regisss/math_qa huggingface dataset:
Problem: the average weight of 10 persons increases by 7.2 kg when a new person comes in place of one of them weighing 65 kg . what might be the weight of the new person ?
Options: a ) 120 kg , b ) 130 kg , c ) 137 kg , d ) 190 kg , e ) none
Rationale: "solution total weight increased = ( 10 x 7.2 ) kg = 72 kg . weight of new person = ( 65 + 72 ) kg = 137 kg . answer c"
Answer: c
Problem: a fold density is 19 times greater than the water and for copper it is 9 times . at what ratio you can mix gold and copper to get 15 times denser than water . ?
Options: a ) 3 : 1 , b ) 3 : 2 , c ) 3 : 5 , d ) 3 : 8 , e ) 3 : 7
Rationale: suppose x units of gold are mixed with y units of copper to make ( x + y ) units of alloy which is 15 times denser than water . then 19 * x + 9 * y = 15 * ( x + y ) = > 19 * x - 15 * x = 15 * y - 9 * y = > 4 * x = 6 * y = > x / y = 3 / 2 so gold and copper should be mixed respectively in the ratio of 3 : 2 answer : b
Answer: b
Problem: r is the set of positive even integers less than 50 , and s is the set of the squares of the integers in r . how many elements does the intersection of r and s contain ?
Options: a ) none , b ) two , c ) three , d ) five , e ) seven
Rationale: "squares < 50 { 1 , 4,9 , 16,25 , 36,49 } s = { 1,4 , 16,36 } r = { 2 , . . . . . 48 } hence c ."
Answer: c
Problem: a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 50 paisa . if the share of y is rs . 45 , what is the total amount ?
Options: a ) 115 , b ) 116 , c ) 195 , d ) 118 , e ) 119
Rationale: "x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 45 39 - - - ? = > 195 answer : c"
Answer: c
Problem: a jar contains a mixture of two liquids acid ( nitric acid ) and base ( ammonium chloride ) in the ratio 4 : 1 . when 10 litres of the mixture is taken out and 10 litres of liquid base is poured into the jar , the ratio becomes 2 : 3 . how many litres of liquid acid was contained in the jar ?
Options: a ) 14 lts , b ) 18 lts , c ) 20 lts , d ) 16 lts , e ) 26 lts
Rationale: % age of liquid base in the original mixture = 1 / 5 x 100 = 20 % in the final mixture % of the liquid base = 3 / 5 x 100 = 60 % now using the rule of allegation hence reduced quantity of the first mixture and the quantity of mixture b which is to be added are the same . total mixture = 10 + 10 = 20 liters and quantity of liquid a = 20 / 5 x 4 = 16 lts answer : d
Answer: d
Problem: if x + y = 4 , x - y = 36 , for integers of x and y , x = ?
Options: a ) 20 , b ) 15 , c ) 25 , d ) 13 , e ) 42
Rationale: "x + y = 4 x - y = 36 2 x = 40 x = 20 answer is a"
Answer: a
Problem: if c is 25 % of a and 15 % of b , what percent of a is b ?
Options: a ) 2.5 % , b ) 15 % , c ) 25 % , d ) 125 % , e ) 250 %
Rationale: "answer = d 25 a / 100 = 20 b / 100 b = 25 a / 20 = 125 a / 100 = 125 %"
Answer: d
Problem: a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 864 in total compensation , how many total hours did he work that week ?
Options: a ) 36 , b ) 40 , c ) 44 , d ) 48 , e ) 52
Rationale: "for 40 hrs = 40 * 16 = 640 excess = 864 - 640 = 224 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 224 / 28 = 56 / 7 = 8 total hrs = 40 + 8 = 48 answer d 48"
Answer: d
Problem: praveen starts business with rs . 3500 and after 5 months , hari joins with praveen as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is hari ’ s contribution in the capital ?
Options: a ) rs . 7500 , b ) rs . 8000 , c ) rs . 8500 , d ) rs . 9000 , e ) rs . 6000
Rationale: "let hari ’ s capital be rs . x . then , 3500 * 12 / 7 x = 2 / 3 = > 14 x = 126000 = > x = 9000 . answer : d"
Answer: d
Problem: harkamal purchased 8 kg of grapes at the rate of 70 per kg and 9 kg of mangoes at the rate of 50 per kg . how much amount did he pay to the shopkeeper ?
Options: a ) a ) 1000 , b ) b ) 1010 , c ) c ) 1065 , d ) d ) 1075 , e ) e ) 1080
Rationale: "cost of 8 kg grapes = 70 × 8 = 560 . cost of 9 kg of mangoes = 50 × 9 = 450 . total cost he has to pay = 560 + 450 = 1010 . b )"
Answer: b
Problem: in a group of 250 readers who read science fiction or literacy works or both , 180 read science fiction and 88 read literacy works . how many read both science fiction and literacy works ?
Options: a ) 268 , b ) 180 , c ) 150 , d ) 18 , e ) 88
Rationale: "consider total number of reader n ( s u l ) = 250 people who read science fiction n ( s ) = 180 people who read literacy works n ( l ) = 88 both science fiction and literacy n ( s â ˆ © l ) = ? n ( s u l ) = n ( s ) + n ( l ) - n ( s â ˆ © l ) 250 = 180 + 88 - n ( s â ˆ © l ) n ( s â ˆ © l ) = 268 - 250 n ( s â ˆ © l ) = 18 so people read both science fiction and literacy works are 18 answer : d"
Answer: d
Problem: in a certain circle there are 10 points . what is the number of the triangles connecting 5 points of the 10 points ?
Options: a ) 80 , b ) 96 , c ) 108 , d ) 120 , e ) 252
Rationale: "imo : e here we have to select 5 points out of 10 points . order is not important so the answer will be 10 c 5 = 252 answer e"
Answer: e
Problem: solution x is 10 percent alcohol by volume , and solution y is 30 percent alcohol by volume . how many milliliters of solution y must be added to 200 milliliters of solution x to create a solution that is 14 percent alcohol by volume ?
Options: a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250
Rationale: "14 % is 4 % - points higher than 10 % but 16 % - points lower than 30 % . thus there should be 4 parts of solution x for 1 part of solution y . we should add 50 ml of solution y . the answer is a ."
Answer: a
Problem: the difference between the place value and the face value of 6 in the numerical 856973 is
Options: a ) 973 , b ) 6973 , c ) 5994 , d ) 9554 , e ) none of them
Rationale: "= ( place value of 6 ) - ( face value of 6 ) = ( 6000 - 6 ) = 5994 answer is c"
Answer: c
Problem: if n = 8 ^ 8 – 5 , what is the units digit of n ?
Options: a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5
Rationale: "8 ^ 8 - 8 = 8 ( 8 ^ 7 - 1 ) = = > 8 ( 2 ^ 21 - 1 ) last digit of 2 ^ 21 is 2 based on what explanation livestronger is saying . 2 ^ 24 - 1 yields 2 - 1 = 1 as the unit digit . now on multiply this with 5 , we get unit digit as 5 answer : e"
Answer: e
Problem: patrick purchased 80 pencils and sold them at a loss equal to the selling price of 16 pencils . the cost of 80 pencils is how many times the selling price of 80 pencils ?
Options: a ) 0.75 , b ) 1.2 , c ) 1.8 , d ) 2.4 , e ) 4.25
Rationale: "say the cost price of 80 pencils was $ 80 ( $ 1 per pencil ) and the selling price of 1 pencil was p . selling at a loss : 80 - 80 p = 16 p - - > p = 5 / 6 . ( cost price ) / ( selling price ) = 1 / ( 5 / 6 ) = 6 / 5 = 1.2 . answer : b ."
Answer: b
Problem: how many odd numbers between 10 and 1,200 are the squares of integers ?
Options: a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16
Rationale: "the square of an odd number is an odd number : 10 < odd < 1,000 10 < odd ^ 2 < 1,000 3 . something < odd < 31 . something ( by taking the square root ) . so , that odd number could be any odd number from 5 to 31 , inclusive : 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19 , 21 , 23 , 25 , 27 , 29 , and 31 . 16 numbers . answer : e ."
Answer: e
Problem: if 6 women take an hour to dig a ditch , then how long should 18 women take a dig to ditch of the same type ?
Options: a ) 25 min , b ) 20 min , c ) 28 min , d ) 15 min , e ) 10 min
Rationale: if 6 women take an hour to dig a ditch , then 18 men will take 6 * 60 / 18 = 20 mins to dig a ditch of the same type . answer : b
Answer: b
Problem: if the price of a book is first decreased by 50 % and then increased by 60 % , then the net change in the price will be ?
Options: a ) 20 % decrease , b ) 10 % increase , c ) 10 % decrease , d ) 15 % increase , e ) 25 % decrease
Rationale: "let the original price be $ 100 new final price = 160 % of ( 50 % of $ 100 ) = 160 / 100 * 50 / 100 * 100 = $ 80 decrease is 20 % answer is a"
Answer: a
Problem: if a farmer sells 15 of his chickens , his stock of feed will last for 4 more days than planned , but if he buys 10 more chickens , he will run out of feed 6 days earlier than planned . if no chickens are sold or bought , the farmer will be exactly on schedule . how many chickens does the farmer have ?
Options: a ) 20 , b ) 50 , c ) 40 , d ) 30 , e ) 60
Rationale: "let x = total feed required for the planned period n = number of chicken t = total time of the planned feed x = nt 1 ) x = ( n - 15 ) * ( t + 4 ) 2 ) x = ( n + 10 ) * ( t - 6 ) equating 1 & 2 ( n - 15 ) * ( t + 4 ) = ( n + 10 ) * ( t - 6 ) or nt + 4 n - 15 t - 60 = nt - 6 n + 10 t - 60 10 n = 25 t n = 5 / 2 * t or t = 2 n / 5 x = n * 2 n / 5 substituting this value in 1 n * 2 n / 5 = ( n - 15 ) * ( 2 n / 5 + 4 ) 2 n = 60 n = 30 d"
Answer: d
Problem: a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability q that all 3 jellybeans removed from the bag are blue ?
Options: a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10
Rationale: "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total outcomes = 10 c 3 / 20 c 3 probability q = 120 / 1140 = 2 / 19 answer : option b method - 2 probability of first jelly bean to be blue = 10 / 20 [ total 10 blue out of total 20 jellybeans ] probability of second jelly bean to be blue = 9 / 19 [ total 9 blue remaining out of total 19 jellybeans remaining ] probability of third jelly bean to be blue = 8 / 18 [ total 8 blue remaining out of total 18 jellybeans remaining ] required probability = ( 10 / 20 ) * ( 9 / 19 ) * ( 8 / 18 ) = 2 / 19 answer : option b"
Answer: b
Problem: find the value of x from this equation ? 7 ( x - 1 ) = 21
Options: a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10
Rationale: 1 . divide both sides by 7 : 2 . simplify both sides : x - 1 = 3 3 . add 1 to both sides : x - 1 + 1 = 3 + 1 4 . simplify both sides : x = 4 b
Answer: b
Problem: a person incurs 10 % loss by selling a watch for $ 100 . at what price should the watch be sold to earn 10 % profit ?
Options: a ) $ 100 , b ) $ 122 , c ) $ 150 , d ) $ 210 , e ) $ 170
Rationale: "let the new selling price be $ x ( 100 - loss % ) : ( 1 st s . p . ) = ( 100 + gain % ) : ( 2 nd s . p . ) ( 100 - 10 ) / 100 = ( 100 + 10 ) / x x = 110 * 100 / 90 = 122 approximately answer is b"
Answer: b
Problem: a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 30 paisa . if the share of y is rs . 27 , what is the total amount ?
Options: a ) 388 , b ) 105 , c ) 288 , d ) 266 , e ) 281
Rationale: x : y : z = 100 : 45 : 30 20 : 9 : 6 9 - - - 27 35 - - - ? = > 105 answer : b
Answer: b
Problem: an association of mathematics teachers has 1,500 members . only 525 of these members cast votes in the election for president of the association . what percent of the total membership voted for the winning candidate if the winning candidate received 40 percent of the votes cast ?
Options: a ) 75 % , b ) 58 % , c ) 42 % , d ) 34 % , e ) 14 %
Rationale: "total umber of members = 1500 number of members that cast votes = 525 since , winning candidate received 40 percent of the votes cast number of votes for winning candidate = ( 40 / 100 ) * 525 = 210 percent of total membership that voted for winning candidate = ( 210 / 1500 ) * 100 = 14 % answer e"
Answer: e
Problem: to fill a tank , 25 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to 3 / 5 of its present ?
Options: a ) 41.66667 , b ) 60.5 , c ) 63.5 , d ) 62.5 , e ) 64.5
Rationale: "let capacity of 1 bucket = x capacity of the tank = 25 x new capacity of the bucket = 3 x / 5 hence , number of buckets needed = 25 x / ( 3 x / 5 ) = ( 25 × 5 ) / 3 = 41.66667 answer is a ."
Answer: a
Problem: salesperson a ' s compensation for any week is $ 300 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of b ' s total sales for that week . for what amount of total weekly sales would both salepeople earn the same compensation ?
Options: a ) $ 21,000 , b ) $ 12,000 , c ) $ 15,000 , d ) $ 4,500 , e ) $ 4,000
Rationale: "sometime , setting up an equation is an easy way to go with : 300 + 0.06 ( x - 1000 ) = 0.08 x x = 12,000 ans : b"
Answer: b
Problem: ritesh and co . generated revenue of rs . 1,800 in 2006 . this was 12.5 % of its gross revenue . in 2007 , the gross revenue grew by rs . 2,500 . what is the percentage increase in the revenue in 2007 ?
Options: a ) 17.36 % , b ) 20 % , c ) 25 % , d ) 50 % , e ) none of these
Rationale: "explanation : given , ritesh and co . generated revenue of rs . 1,800 in 2006 and that this was 12.5 % of the gross revenue . hence , if 1800 is 12.5 % of the revenue , then 100 % ( gross revenue ) is : = > ( 100 / 12.5 ) × 1800 . = > 14,400 . hence , the total revenue by end of 2007 is rs . 14,400 . in 2006 , revenue grew by rs . 2500 . this is a growth of : = > ( 2500 / 14400 ) × 100 . = > 17.36 % . answer : a"
Answer: a
Problem: there are 28 stations between kolkatta and chennai . how many second class tickets have to be printed , so that a passenger can travel from any station to any other station ?
Options: a ) 870 , b ) 970 , c ) 1070 , d ) 1000 , e ) 800
Rationale: the total number of stations = 30 from 30 stations we have to choose any two stations and the direction of travel ( i . e . , kolkatta to chennai is different from chennai to kolkatta ) in 30 p 2 ways . 30 p 2 = 30 * 29 = 870 answer : a
Answer: a
Problem: a contractor is engaged for 30 days on the condition that he receives rs . 25 for each day he works & is fined rs . 7.50 for each day is absent . he gets rs . 490 in all . for how many days was he absent ?
Options: a ) 8 , b ) 10 , c ) 15 , d ) 17 , e ) 19
Rationale: 30 * 25 = 750 490 - - - - - - - - - - - 260 25 + 7.50 = 32.5 260 / 32.5 = 8 a
Answer: a
Problem: what is the area of a rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1 ?
Options: a ) 1 / 9 sq units .
b ) 4 / 9 sq units .
c ) 8 / 9 sq units .
d ) 7 / 9 sq units .
e ) 6 / 9 sq units .
Rationale: area will be 2 / 3 * 4 / 3 = 8 / 9 sq units . answer : c
Answer: c
Problem: albert is 2 times mary ’ s age and 4 times as old as betty . mary is 10 years younger than albert . how old is betty ?
Options: a ) 5 , b ) 8 , c ) 10 , d ) 12 , e ) 15
Rationale: "a = 2 m = m + 10 m = 10 a = 20 a = 4 b , and so b = 5 the answer is a ."
Answer: a
Problem: a worker is paid rs . 20 / - for a full days work . he works 11 / 32 / 31 / 8.3 / 4 days in a week . what is the total amount paid for that worker ?
Options: a ) 56.5 rupees , b ) 58.5 rupees , c ) 57.5 rupees , d ) 59.5 rupees , e ) 55.5 rupees
Rationale: the total days worked = 1 + 0.333 + 0.667 + 0.125 + 0.75 = 2.875 days amount for 2.875 days is ( 2.875 * 20 ) 57.5 rupees answer : c
Answer: c
Problem: how many odd factors does 270 have ?
Options: a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8
Rationale: "start with the prime factorization : 270 = 2 * 3 * 5 for odd factors , we put aside the factor of two , and look at the other prime factors . set of exponents = { 1 , 1 } plus 1 to each = { 2 , 2 } product = 2 * 2 = 4 therefore , there are 4 odd factors of 270 . answer : b ."
Answer: b
Problem: find a number such that when 15 is subtracted from 7 times the number , the results is more than twice the number ?
Options: a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6
Rationale: solution let the number be x . = then , 7 x - 15 = 2 x + 10 = 5 x = 25 ‹ = › x = 5 . hence the required number is 5 . answer d
Answer: d
Problem: having received his weekly allowance , a student spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 1.00 at the candy store . what is this student ’ s weekly allowance ?
Options: a ) $ 2.75 , b ) $ 3.25 , c ) $ 3.75 , d ) $ 4.25 , e ) $ 4.75
Rationale: "let x be the value of the weekly allowance . ( 2 / 3 ) ( 2 / 5 ) x = 100 cents ( 4 / 15 ) x = 100 x = $ 3.75 the answer is c ."
Answer: c
Problem: if 2 | 2 – e | = 5 , what is the sum of all the possible values for e ?
Options: a ) 1 / 2 , b ) 2 , c ) 5 / 2 , d ) 4 , e ) 5
Rationale: if e < 2 , then 2 - e = 5 / 2 ; e = - 1 / 2 if e > 2 , then e - 2 = 5 / 2 ; e = 9 / 2 9 / 2 - 1 / 2 = 8 / 2 = 4 = d
Answer: d
Problem: the market value of a certain machine decreased by 10 percent of its purchase price each year . if the machine was purchased in 1982 for its market value of $ 8,000 , what was its market value two years later ?
Options: a ) $ 8,000 , b ) $ 5,600 , c ) $ 3,200 , d ) $ 6,400 , e ) $ 800
Rationale: d market value in 1982 = $ 8000 market value in 1983 = $ 8000 - ( $ 8000 x 10 / 100 ) = 8000 - 800 = $ 7200 market value in 1984 = market value in 1983 - ( 10 % of $ 8000 ) = 7200 - 800 = $ 6400
Answer: d
Problem: fox jeans regularly sell for $ 15 a pair and pony jeans regularly sell for $ 18 a pair . during a sale these regular unit prices are discounted at different rates so that a total of $ 5 is saved by purchasing 5 pairs of jeans : 3 pairs of fox jeans and 2 pairs of pony jeans . if the sum of the two discounts rates is 18 percent , what is the discount rate on pony jeans ?
Options: a ) 9 % , b ) 10 % , c ) 34.4 % , d ) 12 % , e ) 15 %
Rationale: "you know that fox jeans costs $ 15 , and pony jeans costs $ 18 , you also know that 3 pairs of fox jeans and 2 pairs of pony jeans were purchased . so 3 ( 15 ) = 45 - fox 2 ( 18 ) = 36 - pony the total discount discount is $ 5 and you are asked to find the percent discount of pony jeans , so 45 ( 18 - x ) / 100 + 36 ( x ) / 100 = 5 or 45 * 18 - 45 * x + 36 * x = 5 * 100 or 9 x = - 5 * 100 + 45 * 18 x = 310 / 9 = 34.4 c"
Answer: c
Problem: if | y + 5 | = 8 , what is the sum of all the possible values of y ?
Options: a ) - 14 , b ) - 13 , c ) - 8 , d ) - 12 , e ) - 10
Rationale: "there will be two cases y + 5 = 8 or y + 5 = - 8 = > y = 3 or y = - 13 sum of both the values will be - 13 + 3 = - 10 answer : e"
Answer: e
Problem: let n ~ be defined for all positive integers n as the remainder when ( n - 1 ) ! is divided by n . what is the value of 30 ~ ?
Options: a ) 1 , b ) 0 , c ) 2 , d ) 8 , e ) 31
Rationale: n ~ = ( n - 1 ) ! so 30 ~ = ( 30 - 1 ) ! = 29 ! when 29 ! / 30 we have 16 * 2 inside 29 ! hence 30 gets cancelled and we get remainder as 0 b
Answer: b
Problem: two unbiased coins are tossed . what is the probability of getting two heads ?
Options: a ) 1 / 2 , b ) 1 / 4 , c ) 2 / 3 , d ) 1 / 5 , e ) 2 / 7
Rationale: "s = { hh , ht , th , tt } e = event of getting 2 heads e = { tt } p ( e ) = 1 / 4 answer is b"
Answer: b
Problem: a and b complete a work in 10 days . a alone can do it in 14 days . if both together can do the work in how many days ?
Options: a ) 5.8 days , b ) 4.78 days , c ) 6.25 days , d ) 3.15 days , e ) 2.75 days
Rationale: 1 / 10 + 1 / 14 = 6 / 35 35 / 6 = 5.8 days answer : a
Answer: a
Problem: if jake loses 33 pounds , he will weigh twice as much as his sister . together they now weigh 153 pounds . what is jake ' s present weight , in pounds ?
Options: a ) 131 , b ) 135 , c ) 113 , d ) 147 , e ) 188
Rationale: "j = jake ’ s current weight , in pounds s = sister ’ s current weight , in pounds we are told that “ if jake loses 8 pounds , he will weigh twice as much as his sister . we put this into an equation : j – 33 = 2 s j = 2 s + 33 ( equation 1 ) next , we are told that “ together they now weigh 153 pounds . ” we can also put this into an equation . j + s = 153 ( equation 2 ) to solve this equation , we can substitute 2 s + 8 from equation 1 for the variable j in equation 2 : 2 s + 33 = 153 - s 3 s = 120 s = 40 j + 40 = 153 j = 113 answer : c"
Answer: c
Problem: the value of a machine depreciates at 24 % per annum . if its present value is $ 1 , 50,000 , at what price should it be sold after two years such that a profit of $ 24,000 is made ?
Options: a ) $ 250640 , b ) $ 430640 , c ) $ 120640 , d ) $ 110640 , e ) $ 150640
Rationale: "the value of the machine after two years = 0.76 * 0.76 * 1 , 50,000 = $ 86,640 sp such that a profit of $ 24,000 is made = 86,640 + 24,000 = $ 1 , 10,640 d"
Answer: d
Problem: a vendor buys 10 t - shirts at an average price of $ 14 per t - shirt . he then buys 15 more t - shirts at an average price of $ 11 per t - shirt . what is the average price v per t - shirt that the vendor paid for these purchases ?
Options: a ) $ 12.20 , b ) $ 12.50 , c ) $ 12.55 , d ) $ 12.70 , e ) $ 13.00
Rationale: "correct answer : a explanation : the relevant formula for this problem is average v = ( sum ) / ( number of terms ) . another way to look at the formula is sum = average x number of terms . for the first purchase , the vendor ' s sum ( total cost ) was $ 140 , since 14 x 10 = 140 . for the second purchase , the vendor ' s cost was $ 165 , since 11 x 15 = 165 . the grand sum is then $ 140 + $ 165 , which equals $ 305 . the total number of shirts purchased was 25 , so to get the average price per shirt , we divide 305 by 25 , which equals $ 12.20 . as a result , the correct answer is a . note : a relative understanding of weighted average offers a shortcut to this problem . because the true average of 11 and 14 is 12.5 , but the vendor sells more shirts at the lower price than at the higher price , the weighted average must be less than $ 12.50 ; only answer choice a is a possibility ."
Answer: a
Problem: if 5 machines can produce 20 units in 10 hours , how long would it take 20 machines to produce 120 units ?
Options: a ) 50 hours , b ) 40 hours , c ) 15 hours , d ) 12 hours , e ) 8 hours
Rationale: "here , we ' re told that 5 machines can produce 20 units in 10 hours . . . . that means that each machine works for 10 hours apiece . since there are 5 machines ( and we ' re meant to assume that each machine does the same amount of work ) , then the 5 machines equally created the 20 units . 20 units / 5 machines = 4 units are made by each machine every 10 hours now that we know how long it takes each machine to make 4 units , we can break this down further if we choose to . . . 10 hours / 4 units = 2.5 hours per unit when 1 machine is working . the prompt asks us how long would it take 20 machines to produce 120 units . if 20 machines each work for 2.5 hours , then we ' ll have 20 units . since 120 units is ' 6 times ' 20 , we need ' 6 times ' more time . ( 2.5 hours ) ( 6 times ) = 15 hours final answer : [ reveal ] spoiler : c"
Answer: c
Problem: 66 cubic centimetres of silver is drawn into a wire 1 mm in diameter . the length if the wire in meters will be :
Options: a ) 76 m
b ) 80 m
c ) 84 m
d ) 88 m
e ) none of these
Rationale: explanation : let the length of the wire be h radius = 1 / 2 mm = 120 cm π r 2 h = 66 22 / 7 ∗ 1 / 20 ∗ 1 / 20 ∗ h = 66 = > h = 66 ∗ 20 ∗ 20 ∗ 7 / 22 = 8400 cm = 84 m option c
Answer: c
Problem: a certain sum of money is divided among a , b and c so that for each rs . a has , b has 65 paisa and c 40 paisa . if c ' s share is rs . 48 , find the sum of money ?
Options: a ) 288 , b ) 262 , c ) 246 , d ) 205 , e ) 267
Rationale: "a : b : c = 100 : 65 : 40 = 20 : 13 : 8 8 - - - - 48 41 - - - - ? = > rs . 246 answer : c"
Answer: c
Problem: a is four times as fast as b . if b alone can do a piece of work in 60 days , in what time can a and b together complete the work ?
Options: a ) 10 , b ) 12 , c ) 22 , d ) 28 , e ) 20
Rationale: a can do the work in 60 / 4 i . e . , 15 days . a and b ' s one day ' s work = 1 / 15 + 1 / 60 = ( 4 + 1 ) / 60 = 1 / 12 so a and b together can do the work in 12 days . answer : b
Answer: b
Problem: 8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 10 men , 4 women and 10 children together complete the piece of work ?
Options: a ) 5 days , b ) 15 days , c ) 28 days , d ) 6 days , e ) 7 days
Rationale: explanation : 1 man ’ s 1 day ’ s work = 1 / 8 × 12 = 1 / 96 10 men ’ s 1 day ’ s work = 1 × 10 / 96 = 5 / 48 1 woman ’ s 1 day ’ s work = 1 / 192 4 women ’ s 1 day ’ s work = 1 / 192 × 4 = 1 / 48 1 child ’ s 1 day ’ s work = 1 / 240 10 children ’ s 1 day ’ s work = 1 / 24 therefore , ( 10 men + 4 women + 10 children ) ’ s 1 day ’ s work = 5 / 48 + 1 / 48 + 1 / 24 = 8 / 48 = 1 / 6 the required no . of days = 6 days answer : option d
Answer: d
Problem: if a positive integer n has 211 factors , then how many prime factors does n have ?
Options: a ) 3 , b ) 2 , c ) can not be determined , d ) 4 , e ) 1
Rationale: let the factor be a ^ x * b ^ y . . . . . so ( x + 1 ) ( y + 1 ) . . . . . = 211 . . now 211 is a prime number . . . so only one variable x or y is 210 and the others are 0 . . ( 210 + 1 ) ( 0 + 1 ) . . . so the number becomes a ^ 210 . . thus only one prime factor is there .
Answer: e
Problem: rahim bought 65 books for rs . 1150 from one shop and 50 books for rs . 920 from another . what is the average price he paid per book ?
Options: a ) rs . 17 , b ) rs . 18 , c ) rs . 12 , d ) rs . 11 , e ) rs . 10
Rationale: "average price per book = ( 1150 + 920 ) / ( 65 + 50 ) = 2070 / 115 = rs . 18 answer : b"
Answer: b
Problem: a cistern 10 m long and 6 m wide contains water up to a breadth of 1 m 35 cm . find the total area of the wet surface .
Options: a ) 100 m sqaure , b ) 103 m sqaure , c ) 152 m sqaure , d ) 164 m sqaure , e ) none of these
Rationale: "explanation : area of the wet surface = 2 [ lb + bh + hl ] - lb = 2 [ bh + hl ] + lb = 2 [ ( 6 * 1.35 + 10 * 1.35 ) ] + 10 * 6 = 103 m square option b"
Answer: b
Problem: a city with a population of 173,460 is to be divided into 9 voting districts , and no district is to have a population that is more than 10 percent greater than the population of any other district . what is the minimum possible population that the least populated district could have ?
Options: a ) 17,400 , b ) 17,500 , c ) 17,600 , d ) 17,700 , e ) 17,800
Rationale: "the minimum possible population occurs when all the other districts have a population that is 10 % greater than the least populated district . let p be the population of the least populated district . then 173,460 = p + 8 ( 1.1 ) p 9.8 p = 173,460 p = 17,700 the answer is d ."
Answer: d
Problem: car q and car y traveled the same 80 - mile route . if car q took 2 hours and car y traveled at an average speed that was 50 percent faster than the average speed of car q , how many hours did it take car y to travel the route ?
Options: a ) 2 / 3 , b ) 1 , c ) 4 / 3 , d ) 8 / 5 , e ) 3
Rationale: the speed of car q is ( distance ) / ( time ) = 80 / 2 = 40 miles per hour . the speed of car y = 3 / 2 * 40 = 60 miles per hour - - > ( time ) = ( distance ) / ( speed ) = 80 / 60 = 4 / 3 hours . answer : c . or : to cover the same distance at 3 / 2 as fast rate 2 / 3 as much time is needed - - > ( time ) * 2 / 3 = 2 * 2 / 3 = 4 / 3 hours . answer : c .
Answer: c
Problem: 3 pig ’ s and 10 hens are brought for rs . 1200 . if the average price of a hen be rs . 30 . what is the average price of a pig .
Options: a ) 300 , b ) 350 , c ) 400 , d ) 425 , e ) 275
Rationale: explanation : average price of a hen = rs . 30 total price of 10 hens = 10 * 30 = rs . 300 but total price of 3 pigs and 10 hens = rs . 1200 total price of 3 pigs is = 1200 - 300 = 900 average price of a pig = 900 / 3 = rs . 300 answer : a
Answer: a
Problem: how many factors of 240 are also multiples of 3 ?
Options: a ) 5 , b ) 8 , c ) 9 , d ) 10 , e ) 20
Rationale: "take factors of 240 240 = 2 ^ 4 * 3 ^ 1 * 5 ^ 1 total factors of 240 = ( 4 + 1 ) * ( 1 + 1 ) * ( 1 + 1 ) = 5 * 2 * 2 = 20 out of total factors of 20 , half will have 0 as power of 3 and half will have 1 as power of 3 . it is because we have considered only 0 and 1 as power of 3 to compute total factors of 240 . therefore factors that are multiple of 3 = 20 / 2 = 10 ans - d"
Answer: d
Problem: mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 46800 , how much amount per month would he have spent on clothes and transport ?
Options: a ) 4038 , b ) 8076 , c ) 9691.2 , d ) 4845.6 , e ) 3900
Rationale: "∵ amount , he have spent in 1 month on clothes transport = amount spent on saving per month ∵ amount , spent on clothes and transport = 46800 ⁄ 12 = 3900 answer e"
Answer: e
Problem: a train 605 m long is running with a speed of 60 km / hr . in what time will it pass a man who is running at 6 km / hr in the direction opposite to that in which the train is going ?
Options: a ) 81 , b ) 16 , c ) 33 , d ) 54 , e ) 12
Rationale: "speed of train relative to man = 60 + 6 = 66 km / hr . = 66 * 5 / 18 = 55 / 3 m / sec . time taken to pass the men = 605 * 3 / 55 = 33 sec . answer : c"
Answer: c
Problem: tom opened a shop investing rs . 3000 . jose joined him 2 months later , investing rs . 4500 . they earned a profit of rs . 6300 after completion of one year . what will be jose ' s share of profit ?
Options: a ) 3500 , b ) 2800 , c ) 3251 , d ) 4251 , e ) 3508
Rationale: "sol = ~ s - so anju ’ s share = [ 5 / 9 ] x 6300 = 3500 a"
Answer: a
Problem: the simple form of the ratio 2 / 3 : 2 / 5 is
Options: a ) 5 : 7 , b ) 5 : 2 , c ) 5 : 9 , d ) 5 : 3 , e ) 5 : 4
Rationale: "2 / 3 : 2 / 5 = 10 : 6 = 5 : 3 answer : d"
Answer: d
Problem: rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 83 . how much was lent at 5 % ?
Options: a ) 299 , b ) 566 , c ) 678 , d ) 700 , e ) 8277
Rationale: "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 83 5 x / 100 + 90 – 6 x / 100 = 83 x / 100 = 7 = > x = 700 answer : d"
Answer: d
Problem: a towel , when bleached , was found to have lost 20 % of its length and 10 % of its breadth . the percentage of decrease in area is ?
Options: a ) 25 % , b ) 26 % , c ) 27 % , d ) 28 % , e ) 29 %
Rationale: "explanation : let original length = x and original width = y decrease in area will be = xy − ( 80 x / 100 × 90 y / 100 ) = ( xy − 18 / 25 xy ) = 7 / 25 xy decrease = ( 7 xy / 25 xy × 100 ) % = 28 option d"
Answer: d
Problem: gold is 10 times as heavy as water and copper is 5 times as heavy as water . in what ratio should these be mixed to get an alloy 6 times as heavy as water ?
Options: a ) 3 : 2 , b ) 1 : 4 , c ) 3 : 1 , d ) 5 : 2 , e ) 4 : 3
Rationale: g = 10 w c = 5 w let 1 gm of gold mixed with x gm of copper to get 1 + x gm of the alloy 1 gm gold + x gm copper = x + 1 gm of alloy 10 w + 5 wx = x + 1 * 6 w 10 + 5 x = 6 ( x + 1 ) x = 4 ratio of gold with copper = 1 : 4 = 1 : 4 answer is b
Answer: b
Problem: calculate how many days it will take for 8 boys to paint a 50 m long wall if 6 boys can paint a 60 m long wall in 5 days ,
Options: a ) 4.13 days , b ) 3.13 days , c ) 513 days , d ) 9.13 days , e ) 2 days
Rationale: the length of wall painted by one boy in one day = 60 / 6 * 1 / 5 = 2 m no . of days required to paint 50 m cloth by 8 boys = 50 / 8 * 1 / 2 = 3.13 days . b
Answer: b
Problem: a fair price shopkeeper takes 10 % profit on his goods . he lost 20 % goods during theft . his loss percent is :
Options: a ) 10 , b ) 13 , c ) 12 , d ) 18 , e ) 19
Rationale: "c 12 % suppose he has 100 items . let c . p . of each item be $ 1 . total cost = $ 100 . number of items left after theft = 80 . s . p . of each item = $ 1.10 total sale = 1.10 * 80 = $ 88 hence , loss % = 12 / 100 * 100 = 12 %"
Answer: c
Problem: a customer went to a shop and paid a total of $ 40 , out of which 90 cents was for sales tax on taxable purchases . if the tax rate was 6 % , then what was the cost of the tax free items ?
Options: a ) $ 22.70 , b ) $ 23.20 , c ) $ 24.10 , d ) $ 25.50 , e ) $ 26.90
Rationale: "the total cost was $ 40 . the tax was $ 0.90 let the original price of the taxable items = x given that tax rate = 6 % 0.06 x = 0.90 x = $ 15 the cost of the tax free items was $ 40 - $ 15 - $ 0.90 = $ 24.10 the answer is c ."
Answer: c
Problem: albert buys 4 horses and 9 cows for rs . 13,400 . if he sells the horses at 10 % profit and the cows at 20 % profit , then he earns a total profit of rs . 1880 . the cost of a horse is :
Options: a ) rs . 2007 , b ) rs . 2000 , c ) rs . 2089 , d ) rs . 2067 , e ) rs . 2098
Rationale: "let c . p . of each horse be rs . x and c . p . of each cow be rs . y . then , 4 x + 9 y = 13400 - - ( i ) and , 10 % of 4 x + 20 % of 9 y = 1880 2 / 5 x + 9 / 5 y = 1880 = > 2 x + 9 y = 9400 - - ( ii ) solving ( i ) and ( ii ) , we get : x = 2000 and y = 600 . cost price of each horse = rs . 2000 . answer : b"
Answer: b
Problem: two motor bikes cover the same distance at the speed of 60 and 64 kmps respectively . find the distance traveled by them if the slower bike takes 1 hour more than the faster bike ?
Options: a ) 860 km , b ) 870 km , c ) 960 km , d ) 260 km , e ) 840 km
Rationale: explanation : 60 ( x + 1 ) = 64 x x = 15 60 x 16 = 960 km answer : c
Answer: c
Problem: the edge of a cube is 7 a cm . find its surface ?
Options: a ) 24 a 8
b ) 24 a 4
c ) 24 a 1
d ) 24 a 2
e ) 294 a 2
Rationale: 6 a 2 = 6 * 7 a * 7 a = 294 a 2 answer : e
Answer: e
Problem: if x < y < z and y - x > 5 , where x is an even integer and y and z are odd integers , what is the least possible value s of z - x ?
Options: a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10
Rationale: "x < y < z to find the least possible value for z - x ; we need to find the values for z and x that can be closest to each other . if x is some even number , then what could be minimum possible odd z . if x is some even number y - x > 5 ; y > x + 5 ; minimum value for y = x + 5 + 2 = x + 7 [ note : x + 5 is as even + odd = odd and nearest odd greater than x + 5 is x + 5 + 2 ] minimum value for z = y + 2 = x + 7 + 2 = x + 9 [ note : z = y + 2 because both z and y are odd . difference between two odd numbers is 2 ] s = z - x = x + 9 - x = 9 ans : d"
Answer: d
Problem: if renee earns a raise of between 5 % and 10 % , non - inclusive , by what fraction could her salary have increased ?
Options: a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 10 , d ) 2 / 25 , e ) 1 / 20
Rationale: 5 % is 5 / 100 = 1 / 20 10 % is 10 / 100 = 1 / 10 the increase must be greater than 1 / 20 and less than 1 / 10 . 1 / 4 , 1 / 5 , and 1 / 10 are not less than 1 / 10 . 1 / 20 is not greater than 1 / 20 . the only option left is 2 / 25 , which is in the specified range . answer : d
Answer: d
Problem: a merchant sells an item at a 20 % discount , but still makes a gross profit of 20 percent of the cost . what percent r of the cost would the gross profit on the item have been if it had been sold without the discount ?
Options: a ) 20 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 75 %
Rationale: "let the market price of the product is mp . let the original cost price of the product is cp . selling price ( discounted price ) = 100 % of mp - 20 % mp = 80 % of mp . - - - - - - - - - - - - - - - - ( 1 ) profit made by selling at discounted price = 20 % of cp - - - - - - - - - - - - - - ( 2 ) apply the formula : profit r = selling price - original cost price = > 20 % of cp = 80 % of mp - 100 % cp = > mp = 120 cp / 80 = 3 / 2 ( cp ) now if product is sold without any discount , then , profit = selling price ( without discount ) - original cost price = market price - original cost price = mp - cp = 3 / 2 cp - cp = 1 / 2 cp = 50 % of cp thus , answer should bec ."
Answer: c
Problem: in δ pqs above , if pq = 5 and ps = 6 , then
Options: a ) 9 / 4 , b ) 12 / 5 , c ) 16 / 5 , d ) 30 / 7 , e ) 50 / 3
Rationale: "there are two ways to calculate area of pqs . area remains same , so both are equal . 5 * 6 / 2 = pr * 7 / 2 pr = 30 / 7 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 male students is single ?
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 = 70 married males = 20 then single males = 70 - 20 = 50 we need to find the fraction of male students who are single i . e single male students / total male student = 50 / 70 = 5 / 7 [ e ]"
Answer: e
Problem: johnny travels a total of one hour to and from school . on the way there he jogs at 4 miles per hour and on the return trip he gets picked up by the bus and returns home at 20 miles per hour . how far is it to the school ?
Options: a ) 2 miles , b ) 6.6 miles , c ) 4.8 miles , d ) 8 miles , e ) 10 miles
Rationale: "answer : b ) 6.6 miles . average speed for round trip = 2 * a * b / ( a + b ) , where a , b are speeds so , average speed was = 2 * 4 * 20 / ( 4 + 20 ) = 6.6 m / hr the distance between schoolhome should be half of that . ie . 6.6 miles answer b"
Answer: b
Problem: an engine moves at the speed of 90 kmph without any coaches attached to it . speed of the train reduces at the rate that varies directly as the square root of the number of coaches attached . when 4 coaches are attached speed decreases to 78 kmph . what will be the speed of train when 16 coaches are attached .
Options: a ) 66 , b ) 72 , c ) 80 , d ) 68 , e ) 73
Rationale: "1 . no . of coaches = 4 sqr root = 2 speed decreases by 12 12 = k * 2 k = 6 no . of coaches = 16 swr root = 4 decrease = 4 * 6 = 24 new speed = 90 - 24 = 66 a"
Answer: a
Problem: at what rate percent of simple interest will a sum of money double itself in 77 years ?
Options: a ) 6.54 % , b ) 2.54 % , c ) 8 . 2 % , d ) 4.94 % , e ) 1.29 %
Rationale: let sum = x . then , s . i . = x . rate = ( 100 * s . i . ) / ( p * t ) = ( 100 * x ) / ( x * 77 ) = 100 / 77 = 1.29 % answer : e
Answer: e
Problem: what is the value of 10 ^ 7 - 6 ^ 5 ?
Options: a ) 99 , 91,223 , b ) 99 , 92,224 , c ) 99 , 94,265 , d ) 99 , 95,300 , e ) 99 , 96,307
Rationale: "as 10 ^ n will always have last digit as 0 and 6 ^ n will always as last digit 6 . . hence difference of such sum should always be ending with 4 and there is only on option . . answer b"
Answer: b
Problem: dacid obtained 73 , 69 , 92 , 64 and 82 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ?
Options: a ) 79 , b ) 99 , c ) 88 , d ) 88 , e ) 76
Rationale: "average = ( 73 + 69 + 92 + 64 + 82 ) / 5 = 380 / 5 = 76 answer : e"
Answer: e
Problem: in may mrs lee ' s earnings were 60 percent of the lee family ' s total income . in june mrs lee earned 30 percent more than in may . if the rest of the family ' s income was the same both months , then , in june , mrs lee ' s earnings were approximately what percent of the lee family ' s total income ?
Options: a ) 64 % , b ) 66 % , c ) 72 % , d ) 76 % , e ) 80 %
Rationale: "let in may lee family ' s total income = 100 in may mrs lee ' s income = 60 in may rest of the family ' s income = 40 in june mrs lees income = 60 * 130 / 100 = 78 in june total income = 78 + 40 = 118 % of mrs lee ' s income = 72 / 112 = 66.10 ( b )"
Answer: b
Problem: r and s together can plough a field in 10 hours but by himself r requires 15 hours . how long would s take to plough the same field ?
Options: a ) 24 hours , b ) 5 hours , c ) 10 hours , d ) 15 hours , e ) 20 hours
Rationale: f r and s together can do a piece of work in x days and r alone can do the same work in y days , then s alone can do the same work in x y / y â € “ x days . the no . of hours required by s = 10 ã — 20 / 20 â € “ 10 = 200 / 10 = 20 hours e
Answer: e
Problem: the area of a square is equal to five times the area of a rectangle of dimensions 60 cm * 12 cm . what is the perimeter of the square ?
Options: a ) 289 cm , b ) 800 cm , c ) 829 cm , d ) 288 cm , e ) 240 cm
Rationale: "area of the square = s * s = 5 ( 60 * 12 ) = > s = 60 = 60 cm perimeter of the square = 4 * 60 = 240 cm . answer : e"
Answer: e
Problem: 36 is divided in 2 parts such that 8 times the first part added to 3 times the second part makes 203 . what is the first part ?
Options: a ) 19 , b ) 88 , c ) 267 , d ) 26 , e ) 28
Rationale: explanation : let the first part be x . 8 x + 3 ( 36 – x ) = 203 8 x + 108 – 3 x = 203 5 x + 108 = 203 5 x = 95 x = 19 answer : a
Answer: a
Problem: roy is now 8 years older than julia and half of that amount older than kelly . if in 2 years , roy will be twice as old as julia , then in 2 years what would be roy ’ s age multiplied by kelly ’ s age ?
Options: a ) 160 , b ) 168 , c ) 176 , d ) 184 , e ) 192
Rationale: "r = j + 8 = k + 4 r + 2 = 2 ( j + 2 ) ( j + 8 ) + 2 = 2 j + 4 j = 6 r = 14 k = 10 in 2 years ( r + 2 ) ( k + 2 ) = 16 * 12 = 192 the answer is e ."
Answer: e
Problem: the radius of a cylindrical vessel is 5 cm and height is 3 cm . find the whole surface of the cylinder ?
Options: a ) 251.4 sq cm , b ) 220 sq cm , c ) 440 sq cm , d ) 132 sq cm , e ) 138 sq cm
Rationale: "r = 5 h = 3 2 π r ( h + r ) = 2 * 22 / 7 * 5 ( 8 ) = 251.4 answer : a"
Answer: a
Problem: mike works at a science lab that conducts experiments on bacteria . the population of the bacteria multiplies at a constant rate , and his job is to notate the population of a certain group of bacteria each hour . at 1 p . m . on a certain day , he noted that the population was 500 and then he left the lab . he returned in time to take a reading at 4 p . m . , by which point the population had grown to 62,500 . now he has to fill in the missing data for 2 p . m . and 3 p . m . what was the population at 3 p . m . ?
Options: a ) 13000 , b ) 12500 , c ) 13500 , d ) 14000 , e ) 15000
Rationale: let the rate be x , then population of the bacteria after each hour can be given as 500,500 x , 500 ( x ^ 2 ) , 500 ( x ^ 3 ) now population at 4 pm = 62,500 thus we have 500 ( x ^ 3 ) = 62,500 = 125 thus x = 5 therefore population at 3 pm = 500 ( 25 ) = 12500 answer : b
Answer: b
Problem: the average weight of a group of boys is 20 kg . after a boy of weight 35 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ?
Options: a ) 12 , b ) 14 , c ) 18 , d ) 24 , e ) 10
Rationale: "let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 35 kg joins the group , total weight of boys = 20 x + 35 so 20 x + 35 = 21 ( x + 1 ) = > x = 14 . answer : b"
Answer: b
Problem: at a certain bowling alley , it costs $ 0.50 to rent bowling shoes for the day and $ 1.75 to bowl 1 game . if a person has $ 12.80 and must rent shoes , what is the greatest number of complete games that person can bowl in one day ?
Options: a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
Rationale: "after renting bowling shoes the person is left with $ 12.80 - $ 0.5 = $ 12.30 , which is enough for 12.3 / 1.75 = 7.02 - > ~ 7 . answer : a ."
Answer: a
Problem: the ratio of number of boys and girls in a school is 3 : 5 . if there are 320 students in the school , find the number of girls in the school ?
Options: a ) 150 , b ) 250 , c ) 300 , d ) 200 , e ) 280
Rationale: "let the number of boys and girls be 3 x and 5 x total students = 320 number of girls in the school = 5 * 320 / 8 = 200 answer is d"
Answer: d
Problem: the arithmetic mean and standard deviation of a certain normal distribution are 14.5 and 1.5 , respectively . what value is exactly 2 standard deviations less than the mean ?
Options: a ) 10.5 , b ) 11 , c ) 11.5 , d ) 12 , e ) 12.5
Rationale: "the value which isexactlytwo sd less than the mean is : mean - 2 * sd = 14.5 - 2 * 1.5 = 11.5 . answer : c ."
Answer: c
Problem: a girl sitting in a train which is travelling at 100 kmph observes that a goods train travelling in a opposite direction , takes 6 seconds to pass him . if the goods train is 560 m long , find its speed .
Options: a ) 230 , b ) 232 , c ) 234 , d ) 236 , e ) 238
Rationale: "relative speed = ( 560 / 6 ) m / s = ( 560 / 6 ) * ( 18 / 5 ) = 336 kmph speed of goods train = 336 - 100 = 236 kmph answer is e"
Answer: e
Problem: the average age of 15 students of a class is 15 years . out of these , the average age of 4 students is 14 years and that of the other 9 students is 16 years . tee age of the 15 th student is ?
Options: a ) 20 years , b ) 22 years , c ) 24 years , d ) 25 years , e ) 27 years
Rationale: "age of the 15 th student = [ 15 * 15 - ( 14 * 4 + 16 * 9 ) ] = ( 225 - 200 ) = 25 years . answer : d"
Answer: d
Problem: a can do a piece of work in 10 days and b alone can do it in 15 days . how much time will both take to finish the work ?
Options: a ) 2 days , b ) 7 days , c ) 6 days , d ) 9 days , e ) 3 days
Rationale: "c 6 days time taken to finish the job = xy / ( x + y ) = 10 x 15 / ( 10 + 15 ) = 150 / 25 = 6 days"
Answer: c
Problem: in a class of 25 students in an examination in maths 5 students scored 95 marks each , 3 get zero each and the average of the rest was 45 . what is the average of the whole class ?
Options: a ) a ) 47 , b ) b ) 45.6 , c ) c ) 44 , d ) d ) 48 , e ) e ) 49.6
Rationale: "explanation : total marks obtained by a class of 25 students = 5 * 95 + 3 * 0 + 17 * 45 = 1240 average marks of whole class = 1240 / 25 = 49.6 answer : option e"
Answer: e
Problem: if | 20 x - 10 | = 190 , then find the product of the values of x ?
Options: a ) - 45 , b ) 50 , c ) - 62 , d ) 35 , e ) - 90
Rationale: "| 20 x - 10 | = 190 20 x - 10 = 190 or 20 x - 10 = - 190 20 x = 200 or 20 x = - 180 x = 10 or x = - 9 product = - 9 * 10 = - 90 answer is e"
Answer: e
Problem: how many positive integers between 1 and 400 are there such that they are multiples of 25 ?
Options: a ) 12 , b ) 13 , c ) 16 , d ) 15 , e ) 11
Rationale: "multiples of 25 = 25 , 50,75 , - - - - - 400 number of multiples of 25 = > 25 * 16 = 400 answer is c"
Answer: c
Problem: if the range w of the 6 numbers 4 , 314 , 710 and x is 12 , what is the difference between the greatest possible value of x and least possible value of x ?
Options: a ) 0 , b ) 2 , c ) 12 , d ) 13 , e ) 15
Rationale: the range w of a set is the difference between the largest and smallest elements of a set . without x , the difference between the largest and smallest elements of a set is 14 - 3 = 11 < 12 , which means that in order 12 to be the range of the set x must be either the smallest element so that 14 - x = 12 - - - > x = 2 or x must the largest element so that x - 3 = 12 - - > x = 15 . the the difference between the greatest possible value of x and least possible value of x is 15 - 2 = 13 . answer : d .
Answer: d
Problem: how many 7 digit number contain number 3 ?
Options: a ) 2748472 , b ) 3748472 , c ) 4748472 , d ) 5748472 , e ) 6748472
Rationale: "total 7 digit no . = 9 * 10 * 10 * 10 * 10 * 10 * 10 = 9000000 not containing 3 = 8 * 9 * 9 * 9 * 9 * 9 * 9 = 4251528 total 7 digit number contain 3 = 9000 - 4251528 = 4748472 answer : c"
Answer: c
Problem: the sale price of an article including the sales tax is rs . 616 . the rate of sales tax is 10 % . if the shopkeeper has made a profit of 18 % , then the cost price of the article is :
Options: a ) 500 , b ) 474.6 , c ) 222 , d ) 297 , e ) 111
Rationale: "explanation : 110 % of s . p . = 616 s . p . = ( 616 * 100 ) / 110 = rs . 560 c . p = ( 100 * 560 ) / 118 = rs . 474.6 answer : b"
Answer: b
Problem: how many integers from 10 to 180 , inclusive , are divisible by 3 but not divisible by 7 ?
Options: a ) 45 , b ) 49 , c ) 50 , d ) 52 , e ) 56
Rationale: "we should find # of integers divisible by 3 but not by 3 * 7 = 21 . # of multiples of 21 in the range from 10 to 180 , inclusive is ( 168 - 21 ) / 21 + 1 = 8 ; 57 - 8 = 49 . answer : b ."
Answer: b
Problem: 12 men work 8 hours per day to complete the work in 10 days . to complete the same work in 8 days , working 15 hours a day , the number of men required
Options: a ) 8 days , b ) 3 days , c ) 7 days , d ) 5 days , e ) 6 days
Rationale: "that is , 1 work done = 12 × 8 × 10 then , 12 8 × 10 = ? × 15 × 8 ? ( i . e . no . of men required ) = 12 × 8 × 10 / 15 × 10 = 8 days . a"
Answer: a