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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. |
if hcf of 60 and 96 is 12 ,than their LCM will be |
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Answer» we know the formulaLCM × HCF = product of numbersLCM × 12 = 60 × 96 LCM = (60 ×96)/12 LCM = 480 60 |
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| 2. |
Find :2(iii) 12% of 800(iv) 63% of 87(i) 70% of 500(v) 7.5% of 60 kg(ii) 0.25% of 96(vi) 9% of 6 litres |
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| 3. |
11. If the cost of each packet of biscuits is 23.85, find the cost of 16 such packets |
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Answer» Cost of one packet of biscuit is ₹ 23.85 Cost of 16 packets = ₹ 23.85 × 16=₹381.6 |
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| 4. |
3+62= |
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Answer» 3+62 = 65 (basic additon) |
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| 5. |
बहुपद 62 -3 - 7x का शू |
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Answer» x=-1/3 or x=3/2 is the right answer |
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| 6. |
3 [| .. 62 |
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| 7. |
С(2+3xeX 62°3 |
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Answer» 2 + 3x = 62° (alternate opposite angles)3x = 62-23x = 60x = 60/3 x = 3 (2+3x)=62; . 3x=62-2=60; X=60/3=20 20° is the right answer. 2+3x=623x=62-23x=60x=60/3x=20 degrees correct answer is x = 20 please like and accept as best answer 20 is the best answer 2 + 3x = 62° ( opposite angles )3x = 62 - 23x = 60x = 60/3x = 20 |
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| 8. |
Find the measure of angle 'a' in each figure. Give reason in each case.6020996(iv) |
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Answer» 1) 90+a= 180a= 902) a= 433) a+96+209= 360a= 360-305= 55 Hope it is helpful |
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| 9. |
EXERCISE 14.3A survey conducted by an organisation for the cause of illness and deaththe women between the ages 15- 44 (in years) worldwide, found the follonfigures (in %):ingFemale fatality rate (%)S.No. Causes1. Reproductive health conditions2. Neuropsychiatric conditions3. Injuries4. Cardiovascular conditions5.6. Other causes31.825.412.44.34.122.0Respiratory conditions0) Represent the information given above graphically.(i) Which condition is the major cause of women's ill health and death worl(ii) Try to find out, with the help of your teacher, any two factors which play arole in the cause in (ii) above being the major cause. |
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| 10. |
EXERCISE 14.3A survey conducted by an organisation for the cause of illness and death amonthe women between the ages 15-44 (in years) worldwide, found the followinfigures (in %):1.Female fatality rate (%)31.82541244.34.122.0S.No. CausesReproductive health conditions2 Neuropsychiatric conditions3. Injuries4. Cardiovascular conditions5.6. | Other causesRespiratory conditions(i) Represent the information given above graphically.(i) Which condition is the major cause of women's ill health and death worldwide?(ii) Try to find out, with the help of your teacher, any two factors which play a majorrole in the cause in (ii) above being the major cause |
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| 11. |
200:9x+2220 |
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Answer» x=2 correct answer please like my answer x=2 is right answer.. |
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| 12. |
11. Rohit and Roshan travelled a distance by car at 62 km anhour and by scooter at the rate of 20 metres per secondrespectively. The ratio of their speed is(a) 31:33(b) 31:35(c) 31:10(d) 31: 36 |
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Answer» zddjdjdhdhdjdjdkekdkdjfhdhdhd d) is the correct answer option d is the right answer option d is correct ratio 31:36 option d is the right answer Option D is tbe right answer Option d is right answer a is the correct answer b) is the correct answer a) 31:33 is right answer option d is the right answer option d is correct answer option d is correct answer. 62 km/hr = 62000 m /3600 seconds = 155/9.(m/sec) So the ratio = 155/(9x20) = 31/36 Option D is the answer |
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| 13. |
In a office, there are 65 employees. The number of female employees is 5 more than 1/5th of male employees. find the number of female employees in the office. |
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Answer» M + F = 65 F = M/5 + 5 M/5 + 5 + M = 65 6M/5 = 60 M = 50 F = 15 Number of female employees = 15 If you find this answer helpful then like it. |
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| 14. |
31. A rhombus sheet whose perimeter31.is 40 m and onc diagonal is 12 mlong is painted on both sides at therate of 5 per m. The total cost ofpainting is(1) P 480(2) P960(3) 360(4) P 720 |
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Answer» just see the unitary method used |
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| 15. |
Find the value of x in the following case: |
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| 16. |
A 500 gm packet of tea is sold for R 80 and toothbrush costing 11 is given free with it. The shopkeepthus makes a profit of 15% on it. What did the shopkeeper pay for it? |
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| 17. |
Find the values of n and X in following case:2(x-12)-10 and (x, -3)-62-1i-1 |
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| 18. |
Solve the following equations and check your answer in each case |
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| 19. |
4. Show that each of the following is meaningless. Give reason in each case.(1) VC(1) IL(iii) VVII(iv) IXX |
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Answer» (i) VC is wrong because V, L and D are never subtracted.(ii) IL is wrong because I can be subtracted from V and X only.(iii) VVII is wrong because V, L and D are never repeated.(iv) IXX is wrong because X (ten) must be placed before IX (nine). |
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| 20. |
From the following find selling or cost price as the case may be-(a) CP= 850, Profit= 25(b) SP= 12,250 Loss =1005 |
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| 21. |
A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the samebatch is packed using 20 bottles in each box, how many boxes would be filled?7. |
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| 22. |
9. A batch of bottles were packed in 30 boxes with 10 bottles in each box. If the samebatch is packed using 12 bottles in each box, how many boxes would be filled? |
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| 23. |
(4) If we put 8 biscuits in one packet, how many packets canbe made from 800 biscuits?Ans. |
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Answer» 8 divided by 800 equals to 100 one packet to 8 800biscuits required to 800 divided by 8 equal to 100 |
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| 24. |
cost of 16 cartons each having 24 biscuits packets ishaving 25 packets of biscuits?1536. What is theCost |
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Answer» (16*24) packets cost 1536=> 384 packets cost Rs. 1536=> 1 packet costs Rs. (1536/384) = Rs. 4 COST OF 25 PACKETS = 4*25 = Rs. 100 PLEASE HIT THE LIKE BUTTON Thnku sir |
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| 25. |
A chocolate box costs 2220.85. A box of biscuits is 30.40 cheaper thanthe box of chocolates. How much do the box of chocolates and the box ofbiscuits cost altogether? |
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Answer» cost of biscuits box=220.85-30.40=190.45total cost =220.85+190.45=411.30 Answer 6713.84₹220.85*₹30.40=6713.84 |
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| 26. |
5. The value of (243)'\5 isA. 3B. 0-3C. 5D. 1 |
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| 27. |
\operatorname { lim } _ { x \rightarrow 3 } \frac { x ^ { 5 } - 243 } { x ^ { 2 } - 9 } |
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| 28. |
d. 98, 120h. TILg. 40, 64Find the HCF of the following using long division method.256, 182, 100j. 1000, 625, 375, Kd.c. 216, 124b. 90, 110hg. 60, 8422. 6475, 250f119, 136 |
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Answer» the HCF of the numbers 64 and 92 is 4 Euclid vibhajan algorithm,b.90,110ANS.a=bq+r 110=90×1+20 90=20×40+10 20=10×2+0isiliye HCF=10 answer. the hcf of the numbers 64 and 92 is 4 hcf of 92 and 64 is 4 HCFa. 4b. 10c. 4f. 29g. 24 90)110(1 -9020)90(4 8010)20(2 20 The HCFof the number 64 and 92 is 4 please give only one sum beause i am confunsed which is a is correct anser is 4 the hcf of the number 64 and 92 is 4 HCF a. 4 and b.10 c.is 4 f 29 is and 24 c answer right sahi hoga HCF of the numbers 64,92is4 90,110 hcf ans- 10 hcf"=4 Last answer 92/64thats remaidar is2864/28thats remaindar is828/8thays remaindar is 48/4thats remaindar is 24/2 and remaindar is 2 and ans is 2 hcf of 92and 64 is 192 90and 110 is9900 110=90*1+2090=20*40+10 20=10*2+0HCF=10 Find the hcf of the following using long division method 60,84 |
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| 29. |
Mrs Joshi has organised a picnic for all the members of her society. 578 memberscame for the picnic. She orders biscuits, sweets and chips for the members.(a) There are 389 packets of biscuits in a carton. How many more packetsare needed if each member has to be given one packet? |
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Answer» 578-389=189189 packets she need 189 packets she need 189 is the correct answer to this question total no of members 578total no biscuits packet 389therefore we subtract 578 - 389 = 189need of more biscuit packets = 189 total no of members 578total no of biscuit packet 389therefore we subtract 578 - 389 =189need of more biscuit packets = 189 |
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| 30. |
find the product of hcf and LCM of two no is 119. find the two no's if none of them is 1 |
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Answer» Product of two numbers = Product OF HCF AND LCM=> A*B = 119=> 1*B = 119=> B = 119 PLEASE HIT THE LIKE BUTTON then why answer is 7 and 17 |
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| 31. |
Find the HCF of 68 and 119. |
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| 32. |
two pol12. Two circles of radii 10 cm and 8 cmintersect each other, and the length ofthe common chord is 12 cm. Find thedistance between their centres.DI O' |
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| 33. |
34. Two numbers'a' and 'W are selected sucessively witbonut replacementin o. Anelemdefiritifrom the integers 1 to 10, The probability that i is an integn,17(di |
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Answer» The total number of possibilities is 10∗9 = 90 since we are choosing two numbers without replacement and order matters. Hence the cases are (2,1),(3,1),(4,1),(5,1),(6,1),(7,1),(8,1),(9,1)and(10,1) (4,2),(6,2),(8,2),(10,2) (6,3),(9,3) (8,4) (10,5) For all b greater than 5 the smallest multiple is greater than 10.Hence they dont contribute to the total. Hence a total of 9+4+2+1+1=17 valid possibilities. Hence the probability is 17/90 |
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| 34. |
12. Two circles of radii 10 cm and 8 cmintersect each other, and the length ofthe common chord is 12 cm. Find thedistance between their centres.DI O |
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Answer» no |
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| 35. |
2 m cloth is required to stitch a shirt,find out how many shirts can be madefrom 13 m of cloth |
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Answer» 11/4 meter cloth = 1 shirt1 meter = 4/11 shirt55/4 meter = 4/11 * 55/4 5 shirts 2(3/4)=11/4 meter cloth, 1 meter = 4/11 shirt, 55/4 meter = 4/11 × 55/4= 5 shirts answer for this question is 5 .... 5 is a correct answer |
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| 36. |
1. If 25 men can weave 120 metres of clothin a day, how many metres of cloth can bewoven by 30 men in a day? |
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| 37. |
2× 243 |
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Answer» 486 is the answer for your question. |
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| 38. |
Ny ाएएएए/5 (y)‘\’gcbséerfl’,m‘ Zal L&D DDMID काफी कंकहें) 1167ंठा1 |कब्ठ)ला11 11 16611(9ककी के क(फक(ऊ कर: थी |(i) 5v3 +18v3 — 243 G 435 + 235 — %5 o(ii)) 3775 + 5v48 — 243 (iv) 5340 + 230625 — 33320 छा! i |
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| 39. |
If, (243) +1 (243)-5 then find the value of |
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| 40. |
2in243 |
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Answer» thanks |
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| 41. |
S,m243 |
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Answer» l |
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| 42. |
(ii) cube root 243 |
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Answer» 243 = 3*3*3*3*3 = 3 (cube root of 9) |
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| 43. |
\log _ { 9 } 243 = x |
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Answer» thx |
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| 44. |
Find the length of the height of an equilateral triangle of side 24 cm. |
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| 45. |
Calculate the side of a rhombus if its diagonals are 18 cm and 24 cm. |
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| 46. |
8.Calculate the side of a rhombus if its diagonals are 18 cm and 24 cm |
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| 47. |
लि | taad o equadfonfi”,,,io,) =3 HI 8) s -% s ST |
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Answer» 3x^2 + x -4 =03x^2 +4x -3x -4=0x(3x+4)-1(3x+4)=0(x-1)(3x+4)=0x=1,-4/3 |
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| 48. |
EXERCISE 2.11. The graphs of y = p(x) are given in Fig. 2.10 below.number of zeroes of p(x), in each case.below, for some polynomials plo), Findmethodwhen(iii)(I(iv)(vi) |
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| 49. |
26. A sheet of paper in the form of a rectangle ABCD in which AB40cm and AD- 28 cm. Asemi-circular portion with BC as diameter iscut off. Find the area of the remaining portion.Or |
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| 50. |
2In an equilateral triangle of side 24 cm, a circle is inscribedtouching its sides. Find the area of the remaining portionof the triangle. |
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