regisss/math_qa huggingface dataset:
Problem: the output of a factory is increased by 10 % to keep up with rising demand . to handle the holiday rush , this new output is increased by 50 % . by approximately what percent would the output of the factory now have to be decreased in order to restore the original output ?
Options: a ) 20 % , b ) 24 % , c ) 30 % , d ) 32 % , e ) 39 %
Rationale: "take it as original output = 100 . to meet demand increase by 10 % , then output = 110 . to meet holiday demand , new output increase by 50 % then output equals 165 to restore new holidy demand output to original 100 . final - initial / final * 100 = 65 / 165 * 100 = 39 % approxiamately . option e is correct ."
Answer: e
Problem: the average of first 10 even numbers is ?
Options: a ) 19 , b ) 18 , c ) 16 , d ) 11 , e ) 17
Rationale: "sum of 10 even numbers = 10 * 11 = 110 average = 110 / 10 = 11 answer : d"
Answer: d
Problem: in town p , 60 percent of the population are employed , and 45 percent of the population are employed males . what percent of the employed people in town p are females ?
Options: a ) 20 % , b ) 25 % , c ) 30 % , d ) 35 % , e ) 40 %
Rationale: "the percent of the population who are employed females is 60 - 45 = 15 % the percent of employed people who are female is 15 % / 60 % = 25 % . the answer is b ."
Answer: b
Problem: there are 2 sections a and b in a class , consisting of 26 and 34 students respectively . if the average weight of section a is 50 kg and that of section b is 30 kg , find the average of the whole class ?
Options: a ) 35.29 kg , b ) 37.25 kg , c ) 42.45 kg , d ) 38.66 kg , e ) 29.78 kg
Rationale: "total weight of 26 + 34 students = 26 * 50 + 34 * 30 = 2320 average weight of the class is = 2320 / 60 = 38.66 kg answer is d"
Answer: d
Problem: find the amount on rs . 4000 in 2 years , the rate of interest being 4 % per first year and 5 % for the second year ?
Options: a ) 4368 , b ) 2678 , c ) 5460 , d ) 1976 , e ) 1671
Rationale: "4000 * 104 / 100 * 105 / 100 = > 4368 answer : a"
Answer: a
Problem: a shopkeeper forced to sell at cost price , uses a 900 grams weight for a kilogram . what is his gain percent ?
Options: a ) 10 % , b ) 9 % , c ) 11.11 % , d ) 12 % , e ) none of these
Rationale: "shopkeeper sells 900 g instead of 1000 g . so , his gain = 1000 - 900 = 100 g . thus , % gain = ( 100 * 100 ) / 900 = 11.11 % . answer : option c"
Answer: c
Problem: john want to buy a $ 100 trouser at the store , but he think it â € ™ s too expensive . finally , it goes on sale for $ 50 . what is the percent decrease ?
Options: a ) 20 % , b ) 30 % , c ) 40 % , d ) 50 % , e ) 60 %
Rationale: "the is always the difference between our starting and ending points . in this case , it â € ™ s 100 â € “ 50 = 40 . the â € œ original â €  is our starting point ; in this case , it â € ™ s 100 . ( 50 / 100 ) * 100 = ( 0.5 ) * 100 = 50 % . d"
Answer: d
Problem: what are the last two digits of ( 301 * 402 * 503 * 604 * 646 * 547 * 449 * 349 ) ^ 2
Options: a ) 96 , b ) 76 , c ) 56 , d ) 36 , e ) 24
Rationale: "( ( 301 * 402 * 503 * 604 * 646 ) * ( 547 * 449 * 349 ) ) ^ 2 if you observe above digits , last digit are : 1,2 , 3,4 , 6,7 , 9,9 ; 5 & 9 are missing ; so i have rearranged them so that multiplication will be easy for me as initial 4 digits have last two digits as 01 , 02,03 , 04,46 and final three as 47 * 49 * 49 . solving for only last two digits and multiplying them we get : ( ( 06 * 04 * 46 ) ( 47 * 01 ) ) ^ 2 = ( 44 * 47 ) ^ 2 = 68 ^ 2 = 24 hence answer is e"
Answer: e
Problem: divide rs . 1800 among a , b and c so that a receives 2 / 5 as much as b and c together and b receives 1 / 5 as a and c together . a ' s share is ?
Options: a ) 400 , b ) 500 , c ) 375 , d ) 450 , e ) 600
Rationale: a + b + c = 1800 a = 2 / 5 ( b + c ) ; b = 1 / 5 ( a + c ) a / ( b + c ) = 2 / 5 a = 1 / 6 * 3600 = > 600 answer : e
Answer: e
Problem: the ratio between the number of sheep and the number of horses at the stewar farm is 4 to 7 . if each of horse is fed 230 ounces of horse food per day and the farm needs a total 12880 ounces of horse food per day . what is number sheep in the form ? ?
Options: a ) 18 , b ) 28 , c ) 32 , d ) 56 , e ) 58
Rationale: "et no of sheep and horses are 4 k and 7 k no of horses = 12880 / 230 = 56 now 7 k = 56 and k = 8 no of sheep = ( 4 * 8 ) = 32 answer : c"
Answer: c
Problem: a man purchased 3 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 154 . find the unknown rate of two blankets ?
Options: a ) 420 , b ) 550 , c ) 490 , d ) 450 , e ) 457
Rationale: "10 * 154 = 1540 3 * 100 + 5 * 150 = 1050 1540 – 1050 = 490 answer : c"
Answer: c
Problem: last year , company x paid out a total of $ 1 , 050,000 in salaries to its 21 employees . if no employee earned a salary that is more than 15 % greater than any other employee , what is the lowest possible salary that any one employee earned ?
Options: a ) $ 40,000 , b ) $ 41,667 , c ) $ 42,000 , d ) $ 43,750 , e ) $ 60,000
Rationale: "employee 1 earned $ x ( say ) employee 2 will not earn more than $ 1.15 x therfore , to minimize the salary of any one employee , we need to maximize the salaries of the other 20 employees ( 1.15 x * 20 ) + x = 1 , 050,000 solving for x = $ 43,750 answer d"
Answer: d
Problem: in a graduate physics course , 70 percent of the students are male and 30 percent of the students are married . if two - sevenths of the male students are married , what fraction of the female students is married ?
Options: a ) 2 / 7 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 5 / 7
Rationale: "let assume there are 100 students of which 70 are male and 30 are females if 30 are married then 70 will be single . now its given that two - sevenths of the male students are married that means 2 / 7 of 70 = 20 males are married if 30 is the total number of students who are married and out of that 20 are males then the remaining 10 will be females who are married . total females = 30 married females = 10 then single females = 30 - 10 = 20 we need to find the fraction of female students who are single i . e single female students / total female student = 10 / 30 = 1 / 3 [ b ]"
Answer: b
Problem: the area of a rectangle is 63 sq m . the width is two meters shorter than the length . what is the width ?
Options: a ) 9
b ) 7
c ) - 9
d ) 11
e ) 6
Rationale: a = l x w w = l - 2 l = w + 2 a = ( w + 2 ) x w a = w ^ 2 + 2 x w 63 = w ^ 2 + 2 w 0 = w ^ 2 + 2 w - 63 0 = ( w + 9 ) ( w - 7 ) w = - 9 and w = 7 , width can not be negative so w = 7 answer is b
Answer: b
Problem: when the positive integer x is divided by 9 , the remainder is 5 . what is the remainder when 8 x is divided by 9 ?
Options: a ) 0 , b ) 1 , c ) 3 , d ) 4 , e ) 6
Rationale: "i tried plugging in numbers x = 9 q + 5 x = 14 8 x = 112 8 x / 9 = 9 * 12 + 4 remainder is 4 . answer is d ."
Answer: d
Problem: a man can row upstream at 25 kmph and downstream at 65 kmph , and then find the speed of the man in still water ?
Options: a ) 45 , b ) 86 , c ) 30 , d ) 78 , e ) 38
Rationale: "us = 25 ds = 65 m = ( 65 + 25 ) / 2 = 45 answer : a"
Answer: a
Problem: a train crosses a bridge of length 150 m in 7.5 seconds and a lamp post on the bridge in 2.5 seconds . what is the length of the train in metres ?
Options: a ) 37.5 m , b ) 75 m , c ) 25 m , d ) 80 m , e ) 30 m
Rationale: let length of train = l case - 1 : distance = 150 + l ( while crossing the bridge ) time = 7.5 seconds i . e . speed = distance / time = ( 150 + l ) / 7.5 case - 2 : distance = l ( while passing the lamp post ) time = 2.5 seconds i . e . speed = distance / time = ( l ) / 2.5 but since speed has to be same in both cases so ( 150 + l ) / 7.50 = ( l ) / 2.5 i . e . 3 l = l + 150 i . e . 2 l = 150 i . e . l = 75 answer : option b
Answer: b
Problem: at a special sale , 12 tickets can be purchased for the price of 3 tickets . if 12 tickets are purchased at the sale , the amount saved will be what percent of the original price of the 12 tickets ?
Options: a ) 20 % , b ) 33.3 % , c ) 40 % , d ) 50 % , e ) 75 %
Rationale: "let the price of a ticket be rs . 100 , so 3 tickets cost 300 & 12 tickets cost 1200 12 tickets purchased at price of 3 tickets ie . , for 300 , so amount saved s rs . 900 , % of 5 tickets = ( 900 / 1200 ) * 100 = 75 % answer : e"
Answer: e
Problem: if x > 0 , x / 40 + x / 20 is what percent of x ?
Options: a ) 6 % , b ) 25 % , c ) 37 1 / 2 % , d ) 60 % , e ) 75 %
Rationale: "just plug and chug . since the question asks for percents , pick 100 . ( but any number will do . ) 100 / 40 + 100 / 20 = 2.5 + 5 = 7.5 7.5 is 75 % of 100 = e"
Answer: e
Problem: the average age of students of a class is 15.8 years . the average age of boys in the class is 16.6 years and that of the girls is 15.4 years . the ration of the number of boys to the number of girls in the class is :
Options: a ) 1 : 2 , b ) 2 : 3 , c ) 2 : 4 , d ) 2 : 1 , e ) 2 : 9
Rationale: "let the ratio be k : 1 . then , k * 16.6 + 1 * 15.4 = ( k + 1 ) * 15.8 = ( 16.6 - 15.8 ) k = ( 15.8 - 15.4 ) = k = 0.4 / 0.6 = 1 / 2 required ratio = 1 / 1 : 1 = 1 : 2 . answer : a"
Answer: a
Problem: solution a is 20 % salt and solution b is 60 % salt . if you have 30 ounces of solution a and 60 ounces of solution b , in what ratio could you mix solution a with solution b to produce 50 ounces of a 50 % salt solution ?
Options: a ) 6 : 4 , b ) 6 : 14 , c ) 4 : 4 , d ) 4 : 6 , e ) 1 : 3
Rationale: "forget the volumes for the time being . you have to mix 20 % and 80 % solutions to get 50 % . this is very straight forward since 50 is int he middle of 20 and 80 so we need both solutions in equal quantities . if this does n ' t strike , use w 1 / w 2 = ( a 2 - aavg ) / ( aavg - a 1 ) w 1 / w 2 = ( 60 - 50 ) / ( 50 - 20 ) = 1 / 3 so the volume of the two solutions will be equal . answer has to be 1 : 3 e ."
Answer: e
Problem: if 11.25 m of a uniform steel rod weighs 42.75 kg . what will be the weight of 8 m of the same rod ?
Options: a ) 22.8 kg , b ) 25.6 kg , c ) 28 kg , d ) 30.4 kg , e ) none of these
Rationale: explanation : let the required weight be x kg . then , less length , less weight ( direct proportion ) = > 11.25 : 8 : : 42.75 : x = > 11.25 x x = 8 x 42.75 = > x = ( 8 x 42.75 ) / 11.25 = > x = 30.4 answer : d
Answer: d
Problem: the radius of the wheel of a bus is 250 cms and the speed of the bus is 66 km / h , then the r . p . m . ( revolutions per minutes ) of the wheel is
Options: a ) 70 , b ) 125 , c ) 300 , d ) 500 , e ) none of these
Rationale: "radius of the wheel of bus = 250 cm . then , circumference of wheel = 2 ï € r = 500 ï € = 1571.43440 cm distance covered by bus in 1 minute = 66 â  „ 60 ã — 1000 ã — 100 cms distance covered by one revolution of wheel = circumference of wheel = 1571.45 cm â ˆ ´ revolutions per minute = 6600000 / 60 ã — 1571.43 = 70 answer a"
Answer: a
Problem: 1000 men have provisions for 21 days . if 800 more men join them , for how many days will the provisions last now ?
Options: a ) 12.9 , b ) 12.5 , c ) 12.6 , d ) 11.6 , e ) 12.1
Rationale: "1000 * 21 = 1800 * x x = 11.6 answer : d"
Answer: d
Problem: a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 / 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 40 square feet of the wall , and no tiles overlap , what is the area of the entire wall ?
Options: a ) 160 , b ) 220 , c ) 360 , d ) 440 , e ) 560
Rationale: "the number of jumbo tiles = x . the number of regular tiles = 2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ 2 . the dimensions of a jumbo tile is 3 a : 3 a - - > area = 9 a ^ 2 . the area of regular tiles = 2 x * a ^ 2 = 40 . the area of jumbo tiles = x * 9 a ^ 2 = 4.5 ( 2 x * a ^ 2 ) = 4.5 * 40 = 180 . total area = 40 + 180 = 220 . answer : b ."
Answer: b
Problem: how many positive integers less than 10,000 are such that the product of their digits is 210 ?
Options: a ) 24 , b ) 30 , c ) 48 , d ) 54 , e ) 72
Rationale: "210 = 2 x 5 x 3 x 7 = 5 x 6 x 7 x 1 = 5 x 6 x 7 those are the only sets of digits we can use to for the numbers ( any other combination of factors will have two digit factors ) . numbers using 2,5 , 3,7 = 4 ! numbers using 5,6 , 7,1 = 4 ! numbers using 5 , 6,7 ( 3 - digit numbers ) = 3 ! answer = 24 + 24 + 6 = 54 answer is ( d )"
Answer: d
Problem: find the last term of a g . p whose first term is 9 and common ratio is ( 1 / 3 ) if the sum of the terms of the g . p is ( 40 / 3 )
Options: a ) 1 / 3 , b ) 2 / 5 , c ) 1 / 4 , d ) 2 / 3 , e ) 4 / 5
Rationale: sum of the g . p . = ( first term - r * last term ) / 1 – r 40 / 3 = 9 – 1 / 3 ( last term ) / 2 / 3 last term = ( - 40 / 3 * 2 / 3 + 9 ) * 3 = - 80 / 3 + 27 = 1 / 3 answer : a
Answer: a
Problem: the average of marks obtained by 120 boys was 36 . if the average of marks of passed boys was 39 and that of failed boys was 15 , the number of boys who passed the examination is ?
Options: a ) 105 , b ) 110 , c ) 120 , d ) 130 , e ) 140
Rationale: "let the number of boys who passed = x . then , 39 x x + 15 x ( 120 - x ) = 120 x 36 24 x = 4320 - 1800 = > x = 2520 / 24 x = 105 . hence , the number of boys passed = 105 . answer : a"
Answer: a
Problem: shekar scored 76 , 65 , 82 , 62 and 85 marks in mathematics , science , social studies , english and biology respectively . what are his average marks ?
Options: a ) 74 , b ) 69 , c ) 75 , d ) 85 , e ) 90
Rationale: "explanation : average = ( 76 + 65 + 82 + 62 + 85 ) / 5 = 370 / 5 = 74 hence average = 74 answer : a"
Answer: a
Problem: a rectangular plot measuring 10 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 ?
Options: a ) 46 m , b ) 66 m , c ) 26 m , d ) 24 m , e ) 25 m
Rationale: "perimeter of the plot = 2 ( 10 + 50 ) = 120 m no of poles = 120 / 5 = 24 m answer : d"
Answer: d
Problem: x can finish a work in 21 days . y can finish the same work in 15 days . y worked for 5 days and left the job . how many days does x alone need to finish the remaining work ?
Options: a ) 12 , b ) 13 , c ) 16 , d ) 14 , e ) 18
Rationale: "work done by x in 1 day = 1 / 21 work done by y in 1 day = 1 / 15 work done by y in 5 days = 5 / 15 = 1 / 3 remaining work = 1 – 1 / 3 = 2 / 3 number of days in which x can finish the remaining work = ( 2 / 3 ) / ( 1 / 21 ) = 14 d"
Answer: d
Problem: country c imposes a two - tiered tax on imported cars : the first tier imposes a tax of 16 % of the car ' s price up to a certain price level . if the car ' s price is higher than the first tier ' s level , the tax on the portion of the price that exceeds this value is 8 % . if ron imported a $ 14,000 imported car and ended up paying $ 1440 in taxes , what is the first tier ' s price level ?
Options: a ) $ 1600 , b ) $ 6000 , c ) $ 6050 , d ) $ 7050 , e ) $ 4000
Rationale: "let t be the tier price , p be total price = 14000 per the given conditions : 0.16 t + 0.08 ( p - t ) = 1440 - - - - > t = 8000 . e is the correct answer ."
Answer: e
Problem: what is the remainder when the number r = 14 ^ 2 * 15 ^ 8 is divided by 5 ?
Options: a ) 0 , b ) 1 , c ) 2 , d ) 4 , e ) 5
Rationale: "14 ^ 2 has units digit 6 15 ^ 8 has units digit 5 thus r = 14 ^ 2 * 15 ^ 8 has units digit 0 and will be divisible by 5 . the remainder will be zero answer : ( a )"
Answer: a
Problem: p is able to do a piece of work in 20 days and q can do the same work in 10 days . if they can work together for 2 days , what is the fraction of work left ?
Options: a ) 5 / 10 , b ) 9 / 10 , c ) 7 / 10 , d ) 6 / 10 , e ) 4 / 10
Rationale: "explanation : amount of work p can do in 1 day = 1 / 20 amount of work q can do in 1 day = 1 / 10 amount of work p and q can do in 1 day = 1 / 20 + 1 / 10 = 3 / 20 amount of work p and q can together do in 2 days = 2 × ( 3 / 20 ) = 3 / 10 fraction of work left = 1 – 3 / 10 = 7 / 10 answer : option c"
Answer: c
Problem: carina has 70 ounces of coffee divided into 5 - and 10 - ounce packages . if she has 2 more 5 - ounce packages than 10 - ounce packages , how many 10 - ounce packages does she have ?
Options: a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
Rationale: "lets say 5 and 10 ounce packages be x and y respectively . given that , 5 x + 10 y = 70 and x = y + 2 . what is the value of y . substituting the x in first equation , 5 y + 10 + 10 y = 70 - > y = 60 / 15 . = 4 c"
Answer: c
Problem: the average weight of a , b and c is 45 kg . if the average weight of a and b be 42 kg and that of b and c be 43 kg , then the weight of b is :
Options: a ) 33 kg , b ) 31 kg , c ) 32 kg , d ) 36 kg , e ) 35 kg
Rationale: "let a , b , c represent their respective weights . then , we have : a + b + c = ( 45 x 3 ) = 135 . . . . ( i ) a + b = ( 42 x 2 ) = 84 . . . . ( ii ) b + c = ( 43 x 2 ) = 86 . . . . ( iii ) adding ( ii ) and ( iii ) , we get : a + 2 b + c = 170 . . . . ( iv ) subtracting ( i ) from ( iv ) , we get : b = 35 b ' s weight = 35 kg . e"
Answer: e
Problem: john found that the average of 15 numbers is 40 . if 11 is added to each number then the mean of number is ?
Options: a ) 51 , b ) 45 , c ) 65 , d ) 78 , e ) 64
Rationale: "( x + x 1 + . . . x 14 ) / 15 = 40 51 option a"
Answer: a
Problem: a boat goes 100 km downstream in 8 hours , and 75 km upstream in 15 hours . the speed of the stream is ?
Options: a ) 3 , b ) 6.5 , c ) 5.5 , d ) 4 , e ) 3.75
Rationale: "100 - - - 10 ds = 12.5 ? - - - - 1 75 - - - - 15 us = 5 ? - - - - - 1 s = ( 12.5 - 5 ) / 2 = 3.75 kmph . answer : e"
Answer: e
Problem: what is the number of integers from 1 to 1000 ( inclusive ) that are divisible by neither 11 nor by 30 ?
Options: a ) 884 , b ) 890 , c ) 892 , d ) 910 , e ) 945
Rationale: "normally , i would use the method used by bunuel . it ' s the most accurate . but if you are looking for a speedy solution , you can use another method which will sometimes give you an estimate . looking at the options ( most of them are spread out ) , i wont mind trying it . ( mind you , the method is accurate here since the numbers start from 1 . ) in 1000 consecutive numbers , number of multiples of 11 = 1000 / 11 = 90 ( ignore decimals ) in 1000 consecutive numbers , number of multiples of 35 = 1000 / 35 = 28 number of multiples of 11 * 35 i . e . 385 = 1000 / 385 = 2 number of integers from 1 to 1000 that are divisible by neither 11 nor by 35 = 1000 - ( 90 + 28 - 2 ) { using the concept of sets here ) = 884 think : why did i say the method is approximate in some cases ? think what happens if the given range is 11 to 1010 both inclusive ( again 1000 numbers ) what is the number of multiples in this case ? a"
Answer: a
Problem: two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 26 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ?
Options: a ) 288 , b ) 564 , c ) 877 , d ) 278 , e ) 178
Rationale: "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 26 + 21 = 47 t = 12 d = 47 * 12 = 564 answer : b"
Answer: b
Problem: what is the smallest no . which must be added to 25268 so as to obtain a sum which is divisible by 11 ?
Options: a ) 5 , b ) 10 , c ) 11 , d ) 20 , e ) 30
Rationale: "for divisibility by 11 , the difference of sums of digits at even and odd places must be either zero or divisible by 11 . for 25268 , difference = ( 2 + 2 + 8 ) - ( 5 + 6 ) = 12 - 11 = 1 . the units digit is at odd place . so we add 10 to the number = > 25268 + 10 = 25278 now , ( 2 + 2 + 8 ) - ( 5 + 7 ) = 12 - 12 = 0 = > 25268 is also divisible by 11 b"
Answer: b
Problem: a train 455 m long , running with a speed of 63 km / hr will pass a tree in ?
Options: a ) 22 sec , b ) 16 sec , c ) 17 sec , d ) 39 sec , e ) 12 sec
Rationale: "speed = 63 * 5 / 18 = 35 / 2 m / sec time taken = 455 * 2 / 35 = 39 sec answer : d"
Answer: d
Problem: the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 10 percent , and on day 3 , it is discounted an additional 15 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ?
Options: a ) 28 % , b ) 40 % , c ) 64.8 % , d ) 76.5 % , e ) 72 %
Rationale: "original price = 100 day 1 discount = 10 % , price = 100 - 10 = 90 day 2 discount = 10 % , price = 90 - 9 = 81 day 3 discount = 15 % , price = 81 - 12.15 = 68.85 which is 68.85 / 90 * 100 of the sale price on day 1 = ~ 76.5 % answer d"
Answer: d
Problem: an aeroplane covers a certain distance at a speed of 120 kmph in 4 hours . to cover the same distance in 1 2 / 3 hours , it must travel at a speed of :
Options: a ) 520 , b ) 288 , c ) 820 , d ) 740 , e ) 720
Rationale: "distance = ( 240 x 5 ) = 480 km . speed = distance / time speed = 480 / ( 5 / 3 ) km / hr . [ we can write 1 2 / 3 hours as 5 / 3 hours ] required speed = ( 480 x 3 / 5 ) km / hr = 288 km / hr answer b ) 288 km / hr"
Answer: b
Problem: the present age of a father is 3 years more than 3 times the age of his son . 5 years hence , father ' s age will be 10 years more than twice the age of the son . find the present age of the father .
Options: a ) 33 , b ) 38 , c ) 27 , d ) 40 , e ) 48
Rationale: if the present age be x years . father ' s will be ( 3 x + 3 ) years . . so , ( 3 x + 3 + 5 ) = 2 ( x + 3 ) + 10 or , x = 8 so the fathers present age = ( 3 x + 3 ) = ( 3 * 8 + 3 ) years = 27 years . . answer : option c
Answer: c
Problem: the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 20 , find the other ?
Options: a ) 100 , b ) 360 , c ) 120 , d ) 200 , e ) 150
Rationale: "other number = 12 * 600 / 20 = 360 answer is b"
Answer: b
Problem: what is the next no . 4 12 84
Options: a ) 632 , b ) 732 , c ) 832 , d ) 850 , e ) 902
Rationale: "3 ^ 0 + 3 = 4 3 ^ 2 + 3 = 12 3 ^ 4 + 3 = 84 3 ^ 6 + 3 = 732 answer : b"
Answer: b
Problem: x and y are positive integers . when x is divided by 9 , the remainder is 2 , and when x is divided by 7 , the remainder is 4 . when y is divided by 11 , the remainder is 3 , and when y is divided by 13 , the remainder is 12 . what is the least possible value of y - x ?
Options: a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16
Rationale: "when x is divided by 9 , the remainder is 2 : so , the possible values of x are : 2 , 11 , 21 , 29 , etc . when x is divided by 7 , the remainder is 4 : so , the possible values of x are : 4 , 11,18 , . . . stop . since both lists include 11 , the smallest possible value of x is 11 . when y is divided by 11 , the remainder is 3 : so , the possible values of y are : 3 , 14 , 25 , etc . when y is divided by 13 , the remainder is 12 : so , the possible values of y are : 12 , 25 , . . . stop . since both lists include 25 , the smallest possible value of y is 25 since the smallest possible values of x and y are 11 and 25 respectively , the smallest possible value of y - x is 14 . so , c is the correct answer to the original question ."
Answer: c
Problem: at joes steakhouse the hourly wage for a chef is 22 % greater than that of a dishwasher , and the hourly wage of a dishwasher is half as much as the hourly wage of a manager . if a managers wage is $ 8.50 per hour , how much less than a manager does a chef earn each hour ?
Options: a ) $ 1.40 , b ) $ 2.40 , c ) $ 3.315 , d ) $ 4.40 , e ) $ 5.40
Rationale: "manager wages per hour = $ 8.50 dishwasher wages per hour = half of manager ' s wages . = 1 / 2 ( $ 8.50 ) = = > $ 4.25 chef wages per hour = 22 % greater than dishwasher wages - - > 22 % of $ 4.25 = ( 22 * ( $ 4.25 ) ) / 100 - - > ( $ 93.5 ) / 100 - - > $ 0.935 therefore , chef wages per hour = $ 4.25 + $ 0.935 = = > $ 5.185 difference of wages between manager and chef = $ 8.50 - $ 5.185 = = > $ 3.315 answer : c"
Answer: c
Problem: if xerox paper costs 5 cents a sheet and a buyer gets 10 % discount on all xerox paper one buys after the first 2000 papers and 20 % discount after first 10000 papers , how much will it cost to buy 15000 sheets of xerox paper ?
Options: a ) $ 1250 , b ) $ 700 , c ) $ 1350 , d ) $ 900 , e ) $ 1000
Rationale: 30 sec approach - solve it using approximation 15000 sheet at full price , 5 cent = 750 15000 sheet at max discount price , 4 cent = 600 your ans got to be between these two . ans b it is .
Answer: b
Problem: if the sum of two positive integers is 18 and the difference of their squares is 36 , what is the product of the two integers ?
Options: a ) 108 , b ) 80 , c ) 128 , d ) 135 , e ) 143
Rationale: "let the 2 positive numbers x and y x + y = 18 - - 1 x ^ 2 - y ^ 2 = 36 = > ( x + y ) ( x - y ) = 36 - - 2 using equation 1 in 2 , we get = > x - y = 2 - - 3 solving equation 1 and 3 , we get x = 10 y = 8 product = 10 * 8 = 80 answer b"
Answer: b
Problem: a certain number when divided by 80 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ?
Options: a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 9
Rationale: "explanation : 80 + 25 = 105 / 15 = 7 ( remainder ) d"
Answer: d
Problem: a rectangular circuit board is designed to have a width of w inches , a length of l inches , a perimeter of p inches , and an area of c square inches . which of the following equations must be true ?
Options: a ) 2 w ^ 2 + pw + 2 c = 0
b ) 2 w ^ 2 − pw + 2 c = 0
c ) 2 w ^ 2 − pw − 2 c = 0
d ) w ^ 2 + pw + c = 0
e ) w ^ 2 − pw + 2 c = 0
Rationale: p = 2 ( l + w ) - - - - - - - - - - - - - - - - - 1 ) c = lw - - - - - - - - - - - - - - - - - - - - - - - - 2 ) option a is not possible why ? ? because all the terms are positive . lets try option b , put value of p and a from 1 and 2 we have , 2 w ^ 2 - 2 ( l + w ) w + 2 ( lw ) 2 w ^ 2 - 2 lw - 2 w ^ 2 + 2 lw = 0 . hence answer is b .
Answer: b
Problem: a candidate appearing for an examination has to secure 35 % marks to pass paper i . but he secured only 42 marks and failed by 23 marks . what is the maximum mark for paper i ?
Options: a ) 110 , b ) 120 , c ) 186 , d ) 140 , e ) 150
Rationale: "he secured 42 marks nd fail by 23 marks so total marks for pass the examinatn = 65 let toal marks x x * 35 / 100 = 65 x = 186 answer : c"
Answer: c
Problem: how many odd integers from 1 to 50 ( both inclusive ) have odd number of factors ?
Options: a ) 13 , b ) 4 , c ) 5 , d ) 6 , e ) 7
Rationale: "integers having odd number of factors will be perfect squares . odd numbers will have odd perfect squares . thus , the possible values for the perfect squares are : 1,9 , 25,49 and the corresponding integers are 1,3 , 5,7 ( more than 3 ) . thus b is the correct answer ."
Answer: b
Problem: a rectangular plot measuring 90 metres by 40 metres is to be enclosed by wire fencing . if the poles of the fence are kept 5 metres apart , how many poles will be needed ?
Options: a ) 55 , b ) 56 , c ) 57 , d ) 52 , e ) none of these
Rationale: "solution perimeter of the plot = 2 ( 90 + 40 ) = 260 m . ∴ number of poles = [ 260 / 5 ] = 52 m answer d"
Answer: d
Problem: find the value for x from below equation : x / 3 = - 2 ?
Options: a ) - 6 , b ) 1 , c ) - 2 , d ) - 3 , e ) 4
Rationale: 1 . multiply both sides by 3 : x * 3 / 3 = - 2 / 3 2 . simplify both sides : x = - 6 a
Answer: a
Problem: dan can do a job alone in 12 hours . annie , working alone , can do the same job in just 9 hours . if dan works alone for 4 hours and then stops , how many hours will it take annie , working alone , to complete the job ?
Options: a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10
Rationale: "dan can complete 1 / 12 of the job per hour . in 4 hours , dan completes 4 ( 1 / 12 ) = 1 / 3 of the job . annie can complete 1 / 9 of the job per hour . to complete the job , annie will take 2 / 3 / 1 / 9 = 6 hours . the answer is c ."
Answer: c
Problem: the measurement of a rectangular box with lid is 25 cmx 18 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of π cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere )
Options: a ) 288 , b ) 48 , c ) 72 , d ) 972 , e ) 964
Rationale: "d = 18 , r = 9 ; volume of the largest sphere = 4 / 3 π r 3 = 4 / 3 * π * 9 * 9 * 9 = 972 π cm 3 answer : d"
Answer: d
Problem: the probability that a computer company will get a computer hardware contract is 3 / 4 and the probability that it will not get a software contract is 5 / 9 . if the probability of getting at least one contract is 4 / 5 , what is the probability that it will get both the contracts ?
Options: a ) 11 / 30 , b ) 31 / 60 , c ) 41 / 80 , d ) 51 / 120 , e ) 71 / 180
Rationale: "let , a ≡ event of getting hardware contract b ≡ event of getting software contract ab ≡ event of getting both hardware and software contract . p ( a ) = 3 / 4 , p ( ~ b ) = 5 / 9 = > p ( b ) = 1 - ( 5 / 9 ) = 4 / 9 . a and b are not mutually exclusive events but independent events . so , p ( at least one of a and b ) = p ( a ) + p ( b ) - p ( ab ) . = > 4 / 5 = ( 3 / 4 ) + ( 4 / 9 ) - p ( ab ) . = > p ( ab ) = 71 / 180 . hence , the required probability is 71 / 180 . the answer is e ."
Answer: e
Problem: working alone at its constant rate , machine a produces x boxes in 10 minutes and working alone at its constant rate , machine b produces 3 x boxes in 5 minutes . how many minutes does it take machines a and b , working simultaneously at their respective constant rates , to produce 3 x boxes ?
Options: a ) 3 minutes , b ) 4 minutes , c ) 4.2 minutes , d ) 6 minutes , e ) 12 minutes
Rationale: "rate = work / time given rate of machine a = x / 10 min machine b produces 3 x boxes in 5 min hence , machine b produces 4 x boxes in 10 min . rate of machine b = 6 x / 10 we need tofind the combined time that machines a and b , working simultaneouslytakeat their respective constant rates let ' s first find the combined rate of machine a and b rate of machine a = x / 10 min + rate of machine b = 6 x / 10 = 7 x / 10 now combine time = combine work needs to be done / combine rate = 3 x / 7 x * 10 = 4.2 min ans : c"
Answer: c
Problem: a student chose a number , multiplied it by 4 , then subtracted 138 from the result and got 102 . what was the number he chose ?
Options: a ) 60 , b ) 120 , c ) 130 , d ) 140 , e ) 150
Rationale: solution : let xx be the number he chose , then 4 ⋅ x − 138 = 102 4 x = 240 x = 60 answer a
Answer: a
Problem: what is the speed of the stream if a canoe rows upstream at 4 km / hr and downstream at 12 km / hr
Options: a ) 1 kmph , b ) 4 kmph , c ) 3 kmph , d ) 2 kmph , e ) 1.9 kmph
Rationale: "sol . speed of stream = 1 / 2 ( 12 - 4 ) kmph = 4 kmph . answer b"
Answer: b
Problem: when asked what the time is , a person answered that the amount of time left is 3 / 5 of the time already completed . what is the time .
Options: a ) 2 pm , b ) 9 pm , c ) 3 pm , d ) 8 pm , e ) 6 pm
Rationale: "a day has 24 hrs . assume x hours have passed . remaining time is ( 24 - x ) 24 − x = 3 / 5 x ⇒ x = 15 time is 3 pm answer : c"
Answer: c
Problem: the side of a cube is 15 m , find it ' s surface area ?
Options: a ) 1350 , b ) 1750 , c ) 1150 , d ) 1450 , e ) 1570
Rationale: "surface area = 6 a ( power ) 2 sq . units 6 a ( power ) 2 = 6 × 225 = 1350 m ( power ) 2 answer is a ."
Answer: a
Problem: a water tank is three - fifths full . pipe a can fill a tank in 10 minutes and pipe b can empty it in 6 minutes . if both the pipes are open , how long will it take to empty or fill the tank completely ?
Options: a ) 6 min , b ) 8 min , c ) 7 min , d ) 9 min , e ) 1 min
Rationale: "the combined rate of filling / emptying the tank = 1 / 10 - 1 / 6 = - 1 / 15 since the rate is negative , the tank will be emptied . a full tank would take 15 minutes to empty . since the tank is only three - fifths full , the time is ( 3 / 5 ) * 15 = 9 minutes the answer is d ."
Answer: d
Problem: two vessels p and q contain 62.5 % and 87.5 % of alcohol respectively . if 4 litres from vessel p is mixed with 8 litres from vessel q , the ratio of alcohol and water in the resulting mixture is ?
Options: a ) 19 : 1 , b ) 19 : 4 , c ) 19 : 8 , d ) 38 : 10 , e ) 38 : 2
Rationale: "quantity of alcohol in vessel p = 62.5 / 100 * 4 = 5 / 2 litres quantity of alcohol in vessel q = 87.5 / 100 * 8 = 7 / 1 litres quantity of alcohol in the mixture formed = 5 / 2 + 7 / 1 = 19 / 2 = 9.50 litres as 12 litres of mixture is formed , ratio of alcohol and water in the mixture formed = 9.50 : 2.50 = 38 : 10 . answer : d"
Answer: d
Problem: ( 228 % of 1265 ) ÷ 6 = ?
Options: a ) a ) 125 , b ) b ) 175 , c ) c ) 481 , d ) d ) 375 , e ) e ) 524
Rationale: "explanation : ? = ( 228 x 1265 / 100 ) ÷ 6 = 288420 / 600 = 481 answer : option c"
Answer: c
Problem: the first year , two cows produced 8100 litres of milk . the second year their production increased by 15 % and 10 % respectively , and the total amount of milk increased to 9100 litres a year . how many litres were milked from one cow ?
Options: a ) 2178 lt , b ) 3697 lt , c ) 6583 lt , d ) 4370 lt , e ) 5548 lt
Rationale: "let x be the amount of milk the first cow produced during the first year . then the second cow produced ( 8100 − x ) litres of milk that year . the second year , each cow produced the same amount of milk as they did the first year plus the increase of 15 % 15 % or 10 % so 8100 + 15100 ⋅ x + 10100 ⋅ ( 8100 − x ) = 9100 therefore 8100 + 320 x + 110 ( 8100 − x ) = 9100 120 x = 190 x = 3800 therefore , the cows produced 3800 and 4300 litres of milk the first year , and 4370 and 4730 litres of milk the second year , = > one cow produced 4370 litres correct answer is d ) 4370 lt"
Answer: d
Problem: the average height of 40 girls out of a class of 50 is 169 cm . and that of the remaining girls is 167 cm . the average height of the whole class is :
Options: a ) 138.9 cms , b ) 149.2 cms , c ) 168.6 cms , d ) 159.2 cms , e ) 142.5 cms
Rationale: "explanation : average height of the whole class = ( 40 × 169 + 10 × 167 / 50 ) = 168.6 cms answer c"
Answer: c
Problem: equal amount of water were poured into two empty jars of different capacities , which made one jar 1 / 7 full and other jar 1 / 6 full . if the water in the jar with lesser capacity is then poured into the jar with greater capacity , what fraction of the larger jar will be filled with water ?
Options: a ) 1 / 7 , b ) 7 / 12 , c ) 1 / 2 , d ) 2 / 7 , e ) 2 / 3
Rationale: "same amount of water made bigger jar 1 / 7 full , then the same amount of water ( stored for a while in smaller jar ) were added to bigger jar , so bigger jar is 1 / 7 + 1 / 7 = 2 / 7 full . answer : d ."
Answer: d
Problem: given f ( x ) = 3 x – 5 , for what value of x does 2 * [ f ( x ) ] – 10 = f ( x – 2 ) ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5
Rationale: "2 ( 3 x - 5 ) - 10 = 3 ( x - 2 ) - 5 3 x = 9 x = 3 the answer is c ."
Answer: c
Problem: the population of a town is 10000 . it increases annually at the rate of 40 % p . a . what will be its population after 2 years ?
Options: a ) 14000 , b ) 14400 , c ) 19600 , d ) 14600 , e ) 14700
Rationale: "formula : 10000 × 140 / 100 × 140 / 100 = 19600 answer : c"
Answer: c
Problem: for any integer n greater than 1 , # n denotes the product of all the integers from 1 to n , inclusive . how many prime numbers e are there between # 6 + 2 and # 6 + 6 , inclusive ?
Options: a ) none , b ) one , c ) two , d ) three , e ) four
Rationale: "none is the answer . a . because for every k 6 ! + k : : k , because 6 ! : : k , since k is between 2 and 6 . a"
Answer: a
Problem: the unit digit in the product ( 784 x 618 x 917 x 463 ) is :
Options: a ) 5 , b ) 9 , c ) 16 , d ) 2 , e ) 42
Rationale: "explanation : unit digit in the given product = unit digit in ( 4 x 8 x 7 x 3 ) = ( 672 ) = 2 d"
Answer: d
Problem: the diameter of a cylindrical tin is 8 cm and height is 5 cm . find the volume of the cylinder ?
Options: a ) 33 , b ) 45 , c ) 66 , d ) 80 , e ) 21
Rationale: "r = 4 h = 5 π * 4 * 4 * 5 = 80 π cc answer : d"
Answer: d
Problem: whats the reminder when 54,879 , 856,985 , 421,547 , 895,689 , 874,525 , 826,547 is divided by 2
Options: a ) 1 , b ) 7 . , c ) 2 . , d ) 3 , e ) 9
Rationale: "a number ending in a 0 is divisible by 2 . given the obscene number , you should immediately be convinced that you will need to focus on a very small part of it . 54,879 , 856,985 , 421,547 , 895,689 , 874,525 , 826,547 = 54,879 , 856,985 , 421,547 , 895,689 , 874,525 , 826,540 + 7 the first number is divisible by 16 . you just have to find the remainder when you divide 5287 by 16 . that will be the remainder when you divide the original number by 2 . 7 / 16 gives remainder 1 . answer ( a )"
Answer: a
Problem: each of the dogs in a certain kennel is a single color . each of the dogs in the kennel either has long fur or does not . of the 45 dogs in the kennel , 26 have long fur , 30 are brown , and 8 are neither long - furred nor brown . how many long - furred dogs are brown ?
Options: a ) 26 , b ) 19 , c ) 11 , d ) 8 , e ) 6
Rationale: "no of dogs = 45 long fur = 26 brown = 30 neither long fur nor brown = 8 therefore , either long fur or brown = 45 - 8 = 37 37 = 26 + 30 - both both = 19 answer b"
Answer: b
Problem: a person bought 114 glass bowls at a rate of rs . 13 per bowl . he sold 108 of them at rs . 17 and the remaining broke . what is the percentage gain for a ?
Options: a ) 40 , b ) 3000 / 11 , c ) 5900 / 247 , d ) 2790 / 11 , e ) 2709 / 8
Rationale: "cp = 114 * 13 = 1482 and sp = 108 * 17 = 1836 gain % = 100 * ( 1836 - 1482 ) / 1482 = 5900 / 247 answer : c"
Answer: c
Problem: a boat can travel with a speed of 16 km / hr in still water . if the rate of stream is 5 km / hr , then find the time taken by the boat to cover distance of 147 km downstream .
Options: a ) 4 hours , b ) 5 hours , c ) 6 hours , d ) 7 hours , e ) 8 hours
Rationale: "explanation : it is very important to check , if the boat speed given is in still water or with water or against water . because if we neglect it we will not reach on right answer . i just mentioned here because mostly mistakes in this chapter are of this kind only . lets see the question now . speed downstream = ( 16 + 5 ) = 21 kmph time = distance / speed = 147 / 21 = 7 hours option d"
Answer: d
Problem: the average age of 38 students in a group is 14 years . when teacher â € ™ s age is included to it , the average increases by one . what is the teacher â € ™ s age in years ?
Options: a ) 31 , b ) 36 , c ) 53 , d ) 58 , e ) none
Rationale: "sol . age of the teacher = ( 39 ã — 15 â € “ 38 ã — 14 ) years = 53 years . answer c"
Answer: c
Problem: the area of a square is 4761 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square .
Options: a ) 5 : 28 , b ) 5 : 19 , c ) 15 : 12 , d ) 5 : 13 , e ) 15 : 46
Rationale: "let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 4761 a = 69 l = 2 a and b = a - 24 b : l = a - 24 : 2 a = 45 : 138 = 15 : 46 answer : e"
Answer: e
Problem: a positive integer n is a perfect number provided that the sum of all the positive factors of n , including 1 and n , is equal to 2 n . what is the sum of the reciprocals of all the positive factors of the perfect number 28 ?
Options: a ) 1 / 4 , b ) 56 / 27 , c ) 2 , d ) 3 , e ) 4
Rationale: soln : 28 = 1 * 28 2 * 14 4 * 7 sum of reciprocals = 1 + 1 / 28 + 1 / 2 + 1 / 14 + 1 / 4 + 1 / 7 = 56 / 28 = 2 answer : c
Answer: c
Problem: two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 46 kmph respectively . in how much time will they cross each other , if they are running in the same direction ?
Options: a ) 72 sec , b ) 210 sec , c ) 192 sec , d ) 252 sec , e ) none
Rationale: "solution relative speed = ( 46 - 40 ) kmph = 6 kmph = ( 6 x 5 / 18 ) m / sec = ( 30 / 18 ) m / sec time taken = ( 350 x 18 / 30 ) sec = 210 sec . answer b"
Answer: b
Problem: an employee ’ s annual salary was increased 50 % . if her old annual salary equals $ 80,000 , what was the new salary ?
Options: a ) $ 128,000 , b ) $ 120,000 , c ) $ 110,000 , d ) $ 139,000 , e ) $ 125,000
Rationale: "old annual salary = $ 80,000 salary increase = 50 % . original salary = $ 80,000 * 50 / 100 = $ 40,000 new salary = $ 80,000 + $ 40,000 = $ 120,000 hence b ."
Answer: b
Problem: a can do a job in 15 days and b in 20 days . if they work on it together for 8 days , then the fraction of the work that is left is ?
Options: a ) 2 / 15 , b ) 8 / 15 , c ) 3 / 11 , d ) 1 / 12 , e ) 1 / 15
Rationale: "a ' s 1 day work = 1 / 15 b ' s 1 day work = 1 / 20 a + b 1 day work = 1 / 15 + 1 / 20 = 7 / 60 a + b 8 days work = 7 / 60 * 8 = 14 / 15 remaining work = 1 - 14 / 15 = 1 / 15 answer is e"
Answer: e
Problem: a dishonest milkman wants to make a profit on the selling of milk . he would like to mix water ( costing nothing ) with milk costing rs . 33 per litre so as to make a profit of 20 % on cost when he sells the resulting milk and water mixture for rs . 36 in what ratio should he mix the water and milk ?
Options: a ) 1 : 20 , b ) 1 : 10 , c ) 1 : 8 , d ) 1 : 4 , e ) 6 : 11
Rationale: cost needed to net a 20 % profit : ( 36 - x ) / x = . 2 x = 30 actual cost : 33 solution ( x = liters of water needed to be added to the 1 liter of milk ) : 33 / ( 1 + x ) = 30 x = 1 / 10 so to get the cost down to 30 milk : water 1 : 1 / 10 or ( 10 / 10 ) : ( 1 / 10 ) answer : b
Answer: b
Problem: a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 980 gm . for a kg . his gain is … % .
Options: a ) 2.04 % , b ) 5.36 % , c ) 4.26 % , d ) 6.26 % , e ) 7.26 %
Rationale: "his percentage gain is 100 * 20 / 980 as he is gaining 20 units for his purchase of 980 units . so 2.04 % . . answer : a"
Answer: a
Problem: if 20 % of a number = 400 , then 120 % of that number will be ?
Options: a ) 20 , b ) 120 , c ) 360 , d ) 2400 , e ) 2820
Rationale: "let the number x . then , 20 % of x = 400 x = ( 400 * 100 ) / 20 = 2000 120 % of x = ( 120 / 100 * 2000 ) = 2400 . answer : d"
Answer: d
Problem: at a certain resort , each of the 39 food service employees is trained to work in a minimum of 1 restaurant and a maximum of 3 restaurants . the 3 restaurants are the family buffet , the dining room , and the snack bar . exactly 17 employees are trained to work in the family buffet , 18 are trained to work in the dining room , and 12 are trained to work in the snack bar . if 4 employees are trained to work in exactly 2 restaurants , how many employees are trained to work in all 3 restaurants ?
Options: a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
Rationale: "39 = 17 + 18 + 12 - 4 - 2 x 2 x = 17 + 18 + 12 - 4 - 39 = 43 - 39 = 4 x = 2 a"
Answer: a
Problem: a cistern is filled by pipe a in 20 hours and the full cistern can be leaked out by an exhaust pipe b in 25 hours . if both the pipes are opened , in what time the cistern is full ?
Options: a ) 50 hrs , b ) 60 hrs , c ) 70 hrs , d ) 80 hrs , e ) 100 hrs
Rationale: "time taken to full the cistern = ( 1 / 20 - 1 / 25 ) hrs = 1 / 100 = 100 hrs answer : e"
Answer: e
Problem: the population of a town increased from 1 , 75,000 to 2 , 27,500 in a decade . the average percent increase of population per year is
Options: a ) 4.37 % , b ) 5 % , c ) 3 % , d ) 8.75 % , e ) none
Rationale: "solution increase in 10 years = ( 227500 - 175000 ) = 52500 . increase % = ( 52500 / 175000 ã — 100 ) % = 30 % . required average = ( 30 / 10 ) % = 3 % . answer c"
Answer: c
Problem: - 54 x 29 + 100 = ?
Options: a ) - 1466 , b ) 2801 , c ) - 2801 , d ) - 2071 , e ) none of them
Rationale: given exp . = - 54 x ( 30 - 1 ) + 100 = - ( 54 x 30 ) + 54 + 100 = - 1620 + 154 = - 1466 answer is a
Answer: a
Problem: if 50 % of 100 is greater than 20 % of a number by 47 , what is the number ?
Options: a ) 60 , b ) 30 , c ) 15 , d ) 75 , e ) 100
Rationale: "explanation : 50 / 100 * 100 - 20 / 100 * x = 47 50 - 20 / 100 * x = 47 3 = 20 / 100 * x 3 * 100 / 20 = x 15 = x answer : option c"
Answer: c
Problem: running at the same constant rate , 6 identical machines can produce a total of 270 bottles per minute . at this rate , how many bottles could 14 such machines produce in 4 minutes ?
Options: a ) 2520 , b ) 1800 , c ) 2700 , d ) 10800 , e ) none of these
Rationale: "solution let the required number of bottles be x . more machines , more bottles ( direct proportion ) more minutes , more bottles ( direct proportion ) â ˆ ´ 6 ã — 1 ã — x = 14 ã — 4 ã — 270 â ‡ ” x = 14 x 4 x 270 / 6 = 2520 . answer a"
Answer: a
Problem: a tank holds x gallons of a saltwater solution that is 20 % salt by volume . one fourth of the water is evaporated , leaving all of the salt . when 6 gallons of water and 12 gallons of salt are added , the resulting mixture is 33 1 / 3 % salt by volume . what is the value of x ?
Options: a ) 37.5 , b ) 75 , c ) 100 , d ) 150 , e ) 90
Rationale: nope , 150 . i can only get it by following pr ' s backsolving explanation . i hate that . original mixture has 20 % salt and 80 % water . total = x out of which salt = 0.2 x and water = 0.8 x now , 1 / 4 water evaporates and all salt remains . so what remains is 0.2 x salt and 0.6 x water . now 12 gallons salt is added and 6 gallons of water is added . so salt now becomes - > ( 0.2 x + 12 ) and water - - > ( 0.6 x + 6 ) amount of salt is 33.33 % of total . so amount of water is 66.66 % . so salt is half of the volume of water . so ( 0.2 x + 12 ) = ( 0.6 x + 6 ) / 2 = > 0.4 x + 24 = 0.6 x + 6 - > 0.2 x = 18 solving , x = 90 answer : e
Answer: e
Problem: there are 250 female managers in a certain company . find the total number of female employees in the company , if 2 / 5 of all the employees are managers and 2 / 5 of all male employees are managers .
Options: a ) 325 , b ) 425 , c ) 625 , d ) 700 , e ) none of these
Rationale: as per question stem 2 / 5 m ( portion of men employees who are managers ) + 250 ( portion of female employees who are managers ) = 2 / 5 t ( portion of total number of employees who are managers ) , thus we get that 2 / 5 m + 250 = 2 / 5 t , or 2 / 5 ( t - m ) = 250 , from here we get that t - m = 625 , that would be total number of female employees and the answer ( c )
Answer: c
Problem: income and expenditure of a person are in the ratio 15 : 8 . if the income of the person is rs . 15000 , then find his savings ?
Options: a ) 6999 , b ) 7000 , c ) 7001 , d ) 7002 , e ) 7003
Rationale: "let the income and the expenditure of the person be rs . 15 x and rs . 8 x respectively . income , 15 x = 15000 = > x = 1000 savings = income - expenditure = 15 x - 8 x = 7 x = 7 ( 1000 ) so , savings = rs . 7000 . answer : b"
Answer: b
Problem: ashok secured average of 78 marks in 6 subjects . if the average of marks in 5 subjects is 74 , how many marks did he secure in the 6 th subject ?
Options: a ) 66 , b ) 74 , c ) 78 , d ) 98 , e ) none of these
Rationale: "explanation : number of subjects = 6 average of marks in 6 subjects = 78 therefore total marks in 6 subjects = 78 * 6 = 468 now , no . of subjects = 5 total marks in 5 subjects = 74 * 5 = 370 therefore marks in 6 th subject = 468 – 370 = 98 answer d"
Answer: d
Problem: a man performs 1 / 2 of the total journey by rail , 1 / 3 by bus and the remaining 4 km on foot . his total journey is
Options: a ) 16 km , b ) 10 km , c ) 12 km , d ) 24 km , e ) 25 km
Rationale: "explanation : let the journey be x km then , 1 x / 2 + 1 x / 3 + 4 = x 5 x + 24 = 6 x x = 24 km answer : option d"
Answer: d
Problem: a hat company ships its hats , individually wrapped , in 8 - inch by 10 - inch by 12 - inch boxes . each hat is valued at $ 7.50 . if the company ’ s latest order required a truck with at least 384,000 cubic inches of storage space in which to ship the hats in their boxes , what was the minimum value of the order ?
Options: a ) $ 3,000 , b ) $ 1,350 , c ) $ 1,725 , d ) $ 2,050 , e ) $ 2,250
Rationale: number of boxes = total volume / volume of one box = 384,000 / ( 8 * 10 * 12 ) = 400 one box costs 7.50 , so 400 box will cost = 400 * 7.5 = 3000 a is the answer
Answer: a
Problem: if p / q = 3 / 5 , then 2 p + q = ?
Options: a ) 12 , b ) 11 , c ) 13 , d ) 15 , e ) 16
Rationale: "let p = 3 , q = 5 then 2 * 3 + 5 = 11 so 2 p + q = 11 . answer : b"
Answer: b
Problem: fred and sam are standing 100 miles apart and they start walking in a straight line toward each other at the same time . if fred walks at a constant speed of 5 miles per hour and sam walks at a constant speed of 5 miles per hour , how many miles has sam walked when they meet ?
Options: a ) 5 , b ) 9 , c ) 25 , d ) 30 , e ) 75
Rationale: "relative distance = 100 miles relative speed = 5 + 5 = 10 miles per hour time taken = 100 / 10 = 15 hours distance travelled by sam = 15 * 5 = 75 miles = e"
Answer: e