Std VIII Ganit Pravinya Exam Dec 2016 Test Paper Solution

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Std VIII Ganit Pravinya Eam 2016 4 Dec 2016 Test Paper Solution 1.

1 Std VIII Ganit Pravinya Eam Dec 2016 Test Paper Solution 1. a) Solution - Ratio of length to breath = 1:7 Length = 1 ; Breadth = 7 Area of rectangle = 1 7 = 91 2 Given area = = = 25 = 5 Hence length = 1 =1 1. b) Solution - 1. c) Solution - 5 = 65 cm; Breadth = 7 = 7X 5 = 5 cm = = = = =407 Let the cost price be. After increasing this value by 29% it becomes, = This value is the selling price which is given 2 a) Solution - Factorisation 407 = 999 to be = = 54,242.2 Rs = 2 X 870 = 2 X 2 X 195 = 2 X 2 X X X 215 = 2 X 2 X X X 5 X 4 2. b) Solution To find the number which when divided by given numbers give same remainder (8); the number is of the form k (L.C.M of the number) + 8 k X (L.C.M of 54, 6, 81) + 8 L.C.M of 54 = 9 X 6 6 = 9 X 7 81 = 9 X 9 L.C.M = 9 X X X 2 X 7 = 114. k (114) + 8 is the required number. Since the number has to be smallest 5 digit one- k has to be 9. So that 9 (114) = The required number = = c) Solution of of = Q.. a) Solution- 2.5 liters 45 deciliters 2.5l l deciliters = 10-1 liter. 1

2 A. b) Solution- In PAB PAB = 90 o [Angle subtended by diameter is always right angle] PAB = 7 o PAB + PBA + APB = 180 o 7 o + PBA + 90 o = 180 o PBA = 5 o c) Solution B Given, l(ab) = 12cm l(ad) = 9cm l(bc) = 24cm l(db) l(dc) and m BAD = 90 In ABD, D 9 C O AD 2 + AB 2 = BD = BD = BD = BD 2 l(bd) = 15 = l (DC) ar( ABCD) = ar ( 4 a) Solution - Now, In Draw DL LC = 12cm DLC, BC DL 2 + LC 2 = DC 2 DL 2 =DL 2 LC 2 = = 81 (DL) = 9 DAB) + ar(dbc) = 1/2 X 12 X X 24 X 9 = = 162 cm 2 Ans. (2a-9) - (2a+9) using identity: a b = (a b) (a 2 + b 2 + ab) (2a - 9) (2a + 9) = (2a 9 +2a - 9)[(2a - 9) 2 +(2a+9) 2 + (2a-9) (2a+9)] = (-18) [4a a + 4a a + 4a 2-81] 4 b) Solution - 4 c) Solution Construction Steps- =- 18 [12a ] = - 54 (4a ) Ans 9-1 = (-1) = 5(-1) 2(-1) = 5(-1) 2 2 = 5 5 = = 1 Ans 1. Draw a line segment of length 6.7cm 2. Bisect it and mark middle point. (say M). Draw another line segment of length 5. cm and bisect it. 4. Use the compass to measure the bisected length 2

3 5. Now, taking middle point M as centre and radius as the measure taken with compass, draw arc on perpendicular bisect. 6. Join all the etreme points to make required Rhombus. 5. a) Solution Let, distance covered be km. Time taken be t hr, when average speed is 40 km/hr. As Again Distance Speed = time 40 = t t = ( i) = t - 1 t - 1 = ( ii) 60 Substituting t from equation (i) in eq n (ii) 5 b) Solution t - 1 = 60-1 = = = = 120 km Ans Let total number of students in school be. 5 c) Solutions - 6 a) Solution - 6 b) Solution = = = = = 240 Ans (7a - 6b - 5c) 2 + (7a + 6b - 5c) 2 = [(7a - 5c) - 6b] 2 + [(7a - 5c) + 6b] 2 Using Identity: (a - b) 2 + (a + b) 2 = 2(a 2 + b 2 ) [(7a - 5c) - 6b ] 2 + [(7a-5c) + 6b] 2 = 2 [(7a-5c) 2 + 6b 2 ] = 98 a c ac +72 b 2 2 = - ( + 1) 2 = = - 11 = - = - 11 QTS is a right angle triangle. Hence by Pythgaorous theorem Again QT 2 = QS 2 + TS 2 QT 2 = = 25+9 =4 QT = 4 QTP is a right angled triangle. Hence by Pythgaorous theorem QT 2 = PQ 2 + PT 2 QT 2 = 2 + 2

4 6 c) Solution QT 2 = = =17 = 17 A hare sees a dog 100 m away. Distance = 100m. (between dog and hare) 5 2 Speed of hare = 12 km/hr = 12 5 m/s 18 The hare runs at this speed for 1 min. = 60 Sec distance = Speed X Time 2 = X 5 X 60 = 200 m. Now hare is away from the dog by a distance of total = 00m. Now speed of dog = 18 km/hr = 18 X 5 18 = 5 m/s Hence the distance between dog and hare = 00m. Their relative Speed = 5 m/s - 2 X 5 m/s = 5 m/s. Time taken to cover the distance between Distance Them = = = = 180s. speed a) Solution Part I Le the parallel sides be & y Area of Trapezium 1/2 X 9 (Sum of Parallel Sides) X height = 1/2 ( +y) X 1 = 286 = (+y) = 286 2/1 = 44 cm Part II Let = 20 cm +y= y = 44 y= 24 So two sides are X= 20 cm Y = 24 cm 7 b) Solution - Interest per month = Ts Interest pee Annum = Rs 150 SI = PX R X T / = 500 X R X1 /100 R= 150/5 =0 % Rate = 0 % per annum 7 c) Solution - Let the Interest man has to pay is I 1 & Interest man has to take is I 2 Then I 2 I 1 = (i) Now I 1 = 2500X1.5X4/100 = 150 From Equation (i) I 2 = I1 = = a) Solution- I 2 = 2500X R X1.5/ = R R= /7.50 = 20625/750 = 5.5 R= 5.5 pcpa (6 ) (6) 2(6)( ) ( ) n n n n n b) Solution Average price of 10 chairs = Rs 50 4

Total price of all 10 chairs = Rs 50 X 10 = 500 Average price of 6 of the 10 chairs =20 Total price of the 6 chairs = 20 X 6 = = 1920 = 14m( (m 4 + 9m 2 + 81 + (-6m )- 54 m) + 18m 2 = 14m( (m 4 + 9m

5 Total price of all 10 chairs = Rs 50 X 10 = 500 Average price of 6 of the 10 chairs =20 Total price of the 6 chairs = 20 X 6 = = 1920 = 14m( (m 4 + 9m (-6m )- 54 m) + 18m 2 = 14m( (m 4 + 9m m - 54 m + 18m 2 ) = 14m( (m 4-6m + 9m m 2-54 m + 81) = 14m( (m 4 + m m 2-54 m + 81) Total price of remaining 4 Chairs = Total price of all 10 chairs Total Price of 6 Chairs = = 1580 All the 4 chairs are of same price Price of each of the 4 chairs = Total price of 4 chairs/4 = 1580/4 = 95 Rs. 8 c) Solution = 2X2X2X2 X 2X2X2X2 X 2X2X2X2 X XXX (1776) 1 2 = 2X2X2X = 24 9 (a) Solution - No of lines can be counted by combination of two points from the given 5 points. 5 C 2 5! 5! 54 2!(5 2)! 2!! 2!! Lines can be drawn 9 (b) Solution - (7 m 21 m m) (2 m 2-6 m+18) = 7 m ( m 2 m + 9) (2 m 2-6 m+18) = 7 m ( m 2 m + 9) ( m 2 m + 9) X 2 = 14 m ( m 2 m + 9) 2 = 14 m ( (m 2 )+ (- m) + (9) (m 2 )+ (- m) + 2 (-m) (9)+2 (m 2 ) (9) 9 (c) Solution Line Segment are in the ratio :2 Let the two parts be & 2 By adding the two parts, we will get full line segment +2=5 =9 (Length of Line Segment) =9/5 = 9/5 = 27/5 =5.4 cm 10 (a) Solution 2=.6 cm (14mp-8n) + (12mp+6n)+(mp-11n) =14mp-8n+12mp+6n+mp-11n =14mp+12mp+mp-8n+6n-11n =mp( )+n( ) =27mp-1n 10 (b) Solution 10 c) Solution = (16675) 1 = 5 11 =

6 6

RAO IIT ACADEMY / GANIT PRABHUTWA EXAMINATION / STD - VIII / QP + SOLUTIONS JEE MEDICAL-UG BOARDS KVPY NTSE OLYMPIADS GANIT PRABHUTWA EXAMINATION

RAO IIT ACADEMY / GANIT PRABHUTWA EXAMINATION / STD - VIII / QP + SOLUTIONS JEE MEDICAL-UG BOARDS KVPY NTSE OLYMPIADS GANIT PRABHUTWA EXAMINATION JEE MEDICAL-UG BOARDS KVPY NTSE OLYMPIADS GANIT PRABHUTWA EXAMINATION Date : 0 - - 07 Std - VIII Total Marks : 00 Time : Hours Q. A) Choose the correct alternative and write it against the correct sub