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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 the value of a, b, c and d from the equation:a-b2a + c2a -b 3c+d 0 13 |
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Answer» Thanks |
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| 2. |
triangle formed by the points A (2a, 4a), B (2a, 6a) .(2a + v3a, 5a) isight angledquilateralC(b) isosceles(d) of these |
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| 3. |
the sum of the first 10 term of the AP 2,7,12,............is |
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| 4. |
19. If the sum of the first 14 teruus of an AP is 1050 aud its first term is 10, find the 20th term. |
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Answer» A=10...given S14=1050⇒n/2[2a+(n-1)d]=1050⇒14/2[20+13d]=1050⇒7[20+13d]=1050⇒20+13d=1050/7⇒20+13d=150⇒13d=150-20⇒13d=130⇒d=10 20th term=a+19d =10+19×10 =10+190 =200...Answer |
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| 5. |
S.P. = 121gain = 10 |
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Answer» SP=121 gain=10 so CP=121-10=111 |
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| 6. |
Find the CP if:(a) SP-924, gain-10% |
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| 7. |
findradian7° 21200225 CM330 2 |
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Answer» 1) π/42) π/33) 5π/1π4) 2π/35) 3π/46)5π/37) 11π/6 |
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| 8. |
, 10 Vichy bought a bicicle for Rs.3.000100 land sold it for Rs. 2,700.%.| What was his loss or gain percentgB 10% Lossby 10%.comFlyten's gain d) 11.1 17.15 |
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Answer» correct answer is 10% loss cost price=3000selling price =2700loss = cost price - selling price = 3000-2700 =300profit percent = profit/cost price ×100=300/3000×100=10%there is 10% loss in this transactionoption a is correct answer |
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| 9. |
Ex.2: Find the area of the sector of circlewhich subtends an angle of 1200 at the centre, ifthe radius of the circle is 6 cm. |
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Answer» Area= theeta/ 360*(πr^2)= 120/360*(πx36)1/3*πx36= 12π12*3.1437.68cm^2please like the solution 👍 ✔️👍 |
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| 10. |
Example 12: If the sum of the first 14 terms of an AP is 1050 and its first term is 10,find the 20th term. |
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| 11. |
3. Find four solutions for equation x+2y=6. |
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Answer» 3. x + 2y = 6Solutions arex = 6, y =0x = 0, y = 3x = 2 , y = 2x = 3 , y = 3/2 |
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| 12. |
) a unique solution, (ii) only 1WU 30Write four solutions for each of the following equations:2.(ii) ur +y111) χ=3.Check which of the following are solutions of the equation |
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| 13. |
135. Two supplementary angles are in the ratio 3 : 2. Thesmaller angle measures(a) 108°(c) 72°(c) 81°(d) of these |
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| 14. |
Find the four solutions of 5x-y=10 |
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Answer» Solutions can bex=0then5(0)-y=10y=-10(0,-10)y=05(X)=10x=2(2,0)x=15-y=10y=-5(1,-5)y=15x-1=10x=11/5(11/5,1) 1.x=3y=52.x=0y= -10 |
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| 15. |
EXERCISE 4.2Which one of the following options is true, and why?y=3x + 5 has1.o) aunique solution. (i) onl two solutions, (ii)infiminfinitely many solutiWrite four solutions for each of the following equations(i) 2x y 7 |
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| 16. |
Write four solutions for each of the following equations:(i) 2xty 7(ii) Ttxty9 |
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| 17. |
10. By selling afan for1200, Karin loses200. At what price must he sell it to gain 10%? |
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| 18. |
2010フ443/49. The fourth term of a G.P. is 27 and the 7th term is 729, find the G.P |
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| 19. |
By selling a fan for ₹1200, Karim loses ₹ 200. At what price must he sell it to gain 10%? |
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| 20. |
2x-533. If27, then x = ?(1) 9(2) 86(3) 43(4) 38 |
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Answer» Ans :- Option (3) is correct. |
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| 21. |
By selling a fan for1200, Arun loses? 200. At what price must he sell it to gain 10% |
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| 22. |
The volume of a cuboid is 1200 cm'. The length is 15 cm. and breadth is 10 cm. Find its height. |
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Answer» h=8 CM is the correct answer |
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| 23. |
In an AP of 50 term the sum of first 10 term is 210 and the sum of its 15 term is 2565 .Find the AP |
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Answer» Consider a and d as the first term and the common difference of an A.P. respectively. n th term of an A.P., an= a + ( n – 1)d Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d] Given that the sum of the first 10 terms is 210. ⇒ 10 / 2 [2a + 9d ] = 210 ⇒ 5[ 2a + 9 d ] = 210 ⇒2a + 9d = 42 ----------- (1) 15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning ⇒ a36= a + 35d Sum of the last 15 terms = 15/2 [2a36+ ( 15 – 1)d ] = 2565 ⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565 ⇒ 15 [ a + 35d + 7d ] = 2565 ⇒a + 42d = 171 ----------(2) From (1) and (2), we have d = 4 and a = 3. Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199. Recommend(5) Comment(0) more_horiz person Syeda, SubjectMatterExpert Member since Jan 25 2017 Answer. Consider a and d as the first term and the common difference of an A.P. respectively. n th term of an A.P., an= a + ( n – 1)d Sum of n terms of an A.P., S n = n/ 2 [2a + (n – 1)d] Given that the sum of the first 10 terms is 210. ⇒ 10 / 2 [2a + 9d ] = 210 ⇒ 5[ 2a + 9 d ] = 210 ⇒2a + 9d = 42 ----------- (1) 15 th term from the last = ( 50 – 15 + 1 ) th = 36 th term from the beginning ⇒ a36= a + 35d Sum of the last 15 terms = 15/2 [2a36+ ( 15 – 1)d ] = 2565 ⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565 ⇒ 15 [ a + 35d + 7d ] = 2565 ⇒a + 42d = 171 ----------(2) From (1) and (2), we have d = 4 and a = 3. Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199. |
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| 24. |
Write four solutions for each of thefollowing equations.(i) 2x + y = 7iii) x=4y(ii) nx + y =9 |
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| 25. |
EXERCISE 4.3ing equations:(b) 5t + 28 = 10।==(0x=3 ) 7m419(j) 40 5=3ing equations:2 (b) 3(n-5) = 21=8 (e) 4(2-X) =8ving equations: |
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Answer» I don't know what the answer is b.5t=10-28 5t=-18 t =-18/5 |
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| 26. |
Write four solutions for each of the following equations:(i)2.2x+y =7(ii) TX+y-9(iii) x=4y |
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| 27. |
b) Solve the followinging inequation and write the solution set:13x-5 < 15x +4 <7x +12, x E RRepresent the solution on a real number line. |
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Answer» can you plz explain me?? |
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| 28. |
1) A purse contains 25 paise and 10 paise coins. The total amount in the purse is RS 8.25 if the number of 25paise coins is one third of the number of 10 paise coins in the purse, find the number of coins of eachtype. |
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Answer» Let 10 paise coins=x25 paise coins=1/3*xSo,10x+25*x/3=8.7530x+25x/3=8.7555x=8.75*3=26.25X=26.25/55=0.5So 10 paise coins=.5*10=525 paise coins=.5*25=12 |
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| 29. |
Exercise 2Fing using DMASb. 12+3+3 |
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Answer» Twelve+three+three= 18 the answer of your question is 7 because by using bodmas rule 12 ÷ 3 + 3 so first ÷ (12 ÷3)+34+3 = 7 |
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| 30. |
oten us of a right triangle is 3ă5 cm. If the smaller side is tripled and the larger sideis doubled, the new hypotenuse will be 15cm. Find the length of each side.QR |
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Answer» Like my answer if you find it useful! |
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| 31. |
18.The purse of Sunita contains fifty paise and one rupee coins in the ratio 3:2.The total amount in the purse is Rs. 70. Find the no of 50 paise coins and Rs1 coins. |
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Answer» 3x+2x=705x=70X=14no of 50 paise coin =3*14=42no of 1 rupee coins= 2*14=28 |
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| 32. |
6x3-18otenus of a right triangle is 3Г5 cm. ll the smaller side is tripled and the larger sideThe hypis doubled, the new hypotenuse will be 15cm. Find the length of cach side.OR |
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| 33. |
The hypotenus of a right triangle is 3「5 cm. If the smaller side is tripled and the larger siis doubled, the new hypotenuse will be 15cm. Find the length of each side. |
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| 34. |
The measure of two adjacent sides of arectangular field are in the ratio 17:7. If thesmaller side measures 3.5 m, find the perimeterof the field? |
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Answer» Let the length of Reactangle = 17xLet breadth of Reactangle = 7x Given,7x = 3.5x = 3.5/7 =.5 m Then,Length = 17*.5 = 8.5 mBreadth = 7x = 3.5 m Perimeter of Reactangle = 2(length + breadth)= 2(8.5 + 3.5)= 2(12)= 24 m Perimeter of Reactangle is 24 m |
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| 35. |
चित्र मेंका मान कितना होगा?| 120°12001200120060°(A) 120°(B) 60°(C) 360°(D) 300° |
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Answer» As we know that on a quadrilateral there is the sum of four sides which is 360degrees and on this figure the answer will be 120degrees Chaturbhuj ke Charo kodo ka jod = 360 isliye 120° + 120° + 60° + y = 360° [ chauthe kon ko y maan liya ] Ab yha se y = 60° aayega Ab Ab ek Rekha m kono ka jod = 180° to X° + 60° = 180° yha se X = 120° the correct answer is c 120 is correct answer 120 is the right answer . answer (b) is correct answer (c) is correct A is the correct answer 120 degree is correct answer 120 degree is the right answer a is the right answer.... sum of all internal angle of a quadrilateral = 360°120+120+60+4th angle=3604th angle= 360-300=60°the extended line is a st linex+60=180x= 180-60=120° The right answer is 120 |
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| 36. |
Divi d.e) 6x7-8x6 + 4x5-10x4 + 6x3 by-2x3 |
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| 37. |
, 8x241: 6x4-32; 8x6-43 mnw 4x8nauo27 |
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Answer» 24 is the correct answer |
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| 38. |
( 1200/x + 2) (x-10) - 1200 = 60 |
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Answer» thank u so much to u |
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| 39. |
1200+21(x-10)-1200 = 60. |
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| 40. |
422 +442 +462+ 802市ㄒ牙付5παR |
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| 41. |
(b) 7(d) 2he sum of the series:452-432+442-422+432-412+422-402+till 15 term(a) 1982(b) 2592(c) 2756 |
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Answer» use a^2-b^2 formula or many other method exist, your answer will be (b) 2592 please let me know I will tell you other tactics. like doing its using AP formula |
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| 42. |
50. Simplify : a) (4 +/2) (5-ง2 V2)b)Ns1. In the fig EFB - 60° & DGH 100 State whetherEFB-60" &51. In the fig,DGH- |
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Answer» No AB || CD because corresponding angle are not equal as EFB = 60° ECD = 80° |
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| 43. |
Given a 37.58 b5.56and c 422 findPIf2Cl |
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| 44. |
If the polynomials az +422 +3z-4 and z -4z aleave the same remainder when divided by z 3.Find the value of aINCERT Exemplar) |
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| 45. |
12. In the Fig. 38,ABİİCD:LDCE>60° and<ACD-40Find x, y andz.Fig. 38 |
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| 46. |
14. In Fig. 40 if 22 120 and 5 60° show that mlln.265Fig. 40 |
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Answer» angle 6 + angle 5 = 180°angle 6 = 180° - 60° = 120°as angle 2 = angle 6 which are corresponding angle so m||n |
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| 47. |
Example 5: Observe Fig. 6.30 and then find P6137.63.6012Fig. 6.30 |
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| 48. |
A plot is in the form of a rectangle ABCD having semi-circle on BC as shown inFig. 20.23. If AB 60 m and BC -28 m, find the area of the plot.28 m60 mFig. 20.23 |
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| 49. |
nd the number in repeotedmutiplicalion fromА, 14аC. 72"1. ExpaB. 25D, 22 |
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| 50. |
o A plot is in the form of a rectangle ABCD having semi-cirdle on BC as shown inFig. 20.23. If AB60 m and BC-28 m, find the area of the plot.28 m60 mFig. 20.23n its smaller side |
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Answer» full page |
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