Questions of mensuration, area and volume for aptitude testsPresentation Transcript
Questions of Mensuration, Area and Volume for Aptitude tests by : DR. T.K. JAIN AFTERSCHO ☺ OL centre for social entrepreneurship sivakamu veterinary hospital road bikaner 334001 rajasthan, india FOR – CSE & PGPSE STUDENTS (CSE & PGPSE are free online programmes open for all, free for all) mobile : 91+9414430763
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Two concentric circles form a ring. The inner and outer circumference of the ring are 352/7 m and 528/7m respectively. Find the width of the ring. Solution: Formula of circumference = 2 Pi * radius 2 * 22/7 * radius = 352/7 radius = 8 Formula of circumference = 2 Pi * radius 2 * 22/7 * radius = 528/7 radius = 12 thus width of ring = 12-8 = 4 answer
Four circular cardboard pieces, each of radius 7cm are placed in such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is Solution : area of one circle = (22/7) * 7 * 7 =154 square of on one circle = 14*14 = 196 difference of area : 42 one side 42/4 =10.5 we have 4 cirlces, each has 10.5 cm of space enclosed, so total space enclosed is 42 sq. cm. Answer
A semicircular shaped window has diameter of 63cm. Its perimeter equals Circumference of circle = diameter * 22/7 = 63 * 22/7 = 198 it is semicircle so divide by 2 = 99 add diameter also – to denote one side : 99+63 = 162 cm answer
The length of the room is 5.5m and width is 3.75m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq meter. Total area = 5.5 * 3.75 =20.63 multiply it by 800 =Rs.16500
The no of revolutions a wheel of diameter 49 cm makes in traveling a distance of 176m is Solution : circumference = 22/7 * 49 = 154 17600 / 154 = 114.29 thus the wheel will make 115 revolutions. Answer
.A cow s tethered in the middle of a field with a 14feet long rope.If the cow grazes 100 sq feet per day, then approximately what time will be taken by the cow to graze the whole field? Solution Area 22/7 * 14 * 14 = 616 time required = 616/100 = 6.16 so cow will take little over 6 days to completely graze the whole field. Answer
A man runs round a circular field of radius 49m at the speed of 120 m/hr. What is the time taken by the man to take twenty rounds of the field? Solution : circumference = 2* 22/7 * 49 =308 total distance to be travelled = 308 * 20 = 6160 time required = 6160/120 =51.3 hours.
.The wheel of a motorcycle 70cm in diameter makes 40 revolutions in every 10sec. What is the speed of motorcycle in km/hr? Speed covered in 1 seconds : 4 * 22/7 *70 =880 cm or 8.8 meters. Speed in km per hour : 8.8 * 18/5 = 31.68 km per hour answer
A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will be Solution : its circumference is : 2 * 22/7 *56 = 352 when you make a circle out of it, one side will be : 352/4 = 88 thus its area will be : 88*88 =7744 answer
The area of the largest triangle that can be inscribed in a semicircle of radius 2 is? Area of triangle = Formula = 1/2 * base * height = ½ * (2+2) *2 =4 answer
A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 meters apart. How many poles will be needed? The total boundary = 2(90+50)= 280 280/5 = 56 so we will need 56 poles
The length of a rectangular plot is 20 meters more than its breadth. If the cost of fencing the plot @ 26.50 per meter is Rs. 5300. What is the length of the plot in meter? Total fencing = 5300/26.5 = 200 2(L+B) = 200 2(B+20+B) = 200 2B+20=100 or B=40 and L=60 so length is 60 meters answer
A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq feet, how many feet of fencing will be required? Length = X bredth = 20 20X =680, or X = 34 boundary =2(34+20)=108 but we will not cover one side =108-20=88 feet. Answer
A rectangular paper when folded into two congruent parts had a perimeter of 34cm for each part folded along one set of sides and the same is 38cm. When folded along the other set of sides. What is the area of the paper? When we fold from mid of length = L+2B=34 When we fold from mid of width = 2L+B=38 add them = 3L+3B=72 L+B=24, where L =14, B=10 so area of paper =14*10 =140 answer
A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is? Diagonal = 52*15/60=13 ½ of perimeter=68*15/60=17 X+Y=17 -- 1 st X^2+Y^2 =169 -- 2 nd square of 1 st equation:X^2+Y^2+2XY=289 2XY=120 or XY =60 thus area of field is 60 sq.meter.
The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m Boundary = 10.08/.20 = 50.4 one side is 12.6 area : 158.76 area without pavement : (12.6-6)^2= 43.56 pavement = 115.2 cost = 115.2*.5 = Rs. 57.6 answer
Aman walked diagonally across a square plot. Approximately what was the percent saved by not walking along the edges? Let us assume that the side of square is 1. had he walked along edges, he would have travelled 2. he walked on diagonal so he walked sqrt(2) = 1.41 thus he has saved (2-1.41) =.59 or we can say that he has saved 29.5% answer
.A man walking at the speed of 4 km p.h. crosses a square field diagonally in 30 minutes. The area of the field is The diagonal is 2 km or 2000 meters side of square is : 2000/sqrt(2) =1414 area of field =2000000 sq. m. Or 200 hectares
An error of 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is Instead of 100, we have taken 102 new area : 102*102 = 10404 so error in calculated area is : (10404-10000) = 404 / 10000 * 100 or 4.04% answer
The base of a triangle of 15cm and height is 12cm. The height of another triangle of double the area having the base 20cm is Area of first triangle = ½ 12*15 =90 Area of 2 nd triangle = 180 1/2*20*x=180 X = 18 Height of another traingle is 18 cm.
The sides of a triangle are in the ratio of ½:1/3:1/4. If the perimeter is 52cm, then the length of the smallest side is ½ : 1/3 : ¼ can also be written as : 6:4:3 or the length of longest side is : 6/13*52 = 24 the other two sides are : 16 and 12 cm respectively. Answer
If a square and a rhombus stand on the same base, then the ratio of the areas of the square and the rhombus is: They are of same size. Answer
The area of a circle of radius 5 is numerically what percent its circumference? Circumference of circle =2pi*r =2*22/7*5 =31.43 area = 22/7*5*5 = 78.57 thus area is 250% of the circumference.
One side of a rectangular field is 15m and one of its diagonal is 17m. Find the area of field? The other side of the rectangle is : sqrt(17^2 – 15^2 ) = (289-225) = 64 square root of 64 = 8 so width is 8 area = L * B = 15 * 8=120 sq. m
A lawn is in the form of a rectangle having its sides in the ratio 2:3 The area of the lawn is 1/6 hectares. Find the length and breadth of the lawn. Area of the lawn = 10000/6 sq. Meter 2x*3x = 10000/6 6x^2 = 10000/6 X^2 = 10000 or X =100 length = 200, and bredth = 300 m. Answer
Find the cost of carpeting a room 13m long and 9m broad with a carpet 75cm wide at the rate of Rs 12.40 per sq meter Floor of the room = 13*9 = 117 sq.m. Length of carpet required = 117/.75 =156 meters cost : 156*12.4 = Rs. 1934.4 answer
The length of a rectangle is twice its breadth if its length is decreased by 5cm and breadth is increased by 5cm, the area of the rectangle is increased by 75 sq cm. Find the length of the rectangle. (2x*x) - ( ( 2x-5)*(x+5) ) = -75 2x^2 -(2x^2-25+5x) =-75 -5x=100 or X=20 length = 40, width = 20 answer
In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error percent in the area, calculate from the those measurements. Let us assume length = 200, width = 100 total area : 20000 measurement = 210*96 = 20160 thus there is 160/20000 *100 = .8% variation (increase)
If each side of a square is increased by 25%, find the percentage change in its area? Let us assume present area = 100 * 100 = 10000 new area : 125*125 = 15625 so change in area : 5625 or 56.25%
Find the ratio of the areas of the incircle and circumcircle of a square. The ratio of diameter of incircle and circumcircle is 1:sqrt(2) thus the area (which is in square of radius) will be in ratio : 1 : 2 answer.
If the radius of a circle is decreased by 50% , find the percentage decrease in its area. The area will decrease to square of .5 = .25 so the resulting area will be .75 of the old area or there is 75% decrease in the area. Answer
If the diagonal of a rectangle is 17cm long and its perimeter is 46 cm. Find the area of the rectangle. Area = length * bredth L + b = 23 or L^2+b^2+2lb = 529 L^2+b^2 = 289 or 2lb = 240 L * B = 120 answer
A rectangular grassy plot 110cm by 65cm has a gravel path .5cm (half cm) wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq.mt Plot =110*65 = 7150 net of gravel : (109*64)=6976 cost : (7150-6976)*.8=139.2 answer
The perimeters of two squares are 40cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares. Solution : side of 1 st square : 40/4 = 10 , so, Area of first square = 100, of 2 nd square = 64 difference = 36, one side = 6 so perimeter of 3 rd square = 4*6 = 24 cm answer
A room 5m 44cm long and 3m 74cm broad is to be paved with squre tiles. Find the least number of squre tiles required to cover the floor Find HCF of 544 and 374 first. Divide 544 by 374, remainder is : 170, divide 374 by 170, we get 34, 170 is divisible by 34, so HCF = 34 one side of tile is 34. number of tiles required = (544*374)/(34*34) =176 tiles. Answer
13,16,49,169, 256, .. what will come next, if it will be a repeatition out of this series ? 1+3 = 4 and square of 4 = 16 now 1+6 = 7 and square of 7 = 49 now 4+9 = 13 square of 13 = 169 1+6+9 = 16, square of 16 is 256, 2+5+6 = 13, square of 13 is 169 so the next digit is 169. answer
1,1,4,8,9,27 what will be the next 2 number? 1^2 =1 1^3 = 1 2^2 = 4 2^3 = 8 3^2 = 9 3^3 =27 4^2 = 16 , 4^3 = 64 so next numbers are 16, 64 answer.
What will be the angle in watch at 2.15 pm? Minute hand is at 15, hour hand is at (10*60/55 ) = 10.91 thus there is a gap of 15 – 10.91 = 4.09 the angle is : (4.09/60) *360 =24.54 degree answer
Rain = Drizzle is equal to 1. Boil = Sizzle, 2. Smile = laugh, 3. match = tournament, 4. flow = trickle Rain is bigger form of Drizzle. Thus we are going from bigger to smaller. Similar analogy can be observed in Flow =Trickle
In a family each daughter has the same number of brothers as she has sisters and each son has twice as many sisters as he has brothers. How many daughters and sons does the family have? Options : A 4,3 B. 2,1 C. 3,2 D 5,3 first option gives the answer. For this type of questions, solve through options. In the case of 1 st option, we can see that each daughter has 3 brothers and 3 sisters but each son has 2 brothers and 4 sisters. So this is the answer
If 72 * 23 = 45, 13*12 =12, 23*14 = 25, 64*12 =?? Add 7 and 2 and also 2 and 3 now multiply these : we get : 9*5 = 45 similarly add 1+3 and 1+2 now multiply these we get : 4*3 = 12 similarly, add 2+3 and 1+4, we get : 5*5 = 25, similarly , 6+4 =10 and 1+2 = 3 multiply these : 10*3 = 30 answer
A train passes through a tunnel, whose length is 500 meters, in 1 minute moving at 72 km per hour, the length of the train is ? Solution : - Step 1. convert speed of train in meters per second : 72 * 5/18 = 20 m. p.s. Step 2 : total distance travelled in 1 minute = 20*60 = 1200 step 3. deduct the length of tunnel from this distance : 1200-500 = 700 meters. So answer = 700 meters.
The least number which is a perfect square and is divisible by each of numbers 16,20,24 is ? Look at the factors of these numbers : 16=2^4 20=2^2*5 24=2^3*3 thus taking them all : 2^4*5^2*3^2 = =16*25*9= 3600 answer note : here you have to take all the prime factors in power of multiple of 2.
The least number which should be added to 2497 so that sum is divisible by 5,6,4,3 ? Step 1. take LCM of 5,6,4,3 =product of all prime factors 5*2*3*2 = 60 step 2 : divide 2497 by 60 = remainder is 37 step 3. find difference of 60 and 37 = 23 step 4 : add 23 to 2497 and now it is divisible by 60, so it is divisible by each of the four numbers. Answer=23
.The maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets same number of pens and pencils is ? Solution : - find HCF of 1001 and 910 divide 1001 by 910, remainder is 91, 910 is divisible by 91 so HCF is 91 the number of students is 91 answer
LCM of two numbers is 48..The numbers are in ratio 2:3. The sum of numbers is The two numbers are 2X and 3X 2*3*X = 48 or X = 8 the two numbers are : 16 and 24 the sum of the numbers = 40 answer
LCM and HCFof two numbers are 84 and 21.If ratio of two numbers is 1:4.Then largest of two numbers is ? The numbers are 1X and 4X and their HCF is 21 so the numbers are 21 and 84 respectively. Answer =84
A positive number which when added to 1000 gives a sum , which is greater than when it is multiplied by 1000.The positive integer is? It will be 1. answer
The sum of all possible two digit numbers formed from three different one digit natural numbers when divided by 9 / sum of digits? Let us take 1,2,3 for each digit of the two digits, we have 3 options, so total we have 3*3 =9 options. 9/3 = 3 times – so each digit is repeated 3 times. Thus as a rule it will be 11 * 3 * (1+2+3) = 198 here. Thus divided by 9 gives 22 and divided by 6 gives 33. answer
How many terms are there in 2,4,6,8..........,1024? It is a question on AP (Arithmetic progression). Basic formula is : Tn = a + (n-1)*d a = first term, n= number of terms, d = difference between two terms 1024 = 2 + (n-1)2 2n= 1024 or n =512 answer
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If A+B+C =0, then A^3+B^3+c^3+ABC= ? We know that if A+B+C =0, then : A^3+B^3+C^3=3ABC thus 3ABC+ABC = 4ABC answer
The difference between compound interest and simple interest for 3 years @5% p.a. Can be found out by multiplying the principal by ? Let us assume the principal is 100. simple interest = 15, so the amount will be 115. compound interest = 100 * 1.05*1.05*1.05 = =115.76 Thus the difference between these is : .76 we can find the difference by multiplying the principal with (.76/100) answer
Simple interest on a sum of money is 1/9 th of sum and the number of years and the rate % per annum are equal. What is rate? Formula :( P*R*T)/100 = interest p=principal, r=rate, t=time, here r=t let us assume principal = 9, interest = 1 9 * R^2 / 100 =1 R^2= 100/9 R=10/3 = 3.33 answer
A person invested Rs. 5000 at some rate (simple interest) and 4000 at 1% higher rate of interest. If the interest in both the cases is equal after 4 years. What is the rate of interest in former case? (5000*R*4)/100=(4000*(r+1)*4)/100 5r=4r+4 r=4 answer
On a certain map of India, actual distance is 1450 Km. Between two cities. It is shown as 5 CM on map. What scale is being used? 5Cm represents 1450 Km. Thus 1Cm will represent : 290 Km. Or 1: 29 *10^6
The average price of 10 books is increased by 17 Rupees when one of them whose value is Rs.400 is replaced by a new book. What is the price of new book? Solution : - Total increase in price = 10*17 = 170 price of old item = 400 price of new item = 400+170 = 570 answer
The average marks of girls in a class is 62.5. The average marks of 4 girls among them is 60.The average marks of remaining girls is 63,then what is the number of girls in the class? Solution ...- 62.5X= 240+(X-4)63 62.5X = 240 +63X-252 -.5X = -12 X=24 answer
The average of 2,7,6 and x is 5 and the average of and the average of 18,1,6,x and y is 10 . what is the value of y? Solution : = 2+7+6+X=20, so X = 5 18+1+6+5+Y=50 so Y =20 answer
A team of 8 persons joins in a shooting competition. The best marksman scored 85 points.If he had scored 92 points ,the average score for the team would have been 84.The team scored was? Solution : 84*8= 672 but only 85 was scored instead of 92, so 672-7=665 (team score) average of team:665/8=83.125
16 children are to be divided into two groups A and B of 10 and 6 children. The average percent marks obtained by the children of group A is 75 and the average percent marks of all the 16 children is 76. What is the average percent marks of children of groups B? Solution : 16*76 =1216 – (10*75) = 466 avg. For 6 = 466/6 =77.66 answer shortcut : 10/6 * (76-75) = 1.66, add it to 76= 77.66 answer.
The average temperature of the town in the first four days of a month was 58 degrees. The average for the second ,third,fourth and fifth days was 60 degree .If the temperature of the first and fifth days were in the ratio 7:8 then what is the temperature on the fifth day? (T+W+T+F)-(M+T+W+T) = (60*4)-(58*4)= 8 Difference is just due to 1 st and 5 th day. F-M or 8X-7X = 8 or X =8 temperature on 1 st day : 8*7 = 56 temperature on 5 th day = 8*8 = 64 answer
The average age of a class is 15 years .The average age of boys in the class is 16 years while that of the girls is 14.5 years .What is the ratio of boys to girls in the class. For boys : 16-15 = 1 for girls : 14.5 – 15 = .5 (ignore + - sign) now while calculating final ratio, put these in reverse order : boys : girls = .5:1 (here we have put .5 of girls first and then we have put 1 of boys) thus their ratio is 1:2 answer
A library has an average of 510 visitors on Sundays and 240 on other days .The average number of visitors per day in a month 30 days beginning with a Sunday is? Solution : - We have 5 Sundays : so 5*510=2550 for other days : 25*240=6000 total : 8550, avg=8550/30=285 answer
The difference between C.I and S.I. on a certain sum at 10% per annum for 2 yrs is Rs.631.find the sum Solution : let us assume the sum to be 100. on compound interest it will be : 100*1.1*1.1 = 121 on simple iterest it will be 100+(10*2) = 120 thus difference is 1. sum is : 631*100/1 = 63100 answer
A sum of money doubles itself at C.I in 15yrs.In how many yrs will it become 8 times? Solution : - 1 becomes 2 in 15 years 2 becomes 4 in 30 years 4 becomes 8 in 45 years. Thus it will take 45 years.
Solve the following puzzle : There are six cities A,B,C,D,E and F A is not a hill station. B and E are not historical places. D is not an industrial city. A and D are not historical cities. A and B are not alike. Assuming that there are 2 cities of each type : hill station, historical city and industrial town, Which two cities seem to be industrial centres?
solution... Thus only A and E are industrial towns.
Solve the following puzzle Five friends A,B,C,D,E travelled to five different cities of Chennai,Calcutta,Delhi,Bangalore and Hyderabad by five differnet modes of transport of Bus,Train,Aeroplane,Car and Boat from Mumbai. The person who travelled to Delhi did not travel by Boat. C went to Banalore by Car and B went to Calcutta by Aeroplane. D travelled by Boat whereas E travelled by Train. Mumbai is not connected by Bus to Delhi and Chennai.Person travelling to Delhi went by which mode?
SOLUTION THUS IT IS CLEAR FROM THE PREVIOUS TABLE THAT THE PERSON TRAVELLING TO DELHI TRAVELLED BY TRAIN.
SOLVE THIS PUZZLE Four youngmen A,B,C,D are friendly with four girls E,M,S and L.E and S are friends.C's girlfriend does not like E and S . M does not care for C.B's girlfriend is friendly with E.E does not like A. Who is A's girlfriend?
SOLUTION M IS A'S GIRLFRIEND
SOLVE THIS There are five friends-A,S ,R ,Z and Y .A is shorter than S but taller than Y.R is the tallest. Z is little shorter than S and little taller than A. Who is the shortest? ANSWER : Y<A<Z<S<R thus Y is the shortest. Answer
Solve the following puzzle Among five boys V is taller than M,but not as tall as R,J is taller than D but shorter than M. Who is the tallest in their group? Answer : D<J<M<V<R answer : R is the tallest.
Solve this puzzle S is as much older than K as he is younger than P . N is as old as K .Which of the following is wrong? a)K is younger than P . b)N is younger than P . c)S is older than N . d)P is not the oldest. e)K is younger than S. ANSWER : N=K < S < P SO d IS WRONG
SOLVE THIS PUZZLE From amongst five doctors A,B,C,D and E;four engineers G,H,K and L; six teachers M,N,O,P,Q and R some teams are to be selected.A,B,G,H,O,P and Q are females and the rest are males. Where ever there is a male doctor ,no female teacher. Where ever there is a male engineer, no female doctor. There shall not be more than two male teachers in any team. If the team consists of 3 doctors,2 male engineers and 2 teachers, the members of the team could be: a)A B C K L M R b)B C D K L N R c)C D E K L M N d)C D E K L P R
SOLUTION It is given in question, that we have to take only male engineers. Thus we will be able to take only male doctors. If we take only male doctors, we shall be able to take only male teachers. Thus C is the only option possible (all males). Answer
Solve this puzzle There is a group of six persons A,B,C,D,E and F from a family. They are programmer,manager,lawyer,jeweler,doctor and engineer. The doctor is the grandfather of F who is a programmer. The manager D is married to A. C,the jeweler is married to the lawyer. B is the mother of F and E. There are two married couples in the family. What is the profession of E?
SOLUTION It is clear that the two couples are CB and AD. Thus the profession of A is doctor. Thus only profession left after all these is engineer. Thus profession of E is engineer. Answer
Solve this puzzle In a family of six persons,there are three generations.Each person has seperate profession and also they like different colours.There are two couples in the family.R is a CA and his wife neither is a doctor nor likes green colour. Engineer likes red colour and his wife is a teacher.M is mother-in-law of S and she likes orange colour. V is grandfather of T ,who is principal likes black colour.z is granddaughter of M and she likes blue colour. Z's mother likes white colour. Who is an Engineer?
Solution Thus from the table, it is clear that V is engineer
14 students in a class failed in both QT and Accounts. 20 failed in QT, but 32 failed in Accounts. How many students were there in total ? 20+32 – 14 = 38 students. Answer
Solve the following : All tables are chairs All chairs are almirahs Some almirahs are pens All pens are bikes. Which conclusion can be drawn : 1. all almirahs are tables 2. some bikes are not pens 3.. Some tables are pens 4. some almirahs are bikes..
solution.... You cant be sure that 1 st or 2 nd or 3 rd conclusion is right. There is a possibility that this conclusion is not valid. Thus only 4 th conclusion is a valid conclusion – it is bound to be there. Answer
Solve the following question Statements : 1. some tables are fans 2. some fans are balls 3. Some balls are bats. Which conclusion can be drawn : 1. No table is ball 2. some bats are fans 3. some tables are not bats 4. some bats are tables. SOLUTION : NO CONCLUSION CAN BE DRAWN.
V is 2 years older to T. T is 4 years younger to H. H is of the same age as N. N is 1 year younger to M. What is the age of M, if their total age is 100 years? T seems to be youngest, so let us start with T. V=T+2, H = T+4, N = T+4, M=T+5 Now let us add their age : T+T+2+T+4+T+4+T+5 = 100 5T+15=100, or 5T=85 or T=17 the age of M = 17+5 = 22 answer
Statement v/s argument... Statement : Banks should be privatised. Argument 1 : Yes, they should be privatised, because other countries have also privatised banks. Argument 2: No – they should not be privatised, because government has lots of deposits in Banks. ANSWER : NO ARGUMENT IS STRONG ENOUGH.
Statement v/s argument Statement : Banks should be privatised. Argument 1 : Yes, they should be privatised, because only private sector can be efficient Argument 2: No – they should not be privatised, because they generate employment ANSWER : NO ARGUMENT IS STRONG ENOUGH. IN THE FIRST ARGUMENT THE WORD ONLY CREATES PROBLEMS. IN 2ND ARGUMENT, IT IS NOT CLEAR, WHY EMPLOYMENT IS A PRIORITY FOR BANKS.
There are five friends Lalu, Kalu, Monu, Anuj and raju. Lalu ia shorter than Kalu but taller than raju. Monu is tallest. Anuj is a little shorter than Kalu and a little taller than Lalu. Who is the shortest? R< L <A< K <M answer : Raju
(1)There is a group of five girls. (2)Lata is second in height but younger than Rupa. (3)Pooja is taller than Mem but younger in age. (4)Rupa and Mem are of the same age but Rupa is tallest between them. (5)Niru is taller than Puja and elder to Rupa. Arrange them in age and height AGE : N R M P/ L HEIGHT R L N P M
X knows more than A. Y knows as much as B. Z knows less than C. A nows more than Y. The best knowledge person amongst all is: X>A > Y=B C>Z we are not able to know who knows more Z or A. Thus answer cant be drawn.
Solve the following : Five children were administrated psychological tests to know their intellectual levels. In the report, psychologists pointed out that the child A is less intelligent than the child B. The child C is less intelligent than the child D. The child B is less intelligent than the child C and child A is more intelligent than child E. Which child is most intelligent? D>C>B>A>E thus D is the most intelligent
Solve the following : In an examination, Raj got more marks than Goti but not as many as Neha.Neha got more marks than Jitu and Kamal. Jitu got less marks than Goti but his marks are not the lowest in the group. Who is the second in the descending order of marks? N>R>G>J>K answer = Raj
Solve the following : (A)Gop is shorter than Nanu but taller than Kalu. (B)Lalu is shorter than Kalu. (C)Jitu is taller than Lalu. (D)Nanu is taller than Jitu . Who among them is the tallest? N>G>K / J>L so Nanu is the tallest.
Solve it ?? There are seven students P, Q, R, S, T, U and V (1) they take a series of tests (2)No two students get similar marks. (3)V always scores more than P. (4)P always scores more than Q. (5)Each time either R scores the highest and t gets the least or alternatively S scores the highest and U or Q scores the least. If S is ranked sixth and Q is ranked fifth,solve the questions on next slide ??
Which of these is possible ? a) V is ranked first or fourth (b) R is ranked second or third (c)Pis ranked second or fifth (d) U is ranked third or fourth (e) T is ranked fourth or fifth. Possible solutions : 1 2 3 4 5 6 7 R V P Q S T thus only U can be at 3 rd or 4 th rank. No other options is possible
Who is the oldest ? (1)A, B, C, D and E are five friends. (2)B is elder to E, but not as tall as C. (3)C is younger to A, and is taller to D and E. (4)A is taller to D, But younger to E. (5)D is elder to A but is shorter in the group. B / D>E / D > A>C the oldest can be either B or D.
QUESTIONS... A * B means A and B are the same age. A – B means B is younger than A. A + B means A is younger than B. SOLVE IT : A * b – c means (a) c is youngest (b) c is the oldest (c) b is younger than c (d)None of these ANSWER : c is youngest.
Solve the following Six persons A,B,C,D,E and F are sitting in two rows, three in each. E is not at the end of any row Dis second to the left of F. C the neighbour of E, is sitting diagonally opposite to D. B is the neighbour of F. Who are sitting diagonally opposite? A E C D B F ans : AF and DC are sitting diagonally opposite
Solve it ..... Eight books are kept one over the other counting from the top the second, fifth and sixth books are on plays. Two books on plays are between two books on composition. One book of plays is between two books on poetry. While the book at the top of the book of literature is a book of composition .Which book is fourth from the top? 1 2 3 4 5 6 7 8 P p P C p p C L Thus 4 th book is COMPOSITION.
SOLVE IT ... Four boys are sitting on a bench to be photographed. A is to the left of B. D is to the right of B. C is between B and D. Who would be second from the left in the photograph? A B C D ANS : C In the photo, A will be extreme right and D will be extreme left.
Solve it .... There are five different houses, A to E in a row. A is to the right of B and E is to the left of C and right of A. B is to the right of D. Which of the houses is in the middle? ANSWER : D B A E C THUS A IS IN THE MIDDLE.
SOLVE IT ... In a march past , seven persons are standing in a row. Q is standing left to R but right to P. O is standing right to N and left to P. Similarly , S is standing right to R and left to T. Find out who is standing in middle? ANS : N O P Q R S T the person in MIDDLE Is Q
Solve it ... Five children are sitting in a row. S is sitting next to P but not T who is sitting next to R who is sitting on extreme left and T is not sitting next to K. Who are sitting adjacent to S? Answer : RTPSK answer : P and K are adjacent to S.
Solve it Five Boys are sitting in a row, A is not adjacent to B or C.D is not adjacent to B. A is adjacent to E. E is at the middle in the row. Then, D is adjacent to whom ? DAECB or BCEAD ANSWER : A
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Up to the time the last vote was recorded, it was difficult to decide whether victory lay with the ruling party or the opposition Instead of lay, use lied remember the formats of lie and lay Lay : lay, laid laid lie : lie lay lain lay = to put lie = to rest here we will use lie as an infinitive not as verb
We should always side with those who is true and unselfish and work for others We should always side with those who are true and unselfish and work for others
Before giving the mixture to the child shake it thouroughly. Here it has been used for mixture, but it is coming just after child, thus it seems that child has to be shaked. So correct it : Before giving to the child the mixture, shake it thouroughly.
The short story should not exceed more than two hundred words. Dont use more than exceeds itself means more than, so there is no need of more than or instead of exceed use be of
He looks at everything from their best side. Instead of their, use its everything is singular so use its instead of their.
Page after Page of Gita were read and it gave great consolation to his mind. Instead of were, use was. Page is singular so use was instead of were.
His manners indicate that he has no other intention than to steal his money. Instead of than, use but
I will now deal with him in a manner different from the one I have adopted so far. Instead of I have adopted so far, use : I have been adopting so far (to show continuation).
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