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Class 8th Sample Paper Solution PDF

31 Pages·2016·1.56 MB·English
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Pioneer’s Aspire Scholarship Exam Solution 8th CBSE Examination Centre: Pioneer Education, Sector – 40-D General Instructions: The question paper contains 90 multiple choice questions. There are two sections in the question paper Section: A– MATHEMATICS (1 to 45) Section: B– SCIENCE (46 to 90) Each right answer carries 4 marks and − 1 for every wrong answer. The Maximum marks are 360 and Maximum Time 2.00 hrs. Give your response in the OMR Sheet provided with the Question Paper. Pioneer Education The Best Way To Success IIT – JEE /AIPMT/NTSE/Olympiad Classes Section – A {Mathematics} 2/3 4/3 1 1 1. Simplify : 216 27 3 9 4 4 (a) (b) (c) (d) 4 4 9 3 Ans. (c) Solution: 2/3 4/3 1 1 216 27 2/3 1 1 27 4/3 a m 63 am 1 2 1 62 n am amn 6 33 4/3 34 6 6 4 3 3 3 3 9 2. Find the value of x, if ( 6)x 2 1. (a) x = 3 (b) x = 4 (c) x = 2 (d) x = 1 Ans. (c) Solution: x 2 ( 6)x 2 1 6 1/2 60 [ 60 1] 1 6 2 x 2 60 [Using (am)n = amn] 1 x 2 0 [Using am = an] 2 x 2 0 x 2 Hence, the value of x = 2 3. Find x, if 5x + 5x–1 = 750. (a) x = 4 (b) x = 3 (c) x = 1 (d) x = 2 Ans. (a) Solution: 5x 5x 1 750 Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 1 of 31 www.pioneermathematics.com Pioneer Education The Best Way To Success IIT – JEE /AIPMT/NTSE/Olympiad Classes 5x 5x.5 1 750 1 5x 5x 750 5 5 5x 5x 750 5 5x 5 1 750 5 750 5 53 5 6 5x 5x 6 6 5x 54 x = 4. 4 1 4. Express the answer with positive indices : 3 x4y 3 xy7 y8 x2 x4 y8 (a) (b) (c) (d) x2 y4 y8 x4 Ans. (d) Solution : 4 1 3 x4y 3 xy7 4 1 1/3 1 x4y ma am 1/3 xy7 4 x4/3 y1/3 n = am amn x1/3 y7/3 4 4 1 1 7 = x3 3 y3 3 4 4 = x1 y 6/3 xy 2 = x 4 y8 y8 = x4 Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 2 of 31 www.pioneermathematics.com Pioneer Education The Best Way To Success IIT – JEE /AIPMT/NTSE/Olympiad Classes 5. The difference between compound interest and simple interest at 5% per annum in 2 years is Rs. 30. Find the sum. (a) 14, 000 (b) 12, 000 (c) 13000 (d) 11000 Ans. (b) Solution: Given R = 5%, T = 2 years, C.I. – S.I. = Rs. 30, P = ? P R T P 5 2 P S.I. 100 100 10 2 2 5 21 441P and A P 1 P 100 20 400 441P C.I. A P P 400 A.T.Q., C.I. – S. I. = Rs. 30 41P P 41P 40P 30 30 400 10 400 P 30 P 400 30 Rs.12,000 400 6. At what rate per cent per annum, compound interest will Rs. 10, 000 amount to Rs. 13, 310 in three years? (a) 15% (b) 20% (c) 10% (d) 5% Ans. (c) Solution: Let the rate b R% per annum. We have, P = Rs. 10, 000, A = Rs 13, 310, and n = 3 years. 3 R From formula, A = P 1 100 3 R We get, 13, 310 = 10, 000 1 100 3 3 3 13,310 R 11 R 1 1 10,000 100 10 100 R 11 R 1 1 100 10 100 10 Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 3 of 31 www.pioneermathematics.com Pioneer Education The Best Way To Success IIT – JEE /AIPMT/NTSE/Olympiad Classes 100 R 10 10 Hence, rate = 10% per annum. 7. By selling 288 hens, Malleshwari lost the S.P. of 12 hens. Find her loss percent. (a) 2 % (b) 4% (c) 6% (d) 8% Ans. (b) Solution: Let S.P. of 1 hen = Rs. 1 S.P. of 288 hens = Rs. 288 1 = Rs. 288 and Loss = S.P. of 12 hens = Rs. 12 1 = Rs. 12 C.P. of 288 hens = S.P. + Loss = Rs. 288 + Rs. 12 = Rs. 300 Loss 12 Therefore, loss% = 100 100 4 C.P. 300 Thus, Malleshwari’s loss is 4%. 8. A table with marked price Rs. 2400 was sold to a customer for Rs. 2200. Find the rate of discount allowed on the table. 1 1 1 1 (a) 8 %. (b)10 %. (c) 6 %. (d) 4 %. 3 3 3 3 Ans. (a) Solution: M.P. = Rs. 2400, S.P. = Rs. 2200 Discount = Rs. 2400 – Rs. 2200 = Rs. 200 discount Rate of discount = 100% M.P. 200 1 = 100% 8 . 2400 3 9. Divya bought a set of cosmetic items for Rs. 345 including 15% sales tax and a purse for Rs. 110 including 10% sales tax. What per cent is the sales tax charged on the whole transaction? (a) 15.7% (b) 17.5% (c) 13.7% (d) 13.9% Ans. (c) Solution: Let the LP of cosmetic items = Rs. x Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 4 of 31 www.pioneermathematics.com Pioneer Education The Best Way To Success IIT – JEE /AIPMT/NTSE/Olympiad Classes Given, Rate of sales Tax = 15% 15 Given, Sales price = 345 = x + x 100 115x or 345 100 34500 or x Rs.300 115 Let the list price of purse be y. and Rate of Sales Tax = 10% 10 Given, Sales price = 110 = y + y 100 110 or 110 = y 100 y = Rs. 100 Total sales price of cosmetics and Purse = 345 + 110 = Rs. 455 and = 300 + 100 = Rs. 400 Hence, Discount = Total sales price – total list price = Rs. 455 – Rs. 400 = Rs. 55, i.e., Divya pays Rs. 55 towards the sales tax on purchase of above items. 55 Percentage of sales tax = 100 13.7% 400 10. Price of a DVD player including 12% VAT is 4480. Find the price of DVD player before VAT is added. (a) 4500 (b) 6000 (c) 5000 (d) 4000 Ans. (d) Solution : Let price of DVD player before VAT is added = Rs. x 12 Value added tax (VAT) = x 100 12 Price including VAT = x + x 100 12 x x 4480 100 Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 5 of 31 www.pioneermathematics.com Pioneer Education The Best Way To Success IIT – JEE /AIPMT/NTSE/Olympiad Classes 112x 4480 x 4000 100 Price before VAT in added = Rs. 4000 11. Find the area of rectangle if its length is (3x + 5) and breadth is (2y + 4). (a) 6xy + 12x + 10y + 20 (b) 6xy – 12x + 10y + 20 (c) 6xy + 12x + 8y + 20 (d) 6xy + 12x + 8y + 20 Ans. (a) Solution: Given Length = 3x + 5 and breadth = 2y + 4 Area = length breadth Area = (3x + 5) (2y + 4) = 3x (2y + 4) + 5(2y + 4) = 6xy + 12x + 10y + 20 1 1 2 12. Find the product : a 2b a2 ab 4b2 . 3 9 3 1 1 1 1 (a) 8b3 a3 (b) a3 8b3 (c) a2 8b2 (d) a3 8b3 27 27 27 27 Ans. (d) Solution: 1 1 2 a 2b a2 ab 4b2 3 9 3 2 1 1 1 2 = a 2b a a 2b 2b 3 3 3 3 1 1 = a 2b 3 a3 8b3 3 27 13. Simplify : (4m + 5n)2 + (5m + 4n)2 (a) 41m2 41n2 (b) 41m2 41n2 80mn (c) 16m2 25n2 40mn (d) None of these Ans. (b) Solution: (4m + 5n)2 + (5m + 4n)2 = 16m2 + 25n2 + 40mn + 25m2 + 16n2 + 40mn = 41m2 + 41n2 + 80mn. Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 6 of 31 www.pioneermathematics.com Pioneer Education The Best Way To Success IIT – JEE /AIPMT/NTSE/Olympiad Classes 14. Add : 3(a + b +c) and (a + b) (a – b – c). (a) a2 + b2 + ac + bc + 3a + 3b + 3c (b) a2 – b2 – ac – bc + 3a + 3b + 3c (c) a2 – b2 + bc + ac + 3a + 3b + 3c (d) a2 – b2 – ac – bc – 3a – 3b – 3c Ans. (b) Solution: 3(a + b +c) and (a + b) (a – b – c) = 3a + 3b + 3c + a2 – ab – ac + ab – bc – b2 – c2 = 3a + 3b + 3c + a2 – b2 – c2 – ac – bc 15. By what number (–9)–1 should be multiplied so that the product may be equal to (–12)–1? 3 3 4 4 (a) (b) (c) (d) 4 4 3 3 Ans. (a) Solution: Product = (–12)–1 Given number = (–9)–1 The required number = [Product] (Given number) 1 1 = 12 9 1 1 1 12 12 9 1 9 1 9 1 9 3 = . 12 1 12 1 4 3 Thus, (–9)–1 should be multiplied by to get (–12)–1 as the product. 4 40 5 16. Simplify the following: [5 1 5 19]2 . 2 40 20 1 1 (a) 240 (b) (c) (d) 220 2 2 Ans. (b) Solution: 40 40 5 2 5 [5 1 19 ]2 5 20 2 2 Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 7 of 31 www.pioneermathematics.com Pioneer Education The Best Way To Success IIT – JEE /AIPMT/NTSE/Olympiad Classes 40 40 1 5 = 5 40 40 2 40 1 = 5 40 540 40 2 1 = 5 40 40 [ 1 40 = 1 as 40 is an even number] 240 40 40 1 1 1 = 50 1 . 240 2 2 17. The number 0.000675 can be written as (a) 675 10–6 (b) 67.5 10–5 (c) 6.75 10–4 (d) All of these Ans. (d) Solution: 0.000675 675 675 = 675 10 6 or 67.5 10 5 or 6.75 10 4 1000000 106 18. On a scale map 0.6 cm represents 6.6 km. If the distance between two points on the map is 80.5 cm, what is the actual distance between these points? (a) 885.5 km. (b) 885.7 km (c) 887.7 km (d) 887.5 km Ans. (a) Solution: Distance on map 0.6 80.5 Actual distance 6.6 x Values are directly proportional : Using a a 1 2 we get b b 1 2 0.6 80.5 6.6 x 0.6x 80.5 6.6 x = 885.5 km. Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 8 of 31 www.pioneermathematics.com Pioneer Education The Best Way To Success IIT – JEE /AIPMT/NTSE/Olympiad Classes 19. 120 men had food provision for 200 days. After 5 days, 30 men died due to an epidemic. How long will the remaining food last? (a) 240 days (b) 250 days (c) 260 days (d) 230 days Ans. (c) Solution: Since 30 men die after 5 days. Therefore, the remaining food is sufficient for 120 men for 195 days. Suppose the remaining food lasts for x days for the remaining 90 men. Thus, we have the following table. Number of men 120 90 Number of days 195 x We note that more men will consume the food in less number of days and less number of men will consume the food in more number of days. So, it is a case of inverse variation. Ratio of number of men = Inverse ratio of number of days 120 : 90 = x : 195 120 x 90 195 120 195 x 260 90 Hence, the remaining men will consume the food in 260 days. 20. The price of sugar goes up by 20%. By how much percent must a house wife reduce her consumption so that the expenditure does not increase? 2 2 2 2 (a) 16 % (b) 16 % (c) 16 % (d) 16 % 5 3 7 9 Ans. (b) Solution : Let the consumption of sugar originally be 100 kg and its price be Rs 100. Then, New price of 100 kg sugar = Rs 120 [ Price increases by 20%] Now, Rs 120 can fetch 100 kg sugar 100 250 Rs 100 can fetch = 100 kg sugar = kg sugar 120 3 250 50 2 Reduction in consumption = 100 % % 16 % 3 3 3 Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 9 of 31 www.pioneermathematics.com

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IIT – JEE /AIPMT/NTSE/Olympiad Classes If she wants to reach her school in 15 minutes, what should be her average speed? .. The frequency determines the pitch of a sound. 79. Voice of which of the following is likely to have minimum frequency : (a) baby girl. (b) baby boy. (c) a man. (d) a woman
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