^3/3=2^1=2; 2) (81)^-3/4 = 1/(81)^3/4 = 1/(3)^4^3/4= =1/3^3=1/27 1,_ 211,_ 1/27111,_ 95, _0.8 |
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| 22. |
If 18 binders can bind 900 books in 10 days, how many binders will be required to bind 660 books in12 days?I4. |
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| 23. |
square root 196 |
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Answer» the square root of 196 is 14 |
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| 24. |
Using Cramer Kule, レea+301 |
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| 25. |
141The decimal expansion of 120 willterminate after how many places ofdecimal?7.mhor uhich divides |
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Answer» 141/120=1.175 hence it terminate after 3 decimal places |
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| 26. |
Consider the number 96,385. Using the samedigits, can you write three numbers greaterthan it? Arrange them in ascending order. |
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Answer» The three numbers greater than 96385 in ascending order are 96583, 96853, 98365 |
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| 27. |
Solve the equations Using Cramer Rule:(i) x+2y +4z= 164x +3y-2z 53x-5y+z4 |
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| 28. |
Using prime factorization method find L.C.M and H.C.F of anythree two digit compound numbers. |
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Answer» LCM is least common multiple of any three or two digits. HCF is highest common factor of any two or more than two numbers. LCM of 20 and 12 is 2*2*5*3=60 HCF OF 20 and 12 is 4 becuz we take the most greatest number among common numbers. And we have two numbers 20 and 12 |
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| 29. |
- 1234+10+ 10+- 1274+ 5- 127983Add the following numbers using all the three methods.139 + 52b. 23 +44c. 245 + 26548 +79e. 35+ 184 f 738 + 39 |
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Answer» 1) 139+52 = 139+1+51 = 140+51 = 1912) 23+44 = 20+40+7 = 673) 245+26 = 245+25+1 = 270+1 = 2714) 548+79 = 550+77 = 6275)35+184 = 2196) 738+39 = 737+40 = 777. |
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| 30. |
scMathematics-9.How many three-digit numbers can be formed wihout using the digits 0, 2.3.4, 5 and s |
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Answer» availables are 1,7,8,9so number of three digits=4×4×4=64 |
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| 31. |
Express each of the following as product of prime(D 270write exponential form of 9 × 9 × 9 taking 3 asFind the values of n in each of the following(i) (2 -(2).s.(ii) 768(iii) 64.(ii) 2 2-2(v) 5"1x3"-,-135(iv) 2n-7 ×5". " 1250Ife-en. (, find the value of (q (3Simplify 10 × 25 × 5°3×5"+2+10×5"9"×32 × 3n-(27)^find the value of r27If(3× 23 |
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Answer» thanks.. |
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| 32. |
. Find two consecutive multiples of 3 whose product is 648. |
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| 33. |
iinu the pioduc5154.Express 648 as a product of powers of prime factors. |
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Answer» Please hit the like button if this helped you |
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| 34. |
Three numbers are in AP. If their sthe numbers.um is 27 and their product is 648, find |
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| 35. |
Express each of the following as product of powers of their prime factors:0 648(ii)405(iii)540(iv) 3,600 |
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| 36. |
(118*648)/0.005 |
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Answer» 648.48 ÷ 0.005=129.696 this will give u 129.696 129.696is correct answer of this question. |
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| 37. |
Find the GCF to simplify64 /96 |
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Answer» 64/96 = (4*4*4)/(4*4*2*3)= 4/6= 2/3 Pls I don't understand can u explain more better greatest common factor of 64 and 96 is 64 = 2*2*2*2*2*2 96 = 2*2*2*2*2*3 so, G.C.F = 2×2×2×2×2 = 32 now 64/96 = 32*2/32*3 = 2/3 thank u Pls can u solve dis one here is it -1111-111eeeedfyirjytkudfvj |
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| 38. |
\begin{array} { l } { \text { Evaluate the following } } \\ { \text { (a) } \frac { 27 } { 64 } + \frac { 35 } { 64 } } \end{array} |
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| 39. |
Simplify 216-64V625 196 |
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Answer» (6-4)/(25-14)=2/11=0.18 |
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| 40. |
27, 64. 125 |
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Answer» all are cubes of natural number1^3=12^3=83^3=274^3=645^3=125 so 7 will be at missing place |
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| 41. |
(27*(64*81))/(((256*3^6)*5^2)) |
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| 42. |
\left. \begin{array} { l } { \sqrt [ 3 ] { 27 \times 216 } = \sqrt [ 3 ] { 27 } \times \sqrt [ 3 ] { 216 } } \\ { \frac { \sqrt [ 3 ] { - 64 } } { \sqrt [ 3 ] { 512 } } = \sqrt [ 3 ] { \frac { - 64 } { 512 } } } \end{array} \right. |
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| 43. |
In the given figure, the value of x is |
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| 44. |
The value of Sin A and Cos A are always less than 1. Why? |
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| 45. |
venty-four is divided into two parts such that 7 times the first part added to 5 tsecond part makes 146. Find each part. |
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| 46. |
Twenty-four is divided into two parts such that 7 times the first part added to 5 times the second part makes 146. Find each part. |
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Answer» Thanks😍a🤩lot!!! |
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| 47. |
the value of sinA and cos A are always less than 1why |
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Answer» The value of sin as well as Cos function is always less than 1 because of its definition that for a right triangle, the sine is the opposite side divided by the hypotenuse. And we know Hypotnese is longest side in any right angle triangle.Similarly for cos, therefore the function are less than 1. also we have consider unit circle tu determine the value of sine and cos... then the value of cos and sin is lies between -1 to 1 |
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| 48. |
Prove that the difference of any two sides of a triangle is alwaysless than third side.(a) |
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| 49. |
की कि पर देविश 25 9 5 पद ।७क७ [5 है 5 15फा८क 20 20डे लह ९ पब [७ है ए- हैक 3० 1 el (2 8 kAl |
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| 50. |
Obse(i) Toverify that the difference of any two sides of a triangle is always less than the third side. |
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