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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. |
Two sides of a parallelogram are in the ratio 5: 3. If its perimeter is 64 cm, find the lengthsof its sides. |
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| 2. |
7. The length of the hypotenuse of a right-angledtriangle is 10 cm. If the lengths of the other twosides are in the ratio 3:4, find the length of these sides |
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Answer» suppose the length is x then the length of other two sides would be 3x and 4x using Pythagoras theorem100= 16x^2+9x^2so x^2= 100/25x= 2other sides would be 6 and 8cm |
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| 3. |
2. I cach sido ol a rianagle is doublod hen how many limos tho area ofriangle increased? |
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Answer» area of triangle increases four times. Let the original sides be a, b, c S=(a+b+c) /2 2S=a+b+c Now, Area=S√S(S-a) (s-b) (s-c) New sides are 2a, 2b, 2c New S=(2a+2b+2c) /2 New S=2(a+b+c) /2 New S=a+b+C But a+b+C=2(Original S) (proved above) So, New S=2(original S) So, New area = √New S(New S-2a) (NewS-2b) (newS-2c) Let original S=x New area = √2x(2x-2a)(2x-2b) (2x-2c) New area = √2x*2(x-a)2*(x-b)*2(x-c) New area=4×√(x-a) (x-b) (x-c) But x=original S So, new area = 4*new area |
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| 4. |
Aflooring tStantmed tohis the shape of a parallelogram whose base is 24esponding baght is 10 cm. How many such tilesrafoor of arca 1080 square meters? |
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| 5. |
number ol DA box contains 6 apples. How manyapples in all will seven boxes have? |
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Answer» tnx |
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| 6. |
4. A flooring tile has the shape of a parallelogram whose base is 24 cm and nefioor ofrequired you can split the ules in whatever way you want tofil upcor esponding height is 10cm. How many such tilcs are required to cover aarca 1080 m2? (Ifthe corners). |
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| 7. |
1108. If the length and breadth of arectangular plot are increased by 50%and 20% respectively, then how manytimes will its area be increased?(2) 1g(3) 11(4) 1 |
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| 8. |
120. sweets are distributed between A and B in the ratio 2:3. Find, how many does each get?c ol Ilum itoin 50 litres of netrol. |
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Answer» For A = 2/5*20= 8for B = 3/5*20= 12 |
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| 9. |
The length, breadth and height of a hall are 65cm, 5 dm and 0.15 m, respectively. Find thevolume of the hall in dm^3 |
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Answer» 5dm = 0.5 m65cm = 0.65mnow volume = lbh0.5*0.65*0.150.048m0.48dm^3 |
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| 10. |
32. Determine the longest tape which can be used to measure exactly the lengths 7 m,3 m 85 cm and 12 m 95 cm. |
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Answer» 7 m = 700 cm3 m 85 cm = 385 cm12 m 95 cm = 1295 cm The required length of the tape that can measure the lengths 700 cm, 385 cm and 1295 cm will be given bu the HCF of 700 cm, 385 cm and 1295 cm. Evaluating the HCF of 700, 385 and 1295 using prime factorisation method, we have: 700 = 2 × 2 × 5 × 5 × 7 = 22× 52× 7 385 = 5 × 11 × 7 1295 = 5 × 7 × 37 ∴HCF = 5 ×7 = 35 Hence, the longest tape which can measure the lengths 7 m, 3 m 85 cm and 12 m 95 cm exactly is of 35 cm. |
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| 11. |
Determine the integer values that the length of the third side of a triangle can have if other two sides halengths 3 cm and 7 cm. |
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| 12. |
3.A train is 1 km long. Average speed of the train is 60kmph.How long will it take to pass.(a) An electric post(b) A bridge of length 1 km. |
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Answer» Length of train = 1 kmSpeed of train = 60 km/hr Distance = speed * time 1) Time to pass electric post Distance is 1 km length of train 1 = 60*time time = 1/60 hr or 1 minute 2) Time to pass bridge of length 1km Distance is 2 km 2 = 60*time time = 2/60 hr or 2 minutes |
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| 13. |
What is the area of the triangle having sides of lengths 7 cm, 8 cm and 9 cm? |
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Answer» thanks |
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| 14. |
2 mL of sodium hydroxide solution is added to a few pieces of granulated zinc metaltaken in a test tube. When the contents are warmed, a gas evolves which is bubbledthrough a soap solution before testing. Write the equation of the chemical reactioninvolved and the test to detect the gas. Name the gas which will be evolved when thesame metal reacts with dilute solution of a strong acid.QOR |
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Answer» NAOH + Zn = Na2ZnO2 + H2 This hydrogen gas can be tested by putting matchstick over the solution. The burning matchstick will extinguish by a pop sound. Then, when zn metal reacts with a dilute strong acid such as hcl, it will give zinc chloride and hydrogen gas again. Hcl + Zn= ZnCl2 + H2 Like my answer if you find it useful! |
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| 15. |
A floral design on a floor is made up of 16 tileswhich are triangular, the sides o9 cm, 28 cm and 35 cm (see Fig. 12.18).he cost of polishing the tiles at the rateof 50p per cm2 |
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| 16. |
)A floral design on a floor is made up of 16 tiles which are triangular, the sides of thetriangle being 9 cm, 28 cm and 35 cm. Find the cost of polishing thetiles at the rate of 50p per cm2. |
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| 17. |
45. कोई राशि 5 वर्षों में 50% से बढ़ जाती है । तीन राशियों x, y तथा :को क्रमश: 10, 15 तथा 20 वर्षों के लिए उधार दी जाती है। यदि तीनोंराशियों पर समान मिश्रधन प्राप्त हुए हों, तो ४, y तथा : में क्या अनुपात| [SSC-GL(PT)] |
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Answer» what is your question |
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| 18. |
A floral design on a floor is made up of 16 tileswhich are triangular, the sides of the trianglebeing 9 cm, 28 cm and 35 cm (see Fig. 12.18)Find the cost of polishing the tiles at the rateof 50p per cm2. |
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| 19. |
A foral design on a floor is made up of 16 tileswhich are triangular, the sides of the triangleheing 9 cm, 28 cm and 35 cm (see Fig. 12.18)Find the cost of polishing the tiles at the ratof 50p per cm2. |
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| 20. |
हे ३) ९ YRIES Y1 o 1के... दर |
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Answer» 22/7×16.15×14+(22/7×16)=3.14×16.15×14+(3.14×16)=709.954+ 50.24=760.194 |
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| 21. |
A floral design on a floor is made up of 16 tileswhich are triangular, the sides of the trianglebeing 9 cm, 28 cm and 35 cm (see Fig. 8.18).Find the cost of polishing the tiles at the rateof 50p per cm28. |
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| 22. |
Find the sum of all two digit numbers which when divided by 4,yields 1 as remainder |
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| 23. |
>(sec—172 1-cosY1+QC0SAA - और(20179. _ 1+0059 ) |
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| 24. |
لالالالالالالالاr.SRvery shart Answers Type QuestionIf x=d7 and 5 od 7 lindg. |
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| 25. |
ndodespeively.What must be subtracted from(x" +2x-2r +2r+3).1 +4x + 6).Show that the resultis exactly divisible by |
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| 26. |
yields 1 as remainder.Find the 20th term from the last term of the A.P. 3, 8, 13,.., 253 |
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Answer» We have, l = Last term = 253 and, d = Common difference = 8 - 3 = 5 ∴ 20th term from the end = l - (20 - 1)d = l - 19d = 253 - 19 × 5 = 253 - 95 = 158 |
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| 27. |
251. Adding 4 to twice a number yields. Find the number. |
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Answer» 1÷12 is the number for given data 3/5 is the correct answer of the given question ☆ the answer to this question is 1/12 |
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| 28. |
,esExample 21: EvaluateSolution:169 × vEXERCISE 3.61. Find the square ro7,2251,089 |
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| 29. |
Adding 4 to twice a number yields. Find the number. |
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| 30. |
In the inL6.30the lines a and b parallel? Also, find 41, 42, 23, 44.50gles |
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Answer» ∠2 + 130° = 180°∠2 = 180° - 130° = 50°So, we see that corresponding angles are equal therefore a||b, ∠1 = ∠2 = 50° ( vertically opposite angle)∠3 = 130° ( alternate angles)∠4 = 130° ( corresponding angle) |
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| 31. |
EXERCISE 11.1The length and the breadth of a rectangular piece of land are 500 mrespectively. Find(i) its area1.(i)the cost of the land, if 1 m2 of the land costs t 100m |
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| 32. |
31./The internaland external diameters of a hollow hemispherical vesse are 42 cmi and45.5 em, respectivelyFind its capacity and also its outer curved surface area. |
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| 33. |
collection of good students in class |
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Answer» real life example of vertically opposite and linear pair no in class not good student are there but freature of the class Real life example of vertically opposite and linear pair |
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| 34. |
8. In the adjoining figure, OD is perpendicular tothe chord AB of a circle with centre O. If BC is a Adiameter, show that ACI DO and AC = 2 × OD |
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| 35. |
a) Every resident of a city can speak Hindi or English. If75% of the population speaks Hindi and 60% speaksEnglish, then what percentage can speak both thelanguages? |
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Answer» We consider total population as 100% Resident can speak Hindi = 75% Resident can speak English = 60% Then percentage of resident can speak both language= (75% + 60%) - 100%= 135% - 100= 35% |
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| 36. |
The population of a city doublesevery 6 years. If in 2005, thepopulation is 10.562, bywhich year will the populationincrease to 42,248(A) 2011(C) 201712.(B) 2023(D) 2029 |
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| 37. |
The average weight of 18 boys was calculated to be 40 kg. It was later discovered that oneweight was misread as 36 in place of correct weight 53. Calculate the correct mean. |
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Answer» Average weight of boys = Total weight of boys/ Number of boys Then,40 = Total weight of boys/ 18 Total weight of boys = 40*18= 720 But correct weight 53 was misread as 35 Hence,Actual total weight of boys= 720 - 35 + 53= 720 + 18 = 738 Therefore correct mean= 738/18= 41 kg |
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| 38. |
8. In the adjoining figure, OD is perpendicular tothe chord AB of a circle with centre O. If BC is adiameter, show that ACI DO and AC = 2 x OD.A- |
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| 39. |
Y1OD |
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| 40. |
ताज ODT डा |
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| 41. |
o07od |
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Answer» sum of all angles is 360.so x+120+130+50=360x+300=360so x=60 |
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| 42. |
a^4 - b^4 - 4*a^2*x %2B 4*x^2=0 |
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Answer» 4x^2-4a^2x+( a^4-b^4)=0; 4x^2-{2(a^2+b^2)+2(a^2-b^2)}x+(a^2-b^2)(a^2+b^2)=0, {2x-(a^2-b^2)}{2x-(a^2+ b^2)}=0; 2x=a^2-b^2 or 2x=a^2+b^2; x=a^2-b^2/2/// a^2+ b^2 |
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| 43. |
:x = 15 x 14 = 2110 |
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Answer» 15×14=210÷10=21Sahi hi to hai .. x= 15×14/10x= 21 correct answer for this question x=15×14/10x=210/10x=21HENCE PROVED X=14*15=210210/10 =21So, x=21 X=14 *15 =210210/10=21So,x=21 |
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| 44. |
\frac { 1 } { 2 } \text { of } \frac { 9 } { 10 } \times \frac { 14 } { 15 } + \frac { 2 } { 5 } |
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Answer» 1/2*(9*7/5*15)+2/51/2*63/75+2/563/150+2/563/150+2*30/5*3063+60/150123/15041/50 |
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| 45. |
40Becauseof the budget presented byYashwant Sinha, the price of sugar increasedby 40%. The Verma family reduced itsconsumption so that expenditure on sugar isup by 1 2%. If the total consumption ofsugarbefore the rise in price was 50 kg, what isthe present consumption of sugar (in kg)?(a) 48 kg(c) 36 kg(b 40 kg(d) 32 kg |
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Answer» E = P × CUsing successive rule,Let the c consumption due12= 40 + x + 40x/100x=-20%Thus consumption decreases by 20%Present consumption=(50)(0.8)=40 |
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| 46. |
दी गयी आकृति में, त्रिभुज ABC का अंत: केंद्र 0 है। यदिAO 5 Om CO 32 -7 = = तथा 6 = , है, तो BO/OF का मान क्या है ?ABEC |
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Answer» Solution Construction: Draw : AE ⊥ BC Draw; BF ⊥ AC Draw : DC ⊥ AB O is the intersection point . Given That: AO /OE = 5/4 and OC /OD = 3/2 Let BC = a, CA = b, and AB = c ∴AO/OE = b+c/a = OC/OD = b+c/c ∴ b+c/a = 5/4 ⇒b+c+a/a = 5+4/4 = 9/4 -----(1) and b+a/ c = 3/2 ∴ a+b+c/c = 3+2/2 = 5/2 --(2) on Dividing eq. (2) by (1) we get. a/c = 5/2 Χ 4/9 = 10/9 ⇒ a = 10k and c = 9k ⇒ b = 14k/4 = 7k/2 ∴ BO /OF = c+a / b = 10k+9k / 7k/2 = 19 Х 2/ 7 = 38/7 Hence BO / OF = 38/7 |
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| 47. |
12.The mean weight of thirty bags was calculated as 40kg. Later, on rechecking the calculation, it was observedthat the weight of 50 kg was misread as 32 kg and theweight of 25 kg was misread as 52 kgWhat is the correct mean weight? |
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| 48. |
Construct the following quadrilaterals) Quadrilateral DEARDE 4 cmEA-5 cmAR 4.5 cmLE 60°ZA 90° |
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| 49. |
5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost oftin-plating it on the inside at the rate of 16 per 100 cm2 |
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| 50. |
Construct the following quadrilaterals Quadrilateral DEARDE 4 cmEA 5 cmAR 4.5 cmLE 60ZA 90 |
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