Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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if x is invested in a bank at a rate of simple interest of y % p . a . for two years , then the interest earned is 600 . if x is invested at y % p . a . , for two years when the interest is compounded annually , the interest is 615 . what is the value of x ? | "simple way to solve this question is to use options . from si , we know that x * y = 30,000 . now , put the value of x = 6000 , we will have y = 5 % to calculate ci , now , we know 1 st year amount = 6000 + 5 % of 6000 = 6300 . 2 nd year , amount = 6300 + 5 % of 6300 = 6615 we can see after 2 years interest = 6615 - 6000 = 615 . hence , it satisfies the question . hence b is the correct answer" | a ) 8000 , b ) 6000 , c ) 5000 , d ) 4000 , e ) 3000 | b | divide(power(divide(600, const_2), const_2), subtract(615, 600)) | divide(n0,const_2)|subtract(n1,n0)|power(#0,const_2)|divide(#2,#1)| | gain | B |
what is the 25 th digit to the right of the decimal point in the decimal form of 6 / 11 ? | "explanation : to determine the greatest possible number of members , first recognize that each member had to contribute the lowest amount given . build an inequality for the individual contributions and the total amount collected , with n = the number of members in the club , and solve for n . 12 n β€ 599 n β€ 49 11 / 12 since n represents individual people it must be a whole number . thus the greatest number of people is 49 . answer : option c" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | c | divide(6, 11) | divide(n1,n2)| | general | C |
if ( x + yi ) / i = ( 7 + 9 i ) , where x and y are real , what is the value of ( x + yi ) ( x - yi ) ? | ( x + yi ) / i = ( 7 + 9 i ) ( x + yi ) = i ( 7 + 9 i ) = - 9 + 7 i ( x + yi ) ( x - yi ) = ( - 9 + 7 i ) ( - 9 - 7 i ) = 81 + 49 = 130 correct answer is c 0130 | a ) 110 , b ) 120 , c ) 130 , d ) 140 , e ) 150 | c | multiply(multiply(7, 9), const_2) | multiply(n0,n1)|multiply(#0,const_2) | general | C |
12 : 8 seconds : : ? : 4 minutes | "12 * 4 = 8 * x x = 6 answer : c" | a ) 10 , b ) 15 , c ) 6 , d ) 8 , e ) 7 | c | multiply(4, divide(12, 8)) | divide(n0,n1)|multiply(n2,#0)| | physics | C |
if a rectangular billboard has an area of 120 square feet and a perimeter of 46 feet , what is the length of each of the shorter sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 120 and a perimeter of 46 . we ' re asked for the length of one of the shorter sides of the rectangle . since the answers are all integers , and the area is 120 , the shorter side will almost certainly be less than 10 ( since 10 x 10 = 100 , but we ' re not dealing with a square ) . answer b ( 7 ) does not divide evenly into 120 , let ' s test answer c : 8 if . . . the shorter side = 8 . . . the area = 120 . . . . 120 / 8 = 15 = the longer side perimeter = 8 + 8 + 15 + 15 = 46 c" | a ) 4 , b ) 7 , c ) 8 , d ) 13 , e ) 26 | c | divide(subtract(divide(46, const_2), sqrt(subtract(power(divide(46, const_2), const_2), multiply(const_4, 120)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)| | geometry | C |
x , a , z , and b are single digit positive integers . x = 1 / 8 a . z = 1 / 8 b . ( 10 a + b ) β ( 10 x + z ) could not equal | "a = 8 x , b = 8 z therefore ( 8 x . 10 + 8 z ) - ( 10 x + z ) = ( 8 - 1 ) ( 10 x + z ) = 7 . ( 10 x + z ) number should be divisible by 7 e" | a ) 28 , b ) 35 , c ) 49 , d ) 63 , e ) 55 | e | add(add(subtract(add(multiply(8, 8), multiply(8, 10)), add(multiply(8, 10), 8)), 10), const_3) | multiply(n1,n1)|multiply(n1,n4)|add(#0,#1)|add(n1,#1)|subtract(#2,#3)|add(n4,#4)|add(#5,const_3)| | general | E |
the average age of a group of 12 students is 20 years . if 4 more students join the group , the average age increases by 1 year . the average age of the new students is | "total age of 12 std = 20 * 12 total age of 16 std = 21 * 16 total age of 4 std = ( 21 * 16 ) - 20 * 12 = 96 avg of 4 std = ( 96 ) / 4 = 24 answer : c" | a ) 22 , b ) 23 , c ) 24 , d ) 25 , e ) 26 | c | divide(subtract(multiply(add(20, 1), add(20, 1)), multiply(20, 12)), 4) | add(n1,n3)|multiply(n0,n1)|multiply(#0,#0)|subtract(#2,#1)|divide(#3,n2)| | general | C |
in a 200 meters race a beats b by 56 m or 7 seconds . a ' s time over the course is : | "b runs 56 m in 7 sec . = > b runs 200 m in 7 / 56 * 200 = 25 seconds since a beats b by 7 seconds , a runs 200 m in ( 25 - 7 ) = 18 seconds hence , a ' s time over the course = 18 seconds answer : d" | a ) 22 seconds , b ) 12 seconds , c ) 10 seconds , d ) 18 seconds , e ) 28 seconds | d | subtract(multiply(divide(7, 56), 200), 7) | divide(n2,n1)|multiply(n0,#0)|subtract(#1,n2)| | physics | D |
two numbers are 35 % and 42 % are less than a third number . how much percent is the second number less than the first ? | "explanation : i ii iii 65 58 100 65 - - - - - - - - 7 100 - - - - - - ? = > 10.769 % answer b" | a ) 11.769 % , b ) 10.769 % , c ) 12.769 % , d ) 11.69 % , e ) 11.89 % | b | multiply(divide(subtract(subtract(const_100, 35), subtract(const_100, 42)), subtract(const_100, 35)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|subtract(#0,#1)|divide(#2,#0)|multiply(#3,const_100)| | gain | B |
a boat can travel with a speed of 13 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 68 km downstream . | "speed downstream = ( 13 + 4 ) km / hr = 17 km / hr . time taken to travel 68 km downstream = 68 / 17 = 4 hrs ans - b" | a ) 2 , b ) 4 , c ) 6 , d ) 5 , e ) 3 | b | divide(68, add(13, 4)) | add(n0,n1)|divide(n2,#0)| | physics | B |
2525 * 9 | "explanation : 2525 * ( 10 - 1 ) = 25250 - 2525 = 22725 option a" | a ) 22725 , b ) 25675 , c ) 22655 , d ) 27575 , e ) none of these | a | multiply(divide(2525, 9), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | A |
what is the area of square field whose side of length 8 m ? | "8 * 8 = 64 sq m answer : b" | a ) 225 , b ) 64 , c ) 772 , d ) 882 , e ) 21 | b | square_area(8) | square_area(n0)| | geometry | B |
the ages of two persons differ by 16 years . 6 years ago , the elder one was 3 times as old as the younger one . what are their present ages of the elder person ? | let present age of the elder person = x and present age of the younger person = x β 16 ( x β 6 ) = 3 ( x β 16 β 6 ) x β 6 = 3 x β 66 2 x = 60 x = 60 / 2 = 30 answer : option c | a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | c | add(divide(add(multiply(3, 6), subtract(16, 6)), const_2), 16) | multiply(n1,n2)|subtract(n0,n1)|add(#0,#1)|divide(#2,const_2)|add(n0,#3) | general | C |
timothy leaves home for school , riding his bicycle at a rate of 6 miles per hour . fifteen minutes after he leaves , his mother sees timothy β s math homework lying on his bed and immediately leaves home to bring it to him . if his mother drives at 36 miles per hour , how far ( in terms of miles ) must she drive before she reaches timothy ? i think is a 700 level problem but i tag it as 600 / 700 , let me know . either way i hope in an explanationthanks | in 15 mins , timothy travels = 6 / 4 miles . now , let his mother takes x hours to reach him , traveling at 36 mph . so , 36 x = 6 x + 6 / 4 x = 1 / 20 hrs . thus , the distance traveled by his mother to reach = 36 * 1 / 20 = 9 / 5 miles . ans b | a ) 1 / 3 , b ) 9 / 5 , c ) 4 , d ) 9 , e ) 12 | b | multiply(divide(6, multiply(const_4, subtract(36, 6))), 36) | subtract(n1,n0)|multiply(#0,const_4)|divide(n0,#1)|multiply(n1,#2) | physics | B |
a and b β s salaries together amount to rs . 2,000 . a spends 95 % of his salary and b spends 85 % of his . if now their savings are the same , what is a β s salary ? | "explanation : ( 5 / 100 ) a = ( 15 / 100 ) b a = 3 b a + b = 2000 4 b = 2000 = > b = 500 a = 1500 answer is d" | a ) rs . 500 , b ) rs . 750 , c ) rs . 1250 , d ) rs . 1500 , e ) rs . 1520 | d | divide(multiply(multiply(multiply(const_2, multiply(const_4, add(const_2, const_3))), const_100), subtract(const_1, divide(85, const_100))), add(subtract(const_1, divide(95, const_100)), subtract(const_1, divide(85, const_100)))) | add(const_2,const_3)|divide(n2,const_100)|divide(n1,const_100)|multiply(#0,const_4)|subtract(const_1,#1)|subtract(const_1,#2)|add(#5,#4)|multiply(#3,const_2)|multiply(#7,const_100)|multiply(#8,#4)|divide(#9,#6)| | gain | D |
1397 x 1397 | "1397 x 1397 = ( 1397 ) 2 = ( 1400 - 3 ) 2 = ( 1400 ) 2 + ( 3 ) 2 - ( 2 x 1400 x 3 ) = 1960000 + 9 - 8400 = 1960009 - 8400 = 1951609 . answer : option a" | a ) 1951609 , b ) 1981709 , c ) 18362619 , d ) 2031719 , e ) none of these | a | multiply(divide(1397, 1397), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | A |
john and peter are among the 9 players a volleyball coach can choose from to field a 6 - player team . if all 6 players are chosen at random , what is the probability of choosing a team that includes john and peter ? | the total possible ways of selecting a 6 - member team is 9 c 6 = 84 the possible ways which include john and peter is 7 c 4 = 35 the probability of choosing both john and peter is 35 / 84 = 5 / 12 the answer is c . | a ) 3 / 10 , b ) 4 / 11 , c ) 5 / 12 , d ) 6 / 13 , e ) 7 / 15 | c | divide(divide(factorial(subtract(9, const_2)), multiply(factorial(subtract(6, const_2)), factorial(subtract(subtract(9, const_2), subtract(6, const_2))))), divide(factorial(9), multiply(factorial(6), factorial(subtract(9, 6))))) | factorial(n0)|factorial(n1)|subtract(n0,const_2)|subtract(n1,const_2)|subtract(n0,n1)|factorial(#2)|factorial(#3)|factorial(#4)|subtract(#2,#3)|factorial(#8)|multiply(#1,#7)|divide(#0,#10)|multiply(#6,#9)|divide(#5,#12)|divide(#13,#11) | probability | C |
mary and mike enter into a partnership by investing $ 900 and $ 100 respectively . at the end of one year , they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business . if mary received $ 1600 more than mike did , what was the profit made by their business in that year ? | "explanatory answer let the profit made during the year be $ 3 x therefore , $ x would have been shared equally and the remaining $ 2 x would have been shared in the ratio 9 : 1 . i . e . , 90 % of 2 x would go to mary and 10 % of 2 x would go to mike . hence , mary would get ( 90 - 10 ) % of 2 x more than mike or 80 % of 2 x = $ 1600 i . e . , ( 60 / 100 ) * 2 x = 1600 or 2 x = 2000 . hence , the profit made by the company during the year $ 3 x = $ 3000 . answer : e" | a ) $ 3500 , b ) $ 3400 , c ) $ 3300 , d ) $ 3200 , e ) $ 3000 | e | divide(1600, multiply(subtract(divide(900, const_1000), divide(100, const_1000)), subtract(const_1, divide(const_1, const_3)))) | divide(n0,const_1000)|divide(n1,const_1000)|divide(const_1,const_3)|subtract(#0,#1)|subtract(const_1,#2)|multiply(#3,#4)|divide(n2,#5)| | general | E |
if p ( a ) = 2 / 15 , p ( b ) = 4 / 15 , and p ( a Γ’ Λ Βͺ b ) = 11 / 15 find p ( a | b ) | "p ( a | b ) = p ( a Γ’ Λ Βͺ b ) / p ( b ) p ( a | b ) = ( 11 / 15 ) / ( 4 / 15 ) = 11 / 4 . a" | a ) 11 / 4 , b ) 2 / 3 , c ) 3 / 2 , d ) 4 / 5 , e ) 4 / 7 | a | divide(divide(11, 15), divide(2, 15)) | divide(n4,n1)|divide(n0,n1)|divide(#0,#1)| | general | A |
a no . when divided by 220 gives a remainder 43 , what remainder will beobtained by dividingthe same no . 17 ? | "220 + 43 = 263 / 17 = 8 ( remainder ) b" | a ) 2 , b ) 8 , c ) 9 , d ) 11 , e ) 15 | b | divide(add(220, 43), 17) | add(n0,n1)|divide(#0,n2)| | general | B |
rs . 800 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 800 become in 3 years . | explanation : s . i . = 956 - 800 = rs 156 r = 156 β 100 / 800 β 3 r = 6 1 / 2 % new rate = 6 1 / 2 + 4 = 21 / 2 % new s . i . = 800 Γ 21 / 2 Γ 3100 = 252 now amount will be 800 + 252 = 1052 option a | a ) rupees 1052 , b ) rs 1152 , c ) rs 1252 , d ) rs 1352 , e ) none of these | a | add(800, divide(multiply(multiply(800, add(divide(multiply(subtract(956, 800), const_100), multiply(800, 3)), 4)), 3), const_100)) | multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7) | gain | A |
by selling 8 pencils for a rupee a man loses 60 % . how many for a rupee should he sell in order to gain 60 % ? | "40 % - - - 8 160 % - - - ? 40 / 160 * 12 = 2 answer : c" | a ) 8 , b ) 9 , c ) 2 , d ) 6 , e ) 4 | c | multiply(divide(const_1, multiply(add(const_100, 60), divide(const_1, subtract(const_100, 60)))), 8) | add(n2,const_100)|subtract(const_100,n1)|divide(const_1,#1)|multiply(#0,#2)|divide(const_1,#3)|multiply(n0,#4)| | gain | C |
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 | D |
frank the fencemaker needs to fence in a rectangular yard . he fences in the entire yard , except for one full side of the yard , which equals 40 feet . the yard has an area of 240 square feet . how many feet offence does frank use ? | "area = length x breadth 240 = 40 x breadth so , breadth = 6 units fencing required is - breadth + breadth + length 6 + 6 + 40 = > 52 feet answer must be ( b ) 52" | a ) 14 , b ) 52 , c ) 54 , d ) 180 , e ) 240 | b | add(add(divide(240, 40), divide(240, 40)), 40) | divide(n1,n0)|add(#0,#0)|add(n0,#1)| | geometry | B |
a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 970 gm . for a kg . his gain is β¦ % . | "his percentage gain is 100 * 30 / 970 as he is gaining 30 units for his purchase of 970 units . so 3.09 % . answer : a" | a ) 3.09 % , b ) 5.36 % , c ) 4.26 % , d ) 6.26 % , e ) 7.26 % | a | multiply(subtract(inverse(divide(970, multiply(multiply(add(const_4, const_1), const_2), const_100))), const_1), const_100) | add(const_1,const_4)|multiply(#0,const_2)|multiply(#1,const_100)|divide(n0,#2)|inverse(#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain | A |
the salary of a typist was at first raised by 10 % and then the same was reduced by 5 % . if he presently draws rs . 5225 . what was his original salary ? | "x * ( 110 / 100 ) * ( 95 / 100 ) = 5225 x * ( 11 / 10 ) * ( 1 / 100 ) = 55 x = 5000 answer : b" | a ) 2277 , b ) 5000 , c ) 1000 , d ) 2651 , e ) 1971 | b | divide(5225, multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(5, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|divide(n2,#4)| | gain | B |
find the fraction which has the same ratio to 2 / 3 that 3 / 5 has to 6 / 7 | "p : 2 / 3 = 3 / 5 : 6 / 7 as the product of the means is equal to the product of the extremes . p * 6 / 7 = 2 / 3 * 3 / 5 p * 6 / 7 = 6 / 15 p = 7 / 15 = > p = 7 / 15 answer : d" | a ) 1 / 5 , b ) 2 / 22 , c ) 3 / 4 , d ) 7 / 15 , e ) 5 / 6 | d | multiply(divide(divide(3, 5), divide(6, 7)), divide(2, 3)) | divide(n2,n3)|divide(n4,n5)|divide(n0,n1)|divide(#0,#1)|multiply(#3,#2)| | other | D |
if | x - 12 | = 100 , what is the sum of all the possible values of x ? | "there will be two cases x - 12 = 100 or x - 12 = - 100 = > x = 112 or x = - 88 sum of both the values will be - 88 + 112 = 24 answer is c" | a ) - 12 , b ) - 22 , c ) 24 , d ) 36 , e ) 42 | c | subtract(add(100, 12), subtract(100, 12)) | add(n0,n1)|subtract(n1,n0)|subtract(#0,#1)| | general | C |
given f ( x ) = 3 x β 5 , for what value of x does 2 * [ f ( x ) ] β 1 = f ( 3 x β 6 ) | "explanations we have the function f ( x ) = 3 x β 5 , and we want to some sophisticated algebra with it . let β s look at the two sides of the prompt equation separately . the left side says : 2 * [ f ( x ) ] β 1 β - this is saying : take f ( x ) , which is equal to its equation , and multiply that by 2 and then subtract 1 . 2 * [ f ( x ) ] β 1 = 2 * ( 3 x β 5 ) β 1 = 6 x β 10 β 1 = 6 x β 11 the right side says f ( 3 x β 6 ) β this means , take the algebraic expression ( 3 x β 6 ) and plug it into the function , as discussed above in the section β how a mathematician things about a function . β this algebraic expression , ( 3 x β 6 ) , must take the place of x on both sides of the function equation . f ( 3 x β 6 ) = 3 * [ 3 x β 6 ] β 5 = 9 x β 18 β 5 = 9 x β 23 now , set those two equal and solve for x : 9 x β 23 = 6 x β 11 9 x = 6 x β 11 + 23 9 x = 6 x + 12 9 x β 6 x = 12 3 x = 12 x = 4 answer = b" | a ) 0 , b ) 4 , c ) 6 , d ) 7 , e ) 13 | b | divide(subtract(add(multiply(2, 5), 1), add(multiply(3, 3), 5)), subtract(multiply(2, 3), multiply(3, const_1))) | multiply(n1,n2)|multiply(n4,n0)|multiply(n0,n2)|multiply(n0,const_1)|add(n3,#0)|add(n1,#1)|subtract(#2,#3)|subtract(#4,#5)|divide(#7,#6)| | general | B |
a company has two models of computers , model x and model y . operating at a constant rate , a model x computer can complete a certain task in 60 minutes and a model y computer can complete the same task in 30 minutes . if the company used the same number of each model of computer to complete the task in 1 minute , how many model x computers were used ? | "let ' s say 1 work is processing 60 gb of data . model x : 1 gb per min model y : 2 gb per min working together , 1 x and 1 y = 3 gb per min so , 20 times as many computers would work at 60 gb per min . so no . of x = 20 answer is e" | a ) 18 , b ) 16 , c ) 15 , d ) 17 , e ) 20 | e | divide(multiply(60, 30), add(60, 30)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0)| | gain | E |
a certain college ' s enrollment at the beginning of 1992 was 40 percent greater than it was at the beginning of 1991 , and its enrollment at the beginning of 1993 was 15 percent greater than it was at the beginning of 1992 . the college ' s enrollment at the beginning of 1993 was what percent greater than its enrollment at the beginning of 1991 ? | "suppose enrollment in 1991 was 100 then enrollment in 1992 will be 140 and enrollment in 1993 will be 140 * 1.15 = 161 increase in 1993 from 1991 = 161 - 100 = 61 answer : e" | a ) 63 % , b ) 24 % , c ) 50 % , d ) 55 % , e ) 61 % | e | subtract(multiply(add(const_100, 40), divide(add(const_100, 15), const_100)), const_100) | add(n1,const_100)|add(n4,const_100)|divide(#1,const_100)|multiply(#0,#2)|subtract(#3,const_100)| | gain | E |
pipe p can fill a tank in 3 hours , pipe q in 9 hours and pipe r in 24 hours . if all the pipes are open , in how many hours will the tank be filled ? | "explanation : part filled by ( p + q + r ) in 1 hour = ( 1 / 3 + 1 / 9 + 1 / 24 ) = 35 / 72 all the three pipes together will fill the tank = 72 / 35 = 2.1 hours answer b" | a ) 2 hours , b ) 2.1 hours , c ) 3 hours , d ) 3.5 hours , e ) 4.5 hours | b | inverse(add(divide(const_1, 24), add(divide(const_1, 3), divide(const_1, 9)))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|inverse(#4)| | physics | B |
in a factory there are 3 types of machine m 1 , m 2 and m which produces 25 % , 35 % and 40 % of the total products respectively . m 1 , m 2 and m 3 produces 2 % , 4 % and 5 % defective products , respectively . what is the percentage of non - defective products ? | solution : non - defective products , { [ ( 25 * 0.98 ) + ( 35 * 0.96 ) + ( 40 * 0.95 ) ] / 100 } * 100 = 96.1 % . answer : option c | a ) 89 % , b ) 97.1 % , c ) 96.1 % , d ) 86.1 % , e ) none of these | c | subtract(const_100, add(add(multiply(divide(2, const_100), 25), multiply(divide(4, const_100), 35)), multiply(divide(5, const_100), 40))) | divide(n2,const_100)|divide(n10,const_100)|divide(n11,const_100)|multiply(n3,#0)|multiply(n4,#1)|multiply(n5,#2)|add(#3,#4)|add(#6,#5)|subtract(const_100,#7) | general | C |
a can do a work in 15 days and b in 20 days . if they work on it together then in how many days required to complete the work ? | person ( a ) ( b ) ( a + b ) time - ( 15 ) ( 20 ) ( 300 / 35 ) rate - ( 20 ) ( 15 ) ( 35 ) work - ( 300 ) ( 300 ) ( 300 ) therefore a + b requires ( 300 / 35 ) days to complete entire work = 300 / 35 answer is b | a ) 300 / 31 , b ) 300 / 35 , c ) 300 / 21 , d ) 300 / 15 , e ) 300 / 20 | b | inverse(add(divide(const_1, 15), divide(const_1, 20))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|inverse(#2) | physics | B |
a man can row upstream at 20 kmph and downstream at 28 kmph , and then find the speed of the man in still water ? | "us = 20 ds = 28 m = ( 28 + 20 ) / 2 = 24 answer : a" | a ) 24 , b ) 37 , c ) 30 , d ) 27 , e ) 18 | a | divide(add(20, 28), const_2) | add(n0,n1)|divide(#0,const_2)| | physics | A |
walking with 4 / 5 of my usual speed , i arrive at the bus stop 7 minutes later than normal . how many minutes does it take to walk to the bus stop at my usual speed ? | let t = usual time = distance / usual speed t + 7 = distance / ( 4 * usual speed / 5 ) = ( 5 * distance ) / ( 4 * usual speed ) = 5 t / 4 t = 28 the answer is e . | a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 28 | e | divide(7, subtract(divide(5, 4), const_1)) | divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1) | physics | E |
in a company of 12 employees , 5 employees earn $ 36,000 , 4 employees earn $ 45,000 , and the 3 highest - paid employees earn the same amount . if the average annual salary for the 12 employees is $ 47,500 , what is the annual salary for each of the highest - paid employees ? | 5 * 36,000 + 4 * 45,000 + 3 x = 12 * 47,500 3 x = 570,000 - 180,000 - 180,000 3 x = 210,000 x = 70,000 the answer is c . | a ) $ 60,000 , b ) $ 65,000 , c ) $ 70,000 , d ) $ 75,000 , e ) $ 80,000 | c | subtract(divide(multiply(divide(subtract(subtract(multiply(add(add(multiply(multiply(const_3, const_3), add(const_4, const_1)), const_2), divide(const_1, const_2)), 12), multiply(multiply(12, 3), add(const_4, const_1))), multiply(multiply(multiply(const_3, const_3), add(const_4, const_1)), 4)), 3), const_1000), const_1000), 5) | add(const_1,const_4)|divide(const_1,const_2)|multiply(const_3,const_3)|multiply(n0,n5)|multiply(#0,#2)|multiply(#0,#3)|add(#4,const_2)|multiply(n3,#4)|add(#6,#1)|multiply(n0,#8)|subtract(#9,#5)|subtract(#10,#7)|divide(#11,n5)|multiply(#12,const_1000)|divide(#13,const_1000)|subtract(#14,n1) | general | C |
if the radius of a circle is decreased 40 % , what happens to the area ? | "area of square = pi * radius ^ 2 new radius = 0.6 * old radius so new area = ( 0.6 ) ^ 2 old area = > 4 / 25 of old area = > 36 % old area ans : e" | a ) 10 % decrease , b ) 20 % decrease , c ) 36 % decrease , d ) 40 % decrease , e ) 64 % decrease | e | subtract(const_100, multiply(power(divide(40, const_100), const_2), const_100)) | divide(n0,const_100)|power(#0,const_2)|multiply(#1,const_100)|subtract(const_100,#2)| | geometry | E |
simple interest on a certain sum at a certain annual rate of interest is 1 / 25 of the sum . if the numbers representing rate percent and time in years be equal , then the rate of interest is : | "explanation : let sum = x . then , s . i . = x / 25 let rate = r % and time = r years . [ x * r * r / 100 ] = x / 25 ? r ^ 2 = 100 / 25 = 4 r = 2 hence , rate of interest = 2 % . answer : b" | a ) 5 % , b ) 2 % , c ) 6 % , d ) 8 % , e ) 3 % | b | divide(add(multiply(floor(sqrt(multiply(divide(1, 25), const_100))), const_10), 1), const_3) | divide(n0,n1)|multiply(#0,const_100)|sqrt(#1)|floor(#2)|multiply(#3,const_10)|add(#4,n0)|divide(#5,const_3)| | gain | B |
a fruit seller had some apples . he sells 40 % apples and still has 420 apples . originally , he had : | "suppose originally he had x apples . then , ( 100 - 40 ) % of x = 420 . 60 / 100 x x = 420 x = ( 420 x 100 ) / 60 = 700 . answer e" | a ) 588 apples , b ) 742 apples , c ) 750 apples , d ) 600 apples , e ) 700 apples | e | original_price_before_loss(40, 420) | original_price_before_loss(n0,n1)| | gain | E |
if 6 x ^ 2 + x - 12 = ( ax + b ) ( cx + d ) , then | a | + | b | + | c | + | d | | this is the hard one , definitely a 700 + level question . we need numbers a , b , c , and d such that 6 x ^ 2 + x - 12 = ( ax + b ) ( cx + d ) this means that ac = 6 , bd = β 12 , and ad + bc = 1 . the a & c pair could be ( 1 , 6 ) or ( 2 , 3 ) , in some order . the absolute values of the b & d pair could be ( 1 , 12 ) or ( 2 , 6 ) or ( 3 , 4 ) , and of course , in each case , one of the two would have to be negative . after some trial and error , we find : 6 x ^ 2 + x - 12 = ( 2 x + 3 ) ( 3 x - 4 ) thus , we see : | a | + | b | + | c | + | d | = 2 + 3 + 3 + 4 = 12 answer = b | a ) 10 , b ) 12 , c ) 15 , d ) 18 , e ) 20 | b | add(add(2, const_3), add(const_3, const_4)) | add(n1,const_3)|add(const_3,const_4)|add(#0,#1) | general | B |
walking 6 / 7 of his usual rate , a boy reaches his school 4 min early . find his usual time to reach the school ? | speed ratio = 1 : 6 / 7 = 7 : 6 time ratio = 6 : 4 1 - - - - - - - - 6 4 - - - - - - - - - ? = > 24 m answer : e | a ) 28 m , b ) 20 m , c ) 22 m , d ) 23 m , e ) 24 m | e | multiply(4, 6) | multiply(n0,n2) | gain | E |
if the ratio of a to b is 2 to 3 and the ratio of b to c is 1 to 5 , what is the ratio of a to c ? | "a : b = 2 : 3 - - 1 b : c = 1 : 5 = > b : c = 3 : 15 - - 2 from 1 and 2 , we get a : c = 2 : 15 answer c" | a ) 4 / 15 , b ) 1 / 3 , c ) 2 / 15 , d ) 4 / 5 , e ) 7 / 6 | c | divide(multiply(2, 1), multiply(3, 5)) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)| | other | C |
pipe a can fill a tank in 24 minutes . pipe b can fill the same tank 6 times faster than pipe a . if both the pipes are connected to the tank so that they fill the tank simultaneously , how long will it take for the empty tank to overflow ? | "pipe a can fill a tank in 24 minutes - - > the rate of a = 1 / 24 tank / minute ; pipe b can fill the same tank 6 times fasterthan pipe a - - > the rate of b = 1 / 24 + 6 / 24 = 7 / 24 tank / minute . their combined rate = 1 / 24 + 7 / 24 = 1 / 3 tank / minute ; therefore , a and b can fill the tank in 3 minutes . answer : d" | a ) 4 minutes , b ) 32 / 7 minutes , c ) 192 / 7 minutes , d ) 3 minutes , e ) 28 minutes | d | inverse(add(divide(const_1, 24), add(divide(const_1, 24), divide(6, 24)))) | divide(const_1,n0)|divide(n1,n0)|add(#0,#1)|add(#2,#0)|inverse(#3)| | physics | D |
a line that passes through ( β 2 , β 4 ) and ( 3 , k ) has a slope = k . what is the value of k ? | "slope = ( y 2 - y 1 ) / ( x 2 - x 1 ) = > k = ( k + 4 ) / ( 3 + 2 ) = > 5 k = k + 4 = > k = 1 ans b it is !" | a ) 3 / 4 , b ) 1 , c ) 4 / 3 , d ) 2 , e ) 7 / 2 | b | divide(4, 3) | divide(n1,n2)| | general | B |
kim finds a 3 - meter tree branch and marks it off in thirds and fifths . she then breaks the branch along all the markings and removes one piece of every distinct length . what fraction of the original branch remains ? | 3 pieces of 1 / 5 length and two piece each of 1 / 15 and 2 / 15 lengths . removing one piece each from pieces of each kind of lengths the all that will remain will be 2 pieces of 1 / 5 i . e 2 / 5 , 1 piece of 1 / 15 , and 1 piece of 2 / 15 which gives us 2 / 5 + 1 / 15 + 2 / 15 - - - - - > 3 / 5 answer is b | a ) 2 / 5 , b ) 3 / 5 , c ) 8 / 15 , d ) 1 / 2 , e ) 7 / 5 | b | subtract(const_1, add(add(divide(const_3, multiply(add(const_2, const_3), 3)), divide(const_2, multiply(add(const_2, const_3), 3))), divide(const_1, multiply(add(const_2, const_3), 3)))) | add(const_2,const_3)|multiply(n0,#0)|divide(const_3,#1)|divide(const_2,#1)|divide(const_1,#1)|add(#2,#3)|add(#5,#4)|subtract(const_1,#6) | physics | B |
15 men take 20 days of 8 hours each to do a piece of work . how many days of 6 hours each would 21 women take to do the same . if 3 women do as much work as 2 men ? | 3 w = 2 m 15 m - - - - - - 20 * 8 hours 21 w - - - - - - x * 6 hours 14 m - - - - - - x * 6 15 * 20 * 8 = 14 * x * 6 x = 28.57 answer : b | a ) 27.57 , b ) 28.57 , c ) 29.57 , d ) 30.57 , e ) 32 | b | add(floor(divide(multiply(multiply(20, 8), multiply(15, 3)), multiply(multiply(21, 2), 6))), const_1) | multiply(n1,n2)|multiply(n0,n5)|multiply(n4,n6)|multiply(#0,#1)|multiply(n3,#2)|divide(#3,#4)|floor(#5)|add(#6,const_1) | physics | B |
a train covers a distance of 6 km in 10 min . if it takes 12 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 6 / 10 * 60 ) km / hr = ( 36 * 5 / 18 ) m / sec = 10 m / sec . length of the train = 10 * 12 = 120 m . answer : a" | a ) 120 m , b ) 180 m , c ) 240 m , d ) 220 m , e ) 280 m | a | divide(6, subtract(divide(6, 10), 12)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics | A |
a father purchased dress for his 3 daughters . the dresses are of same color but diff size and they are kept in dark room . what is probability that all the 3 will not choose there own dress ? | explanation : let 1 st girl come and she choose wrong dress so probability of that girl to choose wrong dress out of 3 is = 2 / 3 . now 2 nd girl come nd she choose wrong dress so probability of that girl to choose wrong dress out of 2 is 1 / 2 . now for 3 rd girl probability is 1 to choose wrong dress . so probability tht all the 3 wil not choose der own dress is = 2 / 3 * 1 / 2 * 1 = 1 / 3 . hence ( a ) is the correct answer . answer : a | a ) 1 / 3 , b ) 2 / 3 , c ) 1 / 2 , d ) 3 / 4 , e ) 3 / 7 | a | divide(choose(const_2, const_1), factorial(const_3)) | choose(const_2,const_1)|factorial(const_3)|divide(#0,#1) | probability | A |
what is the least number which when divided by 6 , 9 , 12 and 18 leaves remainder 4 in each care ? | "explanation : lcm of 6 , 9 , 12 and 18 is 36 required number = 36 + 4 = 40 answer : option b" | a ) 30 , b ) 40 , c ) 36 , d ) 56 , e ) 66 | b | add(lcm(lcm(6, 9), lcm(12, 18)), 4) | lcm(n0,n1)|lcm(n2,n3)|lcm(#0,#1)|add(n4,#2)| | general | B |
if price of t . v set is reduced by 10 % , then its sale increases by 85 % , find net effect on sale value | - a + b + ( ( - a ) ( b ) / 100 ) = - 10 + 85 + ( - 10 * 85 ) / 100 = - 10 + 85 - 8.5 = 67 answer : d | a ) 44 , b ) 45 , c ) 46 , d ) 67 , e ) 48 | d | multiply(subtract(multiply(divide(subtract(const_100, 10), const_100), divide(add(const_100, 85), const_100)), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100) | gain | D |
what is the sum of all the multiples of 7 between 30 and 100 ? | "you first have to know all the multiples of 7 between 30 and 100 . they are 35 , 42 , 49 , 56 , 63 , 70 , 77 , 84 , 91 , and 98 . if you add all these numbers together , you get 665 . final answer : e" | a ) 692 , b ) 700 , c ) 677 , d ) 654 , e ) 665 | e | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 7), add(const_2, const_4)) | add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)| | general | E |
a rower can row 6 km / h in still water . when the river is running at 1 km / h , it takes the rower 1 hour to row to big rock and back . how many kilometers is it to big rock ? | "let x be the distance to big rock . time = x / 5 + x / 7 = 1 x = 35 / 12 = 2.92 the answer is e ." | a ) 2.66 , b ) 2.78 , c ) 2.8 , d ) 2.87 , e ) 2.92 | e | multiply(divide(subtract(6, 1), add(add(6, 1), subtract(6, 1))), add(6, 1)) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|divide(#1,#2)|multiply(#0,#3)| | physics | E |
rs . 1500 is divided into two parts such that if one part is invested at 6 % and the other at 5 % the whole annual interest from both the sum is rs . 80 . how much was lent at 5 % ? | "( x * 5 * 1 ) / 100 + [ ( 1500 - x ) * 6 * 1 ] / 100 = 80 5 x / 100 + 90 Γ’ β¬ β 6 x / 100 = 80 x / 100 = 10 = > x = 1000 answer : e" | a ) 228 , b ) 299 , c ) 266 , d ) 500 , e ) 1000 | e | multiply(add(5, 6), const_100) | add(n1,n2)|multiply(#0,const_100)| | gain | E |
5357 x 51 = ? | "5357 x 51 = 5357 x ( 50 + 1 ) = 5357 x 50 + 5357 x 1 = 267850 + 5357 = 273207 e )" | a ) 273232 , b ) 273243 , c ) 273247 , d ) 273250 , e ) 273207 | e | multiply(divide(5357, 51), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | E |
the area of a square is 4356 sq cm . find the ratio of the breadth and the length of a rectangle whose length is twice the side of the square and breadth is 24 cm less than the side of the square . | "let the length and the breadth of the rectangle be l cm and b cm respectively . let the side of the square be a cm . a 2 = 4356 a = 66 l = 2 a and b = a - 24 b : l = a - 24 : 2 a = 42 : 132 = 7 : 22 answer : e" | a ) 5 : 28 , b ) 5 : 19 , c ) 7 : 12 , d ) 5 : 13 , e ) 7 : 22 | e | divide(subtract(sqrt(4356), 24), multiply(sqrt(4356), const_2)) | sqrt(n0)|multiply(#0,const_2)|subtract(#0,n1)|divide(#2,#1)| | geometry | E |
if 20 machine can finish a job in 36 days , then how many more machines would be needed to finish the job in one - third less time ? | "you might think of this in a management context - we can use the principle of ' person - hours ' to solve any problem where we have identical workers . so , using simpler numbers , suppose you know that 6 identical employees , working simultaneously , would finish a job in 5 hours . then that job requires 6 * 5 = 30 total hours of person - work . if instead you wanted the job done in 3 hours , you ' d assign 30 / 3 = 10 employees to do the job , because you want to get a total of 30 hours of work from the employees . we can solve this problem identically . if 20 machines ( identical ones , i assume ) work simultaneously for 36 days , they will do a total of 20 * 36 machine - days of work . so the job requires 20 * 36 days of machine work in total . we instead want the job done in 1 / 3 less time , so in 24 days . so we ' ll need 20 * 36 / 24 = 30 machines , or 10 additional machines . c" | a ) a . 4 , b ) b . 8 , c ) c . 10 , d ) d . 12 , e ) e . 16 | c | subtract(divide(multiply(20, 36), multiply(divide(const_3, const_4), 36)), 20) | divide(const_3,const_4)|multiply(n0,n1)|multiply(n1,#0)|divide(#1,#2)|subtract(#3,n0)| | general | C |
a group of people participate in some curriculum , 25 of them practice yoga , 18 study cooking , 10 study weaving , 4 of them study cooking only , 5 of them study both the cooking and yoga , 4 of them participate all curriculums . how many people study both cooking and weaving ? | "both cooking and weaving = 18 - ( 4 + 4 + 5 ) = 5 so , the correct answer is e ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | e | subtract(subtract(subtract(18, 5), 4), 4) | subtract(n1,n4)|subtract(#0,n5)|subtract(#1,n3)| | other | E |
find the area of a cuboid of length 8 cm , breadth 10 cm . and height 12 cm . | "area of a cuboid = lxbxh = 8 cm x 10 cm x 12 cm = 960 cm cube answer : e" | a ) 900 cm cube , b ) 910 cm cube , c ) 920 cm cube , d ) 930 cm cube , e ) 960 cm cube | e | multiply(multiply(8, 10), 12) | multiply(n0,n1)|multiply(n2,#0)| | physics | E |
three pipes of same capacity can fill a tank in 8 hours . if there are only two pipes of same capacity , the tank can be filled in ? | "the part of the tank filled by three pipes in one hour = 1 / 8 = > the part of the tank filled by two pipes in 1 hour = 2 / 3 * 1 / 8 = 1 / 12 . the tank can be filled in 12 hours . answer : b" | a ) 08 hours , b ) 12 hours , c ) 03 hours , d ) 04 hours , e ) 21 hours | b | inverse(multiply(divide(const_2, const_3), divide(const_1, 8))) | divide(const_2,const_3)|divide(const_1,n0)|multiply(#0,#1)|inverse(#2)| | physics | B |
mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 24624 , how much amount per month would he have spent on clothes and transport ? | "β΅ amount , he have spent in 1 month on clothes transport = amount spent on saving per month β΅ amount , spent on clothes and transport = 24624 β 12 = 2052 answer c" | a ) 4038 , b ) 8076 , c ) 2052 , d ) 4845.6 , e ) none of these | c | multiply(divide(divide(24624, divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)), multiply(const_3, const_4)), divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)) | multiply(const_3,const_4)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|divide(n2,#4)|divide(#5,#0)|multiply(#6,#4)| | general | C |
at a certain company , each employee has a salary grade s that is at least 1 and at most 5 . each employee receives an hourly wage p , in dollars , determined by the formula p = 9.50 + 0.25 ( s β 2 ) . an employee with a salary grade of 5 receives how many more dollars per hour than an employee with a salary grade of 1 ? | oa is definitely wrong . the answer should be e . | a ) $ 0.50 , b ) $ 1.00 , c ) $ 1.25 , d ) $ 1.50 , e ) $ 1.75 | e | add(multiply(0.25, subtract(5, 1)), 0.25) | subtract(n1,n0)|multiply(n3,#0)|add(n3,#1)| | general | E |
the area of a circle is increased by 800 % . by what percent has the diameter of the circle increased ? | "a diameter of 2 it ' s radius = 1 it ' s area = ( 1 ^ 2 ) pi = 1 pi answer a : if we increase that diameter 100 % , we have . . . . a diameter of 4 it ' s radius = 2 it ' s area = ( 2 ^ 2 ) pi = 4 pi this area has increased ( 4 pi - 1 pi ) / 1 pi = 3 pi / 1 pi = 3 = 300 % answer b : if we increase the diameter 200 % , we have . . . a diameter of 6 it ' s area = 3 it ' s area = ( 3 ^ 2 ) pi = 9 pi this area has increased ( 9 pi - 1 pi ) / 1 pi = 8 pi / 1 pi = 8 = 800 % answer : e" | a ) 100 % , b ) 200 % , c ) 300 % , d ) 600 % , e ) 800 % | e | multiply(const_100, divide(const_2, const_2)) | divide(const_2,const_2)|multiply(#0,const_100)| | geometry | E |
working at constant rate , pump x pumped out half of the water in a flooded basement in 2 hours . the pump y was started and the two pumps , working independently at their respective constant rates , pumped out rest of the water in 3 hours . how many hours would it have taken pump y , operating alone at its own constant rate , to pump out all of the water that was pumped out of the basement ? | rate of x = 1 / 8 rate of x + y = 1 / 6 rate of y = 1 / 6 - 1 / 8 = 1 / 24 18 hours d | a ) a . 10 , b ) b . 12 , c ) c . 14 , d ) d . 18 , e ) e . 24 | d | add(add(add(add(add(add(add(add(add(add(add(add(const_1, add(const_3, const_2)), const_1), const_1), const_1), const_1), const_2), const_1), const_1), const_1), const_1), const_1), const_1) | add(const_2,const_3)|add(#0,const_1)|add(#1,const_1)|add(#2,const_1)|add(#3,const_1)|add(#4,const_1)|add(#5,const_2)|add(#6,const_1)|add(#7,const_1)|add(#8,const_1)|add(#9,const_1)|add(#10,const_1)|add(#11,const_1) | physics | D |
the price of a t . v . set worth rs . 10000 is to be paid in 20 installments of rs . 1000 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 . 1000 balance payment = ( 10000 - 1000 ) = rs . 9000 answer : a" | a ) 9000 , b ) 26699 , c ) 96000 , d ) 19000 , e ) 26711 | a | subtract(10000, 1000) | subtract(n0,n2)| | gain | A |
a man purchased 3 blankets @ rs . 100 each , 5 blankets @ rs . 150 each and two blankets at a certain rate which is now slipped off from his memory . but he remembers that the average price of the blankets was rs . 160 . find the unknown rate of two blankets ? | "10 * 160 = 1600 3 * 100 + 5 * 150 = 1050 1600 β 1050 = 550 answer : b" | a ) 420 , b ) 550 , c ) 490 , d ) 450 , e ) 457 | b | subtract(multiply(const_10, 160), add(multiply(3, 100), multiply(5, 150))) | multiply(n4,const_10)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(#0,#3)| | general | B |
a number is doubled and 9 is added . if the resultant is trebled , it becomes 63 . what is that number ? | "let the number be x . then , 3 ( 2 x + 9 ) = 63 2 x = 12 = > x = 6 answer : b" | a ) 3.5 , b ) 6 , c ) 8 , d ) 7 , e ) 4 | b | divide(subtract(63, multiply(const_3, 9)), multiply(const_3, const_2)) | multiply(n0,const_3)|multiply(const_2,const_3)|subtract(n1,#0)|divide(#2,#1)| | general | B |
the salaries of a , b , and c are in the ratio of 1 : 2 : 3 . the salary of b and c together is rs . 6000 . by what percent is the salary of c more than that of a ? | explanation : let the salaries of a , b , c be x , 2 x and 3 x respectively . then , 2 x + 3 x = 6000 = > x = 1200 . a ' s salary = rs . 1200 , b ' s salary = rs . 2400 , and cs salary rs . 3600 . excess of c ' s salary over a ' s = [ ( 2400 / 1200 ) x 100 ] = 200 % . answer : b ) 200 % | a ) 209 % , b ) 200 % , c ) 290 % , d ) 600 % , e ) 100 % | b | multiply(subtract(divide(multiply(3, divide(6000, add(2, 3))), divide(6000, add(2, 3))), const_1), const_100) | add(n1,n2)|divide(n3,#0)|multiply(n2,#1)|divide(#2,#1)|subtract(#3,const_1)|multiply(#4,const_100) | other | B |
x and y began business with rs . 10000 and rs . 8000 after 6 months , x advances rs . 2000 and y advances rs . 2000 more . at the end of the year , their profits amounted to rs . 2000 find the share of x and y respectively . | ( 10 * 6 + 12 * 6 ) : ( 8 * 6 + 10 * 6 ) 11 : 9 x share = 11 / 20 * 2000 = 1100 . y share = 9 / 20 * 2000 = 900 . x and y share respectively is 1100 , 900 . answer : e | a ) 10001000 , b ) 9001100 , c ) 1200800 , d ) 8001200 , e ) 1100 , 900 | e | multiply(divide(divide(add(multiply(10000, 6), multiply(add(10000, 2000), 6)), gcd(add(multiply(10000, 6), multiply(add(10000, 2000), 6)), add(multiply(8000, 6), multiply(add(8000, add(2000, 2000)), 6)))), add(divide(add(multiply(10000, 6), multiply(add(10000, 2000), 6)), gcd(add(multiply(10000, 6), multiply(add(10000, 2000), 6)), add(multiply(8000, 6), multiply(add(8000, add(2000, 2000)), 6)))), divide(add(multiply(8000, 6), multiply(add(8000, add(2000, 2000)), 6)), gcd(add(multiply(10000, 6), multiply(add(10000, 2000), 6)), add(multiply(8000, 6), multiply(add(8000, add(2000, 2000)), 6)))))), 2000) | add(n0,n3)|add(n3,n3)|multiply(n0,n2)|multiply(n1,n2)|add(n1,#1)|multiply(n2,#0)|add(#2,#5)|multiply(n2,#4)|add(#3,#7)|gcd(#6,#8)|divide(#6,#9)|divide(#8,#9)|add(#10,#11)|divide(#10,#12)|multiply(n3,#13) | general | E |
mary works in a restaurant a maximum of 60 hours . for the first 20 hours , she is paid $ 8 per hour . for each overtime hour , she is paid at a rate which is 25 % higher than her regular rate . how much mary can earn in a week ? | mary receives $ 8 ( 20 ) = $ 160 for the first 20 hours . for the 40 overtime hours , she receives $ 8 ( 0.25 ) + $ 8 = $ 10 per hour , that is $ 10 ( 40 ) = $ 400 . the total amount is $ 160 + $ 400 = $ 560 answer c 560 . | a ) 300 , b ) 420 , c ) 560 , d ) 320 , e ) 400 | c | add(multiply(60, 8), multiply(subtract(60, 20), multiply(8, divide(25, const_100)))) | divide(n3,const_100)|multiply(n0,n2)|subtract(n0,n1)|multiply(n2,#0)|multiply(#3,#2)|add(#1,#4) | gain | C |
if x / y = 3 and ( 2 a - x ) / ( 5 b - y ) = 3 , then the value of a / b is ? | "x = 3 y 2 a - 3 y = 3 ( 5 b - y ) 2 a - 3 y = 15 b - 3 y 2 a = 15 b a / b = 15 / 2 answer : a" | a ) 15 / 2 , b ) - 2 , c ) 1 , d ) 2 , e ) 3 | a | divide(multiply(3, 3), 2) | multiply(n0,n0)|divide(#0,n1)| | general | A |
of the people who responded to a market survey , 60 preferred brand x and the rest preferred brand y . if the respondents indicated a preference for brand x over brand y by ratio of 3 to 1 , how many people responded to the survey ? | "ratio = 3 : 1 = > 3 x respondents preferred brand x and x preferred brand y since , no . of respondents who preferred brand x = 60 = > 3 x = 60 = > x = 20 hence total no . of respondents = 60 + 20 = 80 hence a is the answer ." | a ) 80 , b ) 150 , c ) 240 , d ) 360 , e ) 480 | a | add(divide(60, 3), 60) | divide(n0,n1)|add(n0,#0)| | other | A |
virginia , adrienne , and dennis have taught history for a combined total of 102 years . if virginia has taught for 9 more years than adrienne and for 9 fewer years than dennis , for how many years has dennis taught ? | "let number of years taught by virginia = v number of years taught by adrienne = a number of years taught by dennis = d v + a + d = 96 v = a + 9 = > a = v - 9 v = d - 9 = > a = ( d - 9 ) - 9 = d - 18 d - 9 + d - 18 + d = 102 = > 3 d = 102 + 27 = 129 = > d = 43 answer d" | a ) 23 , b ) 32 , c ) 35 , d ) 43 , e ) 44 | d | add(divide(subtract(102, add(add(9, 9), 9)), const_3), add(9, 9)) | add(n1,n2)|add(n1,#0)|subtract(n0,#1)|divide(#2,const_3)|add(#0,#3)| | general | D |
the sum of four consecutive even numbers is 292 . what would be the largest number ? | "let the four consecutive even numbers be 2 ( x - 2 ) , 2 ( x - 1 ) , 2 x , 2 ( x + 1 ) their sum = 8 x - 4 = 292 = > x = 37 smallest number is : 2 ( x + 1 ) = 76 . answer : c" | a ) 33 , b ) 88 , c ) 76 , d ) 123 , e ) 12 | c | add(add(power(add(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(292, const_10), const_2), const_4), const_2), const_2))) | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics | C |
a person can row at 9 kmph and still water . he takes 3 1 / 2 hours to row from a to b and back . what is the distance between a and b if the speed of the stream is 1 kmph ? | "let the distance between a and b be x km . total time = x / ( 9 + 1 ) + x / ( 9 - 1 ) = 3.5 = > x / 10 + x / 8 = 7 / 2 = > ( 4 x + 5 x ) / 40 = 7 / 2 = > x = 16 km . answer : e" | a ) 32 , b ) 28 , c ) 29 , d ) 54 , e ) 16 | e | divide(multiply(multiply(subtract(9, 1), add(9, 1)), 3), add(add(9, 1), subtract(9, 1))) | add(n0,n2)|subtract(n0,n2)|add(#0,#1)|multiply(#0,#1)|multiply(n1,#3)|divide(#4,#2)| | physics | E |
sale of rs 6835 , rs . 9927 , rs . 6855 , rs . 7230 and rs . 6562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs , 6700 ? | "total sale for 5 months = rs . ( 6435 + 6927 + 6855 + 7230 + 6562 ) = rs . 34009 . required sale = rs . [ ( 6700 x 6 ) - 34009 ] = rs . ( 40200 - 34009 ) = rs . 6191 answer : b" | a ) 4966 , b ) 6191 , c ) 2877 , d ) 2676 , e ) 1881 | b | multiply(subtract(divide(add(add(add(add(6835, 9927), 6855), 7230), 6562), 5), 6700), 5) | add(n0,n1)|add(n2,#0)|add(n3,#1)|add(n4,#2)|divide(#3,n5)|subtract(#4,n6)|multiply(n5,#5)| | general | B |
a car is running at a speed of 90 kmph . what distance will it cover in 15 second ? | "explanation : given : speed = 108 kmph = ( 90 x ( 5 / 18 ) ) m / sec = 25 m / sec distance covered in 15 second = ( 25 x 15 ) m = 375 m . answer : c" | a ) 100 m , b ) 255 m , c ) 375 m , d ) can not be determined , e ) none of these | c | multiply(divide(90, const_3_6), 15) | divide(n0,const_3_6)|multiply(n1,#0)| | physics | C |
simplify : 0.3 * 0.8 + 0.1 * 0.5 | "given exp . = 0.3 * 0.8 + ( 0.1 * 0.5 ) = 0.24 + 0.05 = 0.29 answer is b ." | a ) 0.52 , b ) 0.29 , c ) 0.48 , d ) 0.64 , e ) 0.46 | b | add(multiply(0.3, 0.8), multiply(0.1, 0.5)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | general | B |
mariah has decided to hire three workers . to determine whom she will hire , she has selected a group of 17 candidates . she plans to have one working interview with 3 of the 17 candidates every day to see how well they work together . how many days will it take her to have working interviews with all the different combinations of job candidates ? | "420 . answer b" | a ) 720 , b ) 420 , c ) 300 , d ) 30 , e ) 333 | b | subtract(subtract(subtract(divide(divide(divide(factorial(17), factorial(subtract(17, 3))), factorial(3)), const_2), 17), 17), const_10) | factorial(n0)|factorial(n1)|subtract(n0,n1)|factorial(#2)|divide(#0,#3)|divide(#4,#1)|divide(#5,const_2)|subtract(#6,n0)|subtract(#7,n0)|subtract(#8,const_10)| | physics | B |
a rectangular courty 3.78 metres long and 5.25 metres wide is to be paved exactly with square tiles , all of the same size . what is the largest size of the tile which could be used for the purpose ? | solution largest size of the tile . h . c . f of 378 cm and 525 cm = 21 cms . answer b | ['a ) 14 cms', 'b ) 21 cms', 'c ) 42 cms', 'd ) none of these', 'e ) can not be determined'] | b | multiply(divide(divide(divide(divide(multiply(const_100, 3.78), const_3), const_3), const_3), const_2), const_3) | multiply(n0,const_100)|divide(#0,const_3)|divide(#1,const_3)|divide(#2,const_3)|divide(#3,const_2)|multiply(#4,const_3) | geometry | B |
the cost of a one - family home was $ 120,000 in 1980 . in 1988 , the price had increased to $ 192,000 . what was the percent increase in the cost of the home ? | increase = 192000 - 120000 = 72000 % increase = 72000 * 100 / 120000 = 60 % answer : option a | a ) 60 % , b ) 50 % , c ) 55 % , d ) 40 % , e ) 33.3 % | a | multiply(divide(subtract(multiply(multiply(const_12, multiply(const_4, const_4)), const_1000), multiply(multiply(const_12, const_1000), const_10)), multiply(multiply(const_12, const_1000), const_10)), const_100) | multiply(const_4,const_4)|multiply(const_1000,const_12)|multiply(#0,const_12)|multiply(#1,const_10)|multiply(#2,const_1000)|subtract(#4,#3)|divide(#5,#3)|multiply(#6,const_100) | general | A |
in a group of ducks and cows , the total number of legs are 36 more than twice the no . of heads . find the total no . of buffaloes . | "let the number of buffaloes be x and the number of ducks be y = > 4 x + 2 y = 2 ( x + y ) + 36 = > 2 x = 36 = > x = 18 e" | a ) 10 , b ) 12 , c ) 13 , d ) 15 , e ) 18 | e | divide(36, const_2) | divide(n0,const_2)| | general | E |
what number times ( 1 β 2 ) ^ 2 will give the value of 2 ^ 3 ? | x * ( 1 / 2 ) ^ 2 = 2 ^ 3 x = 2 ^ 2 * 2 ^ 3 = 2 ^ 5 = 32 the answer is e . | a ) 2 , b ) 4 , c ) 8 , d ) 16 , e ) 32 | e | multiply(power(2, 2), power(2, 3)) | power(n1,n1)|power(n1,n4)|multiply(#0,#1) | general | E |
a caterer ordered 200 ice - cream bars and 200 sundaes . if the total price was $ 200.00 and the price of each ice - cream bar was $ 0.40 , what was the price of each sundae ? | "let price of a sundae = s price of ice cream bar = . 4 $ 200 * . 4 + 200 * s = 200 = > 200 * s = 200 = > s = 0.6 answer a" | a ) $ 0.60 , b ) $ 0.80 , c ) $ 1.00 , d ) $ 1.20 , e ) $ 1.60 | a | divide(subtract(200.00, multiply(200, 0.40)), 200) | multiply(n0,n3)|subtract(n2,#0)|divide(#1,n1)| | general | A |
two dogsled teams raced across a 300 mile course in wyoming . team a finished the course in 3 fewer hours than team r . if team a ' s average speed was 5 mph greater than team r ' s , what was team r ' s average mph ? | "this is a very specific format that has appeared in a handful of real gmat questions , and you may wish to learn to recognize it : here we have a * fixed * distance , and we are given the difference between the times and speeds of two things that have traveled that distance . this is one of the very small number of question formats where backsolving is typically easier than solving directly , since the direct approach normally produces a quadratic equation . say team r ' s speed was s . then team r ' s time is 300 / s . team a ' s speed was then s + 5 , and team a ' s time was then 300 / ( s + 5 ) . we need to find an answer choice for s so that the time of team a is 3 less than the time of team r . that is , we need an answer choice so that 300 / ( s + 5 ) = ( 300 / s ) - 3 . you can now immediately use number properties to zero in on promising answer choices : the times in these questions will always work out to be integers , and we need to divide 300 by s , and by s + 5 . so we want an answer choice s which is a factor of 300 , and for which s + 5 is also a factor of 300 . so you can rule out answers a and c immediately , since s + 5 wo n ' t be a divisor of 300 in those cases ( sometimes using number properties you get to the correct answer without doing any other work , but unfortunately that ' s not the case here ) . testing the other answer choices , if you try answer d , you find the time for team r is 15 hours , and for team a is 12 hours , and since these differ by 3 , as desired , d is correct ." | a ) 12 , b ) 15 , c ) 18 , d ) 20 , e ) 25 | d | divide(divide(300, 5), 3) | divide(n0,n2)|divide(#0,n1)| | physics | D |
obra drove 120 Ο meters along a circular track . if the area enclosed by the circular track on which she drove is 57,600 Ο square meters , what percentage of the circular track did obra drive ? | "area enclosed by the circular track on which she drove is 57,600 Ο square meters so , Ο ( r ^ 2 ) = 57,600 Ο - - - > ( r ^ 2 ) = 57,600 - - - > r = 240 circumference of the circular track = 2 Ο r = 480 Ο therefore , part of circumference covered = 120 Ο / 480 Ο = 25 % hence , answer is d ." | a ) 6.67 % , b ) 12.5 % , c ) 18.75 % , d ) 25 % , e ) 33.3 % | d | subtract(subtract(multiply(multiply(divide(circumface(120), 120), const_3), const_2), const_4), const_2) | circumface(n0)|divide(#0,n0)|multiply(#1,const_3)|multiply(#2,const_2)|subtract(#3,const_4)|subtract(#4,const_2)| | geometry | D |
a can do a job in 12 days and b can do it in 20 days . a and b working together will finish twice the amount of work in - - - - - - - days ? | "a 1 / 12 + 1 / 20 = 2 / 15 15 / 2 * 2 = 15 days" | a ) 15 days , b ) 13 days , c ) 22 days , d ) 11 days , e ) 19 days | a | add(divide(const_1, 12), divide(const_1, 20)) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)| | physics | A |
find the numbers which are in the ratio 3 : 2 : 4 such that the sum of the first and the second added to the difference of the third and the second is 21 ? | let the numbers be a , b and c . a : b : c = 3 : 2 : 4 given , ( a + b ) + ( c - b ) = 21 = > a + c = 21 = > 3 x + 4 x = 21 = > x = 3 a , b , c are 3 x , 2 x , 4 x a , b , c are 9 , 6 , 12 . answer : d | a ) 4 , 3,22 , b ) 4 , 4,22 , c ) 9 , 3,32 , d ) 9 , 6,12 , e ) 9 , 2,23 | d | divide(multiply(4, const_3), const_3) | multiply(n2,const_3)|divide(#0,const_3) | general | D |
a group of men decided to do a work in 17 days , but 8 of them became absent . if the rest of the group did the work in 21 days , find the original number of men ? | "original number of men = 8 * 21 / ( 21 - 17 ) = 42 answer is c" | a ) 15 , b ) 20 , c ) 42 , d ) 25 , e ) 18 | c | divide(multiply(8, 21), subtract(21, 17)) | multiply(n1,n2)|subtract(n2,n0)|divide(#0,#1)| | physics | C |
the ratio of males to females in a class is 2 : 3 . the career preferences of the students in the class are represented in a circle graph . if the area of the graph allocated to each career preference is proportional to the number of students who have that career preference , how many degrees of the circle should be used to represent a career that is preferred by one - half of the males and one - fourth of the females in the class ? | "1 / 2 * 2 / 5 + 1 / 4 * 3 / 5 = 4 / 20 + 3 / 20 = 7 / 20 the number of degrees is 7 / 20 * 360 = 126 degrees the answer is b ." | a ) 120 , b ) 126 , c ) 132 , d ) 138 , e ) 144 | b | divide(multiply(2, const_360), add(2, 3)) | add(n0,n1)|multiply(n0,const_360)|divide(#1,#0)| | geometry | B |
how many positive integers less than 50 have a reminder 2 when divided by 7 ? | "take the multiples of 7 and add 2 0 x 7 + 2 = 2 . . . . 6 x 7 + 2 = 44 there are 7 numbers answer e" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | e | divide(factorial(subtract(add(const_4, 2), const_1)), multiply(factorial(2), 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 | E |
the simple form of the ratio 4 / 5 : 8 / 9 is ? | 4 / 5 : 8 / 9 = 9 : 10 answer : e | a ) 5 : 8 , b ) 4 : 9 , c ) 5 : 9 , d ) 5 : 3 , e ) 9 : 10 | e | divide(8, 9) | divide(n2,n3) | other | E |
calculate the average of all the numbers between 7 and 49 which are divisible by 6 . | "explanation : numbers divisible by 6 are 12 , 18,24 , 30,36 , 42,48 , average = ( 12 + 18 + 24 + 30 + 36 + 42 + 48 , ) / 7 = 210 / 7 = 30 answer : e" | a ) 20 , b ) 18 , c ) 19 , d ) 15 , e ) 30 | e | multiply(divide(add(add(floor(divide(7, 6)), const_1), floor(divide(49, 6))), const_2), 6) | divide(n0,n2)|divide(n1,n2)|floor(#0)|floor(#1)|add(#2,const_1)|add(#4,#3)|divide(#5,const_2)|multiply(n2,#6)| | general | E |
a rectangular wall is covered entirely with two kinds of decorative tiles : regular and jumbo . 1 / 3 of the tiles are jumbo tiles , which have a length three times that of regular tiles and have the same ratio of length to width as the regular tiles . if regular tiles cover 40 square feet of the wall , and no tiles overlap , what is the area of the entire wall ? | "the number of jumbo tiles = x . the number of regular tiles = 2 x . assume the ratio of the dimensions of a regular tile is a : a - - > area = a ^ 2 . the dimensions of a jumbo tile is 3 a : 3 a - - > area = 9 a ^ 2 . the area of regular tiles = 2 x * a ^ 2 = 40 . the area of jumbo tiles = x * 9 a ^ 2 = 4.5 ( 2 x * a ^ 2 ) = 4.5 * 40 = 180 . total area = 40 + 180 = 220 . answer : b ." | a ) 160 , b ) 220 , c ) 360 , d ) 440 , e ) 560 | b | add(40, multiply(divide(multiply(40, 3), const_2), 3)) | multiply(n2,n1)|divide(#0,const_2)|multiply(n1,#1)|add(n2,#2)| | geometry | B |
there are 4 more women than there are men on a local co - ed softball team . if there are a total of 14 players on the team , what is the ratio of men to women ? | "w = m + 4 w + m = 14 m + 4 + m = 14 2 m = 10 m = 5 w = 9 ratio : 5 : 9 ans : e" | a ) 10 / 16 , b ) 6 / 16 , c ) 4 / 16 , d ) 6 / 10 , e ) 5 / 9 | e | divide(divide(subtract(14, 4), add(const_1, const_1)), add(divide(subtract(14, 4), add(const_1, const_1)), 4)) | add(const_1,const_1)|subtract(n1,n0)|divide(#1,#0)|add(n0,#2)|divide(#2,#3)| | general | E |
if w / x = 5 / 7 and w / y = 4 / 7 , then ( x + y ) / y = | "ratio 1 : 7 w = 5 x ratio 2 : 7 w = 4 y 5 x = 4 y x = 4 y / 3 ( x + y ) / y = ( ( 4 y / 3 ) + y ) / y = y ( 4 / 3 + 1 ) / y = 7 / 3 answer is b" | a ) 1 / 2 , b ) 7 / 3 , c ) 6 / 7 , d ) 1 / 5 , e ) 11 / 13 | b | add(divide(divide(4, 5), divide(7, 7)), const_1) | divide(n2,n0)|divide(n3,n1)|divide(#0,#1)|add(#2,const_1)| | general | B |
the sector of a circle has radius of 18 cm and central angle 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 18 ) + 2 ( 18 ) = 42.4 + 36 = 78.4 cm answer : a" | a ) 78.4 cm , b ) 11.5 cm , c ) 91.8 cm , d ) 92.5 cm , e ) 99.5 cm | a | multiply(multiply(const_2, divide(multiply(subtract(18, const_3), const_2), add(const_4, const_3))), 18) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics | A |
john and jane went out for a dinner and they ordered the same dish . both used a 10 % discount coupon . john paid a 15 % tip over the original price of the dish , while jane paid the tip over the discounted price for the coupon . if john paid $ 1.26 more than jane , what was the original price of the dish ? | the difference between the amounts john paid and jane paid is the deference between 15 % of p and 15 % of 0.9 p : 0.15 p - 0.15 * 0.9 p = 1.26 - - > 15 p - 13.5 p = 126 - - > p = 84 . answer : e . | a ) 24 , b ) 34.8 , c ) 37.8 , d ) 42 , e ) 84 | e | divide(1.26, subtract(divide(15, const_100), multiply(subtract(const_1, divide(10, const_100)), divide(15, const_100)))) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#1)|multiply(#0,#2)|subtract(#0,#3)|divide(n2,#4) | gain | E |
two persons start running simultaneously around a circular track of length 250 m from the same point at speeds of 20 kmph and 40 kmph . when will they meet for the first time any where on the track if they are moving in the opposite direction ? | time taken to meet the first time = length of track / relative speed = 250 / ( 20 + 40 ) ( 5 / 18 ) = 250 / 60 * ( 18 / 5 ) = 15 sec . answer : b | a ) 144 , b ) 15 , c ) 18 , d ) 32 , e ) 38 | b | divide(250, multiply(add(20, 40), const_0_2778)) | add(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1) | physics | B |
linda spent 5 / 6 of her savings on furniture and the rest on a tv . if the tv cost her $ 500 , what were her original savings ? | "if linda spent 5 / 6 of her savings on furniture , the rest 6 / 6 - 5 / 6 = 1 / 6 on a tv but the tv cost her $ 500 . so 1 / 6 of her savings is $ 500 . so her original savings are 6 times $ 500 = $ 3000 correct answer b" | a ) $ 9000 , b ) $ 3000 , c ) $ 6000 , d ) $ 7000 , e ) $ 8000 | b | divide(500, subtract(const_1, divide(5, 6))) | divide(n0,n1)|subtract(const_1,#0)|divide(n2,#1)| | general | B |
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