Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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r = { 2 , 3 , 4 , 5 } b = { 4 , 5 , 6 , 7 , 8 } two integers will be randomly selected from the sets above , one integer from set r and one integer from set b . what is the probability that the sum of the two integers will equal 9 ? | the total number of pairs r , b possible is 4 * 5 = 20 . out of these 20 pairs only 4 sum up to 9 : ( 2 , 7 ) ; ( 3 , 6 ) , ( 4 , 5 ) and ( 5 , 4 ) . the probability thus is 4 / 20 = 0.2 . answer : b . | a ) 0.15 , b ) 0.20 , c ) 0.25 , d ) 0.30 , e ) 0.33 | b | multiply(divide(4, 4), divide(const_1, 5)) | divide(n2,n2)|divide(const_1,n3)|multiply(#0,#1) | general |
a $ 79.95 lawn chair was sold for $ 59.95 at a special sale . by approximately what percent was the price decreased ? | "listed selling price of chair = 79.95 $ discounted selling price of chair = 59.95 $ discount = 79.95 - 59.95 = 20 $ % decrease in price of chair = ( 20 / 79.95 ) * 100 % = 25 % approx answer c" | a ) 15 % , b ) 20 % , c ) 25 % , d ) 60 % , e ) 80 % | c | multiply(divide(subtract(79.95, 59.95), 79.95), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | general |
5 drainage pipes , each draining water from a pool at the same constant rate , together can drain a certain pool in 16 days . how many additional pipes , each draining water at the same constant rate , will be needed to drain the pool in 4 days ? | this is an inverse proportional problem . . . . . . 5 pipes in 16 days ; so for 4 days , it will be = 16 x 5 / 4 = 20 so , 20 - 5 = 15 answer e | a ) 6 , b ) 9 , c ) 10 , d ) 12 , e ) 15 | e | subtract(multiply(4, 5), 5) | multiply(n0,n2)|subtract(#0,n0) | general |
a heap of grapes is divided into groups of 3 , 5 and 7 and each time one coconut is left over . the least number of grapes in the heap is ? a . 31 b . 41 c . 51 d . 61 | lcm = 105 = > 105 + 1 = 106 answer : c | a ) a ) 31 , b ) b ) 41 , c ) c ) 106 , d ) d ) 61 , e ) e ) 71 | c | lcm(lcm(3, 5), 7) | lcm(n0,n1)|lcm(n2,#0) | general |
the diagonals of a rhombus are 12 cm and 10 cm . find its area ? | "1 / 2 * 12 * 10 = 60 answer : e" | a ) 158 , b ) 129 , c ) 150 , d ) 123 , e ) 60 | e | rhombus_area(12, 10) | rhombus_area(n0,n1)| | geometry |
three 6 faced dice are thrown together . the probability that no two dice show the same number on them is | "explanation : no two dice show same number would mean all the three faces should show different numbers . the first can fall in any one of the six ways . the second die can show a different number in five ways . the third should show a number that is different from the first and second . this can happen in four ways .... | a ) 5 / 0 , b ) 5 / 9 , c ) 5 / 1 , d ) 5 / 3 , e ) 5 / 6 | b | multiply(multiply(multiply(divide(const_1, 6), divide(const_1, 6)), divide(const_1, 6)), divide(const_1, 6)) | divide(const_1,n0)|multiply(#0,#0)|multiply(#0,#1)|multiply(#0,#2)| | probability |
find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively . | "required number = h . c . f . of ( 1657 - 6 ) and ( 2037 - 5 ) = h . c . f . of 1651 and 2032 _______ 1651 ) 2032 ( 1 1651 1651 _______ 381 ) 1651 ( 4 1524 _________ 127 ) 381 ( 3 381 0 required number = 127 . answer is c ." | a ) 121 , b ) 124 , c ) 127 , d ) 122 , e ) 129 | c | gcd(subtract(1657, 6), subtract(2037, 5)) | subtract(n0,n2)|subtract(n1,n3)|gcd(#0,#1)| | general |
ram - leela has $ 100 in her piggy bank . how much will she have in her bank 52 weeks from now if she puts $ 1 in the bank next week , $ 2 two weeks from now , $ 3 3 weeks from now , and continues to increase the amount that she puts in by $ 1 each week ? | the dollar deposits are in an a . p . 1,2 , 3,4 . . . 52 with common difference 1 sum of the terms is n ( n + 1 ) / 2 i . e 52 * ( 52 + 1 ) / 2 = 52 * 53 / 2 = 1378 total deposit therefore with chiu - lihas is 100 + 1378 = 1478 $ | a ) 1478 , b ) 1578 , c ) 1678 , d ) 1778 , e ) 1798 | a | add(divide(multiply(add(divide(100, 2), 3), 52), 2), 100) | divide(n0,n3)|add(n4,#0)|multiply(n1,#1)|divide(#2,n3)|add(n0,#3) | general |
a certain mixture of nuts consists of 5 parts almonds to 2 parts walnuts , by weight . what is the number of pounds of almonds in 350 pounds of the mixture ? | almonds : walnuts = 5 : 2 total mixture has 7 parts in a 350 pound mixture , almonds are 5 / 7 ( total mixture ) = 5 / 7 * 350 = 250 pounds answer ( a ) | a ) 250 , b ) 84 , c ) 40 , d ) 28 , e ) 20 | a | divide(multiply(5, 350), add(5, 2)) | add(n0,n1)|multiply(n0,n2)|divide(#1,#0) | general |
if x = - 5 / 4 and y = - 3 / 2 , what is the value of the expression - 2 x β y ^ 2 ? | x = - 5 / 4 and y = - 3 / 2 = = > - 2 ( - 5 / 4 ) - ( 3 / 2 ) ^ 2 = 10 / 4 - 9 / 4 = 1 / 4 ans : a | a ) 1 / 4 , b ) - 1 , c ) 5 / 4 , d ) 3 / 2 , e ) 3 / 4 | a | subtract(multiply(2, divide(5, 4)), power(divide(3, 2), 2)) | divide(n0,n1)|divide(n2,n3)|multiply(n3,#0)|power(#1,n3)|subtract(#2,#3) | general |
which expression is the greatest | "options can be re - written as ( x - 5 ) x = > 1 - ( 5 / x ) a ) 1 - ( 5 / 3257 ) b ) 1 - ( 5 / 3461 ) c ) 1 - ( 5 / 3596 ) d ) 1 - ( 5 / 3656 ) e ) 1 - ( 5 / 3458 ) to get the largest among these second half should be the least and so denominator to be largest . hence ' d ' ." | a ) 3252 / 3257 , b ) 3456 / 3461 , c ) 3591 / 3596 , d ) 3641 / 3656 , e ) 3453 / 3458 | d | divide(subtract(const_3600, multiply(const_3, const_3)), subtract(const_3600, const_4)) | multiply(const_3,const_3)|subtract(const_3600,const_4)|subtract(const_3600,#0)|divide(#2,#1)| | general |
jerry β s average ( arithmetic mean ) score on the first 3 of 4 tests is 78 . if jerry wants to raise his average by 2 points , what score must he earn on the fourth test ? | "total score on 3 tests = 78 * 3 = 234 jerry wants the average to be = 80 hence total score on 4 tests should be = 80 * 4 = 320 score required on the fourth test = 320 - 234 = 86 option c" | a ) 87 , b ) 89 , c ) 86 , d ) 93 , e ) 95 | c | subtract(multiply(4, add(78, 2)), multiply(78, 3)) | add(n2,n3)|multiply(n0,n2)|multiply(n1,#0)|subtract(#2,#1)| | general |
guy drives 60 miles to attend a meeting . halfway through , he increases his speed so that his average speed on the second half is 16 miles per hour faster than the average speed on the first half . his average speed for the entire trip is 30 miles per hour . guy drives on average how many miles per hour during the fir... | let x be the average speed for 1 st half of the distance . then the average speed for 2 nd half of the distance will be x + 16 avg speed = total distance / total time 30 = 60 / { ( 30 / x ) + ( 30 / ( x + 16 ) ) } solving we get x ^ 2 - 14 x - 240 = 0 x = - 10 or 24 x cant be negative hence x = 24 answer : d | a ) 12 , b ) 14 , c ) 16 , d ) 24 , e ) 40 | d | divide(divide(multiply(16, 30), const_2), divide(60, multiply(const_3, const_2))) | multiply(n1,n2)|multiply(const_2,const_3)|divide(#0,const_2)|divide(n0,#1)|divide(#2,#3) | physics |
8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 18 men , 4 women and 10 children together complete the piece of work ? | "explanation : 1 man β s 1 day β s work = 1 / 8 Γ 12 = 1 / 96 18 men β s 1 day β s work = 1 Γ 18 / 96 = 3 / 16 1 woman β s 1 day β s work = 1 / 192 4 women β s 1 day β s work = 1 / 192 Γ 4 = 1 / 48 1 child β s 1 day β s work = 1 / 240 10 children β s 1 day β s work = 1 / 24 therefore , ( 18 men + 4 women + 10 children ... | a ) 5 days , b ) 15 days , c ) 28 days , d ) 4 days , e ) 7 days | d | inverse(add(multiply(10, inverse(multiply(24, 10))), add(multiply(inverse(multiply(12, 8)), 10), multiply(inverse(multiply(48, 4)), 4)))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,n5)|inverse(#0)|inverse(#1)|inverse(#2)|multiply(n4,#3)|multiply(n2,#4)|multiply(n4,#5)|add(#6,#7)|add(#9,#8)|inverse(#10)| | physics |
square a is inscribed in circle b . if the perimeter of a is 32 , what is the circumference of b ? | square forms two right angled triangles . any time we have a right angle triangle inside a circle , the hypotenuse is the diameter . hypotenuse here = diagonal of the square = 8 sqrt ( 2 ) = diameter = > radius = 4 sqrt ( 2 ) circumference of the circle = 2 pi r = 8 pi sqrt ( 2 ) answer is c . | ['a ) 6 β 2 Ο', 'b ) 4 β 2 Ο', 'c ) 8 β 2 Ο', 'd ) 7 β 2 Ο', 'e ) 5'] | c | circumface(divide(sqrt(add(power(square_edge_by_perimeter(32), const_2), power(square_edge_by_perimeter(32), const_2))), const_2)) | square_edge_by_perimeter(n0)|power(#0,const_2)|add(#1,#1)|sqrt(#2)|divide(#3,const_2)|circumface(#4) | geometry |
15 years hence , rohan will be just 4 times as old as he was 15 years ago . how old is rohan at present ? | let the present age of rohan be x years then , given : x + 15 = 4 ( x - 15 ) x = 25 answer : b | a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 40 | b | divide(add(multiply(4, 15), 15), subtract(4, const_1)) | multiply(n0,n1)|subtract(n1,const_1)|add(n0,#0)|divide(#2,#1) | general |
dan β s car gets 32 miles per gallon . if gas costs $ 4 / gallon , then how many miles can dan β s car go on $ 54 of gas ? | 54 / 4 = 13.5 gallons 13.5 * 32 = 432 miles the answer is c . | a ) 236 , b ) 354 , c ) 432 , d ) 512 , e ) 670 | c | divide(multiply(54, 32), 4) | multiply(n0,n2)|divide(#0,n1) | physics |
if the operation ΓΈ is defined for all positive integers x and w by x ΓΈ w = ( 2 ^ x ) / ( 2 ^ w ) then ( 3 ΓΈ 1 ) ΓΈ 1 = ? | "3 ΓΈ 1 = 2 ^ 3 / 2 ^ 1 = 4 4 ΓΈ 1 = 2 ^ 4 / 2 = 8 the answer is c ." | a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | c | divide(power(2, divide(power(2, 3), power(2, 2))), power(2, 1)) | power(n0,n2)|power(n0,n0)|power(n0,n4)|divide(#0,#1)|power(n0,#3)|divide(#4,#2)| | general |
what will be the remainder when ( 67 ^ 67 ) + 67 is divided by 68 ? | "x ^ n + 1 will be divisible by x + 1 only when n is odd 67 ^ 67 + 1 will be divisible by 67 + 1 ( 67 ^ 67 + 1 ) + 66 , when divided by 68 will give 66 as remainder answer is c" | a ) 52 , b ) 62 , c ) 66 , d ) 68 , e ) 72 | c | power(const_2.0, const_2) | power(const_2.0,const_2)| | general |
walking at the rate of 5 kmph a man cover certain distance in 5 hr . running at a speed of 15 kmph the man will cover the same distance in . | distance = speed * time 5 * 5 = 25 km new speed = 15 kmph therefore time = 25 / 15 = 5 / 3 = 36 min answer : b | a ) 12 min , b ) 36 min , c ) 40 min , d ) 48 min , e ) 60 min | b | multiply(divide(15, multiply(5, 5)), const_60) | multiply(n0,n0)|divide(n2,#0)|multiply(#1,const_60) | physics |
subash can copy 50 pages in 10 hrs . subash and prakash together can copy 300 pages in 40 hours . in how much time prakash can copy 18 pages . | subhas ' s 1 hr copy page = 50 / 10 = 5 page ( subhas + prakash ) ' s 1 hr copy page = 300 / 40 = 7.5 page from above prakash ' s 1 hr copy page = 2.5 page so time taken in 30 page ' s copy = ( 5 / 2.5 ) = 2 hrs answer : c | a ) 8 , b ) 10 , c ) 2 , d ) 14 , e ) 16 | c | floor(subtract(divide(300, 40), divide(50, 10))) | divide(n2,n3)|divide(n0,n1)|subtract(#0,#1)|floor(#2) | general |
a pair of articles was bought for $ 720 at a discount of 10 % . what must be the marked price of each of the article ? | "s . p . of each of the article = 720 / 2 = $ 360 let m . p = $ x 90 % of x = 360 x = 360 * 100 / 90 = $ 400 answer is d" | a ) $ 300 , b ) $ 500 , c ) $ 350 , d ) $ 400 , e ) $ 600 | d | divide(multiply(subtract(const_100, 10), divide(720, const_2)), const_100) | divide(n0,const_2)|subtract(const_100,n1)|multiply(#0,#1)|divide(#2,const_100)| | gain |
average expenditure of a person for the first 3 days of a week is rs . 330 and for the next 4 days is rs . 420 . average expenditure of the man for the whole week is : | "explanation : assumed mean = rs . 330 total excess than assumed mean = 4 Γ ( rs . 420 - rs . 350 ) = rs . 280 therefore , increase in average expenditure = rs . 280 / 7 = rs . 40 therefore , average expenditure for 7 days = rs . 330 + rs . 40 = rs . 370 correct option : b" | a ) 350 , b ) 370 , c ) 390 , d ) 430 , e ) none | b | add(330, divide(multiply(4, subtract(420, 330)), add(3, 4))) | add(n0,n2)|subtract(n3,n1)|multiply(n2,#1)|divide(#2,#0)|add(n1,#3)| | general |
set x consists of 10 integers and has median of 30 and a range of 30 . what is the value of the greatest possible integer that can be present in the set ? | "note that both median and range do not restrict too many numbers in the set . range is only concerned with the smallest and greatest . median only cares about the middle . quick check of each option starting from the largest : ( e ) 50 range of 20 means the smallest integer will be 30 . so 20 can not lie in between an... | a ) 32 , b ) 37 , c ) c . 40 , d ) 43 , e ) 50 | e | add(30, 30) | add(n1,n2)| | general |
in a class there are 55 pupil , out of them 10 are in debate only and 18 in singing only . then how many in both ? | explanation : total pupil = 55 debate + singing = 10 + 18 = 28 the intersection for two = 55 Γ’ β¬ β 10 Γ’ β¬ β 28 = 17 play both games . answer : c | a ) 132 , b ) 26 , c ) 17 , d ) 11 , e ) 12 | c | subtract(subtract(subtract(55, 18), 10), 10) | subtract(n0,n2)|subtract(#0,n1)|subtract(#1,n1) | other |
in a hostel , the number of students decreased by 8 % and the price of food increased by 20 % over the previous year . if each student consumes the same amount of food then by how much should the consumption of food be cut short by every student , so that the total cost of the food remains the same as that of the previ... | "cost of food ( c ) = food consumed per student ( f ) * number of students ( n ) * price of food ( p ) originally , c = fnp when number of students decrease by 8 % , and the price of food increases by 20 % , c = f ( new ) * ( 0.92 n ) * ( 1.2 p ) = > f ( new ) = f / ( 0.92 * 1.2 ) = > f ( new ) = 0.906 f therefore the ... | a ) 19 % , b ) 15 % , c ) 25 % , d ) 40 % , e ) 9.4 % | e | multiply(subtract(const_1, divide(multiply(const_100, const_100), multiply(subtract(const_100, 8), add(const_100, 20)))), const_100) | add(n1,const_100)|multiply(const_100,const_100)|subtract(const_100,n0)|multiply(#0,#2)|divide(#1,#3)|subtract(const_1,#4)|multiply(#5,const_100)| | general |
the total circumference of two circles is 88 . if the first circle has a circumference that is exactly twice the circumference of the second circle , then what is the approximate sum of their two radii ? | "let r = radius of smaller circle . let r = radius of larger circle therefore : 2 Ο r + 2 Ο r = 88 where 2 r = r thus : 2 Ο r + 4 Ο r = 88 6 Ο r = 88 r = approx 4.7 Ο r + 2 r Ο = 88 3 Ο r = 88 r = approx 9.3 r + r = approx 14.0 answer : e" | a ) 5.7 , b ) 6.0 , c ) 6.7 , d ) 9.7 , e ) 14.0 | e | divide(add(divide(divide(88, const_3), const_3), divide(multiply(divide(88, const_3), const_2), const_3)), const_2) | divide(n0,const_3)|divide(#0,const_3)|multiply(#0,const_2)|divide(#2,const_3)|add(#1,#3)|divide(#4,const_2)| | general |
how many multiples of 2 are there between 101 and 999 ? | "2 multiples are 102 , 104,106 , - - - - - - - - - , 994 , 996,998 it should be mentioned whether 1 and 89 are inclusive . the answer is ( 998 - 102 ) / 2 + 1 = 449 answer is e" | a ) 250 , b ) 440 , c ) 510 , d ) 575 , e ) 449 | e | add(divide(subtract(999, 101), 2), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
find the smallest number which when divided by 13 and 16 leaves respective remainders of 2 and 5 . | "explanation : let ' n ' is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5 . required number = ( lcm of 13 and 16 ) - ( common difference of divisors and remainders ) = ( 208 ) - ( 11 ) = 197 . answer : b" | a ) 128 , b ) 197 , c ) 127 , d ) 182 , e ) 091 | b | subtract(multiply(13, 16), add(const_10, const_1)) | add(const_1,const_10)|multiply(n0,n1)|subtract(#1,#0)| | general |
how much is 80 % of 40 is greater than 4 / 5 of 20 ? | "( 80 / 100 ) * 40 Γ’ β¬ β ( 4 / 5 ) * 20 32 - 16 = 16 answer : d" | a ) 12 , b ) 27 , c ) 18 , d ) 16 , e ) 81 | d | subtract(multiply(40, divide(80, const_100)), multiply(divide(4, 5), 20)) | divide(n0,const_100)|divide(n2,n3)|multiply(n1,#0)|multiply(n4,#1)|subtract(#2,#3)| | general |
if the price of sugar rises from rs . 10 per kg to rs . 13 per kg , a person , to have no increase in the expenditure on sugar , will have to reduce his consumption of sugar by | "sol . let the original consumption = 100 kg and new consumption = x kg . so , 100 x 10 = x Γ 13 = x = 77 kg . β΄ reduction in consumption = 23 % . answer c" | a ) 15 % , b ) 20 % , c ) 23 % , d ) 30 % , e ) none | c | multiply(subtract(const_1, divide(multiply(const_1, 10), 13)), const_100) | multiply(n0,const_1)|divide(#0,n1)|subtract(const_1,#1)|multiply(#2,const_100)| | general |
a train crosses a bridge of length 800 m in 45 seconds and a lamp post on the bridge in 15 seconds . what is the length of the train in metres ? | "let length of train = l case - 1 : distance = 800 + l ( while crossing the bridge ) time = 45 seconds i . e . speed = distance / time = ( 800 + l ) / 45 case - 2 : distance = l ( while passing the lamp post ) time = 15 seconds i . e . speed = distance / time = ( l ) / 15 but since speed has to be same in both cases so... | a ) 375 m , b ) 750 m , c ) 400 m , d ) 800 m , e ) 300 m | c | multiply(divide(800, subtract(45, 15)), 15) | subtract(n1,n2)|divide(n0,#0)|multiply(n2,#1)| | physics |
the price of an item is discounted 10 percent on day 1 of a sale . on day 2 , the item is discounted another 10 percent , and on day 3 , it is discounted an additional 20 percent . the price of the item on day 3 is what percentage of the sale price on day 1 ? | "let initial price be 1000 price in day 1 after 10 % discount = 900 price in day 2 after 10 % discount = 810 price in day 3 after 20 % discount = 648 so , price in day 3 as percentage of the sale price on day 1 will be = 648 / 900 * 100 = > 72 % answer will definitely be ( e )" | a ) 28 % , b ) 40 % , c ) 64.8 % , d ) 70 % , e ) 72 % | e | add(multiply(divide(divide(20, const_100), subtract(1, divide(1, 10))), const_100), 2) | divide(n5,const_100)|divide(n1,n0)|subtract(n1,#1)|divide(#0,#2)|multiply(#3,const_100)|add(n2,#4)| | gain |
the sum of three consecutive numbers is 87 . the greatest among these three number is : | "let the numbers be x , x + 1 and x + 2 then , x + ( x + 1 ) + ( x + 2 ) = 87 3 x = 84 x = 28 greatest number , ( x + 2 ) = 30 . answer : d" | a ) 26 , b ) 28 , c ) 29 , d ) 30 , e ) 31 | d | divide(add(87, const_1), const_2) | add(n0,const_1)|divide(#0,const_2)| | physics |
a train travels 290 km in 4.5 hours and 400 km in 5.5 hours . find the average speed of train . | "as we know that speed = distance / time for average speed = total distance / total time taken thus , total distance = 290 + 400 = 690 km thus , total speed = 10 hrs or , average speed = 690 / 10 or , 69 kmph . answer : b" | a ) 80 kmph , b ) 69 kmph , c ) 70 kmph , d ) 90 kmph , e ) none of these | b | divide(add(290, 400), add(4.5, 5.5)) | add(n0,n2)|add(n1,n3)|divide(#0,#1)| | physics |
what is the average of 12 , 13 , 14 , 510 , 520 , 530 , 1,115 , 1,120 , and 1 , 125,2140 , 2345 ? | "add 12 , 13 , 14 , 510 , 520 , 530 , 1,115 , 1,120 , and 1,125 , 2140 , 2345 grouping numbers together may quicken the addition sum = 9444 4959 / 11 = 858.54 . e" | a ) 419 , b ) 551 , c ) 601 , d ) 620 , e ) 858.54 | e | divide(add(add(add(add(add(add(12, 13), 14), 510), 520), 530), 12), add(const_3, const_4)) | add(n0,n1)|add(const_3,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|add(n5,#4)|add(n0,#5)|divide(#6,#1)| | general |
when a certain tree was first planted , it was 4 feet tall , and the height of the tree increased by a constant amount each year for the next 6 years . at the end of the 6 th year , the tree was 1 / 3 taller than it was at the end of the 4 th year . by how many feet did the height of the tree increase each year ? | "say , the tree grows by x feet every year . then , 4 + 6 x = ( 1 + 1 / 3 ) ( 4 + 4 x ) or , x = 2 answer b" | a ) 3 / 10 , b ) 2 , c ) 1 / 2 , d ) 2 / 3 , e ) 6 / 5 | b | divide(4, subtract(multiply(subtract(6, 4), 3), 4)) | subtract(n1,n0)|multiply(n4,#0)|subtract(#1,n0)|divide(n0,#2)| | general |
how many integerskgreater than 100 and less than 800 are there such that if the hundreds and the units digits ofkare reversed , the resulting integer is k + 99 ? | "numbers will be like 102 = > 201 = 102 + 99 203 = > 302 = 103 + 99 so the hundereth digit and units digit are consecutive where unit digit is bigger than hundred digit . there will be six pairs of such numbers for every pair there will 10 numbers like for 12 = > 102 , 112,132 , 142,152 , 162,172 , 182,192 . total = 6 ... | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | b | add(multiply(divide(100, const_10), multiply(const_2, const_4)), divide(100, const_10)) | divide(n0,const_10)|multiply(const_2,const_4)|multiply(#0,#1)|add(#0,#2)| | general |
local kennel has cats and dogs in the ratio of 5 : 10 . if there are 40 fewer cats than dogs , how many dogs are in the kennel ? | "lets work with the data given to us . we know that there ratio of cats to dogs is 5 : 10 or cats 5 dogs 10 we can write number of cats as 5 x and number of dogs as 10 x and we know that 10 x - 5 x = 40 ( therefore 5 x = 40 = > x = 8 ) then # of dogs = 10 x 8 = 80 answer is a" | a ) 80 , b ) 70 , c ) 60 , d ) 50 , e ) 45 | a | multiply(40, const_3.0) | multiply(const_3.0,n2)| | other |
for how many unique pairs of nonnegative integers { a , b } is the equation a ^ 2 - b ^ 2 = 45 true ? | "answer b ( a + b ) ( a - b ) = 45 3 cases for ( a + b ) , ( a - b ) 45 , 1 15 , 3 9 , 5 answer b" | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | divide(log(45), log(add(const_4, const_1))) | add(const_1,const_4)|log(n2)|log(#0)|divide(#1,#2)| | general |
if the number 892 , 142,24 x is divisible by 11 , what must be the value of x ? | "multiplication rule of 11 : ( sum of digits at odd places - sum of digits at even places ) should be divisible by 11 given number : 892 , 142,24 x sum of digits at odd places = 8 + 2 + 4 + 2 + x = 16 + x ( i ) sum of digits at even places = 9 + 1 + 2 + 4 = 16 ( ii ) ( i ) - ( ii ) = 16 + x - 16 = x - 0 hence x should ... | a ) 1 , b ) 2 , c ) 3 , d ) 0 , e ) 5 | d | multiply(multiply(multiply(add(multiply(const_3, const_10), const_1), const_2), const_4), 11) | multiply(const_10,const_3)|add(#0,const_1)|multiply(#1,const_2)|multiply(#2,const_4)|multiply(n2,#3)| | general |
a 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2 . 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more . at the end of the two removal and replacement , what is the ratio w of milk and water in the resultant mixture ? | "he 20 litre mixture contains milk and water in the ratio of 3 : 2 . therefore , there will be 12 litres of milk in the mixture and 8 litres of water in the mixture . step 1 . when 10 litres of the mixture is removed , 6 litres of milk is removed and 4 litres of water is removed . therefore , there will be 6 litres of ... | a ) 17 : 3 , b ) 9 : 1 , c ) 3 : 17 , d ) 5 : 3 , e ) 11 : 2 | b | divide(add(multiply(divide(add(multiply(divide(3, add(3, 2)), subtract(20, 10)), 10), 20), subtract(20, 10)), 10), multiply(divide(multiply(divide(2, add(3, 2)), subtract(20, 10)), 20), subtract(20, 10))) | add(n1,n2)|subtract(n0,n3)|divide(n1,#0)|divide(n2,#0)|multiply(#2,#1)|multiply(#3,#1)|add(n3,#4)|divide(#5,n0)|divide(#6,n0)|multiply(#7,#1)|multiply(#8,#1)|add(n3,#10)|divide(#11,#9)| | general |
of the families in city x in 1992 , 60 percent owned a personal computer . the number of families in city x owning a computer in 1993 was 50 percent greater than it was in 1992 , and the total number of families in city x was 3 percent greater in 1993 than it was in 1992 . what percent of the families in city x owned a... | "say a 100 families existed in 1994 then the number of families owning a computer in 1994 - 60 number of families owning computer in 1998 = 60 * 150 / 100 = 90 number of families in 1998 = 103 the percentage = 90 / 103 * 100 = 87.37 % . answer : e" | a ) 68.99 % , b ) 66.55 % , c ) 91.23 % , d ) 77.77 % , e ) 87.37 % | e | multiply(const_100, divide(divide(multiply(add(50, const_100), 60), const_100), add(const_100, 3))) | add(n3,const_100)|add(n5,const_100)|multiply(n1,#0)|divide(#2,const_100)|divide(#3,#1)|multiply(#4,const_100)| | general |
find the value of ( β 1.21 ) / ( β 0.81 ) + ( β 1.00 ) / ( β 0.49 ) is | ( β 1.21 ) / ( β 0.81 ) + ( β 1.00 ) / ( β 0.49 ) 11 / 9 + 10 / 7 = > 2.65 answer is a | a ) 2.65 , b ) 145 / 63 , c ) 155 / 63 , d ) 125 / 63 , e ) 185 / 63 | a | add(divide(sqrt(1.21), sqrt(0.81)), divide(sqrt(1), sqrt(0.49))) | sqrt(n0)|sqrt(n1)|sqrt(n2)|sqrt(n3)|divide(#0,#1)|divide(#2,#3)|add(#4,#5) | general |
louie takes out a 3 - month loan of $ 1000 . the lender charges him 10 % interest per month compounded monthly . the terms of the loan state that louie must repay the loan in 3 equal monthly payments . to the nearest dollar , how much does louis have to pay each month ? | after 1 st month : ( 1000 ) ( 1.1 ) - x = 1100 - x after 2 nd month : ( 1100 - x ) ( 1.1 ) - x = 1210 - 2.21 x after 3 rd month : ( 1210 - 2.21 x ) ( 1.1 ) - x = 1331 - 3.31 x now , the amount after the last payment in 3 rd month must bring the total to 0 . hence : 1331 - 3.31 x = 0 x = 1331 / 3.31 = 402.11 the answer ... | a ) 333 , b ) 383 , c ) 402 , d ) 433 , e ) 483 | c | divide(add(1000, add(add(multiply(multiply(1000, 3), divide(10, const_100)), multiply(multiply(multiply(1000, 3), divide(10, const_100)), divide(10, const_100))), const_1)), divide(add(add(multiply(multiply(1000, 3), divide(10, const_100)), multiply(multiply(multiply(1000, 3), divide(10, const_100)), divide(10, const_1... | divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#0,#2)|add(#2,#3)|add(#4,const_1)|add(n1,#5)|divide(#5,const_100)|divide(#6,#7) | general |
if the perimeter of a rectangular garden is 600 m , its length when its breadth is 200 m is ? | "2 ( l + 200 ) = 600 = > l = 100 m answer : d" | a ) 299 m , b ) 777 m , c ) 200 m , d ) 100 m , e ) 128 m | d | subtract(divide(600, const_2), 200) | divide(n0,const_2)|subtract(#0,n1)| | physics |
rs 3000 is divided into two parts such that one part is put out at 3 % and the other at 5 % . if the annual interest earned from both the investments be rs 144 , find the first part . | explanation : average rate = ( 144 / 3000 ) * 100 = 4.8 ratio = 2 : 18 so , first part = ( 2 / 20 ) * 3000 = rs 300 . answer : c | a ) s 400 , b ) s 280 , c ) s 300 , d ) s 350 , e ) s 310 | c | divide(subtract(multiply(3000, divide(5, const_100)), 144), subtract(divide(5, const_100), divide(3, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3) | gain |
the area of a square is equal to three times the area of a rectangle of dimensions 3 cm * 9 cm . what is the perimeter of the square ? | "area of the square = s * s = 3 ( 3 * 9 ) = > s = 9 cm perimeter of the square = 4 * 9 = 36 cm . answer : option b" | a ) 30 , b ) 36 , c ) 46 , d ) 66 , e ) 76 | b | multiply(sqrt(multiply(rectangle_area(3, 9), divide(9, const_2))), const_4) | divide(n1,const_2)|rectangle_area(n0,n1)|multiply(#0,#1)|sqrt(#2)|multiply(#3,const_4)| | geometry |
a machine , working at a constant rate , manufactures 18 dies in 25 minutes . how many dies does it make in 1 hr 15 min ? | "change 1 hr 15 min to 75 min . for this , we need to set up a simple proportion of dies per time 18 / 25 = s / 75 the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . we can cancel the common factor of 25 in the two denominators . 18... | a ) 55 , b ) 53 , c ) 54 , d ) 52 , e ) 50 | c | multiply(divide(add(multiply(1, const_60), 15), 25), 18) | multiply(n2,const_60)|add(n3,#0)|divide(#1,n1)|multiply(n0,#2)| | physics |
a room 5.44 m long and 3.74 m broad is to be paved with square tiles . the least number of square tiles required to cover the floor is : | area of the room = 544 * 374 sq . cm size of largest square tile = hcf of 544 cm & 374 cm area of 1 tile = 34 * 34 sq . cm therefore , number of tiles = ( 544 * 374 / 34 * 34 ) = 176 answer : a | ['a ) 176', 'b ) 192', 'c ) 184', 'd ) 162', 'e ) 172'] | a | divide(multiply(multiply(5.44, const_100), multiply(3.74, const_100)), multiply(add(add(add(const_10, const_10), const_10), const_4), add(add(add(const_10, const_10), const_10), const_4))) | add(const_10,const_10)|multiply(n0,const_100)|multiply(n1,const_100)|add(#0,const_10)|multiply(#1,#2)|add(#3,const_4)|multiply(#5,#5)|divide(#4,#6) | physics |
find the maximum value of n such that 50 ! is perfectly divisible by 2520 ^ n | 2520 = 2 ^ 3 * 3 ^ 2 * 5 * 7 here , 7 is the highest prime , so find the no . of 7 ' s in 50 ! only no . of 7 ' s in 50 ! = [ 50 / 7 ] + [ 50 / 7 ^ 2 ] = 7 + 1 = 8 answer : b | a ) 7 , b ) 8 , c ) 9 , d ) 6 , e ) 5 | b | add(divide(50, const_10), const_3) | divide(n0,const_10)|add(#0,const_3) | general |
in two triangle , the ratio of the areas is 4 : 3 and that of their heights is 3 : 4 the ratio of their bases is | explanation : given a 1 / a 2 = 4 / 3 , h 1 / h 2 = 3 / 4 also , we know that a 1 / a 2 = ( 1 / 2 Γ b 1 Γ h 1 ) / ( 1 / 2 Γ b 2 Γ h 2 ) on substituting in above equation , we get b 1 / b 2 = 16 / 9 therefore b 1 : b 2 = 16 : 9 answer : option a | ['a ) 16 : 9', 'b ) 9 : 16', 'c ) 9 : 12', 'd ) 16 : 12', 'e ) 4 : 12'] | a | divide(multiply(4, 4), multiply(3, 3)) | multiply(n0,n0)|multiply(n1,n1)|divide(#0,#1) | geometry |
a number exceeds by 30 from its 3 / 8 part . then the number is ? | "x β 3 / 8 x = 30 x = 48 answer : e" | a ) a ) 32 , b ) b ) 35 , c ) c ) 39 , d ) d ) 40 , e ) e ) 48 | e | divide(multiply(30, 8), subtract(8, 3)) | multiply(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | general |
the h . c . f . of two numbers is 12 and their l . c . m . is 520 . if one of the numbers is 480 , then the other is : | "other number = ( 12 x 520 ) / 480 = 13 . answer : a" | a ) 13 , b ) 18 , c ) 21 , d ) 24 , e ) 38 | a | multiply(12, 480) | multiply(n0,n2)| | physics |
the radius of a cylinder is 12 m , height 21 m . the lateral surface area of the cylinder is : | lateral surface area = 2 Ο rh = 2 Γ 22 / 7 Γ 12 Γ 21 = 44 Γ 36 = 1584 m ( power 2 ) answer is a . | ['a ) 1584', 'b ) 1854', 'c ) 1458', 'd ) 1485', 'e ) none of them'] | a | multiply(circumface(12), 21) | circumface(n0)|multiply(n1,#0) | geometry |
in an election between two candidates , the first candidate got 80 % of the votes and the second candidate got 240 votes . what was the total number of votes ? | "let v be the total number of votes . 0.2 v = 240 v = 1200 the answer is c ." | a ) 600 , b ) 900 , c ) 1200 , d ) 1500 , e ) 1800 | c | divide(multiply(240, const_100), subtract(const_100, 80)) | multiply(n1,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain |
how many of the positive factors of 19 are not factors of 29 ? | "factors of 19 - 1 , 19 factors of 29 - 1 , 29 comparing both , we have three factors of 19 which are not factors of 29 - 19 , the answer is b" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | divide(29, 19) | divide(n1,n0)| | other |
a retailer buys a radio for rs 225 . his overhead expenses are rs 30 . he sellis the radio for rs 300 . the profit percent of the retailer is | "explanation : cost price = ( 225 + 30 ) = 255 sell price = 300 gain = ( 45 / 255 ) * 100 = 17.6 % . answer : d" | a ) 10 % , b ) 50 % , c ) 25 % , d ) 17.6 % , e ) none of these | d | subtract(multiply(divide(300, add(225, 30)), const_100), const_100) | add(n0,n1)|divide(n2,#0)|multiply(#1,const_100)|subtract(#2,const_100)| | gain |
the number of boxes in a warehouse can be divided evenly into 6 equal shipments by boat or 27 equal shipments by truck . what is the smallest number of boxes that could be in the warehouse ? | "explanations 1 ) this tells us that the number of boxes is evenly divisible by both 6 and 27 ; in other words , it β s a common multiple of 6 and 27 . the question says : what β s the smallest value it could have ? in other words , what β s the lcm of 6 and 27 ? ( this question is one example of a real - world set - u... | a ) 27 , b ) 33 , c ) 54 , d ) 81 , e ) 162 | c | multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_3), 6) | multiply(const_2,const_2)|multiply(#0,const_2)|multiply(#1,const_3)|multiply(n0,#2)| | general |
if it is assumed that 60 percent of those who receive a questionnaire by mail will respond and 210 responses are needed , what is the minimum number of questionnaires that should be mailed ? | let x be the minimum number of questionnaires to be mailed . 0.6 x = 210 x = 350 the answer is c . | a ) 290 , b ) 320 , c ) 350 , d ) 380 , e ) 410 | c | divide(210, divide(60, const_100)) | divide(n0,const_100)|divide(n1,#0) | gain |
a lemonade stand sold only small and large cups of lemonade on tuesday . 4 / 5 of the cups sold were small and the rest were large . if the large cups were sold for 7 / 6 as much as the small cups , what fraction of tuesday ' s total revenue was from the sale of large cups ? | "a simpler way i guess would be to think that in total 5 cups were sold . out of which 4 are small and 1 is large . now let the small ones cost $ 6 . so the large ones would cost $ 7 . so , 4 * 6 = 24 and 1 * 7 = 7 . total revenue was 24 + 7 = 31 and large cup sales as found above is 7 therefore answer is 7 / 31 d" | a ) ( a ) 7 / 16 , b ) ( b ) 7 / 15 , c ) ( c ) 10 / 21 , d ) ( d ) 7 / 31 , e ) ( e ) 1 / 2 | d | divide(multiply(subtract(const_10, multiply(divide(4, 5), const_10)), multiply(divide(7, 6), multiply(divide(4, 5), const_10))), add(multiply(multiply(divide(4, 5), const_10), multiply(divide(4, 5), const_10)), multiply(subtract(const_10, multiply(divide(4, 5), const_10)), multiply(divide(7, 6), multiply(divide(4, 5), ... | divide(n2,n3)|divide(n0,n1)|multiply(#1,const_10)|multiply(#0,#2)|multiply(#2,#2)|subtract(const_10,#2)|multiply(#3,#5)|add(#4,#6)|divide(#6,#7)| | general |
96 % of the population of a city is 23040 . find the total population of the city ? | "population consider as x x * ( 96 / 100 ) = 23040 x = 240 * 100 = = > 24000 answer c" | a ) 30000 , b ) 28000 , c ) 24000 , d ) 25000 , e ) 25500 | c | multiply(divide(const_100, 96), 23040) | divide(const_100,n0)|multiply(n1,#0)| | gain |
the average weight of 10 men is increased by 2 Β½ kg when one of the men who weighs 58 kg is replaced by a new man . what is the weight of the new man ? | since the average has increased by 2.5 kg , the weight of the man who stepped in must be equal to 58 + 10 x 2.5 58 + 25 = 83 kg ans : ' b ' | a ) 80 kg , b ) 83 kg , c ) 70 kg , d ) 75 kg , e ) 85 kg | b | add(58, multiply(10, add(2, divide(const_1, const_2)))) | divide(const_1,const_2)|add(n1,#0)|multiply(n0,#1)|add(n2,#2) | general |
if 6 log ( 4 * 5 ^ 2 ) = x , find x | 6 ( log 2 ^ 2 * 5 ^ 2 ) = x 6 log ( 5 * 2 ) ^ 2 = x 6 * 2 log ( 5 * 2 ) = x 12 log 10 = x log 10 base 10 = 1 so 12 * 1 = x x = 12 answer : d | a ) 10 , b ) 11 , c ) 7 , d ) 12 , e ) 9 | d | power(multiply(4, power(5, 2)), 6) | power(n2,n3)|multiply(n1,#0)|power(#1,n0) | general |
in the rectangular coordinate system , what is the x - intercept of a line passing through ( 10 , 3 ) and ( β 10 , β 7 ) ? | "slope = rise / run = 10 / 20 = 1 / 2 the equation of the line is y = ( 1 / 2 ) x + b 3 = ( 1 / 2 ) ( 10 ) + b b = - 2 the equation of the line is y = ( 1 / 2 ) x - 2 to find the x - intercept , let y = 0 : 0 = ( 1 / 2 ) x - 2 x = 4 the answer is a ." | a ) 4 , b ) 2 , c ) 0 , d ) β 2 , e ) β 4 | a | divide(subtract(multiply(divide(subtract(negate(7), 3), subtract(negate(10), 10)), 10), 3), divide(subtract(negate(7), 3), subtract(negate(10), 10))) | negate(n3)|negate(n2)|subtract(#0,n1)|subtract(#1,n0)|divide(#2,#3)|multiply(n0,#4)|subtract(#5,n1)|divide(#6,#4)| | general |
the price of a t . v . set worth rs . 40000 is to be paid in 20 installments of rs . 2500 each . if the rate of interest be 6 % per annum , and the first installment be paid at the time of purchase , then the value of the last installment covering the interest as well will be ? | "money paid in cash = rs . 2500 balance payment = ( 40000 - 2500 ) = rs . 37500 . answer : d" | a ) 29997 , b ) 28088 , c ) 27098 , d ) 37500 , e ) 2799 | d | subtract(40000, 2500) | subtract(n0,n2)| | gain |
when greenville state university decided to move its fine arts collection to a new library , it had to package the collection in 20 - inch by 20 - inch by 12 - inch boxes . if the university pays $ 0.50 for every box , and if the university needs 2.4 million cubic inches to package the collection , what is the minimum ... | "the volume of each box is 20 * 20 * 12 = 4800 cubic inches . number of boxes = 2 , 400,000 / 4800 = 500 boxes total cost = 500 Γ $ 0.5 = $ 250 the answer is a ." | a ) $ 250 , b ) $ 275 , c ) $ 510 , d ) $ 1,250 , e ) $ 2,550 | a | multiply(divide(multiply(2.4, multiply(const_1000, const_1000)), multiply(multiply(20, 20), 12)), 0.50) | multiply(const_1000,const_1000)|multiply(n0,n0)|multiply(n4,#0)|multiply(n2,#1)|divide(#2,#3)|multiply(n3,#4)| | general |
boy sells a book for rs . 450 he gets a loss of 10 % , to gain 10 % , what should be the sp ? | "find selling price to gain 10 % . now , we are asked to find selling price to gain 10 % profit . hint : selling price = ( 100 + gain % ) Γ c . p . 100 selling price = ( 100 + 10 ) Γ 500 100 selling price = ( 110 ) Γ 500 100 therefore , selling price = rs . 550 c" | a ) rs . 320 , b ) rs . 450 , c ) rs . 550 , d ) rs . 640 , e ) rs . 680 | c | add(divide(450, subtract(const_1, divide(10, const_100))), multiply(divide(450, subtract(const_1, divide(10, const_100))), divide(10, const_100))) | divide(n1,const_100)|divide(n2,const_100)|subtract(const_1,#0)|divide(n0,#2)|multiply(#3,#1)|add(#3,#4)| | gain |
in country z , 10 % of the people do not have a university diploma but have the job of their choice , and 25 % of the people who do not have the job of their choice have a university diploma . if 20 % of the people have the job of their choice , what percent of the people have a university diploma ? | "setting up a matrix is how i solve this one . diploma no diploma totals job of choice w / diploma job of choice w / o diploma = 10 % job of choice total = 20 % not job of choice with diploma = . 25 x not job of choice w / o diploma = . 75 x total not job of choice = x total with diploma total without diploma total cit... | a ) 30 % , b ) 45 % , c ) 55 % , d ) 65 % , e ) 75 % | a | add(divide(multiply(25, subtract(const_100, 20)), const_100), subtract(20, 10)) | subtract(const_100,n2)|subtract(n2,n0)|multiply(n1,#0)|divide(#2,const_100)|add(#3,#1)| | gain |
a batsman makes a score of 84 runs in the 17 th inning and thus increases his averages by 3 . find his average after 17 th inning ? | "let the average after 17 th inning = x then average after 16 th inning = ( x - 3 ) therefore 16 ( x - 3 ) + 84 = 17 x therefore x = 36 answer : c" | a ) 19 , b ) 29 , c ) 36 , d ) 49 , e ) 59 | c | add(subtract(84, multiply(17, 3)), 3) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
of the 120 passengers on flight 750 , 40 % are female . 10 % of the passengers sit in first class , and the rest of the passengers sit in coach class . if 1 / 3 of the passengers in first class are male , how many females are there in coach class ? | "number of passengers on flight = 120 number of female passengers = . 4 * 120 = 48 number of passengers in first class = ( 10 / 100 ) * 120 = 12 number of passengers in coach class = ( 90 / 100 ) * 120 = 108 number of male passengers in first class = 1 / 3 * 12 = 4 number of female passengers in first class = 12 - 4 = ... | a ) 40 , b ) 48 , c ) 50 , d ) 52 , e ) 56 | a | subtract(multiply(120, divide(40, const_100)), subtract(multiply(120, divide(10, const_100)), divide(multiply(120, divide(10, const_100)), 3))) | divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n0,#1)|divide(#3,n5)|subtract(#3,#4)|subtract(#2,#5)| | gain |
the maximum numbers of students among them 451 pens and 410 toys can be distributed in such a way that each student gets the same number of pens and same number of toys is | "olution required number of students . = h . c . f of 451 and 410 . Γ’ β¬ ΒΉ = Γ’ β¬ ΒΊ 41 . answer a" | a ) 41 , b ) 910 , c ) 1001 , d ) 1911 , e ) none | a | subtract(451, 410) | subtract(n0,n1)| | general |
find the average of all the numbers between 4 and 32 which are divisible by 5 . | "solution average = ( 5 + 10 + 15 + 20 + 25 + 30 ) / 6 ) = 105 / 6 = 17.5 . answer a" | a ) 17.5 , b ) 20 , c ) 24 , d ) 30 , e ) 32 | a | divide(add(add(4, const_4), subtract(32, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general |
a cylinder with base radius of 8 cm and height of 2 cm is melted to form a cone of height 6 cm . the radius of the cone will be : | sol . let the radius of the cone be r cm . then , 1 / 3 β * r Β² * 6 = β * 8 * 8 * 2 β r Β² = [ 8 * 8 * 2 * 3 / 6 ] = 64 β r = 8 cm . answer a | ['a ) 8 cm', 'b ) 9 cm', 'c ) 10 cm', 'd ) 11 cm', 'e ) none'] | a | sqrt(divide(divide(multiply(volume_cylinder(8, 2), const_3), 6), const_pi)) | volume_cylinder(n0,n1)|multiply(#0,const_3)|divide(#1,n2)|divide(#2,const_pi)|sqrt(#3) | geometry |
if @ is a binary operation defined as the difference between an integer n and the product of n and 5 , then what is the largest positive integer n such that the outcome of the binary operation of n is less than 18 ? | "@ ( n ) = 5 n - n we need to find the largest positive integer such that 5 n - n < 18 . then 4 n < 18 and n < 4.5 the largest possible integer is n = 4 . the answer is d ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | floor(divide(18, subtract(5, const_1))) | subtract(n0,const_1)|divide(n1,#0)|floor(#1)| | general |
a runs twice as fast as b and gives b a start of 60 m . how long should the racecourse be so that a and b might reach in the same time ? | ratio of speeds of a and b is 2 : 1 b is 60 m away from a but we know that a covers 1 meter ( 2 - 1 ) more in every second than b the time taken for a to cover 60 m is 60 / 1 = 60 m so the total time taken by a and b to reach = 2 * 60 = 120 m answer : a | a ) 120 m . , b ) 80 m . , c ) 150 m . , d ) 100 m . , e ) none of the above | a | multiply(60, const_2) | multiply(n0,const_2) | physics |
how long does a train 100 m long running at the speed of 80 km / hr takes to cross a bridge 142 m length ? | "speed = 80 * 5 / 18 = 22.2 m / sec total distance covered = 100 + 142 = 242 m . required time = 242 / 22.2 ' = 10.9 sec . answer : b" | a ) 10.7 sec , b ) 10.9 sec , c ) 10.1 sec , d ) 15.1 sec , e ) 12.7 sec | b | divide(add(100, 142), multiply(80, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1)| | physics |
a can complete a project in 20 days and b can complete the same project in 30 days . if a and b start working on the project together and a quits 5 days before the project is completed , in how many days total will the project be completed ? | "a ' s rate is 1 / 20 of the project per day . b ' s rate is 1 / 30 of the project per day . the combined rate is 1 / 12 of the project per day . in the last 5 days , b can do 1 / 6 of the project . thus a and b must complete 5 / 6 of the project , which takes 10 days . the total number of days is 10 + 5 = 15 . the ans... | a ) 12 , b ) 15 , c ) 18 , d ) 21 , e ) 24 | b | add(divide(subtract(const_1, multiply(divide(const_1, 30), 5)), add(divide(const_1, 20), divide(const_1, 30))), 5) | divide(const_1,n1)|divide(const_1,n0)|add(#1,#0)|multiply(n2,#0)|subtract(const_1,#3)|divide(#4,#2)|add(n2,#5)| | physics |
angelo and isabella are both salespersons . in any given week , angelo makes $ 580 in base salary plus 8 percent of the portion of his sales above $ 1,000 for that week . isabella makes 10 percent of her total sales for any given week . for what amount of weekly sales would angelo and isabella earn the same amount of m... | "let the weekly sales of both = x 580 + ( x β 1000 ) 8 / 100 = 10 / 100 x x = 25000 answer : c" | a ) 23,500 , b ) 24,500 , c ) 25,000 , d ) 26,500 , e ) 27,500 | c | floor(divide(divide(subtract(580, multiply(1,000, divide(8, const_100))), subtract(divide(10, const_100), divide(8, const_100))), 1,000)) | divide(n1,const_100)|divide(n3,const_100)|multiply(#0,n2)|subtract(#1,#0)|subtract(n0,#2)|divide(#4,#3)|divide(#5,n2)|floor(#6)| | general |
a dessert recipe calls for 50 % melted chocolate and 50 % raspberry puree to make a particular sauce . a chef accidentally makes 15 cups of the sauce with 30 % melted chocolate and 70 % raspberry puree instead . how many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce t... | "yes , we assume that the mix is homogeneous . otherwise , we will not be able to solve the question . you have 15 cups of sauce with 30 % chocolate . you also have unlimited amount of pure chocolate sauce . now you need to mix these two in such a way that you get total 15 cups of sauce with 50 % chocolate . using scal... | a ) 1.5 , b ) 2.14 , c ) 3 , d ) 4.5 , e ) 5 | b | divide(divide(subtract(add(50, 50), 50), subtract(50, 30)), const_2) | add(n0,n0)|subtract(n0,n3)|subtract(#0,n0)|divide(#2,#1)|divide(#3,const_2)| | gain |
the price of commodity x increases by 45 cents every year , while the price of commodity y increases by 20 cents every year . in 2001 , the price of commodity x was $ 5.20 and the price of commodity y was $ 7.30 . in which year will the price of commodity x be 35 cents less than the price of commodity y ? | "the price of commodity x increases 25 cents each year relative to commodity y . the price difference is $ 2.10 and commodity x needs to be 35 cents less than commodity y . $ 1.75 / 25 cents = 7 years the answer is 2001 + 7 years = 2008 . the answer is c ." | a ) 2006 , b ) 2007 , c ) 2008 , d ) 2009 , e ) 2010 | c | add(2001, divide(add(divide(35, const_100), subtract(7.30, 5.20)), subtract(divide(45, const_100), subtract(7.30, 5.20)))) | divide(n5,const_100)|divide(n0,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#1,#2)|divide(#3,#4)|add(n2,#5)| | general |
if log 1087.5 = 1.9421 , then the number of digits in ( 875 ) 10 is ? | "x = ( 875 ) 10 = ( 87.5 x 10 ) 10 therefore , log 10 x = 10 ( log 1087.5 + 1 ) = 10 ( 1.9421 + 1 ) = 10 ( 2.9421 ) = 29.421 x = antilog ( 29.421 ) therefore , number of digits in x = 30 . answer : a" | a ) 30 , b ) 28 , c ) 27 , d ) 26 , e ) 25 | a | add(multiply(const_4, 1.9421), divide(log(const_100), log(const_10))) | log(const_100)|log(const_10)|multiply(n1,const_4)|divide(#0,#1)|add(#3,#2)| | other |
a river 4 m deep and 65 m wide is flowing at the rate of 6 kmph the amount of water that runs into the sea per minute is ? | "rate of water flow - 6 kmph - - 6000 / 60 - - 100 m / min depth of river - - 4 m width of river - - 65 m vol of water per min - - 100 * 4 * 65 - - - 26000 answer b" | a ) 25000 , b ) 26000 , c ) 27000 , d ) 28000 , e ) 29000 | b | divide(multiply(multiply(4, 65), multiply(6, const_1000)), multiply(const_1, const_60)) | multiply(n0,n1)|multiply(n2,const_1000)|multiply(const_1,const_60)|multiply(#0,#1)|divide(#3,#2)| | physics |
in a garden , there are 10 rows and 17 columns of mango trees . the distance between the two trees is 2 metres and a distance of one metre is left from all sides of the boundary of the garden . the length of the garden is | "explanation : each row contains 17 plants . there are 15 gapes between the two corner trees ( 16 x 2 ) metres and 1 metre on each side is left . therefore length = ( 32 + 2 ) m = 34 m . answer : c" | a ) 20 m , b ) 22 m , c ) 34 m , d ) 26 m , e ) 28 m | c | add(add(multiply(subtract(17, const_1), 2), divide(10, 2)), divide(10, 2)) | divide(n0,n2)|subtract(n1,const_1)|multiply(n2,#1)|add(#0,#2)|add(#3,#0)| | physics |
a person spent rs . 6,040 from his salary on food and 8,000 on house rent . after that he was left with 70 % of his monthly salary . what is his monthly salary ? | "total money spent on food and house rent = 6,040 + 8,000 = 14,040 which is 100 - 70 = 30 % of his monthly salary β΄ his salary = 14040 x 100 / 30 = 46800 answer : a" | a ) 46,800 , b ) 66,800 , c ) 56,800 , d ) 26,800 , e ) 76,800 | a | divide(add(multiply(add(const_4, const_1), const_100), 70), sqrt(const_100)) | add(const_1,const_4)|sqrt(const_100)|multiply(#0,const_100)|add(n2,#2)|divide(#3,#1)| | gain |
a basketball is dropped from a height of 40 feet . if it bounces back up to a height that is exactly half of its previous height , and it stops bouncing after hitting the ground for the fourth time , then how many total feet will the ball have traveled after 3 full bounces . | "initial distance = 40 feet first bounce = 20 feet up + 20 feet down = 40 feet second bouche = 10 feet up + 10 feet down = 20 feet third bounce = 5 feet up and 5 feet down = 10 feet total distance covered = 40 + 40 + 20 + 10 = 110 answer is c" | a ) 50 , b ) 55 , c ) 110 , d ) 75 , e ) 80 | c | subtract(subtract(subtract(subtract(multiply(add(divide(divide(divide(divide(40, 3), 3), 3), 3), add(add(add(add(add(add(40, divide(40, 3)), divide(40, 3)), divide(divide(40, 3), 3)), divide(divide(40, 3), 3)), divide(divide(divide(40, 3), 3), 3)), divide(divide(divide(40, 3), 3), 3))), 3), divide(divide(divide(40, 3),... | divide(n0,n1)|add(n0,#0)|divide(#0,n1)|add(#1,#0)|divide(#2,n1)|add(#3,#2)|divide(#4,n1)|add(#5,#2)|add(#7,#4)|add(#8,#4)|add(#9,#6)|multiply(n1,#10)|subtract(#11,#4)|subtract(#12,#4)|subtract(#13,#4)|subtract(#14,#4)| | general |
80 370 860 1550 ? 3530 | "10 ^ 2 - 20 = 80 20 ^ 2 - 30 = 370 30 ^ 2 - 40 = 860 40 ^ 2 - 50 = 1550 50 ^ 2 - 60 = 2440 60 ^ 2 - 70 = 3530 . answer : b" | a ) 900 , b ) 2440 , c ) 750 , d ) 244 , e ) 960 | b | divide(multiply(add(80, divide(370, 860)), 1550), const_100) | divide(n1,n2)|add(n0,#0)|multiply(n3,#1)|divide(#2,const_100)| | general |
a sum fetched a total simple interest of 4052.25 at the rate of 9 % . p . a . in 5 years . what is the sum ? | principal = ( 100 x 4052.25 ) / ( 9 x 5 ) = 405225 / 45 = 9005 . answer a | a ) 9005 , b ) 8925 , c ) 2345 , d ) 6474 , e ) 8723 | a | divide(divide(multiply(4052.25, const_100), 9), 5) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2) | gain |
a rotameter is a device that measures flow of liquid and gases . when measuring liquid phase flows , 2.5 inches represent 60 liters per minute of liquid . with gas measurements the rotameter moves 50 % of the movement he moves with the liquid phase . how many liters of gas passed through the rotameter if it measured 4 ... | in case of liquid - 2.5 inches represents 60 lit / min . in case of gas - 50 % of 2.5 inches represents 60 lit / min 1.25 inches represents 60 lit / min 4 inches will represent 60 * 4 / 1.25 = 192 b is the answer | a ) 176 , b ) 192 , c ) 202 , d ) 218 , e ) 284 | b | divide(divide(multiply(60, 4), 2.5), divide(50, const_100)) | divide(n2,const_100)|multiply(n1,n3)|divide(#1,n0)|divide(#2,#0) | physics |
if a trader sold two cars each at rs . 325475 and gains 10 % on the first and loses 10 % on the second , then his profit or loss percent on the whole is ? | "sp of each car is rs . 325475 , he gains 10 % on first car and losses 10 % on second car . in this case , there will be loss and percentage of loss is given by = [ ( profit % ) ( loss % ) ] / 100 = ( 10 ) ( 10 ) / 100 % = 1.00 % answer : e" | a ) 1.44 % , b ) 1.74 % , c ) 1.84 % , d ) 1.47 % , e ) 1.00 % | e | divide(multiply(10, 10), const_100) | multiply(n1,n1)|divide(#0,const_100)| | gain |
how many positive integers less than 130 are there such that they are multiples of 13 or multiples of 12 but not both ? | "for 13 : 13 . . . 130 = 13 * 10 for 12 : 12 . . . 130 = 12 * 10 but there is one integer 13 * 12 . so n = ( 10 ) + ( 10 ) = 20 c" | a ) 16 , b ) 18 , c ) 20 , d ) 24 , e ) 28 | c | divide(factorial(subtract(add(const_4, 13), const_1)), multiply(factorial(13), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
the marks obtained by polly and sandy are in the ratio 5 : 6 and those obtained by sandy and willy are in the ratio of 3 : 2 . the marks obtained by polly and willy are in the ratio of . . . ? | "polly : sandy = 5 : 6 sandy : willy = 3 : 2 = 6 : 4 polly : sandy : willy = 5 : 6 : 4 polly : willy = 5 : 4 the answer is b ." | a ) 3 : 2 , b ) 5 : 4 , c ) 7 : 6 , d ) 9 : 8 , e ) 11 : 10 | b | divide(multiply(5, 3), multiply(6, 2)) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)| | other |
what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 7 ? | what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 7 ? 9 ^ 7 = ( 3 ^ 2 ) ^ 7 = 3 ^ 14 b . 14 | a ) 5 , b ) 14 , c ) 10 , d ) 20 , e ) 30 | b | multiply(subtract(9, 7), 7) | subtract(n1,n2)|multiply(n2,#0) | general |
if 70 percent of a class answered the first question on a certain test correctly , 55 percent answered the second question on the test correctly , and 20 percent answered neither of the questions correctly , what percent answered both correctly ? | "70 % answered the first question correctly and 20 % answered neither correctly . then 10 % missed the first question but answered the second question correctly . then the percent who answered both correctly is 55 % - 10 % = 45 % . the answer is a ." | a ) 45 % , b ) 40 % , c ) 35 % , d ) 30 % , e ) 25 % | a | subtract(add(add(70, 55), 20), const_100) | add(n0,n1)|add(n2,#0)|subtract(#1,const_100)| | other |
each machine of type a has 3 steel parts and 2 chrome parts . each machine of type b has 2 steel parts and 4 chrome parts . if a certain group of type a and type b machines has a total of 50 steel parts and 66 chrome parts , how many machines are in the group | "look at the below representation of the problem : steel chrome total a 3 2 50 > > no . of type a machines = 50 / 5 = 10 b 2 4 66 > > no . of type b machines = 66 / 6 = 11 so the answer is 21 i . e b . hope its clear ." | a ) 11 , b ) 21 , c ) 31 , d ) 61 , e ) 51 | b | add(divide(66, add(4, 2)), divide(50, add(3, 2))) | add(n1,n3)|add(n0,n1)|divide(n5,#0)|divide(n4,#1)|add(#2,#3)| | general |
a batsman in his 12 th innings makes a score of 75 and thereby increases his average by 1 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 1 ) + 75 β΄ x = 64 answer d" | a ) 42 , b ) 43 , c ) 44 , d ) 64 , e ) 46 | d | add(subtract(75, multiply(12, 1)), 1) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)| | general |
students at a school were on average 180 cm tall . the average female height was 170 cm , and the average male height was 188 cms . what was the ratio of men to women ? | "we ' re given a few facts to work with : 1 ) the average height of the females is 170 cm 2 ) the average height of the males is 188 cm 3 ) the average of the group is 180 cm we ' re asked for the ratio of men to women . w = number of women m = number of men ( 170 w + 188 m ) / ( w + m ) = 180 170 w + 188 m = 180 w + 1... | a ) 5 : 2 , b ) 5 : 1 , c ) 4 : 3 , d ) 5 : 4 , e ) 3 : 1 | d | divide(subtract(180, 170), subtract(188, 180)) | subtract(n0,n1)|subtract(n2,n0)|divide(#0,#1)| | general |
the average ( arithmetic mean ) of the integers from 100 to 200 , inclusive , is how much lesser than the average of the integers from 150 to 950 , inclusive ? | "for an ap the mean or average of series is average of first and last term . so , average of numbers between 150 to 950 , inclusive = ( 150 + 950 ) / 2 = 550 average of numbers between 100 to 200 , inclusive = ( 100 + 200 ) / 2 = 150 difference = 550 - 150 = 400 answer is e" | a ) 350 , b ) 475 , c ) 300 , d ) 425 , e ) 400 | e | subtract(divide(add(100, 200), const_2), divide(add(150, 950), const_2)) | add(n0,n1)|add(n2,n3)|divide(#0,const_2)|divide(#1,const_2)|subtract(#2,#3)| | general |
the present ages of 3 persons are in the proportion of 4 : 7 : 9 . 8 years ago , the sum of their ages was 116 . find their present ages . | let the present ages of three persons be 4 k , 7 k and 9 k respectively . ( 4 k - 8 ) + ( 7 k - 8 ) + ( 9 k - 8 ) = 116 20 k = 140 k = 7 therefore , then present ages are 28 , 49,63 . answer : b | a ) 20 , 35,45 , b ) 28 , 49,63 , c ) 16 , 28,36 , d ) 16 , 28,46 , e ) none of these | b | add(multiply(4, divide(add(multiply(3, 8), 116), add(add(4, 7), 9))), 8) | add(n1,n2)|multiply(n0,n4)|add(n5,#1)|add(n3,#0)|divide(#2,#3)|multiply(n1,#4)|add(n4,#5) | general |
a car traveling at a certain constant speed takes 5 seconds longer to travel 1 km than it would take to travel 1 km at 90 km / hour . at what speed , in km / hr , is the car traveling ? | "time to cover 1 kilometer at 80 kilometers per hour is 1 / 90 hours = 3,600 / 90 seconds = 40 seconds ; time to cover 1 kilometer at regular speed is 40 + 5 = 45 seconds = 45 / 3,600 hours = 1 / 80 hours ; so , we get that to cover 1 kilometer 1 / 80 hours is needed - - > regular speed 80 kilometers per hour ( rate is... | a ) 70 , b ) 80 , c ) 74 , d ) 75 , e ) 78 | b | divide(1, divide(add(multiply(const_3600, divide(1, 90)), 5), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics |
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