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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. |
capacity of a cuboidal shaped tank is 12 litres. If the area of the base of50 sq cm, then the height of the tank is10 cm (2) 12 cm (3) 16 cm (4) 20 cm |
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Answer» volume=12 litres=12000cm^3volume=πr^2h=50h=12000cmh=240cm |
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| 2. |
25The external and internal diameters of a hollow hemispherical bowl are12 cm and 10 cm respectively. Find the cost of painting it all over at the rate of 2 per sq, cm. |
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| 3. |
3If \frac{\cos x}{\cos y}=2 and cos( x-y)=root 3/2 then tan yA) 3+4в) V3 + 1c) V3-4D) V3-1 |
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| 4. |
aernplene leaves an aispest and is e Nexth at a SpĂŠed of leakh at the iane time, nees aezphne liaves he saine atipext and |
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Answer» Hence, the aeroplanes would be 300√61 km far from each other. |
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| 5. |
and the four walls at the rate of 19 per m2.4. An open tank is in the shape of a cube. Themeasure of its inner edge is 25 cm. Calculatethe cost of painting the inner sides at the rateof 0.15 per sq cm. |
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Answer» Total surface area of cube = 6a² = 6*25² = 3750 cm² Cost = 3750*0.15 = 562.5 rupees |
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| 6. |
(3+V3)+(3- V3) |
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Answer» The answer us 3+√3+3-√3 =6 |
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| 7. |
The base of a triangular field is thrmetre is 6588, find its base and height. 3lom, 12mee times its altitude. If the cost of leveling the field at3050per sa |
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| 8. |
V3।46. यदि ।तथा ।' = 5, तब (।।।) कामान, V3) तथा |
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Answer» 14 x = root(3) + 1/ root(3) - 1x = (root(3) + 1)(root(3) + 1)/ (root(3) - 1)(root(3) + 1) = (3 + 1 + 2root(3))/ (3 - 1) = (4 + 2root(3))/2 = 2 + root(3) y = root(3) - 1/ root(3) + 1y = (root(3) - 1)(root(3) - 1)/ (root(3) + 1)(root(3) - 1) = (3 + 1 - 2root(3))/ (3 - 1) = (4 - 2root(3))/2 = 2 - root(3) Therefore,Value of x^2 + y^2= (2 + root(3))^2 + (2 - root(3))^2= 4 + 3 + 4root(3) + 4 + 3 - 4root(3)= 7 + 7 = 14 |
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| 9. |
The cost of 25 metres of cloth is 800. Find the cost of 16 metres of cloth |
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| 10. |
Az N(3+'i\/_5) (3-iÂĽ5) .....(V3 +42i)-(V3-i2) |
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Answer» (3 + i√5) ( 3 - i√5)---------------------------------(√3 + √2i) - (√3 -i√2) 3² - (i√5)²---------------- 2√2i 9 + 5----------- 2√2i 7----------- √2 i 7√2i-----------√2×√2i² -7√2i--------- 2 |
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| 11. |
(vii)f(x) = x2-(V3 + 1) x + V3(iv) |
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| 12. |
how that 5-v3 is irrational, given v3 is irrational. |
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Answer» To prove :5-√3 is irrational Proof:Let us assume that5-√3 is rational. Let ,5 -√3 = r , where "r" is rational 5 - r =√3 Here,LHS is purely rational.But,on the other hand ,RHS is irrational.This leads to a contradiction.Hence,5-√3 is irrational |
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| 13. |
LAWERANCE PUBLIC SR. SEC. SCHOOL uth-Class -X/ Maths AssignmentChapter-4/Quadratic EquationsSORRY2*1Q.1Solve for x =--+x-112x+1 2-1Solvex-1 x+1 6XIX-31Solvex-2x-403.3في {;* 0,5* 0,**004.Solvea+b+Xabx25. Solve for x = 2 -3(+3) = 506. Solve for x = 2 (#73 - (25Q7.Solve 4x? -2(a + bºx + a2b2 = 09x? Ga? x + (9-6% =0(a + b)2 x2 +8 (a? -- 62) x + 16 (a - b) = 009.or the following |
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Answer» x/x-1 + x+1/x=4;, x^2+( x-1)( x+1)=4(x)(x+1); x^2+x^2-1=4(x^2+x); 2x^2-1= 4x^2+4x;; 4x^2-2x^2+4x-1=2x^2+4x-1 2)x+1/ x-1 + x-1/ x+1=5/6; (x+1)(x+1)+( x-1)(x-1)=( x-1)(x+1)=(x^2+2x+1)+(x^2-2x+1)=(x^2-1); ( 2x^2+2)=(x^2-1); |
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| 14. |
Write the polynomial whose zeroes are 2 +V3 and 2 - V3. |
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Answer» x^2-4x+1=0 is the correct answer of the given question yes it is correct answer x2-4x+1 is the answer (2+V3)(2-V3)=2(2)-(V3)^2=4-3=1 |
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| 15. |
A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm.Find the volume and the curved surface of the cone so formed. (Take π=3.14) |
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Answer» When a right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm, then solid formed is a cone whose height of a cone, h = 8 cm and radius of a cone, r = 6 cm. Slant height of a cone, l = 10 cmVolume of a cone = 1/3 πr2h = (1/3) x (22/7) x 6 x 6 x 8⇒ 6336/21 = 301.7 cm3and curved surface of the area of cone = πrl⇒ (22/7) x 6 x 10 = 1320/7 = 188.5 cm2Hence, the volume and surface area of a cone are 301.7 cm3 and 188.5 cm2, respectively. |
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| 16. |
Write the polynomial whose zeroes are-2 -v3 and 2 V3 |
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| 17. |
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar toitvsides are of the corresponding sides of the first triangle. |
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| 18. |
Solve theSolve the following equations :-12 m |
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| 19. |
Consructa triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to itwsods are of the corresponding sides of the first triangle.sides are - 0 |
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| 20. |
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whosesides areof the corresponding sides of the first triangle.3 |
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| 21. |
If 20 metres of cloth cost 3600, find the cost of 16 m of cloth |
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Answer» 20 metres costs= 36001metre = 3600/20= 180 cost of 16 metre= 180*162880 rupees |
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| 22. |
1)If20metresofclothcost# 3600, find the cost of 16 m of cloth. |
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Answer» 20 m of cloth = 3600 1 m of cloth = 3600/20 1 m of cloth = Rs: 180 16 m of cloth = 180 * 16 = Rs: 2880 |
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| 23. |
olve the following.1) If 20 metres of cloth cost 3600, find the cost of 16 m of cl |
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| 24. |
Solve the following.(1) If 20 metres of cloth cost 3600, find the cost of 16 m of cloth. |
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Answer» Thanks |
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| 25. |
The cost of 5m cloth is 82.50.Find the cost 12m of such cloth |
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Answer» Cost of 5m cloth is ₹82.5 Cost of 1 m cloth is ₹82.5/5 = ₹16.5 Cost of 12 m cloth is ₹16.5 × 12 = ₹198 |
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| 26. |
nees ul J ui283t3 R3.3725, how much cloth can be purchased8. If 18 dolls cost Rs.630, how many dolls can be bought for Rs.455? |
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| 27. |
Find gain or loss. C P = Rs .1100 S .P = Rs.950 |
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Answer» CP + Gain = SP1100 + Gain = 950Gain = -150Hence this is a loss of Rs. 150 |
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| 28. |
(1) mtP (2) 2mp (3) m-p+ pm -pm + p |
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| 29. |
41. A girl's height increased by 20% last year and 10% this year. What is the total percentage increase inher height in 2 years? |
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Answer» let the height of girl be x now , after 2 years , the height of girl will be x+20%of x = 1.2x and again after 1 year , the height of girl is 1.2x + 10%of(1.2x) = 1.32x now, actually increase in height over two years is. = (1.32-1)x/x *(100%) = 32% |
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| 30. |
the length of a rectangular field is 18 m and width is 12 m. find the total cost of leveling the field at ₹ 1.50 per sq. m . |
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Answer» Area of rectangular field=length×breadth=18×12 m²=216 m² Cost of levelling 1 m²=₹1.50Cost of levelling 216 m²=₹1.50×216=₹324 |
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| 31. |
41 . A girl's height increased by 20% last year and 10% this year. What is the total percentage increase inher height in 2 years? |
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Answer» let the height of girl be x now , after 2 years , the height of girl will be x+20%of x = 1.2x and again after 1 year , the height of girl is 1.2x + 10%of(1.2x) = 1.32x now, actually increase in height over two years is. = (1.32-1)x/x *(100%) = 32%. |
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| 32. |
10.Calculate the cost of carpeting a rectangular floor at the rate ofて225 per sq m, if the length andbreadthofthe floor are 16 m and 12 m respectively |
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| 33. |
6.Check whether-150 is a term ofthe AP : 1 1 , 8, 5, 2、、 |
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| 34. |
The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden..What is the cost of tiling a rectangular plot of land 500 m long and 200 m wideat the rate of Rs 8 per hundred sq m.? |
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| 35. |
Write the nth term of the AP1 1+m 1 + 2m7l |
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| 36. |
Find the nth term of the AP 1,5 2 16' 3' 2 |
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| 37. |
If you borrow $500for 6 years at AnAnnual interest rateof 7%, how muchwill you pay Altogether |
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Answer» SI= PRT/100= 500*6*7/100= 42*5= 210 $You will pay 210$ all together |
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| 38. |
d ouut the profit or loss in each of theseO PS. PProli Loss.Rs 2,090 Rs 2, 100Rs 8395 Rs 8,935Rs 14,060 Rs 14,600Rs 9,319 Rs 9,139s 6,250 Rs 6,175s 11,190 Rs 11,86515,000 Rs 14,905С.Pe talble |
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| 39. |
5 2 16' 3' 2'7. Find the nth term of the AP 1, |
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Answer» My question are not a same your answer My question is Find the nth term of the AP 1,5/6,2/3,1/2.... |
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| 40. |
2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whos2sides areof the corresponding sides of the first triangle. |
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| 41. |
2202. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose2sides areof the corresponding sides of the first triangle. |
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| 42. |
mplete the table.S. PRs 2,385Rs 1,900Rs 8,630Rs 11,391Profit LatsC.PRs 195Rs 628Rs 1,020Rs 4,060Rs 74,365 Rs 2,315lete the table |
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| 43. |
MCQ SINGLE CORRECT1: Nine percent of Rs.700 is:(a) Rs.63 (b) Rs.630 (c) Rs.6.3 (d) Rs.0.632: A radio marked Rs.1000 is given away for Rs.850.The discount is:(a) Rs.50 (b) Rs.100 (c) Rs.150 (d) Rs.2003: The marked price of an article is Rs.200. If 15% ofdiscount is allowed on it, its selling price is:(a) Rs.185 (b) Rs.170 (c) Rs.215 (d) Rs.175The simple interest on Rs.5 ,000 at 2% per monthfor 3 months is:n oT 300 R 400 |
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| 44. |
The incomes of 7 months of a man are Rs. 7250 Rs. 6000, RS. 3500Rs. 2750, Rs. 6500, Rs. 7150 and Rs. 5200. Find his mean income |
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| 45. |
for the A.P. 5,11/2, 6, 13/2 ..., T40-T20 =...A.15B.20C.5D.10 |
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| 46. |
ind the curved surface area and total surface area of a right dircular cylinderhose height is 15 cm and the radius of the base is 7 cm. |
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| 47. |
17. 336 और 54 का एक तथा. 1.04 ज्ञात कीजिएकद .. मे...“ ># मी 31525: ¢ |
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Answer» Let a = 336 , b = 54 Expressing a and b as a product of prime factors 336 = 2 × 2 × 2× 2 × 3 × 7 = 2^4 × 3 × 7 54 = 2 × 3 × 3 × 3 = 2 × 3^3 HCF ( 336 , 54 ) = 2 × 3 = 6 LCF( 336 , 54 ) = 2^4 × 3^3 × 7 = 3024 any two positive integers a and b . HCF( a , b ) × LCM( a , b ) = a × b_____________________________ Verification :------------------ HCF( 336 , 54 ) × LCM( 336 , 54 ) = 6 ×3024 = 18144 -----( 1 ) a × b = 336 × 54 = 18144 ----( 2 ) Therefore , ( 1 ) = ( 2 ) |
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| 48. |
increased by 17 becomes 54. Find the number |
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Answer» Let the unknown number be X. Then, X + 17 = 54=> X = 54-17=> X = 37. Please hit the like button if this helped you |
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| 49. |
17.The amount of what surm becomes? 400 in years at 12% simple annual interest? |
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Answer» Given R=12%, T = 5, A = 400P =? We know,SI = P*R*T/100SI = A - P Then,A - P = P*R*T/100400 - P = P*12*5/1003P/5 + P = 4008P/5 = 4008P = 2000P = 2000/8 = 250 Therefore amount of Rs 250 becomes Rs 400 in 5 years at 12% simple Interest |
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| 50. |
1. After paying 30 out of 40 installments of adebt of Rs. 3600/-, one third of the dept isunpaid. If the installments are forming an A.P.Then what is the first installment.(a) Rs. 50/- (b) Rs. 51/.(c) Rs. 105/- (d) Rs. 110/- |
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Answer» Let the first term/instalment = Rsa. Let the common difference between successive instalment = Rsd Istcase:A man pays off a debt of Rs 3600 in 40 annual instalments ∴n= 40 and Sn= S40= Rs 3600Sn=n/2(2a+(n-1)d)S40=40/2(2a+39d)3600=20(2a+39d)2a+39d=180...(1)IIndcase:Man dies, leaving one-third of the debt unpaid. Now debt unpaid1/3*3600=1200∴ Debt paid by Man in 30 instalments = Total debt paid – Debt unpaid = 3600 – 1200 = Rs 2400 Heren= 30 and Sn= S30= 2400 S30=30/2(2a+29d)160=2a+29d2a+28d=60 Subtracting equation (2) from equation (1), we get (2a+ 39d) – (2a+ 29d) = 180 – 160 ⇒39d– 29d= 20 ⇒10d= 20 ⇒d= 2 Putting value ofdin (1), we get 2a+ 39d= 180 ⇒2a+ 39 × 2 = 180 ⇒2a= 180 – 78 2a= 102 ⇒a= 51 Rs Therefore, the first instalment paid by man = Rs 51 Th Thanks take weights of five ofyour friends. find out what their weights will be on the moon and the mars. |
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