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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. |
Pege No. |
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| 2. |
(a) Sheetal had 2500. She spent 178.50 and she earned 249. How much moneydoes she now have? |
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| 3. |
(2) In the figure, AC- 24 cm, BC 10 cm and O is thecentre of the circle. Find the area of shaded region(n=314) |
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| 4. |
Sabtractthesumof2a+ Sab-b and -3a-ab + 2from a+4ab+b. Find x, if314-1 |
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Answer» Don't study.Study is waste of time |
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| 5. |
(ii) (a + 2ab+b') a+2ac)(2a + b+c) |
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| 6. |
The inner circumference of a circular track is 264 m and the width of the track is7m. Calculate the cost of putting up fence along the outer circle of the track atthe rate of Rs 3. 50 per meter. |
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| 7. |
Page NoDete |
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Answer» 12 is the answer to this 5+7=12 so the answer is 12 5 + 7 = 12 12 is the correct answer 7+5=12 this is addition correct answer is 12 the correct answer is 12 5+7=12 is the right answer 5+7=12 is the right answer. 12 is the best answer 7 + 5 = 12 |
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| 8. |
SBBDete:Page No. : |
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Answer» Given quadratic equation: 100x²-20x+1=0 => (10x)² -2*(10x)+1² = 0 => (10x-1)² = 0 /* By algebraic identity a²-2ab+b² = (a-b)² */ => 10x-1 = 0 or 10x-1 = 0 => 10x = 1 or 10x = 1 => x = 1/10 or x = 1/10 [ equal roots ] Therefore, 1/10, 1/10 are two roots of given Quadratic equation. 100x^2 -20x +1=0100x^2-10x-10x+1=010x(10x-1)-(10x-1)=0(10x-1)(10x-1)=0either 10x-1=0, x= 1/10or 10x-1=0, x= 1/10 100x² - 20x + 1 = 0100x² - 10x - 10x + 1 = 010x(10x - 1) -1(10x - 1) = 0(10x - 1) (10x - 1) = 0x =1/10 , x = 1/10 PLEASE LIKE AND ACCEPT AS BEST ANSWER (10x-1) (10x-1) =0x=1/10 is answer |
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| 9. |
DATE: I1SheetalPege NoDete4Pio value |
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| 10. |
Dete4. |
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Answer» Tan(pi/4 + x) + tan(pi/4 - x) =(1+tan x)/(1-tan x)+ (1- tan x)/(1+tan x) =[(1+tan x)^2 +(1-tan x)^2]/[(1-tan^2 x] =2(1+tan^2 x)/(1-tan^2 x) =2(cos^2 x+sin^2 x)/(cos^2x-sin^2 x) =2/cos2x = 2sec2x |
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| 11. |
8. In a AABC, ADL BC and BD-3CD. Prove that 2AB2AC+ BC |
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Answer» Given that in ΔABC, we haveAD ⊥BC and BD = 3CDIn right angle triangles ADB and ADC, we haveAB°2 = AD^2 + BD^2 ...(i)AC^2 = AD^2 + DC^2 ...(ii) [By Pythagoras theorem]Subtracting equation (ii) from equation (i), we getAB^2 - AC^2 = BD^2 - DC^2= 9CD^2 - CD^2 [∴ BD = 3CD] = 9CD2 = 8(BC/4)^2 [Since, BC = DB + CD = 3CD + CD = 4CD]Therefore, AB^2 - AC^2 = BC^2/2⇒ 2(AB^2 - AC^2) = BC^2⇒ 2AB^2 - 2AC^2 = BC^2∴ 2AB^2 = 2AC^2 + BC^2 thanks |
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| 12. |
11. In figure zBAC = 90°, ADL BC.Prove that AB+CD2-BD+A2 |
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| 13. |
# AB = AC i Adl szfia त्रिधेश 880 हे, तो |
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Answer» AB=AC so angle B and C are equal.so A+B+C=180⁰=>90⁰+B+C=90⁰=>B+C=90⁰=>2B=90⁰ (B=C)=>B=C=45⁰ |
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| 14. |
l 10999992. Find the common factors of 20 anszG3 ind (7)-8-(26)nd the common factors of 20 and 28h oo |
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Answer» Q2. Common factor of 20 and 28 20 = 1 × 2² × 5 28 = 1 × 2² × 7 Common factors are 1, 2 , 4 Q3. (-7) -8 - (-25) , beacuse - × - = + -7 - 8 + 25 -15 + 25 10 Hit like and BeScholr |
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| 15. |
8x, 24 common factors |
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Answer» 8x = 2³ × x24 = 2³ × 3 Common factor are1,2,4,8 |
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| 16. |
Find the common factors:(i) 12x, 36 |
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| 17. |
17.InÎABC, ADzBCprovethat AB2+CD2=BD2+AC2 |
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Answer» ABD is a right angled traingle and AB is the hypotaneous (AB)^2 = (AD)^2 + (BD)^2 (AB)^2 - (BD)^2 = (AD)^2 ---------(i) now,ACD is also a right angled traingle (AC)^2 = (AD)^2 + (DC)^2 (AC)^2 - (DC)^2 = (AD)^2 ---(ii) (i) = (ii) (AB)^2 - (BD)^2 =(AC)^2 - (DC)^2(AB)^2 +(DC)^2 = (AC)^2 + (BD)^2 |
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| 18. |
If in ΔABC, AD is median and AE BC, then prove thatAB2 +AC2 2AD2+1/2 BC2, |
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| 19. |
7. Dheeraj has twice as many sisters as he hasbrothers. If Deepa, Dheeraj's sister has the samenumber of brothers as she has sisters, then Deepahas how many brothers? |
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Answer» SMTP Subho Chung vino beso CDL deepa has 2brothers. .... deepa has 1 brother is answer condition 1 deepa has 1 brother 1 sister equal satisies condition 2 dheeraj has 2 sisters and 1 brother ( here deepa is included) 1 brother is the right answer. 1 brother is the best answer 1 is the correct answer of the given question |
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| 20. |
In the given figure < ACB = 90°andCD LAB Prove thatBC2 BDAC2 AD |
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| 21. |
18.>In Fig. 3, ADL BC. Prove thatAB2 + CD2-BD2 + AC2. |
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Answer» Like if you find it useful |
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| 22. |
In figure ZBAC-90, ADL BCProve that AB2+CD2-BD?+AC |
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| 23. |
241CD3. In the given figure, AD LBC and BD =Prove that: 2CA2-2AB2+ BC2 |
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| 24. |
7. Dheeraj has twice as many sisters as he hasbrothers. If Deepa, Dheeraj's sister has the sâmenumber of brothers as she has sisters, then Deepahas how many brothers? |
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Answer» There are 3 brothers and 4 sisters in total. s=sister, b=brother Dheeraj has twice as many sisters as he has brothers2(b-1)=s...............A Dheeraj 's sister has the same number of brothers as she has sisterss-1=b.....................B Solving equation A and B, we get, s=4, b=3 |
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| 25. |
\left(\sin ^{4} \theta-\cos 40+1\right) \cos \theta |
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| 26. |
4. After coveringof my journey, I find that 15 km is still left. How much distance is the total journey? |
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| 27. |
7 UL.of my journey, I find that I have covered 16 km. How much journey20. After goingstill left? |
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Answer» aftwr going 3 sjhs high speed 62 by the hours |
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| 28. |
2. Bharat covered a distance of 210 km. He found that two-fifths of his journey was stillleft. Find the total length of the journey. |
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| 29. |
A boat whose speed in 15 km/hr in stillwater goes 30 km downstream andcomes back in a total of 4 hours 30minutes. The speed of the stream (inkm/hr) is: |
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Answer» thanks |
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| 30. |
5.In the figure, AD is a median of triangle ABCand AM I BC. Prove that(i) AC2 = AD2 + BC . DM-(i) AB2- AD2 - BC.DM(i) AC2+ AB2 -2AD2+ Bc2222DM D1i |
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Answer» ABC is an isosceles right triangle, right -angled at C. prove that AB²=2AB² |
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| 31. |
BC2 = AB2 + AC2 |
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| 32. |
32CA2 =AB2 + BC2 |
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| 33. |
0 Prove that |
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| 34. |
[0+1 4 40 COS OU(e) Prove that (tapet ) 2 cosec = tan+cot |
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| 35. |
25.co0s52°(A) 3(0) 2का मान हैं-:.(8) 14L ml e st दे TR T |
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Answer» 2 sin(38°)--------------- cos(58°) 2 × sin(90- 58°)------------------------- cos(58°) 2 × cos(58°)------------------- cos(58°) 2 Option (c) is correct |
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| 36. |
S eRl st 14 204 ाच,1+sinA 1+ sec A |
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Answer» cot²A(secA-1/1+sinA)+sec²A(sinA-1/1+secA) =cot²A[(secA-1)(1-sinA)/(1+sinA)(1-sinA)]+sec²A[(sinA-1)(secA-1)/(secA+1)(secA-1)] =cot²A[(secA-secA.sinA-1+sinA)/(1-sin²A)]+sec²A[(sinA.secA-sinA-secA+1)/(sec²A-1)] =cot²A[(secA-secA.sinA-1+sinA)/cos²A]+sec²A[(sinA.secA-sinA-secA+1)/tan²A] =(cos²A/sin²A)[(secA-secA.sinA-1+sinA)/cos²A]+[sec²A(sinA.secA-sinA-secA+1)](cos²A/sin²A) =[(secA-secA.sinA-1+sinA)/sin²A]+[(sinA.secA-sinA-secA+1)/sin²A]=(secA-secA.sinA-1+sinA+sinA.secA-sinA-secA+1)/sin²A =0/sin²A =0 R.H.S=L.H.S Hence proved… |
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| 37. |
才17) if(5xt 1) + E -1, then what is the value of x ?st oar what is the |
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| 38. |
2asquareb-9abc |
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Answer» 2a^2b-9abcab(2a-9c) |
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| 39. |
6. Find the product of (abc) (9abc) (-4abcf) and verify the result forC = 1. |
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Answer» A = 1/2, b=-1; c=1; ( a^2bc^2)(9ab^2c^2)(-4ab^2c^4);; (a^2bc^2)=(1/2)^2(-1)(1)^2 = -1/4_______________(1), (9ab^2c^2)= (9(1/2)(-1)^2(1)^2=-9/2;_(2) (-4ab^2c^4)= (-4(1/2)(-1)^2(1)^3= -2/4=1/-2; (-1/4)(-9/2)(-1/2)= (-1/4)(-9/2)(-1/2) = (-1/4)(18/4) = (-18/4)= (-9/2) |
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| 40. |
in0—2sin’ 0Q. 14 Prove that sec’ 0 — | -2cos*0—cos? 0 |
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| 41. |
1.Bob has 36 candy bars. He eats 29.What does he have now? |
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Answer» He has 7 candies now. Explanation:36 - 29 = 7 Given : BOB HAS 36 CANDY BARS : BOB ATE 29 SO, IF WE WILL SUBTRACT TOTAL NUMBERS OF BAR FROM HE ATE SO WE CAN GET REMAINING CANDY BAR 36-29 7 THEREFORE NOW HE HAVE ONLY 7 CANDY BARS |
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| 42. |
A open box is made of wood 3 cm thick. Its external length,breadth and height are 1.48 m, 1.16 m and 8.3 dm. Findthe cost of painting the inner surface at 50 per squametre. |
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| 43. |
The speed of a boat in still water is15 km/hr. It can go 30 km upstream and returndownstream to the original point in 4 hours30 minutes. Find the speed of the stream. |
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| 44. |
2.A box of candy costs 35. Each box has 5 pieces of candy.What is the cost of 20 pieces of candy? |
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Answer» 700 is the right answer of the given question 700 is the right answer 140 is the correct answer to this question. cost of 1 box = 35 rs. number of candy in each box = 5 Cost of each candy = 35/5 = 7 rs. cost of 20 candy = 20 ✖ 7 = 140 rs.. (ANS) price of 1 candy = 35/5 =7₹ then price of 20 = 20 x 7 = 140 ₹ 140 rs. is the right answer. Are You Send me You D.P The right ans is Rupees 140 coast of 20 pieces candy(🍬)=140 is correct answer cost of 20 pieces of candy =120₹ |
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| 45. |
(i) compoĂźnueu lllMaria invested 8,000 in a business. She would be paid intecompounded annually. Find0 The amount credited against her name at the end of the seco(i) The interest for the 3rd year7.rest at 50yea |
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| 46. |
6. A bag contains lemon flavoured candies only. Malini takes out one candy withoutlooking into the bag. What is the probability that she takes out(0 an orange flavoured candy?(i) a lemon flavoured candy? |
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| 47. |
3abc from -9abc |
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Answer» subtract -9abc -(3abc)= -12abc |
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| 48. |
21. Prove that:-3abc =-(a + b + c)(a-b |
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| 49. |
a+b+c=0 prove that a³+b³+c³=3abc |
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| 50. |
b+c a-b a= 3abc-a"-a +b c -a c |
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Answer» this is first part of soln second part of soln |
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