Raashid Shah

Q.1 A certain type of board is sold only in length of multiples of 2ft. The shortest board sold is 6ft & the largest is 24 ft. A builder needs a large quantity of this type of board in 5 ½ ft length. For minimum wastage the lengths to be ordered should be : A) 24 ft B) 26 ft C) 12 ft D) 22 ft Ans. D Just see Options, & according to the options just put values & then find the wastage: Each 6ft board yields one 5 ½ board with ½ waste Each 12 ft board yields two 5 ½ board with 1 ft waste (2*5 ½ =11, 12-11 = 1 ft waste) Each 24 ft board yields four 5 ½ boards with 2ft waste (4*5 ½ =22, 24-22 = 2 ft waste) Each 22ft board may be divided into four 5 ½ ft boards with no waste (4*5 ½ = 22) Q.2 A gardener plants 17,956 trees in such a way that there are many rows as are trees in a row. The number of trees in a row is : A) 144 B) 154 C) 134 D) 136 Ans. C v 17,956 = 134 Q.3 A men went to shop & was shown some shirts costing Rs. 100 each & some trousers costing Rs. 150 each. If he had Rs.1200 with him, what was the maximum number of trousers that he could have purchased , if he had also wanted to purchase some shirts ? A) 5 B) 8 C) 2 D) 6 Ans. D We have to find greatest multiple of 150 contained in 1200, such that there is a remainder which is a multiple of 100 (i.e. Just divide 1200/150 in such a way that it leaves a remainder which is divisible by 100) Therefore, 1200 = 150*6+3*100 Therefore he can buy 6 trousers along with the 3 shirts Q.4 Karan ranked thirteen from the top & twenty six from the bottom among those who have passed in the annual examination in a class. If six students have failed in the annual examination, how many students appeared ? A) 45 B 38 C) 44 D) 50 Ans. C No. of students who pass= (12+25) + 1= 38 Total number of students= 38 + 6=44 Q.5 John ranked 21st in a class of fifty Students. What is his rank from the end ? A) 30th B) 32nd C) 20th D) 31st Ans. A 50-21=29 29+1=30

Q.1 Diameters of two circular coins are in the ratio of 1:3. The smaller coin is to roll around the bigger coin till it returns to the position from where the process of rolling started. How many times smaller coin rolled around the bigger coin ? A) 6 B) 9 C) 3 D) 4 Ans. C Since, Diameter of small coin/Diameter of big coin = 1/3 Radius of small coin/Radius of big coin = 1/3 Therefore, circumference of small coin = 2pk Circumference of big coins = 2p (3 k) = 6pk = 3 (2pk) Q.2 A ray of light is incident towards a plane mirror at an angle of 60O with the mirror surface. What will be the value of angle of reflection ? 600 Mirror A) 300 B) 00 C) 600 D) 900 Ans. A Since, Normal to the surface of mirror is 900, Therefore angle of incidence will be 90 – 60 = 300 As angle of reflection must be equal to the angle of incidence according to the laws of reflection. Q.3 A man is facing west turns through 450 clockwise, again 1800 clockwise and then turns through 2700 anti clockwise. Which direction is he facing now? A) North–East B) North-West C) South–East D) South–West Ans. D Clearly, the man initially faces in the  west direction. On moving 45o clockwise, he faces in the direction North. On further moving 180o clockwise, he faces in the direction South.  Finally, on moving 270o anti–clockwise, he faces in the direction, which is South–west. Q.4  I am facing east. I turn 1000 in the clockwise direction and then 1450 in the anticlockwise direction. Which direction am I facing now? A) East B) Northeast C) North D) South-west Ans. B Q.5 A bus starts from city A. The no. of women in the bus is half of the no. of men. In city B, 10 men left bus & 5 women entered. Now no. of men & women is equal. In the beginning how many passengers entered the bus? A) 15 B) 30 C) 36 D) 45 Ans. D Suppose in city A the no of men is 2x Then the no. of women is =x In city B: 2x-10 = x+5 X=15 So, A= 30 Thus , the no. of passengers in the beginning =30+15 =45

Q.1 Doctors have advised sheetal, a chocolate freak, not to take more than 20 chocolates in one day. When she went to the market to buy her daily quota, she found that if she buys chocolates from a market complex she would have 3 Rs more for the same no. of chocolates than she would have spent. Had she bought them from her uncle scrooge’s shop , getting two sweets less per rupee. she finally decided to get them from uncle scrooge shop paying only in one rupee coins. how many chocolates did she buy? A) 12 B) 9 C) 18 D) 15 Ans. A Let n be the number of chocolates she bought. Price at market = 1/(x-2) Price at Scrooge’s shop = 1/(x) n/(x-2) = n/(x) + 3 2n = 3x(x-2) n = 3x(x-2)/2 Now, through hit &trial method, (Just put value of X starting from 1,,,then 2,,,then 3 as so on…& you will find after putting x=4 it will calculate & comes as 12) n=12 Q.2 How much would she have spent at the market complex? A) 6 B) 9 C) 12 D) 5 Ans. A At the market complex n/(x-2) = 12/2 = 6 Q.3 Five farmers have 7, 9, 11, 13 & 14 apple trees, respectively in their orchards. Last year, each of them discovered that every tree in their own orchard bore exactly the same number of apples. Further, if the third farmer gives one apple to the first, and the fifth gives three to each of the second and the fourth, they would all have exactly the same number of apples. What were the yields per tree in the orchards of the third and fourth farmers? A) 12 & 8 Apples per tree B) 11 & 9 Apples per tree C) 10 & 14 Apples per tree D) 17 & 21 Apples per tree Ans. B Explanation: Let a, b, c, d & e be the total number of apples bored per year in A, B, C, D & E ‘s orchard. Given that a + 1 = b + 3 = c – 1 = d + 3 = e – 6 But the question is to find the number of apples bored per tree in C and D ‘s orchard. If is enough to consider c – 1 = d + 3. Since the number of trees in C’s orchard is 11 and that of D’s orchard is 13. Let x and y be the number of apples bored per tree in C & d ‘s orchard respectively. Therefore 11 x – 1 = 13 y + 3 By trial and error method, we get the value for x and y as 11 and 9 Q.4 If a light flashes every 6 seconds, how many times will it flash in ¾ of an hour? A) 451 times B) 651 times C) 751 times D) 851 times Ans. A Explanation: There are 60 minutes in an hour. In ¾ of an hour there are (60 * ¾) minutes = 45 minutes. In ¾ of an hour there are (60 * 45) seconds = 2700 seconds. Light flashed for every 6 seconds. In 2700 seconds 2700/6 = 450 times. The count start after the first flash, the light will flashes 451 times in ¾ of an hour Q.5 There are 3 persons Sudhir, Arvind, and Gauri. Sudhir lent cars to Arvind and Gauri as many as they had already. After some time Arvind gave as many cars to Sudhir and Gauri as many as they have. After sometime Gauri did the same thing. At the end of this transaction each one of them had 24. Find the cars each originally had ? A) Sudhir had 39 cars, Arvind had 21 cars and Gauri had 12 cars. B) Sudhir had 40 cars, Arvind had 30 cars & Gauri had 10 cars C) Sudhir had 41 cars, Arvind had 40 cars & Gauri had 8 cars D) Sudhir had 42 cars, Arvind had 50 cars & Gauri had 6 cars Ans. A Explanation: Sudhir Arvind Gauri Finally 24 24 24 Before Gauri’s transaction 12 12 48 Before Arvind’s transaction 6 42 24 Before Sudhir’ s transaction 39 21 12

Q.1 There are 31 kgs of rice. With a 1 kg weight,Find out the minimum number of weighings required to weigh 31 kgs. A) 5 weighings B) 6 weighings C) 7 weighings D) 8 weighings Ans. A 5 weighings ie., 1-1, 2-2, 4-4, 8-8, 16-16. Q.2 A green grocer was selling apple at a penny each, chikoos at 2 for a penny, peanuts at 3 for a penny. A father spent 7 pennies and got same amount of each type of fruit for each of his three children. What did each children get? Options: A) 1 apple, 2 chikoos, 2 peanuts B) 1 apple, 2 chikoos, 1 peanut C) 1 apple, 3chikoos, 2 peanuts D) 1 apple, 1 chikoos, 1 peanut Ans. B Apple=Rs.1, 2 Chikoos are of 1 penny or 1 chikoo= ½ =0.5 In same way, 3 peanuts for 1 penny ,or 1 peanut = 1/3 =0.33 Now see options & determine where 7 pennies are consuming: A) 1 apple i.e. 1*3 (since 3 students are there) 2 chikoos,i.e. 2*3=6 Or 6*0.5 = 3 2 peanuts,i.e. 0.33*6 Now,3+6*0.5+0.33*6 =12 Q.3 After gathering 770 chestnuts, three girls divided them up so that amounts were in the same proportion as their ages. As often as Mary took four chestnuts, Nelli took three, and for every six that Mary received, Susie took seven. How many chestnuts did each girl get? A) Mary-264,Nelli-198,Susie-298 B) Mary-198, Nelli-264,Susie-298 C) Mary-200,Nelli-275,Susie-300 D) Mary-255,Nelli-280,Susie-310 Ans. A Let mary take m , nellie take n and susie take s chestnuts … then , m : n =4:3 = 12:9 and , s : m = 7:6 = 14:12 therefore, s : m : n = 14 : 12 :9 so susie got (14/35) *770 =308 mary got =264 and nellie =198 Q.4 A king gave each of his sons a few gold coins such that no pair of sons has the same number of gold coins as any other pair of sons. If the king has six sons, What is the minimum number of gold coins that the king could have given his sons? A) 21 B) 28 C) 30 D) 32 Ans. D The series is 1,2,3,5,8,13… ( addition of two consecutive numbers gives the next number). no pair of sons will have equal number of gold coins Q.5 Maria buys one clown fish and 3 guppies for $5 and 2 clown fish and 5 guppies for $9. Find the price of 1 guppy. A) $1 B) $2 C) $3 D) $4 Ans. A Let c be price of clown fish and g be that of guppy. c + 3g = 5 …….(i) and 2c + 5g = 9…………(ii) Solving (i) and (ii), we get g = 1, c = 2