49 kg of blended tea contain Assam and Darjeeling tea in the ratio 5: 2. ... of Darjeeling tea that is to be added to th
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www.oliveboard.in 1. Let 0 < x < 1, then the correct inequality is (a) x < √x < x2 (b) √x < x < x2 (c) x2 < x < √x (d) √x < x2 < x 2. Three bells ring at interval of 36 seconds, 40 seconds and 48 seconds respectively. They start ringing together at a particular time. They will ring together after every (a) 6 minutes (b) 12 minutes (c) 18 minutes (d) 24 minutes 3. If the sum of the digits of a three-digit number is subtracted from that number, then it will always be divisible by, (a) 3 only (b) 9 only (c) both 3 and 9 (d) all of 3, 6 and 9 4. Which of the following is correct? (a) 2/3 < 3/5 < 11/15 (b) 3/5 < 2/3 < 11/15 (c) 11/15 < 3/5 < 2/3 (d) 3/5 < 11/15 < 2/3 5. The greater of two numbers whose product is 900 and sum exceeds their difference by 30 is (a) 60 (b) 75 (c) 90 (d) 100 6. The smallest fraction, which should be added to the sum of 2 1/2, 3 1/3, 4 1/4 and 5 1/5 to make the result a whole number is (a) 13/60 (b) 1/4 (c) 17/60 (d) 43/60 7. Find the cube root of (-13824) Or Find the value of (a) 38 (b) -38 (c) 24 (d) -24
3
13824
8. The sum of three positive numbers is 18 and their product is 162. If the sum of two numbers is equal to the third, then the sum of squares of the numbers is (a) 120 (b) 126 (c) 132 (d) 138 9. The sum of three consecutive even numbers is 28 more than the average of these three numbers. Then the smallest of these three numbers is (a) 6 (b) 12 (c) 14 (d) 16
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www.oliveboard.in 10. In a division sum, the divisor ‘d’ is 10 times the quotient ‘q’ and 5 times the remainder ‘r’, if r = 46, the dividend will be (a) 5042 (b) 5328 (c) 5336 (d) 4276 11. A man can do a piece of work in 30 hours, if he works with his son then the same piece of work finished in 20 hours, if the son works alone he can do the work in (a) 60 hours (b) 50 hours (c) 25 hours (d) 10 hours 12. A water tap fills a tub in ‘p’ hours and a sink at the bottom empties it in ‘q’ hours. If pP) at y% per annum at simple interest at the same time, then the amount of their debts will be equal after (a) 100(Q-P)/(Px-Qy) years (b) 100(Px-Qy)/(Q-P) years (c) 100(Px-Qy)/(P-Q) years (d) 100(P-Q)/(Px-Qy) years 49. A man invested a sum of money at compound interest. It amounted to Rs.2420 in 2 years and to Rs.2662 in 3 years. Find the sum. (a) Rs.1000 (b) Rs.2000 (c) Rs.5082 (d) Rs.3000
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www.oliveboard.in 50. If a sum of money becomes 4000 in 2 years and 5500 in 4 years 6 months at the same rate of simple interest per annum. Then the rate of simple interest is (a) 21 3/7 % (b) 21 2/7 % (c) 21 1/7 % (d) 21 5/7 % 51. A hollow cylindrical tube 20cm along is made of iron and its external and internal diameters are 8 cm and 6 cm respectively. The volume (in cubic cm) of iron used in making the tube is (take π=22/7) (a) 1760 (b) 440 (c) 220 (d) 880 52. If the areas of three adjacent faces of a rectangular box which meet in a corner are 12cm 2, 15cm2, and 20cm2 respectively. Then the volume of the box is (a) 3600cm3 (b) 300cm3 (c) 60cm3 (d) 180cm3 53. The ratio between the length and the breadth of a rectangular park is 3:2. If a man cycling along the boundary of the park at the speed of 12km/hr completes one round in 8 minutes, then the area of the park is (a) 153650 m2 (b) 135600m2 (c) 153600m2 (d) 156300m2 54. If the radius of a right circular cylinder open at both the ends, is decreased by 25% and the height of the cylinder is increased by 25%. Then the curved surface area of the cylinder thus formed. (a) remains unaltered (b) is increased by 25% (c) is increased by 6.25% (d) is decreased by 6.25% 55. A cylindrical pencil of diameter 1.2cm has one of its end sharpened into a conical shape of height 1.4cm. the volume of the material removed is (a) 1.056 cm3 (b) 4.224 cm3 (c) 10.56 cm3 (d) 42.24 cm3 56. A rectangular park 60m long and 40m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 m2 then the width of the road is (a) 3m (b) 5m (c) 6m (d) 2m 57. Four circles of equal radii are described about the four corners of a square so that each touches two of the other circles. If each side of the square is 140 cm then area of the space enclosed between the circumference of the circle is (taken π = 22/7) (a) 4200 cm2 (b) 2100 cm2 (c) 7000 cm2 (d) 2800 cm2
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www.oliveboard.in 58. The amount of concrete required to build a concrete cylindrical pillar whose base has a perimeter 8.8 metre and curved surface area 17.6 sq.metre, is (Take π = 22/7) (a) 8.325 m3 (b) 9.725 m3 (c) 10.5 m3 (d) 12.32 m3 59. A hemispherical bowl of internal radius 9 cm, contains a liquid. This liquid is to be filled into small cylindrical bottles of diameter 3 cm and height 4 cm. Then the number of bottles necessary to empty the bowl is (a) 18 (b) 45 (c) 27 (d) 54 60. A rectangular water tank is 80m × 40 m. Water flows into it through a pipe of 40 sq.cm. at the opening at a speed of 10 km/hr. The water level will rise in the tank in half an hour is (a) 3/2 cm (b) 4/9 cm (c) 5/9 cm (d) 5/8 cm 61. A square and a regular hexagon are drawn such that all the vertices of the square and the hexagon are on a circle of radius r cm. the ratio of area of the square and the hexagon is (a) 3:4 (b) 4:3√3 (c) √2:√3 (d) 1:√2 62. A solid cylinder has the total surface area 231 sq.cm. If its curved surface area is 2/3 of the total surface area, then the volume of the cylinder is (a) 154 cu.cm (b) 308 cu.cm (c) 269.5 cu.cm (d) 370 cu.cm 63. The lateral surface area of frustum of a right circular cone, if the area of its base is 16πcm 2 and the diameter of circular upper surface is 4 cm and slant height 6 cm, will be (a) 30 πcm2 (b) 48 πcm2 (c) 36 πcm2 (d) 60 πcm2 64. The diameter of a sphere is twice the diameter of another sphere. The surface area of the first sphere is equal to the volume of the second sphere. The magnitude of the radius of the first sphere is (a) 12 (b) 24 (c) 16 (d) 48 65. A right circular cylinder having diameter 21 cm and height 38 cm is full of ice cream. The ice cream is to be filled in cones of height 12 cm and diameter 7 cm having a hemispherical shape on the top. The number of such cones to be filled with ice cream is (a) 54 (b) 44 (c) 36 (d) 24
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www.oliveboard.in 66. The simplified value of [1 – (2xy/(x2+y2))] [(x3-y3)/(x-y) – 3xy] is (a) 1/(x2-y2) (b) 1/(x2+y2) (c) 1/(x-y) (d) 1/(x+y) 67. If a + b + c = 0 then the value of [1/(a+b)(b+c)] + [1/(b+c)(c+a)] + [1/(c+a)(a+b)] (a) 0 (b) 1 (c) 3 (d) 2 68. If x2+y2+2x+1 = 0, then the value of x 31 + y35 is (a) -1 (b) 0 (c) 1 (d) 2 69. If x = (√5+1)/(√5-1) and y = (√5-1)/(√5+1) , then the value of (x 2+xy+y2)/(x2-xy+y2) (a) 3/4 (b) 4/3 (c) 3/5 (d) 5/3 70. If (x-(1/x))2 = 3, then the value of x 6+(1/x6) equals (a) 90 (b) 100 (c) 110 (d) 120 71. If x4+2x3+ax2+bx+9 is a perfect square, where a and b are positive real numbers, then the value of a and b are (a) a = 5, b = 6 (b) a = 6, b = 7 (c) a = 7, b = 6 (d) a = 7, b = 8 72. If a2 + b2 + c2 = 16, x2 + y2 + z2 = 25 and ax +by +cz = 20, then the value of (a+b+c)/(x+y+z) (a) 3/5 (b) 5/3 (c) 4/5 (d) 5/4 73. The value of x which satisfies the equation ((x+a2+2c2)/b+c) + ((x+b2+2a2 )/c+a) + ((x+c+2b2 )/a+b) = 0 is (a) (a2+b2+c2) (b) –(a2+b2+c2) (c) (a2+2b2+c2) (d) –(a2+b2+2c2) 74. If a3 = 117+b3 and a = 3+b, then the value of a+b is (a) ±7 (b) ±49 (c) ±13 (d) 0
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www.oliveboard.in 75. If a + 1/a = -2 then the value of a1000 + a-1000 is (a) 2 (b) 0 (c) 1 (d) 1/2 76. ABC is similar to DEF. If area of ABC is 9 sq.cm. and area of DEF is 16 sq.cm. and BC = 2.1 cm. Then the length of EF will be (a) 5.6 cm (b) 2.8 cm (c) 3.7 cm (d) 1.4 cm 77. A chord of a circle is equal to its radius. The angle subtended by this chord at a point on the circumference is (a) 80° (b) 60° (c) 30° (d) 90° 78. Let two chords AB and AC of the larger circle touch the smaller circle having same centre at X and Y. then XY =? (a) BC (b) (1/2)BC (c) (1/3)BC (d) (1/4)BC 79. Let G be the centroid of the equilateral triangle ABC of perimeter 24 cm. then the length of AG is (a) 2√3 cm (b) 8/√3 cm (c) 8√3 (d) 4√3 80. A and B are the centres of two circles with radii 11 cm and 6 cm respectively. A common tangent touches these circles at P and Q respectively. If AB = 13 cm, then the length of PQ is (a) 13 cm (b) 17 cm (c) 8.5 cm (d) 12 cm 81. ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12√5 and BC = 24 cm then radius of circle is (a) 10 cm (b) 15 cm (c) 12 cm (d) 14 cm 82. ABC is an isosceles triangle where AB = AC which is circumscribed about a circle. If P is the point where the circle touches the side BC, then which of the following is true? (a) BP = PC (b) BP > PC (c) BP < PC (d) BP = (1/2)PC 83. If D and E are the mid points of AB and AC respectively of ABC, then the ratio of the areas of ADE and BCED is? (a) 1:2 (b) 1:4 (c) 2:3 (d) 1:3
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www.oliveboard.in 84. O is the circumcentre of the isosceles ABC. Given that AB = AC = 17 cm and BC = 6 cm. The radius of the circle is (a) 2.115 cm (b) 40/280 cm (c) 42/298 cm (d) 42/280 cm 85. B1 is a point on the side AC of ABC and B1B is joined. A line is drawn through A parallel to B 1B meeting BC at A1 and another line is drawn through C parallel to B 1B meeting AB produced at C 1. Then (a) [(1/CC1) - (1/AA1)] = 1/BB1 (b) [(1/CC1) + (1/AA1)] = 1/BB1 (c) [(1/BB1) – (1/AA1)] = 2/CC1 (d) [(1/AA1) – (1/CC1)] = 2/BB1 86. The value of the expression (1+sec22°+cot68°)(1-cosec22°+tan68°) is (a) o (b) 1 (c) -1 (d) 2 87. If Xsin3𝜃 + Ycos3𝜃 = sin 𝜃cos𝜃 and Xsin𝜃 - Ycos𝜃 = 0, then the value of x2+y2 equals (a) 1 (b) 1/2 (c) 3/2 (d) 2 88. If sec𝜃 + tan𝜃 = m (>1), then the value of sin 𝜃 is (0°< 𝜃 10/15 > 9/15, which means 11/15 > 2/3 > 3/5 or we can say 3/5 < 2/3 < 11/15 5. A Let the two numbers be a and b, where a>b a*b = 900 ……. (i) (a + b) – (a – b) = 30 2b = 30 b = 15 From equation (i) a*15 = 900 a = 60 Hence, greater number = 60 6. D 2 1/2 + 3 1/3 + 4 1/4 + 5 1/5 = 5/2 + 10/3 + 17/4 + 26/5 = (150 + 200 + 255 + 312)/60 = 917/60 = 15 + 17/60 To make this number a whole number, we need to add (1 – 17/60) to it = (60 – 17)/60 = 43/60
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www.oliveboard.in 7. D -13824 = -24 * -24 * -24 Hence,
3
13824 = -24
8. B Let the three positive numbers be x, y and z. x + y + z = 18 ………… (i) x*y*z = 162 …………. (ii) Also, x + y = z ……….. (iii) From (i) and (iii) z + z = 18 2z = 18 z = 9 …………… (iv) From (ii) and (iv) x*y = 18 Also, x + y = z = 9 From above, (x, y) = (6, 3) or (3, 6) Sum of squares of the numbers = 3 2 + 62 + 92 = 9 + 36 + 81 = 126 9. B Let the three consecutive even numbers be x, x+2 and x+4. Sum of three consecutive even numbers = (x) + (x+2) + (x+4) = 3x + 6 Average of three consecutive even numbers = (3x + 6)/3 = x + 2 Now, 3x + 6 = 28 + (x + 2) 2x = 24 x = 12 Hence, the smallest number is 12. 10. C divisor (d) = 10*quotient(q) = 5*remainder(r) r = 46 Hence, d = 5 * 46 = 230 q = 46/2 = 23 We know that Dividend (D) = divisor (d) * quotient(q) + remainder(r) D = 230 * 23 + 46 = 5290 + 46 = 5336 11. A A man can finish the work by working alone in 30 hours Hence, work done by man in 1 hour = 1/30 Let his son can finish the work by working alone in x hours Hence, work done by his son in 1 hour = 1/x While working together, they can finish the work in 20 hours 20 * (1/30 + 1/x) = 1 1/30 + 1/x = 1/20 1/x = 1/20 – 1/30 = 10/600 = 1/60 x = 60 hours 12. B Let the volume of tub be 1 unit. Volume of tub filled by tap in 1 hour by working alone = 1/p Volume of tub emptied by sink in 1 hour by working alone = 1/q When both tap and sink are opened simultaneously for r hours, where p 5.5x=330 => x=60 then, Mukesh has 3*60 = Rs.180 32. C Total daily income = 257.50 × 20 = Rs.5150 Average Daily income of a boy = Rs.x Average Daily income of a woman = Rs.(x + 10) Average Daily income of a man = Rs.(x + 20) So, 7(x+20)+11(x+10)+2x = 5150 20x = 4900 x = Rs.245 Hence, average daily income of a man = 245+20=Rs.265 33. D let the cost price of an article=Rs. x selling price of an article=Rs. 425 profit=selling price-cost price=425-x in 2nd case, selling price of an article=Rs. 355 loss=cost price- selling price=x-355 according to the condition, => 425-x=x-355 => x+x=355+425 => 2x=780 => x=Rs. 390 cost price of an article=Rs. 390
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www.oliveboard.in 34. B Let the profit share of A = x Profit share of B = 1650-x According to the condition, x/3=2(1650-x)/5 => 5x=9900-6x => x=900 then profit of B = 1650-900=Rs.750 35. C Let the cost price of the article = x Selling price of the article = y According to the 1st condition, 4y/100=5x/100 => x/y=4/5 required ratio = 4:5 36. C Let the original rate per dozen = P And initially number of dozens of eggs that can be bought with this rate = Q So, PQ= 162 ---------(i) Now according to the condition, 0.75P(Q+2)=162 --------(ii) On equating (i) and (ii), 0.75PQ+6P/4 = PQ 6P=PQ Q=6 So, P = 162/6 = 27 Original rate per dozen = Rs. 27 37. A For first painting SP = Rs. 20000, profit = 25% Let CP = Rs. X So, => 25/100 = (20000-x)/x => x=Rs. 16000 For 2nd painting SP= Rs. 20000, loss = 25% Let CP = Rs. y So, => 25/100 = (y-20000)/y => y= Rs. 80000/3 total SP = 2*20000 = Rs. 40000 total CP = 16000 + 80000/3 = Rs. 42666.67 net loss = 42666.67-40000 = Rs.2666.67 which is more than Rs. 2000. 38. B Let the actual price of per kg of wheat = Rs. 100 Then, selling price = 110% of 100 = ₹ 110 Now, shopkeeper sell 30% less than actual weight so cost price to shopkeeper will be = 70% of 100 = (70 x 100)/100 = Rs. 70 Hence, required profit gained by shopkeeper = [(110 - 70)/70] x 100 = (40/70) x 100 = 57 1/7 % 39. B In a day, total minutes = 24*60 = 1440 Required percentage = 30*100/1440 = 2.083 %
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www.oliveboard.in 40. C Let the present earning = Rs. x So, According to the question, x*1.25*0.96*1.25*0.96*1.25 =72000 => 1.8x = 72000 => x = Rs. 40000 41. C Let the total candidates = 100x Candidate passed in qualitative aptitude test = 73x Candidate passed in general awareness = 70x Candidate passed in both = 64x Candidates who passed in only quantitative aptitude = 73x-64x = 9x Candidate who passed in only general awareness = 70x -64x=6x So, candidates who failed in both subject = 100x-(64x+9x+6x) = 21x But, 21x = 6300 So, 100x = 30000 Total number of candidates = 30000 42. B Let the initial income = 100x Expenditure = 75x Saving = 100x-75x = 25x Increased income = 1.2*100x = 120x Increased expenditure = 1.1*75x = 82.5x Saving2=120x-82.5x = 37.5x Percentage increase in savings = (37.5x-25x)*100/25x=50% 43. D
Let the speed of boat in still water = x kmph Let the speed of the stream = y kmph And distance between P to R= 2D km If river is flowing from R to P then, According to the condition, D/(x+y)+D/(x-y) = 12 --------(i) And 2D/(x-y) = 16 hr 40 min = 16(2/3) ---------(ii) Replacing the value of equation (ii) in equation (i), 2D/(x+y)+ 16(2/3) = 24 => 2D/(x+y)=24-16(2/3) => 2D/(x+y)=22/3 hrs Time taken to cover distance R to P=2D/(x+y) = 22/3 hrs = 7 1/3 hr If river is flowing from P to R then, According to the condition, D/(x+y)+D/(x-y) = 12 --------(i) And 2D/(x+y) = 16 hr 40 min = 16(2/3) ---------(ii) Replacing the value of equation (ii) in equation (i), 2D/(x-y)+ 16(2/3) = 24 => 2D/(x-y)=24-16(2/3) => 2D/(x-y)=22/3 hrs Time taken to cover distance R to P=2D/(x-y) = 22/3 hrs = 7 1/3 hr
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www.oliveboard.in 44. C With speed of 42kmph and time 10hrs Distance covered = 42×10=420km person wants to cover the distance in 7 hrs using formula speed=distance/time Required speed= 420/7= 60km/hrs so speed of the car should be increase by 60-42=18kmph in order to cover distance in 7 hrs. 45. A Let the speed of train = x km/hr and speed of car= y km/hr Case 1 Distance travelled by train = 240 Km Distance travelled by car = 450-240 = 210 km Time taken by train = 240/x hr Time taken by car = 210/y hr Total time taken = 240/x+210/y hr 240/x+210/y = 8(2/3) ---------------(i) Case 2 Distance travelled by train = 180 Km Distance travelled by car = 450-180 = 270 km Time taken by train = 180/x hr Time taken by car = 270/y hr Total time taken = 180/x+270/y hr 180/x+270/y = 9 180/x+270/y = 9 --------------(ii) on solving (i) and (ii), x=60,y=45 Speed of Car =45 km/hr Speed of train = 60 km/hr 46. B Let the speed of train C is ‘c’ kmph. As both trains are travelling in same directions, so relative speed = (100-c) kmph So, (0.150+0.250)/(100-c)=2/60 0.4*30=100-c c=88 kmph 47. B Principal = Rs.30000 Interest = Rs. 4347 Compounded amount = 30000+4347 = Rs.34347 Now, A=P(1+r/100)n => 34317=30000(1+7/100) n => 11449/10000=(1+7/100) n => (107/100)2=(1+7/100)n => n=2 years
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www.oliveboard.in 48. A Let the amount be equal in T years. SI=PRT/100 Amount = P+SI Case 1: Amount of debt meet by A, A1 = P + PxT/100 Case 2: Amount of debt meet by B, A2 = Q + QyT/100 It is given that A1=A2 So, => P + PxT/100= Q + QyT/100 => (PxT-QyT)/100 = Q-P => T(Px-Qy)=100(Q-P) => T=100(Q-P)/(Px-Qy) years 49. B Amount after 2 years will be the principal for the next year. So, A=P(1+R/100) => 2662=2420(1+R/100) => (1+R/100)=1.1 => R=10% Now let the initial sum = Z So, 2420=Z(1+10/100)2 => Z=Rs.2000 50. A Let the initial sum =P Rate of interest =R % SI=PRT/100 Case 1: 4000-P=PR*2/100 ----------(i) Case 2: 5500-P=PR*4.5/100 ----------(ii) By (i)/(ii), (4000-P)/(5500-P)=2/4.5 => 36000-9P=22000-4P => 5P=14000 => P=2800 Substituting the value of P in equation (i), 4000-2800=4000*R*2/100 => R=300/14 => R=21 3/7 % 51. B Volume of iron used to make a hollow cylindrical tube = π(r 22-r12)h where r1 and r2 internal and external radius of the hollow cylinder respectively. r1=6/2=3 cm r2 = 8/2 =4 cm h=20 cm volume = 22*(42-32)*20/7 = 440 cubic cm
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www.oliveboard.in 52. C If length, breadth and height of the rectangular box are l, b and h respectively. So, It is given that, lb=12,bh=15 and lh=20 now, on multiplying these three values, (lbh)2=12*15*20 => (lbh)2 =3600 => lbh = 60 so, volume of the box is = 60 cm3 53. C Speed of the man = 12 kmph = 12*5/18=10/3 m/sec Man completes one round in =8 min = 8*60=480 sec Area covered in one round will be the perimeter of the park. So, perimeter of the park = 480*10/3=1600 m Now, Let the length and breadth of the park are 3x and 2x respectively. So, Perimeter = 2(L+B) => 1600=2(3x+2x) => x=160 so, area of the park = L*B=3x*2x=6x 2 = 6*(160)2=153600 m2 54. D Let the radius of the initial cylinder be r and height be h. New radius = r – 25r/100 = 3r/4 New height = h + 25h/100 = 5h/4 So, the percentage change in the surface area can be calculated as: => % change = [2π(3r/4)(5h/4)-2πrh]*100/2πrh => % change = (-rh/16)*100/rh => % change = -6.25 % So, surface area is decreased by 6.25%. 55. A
Given : h=1.4 and r=0.6 Volume of material removed = volume of cylinder ABCD – volume of cone APB = πr2h – πr2h/3 = 2(πr2h)/3 = (2*22*0.6*0.6*1.4)/(3*7) = 1.056 cm3
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www.oliveboard.in 56. A
Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 x2 – 100x + 291 = 0 (x - 97)(x - 3) = 0 x = 3. Width of the road = 3 m 57. A
Radius of each circle = side/2 = 70 cm The area of the circles in the square makes a complete circle so, Area of four sectors = 22*70*70/7 = 15400 cm 2 Area of the square = 140*140=19600 cm2 So, area of enclosed space = 19600-15400 = 4200 cm2 58. D Perimeter of the base = 8.8 m Radius of the base = 8.8/(2π) =1.4 m Curved surface area of the cylinder = 2πrh =17.6 So h = 2 m So amount of concrete required=volume of the cylindrical pillar= πr2h=12.32 m3 59. D R= 9cm and r = 3/2 cm, h = 4 cm Let the number of bottles required = n Volume of hemispherical bowl = n * volume of one cylindrical bottle 2πR3/3 = n * πr2h n= 2R3/(3r2h) n=2*93/(3*1.52*4) n=54 60. D Length of water column flown in 1 min = (10*1000/60)m = 500/3 m Volume flown per minute = (500/3)*(40/100 2) = 2/3 m3 Volume flown in half an hour = (2/3)*30 = 20 m 3 So, rise in water level =20/(40*80) m = (1*100/160)cm = 5/8 cm
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www.oliveboard.in 61. B Square: -
Diagonal of the square = diameter of the circle = 2r Side2 + Side2 = Diagonal 2 2 x Side2 = 4r2 Side = 2r Area of the square = 2r2 Regular hexagon: -
If we join all the vertices with the centre, then we will get 6 equilateral triangles. Area of Hexagon = 6 x area of triangle = 6 x 3/4 x r x r = (33 r2/2) Ratio of area of square to area of Hexagon = 2r 2/(33 r2/2) = 4/33 62. C Let, r be the radius of the cylinder and l be the height. Total surface area = 2πr2 + 2πrl = 2πr (r + l) Curved surface area = 2πrl Curved surface area = 2/3 x Total surface area 2πrl = 2/3 x 2πr (r + l) 3l = 2r + 2l l = 2r … (1) Cylinder has the total surface area 231 sq.cm So, 2πr2 + 2πrl = 231 From equation (1) 6πr2 = 231 πr2 = 231/6 … (2) Volume of the cylinder = πr2l = 7πr2 = 7 x 231/6 = 269.5 cu.cm
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www.oliveboard.in 63. C Area of the base = πr2 = 16πcm2 Radius of the base = 4 Diameter of the base = 8 cm The given frustum can be drawn as:
In the above diagram, Triangle ABC ≈ Triangle CED So, AC = CD = 6 cm Total slant height = 12 cm Lateral surface area of ADG = rl = 48 Lateral surface area of ACF = rl = 12 Lateral surface area of frustum = 48 - 12 = 36 64. B Let, radius of first sphere is 2R and that of another sphere is R. Surface area of first sphere = 4(2R)2 Volume of second sphere = 4/3 R3 Given; 4(2R)2 = 4/3 R3 16R2 = 4/3 R3 12 = R So, radius of first sphere = 2R = 24 units 65. A Volume of right circular cylinder = r2h = 22 x (21/2)2 x 38/7 = 13167 cm3 Volume of cone = 1/3r2h = (22 x (7/2) 2 x 12)/(7 x 3) = 154 cm3 Volume of hemisphere on the top = 2/3 x r3 = 2/3 x (22/7) x (7/2)3 = 89.833 cm3 Total volume of the cone = 154 + 89.833 = 243.833 cm 3 Number of cones to be filled = 13167/243.833 = 54 66. B The given equation is: [1 – (2xy/(x2+y2))] [(x3-y3)/(x-y) – 3xy] = [(x2+y2 - 2xy)/(x2+y2)] [(x-y)( x2+y2 + xy)/(x-y) – 3xy] = (x-y)2/(x2+y2) (x-y) = 1/(x2+y2)
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www.oliveboard.in 67. A [1/(a+b)(b+c)] = [1/(ab + ac + b2 + bc)] = 1[b(a + b + c) + ac] = 1/(0 + ac) = 1/ac … (I) Similarly, [1/(b+c)(c+a)] = 1/ab … (II) And, [1/(c+a)(a+b)] = 1/bc … (III) So, [1/(a+b)(b+c)] + [1/(b+c)(c+a)] + [1/(c+a)(a+b)] = 1/ab + 1/ac + 1/bc … (IV) Multiplying by abc on both sides; abc x [1/(a+b)(b+c)] + [1/(b+c)(c+a)] + [1/(c+a)(a+b)] = a + b + c = 0 [1/(a+b)(b+c)] + [1/(b+c)(c+a)] + [1/(c+a)(a+b)] = 0/abc = 0 68. A x2+y2+2x+1 = 0 (x+1) 2 + y2 = 0 … (I) So, x = -1 and y = 0 x31 + y35 = (-1)31 + (0)35 = -1 + 0 = -1 69. B (√5+1) 2 = 6 + 2√5 (√5-1) 2 = 6 - 2√5 Now, (x2+xy+y2) = [(6 + 2√5)/(6 - 2√5) + 1 + (6 - 2√5)/(6 + 2√5)] = [(6 + 2√5)2 + (6 - 2√5) 2 + (6 + 2√5)(6 - 2√5)]/[(6 - 2√5)/(6 + 2√5)]] = 128/16 Similarly, (x2-xy+y2) = [(6 + 2√5)/(6 - 2√5) - 1 + (6 - 2√5)/(6 + 2√5)] = [(6 + 2√5)2 + (6 - 2√5) 2 - (6 + 2√5)(6 - 2√5)]/[(6 - 2√5)/(6 + 2√5)]] 96/16 (x2+xy+y2)/(x2-xy+y2) = 128/96 = 4/3 70. C (x-(1/x))2 = 3 x2 + 1/x2 - 2 = 3 x2 + 1/x2 = 5 … (1) (x2 + 1/x2)3 = 125 x6 + 1/x6 + 3x4/x2 + 3x2/x4 = 125 x6 + 1/x6 + 3x2 + 3/x2 = 125 x6 + 1/x6 + 3(x2 + 1/x2) = 125 x6 + 1/x6 = 125 – 15 = 110 71. C x4+2x3+ax2+bx+9 = (x2+nx+3)2 = x4+n2x2+ 9 + 2x2nx+ 6x2 + 6nx = x4+2nx3 + (n2 + 6)x2 + 6nx + 9 … (I) Comparing both sides of the equation; We get, 2n = 2, a = (n2 + 6) and b = 6n So, n = 1, a = 7 and b = 6 72. C As, 16 and 25 are perfect squares of 4 and 5. So, let a = 4 and x = 5. b=c=y=z=0 (a+b+c)/(x+y+z) = (4+0+0)/(5+0+0) = 4/5
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www.oliveboard.in 73. B By putting values from options; Option 2: ((x+a2+2c2)/(b+c) + ((x+b2+2a2 )/(c+a) + ((x+c+2b2 )/(a+b) = (b2-c2)/ (b+c) + (c2-a2)/ (c+a) + (a2-b2)/ (a+b) = (b-c)(b+c)/(b+c) + (c-a)(c+a)/(c+a) + (a-b)(a+b)/(a+b) =b–c+c–a+a–b=0 Hence option 2 satisfies the given equation. Hence, option 2 is the correct answer. 74. A a=b+3 taking cube from both the sides; a3 = b3 + 9b2 + 27b + 27 117 + b3 = b3 + 9b2 + 27b + 27 9b2 + 27b – 90 = 0 b2 + 3b – 10 = 0 b = -5, 2 When, b = 2; a=3+2=5 a+b=7 When, b = -5; a = 3 + (-5) = -2 a + b = -2 + (-5) = -7 So, a + b = ±7 75. A a + 1/a = -2 … (I) (a + 1/a)2 = 4 a2 + 1/a2 + 2 = 4 a2 + 1/a2 = 2 … (II) (a + 1/a)3 = -8 a3 + 1/a3 + 3a + 3/a = -8 a3 + 1/a3 + 3(a + 1/a) = -8 a3 + 1/a3 + (-6) = -8 a3 + 1/a3 = -2 … (III) taking square of equation (II); (a2 + 1/a2) 2 = 2 a4 + 1/a4 + 2 = 4 a4 + 1/a4 = 2 … (IV) From above equations, it is evident that; If, a + 1/a = -2 then, an + 1/an = -2 (for n = odd) / 2 (for n = even) So, a1000 + a-1000 = 2 76. B For similar triangles, area of the triangles and square of length of the corresponding sides are in proportion. So, Area of ABC/Area of DEF = BC2/EF2 9/16 = (2.1)2/(EF)2 3/4 = 2.1/EF EF = 2.1 * 4/3 = 2.8 cm
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www.oliveboard.in 77. C
Chord AB = Radius AC = Radius BC So, in equilateral triangle ABC; ∠BAC = 600 Radius is always perpendicular to the radius. So, ∠DAC = 900 At the point of contact (i.e. A), circumference of the circle is parallel to the tangent AD. So, required angle = ∠DAB = 90 – 60 = 300 78. B
XZ(radius) is perpendicular to AB (tangent). In triangle AXZ, (AX)2 = (AZ)2 - (XZ)2 … (I) Similarly, in triangle BXZ, (BX)2 = (BZ)2 - (XZ)2 … (II) But, AZ = BZ … [radius of the same circle] So, AX = BX … (III) AB = AX + BX = 2AX Similarly, AC = 2AY As, 2 sides of both the triangles are in proportion and angles subtended by both the sides are equal. So, both the triangles are similar triangles. Therefore, AX/AB = AY/AC = XY/BC 1/2 = XY/BC XY = 1/2 BC
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www.oliveboard.in 79. B
Perimeter of the equilateral triangle = 24 cm AB = 24/3 = 8 cm Centroid is the common point of three medians of the triangle. Medians are also perpendicular bisectors in the equilateral triangle. AD is perpendicular to BC and it equally divides BC. Therefore, BC = 4 cm In triangle ABD; AD2 = 82 - 42 AD = 43 … (I) Centroid divides the median in the ratio 2: 1. So, AG = 2/3 x AD = 83/3 = 8/3 cm 80. D
Two circles with centre A and B and radius 6 and 11 respectively are drawn. Line BC drawn parallel to the common tangent PQ. So, PQBC is a rectangle. CQ = 6 cm and AC = 5 cm In triangle ABC, BC2 = 132 - 52 BC = 12 cm As, PQBC is a rectangle. PQ = BC = 12 cm
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www.oliveboard.in 81. B
As, ABC is isosceles triangle, the line passing from AO will be perpendicular to BC. So, in triangle ADC, AD2 = (12√5)2 - 122 AD = 24 cm AO = OC = r OD = 24 – r In triangle ODC, OC2 = (OD)2 + 122 r2 = (12)2 + (24 - r)2 r2 = 144 + r2 + 576 – 48r 48r = 720 R = 720/48 = 15 cm 82. A
Tangents drawn from the same point are equal in the length which are denoted by X, Y and Z in the diagram. Given that; AB = AC X+Y=X+Z Y = Z … (I) So, BP = PC
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www.oliveboard.in 83. D
D and E are midpoints of AB and AC respectively. AD/AB = AE/AC = 1/2 … (I) In triangle ADE and ABC; 2 sides are in proportion and angle included by both the sides is equal. Therefore, ADE and ABC are similar triangles. Let, AP is perpendicular to DE and BC. So, in similar triangle; AO/AP = AE/AC = DE/BC = 1/2 Area of ABC = ½ x BC x AP … (I) Area of ADE = ½ x DE x AO = ½ x BC/2 x AP/2 = 1/8 x BC x AP … (II) Area of DECB = Area of ABC - Area of ADE = 3/8 x BC x AP … (III) Required ratio = (1/8 x BC x AP)/(3/8 x BC x AP) = 1/3 84. D
We have; AB = AC = 17 cm BC = 6 cm OR will divide BC in 2 equal parts. BR = RC = 3 cm Also, BP = BR = 3 cm and RC = CQ = 3 cm … (tangents to a circle from a common point are equal in magnitude) So, AP = AQ = 14 cm In triangle ARB; AR2 = AB2 – BR2 = 289 – 9 = 280 AR = 280 cm … (I) Let, radius of the circle = r cm Now, In triangle APO; AP2 + PO2 = AO2 (Since, OP is perpendicular to AB) 196 + r2 = (280 - r)2 196 + r2 = 280 + r2 – 2r280 r = 42/280
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www.oliveboard.in 85. B The diagram can be drawn as follows:
Where, AA1 || BB1 || CC1 As we know, all lines intersecting 3 parallel lines will gets divided into the same ratio. Let, AC, AC1 and A1C gets divided into the ratio x : y. So, AB/BC1 = AB1/B1C = x/y AB/AC1 = AB1/AC = x/(x+y) In triangle ABB1 and AC1C; 2 sides are in the same proportion and angle included by both the sides is equal. Therefore, both the triangles are similar triangles. AB/AC1 = AB1/AC = BB1/C1C = x/(x+y) … (I) So, BB1/C1C = x/(x+y) … (II) Similarly, If we consider the triangles CBB1 and CA1A; We get, BB1/A1A = y/(x + y) … (III) Adding equation (II) and (III); BB1 (1/AA1 + 1/CC1) = x/(x+y) + y/(x+y) [(1/AA1) + (1/CC1)] = 1/BB1 86. D (1+sec22°+cot68°)(1-cosec22°+tan68°) = (1 + sec22 + tan22)(1 – cosec22 + cot22) = (cos22 + 1 + sin22)/cos22 x (sin22 – 1 + cos22)/sin22 = ((Cos22 + sin22)2 – 1)/sin22.cos22 = (Cos222 + sin222 + 2sin22.cos22 -1)/sin22.cos22 = (1 + 2sin22.cos22 -1)/ sin22.cos22 = 2 87. A Xsin3𝜃 + Ycos3𝜃 = sin 𝜃cos𝜃 … (I) Xsin𝜃 - Ycos𝜃 = 0 … (II) So, X = Y cot𝜃 Putting this values in equation (I); Xsin3𝜃 + Ycos3𝜃 = Ycos𝜃sin2𝜃 + Ycos3𝜃 = Ycos𝜃(sin2𝜃 + cos2𝜃) Y = sin𝜃 … (III) Similarly, putting Y = X tan𝜃; We will get, X = cos𝜃 … (IV) x2+y2 = (sin𝜃)2+(cos𝜃)2 = 1
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www.oliveboard.in 88. B Sec𝜃 + tan𝜃 = m Sec𝜃 = m – tan𝜃 Sec2𝜃 = m2 + tan2𝜃 – 2mtan𝜃 Sec2𝜃 – tan2𝜃 = m2 – 2mtan𝜃 1-m2 = -2mtan𝜃 (m2 -1)/2m = tan𝜃
So, AC2 = (2m)2 + (m2 -1)2 = m4 -2m2 + 1 + 4m2 = m4 + 2m2 + 1 = (m2 + 1) 2 AC = m2 + 1 Sin𝜃 = BC/AC = (m2 - 1)/(m2 + 1) 89. B (a2-b2)sin𝜃 + 2abcos 𝜃 = a2 + b2 Let, a = kcosx and b = ksinx. So, a2 + b2 = k2 k2 (cos2x – sin2x)sin𝜃 + 2k2cosx.sin𝜃 = k2 cos2x.sin𝜃 + sin2x.cos𝜃 = 1 sin (2x+𝜃) = 1 So, it is clear that; 𝜃 = /2 - 2x So, tan𝜃 = cot2x = (1 + tan2x) /2.tanx = (a2 + b2)/2ab 90. C
Let, angle ABE = 450 and angle ABD = 600 So, angle EBC = 450 and angle DBC = 300 In triangle EBC; EC = BC = x In triangle BDC; DC = x/3 Distance covered in 10 minutes = EC – DC = x (1-1/3) = x (3-1)/ 3 Time required to travel remaining distance (t) is given by; ED/DC = 10/t [x(3-1)/ 3]/[ x/3] = 10/t t = 10/(3-1) = 10/0.732 = 13.66 = 13 minutes 40 seconds
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www.oliveboard.in 91. D 2ycos𝜃 = xsin𝜃 2ycosec𝜃 – xsec𝜃 = 0 4ycosec𝜃 – 2xsec𝜃 = 0 … (I) 2xsec𝜃 – ycosec𝜃 = 3 … (II) Adding equation, I and II; 3ycosec𝜃 = 3 y = sin𝜃 … (III) Putting value of y in equation I; x = 2 cos𝜃 … (IV) x2+4y2 = 4cos2𝜃 + 4sin2𝜃 = 4 (cos2𝜃 + sin2𝜃) = 4 92. A
In above figure; dbz = 450 dby = 600 in triangle bmz ; bm = zm = l … (I) In triangle bay; ba = √3/2 x by by = 200/√3 ay = ½ x by ay = 100/√3 … (II) Height of pole = yz = ba – bm = ba – zm = ba = ay = 100 - 100/√3 = (100/3)(3-√3) metre 93. B
Given, DC = h Tan⍺ = BC/AB Tan 𝛽 = (h +BC)/AB Tan 𝛽 - tan⍺ = h/AB … (I) tan⍺ = BC/AB h.tan⍺ = h.BC/AB … (II) Divinding equation (II) by (I); [h tan⍺/(tan 𝛽 -tan⍺)] = BC
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www.oliveboard.in 94. A
Angle EDC = 300 and angle EDA = 600 In triangle BCD, Angle CDB = 600 Tan60 = 3 BC = BD. 3 In triangle BAD, Angle ADB = 300 Tan60 = 1/3 AB = BD.(1/3) Distance covered in 10 minutes; AC = BC – AB = BD (3 – 1/3) = 2/3. BD Let time taken to reach the shore is ‘t’. AC/AB = 10/t (2/3)/(1/3) = 10/t t = 5 minutes 95. A (cot𝜃 + cosec𝜃 - 1)/(cot𝜃 - cosec𝜃 + 1) = [(cotA + cosecA) – (cosec2A – cot2A)]/(cotA – cosecA + 1) = cotA + cosecA = (1 + cosA)/sinA 96. B Monthly expenditure on clothing = 10% of the income = Rs.825 Total monthly income = 100% = Rs.8250 97. A From given pie chart, it is clearly apparent that 15% of total income family saves. 98. C Expenditure on food = 30% = Rs.2475 Expenditure on miscellaneous = 20% = Rs.1650 Required ratio = 2475/1650 = 3/2 99. A Expenditure on clothing = 10% = Rs.825 Expenditure on rent = 25% = Rs.2062.5 Average expenditure = 2887.5/2 = Rs.1443.75 100. D Expenditure on food = 30% = Rs.2475 Expenditure on miscellaneous = 20% = Rs.1650 Expenditure on clothing = 10% = Rs.825 Average expenses = (2475+1650+825)/3 = Rs.1650 Expenditure on rent = 25% = Rs.2062.5 Expenditure on saving = 15% = Rs.1237.5 Average expenses = (2062.5+1237.5)/2 = Rs.1650 Require ratio = 1650/1650 = 1: 1
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