regisss/math_qa huggingface dataset:
Problem: a lady grows broccoli in her garden that is in the shape of a square . each broccoli takes 1 square foot of area in her garden . this year , she has increased her output by 79 broccoli when compared to last year . the shape of the area used for growing the broccoli has remained a square in both these years . how many broccoli did she produce this year ?
Options: a ) 1600
b ) 1500
c ) 1650
d ) 1700
e ) can not be determined
Rationale: explanatory answer the shape of the area used for growing broccoli has remained a square in both the years . let the side of the square area used for growing broccoli this year be x ft . therefore , the area of the ground used for cultivation this year = x 2 sq . ft . let the side of the square area used for growing broccoli last year be y ft . therefore , the area of the ground used for cultivation last year = y 2 sq . ft . as the number of broccoli grown has increased by 79 , the area would have increased by 79 sq ft because each broccoli takes 1 sq ft space . hence , x 2 - y 2 = 79 ( x + y ) ( x - y ) = 79 . 79 is a prime number and hence it will have only two factors . i . e . , 79 and 1 . therefore , 79 can be expressed as product of 2 numbers in only way = 79 * 1 i . e . , ( x + y ) ( x - y ) = 79 * 1 so , ( x + y ) should be 79 and ( x - y ) should be 1 . solving the two equations we get x = 40 and y = 39 . therefore , number of broccoli produced this year = x 2 = 402 = 1600 . alternative approach : use answer choices the area in both the years are squares of two numbers . that rules out choice b , c and d . as 1500,1650 and 1700 are not the square of any number . check choice a : if this year ' s produce is 1600 , last year ' s produce would have been 1600 - 79 = 1521 1521 is the square of 39 . so , 1600 is the answer . choice a
Answer: a
Problem: a train 120 m long is running at 60 kmph . in how much time will it pass a platform 240 m long ?
Options: a ) 15.6 seconds , b ) 24.6 seconds , c ) 21.6 seconds , d ) 30.6 seconds , e ) 35.6 seconds
Rationale: distance travelled = 120 + 240 m = 360 m speed = 60 * 5 / 8 = 50 / 3 m time = 360 * 3 / 50 = 21.6 seconds answer : c .
Answer: c
Problem: a is twice as good a workman as b and together they complete a work in 12 days . in how many days a alone can do the work ?
Options: a ) 32 , b ) 33 , c ) 34 , d ) 35 , e ) 36
Rationale: a ’ s one day ’ s work : b ’ s one day ’ s work = 2 : 1 ( a + b ) ’ s one day ’ s work = 1 / 12 b ’ s one day ’ s work = 1 / 12 * 1 / 3 = 1 / 36 b alone can do the work in 36 days answer : e
Answer: e
Problem: 10 women can complete a work in 5 days and 10 children take 10 days to complete the work . how many days will 5 women and 10 children take to complete the work ?
Options: a ) 4 , b ) 5 , c ) 7 , d ) 8 , e ) 2
Rationale: "1 women ' s 1 day work = 1 / 50 1 child ' s 1 day work = 1 / 100 ( 5 women + 10 children ) ' s 1 day work = ( 5 / 50 + 10 / 100 ) = 1 / 5 5 women and 10 children will complete the work in 5 days . b"
Answer: b
Problem: if p # q denotes the least common multiple of p and q , then e = ( ( 12 # 16 ) # ( 18 # 24 ) ) = ?
Options: a ) 216 , b ) 180 , c ) 144 , d ) 108 , e ) 72
Rationale: there are several ways to find the least common multiple of two numbers . in this case , the most efficient method is to use the greatest common factor : ( a * b ) / ( gcf ab ) = lcm ab the greatest common factor of 12 and 16 is 4 . so , 12 # 16 = 12 * 16 / 4 = 48 . the greatest common factor of 18 and 24 is 6 . so , 18 # 24 = 18 * 24 / 6 = 72 finally , the greatest common factor of 48 and 72 is 24 . so , e = ( ( 12 # 16 ) # ( 18 # 24 ) ) = 48 # 72 = ( 48 * 72 ) / 24 = 2 * 72 = 144 the correct answer is c .
Answer: c
Problem: if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 6 - m ) then m = ?
Options: a ) 1 , b ) 2 , c ) 3 , d ) 5 , e ) 6
Rationale: ( - 2 ) ^ ( 2 m ) = 4 ^ m and 2 ^ ( 6 - m ) = 4 ^ ( ( 6 - m ) / 2 ) therefore , m = ( 6 - m ) / 2 2 m = 6 - m m = 2 answer b
Answer: b
Problem: a train passes a platform in 44 seconds . the same train passes a man standing on the platform in 20 seconds . if the speed of the train is 54 km / hr , the length of the platform is
Options: a ) 240 , b ) 250 , c ) 360 , d ) 230 , e ) 220
Rationale: "speed of the train = 54 km / hr = ( 54 × 10 ) / 36 m / s = 15 m / s length of the train = speed × time taken to cross the man = 15 × 20 = 300 m let the length of the platform = l time taken to cross the platform = ( 300 + l ) / 15 = > ( 300 + l ) / 15 = 36 = > 300 + l = 15 × 44 = 660 = > l = 660 - 300 = 360 meter answer is c ."
Answer: c
Problem: exactly 35 % of the reporters for a certain wire service cover local politics in country x . if 30 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ?
Options: a ) 20 % , b ) 42 % , c ) 44 % , d ) 50 % , e ) 84 %
Rationale: "you are correct . people who cover local politics are a subset of people who cover politics . 30 % of reporters who cover politics do not cover local politics so 70 % do cover local politics . reporters covering local politics = 70 % of reporters covering politics = 35 % ofall reporters reporters covering politics / all reporters = 35 / 70 = 1 / 2 reporters not covering politics / all reporters = 1 / 2 = 50 % = d"
Answer: d
Problem: if rs . 10 be allowed as true discount on a bill of rs . 110 due at the end of a certain time , then the discount allowed on the same sum due at the end of double the time is :
Options: a ) rs . 20 , b ) rs . 21.81 , c ) rs . 22 , d ) rs . 18.33 , e ) none
Rationale: "solution s . i . on rs . ( 110 - 10 ) for a certain time = rs . 10 . s . i . on rs . 100 for double the time = rs . 20 . t . d . on rs . 120 = rs . ( 120 - 100 ) = rs . 20 . t . d . on rs . 110 = rs . ( 20 / 120 x 120 ) = rs . 18.33 . answer d"
Answer: d
Problem: the speed at which a man can row a boat in still water is 150 kmph . if he rows downstream , where the speed of current is 48 kmph , what time will he take to cover 390 metres ?
Options: a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14
Rationale: "speed of the boat downstream = 150 + 48 = 198 kmph = 198 * 5 / 18 = 30 m / s hence time taken to cover 390 m = 390 / 30 = 13 seconds . answer : d"
Answer: d
Problem: a shopkeeper gave an additional 20 per cent concession on the reduced price after giving 30 per cent standard concession on an article . if arun bought that article for 1120 , what was the original price ?
Options: a ) 3000 , b ) 4000 , c ) 2400 , d ) 2000 , e ) none of these
Rationale: original price = 1120 × 100 ⁄ 70 × 100 ⁄ 80 = 2000 answer d
Answer: d
Problem: if x , y , and z are positive integers and 3 x = 4 y = 7 z , then the least possible value of x - y + z is
Options: a ) 33 , b ) 40 , c ) 49 , d ) 19 , e ) 61
Rationale: x - y + z = - 3 x / 4 + x + 3 x / 7 = ( - 21 + 28 + 12 ) x / 28 = 19 x / 28 19 is not divisible by 28 ( it ' s a prime # ) , so for least value , x = 28 answer - d
Answer: d
Problem: a man has some hens and cows . if the number of heads be 48 and the number of feet equals 140 , then the number of cows will be :
Options: a ) 30 , b ) 23 , c ) 24 , d ) 26 , e ) 22
Rationale: "explanation : let the number of hens be x and the number of cows be y . then , x + y = 48 . . . . ( i ) and 2 x + 4 y = 140 x + 2 y = 70 . . . . ( ii ) solving ( i ) and ( ii ) we get : x = 26 , y = 22 . the required answer = 22 . answer : e"
Answer: e
Problem: a man buys an article for $ 100 . and sells it for $ 130 . find the gain percent ?
Options: a ) 10 % , b ) 15 % , c ) 25 % , d ) 20 % , e ) 30 %
Rationale: "c . p . = $ 100 s . p . = $ 130 gain = $ 30 gain % = 30 / 100 * 100 = 30 % answer is e"
Answer: e
Problem: how many cubes of 20 cm edge can be put in a cubical box of 1 m edge .
Options: a ) 177 cm , b ) 125 cm , c ) 786 cm , d ) 617 cm , e ) 187 cm
Rationale: "number of cubes = 100 â ˆ — 100 â ˆ — 100 / 20 â ˆ — 20 â ˆ — 20 = 125 note : 1 m = 100 cm answer : b"
Answer: b
Problem: 3 houses are available in a new flat . 3 persons apply for the houses . each applies for 1 house without consulting others . the probability that all the 3 apply for the same house is ?
Options: a ) 1 / 7 , b ) 1 / 9 , c ) 2 / 7 , d ) 3 / 7 , e ) 2 / 9
Rationale: one person can select one house out of 3 = \ inline { \ color { black } 3 c _ { 1 } } ways = 3 . hence , three persons can select one house out of 3 in 3 x 3 x 3 = 9 . therefore , probability that all thre apply for the same house is 1 / 9 b
Answer: b
Problem: if log 10 5 + log 10 ( 5 x + 1 ) = log 10 ( x + 5 ) + 1 , then x is equal to :
Options: a ) 1 , b ) 5 , c ) 3 , d ) 2 , e ) 6
Rationale: log 10 5 + log 10 ( 5 x + 1 ) = log 10 ( x + 5 ) + 1 log 10 5 + log 10 ( 5 x + 1 ) = log 10 ( x + 5 ) + log 10 10 log 10 [ 5 ( 5 x + 1 ) ] = log 10 [ 10 ( x + 5 ) ] 5 ( 5 x + 1 ) = 10 ( x + 5 ) 5 x + 1 = 2 x + 10 3 x = 9 = > x = 3 answer c
Answer: c
Problem: the sum of 4 consecutive even numbers is 140 . what would be the smallest number ?
Options: a ) 32 , b ) 88 , c ) 70 , d ) 123 , e ) 12
Rationale: let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 140 = > x = 18 smallest number is : 2 ( x - 2 ) = 32 answer : a
Answer: a
Problem: a train speeds past a pole in 15 seconds and a platform 100 m long in 30 seconds . its length is ?
Options: a ) 188 m , b ) 876 m , c ) 251 m , d ) 200 m , e ) 145 m
Rationale: "let the length of the train be x meters and its speed be y m / sec . they , x / y = 15 = > y = x / 15 x + 100 / 30 = x / 15 x = 200 m . answer : d"
Answer: d
Problem: an article is sold at a certain price . by selling it at 2 / 3 of that price one loses 10 % . find the gain percent at original price ?
Options: a ) 25 % , b ) 20 % , c ) 35 % , d ) 40 % , e ) 50 %
Rationale: let the original s . p . be $ x then , new s . p . = $ 2 / 3 x loss = 10 % c . p . = 100 / 90 * 2 / 3 x = 20 x / 27 c . p . = 20 x / 27 gain = x - 20 x / 27 = 7 x / 27 gain % = 7 x / 27 * 27 / 20 x * 100 = 35 % answer is c
Answer: c
Problem: in a sports club with 30 members , 17 play badminton and 19 play tennis and 3 do not play either . how many members play both badminton and tennis ?
Options: a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11
Rationale: "let x play both badminton and tennis so 17 - x play only badminton and 19 - x play only tennis . 2 play none and there are total 30 students . hence , ( 17 - x ) + ( 19 - x ) + x + 3 = 30 39 - 2 x + x = 30 39 - x = 30 x = 9 so 9 members play both badminton and tennis . c"
Answer: c
Problem: the local recycling plant had a contract requiring anyone who obtained recycled bottles for storage for them , to bring them back to be recycled again . the plant could make one new bottle from every 7 bottles returned . one week , on monday , they got 343 bottles to recycle . assuming that everybody brought back all the empties , how many could they eventually remake from the 343 ?
Options: a ) 53 , b ) 54 , c ) 55 , d ) 56 , e ) 57
Rationale: in 1 st level they will make 343 / 7 = 49 bottles . in 2 nd level they will make 49 / 7 = 7 bottles . in 3 rd level they will make 7 / 7 = 1 bottle . totally they can remake 49 + 7 + 1 = 57 bottles answer : e
Answer: e
Problem: how many numbers between 400 and 1000 can be made with the digits 2 , 3,4 , 5,6 and 0 ?
Options: a ) 30 , b ) 60 , c ) 90 , d ) 120 , e ) 150
Rationale: "required number = 3 * 5 p 2 = 3 * 20 = 60 answer is b"
Answer: b
Problem: the difference between the heights of two trees is 20 cm . 10 years ago , the taller one was thrice as taller as the shorter one . the present height of the taller tree is
Options: a ) 300 cm , b ) 15 cm , c ) 20 cm , d ) 10 cm , e ) 40 cm
Rationale: let their heights of x cm and ( x + 20 ) cm respectively . then , ( x + 20 ) - 10 = 3 ( x - 10 ) ⇔ x + 10 = 3 x - 30 ⇔ 2 x = 40 ∴ present height of taller tree = ( 20 + 10 ) = 30 cm . answer : 30 cm
Answer: a
Problem: water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 108 grams of water ?
Options: a ) 16 , b ) 96 , c ) 112 , d ) 128 , e ) 142
Rationale: "solution : we are given that the ratio of hydrogen to oxygen in water , by mass , is 2 : 16 . using our ratio multiplier we can re - write this as 2 x : 16 x . we can now use these expressions to determine how much oxygen is in 108 grams of water . 2 x + 16 x = 108 18 x = 108 x = 6 since x is 6 , we know that there are 16 x 6 = 96 grams of oxygen in 108 grams of water . answer b ."
Answer: b
Problem: in an examination , 35 % of total students failed in hindi , 45 % failed in english and 20 % in both . the percentage of these who passed in both the subjects is :
Options: a ) 10 % , b ) 20 % , c ) 30 % , d ) 40 % , e ) 50 %
Rationale: "pass percentage = 100 - ( 35 + 45 - 20 ) = 100 - 60 = 40 answer : d"
Answer: d
Problem: 8 people went to a restaurant for dinner . 7 of them spent $ 10 each on their meals and the eighth spent $ 7 more than the average expenditure of all 8 people . what was the total amount of money that the 8 people spent ?
Options: a ) $ 85 , b ) $ 88 , c ) $ 91 , d ) $ 94 , e ) $ 97
Rationale: let t be the total amount of money spent . 7 * 10 + ( t / 8 + 7 ) = t 77 = 7 t / 8 t = ( 8 * 77 ) / 7 = 88 the answer is b .
Answer: b
Problem: a person crosses a 1080 m long street in 12 minutes . what is his speed in km per hour ?
Options: a ) 4.1 , b ) 4.5 , c ) 4.8 , d ) 5.4 , e ) 5.5
Rationale: "speed = 1080 / ( 12 x 60 ) m / sec = 1.5 m / sec . converting m / sec to km / hr = 1.5 x ( 18 / 5 ) km / hr = 5.4 km / hr . answer : d"
Answer: d
Problem: there were two candidates in an election . winner candidate received 62 % of votes and won the election by 384 votes . find the number of votes casted to the winning candidate ?
Options: a ) 288 , b ) 744 , c ) 992 , d ) 298 , e ) 177
Rationale: "w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 384 62 % - - - - - - - - ? = > 992 answer : c"
Answer: c
Problem: the compound ratio of 1 : 2 , 2 : 5 and 3 : 10 ?
Options: a ) 1 : 2 , b ) 1 : 5 , c ) 3 : 5 , d ) 1 : 50 , e ) 3 : 50
Rationale: 1 / 2 * 2 / 5 * 3 / 10 = 3 / 50 3 : 50 answer : e
Answer: e
Problem: if the product of the integers from 1 to n is divisible by 210 , what is the least possible value of n ?
Options: a ) 7 , b ) 14 , c ) 21 , d ) 28 , e ) 35
Rationale: "210 = 2 x 3 x 5 x 7 n must include at least up to the number 7 . the answer is a ."
Answer: a
Problem: a grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches . if the measurement of each soap box is 7 inches by 6 inches by 10 inches , then what is the maximum number of soap boxes that can be placed in each carton ?
Options: a ) 210 , b ) 252 , c ) 280 , d ) 150 , e ) 420
Rationale: "however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers dimensions of the box = 25 * 42 * 60 dimensions of the soap = 10 * 6 * 7 we get = 5 x 6 x 5 = 150 so the question is why this particular arrangement , in order to maximize number of soaps we need to minimize the space wasted and this is the only config where we dont waste any space so we can expect the maximum number the answer is ( d )"
Answer: d
Problem: selling an kite for rs . 30 , a shop keeper gains 25 % . during a clearance sale , the shopkeeper allows a discount of 10 % on the marked price . his gain percent during the sale is ?
Options: a ) 8 % , b ) 10 % , c ) 11 % , d ) 12.5 % , e ) 20 %
Rationale: "explanation : marked price = rs . 30 c . p . = 100 / 125 * 30 = rs . 24 sale price = 90 % of rs . 30 = rs . 27 required gain % = 3 / 24 * 100 = 12.5 % . answer : d"
Answer: d
Problem: in a university the average age in an arts class is 21 and average age in technical class is 18 . what is the age average of university if there are 8 arts classes and 5 technical classes ?
Options: a ) 18.75 , b ) 19.85 , c ) 20.32 , d ) 20.9 , e ) 21.0
Rationale: average age of university = ( ( 8 x 21 ) + ( 5 x 18 ) ) / 13 = 19.85 answer : b
Answer: b
Problem: a theater box office sold an average ( arithmetic mean ) of 65 tickets per staff member to a particular movie . among the daytime staff , the average number sold per member was 75 , and among the evening staff , the average number sold was 60 . if there are no other employees , what was the ratio of the number of daytime staff members to the number of evening staff members ?
Options: a ) 1 : 2 , b ) 1 : 4 , c ) 5 : 11 , d ) 1 : 3 , e ) 13 : 15
Rationale: "deviation from the mean for the daytime staff = 75 - 65 = 10 . deviation from the mean for the evening staff = 65 - 60 = 5 . thus , the ratio of the number of daytime staff members to the number of evening staff members is 5 : 10 = 1 : 2 . the answer is a ."
Answer: a
Problem: a milk man has 40 liters of milk . if he mixes 5 liters of water , which is freely available , in 20 liters of pure milk . if the cost of pure milk is rs . 18 per liter , then the profit of the milkman , when he sells all the mixture at cost price is :
Options: a ) 20 % , b ) 25 % , c ) 45 % , d ) 18 % , e ) none of these
Rationale: "explanation : when the water is freely available and all the water is sold at the price of the milk , then the water gives the profit on the cost of 40 liters of milk . therefore , profit percentage = 45 % . answer : c"
Answer: c
Problem: the sum of ages of 5 children born 1 years different each is 30 years . what is the age of the elder child ?
Options: a ) 8 , b ) 9 , c ) 10 , d ) 16 , e ) 17
Rationale: "let the ages of children be x , ( x + 1 ) , ( x + 2 ) , ( x + 3 ) and ( x + 4 ) years . then , x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) = 30 5 x = 20 x = 4 . x + 4 = 4 + 4 = 8 answer : a"
Answer: a
Problem: in a kilometer race , if abhishek gives bharti a 40 m start , abhishek wins by 19 sec . but if abhishek gives bharti a 30 sec start , bharti wins by 40 m . find the time taken by bharti to run 5,000 m ?
Options: a ) 150 sec , b ) 750 sec , c ) 450 sec , d ) 825 sec , e ) none of these
Rationale: explanation : if abhishek takes x seconds and bharti takes y seconds to run 1 km , then : = > y = 150 sec and x = 125 sec = > x + 19 = 960 y / 1000 and = ( 960 x / 1000 ) + 30 = y = > y = 150 sec and x = 125 sec answer = ( 150 / 1000 ) × 500 = 750 sec answer : b
Answer: b
Problem: every disk in a bag is either blue , yellow or green . the ratio of blue disks to yellow disks to green disks in this bag is 3 : 7 : 8 . if the total number of disks in the bag is 144 , how many more green disks than blue disks are in the bag ?
Options: a ) 25 , b ) 28 , c ) 30 , d ) 35 , e ) 40
Rationale: "let b : y : g = 3 x : 7 x : 8 x . 3 x + 7 x + 8 x = 18 x = 144 - - > x = 8 . g - b = 8 x - 3 x = 5 x = 40 . the answer is e ."
Answer: e
Problem: if the sum of two numbers is 22 and the sum of their squares is 404 , then the product of the numbers is
Options: a ) 40 , b ) 44 , c ) 80 , d ) 88 , e ) 48
Rationale: "according to the given conditions x + y = 22 and x ^ 2 + y ^ 2 = 404 now ( x + y ) ^ 2 = x ^ 2 + y ^ 2 + 2 xy so 22 ^ 2 = 404 + 2 xy so xy = 80 / 2 = 40 answer : a"
Answer: a
Problem: there are 200 female managers in a certain company . find the total number w 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 ) 300 , b ) w = 400 , c ) w = 500 , d ) w = 600 , e ) none of these
Rationale: "{ managers } = { female managers } + { male managers } ; we are told that the total number of managers in the company is 2 / 5 of all the employees , thus { managers } = 2 / 5 ( m + f ) , where m and f are number of female and male employees , respectively . also , we know that 2 / 5 of all male employees are managers : { male managers } = 2 / 5 * mas well as there are total of 200 female managers : { female managers } = 200 ; thus : 2 / 5 ( m + f ) = 200 + 2 / 5 * m - - > f = 500 . answer : c ."
Answer: c
Problem: a store reduced the price of all items in the store by 10 % on the first day and by another 10 % on the second day . the price of items on the second day was what percent of the price before the first reduction took place ?
Options: a ) 80.0 , b ) 80.9 , c ) 81.0 , d ) 81.1 , e ) 81.9
Rationale: "consider price of the all items as $ 100 after a initial reduction of 10 % price becomes = 0.9 * 100 = $ 90 after the final reduction of 10 % price becomes = 0.9 * 90 = $ 81 price of all items on second day is 81 % of price on first day correct answer option c"
Answer: c
Problem: on the first of the year , james invested x dollars at proudstar bank in an account that yields 0.5 % in interest every quarter year . at the end of the year , during which he made no additional deposits or withdrawals , he had y dollars in the account . if james had invested the same amount in an account which pays interest on a yearly basis , what must the interest rate be for james to have y dollars at the end of the year ?
Options: a ) 2.04 % , b ) 4.12 % , c ) 5 % , d ) 8.25 % , e ) 10 %
Rationale: per quarter , interest = 0.5 % so for a year , interest = 2 % due to quarter cumulation , effective yield ( ytm ) would be slight higher than 2 % answer = 2.04 % = answer : a
Answer: a
Problem: the chance of rain on any given day in tel - aviv is 50 % . what is the probability w that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ?
Options: a ) 15 / 64 , b ) 30 / 64 , c ) 1 / 2 , d ) 1 / 4 , e ) 52 / 64
Rationale: chances of rain on exactly 4 days and not rain on 2 days = ( 1 / 2 ) ^ 4 * ( 1 / 2 ) ^ 2 = 1 / 64 chosing 4 days out of 6 = 6 ! / ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days w = 15 / 64 ans a it is .
Answer: a
Problem: the sum of two numbers is 77 . 5 times one number is equal to 6 times the other . the bigger of the two numbers is
Options: a ) 10 , b ) 42 , c ) 14 , d ) 15 , e ) 60
Rationale: a : b 5 : 6 a + b = 77 5 x + 6 x = 77 x = 7 then two numbers are 35,42 . the bigger number is 42 answer : b
Answer: b
Problem: income and expenditure of a person are in the ratio 7 : 6 . if the income of the person is rs . 14000 , then find his savings ?
Options: a ) 500 , b ) 1000 , c ) 1500 , d ) 2000 , e ) 2500
Rationale: "let the income and the expenditure of the person be rs . 7 x and rs . 6 x respectively . income , 7 x = 14000 = > x = 2000 savings = income - expenditure = 7 x - 6 x = x so , savings = rs . 2000 . answer : d"
Answer: d
Problem: 3 people took gmat practice tests in which the minimum and maximum possible scores are 200 and 800 , respectively . they each took a test 3 times , and no one scored below 400 or over 700 . if the individual ranges of the 3 people ' s scores in those 3 practice tests were 40 , 70 and 100 , what is the difference between the maximum and minimum possible ranges of all their scores put together .
Options: a ) 70 , b ) 80 , c ) 100 , d ) 140 , e ) 200
Rationale: according to the information in the question the maximum range would be between 400 and 700 ( 300 ) . the minimum range would be the greatest range for any one individual , which is listed in the problem as 100 . so the difference between these would be 300 - 100 or 200 . answer : e
Answer: e
Problem: if p / q = 3 / 11 , then 2 p + q = ?
Options: a ) 12 , b ) 14 , c ) 17 , d ) 15 , e ) 16
Rationale: let p = 3 , q = 11 then 2 * 3 + 11 = 17 so 2 p + q = 17 . answer : c
Answer: c
Problem: what is the probability of rolling a total of 5 with a single roll of two fair 6 - sided dice , each with the distinct numbers 1 through 6 on each side ?
Options: a ) 1 / 12 , b ) 1 / 6 , c ) 2 / 7 , d ) 1 / 3 , e ) 1 / 9
Rationale: totalsample spacefrom two fair six - sided dice is 36 . there are six possible ways to be 5 i . e . ( 1 + 4 ) , ( 2 + 3 ) , ( 3 + 2 ) , ( 4 + 1 ) so , total probability is 4 / 36 or 1 / 9 . answer is e .
Answer: e
Problem: if the simple interest on a certain sum of money for 6 years is one – fifth of the sum , then the rate of interest per annum is
Options: a ) 3.3 % , b ) 7.2 % , c ) 6.1 % , d ) 5.0 % , e ) 3.1 %
Rationale: "explanation : let the principal ( p ) be x then , simple interest ( si ) = x / 5 time ( t ) = 6 years rate of interest per annum ( r ) = ( 100 × si ) / pt = ( 100 × ( x / 5 ) / ( x × 6 ) = 20 / 6 = 3.3 % answer : option a"
Answer: a
Problem: the population of a city is 415600 . it increased by 25 % in the first year and decreased by 30 % in the second year . what is the population of the city at the end of second year ?
Options: a ) 367300 , b ) 363580 , c ) 363650 , d ) 373650 , e ) 386750
Rationale: as per given data , p = 415600 , r 1 = 25 % incresed , r 2 = 30 % decreased population of the city at the end of the second year = p ( 1 + r 1 / 100 ) ( 1 - r 2 / 100 ) = 415600 ( 1 + 25 / 100 ) ( 1 - 30 / 100 ) = 363650 answer : c
Answer: c
Problem: what is the least value of x , so that 2 x 5472 is divisible by 9
Options: a ) 7 , b ) 8 , c ) 9 , d ) 3 , e ) 2
Rationale: "explanation : the sum of the digits of the number is divisible by 9 . then the number is divisible by 9 . 2 + x + 5 + 4 + 7 + 2 = 20 + x least value of x may be ' 7 ' , so that the total 20 + 7 = 27 is divisible by 9 . answer : option a"
Answer: a
Problem: a thief is spotted by a policeman from a distance of 150 meters . when the policeman starts the chase , the thief also starts running . if the speed of the thief be 8 km / hr and that of the policeman 10 km / hr , how far the thief will have run before he is overtaken ?
Options: a ) 350 m , b ) 200 m , c ) 400 m , d ) 600 m , e ) none of them
Rationale: "relative speed of the policeman = ( 10 - 8 ) km / hr = 2 km / hr . time taken by police man to cover ( 150 m / 1000 ) x 1 / 2 hr = 3 / 40 hr . in 3 / 40 hrs , the thief covers a distance of 8 x 3 / 40 km = 3 / 5 km = 600 m answer is d"
Answer: d
Problem: car x began traveling at an average speed of 35 miles per hour . after 72 minutes , car y began traveling at an average speed of 70 miles per hour . when both cars had traveled the same distance , both cars stopped . how many miles did car x travel from the time car y began traveling until both cars stopped ?
Options: a ) 15 , b ) 20 , c ) 40 , d ) 42 , e ) 68
Rationale: "car y began travelling after 72 minutes or 1.2 hours . let t be the time for which car y travelled before it stopped . both cars stop when they have travelled the same distance . so , 35 ( t + 1.2 ) = 70 t t = 1.2 distance traveled by car x from the time car y began traveling until both cars stopped is 35 x 1.2 = 42 miles answer : - d"
Answer: d
Problem: a fruit juice company earns $ 150 per 100 gallons of apple juice sold , $ 170 per 100 gallons of guava juice sold , $ 130 per 100 gallons of grape juice sold , $ 100 per 100 gallons of orange juice . if the company manages to sell 230 gallons of apple , 210 gallons of guava , 310 gallons of grape , 270 gallons of orange on a particular day . what is the average earning of the company per 100 gallon of the juice sold ?
Options: a ) 134.8 , b ) 141.2 , c ) 147.2 , d ) 152.3 , e ) 160.1
Rationale: earning of the company from apple = 150 / 100 x 230 = 345 earning of the company from guava = 170 / 100 x 210 = 357 earning of the company from grape = 130 / 100 x 310 = 403 earning of the company from orange = 100 / 100 x 270 = 270 total earning of the company = 345 + 357 + 403 + 270 = 1375 total juice sold = 1020 average earning per 100 gallon of juice sold = ( 1375 / 1020 ) x 100 = $ 134.8 answer : a
Answer: a
Problem: if a mixture is 1 ⁄ 7 alcohol by volume and 2 ⁄ 7 water by volume , what is the ratio of the volume of alcohol to the volume of water in this mixture ?
Options: a ) 4 / 2 , b ) 3 / 2 , c ) 1 / 2 , d ) 5 / 2 , e ) 7 / 2
Rationale: "should be a sub - 600 level q . . volume = { 1 / 7 } / { 2 / 7 } = 1 / 2 c"
Answer: c
Problem: a man sitting in a train which is travelling at 50 kmph observes that a goods train , travelling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed ?
Options: a ) 52 kmph . , b ) 62 kmph . , c ) 72 kmph . , d ) 80 kmph . , e ) none
Rationale: "solution relative speed = ( 280 / 9 ) m / sec = ( 280 / 9 x 18 / 5 ) = 112 kmph . speed of the train = ( 112 - 50 ) kmph = 62 kmph . answer b"
Answer: b
Problem: the average of 10 numbers is 7 . if each number is multiplied by 12 , then the average of new set of numbers is
Options: a ) 28 , b ) 236 , c ) 288 , d ) 84 , e ) 26
Rationale: explanation : if each number is multiplied by k , then the new average increases by k times . so new average = 84 answer : d
Answer: d
Problem: a train 500 m long can cross an electric pole in 50 sec and then find the speed of the train ?
Options: a ) 76 kmph , b ) 54 kmph , c ) 72 kmph , d ) 34 kmph , e ) 36 kmph
Rationale: "length = speed * time speed = l / t s = 500 / 50 s = 10 m / sec speed = 10 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 36 kmph answer : e"
Answer: e
Problem: the length of a rectangle is increased by 40 % while its width is halved . what is the % change in area ?
Options: a ) 25 %
b ) 30 %
c ) 50 %
d ) 65 %
e ) 70 %
Rationale: the original area is l * w the new area is 1.4 l * 0.5 w = 0.7 * l * w = l * w - 0.3 * l * w the area decreased by 30 % . the answer is b .
Answer: b
Problem: if 5 x = 3 y and xy ≠ 0 , what is the ratio of 1 / 5 * x to 1 / 6 * y ?
Options: a ) 25 / 6 , b ) 36 / 25 , c ) 18 / 25 , d ) 5 / 6 , e ) 25 / 36
Rationale: "5 x = 3 y = > x / y = 3 / 5 1 / 5 * x to 1 / 6 * y = x / y * 6 / 5 = ( 3 / 5 ) * ( 6 / 5 ) = 18 / 25 ans : c"
Answer: c
Problem: a family pays $ 800 per year for an insurance plan that pays 70 percent of the first $ 1,000 in expenses and 100 percent of all medical expenses thereafter . in any given year , the total amount paid by the family will equal the amount paid by the plan when the family ' s medical expenses total how much ?
Options: a ) $ 1,000 , b ) $ 1,200 , c ) $ 1,400 , d ) $ 1,800 , e ) $ 2,200
Rationale: "assuming the medical expenses are $ 1000 or more , the family pays $ 800 + $ 300 = $ 1100 . the total amount paid by insurance plan for the first $ 1000 of expenses is $ 700 . the insurance will pay another $ 400 when the medical expenses are $ 1400 . the answer is c ."
Answer: c
Problem: for every integer n ≥ 3 , the function g ( n ) is defined as the product of all the odd integers from 1 to n , inclusive . what is the value of g ( 101 ) – g ( 99 ) ?
Options: a ) 0 , b ) 98 , c ) 99 , d ) 100 , e ) 90
Rationale: "g ( 101 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 * 101 g ( 99 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 g ( 101 ) - g ( 99 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 * 101 - 1 * 3 * 5 * 7 * 9 * . . . * 99 = 1 * 3 * 5 * 7 * 9 * . . . * 99 * ( 101 - 1 ) = 1 * 3 * 5 * 7 * 9 * . . . * 99 * 100 hence : d"
Answer: d
Problem: two trains of equal length are running on parallel lines in the same directions at 46 km / hr . and 36 km / hr . the faster trains pass the slower train in 72 seconds . the length of each train is :
Options: a ) 82 m , b ) 50 m , c ) 72 m , d ) 100 m , e ) none of these
Rationale: "explanation : the relative speed of train is 46 - 36 = 10 km / hr = ( 10 x 5 ) / 18 = 25 / 9 m / s 10 × 518 = 259 m / s in 72 secs the total distance traveled is 72 x 25 / 9 = 200 m . therefore the length of each train is = 200 / 2 = 100 m . answer d"
Answer: d
Problem: a vendor bought toffees at 6 for a rupee . how many for a rupee must he sell to gain 30 % ?
Options: a ) 1 / 2 , b ) 2 / 6 , c ) 3 / 6 , d ) 9 / 6 , e ) 15 / 2
Rationale: c . p . of 6 toffees = re . 1 s . p . of 6 toffees = 180 % of re . 1 = rs . 9 / 5 for rs . 9 / 5 , toffees sold = 6 . for re . 1 , toffees sold = 9 x 5 / 6 = 15 / 2 . answer e
Answer: e
Problem: how many seconds will a train 140 meters long take to cross a bridge 150 meters long if the speed of the train is 36 kmph ?
Options: a ) 28 sec , b ) 29 sec , c ) 24 sec , d ) 25 sec , e ) 26 sec
Rationale: "d = 140 + 150 = 290 s = 36 * 5 / 18 = 10 mps t = 290 / 10 = 29 sec b"
Answer: b
Problem: a man swims downstream 96 km and upstream 64 km taking 4 hours each time ; what is the speed of the current ?
Options: a ) 6 , b ) 4 , c ) 2 , d ) 8 , e ) 5
Rationale: "96 - - - 4 ds = 24 ? - - - - 1 64 - - - - 4 us = 16 ? - - - - 1 s = ? s = ( 24 - 16 ) / 2 = 4 answer : b"
Answer: b
Problem: sophia finished 2323 of a book . she calculated that she finished 90 more pages than she has yet to read . how long is her book ?
Options: a ) 140 , b ) 320 , c ) 745 , d ) 270 , e ) 350
Rationale: let x be the total number of pages in the book , then she finished 23 ⋅ x pages . then she has x − 23 ⋅ x = 13 ⋅ xpages left . 23 ⋅ x − 13 ⋅ x = 90 13 ⋅ x = 90 x = 270 so the book is 270 pages long . correct answer is d ) 270
Answer: d
Problem: on her annual road trip to visit her family in seal beach , california , traci stopped to rest after she traveled 1 ⁄ 3 of the total distance and again after she traveled 1 ⁄ 4 of the distance remaining between her first stop and her destination . she then drove the remaining 300 miles and arrived safely at her destination . what was the total distance , in miles , from traci ’ s starting point to seal beach ?
Options: a ) 600 , b ) 300 , c ) 350 , d ) 400 , e ) 550
Rationale: let d = total distance traci traveled 1 / 3 = d / 3 i . e . remaining distance = 2 d / 3 she traveled 1 / 4 th of 2 d / 3 = d / 6 thus : d = ( d / 3 ) + ( d / 6 ) + 300 d = 600 answer : a
Answer: a
Problem: if 5 x = 6 y and xy ≠ 0 , what is the ratio of 1 / 4 * x to 1 / 6 * y ?
Options: a ) 25 / 6 , b ) 36 / 25 , c ) 9 / 5 , d ) 5 / 6 , e ) 25 / 36
Rationale: "5 x = 6 y = > x / y = 6 / 5 1 / 4 * x to 1 / 6 * y = x / y * 3 / 2 = ( 6 / 5 ) * ( 3 / 2 ) = 9 / 5 ans : c"
Answer: c
Problem: the marks obtained by vijay and amith are in the ratio 4 : 5 and those obtained by amith and abhishek in the ratio of 3 : 4 . the marks obtained by vijay and abhishek are in the ratio of ?
Options: a ) 3 : 4 , b ) 3 : 5 , c ) 6 : 5 , d ) 3 : 6 , e ) 3 : 8
Rationale: "4 : 5 3 : 4 - - - - - - - 12 : 15 : 20 12 : 20 - - - - - - > 6 : 10 3 : 5 answer : b"
Answer: b
Problem: the length of a room is 5.5 m and width is 3.75 m . find the cost of paying the floor by slabs at the rate of rs . 1200 per sq . metre .
Options: a ) rs . 15,000 , b ) rs . 24,750 , c ) rs . 15,600 , d ) rs . 16,500 , e ) none
Rationale: "solution area of the floor = ( 5.5 x 3.75 ) m ² = 20.635 m ² cost of paying = rs . ( 1200 x 20.625 ) = rs . 24,750 . answer b"
Answer: b
Problem: a man sitting in a train which is travelling at 80 kmph observes that a goods train , travelling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed ?
Options: a ) 52 kmph . , b ) 62 kmph . , c ) 72 kmph . , d ) 80 kmph . , e ) 32 kmph .
Rationale: solution relative speed = ( 280 / 9 ) m / sec = ( 280 / 9 x 18 / 5 ) = 112 kmph . speed of the train = ( 112 - 80 ) kmph = 32 kmph . answer e
Answer: e
Problem: three pipes , a , b , & c are attached to a tank . a & b can fill it in 20 & 30 minutes respectively while c can empty it in 20 minutes . if a , b & c are kept open successively for 1 minute each , how soon will the tank be filled ?
Options: a ) 2 hours , b ) 4 hours , c ) 1 hour 30 min , d ) 5 hours , e ) 6 hours
Rationale: "in three minute 1 / 20 + 1 / 30 - 1 / 20 = 2 / 60 = 1 / 30 part is filled 3 min - - - - - - - - 1 / 30 parts x min - - - - - - - - - 1 part ( full ) x = 90 min = 1 hour 30 min answer : c"
Answer: c
Problem: how many zeroes are there at the end of the number n , if n = 70 ! + 140 ! ?
Options: a ) 8 , b ) 12 , c ) 16 , d ) 20 , e ) 24
Rationale: "the number of zeroes at the end of 70 ! will be less than the number of zeroes at the end of 140 ! hence it is sufficient to calculate the number of zeroes at the end of 70 ! the number of zeroes = [ 70 / 5 ] + [ 70 / 25 ] + [ 70 / 125 ] = 14 + 2 + 0 = 16 the answer is c ."
Answer: c
Problem: x can finish a work in 25 days . y can finish the same work in 20 days . y worked for 12 days and left the job . how many days does x alone need to finish the remaining work ?
Options: a ) 2 , b ) 5 , c ) 6 , d ) 7 , e ) 8
Rationale: "work done by x in 1 day = 1 / 25 work done by y in 1 day = 1 / 20 work done by y in 12 days = 12 / 20 = 3 / 5 remaining work = 1 – 3 / 5 = 2 / 5 number of days in which x can finish the remaining work = ( 2 / 5 ) / ( 1 / 20 ) = 8 e"
Answer: e
Problem: a , b and c invested rs . 6000 , rs . 7200 and rs . 10800 respectively , in a partnership business . find the share of c in profit of rs . 19000 after a year ?
Options: a ) 3630 , b ) 5580 , c ) 19000 , d ) 8550 , e ) 6000
Rationale: "explanation : 6000 : 7200 : 10800 5 : 6 : 9 9 / 20 * 19000 = 8550 answer : d"
Answer: d
Problem: if 15 percent of the students at a certain school went to a camping trip and took more than $ 100 , and 75 percent of the students who went to the camping trip did not take more than $ 100 , what percentage of the students at the school went to the camping trip ?
Options: a ) 95 , b ) 90 , c ) 85 , d ) 80 , e ) 60
Rationale: "let x be the number of students in the school . 0.15 x students went to the trip and took more than 100 $ . they compose ( 100 - 75 ) = 25 % of all students who went to the trip . therefore the toal of 0.15 x / 0.25 = 0.6 x students went to the camping which is 60 % . the answer is e"
Answer: e
Problem: an object thrown directly upward is at a height of h feet after t seconds , where h = - 14 ( t - 3 ) ^ 2 + 150 . at what height , in feet , is the object 2 seconds after it reaches its maximum height ?
Options: a ) 80 , b ) 88 , c ) 94 , d ) 112 , e ) 124
Rationale: "we see that h will be a maximum h = 150 when t - 3 = 0 , that is when t = 3 . at t = 5 , h = - 14 ( 5 - 3 ) ^ 2 + 150 = - 14 ( 4 ) + 150 = 94 the answer is c ."
Answer: c
Problem: a man spends 2 / 5 of his salary on house rent , 3 / 10 of his salary on food and 1 / 8 of his salary on convenience he has rs . 1400 left with him , find his expenditure on food and conveyance .
Options: a ) 1000 , b ) 3000 , c ) 4000 , d ) 2000 , e ) 5000
Rationale: "part of salary left = 1 - ( 2 / 5 + 3 / 10 + 1 / 8 ) let the monthly salary be rs . x then , 7 / 40 of x = 1400 x = ( 1400 * 40 / 7 ) = 8600 expenditure on food = rs . ( 3 / 10 * 800 ) = rs . 2400 expenditure on conveyance = rs . ( 1 / 8 * 8000 ) = rs . 1000 answer is a ."
Answer: a
Problem: how many cones of radius 4 m , height 2 m can be formed from a cylinder of 12 m radius , 10 m height :
Options: a ) 135
b ) 125
c ) 145
d ) 115
e ) 155
Rationale: ( cylinder volume ) / ( cone volume ) ( π r ( power 2 ) h ) / ( 1 / 3 π r ( power 2 ) h ) = ( π × 12 × 12 × 10 ) / ( 1 / 3 π × 4 × 4 × 2 ) = 135 answer is a .
Answer: a
Problem: the denominator of a fraction is 7 greater than the numerator . if the numerator and the denominator are increased by 4 , the resulting fraction is equal to 5 â  „ 6 . what is the value of the original fraction ?
Options: a ) 23 / 30 , b ) 25 / 32 , c ) 31 / 38 , d ) 37 / 44 , e ) 47 / 54
Rationale: "let the numerator be x . then the denominator is x + 7 . x + 4 / x + 11 = 5 / 6 . 6 x + 24 = 5 x + 55 . x = 31 . the original fraction is 31 / 38 . the answer is c ."
Answer: c
Problem: the average of 11 numbers is 500 . out of 11 numbers the average of first 6 no . is 50 , and last 6 numbers is 60 then find 6 th number ?
Options: a ) 95 , b ) 80 , c ) 110 , d ) 120 , e ) 72
Rationale: "6 th number = sum of 1 st 6 no . s + sum of last 6 no . s - sum of 11 no . s answer = 6 * 50 + 6 * 60 - 11 * 50 = 110 answer is c"
Answer: c
Problem: 70 is increased by 50 % . find the final number .
Options: a ) 100 , b ) 110 , c ) 105 , d ) 30 , e ) 160
Rationale: "final number = initial number + 50 % ( original number ) = 70 + 50 % ( 70 ) = 70 + 35 = 105 . answer c"
Answer: c
Problem: in a group of 8 semifinalists , all but 2 will advance to the final round . if in the final round only the top 3 will be awarded medals , then how many groups of medal winners are possible ?
Options: a ) 20 , b ) 56 , c ) 120 , d ) 560 , e ) 720
Rationale: 1 ) is it important here that how many go into finals . . . no . . . . what is important is how many get medals . . 3 finally these 3 can be any of the 8 - 8 c 3 = 8 ! 5 ! 3 ! = 8 ∗ 7 ∗ 63 ∗ 2 = 568 c 3 = 8 ! 5 ! 3 ! = 8 ∗ 7 ∗ 63 ∗ 2 = 56 2 ) is order important . . . no we are looking for groups only . . ans 56 answer : b
Answer: b
Problem: martha has the unique talent of being able to guess other people ’ s height and weight . for every 6 people that martha meets , she consistently guesses the people ’ s correct height 5 times , and for every 8 people that she meets , she consistently guesses the people ’ s correct weight 6 times . if martha meets 3 people and her success rate remains constant , what is the probability that martha correctly guesses a person ’ s weight and height at least once ?
Options: a ) 8 / 27 , b ) 2 / 5 , c ) 485 / 512 , d ) 98 / 125 , e ) 625 / 952
Rationale: probability that she guesses the height correctly p ( h ) = 5 / 6 probability that she guesses the weight correctlyp ( w ) = 6 / 8 probability that she guesses both weight and height correctly ( p ( h ) * p ( w ) ) = 5 / 6 * 6 / 8 = 30 / 48 now the q asks about the probability of this happening atleast once . we calculate it by finding the probability of not being able to guess in any of the three occasions . probability of not being able to guess any no . of times = 1 - 30 / 48 = 18 / 48 = 3 / 8 for all three occasions p ( a ) = 3 / 8 * 3 / 8 * 3 / 8 = 27 / 512 probability of the event happening atleast once = 1 - p ( a ) = 1 - 27 / 512 = 485 / 512 ans should be c
Answer: c
Problem: the ratio of the number of red cars in a certain parking lot to the number of black cars is 3 to 8 . if there are 90 black cars in the lot , how many red cars are there in the lot ?
Options: a ) 11 , b ) 33 , c ) 24 , d ) 27 , e ) 32
Rationale: b is correct r / b = 3 / 8 and b = 90 r = 90 * 3 / 8 = 33
Answer: b
Problem: ages of two persons differ by 20 years . if 6 year ago , the elder one be 3 times as old the younger one , find their present age
Options: a ) 12,28 , b ) 14,30 , c ) 16,36 , d ) 18,34 , e ) 19,34
Rationale: "explanation : let the age of younger person is x , then elder person age is ( x + 20 ) = > 3 ( x - 6 ) = ( x + 20 - 6 ) [ 6 years before ] = > 3 x - 18 = x + 14 = > x = 16 so other person age is x + 20 = 36 answer : option c"
Answer: c
Problem: a bag contains 12 red marbles . if someone were to remove 2 marbles from the bag , one at a time , and replace the first marble after it was removed , the probability that neither marble would be red is 25 / 36 . how many marbles are in the bag ?
Options: a ) 24 , b ) 48 , c ) 60 , d ) 72 , e ) 84
Rationale: "ok let me see if i can explain what went on in the previous post lets say i have x marbles in the bag in total - - > out of them 12 are red so the probability of pulling a non - red marble is ( x - 12 ) / x now the marble is placed back in the bag and we have x marbles again , of which again 12 are red . so the probability of pulling a non - red marble out is ( x - 12 ) / x probability theorm states that if the probability of event a occuring is m and the probability of event b occuring is n then the probability of both a and b occuring is m * n so therefore the probability of 2 non - red marbles getting pulled out is [ ( x - 12 ) / x ] * [ ( x - 12 ) / x ] this is given as 25 / 36 - - > ( x - 12 ) ^ 2 = 25 / 36 x ^ 2 square rooting u have x - 12 / x = 5 / 6 or 6 x - 72 = 5 x therefore x = 72 b"
Answer: b
Problem: a bowl was filled with 10 ounces of water , and 0.004 ounce of the water evaporated each day during a 50 - day period . what percent of the original amount of water evaporated during this period ?
Options: a ) 0.004 % , b ) 0.04 % , c ) 0.40 % , d ) 2 % , e ) 40 %
Rationale: "total amount of water evaporated each day during a 50 - day period = . 004 * 50 = . 004 * 100 / 2 = . 4 / 2 = . 2 percent of the original amount of water evaporated during this period = ( . 2 / 10 ) * 100 % = 2 % answer d"
Answer: d
Problem: of the families in city x in 1994 , 35 percent owned a personal computer . the number of families in city x owning a computer in 1998 was 20 percent greater than it was in 1994 , and the total number of families in city x was 4 percent greater in 1998 than it was in 1994 . what percent of the families in city x owned a personal computer in 1998 ?
Options: a ) 50.12 % , b ) 40.52 % , c ) 40.56 % , d ) 40.38 % , e ) 40.14 %
Rationale: "say a 100 families existed in 1994 then the number of families owning a computer in 1994 - 40 number of families owning computer in 1998 = 35 * 120 / 100 = 42 number of families in 1998 = 104 the percentage = 42 / 104 * 100 = 40.38 % . option : d"
Answer: d
Problem: the surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each . the radius of the sphere is
Options: a ) 3 cm , b ) 4 cm , c ) 6 cm , d ) 8 cm , e ) none
Rationale: "solution 4 π r 2 = 2 π 6 x 12 ⇒ r 2 = ( 6 x 12 / 2 ) ⇒ 36 ⇒ r = 6 cm . answer c"
Answer: c
Problem: the area of a parallelogram is 72 sq m and its base is 12 m . then the length of the height is ?
Options: a ) 6 m , b ) 7 m , c ) 60 m , d ) 13 m , e ) none
Rationale: b * h = 72 . h * 12 = 72 = > h = 6 m . answer : ( a )
Answer: a
Problem: a fort had provision of food for 150 men for 45 days . after 15 days , 10 men left the fort . the number of days for which the remaining food will last , is :
Options: a ) 32 , b ) 37 1 / 4 , c ) 42 , d ) 54 , e ) 45
Rationale: "we have food for 150 men for 45 days . after 15 days food left for 150 men for 30 days . so 150 : 30 now we have 140 men and x days 140 : 150 : : 30 : x x = ( 150 * 30 ) / 140 = 32 days . answer : a"
Answer: a
Problem: the sum of the ages of 5 children born at the intervals of 3 year each is 60 year . what is the age of the youngest child ?
Options: a ) 4 year , b ) 6 year , c ) 10 year , d ) none of these , e ) 9 year
Rationale: "solution let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) year . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 60 â ‡ ” 5 x = 30 â ‡ ” x = 6 . â ˆ ´ age of the youngest child = x = 6 years . answer b"
Answer: b
Problem: a set of pictures of butterflies contains 10 pictures . jim bought 3 of the pictures . if 2 pictures are to be picked out from the 10 pictures , what is the probability that both pictures are not those that were already bought by jim ?
Options: a ) 14 / 15 , b ) 3 / 5 , c ) 6 / 13 , d ) 7 / 15 , e ) 7 / 10
Rationale: probability of first pick not one of the jim ' s picture is = 7 / 10 ( except for 3 which jim purchased ) probability of second pick not one of the jim ' s picture is = 6 / 9 total = 7 / 10 ∗ 6 / 9 = 7 / 15 answer is d option d
Answer: d
Problem: maths , physics and chemistry books are stored on a library shelf that can accommodate 25 books . currently , 20 % of the shelf spots remain empty . there are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books . among all the books , 12 books are soft cover and the remaining are hard - cover . if there are a total of 7 hard - cover books among the maths and physics books . what is the probability q , that a book selected at random is either a hard cover book or a chemistry book ?
Options: a ) 1 / 10 , b ) 3 / 20 , c ) 1 / 5 , d ) 1 / 4 , e ) 9 / 20
Rationale: first phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf . to do so , you have the equations : m + p + c = 20 ( since 4 / 5 of the 25 spots are full of books ) m = 2 p p = 4 + c from that , you can use substitution to get everything down to one variable . c = p - 4 m = 2 p p = p then ( p - 4 ) + 2 p + p = 20 , so 4 p = 24 and p = 6 . that means that there are 12 math , 6 physics , and 2 chemistry books on the shelf . with those numbers , you also know that there are 8 total hardcovers , 1 of which is chemistry . so if your goal is to get either a hardcover or a chemistry , there are 9 ways towin - either one of the 7 hardcovers that are n ' t chemistry or the two chemistry books . so out of the 20 total , q = 9 provide the desired outcome , making the answer e .
Answer: e
Problem: a man started driving at a constant speed , from the site of a blast , the moment he heard the blast . he heard a second blast after a time of 30 mins and 15 seconds . if the second blast occurred exactly 30 mins after the first , how many meters was he from the site when he heard the second blast ? ( speed of sound = 330 m / s )
Options: a ) 1650 , b ) 2750 , c ) 3850 , d ) 4950 , e ) 6050
Rationale: "the distance the sound traveled to the man is 15 * 330 = 4950 meters the answer is d ."
Answer: d