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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. |
NumberPut commes to separate the penuunnumber namesNumbersNumbers with commas253178613275196b8203005d.e10532755083005389Write the numbers:a. Six hundred eighty-four thousand two hundred sixty-one. |
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Answer» 6,84,261 is the awnser according to indian system and 684,261 is according to international system ,please give me like |
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| 2. |
12. In a godown one-fifth of the acid is sulphuric and one-thirdhydrochloric; besides that is contains 15 dozen bottles of nitric and 30of carbolic: how much sulphuric and hydrochloric does it contain? |
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Answer» Let total bottles in godown be n Then,Bottles of sulphuric acid = n/5Bottles of hydrochloric acid = n/3Bottles of nitric acid = 5*12 = 60Bottles of carbolic = 30 As per given conditionn/5 + n/3 + 60 + 30 = nn - n/5 - n/3 = 90(15n - 3n - 5n) = 90*157n = 1350n = 1350/7 = 192.85 Then quantity of sulphuric acid= n/5 = 192.85/5 = 38.57 Quantity of hydrochloric acid= n/3 = 192.85/3 = 64.2 |
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| 3. |
J36a% + 124* |
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| 4. |
124: 7::?: 8 |
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Answer» 124:7::x:8so 124/7=x/8so x=(124×8)/7=141.71 thanks jii |
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| 5. |
124+5268 |
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Answer» right answer is is 5392 |
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| 6. |
23. The totalsurface area of a cube is 54 cm2. What is the length of its sides? |
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Answer» Total surface area of cube = 6 × side² So, 6 × side² = 54 side² = 9 Side = √(9) = 3 |
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| 7. |
The total surface area of a cube is 54 cm2. What is the length of its sides ? |
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| 8. |
The area of base of right circular cylinder is 54 cm2 and itsheight is 10 cm. Find its volume. |
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| 9. |
16. Maria and Tanya are reading the samebook. Tanya has read half of the book. Mariahas read less than Tanya. Which could be theamount that Maria has read?4A.B.. 1214610 |
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Answer» 4/9 < 6/12 < 8/14 < 6/10 6/12 = 1/2 Since Maria has read less than Tanya, the amount that Maria read is 4/9 A. 4/9 |
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| 10. |
2000=n÷2 {2×32+(n-1)4} |
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| 11. |
यदि Aऔर B दो ऐसे समुच्चय है कि n(A) = 32. n(B) = 28 तथा n(AOB) = 50| तो n(ADB) ज्ञात करें- |
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Answer» Given,n(A) = 32, n(B) = 28, n(AuB) = 50 Then, n(AnB) = n(A) + n(B) - n(AuB) = 32 + 28 - 50 = 60 - 50 = 10 |
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| 12. |
124*(text*(a*(d*n))) - 32 |
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Answer» 92 is the right ans...... 0000000000000000000000000000000000000000092 92 is the right answer Its half question ????? 92 is the correct answer 92 is the correct answer 92 is the right answer 92 is the correct answer. 92 is the right answer 92 is right answer........ correct answer is 92 |
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| 13. |
If each edge of cube is 1 by so%find en in surface area |
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Answer» Initial Surface Area,A1=6x2A1=6x2 The edge increased by 50%, so, the final edge length is,3x/23x/2 So, final Surface Area,A2=6(3x/2)2=6(9x2/4)=6x2(9/4)=9/4(A1)A2=6(3x/2)2=6(9x2/4)=6x2(9/4)=9/4(A1) So, Increase in Surface Area,dA=A2−A1=9/4(A1)−A1=A1(9/4−1)=A1(5/4)dA=A2−A1=9/4(A1)−A1=A1(9/4−1)=A1(5/4) So, the area increased by5/45/4 Now, percentage increase will be((5/4)∗100=1255/4)∗100=125% So, the percentage increase in the Surface Area of the cube will be 125%. |
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| 14. |
4 If A and B are two sets such that n(A)32n(B) 28 and n(A UB) 50, then n(An B) will b |
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| 15. |
If A and B are two sets suchn(AuB) 50,then n(An B) will bethatn(A)-32, n(B) = 28and |
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| 16. |
If each edge of cube is 1 by 50%find en in surface area. |
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| 17. |
5. The TSA of cube is 54 cm2. What is the length of its side? |
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Answer» Total surface area= 54 cm²∴ 6(a)²=54 [a means edge] (a)²=54/6 a=√9 a=3Hence, length of its side are 3 cm |
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| 18. |
Q8. A statue 1.5m tall stands on the top of a pedestal. From a point on a ground,the angle of elevation of the statue is 45째 and from the same point the angle ofelevation of the top of the pedestal is 30째. Find the height of the pedestal. |
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| 19. |
Q18.A statue 1.46m tall, stands on the pedestal. From a point on the ground, theangle of elevation of the top of the statue is 60° and from the same point, the anglkof elevation of the top of the pedestal is 450. Find the height of the pedestal. |
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| 20. |
a) Name the three control structures in Java. |
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Answer» Sequential: default mode. Selection: used for decisions, branching -- choosing between 2 or more alternativepaths. Repetition: used for looping, i.e.repeatinga piece of codemultipletimes in a row. |
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| 21. |
(ă¤nch |
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Answer» 71/150=0.47333.. 0.47333333...is the answer bt in round off we can write 0.47 |
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| 22. |
classmateDatePagewith finding the cubefactorise en 54 ) +by - 32 ) - 32 |
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Answer» (2x-5y)^3+(5y-3z)^3+(3z-2x)^3=0 =3(2x-5y)(5y-3z)(3z-2x) =90xyz-54xz^2-225zy^2+135z-60x^2y+36x^2z+150xy^2-90xz |
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| 23. |
Date ___ //.Page No.i n32 |
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Answer» -9 22----- + ------10 15 Now take LCM of 10 and 15 as denominator -27+ 44-------------- 30 17-----30 |
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| 24. |
6 Write the supplements of the7 The measure of one angle isFind the measure of x in the |
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Answer» angle DOC = 93° ( vertically opposite angle) AOD + AOB = 180° ( linear pair) AOD = 180° - 93° = 87° angle BOC = 87° ( vertically opposite angle) |
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| 25. |
Define: Normality and Molarity |
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Answer» Normality (N) is defined as the number of mole equivalents per liter of solution:normality = number of mole equivalents/1 L of solutionThe solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution. Molarity (M) is defined as the number of moles of solute per liter of solution.molarity = moles of solute/liters of solutionMolarity is a measurement of the moles in the total volume of the solution. |
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| 26. |
1. The adjacent sides of a parallelogram are in the ratio 3:7 and its perimeter is 100 cm. Findsides of the parallelogram.2. The angles of a quadrilateral are in the ratio 2:3:4:6. Find the measure of the angles.3.Diagonals AC and BD of a rectangle ABCD intersect each other at point O. If OA=3cm.AC and BD. |
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Answer» third question is not clear |
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| 27. |
6. The angles of a triangle are in the ratio 2:4:4. Find the measure of each angle and also givename to the triangle. |
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Answer» sum of all 3 angles of a triangle = 180 degree let the ratio factor be X. angles are 2X, 4X, 4X A/Q 2X + 4X + 4X = 180 10X = 180 X = 180/10 X = 18 angles are 2X = 2 ✖ 18 = 36 degree 4X = 4 ✖ 18 = 72 degree 4X = 4 ✖ 18 = 72 degree since 2 angles are equal, triangle is an isosceles triangle. |
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| 28. |
9 x ^ { 2 } y ^ { 2 } ( 3 z - 24 ) + 27 x y ( z - 8 ) |
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Answer» 9x^2 y^2 (3z-24)÷27xy(z-8)=9x^2 y^2 ×3 (z-8)÷(27xy (z-8))=27x^2 y^2 (z-8)÷27xy (z-8)=xy |
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| 29. |
learning outcomes inmathematics are developed-(1) to prepare children for year-endexaminations(2) such that children may be toldsmall steps for calculations(3 to increase the achievementchildren in various educationalsurveys4defineefine classwise compe-tencies and skills to be acby children |
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Answer» to increase the achievement of children in various educational surveys |
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| 30. |
3 = 5sinf —3cosOनए 550 0+ 3९०५9 |
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Answer» We are given 3 cot x = 4, or cot x = 4/3. Now consider a right angled triangle, ABC, with BC as the base , right angle at and A the upper vertex. Thus <ABC = x. So BC=4, AC=3 and AB=5. Then sin x = 3/5 and cos x = 4/5 (1) (5 sin x + 3 cos x)/(5 sin x - 3 cos x) … (2). Put the values of sin x and cos x from (1) it (2) to get [(5*3/5) +(3*4/5)]/[(5*3/5) -(3*4/5)], or = (15+12)/(15–12) [As 5 is common both in the numerator and denominator it can be ignored] = 27/3 = 9 Like my answer if you find it useful! |
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| 31. |
| 675 x5.5 + 5.5 x8.25 =9(1155 (2) 1100 (3) 550 (4) 5500реж |
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| 32. |
Arvind began a business with 550 and wasjoined afterwards by Brij with 330. When didBrij join, if the profits at the end of the yearwere divided in the ratio 10:3?अरविंद ने 550 रूपए के साथ एक कारोबार शुरू कियाऔर बाद में ब्रिज 330 रूपए के साथ जुड़ जाता है. यदिवर्ष के अंत में लाभ को 10:3 के अनुपात में बांटा जाता है,तो ब्रिज कब जुड़ा था? |
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Answer» after six months and last answer is gvien t1 12 months Total Arvind business = 550x12Total Brij business= 330 x XAccording to question(550 x 12)/(330 x X) = 10/36600/330X = 10/3Solving we get X = 6So,Brij joined Arvind after = 12-6 =33 months Total Arvind business = 550 x 12Total Brij business= 330 x XAccording to question(550 x 12)/(330 x X)= 10/3Solving we get X = 6So,Brij joined Arvind after = 12-6 = 6 months |
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| 33. |
ही 92 1! (0. (0. &0051020. of eMA'”W५. ध्व * दि 0कन- ८-4. |
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| 34. |
In the given figure. <3-550, then fof all other anglesInd(measure2467 |
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| 35. |
(3) The probability of scoring 70 marks in a 100 marktest is.. |
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Answer» Probability of scoring 70 in 100 mark test isP(70)=1/100 p(70)=1/100possibe outcomes=1total outcomes=100 probability of scoring 70 in 100 mark test is p(70)=1/100 Probability of scoring 70 in 100 mark test is P(70) =1/100 1/100 is the right answer p(70)=1/100 is the probability of scoring 70 marks in 100 mark test 1/100 is the correct answer |
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| 36. |
A copper rod of diameter 2 cm and length 10 cmis drawn into a wire of uniform thickness andlength 10 m. Find the thickness of the wire. |
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| 37. |
(z A copper rod of diameter 2 cm and length 10 cm is drawn into a wire ofuniform thickness and length 10 m. Find the thickness of the wire.ICBSE 2012 |
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| 38. |
A copper rod of diameter 2 cm and length 10 cm is drawn into a wire ofuniform thickness and length 10 m. Find the thickness of the wire.[CBSE 2012] |
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| 39. |
7. A copper rod of diameter 2 cm and length 10 cm is drawn into a wire ofuniform thickness and length 10 m. Find the thickness of the wire.[CBSE 2012 |
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Answer» thanqs |
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| 40. |
of diameter 2 cm and length 10 cm is drawn into a wire of7. A copper rod dia. A thickness and length 10 m. Find the thickness of the wire.rodICBSE 2012 |
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| 41. |
oru(1 + sin θ) (1-sin θ)( + cos 0)( - cos 6)1 If cotevaluate: (i)(ii) cot 6 |
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| 42. |
Mathematics for Class 8ameter is228Inder? Also,e of 1.5-cm-thick metal. Its external diameteof metal used in making the cylinder? A20. A cylinder is open at both ends and is made of12 cm and height is 84 cm. What is the volume of mof the metal weighs 7.5 g.find the weight of the cylinder if 1 cmmal radius=4.5 cm |
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| 43. |
4. The speed of a car on a highway isS l ur. On entering the city, the speed of the car sloweddown to 50 t/hr. Calculate the per cent decrease in the speed of the car5. IF Lipi's salary is 120% of the sale of |
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Answer» bhai accept it best maine ek que daala hai use tm ans do mai best accept kar lunga dono ko 100 points milenge |
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| 44. |
Find an irrational number lying between v2 and v3 |
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| 45. |
A die is thrown once. Find the probability of gettinyg(i) a prime number(i) a number lying between 2 and 6.(iii) an odd number.22. |
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| 46. |
The product ofapproximately.12.34 x 43.21 is1 |
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Answer» 533.2114 is the write answer 533.2114 is the answer for the product of 12.34 and 43.21 |
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| 47. |
A number lying between 10 and 100 is seven times the sum of its digits.If 9 is substracted from it,the digis of tigits of thenumber are reversed. Then the number is : |
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Answer» suppose number is 10x+yso 10x+y=7(x+y)so 3x=6yso x=2y10x+y-9=10y+x20y+y-9=10y+2yso 9y=9so y=1 and x=2so number is 21 |
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| 48. |
Write the whole number part and decimal part of given numbers.(a) 14.25(b) 99.09(c) 178.752(d) 299.990 |
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Answer» a) 14.25 whole number : 14 decimal part : 25 b) 99.09 whole number : 99 decimal part : 09 c) 178.752 whole number : 178 decimal part : 752 d) 299.990 whole number : 299 decimal part : 990 |
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| 49. |
\begin{array} { l } { \text { The thickness of a hollow metallic cylinder } } \\ { \text { is } 2 \mathrm { cm } . \text { It is } 70 \mathrm { cm } \text { long with outer radius } } \\ { \text { of } 14 \mathrm { cm } \text { . Find the volume of the metal } } \\ { \text { used in making the cylinder, assuming that } } \\ { \text { it is open at both the ends. Also find its } } \\ { \text { weight if the metal weighs } 8 \text { g per } \mathrm { cm } ^ { 3 } \text { . } } \end{array} |
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| 50. |
\begin{array} { l } { \text { The thickness of a hollow metallic cylinder } } \\ { \text { is } 2 \mathrm { cm } \text { . It is } 70 \mathrm { cm } \text { long with outer radius } } \\ { \text { of } 14 \mathrm { cm } . \text { Find the volume of the metal } } \\ { \text { used in making the cylinder, assuming that } } \\ { \text { it is open at both the ends. Also find its } } \\ { \text { weight if the metal weighs } 8 \text { g per } \mathrm { cm } ^ { 3 } \text { . } } \end{array} |
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