MISCELLANEOUS MCQS

Q.1 The following figure shows the displacement time graph of a body in motion. The ratio of the speed in first second & that in next two seconds is:    A) 1 : 2B) 1 : 3C) 3 : 1D) 2 :1Ans. DRequired ratio : Speed in first second / Speed in next two seconds= Displacement in first second/t    /  Displacement in next two seconds/2t30/1  / 30(3-1) = 30/15 = 2 : 1 Q. 2 Seven players A,B,C,D,E,F & G are to be seated such that :1. B must always occupy second position2. F must always occupy sixth position3. D & E must be seated adjacent to each positionIn how many different ways can they be seated ?A) 8B) 12C) 24D) None of the aboveAns. C Position : 1 2 3 4 5 6 7B F As D & E are seated to each other, they can only occupy at position 3,4 & 5 in 2 * 2 = 4 waysRemaining three A,C & G can occupy in 3! = 6 waysTotal no. of required ways = 4 * 6 = 24 Q. 3 When 60% of a no. A is added to another no. B, B becomes 175 % of its previous value. Which one of the following is correct ?A) A > BB) B > AC) Either of the above can be true depending upon the values of A & bD) None of the above three can be predicted without more dataAns. A 60/ 100 * A+B = 175/100 * B3/5 A = 7/4 B-B3/5A = 3/4BA/B = 5/4 Q.4 A cube of length 1 cm is taken out from a cube of length 8 cm. What is the weight of the remining portion ?A) 7/8 of the weight of the original cubeB) 8/9 of the weight of the original cubeC) 63/64 of the weight of the original cubeD) 511/512 of the weight of the original cubeAns. D Volume of remaining portion = 83 – 13512-1 = 511Since Volume is proportional to weight Then, weight of remaining portion = 511/ 512 of the weight of the original cube Q.5 If Saturday falls four days after today which is 6th January, on which day did the first of December of the previous year fall ?A) TuesdayB) FridayC) SundayD) MondayAns. D Today = 6 JanuaryThe date after 4 days = 10 January = SaturdayThe date before 7 days = 3 January = SaturdayLast years, the date of 31 December = Saturday – 3 days = WednesdayTherefore, the day on December 1 of last year = Wednesday – 2 days = Monday Q.6 Five friends F1, F2, F3, F4, F5 have Rs.100 each. F1 gives Rs. 12 to F3 who also borrows Rs.10 from F2 who in turn takes Rs.15 from F5 while F5 gives Rs.18 to F4 & Rs.12 to F1. F1 borrows Rs.20 from F4. What is the amount of money left with F2 ?A) Rs. 120B) Rs. 122C) Rs. 105D) Rs. 97Ans.  C F1 F2 F3 F4 F5100 100 100 100 100-12 -10 +12 +18 -15+12 +15 +10 -20 -18+20 -12105 Q.7 If 2nd & 4th Saturdays & all the Sundays are taken as holidays for an office, what would be the minimum number of possible working days of any month of any year ?A) 20B) 21C) 22D) 23Ans.  C Let February of 28 days& 1st day is SundayNumber of Sunday in that month = 4Number of Staurday = 4No. of holiday = 4+2 = 6 daysNo. of working days = 28-6 = 22 days Q.8 The average of five consecutive nos. is x. If the next two numbers are also included, how shall the average vary ?A) It shall increase by 1B) It shall remain the sameC) It shall increase by 1.4D) It shall increase by 2Ans. A Let 5 consecutive nos. are (x-2), (x-1), x, (x+1), (x+2), (x+3), (x+4)Their average will be (x+1)Hence , average vary by, x+1-x = 1  

Q.1 Which will be the next term in the following?KPA, LQB, MRC, NSD……A) OETB) OTEC) TOED) EOTAns. B Q.2 Consider the following:1. Every square is a rectangle.2. Every rectangle is a parallelogram.3. Every parallelogram is not necessarily a square.Which one of the following conclusions can be drawn on the basis of the above statements?A) All parallelograms are either squares or rectanglesB) A non-parallelogram figures cannot be either a square or a rectangleC) All rectangles are either squares or parallelogramsD) Squares and rectangles are the only parallelograms.Ans. B Q.3 Abhishek had a certain number of Re1 coins, Rs 2 coins and Rs 10 coins. If the number of Re 1 coins he had is six times the number of Rs 2 coins Abhishek had, and the total worth of his coins is Rs 160, find the maximum number of Rs 10 coins Abhishek could have had.A) 12B) 10C) 8D) 6Ans. AIf the Abhishek had x Re 1, y Rs 2 coins and z Rs 10 coins, the total value of coins he had: =x(1)+y(2)+z(10)=x+2y+10z=160Since, 6y=xThus, 8y+10z=160 i.e 8y is a multiple of 10 i.e. y=5 or y=10 i.e. (x,y,z)=(30,5,12) or (60,10,18)Thus, the maximum value of ‘z’ is 12 Q.4 Select a suitable figure from the four alternatives that would complete the figure matrix. miscset19 A) 1B) 2C) 3D) 4Ans. AIn each row, the third figure comprises of a black circle and only those line segments which are not common to the first and the second figures. Q.5 The following table shows he marks obtained by two students in different subjects:Student A   Maximum Marks Student B Maximum MarksEnglish   60 100     80 150Psychology70 100     70 100History   50 100     60 100Sanskrit  30 50     15 25 The difference in the mean aggregate %ge marks of the students is-A) 2.5%B) 13.75%C) 1.25%D) ZeroAns. DExplanation: Mean %ge of A==(60+70+60+15/150+100+100+25)*100%=225/375 *100%=60%Reqd. mean aggregate %ge marks of the students=60%-60%= Zero Q.6 A bell rings every 18 minutes. A second bell rings every 24 minutes. A third bell rings every 32 minutes. If all the three bells ring at the same time at 3 O’ clock in the morning,at what other time will they all ring together?A) 12:40 hrsB) 12:48 hrsC) 12:56 hrsD) 13:04 hrsAns. BExplanation: L.C.M. of 18, 24 & 32=288Time in hrs=288/60=4:48So, reqd. time = 8+ 4:48=12:48 hrs Q.7 ‘Price is not the same thing as the value. Suppose that on a day the price of everything viz, coal, bread, postage stamps, a day’s labour, the rent of houses, etc. were to double. Prices then would  certainly rise, but values of all things except one would not.The writer wants to say that if prices of all things were doubled-A) The values of all things would remain constant.B) The values of the things sold would be doubledC) The values of the things bought would be halvedD) The value of the money only would be halved.Ans. D Q.8 A & B decide to travel from place X to place Y by bus. A has Rs. 10 with him & he finds that it is 80% of the bus fare for two persons. B finds that he has Rs. 3 with him & hands it over to A. in this context, which one is correct?A) Now the money A has is just enough to buy two tickets.B) A still needs Rs. 2 for buying the tickets.C) After buying the two tickets A will be left with 50 paiseD) The money A now has is still not sufficient to buy two tickets.Ans. CExplanation: Let the reqd.amount is= Rs. XAs per question,80% of x = Rs. 10X= 10*100/80=Rs. 12.5(A+B)’s total amount= Rs. (10+3)=Rs. 13Remaining amount=Rs.(13-12.5)=50 paise Q.9 As per aggrement with a bank, a businessman had to refund a loan in some equal instalments without interest. After paying 18 nstalments he found that 60% of hi loan was refunded. How many instalments were there in the agreement?A) 22B) 24C) 30D) 33Ans. CExplanation: Let the no. of reqd. instalments would be x60% of x= 1860/100 * x= 18X= 30 Q.10 A worker reaches his factory 3 minutes late if his speed from house to factory is 5km/hr. if he walks at a speed of 6km/hr, then he reaches the factory 7 minutes early. The distance of the factory from his house is-A) 3 kmB) 4 kmC) 5 kmD) 6 kmAns. CExplanation: Reqd. distance==5*6/6-5*(3/60+7/60)=5 km  

Q.1 Five persons fire a bullets at a target at an interval of 6,7,8,9,& 12 seconds respectively. The no. of times they would fire the bullets together at the target in an hour is A) 6 B) 7 C) 8 D) 9 Ans. B Explanation:L.C.M. of 6,7,8,9 & 12 = 504Reqd. no. = 60*60/504=7 times Q.2 Assume that-1. the hour & minute hands of a clock move without jerking.2. the clock shows a time between 8 O’ clock & 9 O’clock.3. the two hands of the clock are one above the other.After how many minutes (nearest integer) will the two hands be again lying one above the other? A) 60 B) 62 C) 65 D) 67 Ans. C Explanation: Reqd. time=60*60/55=65 5/11= 65 Q.3 A gardener increased the area of his rectangular garden by increasing its length by 40% & decreasing its width by 20%.The area of the new garden A) has increased by 20% B) has increased by 12% C) has increased by 8% D) Same as the old area Ans. B Explanation:The area of the new garden==(40-20- 40*20/100)%=(20-80)%=12% Q.4 If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non blue sector? A) Green = 3/5 ; Non-blue = 4/5 B) Green = 2/5 ; Non-blue = 3/5 C) Green = 4/5 ; Non-blue = 3/5 D) Green = 3/5 ; Non-blue = 2/5 Ans. A Total number of events = 5 Q.5 A bag has 4 red balls and 2 yellow balls. (The balls are identical in all respects other than colour). A ball is drawn from the bag without looking into the bag. What is probability of getting a red ball? A) 2/3 B) 2/4 C) 2/5 D) 2/6 Ans. A There are in all (4 + 2 =) 6 outcomes of the event. Getting a red ballconsists of 4 outcomes. Therefore, the probability of getting a red ball is4/6 = 2/3 Q.6 There are 2401 students in a school. P.T. teacher wants them to stand in rows and columns such that the number of rows is equal to the number of columns. Findthe number of rows. A) 45 B) 47 C) 49 D) 51 Ans. C Let the number of rows be xSo, the number of columns = xTherefore, number of students = x × x = x2Thus, x2 = 2401 gives x = Sq. root of 2401 = 49The number of rows = 49 Q.7 The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there? A) 40 & 24 B) 42 & 24 C) 42 & 20 D) None of the above Ans. B Let the ten’s and the unit’s digits in the first number be x and y, respectively.So, the first number may be written as 10 x + y in the expanded form (for example,56 = 10(5) + 6).When the digits are reversed, x becomes the unit’s digit and y becomes the ten’sdigit. This number, in the expanded notation is 10y + x (for example, when 56 isreversed, we get 65 = 10(6) + 5).According to the given condition.(10x + y) + (10y + x) = 66i.e., 11(x + y) = 66i.e., x + y = 6 (1)We are also given that the digits differ by 2, therefore,either x – y = 2 (2)or y – x = 2 (3)If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2.In this case, we get the number 42.If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4.In this case, we get the number 24.Thus, there are two such numbers 42 and 24. Q.8 Satpal walks 2/3 km from a place P, towards east and then from there 1 5/7 km towards west. Where will he be now from P? A) Distance of 1 1/21 km towards west of P B) Distance of 1 7/21 km towards west of P C) Distance of 1 1/21 km towards east of P D) Distance of 1 7/21 km towards east of P Ans. A 2/3 + (-1 5/7) = 2/3 + (-12)/7= 2 * 7 / 3 * 7 + (-12) * 3 / 7 * 314-36 / 21 = 1 1/21 Since it is negative, it means Satpal is at a distance 1 1/21km towards west of P. Q.9 Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is ¾ m. Find the distance between the first and the last sapling A) 3 m B) 4 m C) 5 m D) 6 m Ans. A With 4 saplings there are 3 gaps: between 1 and 2, 2 and 3, and 3 & 43(3/4 m) = 9/4 m = 2.25 m Q.10 In a class of 40 students 1/5 of the total number of students like to study English, 2/5 of the total number like to study mathematics and the remaining students like to study Science. (i) How many students like to study English? A) 6 B) 7 C) 8 D) 9 Ans. C Total number of students in the class = 40.(i) Of these 1/5 of the total number of students like to study EnglishThus, the number of students who like to study English =1/5 of 40 = 8. (ii) How many students like to study Mathematics? A) 14 B) 15 C) 16 D) 17 Ans. C Total students = 40Students who like English = 1/5 * 40 = 8Students who like to study maths = 2/5 * 40 = 16 (iii) What fraction of the total number of students like to study Science? A) 1/5 B) 2/5 C) 3/5 D) 4/5 Ans. B Students who like

Q.1 Salil wants to put a picture in a frame. The picture is 7 3/5 cm wide. To fit in the frame the picture cannot be more than 7 3/10 cm wide. How much should the picture be trimmed ? A) 3/9 B) 2/10 C) 3/10 D) 2/9 Ans. C Therefore, picture to be trimmed Thus picture should be trimmed by 3/10 cm Q.2 Suman studies for 5 2/3 hours daily. She devotes 4 2/5 hours of her time for Science and Mathematics. How much time does she devote for other subjects? A) 1 13/15 B) 2 13/15 C) 1 12/15 D) 2 12/15 Ans. B Total time of Suman’s study =5 2/3 hr = 17/3 hr.Time devoted by her for Science and Mathematics = 2 4/5 = 14/5 hr Thus, time devoted by her for other subjects =17/3 – 14/5 hr17 * 5 / 15 – 14 * 3/ 15= 43/15 hr = 2 13/15 hr Q.3 Ritu ate 3/5 part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? A) 2/4 part B) 2/5 part C) 2/6 part D) 2/7 part Ans. BThe part eaten by her brother Somu =?Who eat the larger part of apple?The larger part of apple was how much larger?The part of apple left after eaten by RituTherefore, Somu ate 2/5 parts of apple.Now compare the part eaten by them Since, 3>2. Therefore, That means Ritu ate larger part.The difference in both parts Thus,Somu ate 2/5 part of apple.Ritu ate the larger part of appleRitu ate 1/5 part more apple than her brother Somu. Q.4 The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2.Find the rational number ? A) 13/ 21 B) 13/20 C) 11/21 D) 11/ 20 Ans. A Acc. to the question, 2(x + 17) = 3(x + 7)⇒ 2x + 34 = 3x + 21⇒ 34 − 21 = 3x − 2x⇒13 = xNumerator of the rational number = x = 13Denominator of the rational number = x + 8 = 13 + 8 = 21

Q.1 There are 2401 students in a school. P.T. teacher wants them to stand in rows and columns such that the number of rows is equal to the number of columns. Find the number of rows. A) 45 B) 47 C) 49 D) 51 Ans. C Let the number of rows be xSo, the number of columns = xTherefore, number of students = x × x = x2Thus, x2 = 2401 gives x = Sq. root of 2401 = 49The number of rows = 49 Q.2 Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is ¾ m. Find the distance between the first and the last sapling A) 3 m B) 4 m C) 5 m D) 6 m Ans. A With 4 saplings there are 3 gaps: between 1 and 2, 2 and 3, and 3 & 43(3/4 m) = 9/4 m = 2.25 m Q.3 Salil wants to put a picture in a frame. The picture is 7 3/5 cm wide. To fit in the frame the picture cannot be more than 7 3/10 cm wide. How much should the picture be trimmed ? A) 3/9 B) 2/10 C) 3/10 D) 2/9 Ans. C Therefore, picture to be trimmed Thus picture should be trimmed by 3/10 cm Q.4 Ritu ate 3/5 part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? A) 2/4 part B) 2/5 part C) 2/6 part D) 2/7 part Ans. BThe part eaten by her brother Somu =?Who eat the larger part of apple?The larger part of apple was how much larger?The part of apple left after eaten by RituTherefore, Somu ate 2/5 parts of apple.Now compare the part eaten by them Since, 3>2. Therefore, That means Ritu ate larger part.The difference in both parts Thus,Somu ate 2/5 part of apple.Ritu ate the larger part of appleRitu ate 1/5 part more apple than her brother Somu. Q.5 Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd. A) 70 B) 71 C) 72 D) 73 Ans. C let x be total no. of deer, x/2 be half of the herd, 3/4{x/2} be three fourth of remaining half, so, x/2 + 3/4{(x/2} + 9 = x x/2 + 3x/8 + 9 = x 8x/8 – 4x/8 – 3x/8 = 9 x/8 = 9 x = 72

Q.1 Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him Rs 50 per metre and trouser material that costs him Rs 90 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,600. How much trouser material did he buy? A) 198 m B) 199 m C) 200 m D) 201 m Ans. C Let 2x m of trouser material and 3x m of shirt material be bought by him.Per metre selling price of trouser material = = Rs 100.80Per metre selling price of shirt material = = Rs 55Given that, total amount of selling = Rs 36660100.80 × (2x) + 55 × (3x) = 36660201.60x + 165x = 36660366.60x = 36660Dividing both sides by 366.60, we obtainx = 100Trouser material = 2x m = (2 × 100) m = 200 m Q.2 The organisers of an essay competition decide that a winner in the competition gets a prize of Rs 100 and a participant who does not win gets a prize of Rs 25. The total prize money distributed is Rs 3,000. Find the number of winners, if the total number of participants is 63. A) 16 B) 17 C) 18 D) 19 Ans. D Let the number of winners be x. Therefore, the number of participants who did not win will be 63 – x.Amount given to the winners = Rs (100 × x) = Rs 100xAmount given to the participants who did not win = Rs [25(63 – x)]= Rs (1575 – 25x)According to the given question,100x + 1575 – 25x = 3000On transposing 1575 to R.H.S, we obtain75x = 3000 – 157575x = 1425On dividing both sides by 75, we obtainx = 19Hence, number of winners = 19 Q.3 Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4,00,000. How many notes of each denomination does she have? A) Rs. 10 notes= 5000 , Rs.50 notes = 3000 , Rs. 100 notes = 2000 B) Rs. 10 notes = 5000 , Rs. 50 notes = 3000 , Rs. 100 notes = 1000 C) Rs. 10 notes = 5000 , Rs. 50 notes = 3000 , Rs. 100 notes = 2500 D) Rs. 10 notes = 5000 , Rs. 50 notes = 3000 , Rs. 100 notes = 1500 Ans. A Let the common ratio between the numbers of notes of different denominations be x. Therefore, numbers of Rs 100 notes, Rs 50 notes, and Rs 10 notes will be2x, 3x, and 5x respectively.Amount of Rs 100 notes = Rs (100 × 2x) = Rs 200xAmount of Rs 50 notes = Rs (50 × 3x)= Rs 150xAmount of Rs 10 notes = Rs (10 × 5x) = Rs 50xIt is given that total amount is Rs 400000.? 200x + 150x + 50x = 400000? 400x = 400000On dividing both sides by 400, we obtainx = 1000Number of Rs 100 notes = 2x = 2 × 1000 = 2000Number of Rs 50 notes = 3x = 3 × 1000 = 3000Number of Rs 10 notes = 5x = 5 × 1000 = 5000 Q.4 A person has only Rs. 1 and Rs. 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs. 75, then the number of Rs. 1 and Rs. 2 coins are, respectively A) 15 and 35 B) 35 and 15 C) 30 and 20 D) 25 and 25 Ans. D 25 and 25Assume no. of Rs 1 coins = x and no. of Rs 1 coins = yx + y = 50 (given)x + 2y = 75 (total money)hence x = 25 and y = 25 Q.5 In a queue, Mr. X is fourteenth from the front and Mr. Y is seventeenth from the end, while Mr. Z is exactly in between Mr. X and Mr. Y. If Mr. X is ahead of Mr. Y and there are 48 persons in the queue, how many persons are there between Mr. X and Mr. Z ? A) 6 B) 7 C) 8 D) 9 Ans. C +*******************+16X(14th) 8 Z(in Middle) 8 (17th)Required no. of persons between Mr. X & Mr. Z= 8

Q.1 In a rare coin collection, there is one gold coin for every three non-gold coins. 10 more gold coins are added to the collection and the ratio of gold coins to non-gold coins would be 1: 2. Based on the information; the total number of corns in the collection now becomes A) 90 B) 80 C) 60 D) 50Ans. ALet the no,of Gold coins be X & non-gold coins be 3x.As per the question-X+10/3x = ½2(x+10)= 3x2x+20 = 3xTherefore, x= 20Now, total no. of coins in the collection:=(x+10)+3x=20+10+3*2030+60=90 Q.2 A gardener has 1000 plants: He wants to plant them in such a way that the number of rows and the number of columns remains the same. What is the minimum number of plants that he needs more for this purpose?A) 14 B) 24 C) 32 D) 34Ans. BMinimum no. of plants=(32)2 – 10001024-1000=24 Q.3 A sum of RS. 700 has to be used to give seven cash prizes to the students of a school for their overall academic performance. If each prize is RS. 20 less than its preceding prize, what is the least value of the prize? A) RS. 30 B) RS. 40 C) RS. 60 D) RS. 80Ans. BLet the least value of the prize will be Rs. XThen as per question,X+(x+20) +(x+40) + (x+60) + (x+80) + (x+100) + (x+120)=7007x + 420=7007x=700-4207x= 280X=280/7 = 40 Ans. Q.4 Four friends, A, B, C and D distribute some money among themselves in such a manner that A gets one less than B, C gets 5 more than D, D gets 3 more than B. Who gets the smallest amount?A) A B) B C) C D) DAns. ALet the amount distribute in four friends A,B,C & Di.e.A–? XB—> x+1D–? (x+1) + 3 -? (x+4)C–? (x+4) +5 —? (x+9)Hence, A gets the smallest amount Q.5 A, B, C, D and E belong to five different cities P, Q, R, S and T (not necessarily in that order). Each one of them comes from a different city. Further it is given that: 1. Band C do not belong to Q. 2. Band E do not belong to P and R. 3. A and C do not belong to R, Sand T. 4. D and E do not belong to Q and T. Which one of the following statements is not correct? A) C belongs to P B) D belongs to R C) A belongs to Q D) B belongs to SAns. D

Q.1 There are some balls of red, green and yellow colour lying on a table. There are as many red balls as there are yellow balls. There are twice as many yellow balls as there are green ones. The number of red balls A) is equal to the sum of yellow and green balls. B) is double the number of green balls. C) is equal to yellow balls minus green balls. D) cannot be ascertained.Ans. BLet the no. of green balls are X.Then yellow balls will be 2x & also the red balls.Hence, the no. of red balls is double the no. of green ballds. Q.2 In a class of 45 students, a boy is ranked 20th. When two boys joined, his rank was dropped by one. What is his new rank from the end? A) 25th B) 26th C) 27th D) 28thAns. C(45+2) – (20+1) + 1= 47-21+1=48-21 = 27th Q.3 A factory employes skilled workers, unskilled workers & clerks in the ratio of 8 : 5 : 1 & the wages of a skilled worker , an unskilled worker & a clerk are in the ratio of 5 : 2 : 3. When 20 unskilled workers are employed the total daily wages of all amount to Rs. 318. Find out the daily wages paid to each category of employees ?A) Unskilled worker= Rs.3 , Skilled worker= Rs.7.50 , Clerk= Rs. 4.50B) Unskilled worker= Rs.4 , Skilled worker= Rs. 8.50 , Clerk= Rs.5.50C) Unskilled worker= Rs.5 , Skilled worker= Rs.9.50 , Clerk= Rs.6.50D) Unskilled worker= Rs.6 , Skilled worker= Rs.10.50 , Clerk= Rs.7.50Ans. A Q.4 Six faces of a cube numbered from 1 to 6, each face carrying one different number.1) Face 2 is opposite to the face 62) Face 1 is opposite to face 53) Face 3 is opposite to face 1 & face 54) face 4 is adjacent to face 2Which of the following is correct ?A) Face 6 is between face 2 & face 4B) Face 2 is adjacent to face 3C) face 1 is between the face 5 & face 6D) None of the aboveAns. B3 & 4 are opposite to each other.Since 2 is adjacent of 4, hence it must be adjacent of 3 Q.5 A college starts from 10 a.m. & continues till 1:30 p.m. In this duration there are 5 periods . If 5 min are provided before each period to leave the room & enter the other, then what is the duration of each period ? A) 41 minB) 42 min.C) 40 minD) 38 min.Ans. DDuration of college time=1:30 pm – 10:00 am= 13:30 – 10:003:30 hourOr 210 min.Time left= 5 * 4 = 20minTherefore,Time for 5 periods = 210 – 20 = 190 min.Therefore,Duration of 1 period= 190/5 = 38 min

PASASAGE:An airline has a certain free luggage allowance & charges for excess luggage at a fixed rate per kg. Two passengers Raja & Praja have 60 kg of luggage between them, and are charged Rs.1200 & Rs.2400 for excess luggage. Had the entire luggage belonged to one of them , the excess luggage charge would have been Rs. 5400. Q.1 What is the weight of the Praja’s luggage ?A) 20 kgB) 35 kgC) 40 kgD) 30 kgAns. BLet free luggage allowed be x kgRaja & Praja are allowed free luggage of 2xkgLet charge for excess luggage= Rs. Y /kgRaja & Praja pay Rs. 1200 + Rs. 2400 = 3600 excess luggageY(60-2x) = 3600–iIf entire luggage had belonged to one of them , excess luggage charges would have been Rs.5400Y (60-x) = 5400—iiFrom these two equations:60y- 2xy = 360060y – xy =5400Now. Solve these two equations;Y = 120X = 15 kgSince Praja is charged Rs.2400 for excess luggage is 2400/120 = 20kgHence weight of Praja’s luggage + her excess luggage= 15 + 20kg=35 kg Q.2 What is the free luggage allowance per person ?A) 10 kgB) 15 kgC) 20 kgD) 30 kgAns. B Q.3 Two boys begin together to write out a booklet containing 8190 lines. The first boy starts with the first line , writing at the rate of 200 lines an hour & the second boy starts with the last line, then writes 8189th line & so on, proceeding backward at the rate of 150 lines an hour. At what line will they meet ?A) 4910B) 4680C) 4280D) 5200Ans. BRelative speed = 200 + 150 = 350 lines/hrSo, they will meet after =8190 / 350 = 117/5 hrsTherefore, they will meet at 117/5*200 = 4680th line from the beginning Q.4 If 20% of P = 30% of Q = 1/6 of R, then P:Q:R is:A) 3 : 2: 16B) 2 :3 : 16C) 15 : 10 : 18D) 10 : 15 : 18Ans. CPQR=1/5:3/10:1/6=15:10:18 Q.5 A local train is going from Delhi to Saharanpur & it has 4 stoppages in between. If two passengers enter in the train during the journey , how many different sets of tickets they may had ? A) 90 B) 120 C) 45 D) 180 Ans. C 1st station = 4tickets 2nd station = 3 tickets 3rd station = 2 tickets 4th station = 1 ticket Total no. of tickets = 4+3+2+1 = 10 Therefore, sets of ticket they may had = 10C2 = 45

Q.1 If the king gave less than 100 gold coins on the whole to this sons, what is the maximum number of sons he could have had ? A) 7 B) 10 C) 11 D) 8 Ans. D On similar lines as in Q1, we add on the series with the increasing differenceThe sum of the series = 1+2+4+7+11+16+22+29 = 92 which is less than 100So, a maximum of 8 sons Q.2 A man engaged a servant on a condition that he will pay Rs 90 and also give him a bag at the end of the yr. He served for 9 months and was given a turban and Rs 65. So the price of turban is. A) 10 B) 9 C) 19 D) 50 Ans. A (90+B)/12=(65+T)/9=>270+3B=260+4T=>3B+10=4T=>T=10 Q.3 A man fills a basket with eggs in such a way that the number of eggs added on each successively day is same as the number already present in the basket. This way the basket gets completely filled in 24 days. After how many days basket was 1/4th full ? A) 6 B) 17 C) 22 D) 12 Ans. C It is written in 23rd day it was half fill & on 22nd day it was ¼ filled Q.4 A lady buys goods worth Rs.200 from a shop (shopkeeper sells the goods with zero profit). The lady gives him Rs.1000 note. The shopkeeper gets the change from the next shop and keeps 200 for himself and returns 800 to the lady. Later the shopkeeper of the next shop comes with 1000 rs note saying that it’s a duplicate and takes his money back. How much loss did the shopkeeper face? A) 1200 B) 1800 C) 1000 D) 2000 Ans. D Shopkeeper gives Rs 800 and goods which worths Rs 200 to that lady And he gives Rs1000 to his next shopkeeper So, Rs2000 Loss Q.5 A schoolyard contains only bicycles and 4 wheeled wagons. On Tuesday, the total number of wheels in the schoolyard was 90. What could be possible number of bicycles? A) 11 B) 12 C) 14 D) 10 Ans. A