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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. |
15 iron ball of the same size weigh 10 kg 50 g.how many of them will weight 4 kg 20 g. |
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| 2. |
5+x=62 |
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Answer» 5+X=62 X=62-5=57 ok thanks 11s P |
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| 3. |
x-62=12, find the value of x |
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Answer» 50 is the answer of the following 50 is a correct answer 50 is the right answer. X - 62 = 12 X = 62 -12 X = 5050 is the correct answer 74 is the right answer |
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| 4. |
equatios... If△DEF ~ ΔΑBC such that 2AB = DE and BC = 8cm, thenfind the length of EF.nnewer to a certain |
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Answer» DEF ~ ABC AB/DE = BC/EF AB/2AB = 8/EF EF = 8×2 = 16m If you find this answer helpful then like it. |
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| 5. |
evaluate (419)0 |
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| 6. |
ltself in 10 yearsomes of itself in 2 years. The rate of6(iv) At simple interest a sum bec5( )% per annum.T for true and pfor false for each of the follnuiWrite |
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Answer» SI = P*R*T/100Let sum is P6/5P - P = P*5/2*R/1001/5P = 5*P*R/200R = 40/5R = 8% Rate of Interest is 8% |
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| 7. |
In how many years will a sum of money double itself at 8% per annum? |
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Answer» Let the sum of money = Rs x Therefore amount = Rs 2x Interest = sum*time*rate of interest/100 = x*time*8/100 Interest = 8x*time/100 Amount = sum + interest 2x = x + 8x*time/100 2x = (100x + 8x*time)/100 200x = 100x + 8x*time 8x*time = 100x time = 100x/8x = 12.5 years |
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| 8. |
4. IA be the area of a right triangle and b be one of the sides containing the rnght angle, prve thast the2Ablength of the altitude on the hypotenuse is |
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Answer» Like if you find it useful |
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| 9. |
. If A be the area of a right triangle and b be one of the sides containing the right angle, prove that the2Ablength of the altitude on the hypotenuse isb4 +4A2 |
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| 10. |
(c) 10 cmlength of chord AC is(a) /2ABa circle with centre O, AB and CD are two diameters perpendicular to each other. The8.(b) 2 AB(c)ABV2(d) ABm and CE |
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| 11. |
NUDIU UU UUIUCHUI U PULLULTICONIf the length of the perpendicular from the point (1,1) to the line ax --by + c = 0 be unity. Show.1 1 1 cthat - +---Cab 2abin annarrant |
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| 12. |
108 The Sum 4 tro integer is -16. If one or them is 20then find the other. |
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Answer» Let 2nd intezer be x20+x=-16X=-16-20=-36 suppose other integer be x . then we have, 20+x=_16. x=_16_20=-36 |
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| 13. |
20t tro cubes of 10 cm edge are joined end to end, then the surface area of the resulting cuboid isai 1200() 1000(c) 800(d) 1400 |
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Answer» When edge of 10 cm Cube join together the dimensions of formed cuboid areLength (l) = 20 cmBreadth(b)=10 cmHeight (h)=10 cm______________________________________________________________Total Surface Area Of cuboid = 2(lb+bh+hl) = 2(20X10+10X10+10X20) = 2(200+100+200) = 2(500) = 1000 cm² |
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| 14. |
IS28ă. The domain of f(x) =x-2)17-X1)(2, oc)2)(7, cs)3)(2, 7)4)(-oc,2) |
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| 15. |
If 45 iron rods of the same size weigh 12.6 kg, how much is the weight of 24 such rods? |
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| 16. |
x, '):xis a prime number less thanLet A = {x, y, z} and BLet R be the relati', '{1,2). Findthenumber of relations tro07 hr,A to B.B.n integer |
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| 17. |
328o [23 x 62}2 |
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| 18. |
Oc. Why 17 +11*13 4 17 419 is acomposite no. Explain. |
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Answer» Any number which has more than two factors is called composite number.17+11×13×17×19=17(1+11×13×19)=17×k×1here k =1+11×13×19=composite number it has more than two factors17,k,1 |
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| 19. |
4. Simplify: (4 - 5) - (13 - 18+ 2)(d) 2. |
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Answer» (4-5)-(13-18+2)= (-1) - (-3)= -1+3= 2 thanks bro I hope my ans is right |
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| 20. |
7 The median of the following observations arranged in ascending order is 25 findind Ď11, 13, 15, 19, x +2, x+ 4,30, 35, 39, 46. Also find the mean |
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Answer» Numbers are in ascending order 11,13,15,19,x+2,x+4,30,35,39,46 10 terms so median = (5th term+6th term)/2 = 25 x+2+x+4 = 50 2x = 40 x = 20 numbers 11,13,15,19,22,24,30,35,39,46 Mean = (11+13+15+19+22+24+30+35+39+46)/10 = 254/10 = 25.4 If you find this answer helpful then like it. |
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| 21. |
13. यदि (26 - 1, 10 समीकरण 10: - 92 का हल हो. हो : — _) 1 ) 2 (ग) 3 घी |
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| 22. |
(a)Why is the air in a region dry? |
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Answer» The air in a particular region must be dry due to absence oh humidity i.e moisture content in the atmosphere. |
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| 23. |
26. The fifth term of the H.P., 2, 2, 3will be(2) 3510(4) 10 |
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Answer» thanks |
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| 24. |
_ कचलकड . € (चना e >bt A e T Ao e 2/24— |
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Answer» 49(a - 3b)^2 - 16(3a + b)^2= (7a - 21b)^2 - (12a + 4b)^2= (7a - 21b + 12a + 4b)(7a - 21b - 12a - 4b)= (19a - 17b)(-5a - 25b)= - 5(19a - 17b)(a + 5b)= - 5(19a^2 + 95ab - 17ab - 85b^2)= - 5(19a^2 + 78ab - 85b^2) |
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| 25. |
B e e Ao o bl R O S किU7 snlbo-< )2l Her ८:22 2 8 |
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Answer» √2 sin ( 60 - a) = 1 sin(60 - a ) = 1/(√2) sin(60 - a) = sin(45°) Comparing both sides gives 60° - a = 45° a = 60° - 45° = 15° please hit like if you find my solution useful |
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| 26. |
Ao328E39 |
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Answer» x= 39 is the best answer x=39 is the right answer |
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| 27. |
र्द्र/ 2८0 आर 3 2 2:22: #८ s Ao valel की (८ प्रजि न न e e |
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| 28. |
P e senibons oA Ao I RB हैक सामरिक है कि नइलमिनकर |
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Answer» Two Angles are Complementary when they add up to 90 degrees Let first angle be x and its complement angle be 4x Then,x + 4x = 905x = 90x = 18 Angles are 18 degree and 72 degree |
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| 29. |
1. एक महाजन 9 रु. इस शर्त पर उधार देता है कि1 रु. प्रतिमाह की दर से 10 माह में वापसकरना है। ब्याज की दर प्रति वर्ष क्या है ?(1) 26%(2) 26-%(3) 25% |
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Answer» 25℅ is the answer of the quetion 25 % is the right answer 25 % is the right answer option 3 is the right answer option 3 is the right answer option D is the correct answer option d is the right answer D , option is right answer 25% is the right answer of following questions the correct answer is 3) 25% 1 is the correct answer 25% is the right answer 25% is the correct answer 25℅ is the right answer 1 )26% is correct answer c is the correct answer option 4 is write ans.. 25% is the right answer answer (3) 25 is correct answer 25% is the right answer of question |
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| 30. |
18 - 32.5 x 2 / 13 +8 |
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| 31. |
14.Find the missing number, if same rule is followed inall the three figures.26A10Figure (2)Figure (3)A.B.CD.Figure (1)39263525 |
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Answer» 6*8*5=240÷10=245*7*4=140÷10=143*13*10=390÷10=39 39 is the correct answer of this question 39 is the correct answer 1) 6x8x5=240÷10=24, 2)5×7×4=140÷10=14;, 3) 3x13x10=390÷10=39 39 is the right answee.... 3×13×10=390÷10=39 so 39 is the answer 3×13×10=390÷10=39 so 39 is the answer 5*6*7=240/10=244*5*7=140/10=14similarly 10*3*13=390/10=39so A is right option 39 is the right answer figure(3)answer is b. 26 39 is the correct answer 39 is the correct answer of the given question 39 is the best answer of the question 39 is the correct answer 39 is the best answer 39 is the right answer |
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| 32. |
Find the median from the following data:MarksBelow 10Below 20Below 30Below 40Below 50Below 60Below 70Below 80No. of students1232578092116164200 |
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| 33. |
R27won. WeZACB = 90°meter of the circle. Ifknow that angle in a semicircle is 90°ACBAA + CAB + ZACB = 180°sum of angles in a triangle = 1807)BA +27° + 90° = 180°BA = 180° - 27° - 90° = 63Exercise 151. Draw a circle wZAOB = 60°. Me2. Draw a circle of[Hint. To draw chaand radius 5 cm, di60°. Measure the length of the chord AB.le with centre O and radius 2.5 cm. Draw two Tadil OA and OB such theShade the minor segment of the circle.vele of radius 3.2 cm. Draw a chord AB of this circle such that AB 5radius 5 cm, draw an arc to meet the circle at B. Join AB.)To draw chord AB of length 5 cm, take a point A on the circle. With A as cenydistance 5 cm from the centre.the length of the tangent drawn to a circle of radius 3 cm. from a pointIn the adjoining figure, PT is a tangent to the circle withentre C. Given CP = 20 cm and PT = 16 cm, find the radiusof the circle.in each of the following figure, O is the centre of the circle. Find the size ofettered angle :0500 |
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Answer» dont knowggdjjndhdhdhdhdhdhfhdhdhfjjsjsjdhfhhfhdhdhrhdh6equrssurwrueryaharthsshntsetheethsthrshrthsethethdjtsthrhsrthsethdthdtjrjyfkyyfumuyyfmfykku |
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| 34. |
sin 6 —cos0+1 1' (b) Prove that : sin0+cosO—1 secO—tan0 |
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| 35. |
(1 + sinθ)(1-sina)cosig-I-sinzgcos, θ1-coseV1-cose(1 + cos0) (1 + cost)Vcos6) (+cos×1+cos |
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Answer» √1+cosx/√1-cosx= √(1+cosx)*(1+cosx)/√(1-cos²x)= √(1+cosx)²/√sin²x= (1+cosx)/sinx= 1/sinx +cosx/sinx= cosecx + cotx |
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| 36. |
4sine-cos0+1.19 If 4tane-3, Evaluate Asine+cos0-1 |
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Answer» 4 tan theta = 3tan thet = 3/4=P/Bperpendicular =3base =4hypentonuese =5 (by pythagoras thrm)sin theta = P/H=3/5cos theta =B/H=4/5 4sin theta-cos theta/ 4 Sin theta+ Cos theta=4*3/5-4/5÷4*3/5+4/5=144-20+48/60=472/60=118/15=7.86 |
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| 37. |
The cost of fencing a square field atäš14 per metre is?28000. Find the area of the field. |
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Answer» Thanks for giving idea |
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| 38. |
WIthoitgle are in the ratio 11:19:24 and its perimeter is 540 cm. Find the area of the triangle. |
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Answer» Ratio of the sides of the triangle = 11 : 19 : 24 Let the common ratio be x then sides are 11x, 19x and 24x Perimeter of the triangle = 540cm11x + 19x + 24x = 540 cm⇒ 54x = 540cm⇒ x = 10 Sides of triangle are11x = 11 × 10 = 110cm19x = 19 × 10 = 190cm24x = 24 × 10 = 240cmSemi perimeter of triangle(s) = 540/2 = 270cm Using heron's formula,Area of the triangle = √s (s-a) (s-b) (s-c) = √270(270 - 110) (270 - 190) (270 - 240)cm2 = √270 × 160 × 80 × 30 = 7200√2 cm2 |
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| 39. |
There are also some given figure which have one or more thangle: line of symmetry is angle bisector i.e., AO |
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Answer» please mention clearly about what you need in this question.this is an incomplete question |
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| 40. |
Consider a triangle PQR, right angled at Pin which PQ-29Îź.QR-21Îź andPOR-0, then find the value of cos?e+ sin0, cos0-sin0? |
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| 41. |
Simplify:(0) [12 + (18+ 3) - 11(v) 200+ [4 (6+3)](1) [16-(9+3) 2 () 13-(8-3)] +6 (iv) 16-02*(vi) [(123) + (12+3)]*2 |
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Answer» Give me answer |
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| 42. |
sin0°+cos90°=? |
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Answer» sin0°+cos 90°=0+0=0 0 is your right answer. |
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| 43. |
0056090oot T R % fsin0=â7 201 |
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Answer» please like my answer if you find it useful |
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| 44. |
sin0=? |
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Answer» The value of sin(0) = 0 |
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| 45. |
Prove that cot0-tan0-2cos 6-1sin0 cos0 |
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| 46. |
3. If Sin0+ Sinzg1 , prove that Casp + Casg-| |
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Answer» sin theeta= 1- sin ^2 theeta= cos^2 theeta how you solve explain |
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| 47. |
Prove that sin0(1 + tan0) + cos6(1 + cote)secB + cosecB |
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Answer» please like the solution 👍 ✔️👍 |
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| 48. |
रणgleZlool |
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Answer» 2 is the correct answer of the given question 2 is the correct answer 2 is the right answer. plz like my answer 2 is Right answer a it is also 2 is the right answer 2 is correct answer of this question 2 is the correct answer |
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| 49. |
Prove that sin0(1 + tan0) + cos0 (1 + cot0) =sec0+cosec0 |
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Answer» PLEASE LIKE THE SOLUTION |
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| 50. |
OO | -S|~gle |
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Answer» LHS = (1/8 + 1/10) Take LCM of 8 and 10 LCM of 8 and 10 = 40 = (5 + 4)/40 = 9/40 = RHS Hence proved |
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