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a ) 3 : 1 , b ) 4 : 3 , c ) 5 : 4 , d ) 6 : 5 , e ) 17 : 4 | a | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | p and q started a business investing rs . 54,000 and rs . 18,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 54000 : 18000 = 3 : 1 . answer : a" | a = 2 + 3
b = a + 3
c = 2 + 3
d = c * 2
e = b * d
f = 2 + 3
g = e + f
h = 2 + 3
i = h * 2
j = 3 * i
k = 2 + 3
l = j + k
m = g / l
|
a ) 25 , b ) 36 , c ) 10 , d ) 48 , e ) 45 | d | divide(add(multiply(subtract(const_1, divide(1, 3)), 10), 20), subtract(const_1, multiply(subtract(const_1, divide(1, 3)), subtract(const_1, divide(1, 3))))) | a man loses 1 / 3 rd of his money , then wins rs . 10 loses 1 / 3 rd of what he has and wins rs . 20 and finds that he has exactly what be had at the beginning . what was the amount he originally bad ? | let x be that amount so accordind to question , eqn will be like : ( ( ( x - x / 3 ) + 10 ) - ( ( ( x - x / 3 ) + 10 ) / 3 ) ) + 20 = x = > ( ( 2 x / 3 + 10 ) - ( ( 2 x / 3 + 10 ) / 3 ) ) + 20 = x = > 3 x - 4 x / 3 = 80 = > x = 48 answer : d | a = 1 / 3
b = 1 - a
c = b * 10
d = c + 20
e = 1 / 3
f = 1 - e
g = 1 / 3
h = 1 - g
i = f * h
j = 1 - i
k = d / j
|
a ) 28 / 216 , b ) 27 / 216 , c ) 29 / 216 , d ) 26 / 216 , e ) 25 / 216 | b | divide(const_2, choose(add(const_3, const_3), const_3)) | what is the probability of getting sum of 10 on rolling three dices ? | "the no of ways to roll three dies is 6 ^ 3 = 216 the possibles of getting sum as 10 is 631 = 3 ! = 6 622 = 3 ! / 2 ! = 3 541 = 3 ! = 6 523 = 3 ! = 6 442 = 3 ! / 2 ! = 3 433 = 3 ! / 2 ! = 3 total possibles for getting sum as 10 is 27 the probabilty is 27 / 216 answer : b" | a = 3 + 3
b = math.comb(a, 3)
c = 2 / b
|
a ) none , b ) one , c ) two , d ) three , e ) four | a | add(1, 1) | for any integer n greater than 1 , # n denotes the product of all the integers from 1 to n , inclusive . how many prime numbers are there between # 6 + 2 and # 6 + 6 , inclusive ? | "given that n * denotes the product of all the integers from 1 to n , inclusive so , 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 . now , notice that we can factor out 2 our of 6 ! + 2 so it can not be a prime number , we can factor out 3 our of 6 ! + 3 so it can not be a prime number , we can factor out 4 our of 6 ! + 4 so it can not be a prime number , . . . the same way for all numbers between 6 * + 2 = 6 ! + 2 and 6 * + 6 = 6 ! + 6 , inclusive . which means that there are no primes in this range . answer : a ." | a = 1 + 1
|
a ) 4.5 , b ) 7.5 , c ) 9.5 , d ) 14.5 , e ) 24.5 | e | divide(multiply(7, add(1, 6)), const_2) | a carpenter worked alone for 1 day on a job that would take him 6 more days to finish . he and another carpenter completed the job in 7 more days . how many days would it have taken the second carpenter to do the complete job working alone ? | "a carpenter worked only 1 day on something that takes him 6 more days . means ; carpenter finishes his work in 7 days . let his buddy finish the same task in x days . respective rates per day : 1 / 7 and 1 / x to complete 1 work : first guy worked for 5 days @ rate = 1 / 7 per day . second one worked for 7 days @ rate = 1 / x per day expression : days * rate = work 5 * 1 / 7 + 7 * 1 / x = 1 5 x + 49 = 7 x 2 x = 49 x = 24.5 days . ans : e" | a = 1 + 6
b = 7 * a
c = b / 2
|
a ) 12 % , b ) 26 % , c ) 29 % , d ) 30 % , e ) 60 % | c | multiply(divide(add(multiply(divide(25, const_100), 300), multiply(divide(35, const_100), 200)), add(300, 200)), const_100) | for an agricultural experiment , 300 seeds were planted in one plot and 200 were planted in a second plot . if exactly 25 percent of the seeds in the first plot germinated and exactly 35 percent of the seeds in the second plot germinated , what percent of the total number of seeds germinated ? | in the first plot 25 % of 300 seeds germinated , so 0.25 x 300 = 75 seeds germinated . in the second plot , 35 % of 200 seeds germinated , so 0.35 x 200 = 70 seeds germinated . since 75 + 70 = 145 seeds germinated out of a total of 300 + 200 = 500 seeds , the percent of seeds that germinated is ( 145 / 500 ) x 100 % , or 29 % . answer : c . | a = 25 / 100
b = a * 300
c = 35 / 100
d = c * 200
e = b + d
f = 300 + 200
g = e / f
h = g * 100
|
a ) 65 , b ) 50 , c ) 55 , d ) 60 , e ) 58 | d | divide(subtract(420, 240), 3) | on the first day of her vacation , louisa traveled 240 miles . on the second day , traveling at the same average speed , she traveled 420 miles . if the 240 - mile trip took 3 hours less than the 420 - mile trip , what was the average speed , in miles per hour ? | "( time ) * ( rate ) = ( distance ) - - > ( rate ) = ( distance ) / ( time ) - - > given : ( rate ) = 240 / t = 420 / ( t + 3 ) - - > 4 / t = 7 / ( t + 3 ) - - > 4 t + 12 = 7 t - - - - > 3 t = 12 . t = 4 - - - - > ( rate ) = 240 / 4 = 60 answer : d" | a = 420 - 240
b = a / 3
|
a ) 5 , b ) 7 , c ) 13 , d ) 14 , e ) 16 | b | add(add(3, 2), 2) | the number of positive integers n in the range 12 β€ n β€ 40 such that the product ( n β 1 ) ( n β 2 ) . . . 3 . 2.1 is not divisible by n is | explanation : ( n - 1 ) ! is always divisible by n unless and untill n is a prime number e . g . ( 5 - 1 ) ! = 4 ! which is not divisible by 5 bcuz 5 is a prime number therefore , find all prime numbers in the range of 12 to 40 i . e 13 , 17 , 19 , 23 , 29 , 31 and 37 hence , the number of positive integers n is 7 . answer : b | a = 3 + 2
b = a + 2
|
a ) 10 ^ 7 , b ) 10 ^ 9 , c ) 10 ^ 8 , d ) 10 ^ 10 , e ) 10 ^ 11 | c | multiply(multiply(79, 48), add(add(multiply(multiply(const_0_25, const_1000), const_100), multiply(add(const_3, const_4), const_10)), const_3)) | 79 laboratories raise the bacterium , the laboratory have 48 culture dishes on average , which has about 25,073 bacteria each . how many bacteria are there approximately ? | 79 laboratories raise the bacterium , the laboratory have 48 culture dishes on average , which has about 25,073 bacteria each . how many bacteria are there approximately ? a . 10 ^ 7 b . 10 ^ 8 c . 10 ^ 9 d . 10 ^ 10 e . 10 ^ 11 - > due to approximately , 79 = 80 , 48 = 50 , 25,073 = 25,000 are derived , which makes ( 79 ) ( 48 ) ( 25,075 ) = ( 80 ) ( 50 ) ( 25,000 ) = 10 ^ 8 . the answer is c . | a = 79 * 48
b = const_0_25 * 1000
c = b * 100
d = 3 + 4
e = d * 10
f = c + e
g = f + 3
h = a * g
|
a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 5 | d | subtract(15, 10) | a set consists of 15 numbers , all are even or multiple of 5 . if 8 numbers are even and 10 numbers are multiple of 5 , how many numbers is multiple of 10 ? | "{ total } = { even } + { multiple of 5 } - { both } + { nether } . since { neither } = 0 ( allare even or multiple of 5 ) then : 15 = 8 + 10 - { both } + 0 ; { both } = 3 ( so 1 number is both even and multiple of 5 , so it must be a multiple of 10 ) . answer : d ." | a = 15 - 10
|
a ) 74 , b ) 78 , c ) 79 , d ) 81 , e ) 85 | c | divide(subtract(3240, 1), gcd(2928, 3240)) | the greatest number that divides 2928 and 3240 leaving remainders 5 and 1 respectively is : | "explanation : 2928 - 5 = 2923 , 3240 - 1 = 3239 highest number that can divide 2923 and 3239 is hcf of numbers . hcf of 2923 and 3239 = 79 answer : c" | a = 3240 - 1
b = math.gcd(2928, 3240)
c = a / b
|
a ) 1236 , b ) 1356 , c ) 1470 , d ) 1556 , e ) 1635 | e | multiply(divide(subtract(2, 6), subtract(1365, const_1)), 1365) | find large number from below question the difference of 2 no ' s is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 15 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 15 5 x = 1350 x = 270 large number = 270 + 1365 = 1635 e" | a = 2 - 6
b = 1365 - 1
c = a / b
d = c * 1365
|
a ) 2000 , b ) 2100 , c ) 2200 , d ) 2300 , e ) 2400 | e | multiply(divide(72, 3), const_100) | a sum was put at simple interest at certain rate for 3 years . had it been put at 1 % higher rate it would have fetched rs . 72 more . the sum is : a . rs . 2,400 b . rs . 2,100 c . rs . 2,200 d . rs . 2,480 | 1 percent for 3 years = 72 1 percent for 1 year = 24 = > 100 percent = 2400 answer : e | a = 72 / 3
b = a * 100
|
a ) 10 , b ) 99 , c ) 77 , d ) 55 , e ) 6 | e | add(inverse(subtract(divide(const_1, 3), divide(const_1, 6))), divide(const_2, add(const_2, const_3))) | a and b together can do a work in 3 days . if a alone can do it in 6 days . in how many days can b alone do it ? | "1 / 3 β 1 / 6 = 1 / 6 = > 6 answer : e" | a = 1 / 3
b = 1 / 6
c = a - b
d = 1/(c)
e = 2 + 3
f = 2 / e
g = d + f
|
a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | e | subtract(26, const_2) | a man is 26 years older than his studen . in two years , his age will be twice the age of his student . the present age of his student is : | let the student ' s present age be x years . then , man ' s present age = ( x + 26 ) years . ( x + 26 ) + 2 = 2 ( x + 2 ) x + 28 = 2 x + 4 x = 24 . answer : e | a = 26 - 2
|
a ) 0.15 , b ) 0.20 , c ) 0.25 , d ) 0.30 , e ) 0.33 | b | divide(4, multiply(4, 5)) | a = { 2 , 3 , 4 , 5 } b = { 4 , 5 , 6 , 7 , 8 } two integers will be randomly selected from the sets above , one integer from set a and one integer from set b . what is the probability t that the sum of the two integers will equal 9 ? | rearrange the first set : a = { 54 , 32 } b = { 45 , 67 , 8 } as you can see numbers in each column ( the numbers of the same color ) give the sum of 9 . so there are 4 such pares possible , total # of pairs is 4 * 5 = 20 . t = favorable / total = 4 / 20 = 0.2 . answer : b . or : we can select any number from set a ( 4 / 4 = 1 ) but in this case we must select its matching pair from set b ( the number of the same color ) and since there are only one matching pair of this particular number in b then the probability of this is 1 / 5 . so , overall : t = 1 * 1 / 5 . answer : b . | a = 4 * 5
b = 4 / a
|
a ) 10 , b ) 12 , c ) 14 , d ) 36 , e ) 18 | d | multiply(log(divide(multiply(multiply(add(const_4, const_1), 1,000), const_100), 1,000)), 4) | the population of a bacteria culture doubles every 4 minutes . approximately how many minutes will it take for the population to grow from 1,000 to 500,000 bacteria | "this one ' s easy . 1000 * 2 ^ t = 500,000 2 ^ t = 500 now gauging , since 2 ^ 8 = 256 , then 2 ^ 9 = 512 so t = 9 but be careful , ' t ' is in time intervals of 4 minutes so answer is 9 * 4 = 36 minutes answer ( d )" | a = 4 + 1
b = a * 1
c = b * 100
d = c / 1
e = math.log(d)
f = e * 4
|
['a ) 10', 'b ) 11', 'c ) 12', 'd ) 13', 'e ) 14'] | b | subtract(multiply(4, const_3), const_1) | how many pieces can a square pizza be cut into by making 4 linear cuts ? | 1 cut makes two pieces a second cut can make 4 pieces by cutting through 2 of the pieces a third cut can make 7 pieces by cutting through 3 of the pieces a fourth cut can make 11 pieces by cutting through 4 of the pieces b | a = 4 * 3
b = a - 1
|
a ) 28 , b ) 266 , c ) 990 , d ) 25 , e ) 11 | d | divide(add(220, 280), multiply(add(42, 30), const_0_2778)) | two trains of length 220 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 220 + 280 = 500 m . the time required = d / s = 500 / 20 = 25 sec . answer : d" | a = 220 + 280
b = 42 + 30
c = b * const_0_2778
d = a / c
|
a ) 32 , b ) 36 , c ) 38 , d ) 42 , e ) 48 | d | divide(subtract(multiply(46.2, const_100), multiply(60, 70)), subtract(70, 60)) | we bought orange juice and apple juice at the store . a bottle of orange juice costs 70 cents and a bottle of apple juice costs 60 cents . we bought a total of 70 bottles for $ 46.20 . how many bottles of orange juice did we buy ? | let oj be the bottles of orange juice and let aj be the bottles of apple juice . ( oj ) + aj = 70 . aj = 70 - ( oj ) . 70 ( oj ) + 60 aj = 4620 . 70 ( oj ) + 60 ( 70 - oj ) = 4620 . 10 ( oj ) + 4200 = 4620 . oj = 42 . the answer is d . | a = 46 * 2
b = 60 * 70
c = a - b
d = 70 - 60
e = c / d
|
a ) 3 / 5 , b ) 4 / 7 , c ) 5 / 7 , d ) 18 / 35 , e ) 23 / 35 | d | divide(add(multiply(const_2, const_2), multiply(const_3, const_2)), multiply(7, subtract(7, 1))) | if x is to be chosen at random from the integers between 1 to 7 , inclusive , and y is to be chosen at random from the integers between 7 and 11 , inclusive , what is the probability that x + y will be even ? | "x + y will be even if x and y are both even or both odd . p ( x and y are both even ) = 3 / 7 * 2 / 5 = 6 / 35 p ( x and y are both odd ) = 4 / 7 * 3 / 5 = 12 / 35 p ( x + y is even ) = 6 / 35 + 12 / 35 = 18 / 35 the answer is d ." | a = 2 * 2
b = 3 * 2
c = a + b
d = 7 - 1
e = 7 * d
f = c / e
|
a ) 60,000 , b ) 6,000 , c ) 2,500 , d ) 250 , e ) 0 | b | multiply(divide(12000, 12), 12) | a glucose molecule contains 6 carbon , 12 hydrogen and 6 oxygen atoms . during photosynthesis , green plants create glucose daily . if , over a period of 12 days , an eucalyptus plant ( a green plant ) creates 12000 molecules of glucose then approximately how many more atoms of hydrogen than carbon are created on average per day ? | solution : we know that 12,000 glucose molecules are created over a period of 12 days . therefore 12,000 / 12 = 1,000 glucose molecules are created on average per day . each glucose molecule contains 6 carbon atoms and 12 hydrogen atoms . so 1,000 glucose molecules contain 6 Γ 1,000 = 6,000 carbon atoms and 12 Γ 1,000 = 12,000 hydrogen atoms . the difference is 12,000 β 6,000 = 6,000 . the correct answer is b . | a = 12000 / 12
b = a * 12
|
a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | d | subtract(multiply(3, 7), multiply(add(2, 3), 2)) | a basketball player practices 2 hours a day monday through friday . how many hours must he practice saturday and sunday combined if he wants to practice an average ( arithmetic mean ) of 3 hours a day for the 7 day week ? | total number of hours practiced mon through fri : 5 * 2 = 10 total number of hours that will be practiced in a 7 day week considering 3 hours a day average : 7 * 3 = 21 thus 21 - 10 = 11 hours still need to be practiced ( on sat and sun combined ) d | a = 3 * 7
b = 2 + 3
c = b * 2
d = a - c
|
a ) 9 , b ) 20 , c ) 55 , d ) 70 , e ) 81 | e | divide(add(multiply(divide(subtract(multiply(83.2, const_10), const_2), const_10), divide(11, divide(const_2, const_10))), 11), divide(11, divide(const_2, const_10))) | when the positive integer k is divided by the positive integer n , the remainder is 11 . if k / n = 83.2 , what is the value of n ? | here ' s an approach that ' s based on number properties and a bit ofbrute forcemath : we ' re told that k and n are both integers . since k / n = 81.2 , we can say that k = 81.2 ( n ) n has tomultiply outthe . 2 so that k becomes an integer . with the answers that we have to work with , n has to be a multiple of 5 . eliminate a and e . with the remaining answers , we can test the answers and find the one that fits the rest of the info ( k / n = 81.2 and k / n has a remainder of 11 ) answer b : if n = 20 , then k = 1624 ; 1624 / 20 has a remainder of 4 not a match answer c : if n = 55 , then k = 4466 ; 4466 / 55 has a remainder of 11 match . final answer : e | a = 83 * 2
b = a - 2
c = b / 10
d = 2 / 10
e = 11 / d
f = c * e
g = f + 11
h = 2 / 10
i = 11 / h
j = g / i
|
a ) rs . 550 , b ) rs . 650 , c ) rs . 600 , d ) rs . 500 , e ) rs . 700 | d | divide(90, divide(multiply(4, add(4, divide(1, 2))), const_100)) | what sum of money will produce rs . 90 as simple interest in 4 years at 4 1 / 2 percent ? | "explanation : 90 = ( p * 4 * 9 / 2 ) / 100 90 = 36 p / 100 p = 500 answer : option d" | a = 1 / 2
b = 4 + a
c = 4 * b
d = c / 100
e = 90 / d
|
a ) 70 , b ) 20 , c ) 30 , d ) 75 , e ) 15 | a | multiply(divide(subtract(72, 36), const_3_6), 7) | two trains are moving in the same direction at 72 kmph and 36 kmph . the faster train crosses a man in the slower train in 7 seconds . find the length of the faster train ? | "relative speed = ( 72 - 36 ) * 5 / 18 = 2 * 5 = 10 mps . distance covered in 27 sec = 7 * 10 = 70 m . the length of the faster train = 70 m . answer : a" | a = 72 - 36
b = a / const_3_6
c = b * 7
|
a ) β 220 , b ) β 200 , c ) 100 , d ) 135 , e ) it can not be determined from the information given | b | subtract(multiply(180, const_2), multiply(80, const_2)) | if the average ( arithmetic mean ) of a and b is 80 , and the average of b and c is 180 , what is the value of a β c ? | "a + b = 160 b + c = 360 a - c = - 200 . imo option b ." | a = 180 * 2
b = 80 * 2
c = a - b
|
a ) 24 , b ) 35 , c ) 39 , d ) 60 , e ) 65 | a | divide(multiply(multiply(7, 8), 3), 7) | exactly 3 / 7 of the people in the room are under the age of 21 , and exactly 5 / 8 of the people in the room are over the age of 65 . if the total number of the people in the room is greater than 50 and less than 100 , how many people in the room are under the age of 21 ? | "the total number of the people in the room must be a multiple of both 7 and 8 ( in order 3 / 7 and 5 / 8 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 7 and 8 , which is 56 . since , the total number of the people in the room is greater than 50 and less than 100 , then there are 56 people in the room . therefore there are 3 / 7 * 56 = 24 people in the room under the age of 21 . answer : a ." | a = 7 * 8
b = a * 3
c = b / 7
|
a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | a | divide(42, add(const_1, add(divide(50, const_100), multiply(divide(50, const_100), add(const_1, divide(20, const_100)))))) | alice picked a number of pears , each of which she sold , canned , or poached . she canned 20 percent more than she poached , and she poached 50 percent less than she sold . if there were a total of 42 pears , then how many pears did she sell ? | let x be the amount of pears that she sold . then she poached 0.5 x . then she canned 1.2 * 0.5 x = 0.6 x . the total amount of pears is x + 0.5 x + 0.6 x = 2.1 x the fraction of pears that she sold is 1 / 2.1 = 10 / 21 . the number of pears she sold is ( 10 / 21 ) * 42 = 20 . the answer is a . | a = 50 / 100
b = 50 / 100
c = 20 / 100
d = 1 + c
e = b * d
f = a + e
g = 1 + f
h = 42 / g
|
a ) 100000000010 , b ) 10000000010 , c ) 10000000000 , d ) 1200000000 , e ) 12000000010 | a | add(multiply(10, power(10, 10)), 10) | the value of x + x ( x ^ x ) when x = 10 is | solution : x + x ( x ^ x ) put the value of x = 10 in the above expression we get , 10 + 10 ( 10 ^ 10 ) = 10 + 10 ( 10 Γ£ β 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 ) = 10 + 10 ( 10000000000 ) = 10 + 100000000000 = 100000000010 answer : ( a ) | a = 10 ** 10
b = 10 * a
c = b + 10
|
a ) 0 , b ) 1 , c ) 45 , d ) 130 , e ) 520 | b | divide(3, 3) | if k is a non - negative integer and 30 ^ k is a divisor of 929260 then 3 ^ k - k ^ 3 = | 9 + 2 + 9 + 2 + 6 + 0 = 28 , so this number is not divisible by 3 and thus not divisible by 30 . therefore , k = 0 3 ^ k - k ^ 3 = 1 - 0 = 1 the answer is b . | a = 3 / 3
|
a ) 500 , b ) 600 , c ) 700 , d ) 800 , e ) none of these | c | divide(56, multiply(divide(2, const_100), 4)) | a sum was put at simple interest at a certain rate for 4 years had it been put at 2 % higher rate , it would have fetched 56 more . find the sum . | "difference in s . i . = p Γ t / 100 ( r 1 β r 2 ) β 56 = p Γ 4 Γ 2 / 100 ( β΅ r 1 - r 2 = 2 ) β p = 56 Γ 100 / 4 Γ 2 = 700 answer c" | a = 2 / 100
b = a * 4
c = 56 / b
|
a ) $ 8.30 , b ) $ 20.75 , c ) $ 22.50 , d ) $ 42.75 , e ) $ 45.00 | a | multiply(0.1, add(add(add(multiply(divide(90, add(4, divide(const_1, const_2))), const_2), divide(90, add(4, divide(const_1, const_2)))), divide(90, add(4, divide(const_1, const_2)))), 3)) | a vending machine randomly dispenses 4 different types of fruit candy . there are twice as many apple candies as orange candies , twice as many strawberry candies as grape candies , and twice as many apple candies as strawberry candies . if each candy cost $ 0.1 , and there are exactly 90 candies , what is the minimum amount of money required to guarantee that you would buy at least 3 of each type of candy ? | let number of apple , orange , strawberry and grape candies be a , o , s and g respectively . a = 2 o s = 2 g a = 2 s a = 4 g a + o + s + g = 90 = > a + a / 2 + a / 2 + a / 4 = 90 = > 9 a / 4 = 90 = > a = 40 o = 20 s = 20 g = 10 cost of each candy = . 1 $ mininum amount of money required to guarantee that you would buy at least three of each type of candy we can buy 40 apple candies , 20 orange candies , 20 strawberry candies and 3 grape candies to ensure atleast 3 of each type of candies . total = 83 candies amount required = 83 * . 1 = 8.30 $ answer a | a = 1 / 2
b = 4 + a
c = 90 / b
d = c * 2
e = 1 / 2
f = 4 + e
g = 90 / f
h = d + g
i = 1 / 2
j = 4 + i
k = 90 / j
l = h + k
m = l + 3
n = 0 * 1
|
a ) 0.1 , b ) 0.2 , c ) 0.3 , d ) 0.4 , e ) 0.5 | e | add(divide(4, 6), divide(4, 8)) | a can do a work in 6 days , b can do a work in 8 days and c can do it in 12 days . b left work after 4 days . for how many number of days should a and c should work together to complete the remaining work ? | "b work 1 / 8 * 4 = 1 / 2 remaining work = 1 - 1 / 2 = 1 / 2 a and c work together = 1 / 6 + 1 / 12 = 3 / 12 = 1 / 4 take reciprocal 4 * remaining work = 2 * 1 / 4 = 0.5 answer : e" | a = 4 / 6
b = 4 / 8
c = a + b
|
a ) 2 % decrease , b ) 8 % decrease , c ) 9 % decrease , d ) 0 % no change , e ) 2 % decrease | d | subtract(const_100, multiply(multiply(add(const_1, divide(25, const_100)), subtract(const_1, divide(20, const_100))), const_100)) | the tax on a commodity is diminished by 20 % and its consumption increased by 25 % . the effect on revenue is ? | "100 * 100 = 10000 80 * 125 = 10000 - - - - - - - - - - - 10000 - - - - - - - - - - - 0 100 - - - - - - - - - - - ? = > 0 % no change answer : d" | a = 25 / 100
b = 1 + a
c = 20 / 100
d = 1 - c
e = b * d
f = e * 100
g = 100 - f
|
a ) 0 , b ) 2 , c ) 1 , d ) 3 , e ) 5 | c | subtract(power(add(4, 2), 2), multiply(12, const_4)) | if n is a prime number greater than 4 , what is the remainder when n ^ 2 is divided by 12 ? | "there are several algebraic ways to solve this question including the one under the spoiler . but the easiest way is as follows : since we can not have two correct answersjust pick a prime greater than 4 , square it and see what would be the remainder upon division of it by 12 . n = 7 - - > n ^ 2 = 49 - - > remainder upon division 49 by 12 is 1 . answer : c ." | a = 4 + 2
b = a ** 2
c = 12 * 4
d = b - c
|
a ) s . 380 , b ) s . 600 , c ) s . 460 , d ) s . 400 , e ) s . 480 | c | multiply(multiply(3, subtract(divide(const_1, 3), add(divide(const_1, 6), divide(const_1, 8)))), 3680) | a can do a particular work in 6 days . b can do the same work in 8 days . a and b signed to do it for rs . 3680 . they completed the work in 3 days with the help of c . how much is to be paid to c ? | "explanation : amount of work a can do in 1 day = 1 / 6 amount of work b can do in 1 day = 1 / 8 amount of work a + b can do in 1 day = 1 / 6 + 1 / 8 = 7 / 24 amount of work a + b + c can do = 1 / 3 amount of work c can do in 1 day = 1 / 3 - 7 / 24 = 1 / 24 work a can do in 1 day : work b can do in 1 day : work c can do in 1 day = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 amount to be paid to c = 3680 Γ ( 1 / 8 ) = 460 answer : option c" | a = 1 / 3
b = 1 / 6
c = 1 / 8
d = b + c
e = a - d
f = 3 * e
g = f * 3680
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a ) 8.5 seconds , b ) 10 seconds , c ) 3.5 seconds , d ) 2.5 seconds , e ) 2.6 seconds | b | divide(100, multiply(36, const_0_2778)) | in what time will a train 100 meters long cross an electric pole , if its speed is 36 km / hr | "first convert speed into m / sec speed = 36 * ( 5 / 18 ) = 10 m / sec time = distance / speed = 100 / 10 = 10 seconds answer : b" | a = 36 * const_0_2778
b = 100 / a
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a ) 22 , b ) 33 , c ) 44 , d ) 66 , e ) 72 | e | multiply(multiply(const_3, 12), const_2) | if x / ( 12 p ) is an odd prime number , where x is a positive integer and p is a prime number , what is the least value of x ? | "x / ( 12 p ) = odd prime number x = odd prime number * 11 p least value of x = lowest odd prime number * 11 * lowest value of p = 3 * 12 * 2 = 72 answer e" | a = 3 * 12
b = a * 2
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a ) 1 , b ) 4 , c ) 6 , d ) 7 , e ) 13 | a | divide(subtract(add(multiply(2, 5), 16), add(multiply(3, 6), 5)), subtract(multiply(2, 3), multiply(3, const_1))) | given f ( x ) = 3 x β 5 , for what value of x does 2 * [ f ( x ) ] β 16 = f ( x β 6 ) ? | "2 ( 3 x - 5 ) - 16 = 3 ( x - 6 ) - 5 3 x = 3 x = 1 the answer is a ." | a = 2 * 5
b = a + 16
c = 3 * 6
d = c + 5
e = b - d
f = 2 * 3
g = 3 * 1
h = f - g
i = e / h
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a ) 42 , b ) 43 , c ) 44 , d ) 45 , e ) 47 | e | add(subtract(80, multiply(12, 3)), 3) | a batsman in his 12 th innings makes a score of 80 and thereby increases his average by 3 runs . what is his average after the 12 th innings if he had never been β not out β ? | let β x β be the average score after 12 th innings β 12 x = 11 Γ ( x β 3 ) + 80 β΄ x = 47 answer e | a = 12 * 3
b = 80 - a
c = b + 3
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a ) 9 / 20 , b ) 11 / 20 , c ) 11 / 32 , d ) 13 / 40 , e ) 17 / 40 | d | add(divide(1, 5), divide(1, 8)) | each of the three people individually can complete a certain job in 4 , 5 , and 8 hours , respectively . what is the lowest fraction of the job that can be done in 1 hour by 2 of the people working together at their respective rates ? | "the two slowest people work at rates of 1 / 5 and 1 / 8 of the job per hour . the sum of these rates is 1 / 5 + 1 / 8 = 13 / 40 of the job per hour . the answer is d ." | a = 1 / 5
b = 1 / 8
c = a + b
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a ) 3 : 00 , b ) 4 : 00 , c ) 5 : 00 , d ) 6 : 00 , e ) 7 : 00 | d | subtract(multiply(add(7, const_1), const_3), divide(840, 60)) | a train travels from new york to chicago , a distance of approximately 840 miles , at an average rate of 60 miles per hour and arrives in chicago at 7 : 00 in evening , chicago time . at what hour in the morning , new york time , did the train depart for chicago ? ( note : chicago time is one hour earlier than new york time ) | "7 : 00 in evening in chicago = 8 : 00 in evening in new york . so , the train was in chicago 8 : 00 in the evening , new york time . the trip took t = d / r = 840 / 60 = 14 hours . therefore , the train depart from new york at 8 : 00 - 14 hours = 6 : 00 in the morning , new york time . answer : d ." | a = 7 + 1
b = a * 3
c = 840 / 60
d = b - c
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a ) 4 / 15 , b ) 1 / 3 , c ) 2 / 5 , d ) 4 / 5 , e ) 7 / 6 | a | divide(multiply(4, 1), multiply(3, 5)) | if the ratio of a to b is 4 to 3 and the ratio of b to c is 1 to 5 , what is the ratio of a to c ? | a : b = 4 : 3 - - 1 b : c = 1 : 5 = > b : c = 3 : 15 - - 2 from 1 and 2 , we get a : c = 4 : 15 answer a | a = 4 * 1
b = 3 * 5
c = a / b
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a ) 3421 , b ) 2132 , c ) 1111 , d ) 4323 , e ) 1071 | e | multiply(63, 17) | the h . c . f . of two numbers is 63 and the other two factors of their l . c . m . are 11 and 17 . the larger of the two numbers is : | ": explanation : clearly , the numbers are ( 63 x 11 ) and ( 63 x 17 ) . { \ color { blue } \ therefore } larger number = ( 63 x 17 ) = 1071 . answer : e ) 1071" | a = 63 * 17
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a ) 6 / 7 , b ) 5 / 7 , c ) 4 / 7 , d ) 3 / 7 , e ) 2 / 7 | c | divide(const_2, add(const_3, const_4)) | find the probability that a leap year selected at random will not have 53 sundays or mondays . | "there are 366 days in a leap year : 52 weeks and 2 more days . so , 52 sundays , 52 mondays and 2 days . these 2 days can be : { mon , tue } , { tue , wed } , { wed , thu } , { thu , fri } , { fri , sat } , { sat , sun } and { sun , mon } ( 7 cases ) . in order to have 53 sundays or 53 mondays we should have either { mon , tue } { sat , sun } or { sun , mon } case . the probability of that is 3 / 7 . so probability of not having 53 sundays or mondays , probability is 1 - 3 / 7 = 4 / 7 answer : c ." | a = 3 + 4
b = 2 / a
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a ) 90 , b ) 2,400 , c ) 4,500 , d ) 5,400 , e ) 324,000 | b | divide(multiply(add(25, 55), const_60), const_2) | a copy machine , working at a constant rate , makes 25 copies per minute . a second copy machine , working at a constant rate , makes 55 copies per minute . working together at their respective rates , how many copies do the two machines make in half an hour ? | "together the two machines make 25 + 55 = 80 copies per minute . so , in half an hour they will make 80 * 30 = 2,400 copies . answer : b ." | a = 25 + 55
b = a * const_60
c = b / 2
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a ) 64 % , b ) 65 % , c ) 66 % , d ) 67 % , e ) none of these | a | multiply(divide(subtract(divide(5, 3), divide(3, 5)), divide(5, 3)), const_100) | a student multiplied a number by 3 / 5 instead of 5 / 3 . what is the percentage error . | "explanation : let the number be x , then , 5 / 3 β 3 / 5 = 16 / 15 x error % = ( 16 / 15 x β 3 / 5 β 100 ) option a" | a = 5 / 3
b = 3 / 5
c = a - b
d = 5 / 3
e = c / d
f = e * 100
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a ) 1 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | e | divide(12, divide(log(9), log(3))) | solve the given equation 9 ^ y = 3 ^ 12 , what is y ? | 3 ^ 2 y = 3 ^ 12 2 y = 12 , therefore y = 6 e | a = math.log(9)
b = math.log(3)
c = a / b
d = 12 / c
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a ) 2277 , b ) 2977 , c ) 2677 , d ) 4530 , e ) 1971 | d | multiply(subtract(subtract(multiply(multiply(const_2, 12), add(15, 25)), multiply(6, 3)), multiply(3, multiply(4, 3))), 5) | the dimensions of a room are 25 feet * 15 feet * 12 feet . what is the cost of white washing the four walls of the room at rs . 5 per square feet if there is one door of dimensions 6 feet * 3 feet and three windows of dimensions 4 feet * 3 feet each ? | "area of the four walls = 2 h ( l + b ) since there are doors and windows , area of the walls = 2 * 12 ( 15 + 25 ) - ( 6 * 3 ) - 3 ( 4 * 3 ) = 906 sq . ft . total cost = 906 * 5 = rs . 4530 answer : d" | a = 2 * 12
b = 15 + 25
c = a * b
d = 6 * 3
e = c - d
f = 4 * 3
g = 3 * f
h = e - g
i = h * 5
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a ) 70000 , b ) 60000 , c ) 80000 , d ) 90000 , e ) 50000 | d | add(add(multiply(divide(30000, 50000), 36000), multiply(divide(45000, 50000), 36000)), 36000) | a , b and c started a partnership business by investing rs . 30000 , rs . 45000 , rs . 50000 respectively . at the end of the year , the profit were distributed among them . if c ' s share of profit is 36000 , what is the total profit ? | "a : b : c = 30000 : 45000 : 50000 = 6 : 9 : 10 let total profit = p then p Γ£ β 10 / 25 = 36000 p = ( 36000 Γ£ β 25 ) / 10 = 90000 answer is d ." | a = 30000 / 50000
b = a * 36000
c = 45000 / 50000
d = c * 36000
e = b + d
f = e + 36000
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a ) 4 , b ) 7 , c ) 8 , d ) 5 , e ) 3 | d | divide(subtract(add(multiply(7, 7), 5), 4), 10) | if the number is decreased by 5 and divided by 7 the result is 7 . what would be the result if 4 is subtracted and divided by 10 ? | "explanation : let the number be x . then , ( x - 5 ) / 7 = 7 = > x - 5 = 49 x = 54 . : ( x - 4 ) / 10 = ( 54 - 4 ) / 10 = 5 answer : option d" | a = 7 * 7
b = a + 5
c = b - 4
d = c / 10
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a ) 4 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | add(4, const_1) | integer m has 4 different prime factors and n has 3 different prime factors . if m and n has the greatest common factor of 15 , how many different prime factors does mn have ? | as m and n has gcf of 15 , 3 * 5 should be common factor for both m and n . m = 3 * 5 * p 1 * p 2 n = 3 * 5 * p 3 mn = 3 ^ 2 * 5 ^ 2 * p 1 * p 2 * p 3 mn has prime numbers of 3,5 , p 1 , p 2 p 3 b | a = 4 + 1
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a ) 23 , b ) 11 , c ) 12 , d ) 14 , e ) 15 | a | subtract(power(2, 2), 2) | if x ^ 2 + ( 1 / x ^ 2 ) = 5 , x ^ 4 + ( 1 / x ^ 4 ) = ? | "- > x ^ 4 + ( 1 / x ^ 4 ) = ( x ^ 2 ) ^ 2 + ( 1 / x ^ 2 ) ^ 2 = ( x ^ 2 + 1 / x ^ 2 ) ^ 2 - 2 x ^ 2 ( 1 / x ^ 2 ) = 5 ^ 2 - 2 = 23 . thus , the answer is a ." | a = 2 ** 2
b = a - 2
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a ) 2 % , b ) 3 % , c ) 4 % , d ) 5 % , e ) 6 % | b | divide(multiply(const_100, subtract(609, 600)), divide(600, 2)) | on a sum of money , the s . i . for 2 years is $ 600 , while the c . i . is $ 609 , the rate of interest being the same in both the cases . the rate of interest is ? | "difference in c . i . and s . i for 2 years = $ 609 - $ 600 = $ 9 s . i for one year = $ 300 s . i . on $ 300 for 1 year = $ 9 rate = ( 100 * 9 ) / ( 300 ) = 3 % the answer is b ." | a = 609 - 600
b = 100 * a
c = 600 / 2
d = b / c
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a ) 86.5 kg , b ) 63.5 kg , c ) 46.5 kg , d ) 26.5 kg , e ) 16.5 kg | b | divide(add(65, add(62, const_1)), const_2) | in arun ' s opinion , his weight is greater than 62 kg but leas than 72 kg . his brother does not agree with arun and he thinks that arun ' s weight is greater than 60 kg but less than 70 kg . his mother ' s view is that his weight can not be greater than 65 kg . if all of them are correct in their estimation , what is the average of diferent probable weights of arun ? | "let arun ' s weight be x kg . according to arun , 62 < x < 72 . according to arun ' s brother , 60 < x < 70 . according to arun ' s mother , x < 65 . the values satisfying all the above conditions are 63 and 64 . required average = ( 63 + 64 ) / 2 = 63.5 kg answer : b" | a = 62 + 1
b = 65 + a
c = b / 2
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a ) 45 , b ) 43 , c ) 40 , d ) 38 , e ) 29 | a | divide(add(add(subtract(60, const_10), const_100), add(subtract(60, const_10), const_100)), add(divide(add(subtract(60, const_10), const_100), 60), divide(add(subtract(60, const_10), const_100), 36))) | x and y are two towns . ganesh covers the distance from x to y at an average speed of 60 km / hr . however , he covers the distance from y to x at an average speed of 36 km / hr . his average speed during the whole journey in km / hr . is : | solution : average speed = 2 xy / x + y = 2 * 60 * 36 / 60 + 36 = 45 answer : a | a = 60 - 10
b = a + 100
c = 60 - 10
d = c + 100
e = b + d
f = 60 - 10
g = f + 100
h = g / 60
i = 60 - 10
j = i + 100
k = j / 36
l = h + k
m = e / l
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a ) 5184 , b ) 5060 , c ) 5148 , d ) 5084 , e ) none of them | e | divide(multiply(power(12, 3), power(6, 4)), 432) | ( 12 ) ^ 3 x 6 ^ 4 Γ· 432 = ? | = ( 12 ) ^ 3 x 6 ^ 4 / 432 = ( 12 ) ^ 3 x 6 ^ 4 / 12 x 6 ^ 2 = ( 12 ) ^ 2 x 6 ^ 2 = ( 72 ) ^ 2 = 70 answer is e | a = 12 ** 3
b = 6 ** 4
c = a * b
d = c / 432
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['a ) 360 cc', 'b ) 60 cc', 'c ) 300 cc', 'd ) 36 cc', 'e ) 39 cc'] | a | multiply(40, multiply(3, 3)) | two cylinders are of the same height . their radii are in the ratio 1 : 3 . if the volume of the first cylinder is 40 cc . find the volume of the second cylinder ? | r 1 = x r 2 = 3 x Ο * x 2 * h = 40 Ο 9 x 2 h = 40 * 9 = 360 answer : a | a = 3 * 3
b = 40 * a
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a ) rs . 3500 , b ) rs . 1520 , c ) rs . 4996 , d ) rs . 4000 , e ) rs . 900 | b | subtract(subtract(subtract(multiply(2890, const_4), 2500), 4000), 3540) | a grocer has a sale of rs . 2500 , rs . 4000 , rs . 3540 for 3 consecutive months . how much sale must he have in the fourth month so that he gets an average sale of rs . 2890 ? | total sale for 5 months = rs . ( 2500 + 4000 + 3540 ) = rs . 10040 . required sale = rs . [ ( 2890 x 4 ) - 10040 ] = rs . ( 11560 - 10040 ) = rs . 1520 . b ) | a = 2890 * 4
b = a - 2500
c = b - 4000
d = c - 3540
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a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | e | multiply(782,969, const_1) | if 3 ^ x * 4 ^ y = 4 , 782,969 and x β y = 14 , then x = ? | "since 4 , 782,969 is an odd number , it is not a multiple of 4 . this means that y must equal zero . since x - y = 14 and y = 0 , then x = 14 . the answer is e ." | a = 782 * 969
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a ) 100 , b ) 120 , c ) 150 , d ) 200 , e ) 160 | b | subtract(50, subtract(120, 100)) | if x , y , and z are positive real numbers such that x + y - z = 100 , y - z + x = 120 , and z - x + y = 50 , then find the x + y + z value ? | "x + y - z = 100 , - - - - - - 1 ) y - z + x = 120 , - - - - - - 2 ) z - x + y = 50 - - - - - - 3 ) from ( 1 ) & ( 2 ) 2 z = 20 z = 10 from ( 2 ) & ( 3 ) 2 y = 170 y = 85 then x = 100 - 75 = 25 x + y + z = 120 correct option is b" | a = 120 - 100
b = 50 - a
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a ) 25 % loss , b ) 25 % profit , c ) 20 % loss , d ) 20 % profit , e ) 50 % loss | e | multiply(const_100, divide(subtract(const_100, divide(multiply(const_100, 20), 10)), divide(multiply(const_100, 20), 10))) | if the cost price of 10 articles is equal to the selling price of 20 articles , what is the % profit or loss made by the merchant ? | "let the cost price of 1 article be $ 1 . therefore , cost price of 10 articles = 10 * 1 = $ 10 the selling price of 20 articles = cost price of 10 articles = $ 10 . now , we know the selling price of 20 articles . let us find the cost price of 20 articles . cost price of 20 articles = 20 * 1 = $ 20 . therefore , profit made on sale of 20 articles = selling price of 20 articles - cost price of 20 articles = 10 - 20 = - $ 10 . as the profit is in the negative , the merchant has made a loss of $ 10 . therefore , % loss = loss / cp * 100 % loss = - 10 / 20 * 100 = 50 % loss . e" | a = 100 * 20
b = a / 10
c = 100 - b
d = 100 * 20
e = d / 10
f = c / e
g = 100 * f
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a ) 0.66 , b ) 0.68 , c ) 0.96 , d ) 0.69 , e ) 0.76 | a | divide(subtract(power(0.76, 3), power(0.1, 3)), add(add(power(0.76, 2), 0.076), power(0.1, 2))) | ( 0.76 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.76 ) ( power 2 ) + 0.076 + ( 0.1 ) ( power 2 ) is : | "given expression = ( 0.76 ) ( power 3 ) - ( 0.1 ) ( power 3 ) / ( 0.76 ) ( power 2 ) + ( 0.76 x 0.1 ) + ( 0.1 ) ( power 2 ) = a ( power 3 ) - b ( power 3 ) / a ( power 2 ) + ab + b ( power 2 ) = ( a - b ) = ( 0.76 - 0.1 ) = 0.66 answer is a ." | a = 0 ** 76
b = 0 ** 1
c = a - b
d = 0 ** 76
e = d + 0
f = 0 ** 1
g = e + f
h = c / g
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a ) 3 , b ) 30 , c ) 300 , d ) none of these , e ) can not be determined | c | multiply(const_100, divide(divide(1206, const_3), 134)) | one - third of 1206 is what percent of 134 ? | "answer let one - third of 1206 is n % of 134 . β΅ 1206 / 3 = ( n x 134 ) / 100 β΄ n = ( 402 x 100 ) / 134 = 300 correct option : c" | a = 1206 / 3
b = a / 134
c = 100 * b
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a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | d | add(9, 7) | the digit sum of the year 2007 is 2 + 0 + 0 + 7 = 9 , which is a square number . how many years during the 20 - first century have a square digital sum ? | the twenty - first century runs from the year 2001 to 2100 inclusive , so with the exception of the year 2100 , for which the digital sum is three anyway , the first two digits of all the other years are fixed as 20 . hence for the sum of all the digits to be square the sum of the last two digits number be two less than a perfect square . as the maximum sum of the last two digits is 18 , the greatest sum will be 20 . therefore we can obtain digital sums of 4 , 9 , or 16 , with the last two digits adding to 2 , 7 , or 14 respectively . 4 : 2002 , 2011 , 2020 9 : 2007 , 2016 , 2025 , 2034 , 2043 , 2052 , 2061 , 2070 16 : 2059 , 2068 , 2077 , 2086 , 2095 that is , there are sixteen years during the twenty - first century for which the digital sum is square . answer : d | a = 9 + 7
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a ) 30 , b ) 64 , c ) 60 , d ) 70 , e ) 90 | b | multiply(multiply(divide(40, subtract(const_1, divide(2, 3))), divide(2, 3)), divide(4, 5)) | a certain automobile company β s best - selling model is the speedster . the speedster , like all of their other models , comes in coupe and convertible styles . 2 / 3 of the current inventory is speedsters , of which 4 / 5 are convertibles . if there are 40 vehicles that are not speedsters , how many speedster convertibles are there ? | total vehicle = 2 / 3 of speedster + 1 / 3 of others . speedster convertibles = 2 / 3 total vehicle * 4 / 5 given : 1 / 3 constitutes 40 vehicles . hence 2 / 3 constitutes 80 speedster convertibls = 80 * 4 / 5 = 64 b | a = 2 / 3
b = 1 - a
c = 40 / b
d = 2 / 3
e = c * d
f = 4 / 5
g = e * f
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a ) 6 minutes , b ) 5 minutes , c ) 7 minutes , d ) 3 minutes , e ) 4 minutes | d | divide(180, add(36, divide(36, divide(const_3, const_2)))) | earl can stuff advertising circulars into envelopes at the rate of 36 envelopes per minutes and ellen requires a minutes and half to stuff the same number of envelops . working together , how long will it take earl and ellen to stuff 180 envelopes | earl takes 1 min . for 36 envelopes . ellen takes 3 / 2 mins for the same . so ellen can stuff ( ( 36 ) / ( 3 / 2 ) ) in 1 min . i . e . , 24 envelopes a min . so both of them when work together can stuff 36 + 24 = 60 envelopes in 1 min . for 180 envelopes they will take 180 / 60 mins . i . e . , 3 mins . answer : d | a = 3 / 2
b = 36 / a
c = 36 + b
d = 180 / c
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a ) 430 , b ) 990 , c ) 550 , d ) 590 , e ) 600 | b | add(divide(810, subtract(const_1, divide(10, const_100))), multiply(divide(810, subtract(const_1, divide(10, const_100))), divide(10, const_100))) | boy sells a book for rs . 810 he gets a loss of 10 % , to gain 10 % , what should be the sp ? | "cost price = 810 / 90 x 100 = 900 to gain 10 % = 900 x 10 / 100 = 90 sp = cp + gain = 900 + 90 = 990 answer : b" | a = 10 / 100
b = 1 - a
c = 810 / b
d = 10 / 100
e = 1 - d
f = 810 / e
g = 10 / 100
h = f * g
i = c + h
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a ) $ 100.00 , b ) $ 40.00 , c ) $ 50.00 , d ) $ 30.00 , e ) $ 10.00 | b | multiply(100, divide(add(50, 20), const_100)) | a discount electronics store normally sells all merchandise at a discount of 2 percent to 50 percent off the suggested retail price . if , during a special sale , an additional 20 percent were to be deducted from the discount price , what would be the lowest possible price of an item costing $ 100 before any discount ? | "original price : 100 $ max first discount = - 50 % thus : 100 Γ’ Λ β ( 1 Γ’ Λ β 50 / 100 ) = 50 second discount on the discounted price = - 20 % thus : 50 Γ’ Λ β ( 1 Γ’ Λ β 20 / 100 ) = 40 answer b ." | a = 50 + 20
b = a / 100
c = 100 * b
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a ) 50 , b ) 40 , c ) 30 , d ) 20 , e ) 10 | b | divide(multiply(34, const_100), 85) | how many pieces of 85 cm length can be cut from a rod of 34 meters long ? | "number of pieces = 3400 / 85 = 40 the answer is b ." | a = 34 * 100
b = a / 85
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a ) - 1 , b ) 0 , c ) 3 , d ) 4 , e ) 5 | e | add(const_2, const_3) | if 525 / 999 = 0.525 , what is the 81 st digit to the right of the decimal point in the decimal equivalent of the fraction ? | "0 . [ u ] 125 [ / u = 0.525525525 . . . . . every 3 rd digit is 5 and every multiple of 3 will be the digit 5 . since 81 is multiple of 3 , the 81 st digit is 5 . answer e" | a = 2 + 3
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a ) 30000 , b ) 50000 , c ) 40000 , d ) 20000 , e ) 65000 | e | multiply(divide(52000, 4), 5) | p and q invested in a business . the profit earned was divided in the ratio 4 : 5 . if p invested rs 52000 , the amount invested by q is | "let the amount invested by q = q 52000 : q = 4 : 5 β 52000 Γ 5 = 4 q β q = ( 52000 Γ 5 ) / 4 = 65000 answer is e ." | a = 52000 / 4
b = a * 5
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a ) 8 , b ) 15 , c ) 21 , d ) 27 , e ) 39 | b | multiply(floor(power(divide(152, const_2), divide(const_1, const_3))), power(subtract(152, power(floor(power(divide(152, const_2), divide(const_1, const_3))), const_3)), divide(const_1, const_3))) | the number 152 is equal to the sum of the cubes of two integers . what is the product of those integers ? | "actually , decomposition into factors can easily give you the answer here . you should just do the decomposition of the right thing i . e . the options since they represent the product of those integers . since the sum of cubes is 152 , the numbers can not be larger than 5 since 6 ^ 3 itself is 216 . 21 , 27 , 39 - the factors are too large so ignore 8 - ( 2 , 4 ) does not satisfy 15 - ( 3 , 5 ) yes . 3 ^ 3 + 5 ^ 3 = 152 - answer answer : b" | a = 152 / 2
b = 1 / 3
c = a ** b
d = math.floor(c)
e = 152 / 2
f = 1 / 3
g = e ** f
h = math.floor(g)
i = h ** 3
j = 152 - i
k = 1 / 3
l = j ** k
m = d * l
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a ) 240 , b ) 277 , c ) 276 , d ) 270 , e ) 281 | a | subtract(multiply(multiply(divide(72, const_3600), const_1000), 26), 280) | a goods train runs at the speed of 72 km / hr and crosses a 280 m long platform in 26 sec . what is the length of the goods train ? | "speed = 72 * 5 / 18 = 20 m / sec . time = 26 sec . let the length of the train be x meters . then , ( x + 280 ) / 26 = 20 x = 240 m . answer : a" | a = 72 / 3600
b = a * 1000
c = b * 26
d = c - 280
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a ) 1276 , b ) 1560 , c ) 2832 , d ) 1299 , e ) 1236 | b | multiply(add(add(30, divide(1200, 30)), sqrt(add(power(30, 2), power(divide(1200, 30), 2)))), 13) | a rectangular farm has to be fenced one long side , one short side and the diagonal . if the cost of fencing is rs . 13 per meter . the area of farm is 1200 m 2 and the short side is 30 m long . how much would the job cost ? | "explanation : l * 30 = 1200 Γ¨ l = 40 40 + 30 + 50 = 120 120 * 13 = 1560 answer : option b" | a = 1200 / 30
b = 30 + a
c = 30 ** 2
d = 1200 / 30
e = d ** 2
f = c + e
g = math.sqrt(f)
h = b + g
i = h * 13
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a ) 34 , b ) 40 , c ) 25 , d ) 88 , e ) 78 | c | add(multiply(divide(50, 20), const_2), 20) | a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 50 sq . feet , how many feet of fencing will be required ? | "we have : l = 20 ft and lb = 50 sq . ft . so , b = 2.5 ft . length of fencing = ( l + 2 b ) = ( 20 + 5 ) ft = 25 ft . answer : c" | a = 50 / 20
b = a * 2
c = b + 20
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a ) 48.975 , b ) 42.255 , c ) 50 , d ) 51.255 , e ) 52.253 | a | divide(add(multiply(24, 50.25), multiply(8, 45.15)), add(24, 8)) | the average weight of 24 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg . find the average weights of all the boys in the class . | "explanation : average weight of 24 boys = 50.25 total weight of 24 boys = 50.25 Γ 24 average weight of remaining 8 boys = 45.15 total weight of remaining 8 boys = 45.15 Γ 8 total weight of all boys in the class = ( 50.25 Γ 24 ) + ( 45.15 Γ 8 ) total boys = 24 + 8 = 32 average weight of all the boys = ( 50.25 Γ 24 ) + ( 45.15 Γ 8 ) / 32 = 48.975 answer : option a" | a = 24 * 50
b = 8 * 45
c = a + b
d = 24 + 8
e = c / d
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a ) 40 , b ) 80 , c ) 20 , d ) 24 , e ) 36 | c | multiply(4, 5) | if ' y ' is a positive integer exactly divisible by 4 or 5 but not divisible by 8 . what could possibly be the value of ' y ' ? | 40 and 80 are both divisible by 4 and 5 but also by 8 . so they are not the right answer . 24 and 36 are both clearly not divisible by 5 ( not correct ) 20 is both divisible by 4 and 5 but not by 8 . answer : ( c ) | a = 4 * 5
|
a ) 40 , b ) 45 , c ) 48 , d ) 60 , e ) 44 | d | add(45, 30) | two goods trains each 625 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 625 + 625 = 1250 m . required time = 1250 * 6 / 125 = 60 sec . answer : option d" | a = 45 + 30
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a ) 2793 , b ) 2333 , c ) 2993 , d ) 2893 , e ) 2693 | d | add(add(subtract(divide(divide(1000, const_10), const_10), const_1), subtract(subtract(divide(1000, const_10), const_1), subtract(divide(divide(1000, const_10), const_10), const_1))), multiply(subtract(subtract(1000, const_1), subtract(divide(1000, const_10), const_1)), const_3)) | how many figures are required to number the pages the pages of a book containing 1000 pages ? | 1 to 9 = 9 * 1 = 9 10 to 99 = 90 * 2 = 180 100 to 999 = 900 * 3 = 2700 1000 = 4 - - - - - - - - - - - 2893 answer : d | a = 1000 / 10
b = a / 10
c = b - 1
d = 1000 / 10
e = d - 1
f = 1000 / 10
g = f / 10
h = g - 1
i = e - h
j = c + i
k = 1000 - 1
l = 1000 / 10
m = l - 1
n = k - m
o = n * 3
p = j + o
|
a ) 20 % , b ) 40 % , c ) 87.5 % , d ) 60 % , e ) 75 % | c | subtract(const_100, subtract(subtract(const_100, 20), 50)) | a merchant sells an item at a 20 % discount , but still makes a gross profit of 50 percent of the cost . what percent of the cost would the gross profit on the item have been if it had been sold without the discount ? | "original sp = x cost = c current selling price = . 8 x ( 20 % discount ) . 8 x = 1.5 c ( 50 % profit ) x = 1.5 / . 8 * c x = 15 / 8 c original selling price is 1.875 c which is 87.5 % profit answer c" | a = 100 - 20
b = a - 50
c = 100 - b
|
a ) 4 miles , b ) 4 1 / 4 miles , c ) 4 3 / 4 miles , d ) 5 1 / 2 miles , e ) 6 miles | e | divide(divide(multiply(subtract(4.90, 1.40), const_100), const_3), const_4) | a taxi company charges $ 1.40 for the first quarter of a mile and fifteen cents for each additional quarter of a mile . what is the maximum distance someone could travel with $ 4.90 ? | "if we start out with $ 4.90 and have to spend $ 1.40 for the first quarter - mile , we will have $ 3.50 left to spend on quarter - mile intervals . since $ 3.50 / $ 0.15 = 23.33 , we can buy 23.33 more quarter - miles , and will travel 24.33 quarter miles in all : 24.33 Γ 1 / 4 = 6 miles . the correct answer is choice ( e ) ." | a = 4 - 90
b = a * 100
c = b / 3
d = c / 4
|
a ) 75 , b ) 100 , c ) 125 , d ) can not be determined , e ) none of these | c | power(add(50, const_4), const_4) | the difference between a number and its three - fifths is 50 . what is the number ? | "suppose the number is n . then n - 3 β 5 n = 50 β 2 n β 5 = 50 β΄ n = 50 Γ 52 = 125 answer . c" | a = 50 + 4
b = a ** 4
|
a ) 35 % , b ) 45 % , c ) 25 % , d ) 30 % , e ) 20 % | c | multiply(subtract(6, inverse(8)), const_100) | john and david can finish a job together in 6 hours . if john can do the job by himself in 8 hours , what percent of the job does david do ? | "you can also plug in numbers . for example , bob and alice work at a donut factory and make 24 donuts which is the job ( i picked this as a smart number ) . john on his own works 24 / 8 = 3 donuts per hour . john and david work 24 / 6 = 4 donuts per hour so david works 1 donuts / hour to find out the percentage , david works 1 donuts / hr x 6 hours = 6 donuts per hour . therefore 6 donuts / 24 donuts = 1 / 4 = 25 % answer : c" | a = 1/(8)
b = 6 - a
c = b * 100
|
a ) 40 days , b ) 30 days , c ) 60 days , d ) 70 days , e ) 50 days | b | divide(multiply(10, 15), subtract(15, 10)) | a and b finish the job in 15 days . while a , b and c can finish it in 10 days . c alone will finish the job in | "explanation : 10 = ( 15 * x ) / ( 15 + x ) 150 + 10 x = 15 x 5 x = 150 x = 30 answer : option b" | a = 10 * 15
b = 15 - 10
c = a / b
|
a ) 2 : 1 , b ) 1 : 2 , c ) 5 : 1 , d ) 4 : 1 , e ) 3 : 1 | a | divide(divide(subtract(13.25, 10.75), subtract(13.25, 10)), subtract(const_1, divide(subtract(13.25, 10.75), subtract(13.25, 10)))) | in what ratio should a variety of rice costing rs . 10 per kg be mixed with another variety of rice costing rs . 13.25 per kg to obtain a mixture costing rs . 10.75 per kg ? | "let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x / y = ( 13.25 - 10.75 ) / ( 10.75 - 10 ) = 6 / 3 = 2 / 1 answer : a" | a = 13 - 25
b = 13 - 25
c = a / b
d = 13 - 25
e = 13 - 25
f = d / e
g = 1 - f
h = c / g
|
a ) 1 / 5 , b ) 2 / 7 , c ) 2 / 9 , d ) 3 / 10 , e ) 4 / 11 | c | divide(square_perimeter(4), 9) | a 9 - meter long wire is cut into two pieces . if the longer piece is then used to form a perimeter of a square , what is the probability that the area of the square will be more than 4 if the original wire was cut at an arbitrary point ? | "the longer wire will form a square with an area more than 4 if the wire is cut at a point within one meter of either end . the probability of this is 2 / 9 . the answer is c ." | a = square_perimeter / (
|
a ) 3277 , b ) 2977 , c ) 30000 , d ) 6077 , e ) 17112 | c | multiply(multiply(4, multiply(const_100, const_100)), power(divide(3, 4), 1)) | the value of a scooter depreciates in such a way that its value of the end of each year is 3 / 4 of its value of the beginning of the same year . if the initial value of the scooter is rs . 40,000 , what is the value at the end of 1 year ? | "explanation : 40,000 * ( 3 / 4 ) ^ 1 answer : c" | a = 100 * 100
b = 4 * a
c = 3 / 4
d = c ** 1
e = b * d
|
a ) 54 , b ) 76 , c ) 152 , d ) 384 , e ) none of these | d | multiply(multiply(4, multiply(4, const_2)), multiply(4, 3)) | the greatest number by which the product of 3 consecutive multiples of 4 is always divisible is | solution required number = product of first three multiplies of 3 = ( 4 Γ£ β 8 Γ£ β 12 ) = 384 . answer d | a = 4 * 2
b = 4 * a
c = 4 * 3
d = b * c
|
a ) 25 , b ) 27 , c ) 45 , d ) 75 , e ) 125 | d | multiply(power(5, 2), 3) | if both 5 ^ 2 and 3 ^ 3 are factors of n x ( 2 ^ 5 ) x ( 6 ^ 2 ) x ( 7 ^ 3 ) , what is the smallest possible positive value of n ? . | if both 5 ^ 2 & 3 ^ 3 are factors , then they must be present in the number . leaving rest of the prime factors and splitting 6 ^ 2 into 3 ^ 2 * 2 ^ 3 . the number is lacking 5 ^ 2 & a 3 , so that 5 ^ 2 and 3 ^ 3 is a factor . hence the smallest number is 5 ^ 2 * 3 = 75 answer : d | a = 5 ** 2
b = a * 3
|
a ) rs . 1000 , b ) rs . 2596 , c ) rs . 4320 , d ) rs . 6489 , e ) rs . 4789 | c | multiply(multiply(multiply(divide(60, const_100), divide(subtract(const_100, 40), const_100)), divide(subtract(const_100, 20), const_100)), 15000) | rajesh spends 40 % of his monthly salary on food , 20 % of the medicines and saves 60 % of the remaining amount . if his monthly salary is rs . 15000 , how much money does he save every month ? | explanation : ramesh ' s monthly income = rs . 15000 he spends 40 % on food . the total money spent on food = 40 / 100 * 15000 = rs . 6000 now , his monthly remaining income = rs . 15000 β rs . 6000 = rs . 9000 out of rs . 9000 , he spends 20 % on medicines . the total money spent on medicines = 20 / 100 * 9000 = rs . 1800 now , his monthly remaining income = rs . 9000 β rs . 1800 = rs . 7200 out of rs . 7200 , he saves 600 % . amount saved = 60 / 100 * 7200 = rs . 4320 answer : c | a = 60 / 100
b = 100 - 40
c = b / 100
d = a * c
e = 100 - 20
f = e / 100
g = d * f
h = g * 15000
|
a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | c | subtract(add(add(14, 32), 53), add(add(multiply(3, const_3), 21), 47)) | the average ( arithmetic mean ) of 14 , 32 , and 53 is 3 more than the average of 21 , 47 , and x . what is x ? | "the average of 14 , 32 , and 53 is 33 . the average of 21 , 47 and x is 30 . then 21 + 47 + x = 90 . x = 22 . the answer is c ." | a = 14 + 32
b = a + 53
c = 3 * 3
d = c + 21
e = d + 47
f = b - e
|
a ) 33 , b ) 72 , c ) 55 , d ) 82.5 , e ) 62 | d | divide(add(165, 660), multiply(36, const_0_2778)) | how long does a train 165 meters long running at the rate of 36 kmph take to cross a bridge 660 meters in length ? | "explanation : t = ( 660 + 165 ) / 36 * 18 / 5 t = 82.5 answer : option d" | a = 165 + 660
b = 36 * const_0_2778
c = a / b
|
a ) 6 , b ) 12 , c ) 24 , d ) 48 , e ) none | b | multiply(log(64), const_2) | if β 2 n = 64 , then the value of n is : | sol . β 2 n = 64 β ( 2 n ) 1 / 2 β 2 n / 2 = 26 β n / 2 = 6 β n = 12 . answer b | a = math.log(64)
b = a * 2
|
a ) 28000 , b ) 30000 , c ) 31500 , d ) 34000 , e ) none of these | c | divide(4500, divide(multiply(const_1, const_1), add(multiply(const_3, const_2), multiply(const_1, const_1)))) | a & b started a partnership business . a ' s investment was thrice the investment of b and the period of his investment was two times the period of investments of b . if b received rs 4500 as profit , what is their total profit ? | "explanation : suppose b ' s investment = x . then a ' s investment = 3 x suppose bs period of investment = y , then a ' s period of investment = 2 y a : b = 3 x * 2 y : xy = 6 : 1 total profit * 1 / 7 = 4500 = > total profit = 4500 * 7 = 31500 . answer : option c" | a = 1 * 1
b = 3 * 2
c = 1 * 1
d = b + c
e = a / d
f = 4500 / e
|
a ) 10 % , b ) 9 % , c ) 25 % , d ) 30 % , e ) 50 % | b | multiply(divide(subtract(multiply(const_100, const_100), multiply(subtract(const_100, 30), add(const_100, 30))), multiply(const_100, const_100)), const_100) | robert ' s salary was decreased by 30 % and subsequently increased by 30 % . how much percentage does he lose ? | "let original salary be $ 100 salary after decreasing 30 % = 100 - 100 x 30 / 100 = $ 70 salary after increasing 30 % on $ 70 = 70 + 70 x 30 / 100 = $ 91 percentage of loss = 100 - 91 = 9 % answer : b" | a = 100 * 100
b = 100 - 30
c = 100 + 30
d = b * c
e = a - d
f = 100 * 100
g = e / f
h = g * 100
|
a ) 8 , b ) 2 , c ) 1 , d ) 5 , e ) 2 | c | multiply(divide(4, 4), const_100) | 4 + 4 | c | a = 4 / 4
b = a * 100
|
a ) 20 , b ) 12 , c ) 15 , d ) 16 , e ) 10 | a | multiply(20, const_1) | the total age of a and b is 20 years more than the total age of b and c . c is how many years younger than a . ? | "( a + b ) - ( b - c ) = 20 a - c = 20 answer is a" | a = 20 * 1
|
a ) 1 minutes , b ) 2 minutes , c ) 3 minutes , d ) 4 minutes , e ) 6 minutes | e | multiply(6, const_1) | if 6 cats can kill 6 rats in 6 minutes , how long will it take 100 cats to kill 100 rats ? | "it will take 6 minutes for 100 cats to kill 100 rats . 1 cat can kill 1 rat in 6 minutes , so 100 cats can kill 100 rats in 6 minutes answer e" | a = 6 * 1
|
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