1796 words - 8 pages

Unproctored Mock-1 2013 Answers and Explanations

1 11 21 31 41 51 b d b b d b 2 12 22 32 42 52 b b a b c a 3 13 23 33 43 53 c c d a b a 4 14 24 34 44 54 a b d d c a 5 15 25 35 45 55 a a d b a b 6 16 26 36 46 56 c a b c b a 7 17 27 37 47 57 a b c b a d 8 18 28 38 48 58 b c a b d d 9 19 29 39 49 59 c d b c d c 10 20 30 40 50 60 d b c c a d

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Unproctored Mock-1 2013

1. b

1! + 2 × 2! + 3 × 3! + 4 × 4! +…+ 12 × 12! = (2! – 1!) + (3! – 2!) + (4! – 3!) +…+ (13! – 12!) = 13! – 1 So the remainder = –1 i.e. 12 Let the smallest number be x and the three equal numbers be y. Then, x + 3y + x + 100 = 150 5

4. a

A

2. b

3 y 2 Hence, y can take only even values.

⇒ 2x + 3y = ...view middle of the document...

+ = 33 33 33

6. c

m× d×h = cons tan t, where w m = number of men d = number of days h = number of hours w = amount of work

Let the number of additional men required be x.

75 × 90 × 8 (75 + x) × 60 × 10 or x = 150 = 2 5 Hence, 150 additional men would be required. ∴

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Unproctored Mock-1 2013

7. a

The number of ways of picking two small cubes

= 64C2 = 32 × 63

The number of small cubes with exactly two faces painted red = 2 × 12 = 24 (Since two such cubes will be obtained from each edge of the large cube.) The number of ways of picking two such cubes

= 24C2 = 23 × 12

Now, 32 + 1 when divided by 22, always gives a remainder of 2. Therefore, the highest power of 2 which can divide each of the terms

29 28 21 3 + 1 , 3 + 1 ,..., 3 + 1 is 1.

n

So the required probability =

23 × 12 23 . = 63 × 32 168

Hence, 31024 − 1 is divisible by 29 × 22 × 21 and so the highest power of 2 by which it is divisible = 9 + 2 + 1 = 12 12. b Let the speeds of Prakash and Arpit be 2x and 5x respectively. Let’s assume that Arpit turned back from point Q as shown in the figure given below, ran at a speed of x after turning back and met Prakash at point R, ‘t’ hours after they started running.

P 1 0 km a Q

For questions 8 to 10: The data given in the bar graph can be tabulated as shown below. Let the total number of visitors in the year 2006 be 100x.

Year 2007 2008 2009 2010 2011

Total number of visitors 120x 108x 144x 180x 160x

Number of male visitors 60x 81x 96x 108x 104x

Number of female visitors 60x 27x 48x 72x 56x

13. c

R From the given conditions, 2x × t = 10 = 5x × 2 – x × (t – 2) ⇒t=4 So the running speed of Prakash

= 2x =

10 = 2.5 km / hr. 4

It is given that (0, 0) is one of the ends of the diagonal. The point of intersection of the other two lines

2 3 x + y = 1 and 6x + y = 3 is , , which should be the 5 5 other end of the diagonal. The required equation is given by: 3 −0 y−0 5 i.e. 3x − 2y = 0. = x−0 2 −0 5

8. b

The total number of female visitors in the year 2011 = 56x = 56 × 468.50 = 26236 Percentage growth in the number of female visitors from 2009 to 2010

9. c

72x − 48x = × 100 = 50% 48x

10. d Gender Gap in the year 2009 = 96x – 48x = 48x the year 2011 = 104x – 56x = 48x Hence, it is equal in 2009 and 2011.

14. b

cm

B E

5 .6 cm A 9 0° 9 0° D F G H

11. d

31024 − 1 = 32 9 = 32 9 = 32 9 = 32 9 = 32

9 9 − 1 = 32 + 1 32 − 1 28 28 + 1 3 + 1 3 − 1 8 0 0 + 1 32 + 1 × ... × 32 + 1 32 − 1 28 21 20 0 + 1 3 + 1 × ... × 3 + 1 × 3 + 1 × 32 − 1 21 2 1 28 + 1 3 + 1 × ... × 3 + 1 × 2 × 2

10

Let’s draw CH parallel to AE intersecting DF at G. As CE || AH and AE || CH, AECH is also a parallelogram. Hence, CE = AH and H is the midpoint of AD. In ∆ADF,GH || AF,...

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