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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find the sum the difference between the compound and s . i . on a certain sum of money for 2 years at 10 % per annum is rs . 100 of money ? | "p = 100 ( 100 / 10 ) 2 = > p = 2000 answer : e" | a ) 1500 , b ) 6000 , c ) 2500 , d ) 1400 , e ) 2000 | e | multiply(multiply(divide(100, multiply(10, 2)), const_100), multiply(10, 2)) | multiply(n0,n1)|divide(n2,#0)|multiply(#1,const_100)|multiply(#2,#0)| | general | E |
what is the minimum value of | x - 4 | + | x + 7 | + | x - 5 | ? | "a can not be the answer as all the three terms are in modulus and hence the answer will be non negative . | x - 4 | > = 0 - - > minimum occurs at x = 4 | x + 7 | > = 0 - - > minimum occurs at x = - 7 | x - 5 | > = 0 - - > minimum occurs at x = 5 x = - 7 - - > result = 11 + 0 + 12 = 23 . also any negative value will push the combined value of | x - 4 | + | x - 5 | to a value > 9 . x = 4 - - > result = 0 + 11 + 1 = 12 x = 5 - - > result = 1 + 12 + 0 = 13 x = 7 - - > result = 3 + 14 + 2 = 19 so minimum value of the expression occurs at x = 4 and the resultant value = 12 answer : e" | a ) - 3 , b ) 3 , c ) 5 , d ) - 12 , e ) 12 | e | add(7, 5) | add(n1,n2)| | general | E |
45 x ? = 60 % of 900 | "answer let 45 x a = ( 60 x 900 ) / 100 ∴ a = ( 60 x 9 ) / 45 = 12 correct option : b" | a ) 16.2 , b ) 12 , c ) 5 , d ) 500 , e ) none | b | divide(multiply(divide(60, const_100), 900), 45) | divide(n1,const_100)|multiply(n2,#0)|divide(#1,n0)| | general | B |
the area of a square field is 900 km 2 . how long will it take for a horse to run around at the speed of 12 km / h ? | explanation area of field = 900 km 2 . then , each side of field = √ 900 = 30 km distance covered by the horse = perimeter of square field = 30 × 4 = 120 km ∴ time taken by horse = distances / peed = 120 / 12 = 10 h answer b | ['a ) 12 h', 'b ) 10 h', 'c ) 8 h', 'd ) 6 h', 'e ) none of these'] | b | divide(multiply(const_4, sqrt(900)), 12) | sqrt(n0)|multiply(#0,const_4)|divide(#1,n2) | geometry | B |
cereal a is 11 % sugar by weight , whereas healthier but less delicious cereal b is 2 % sugar by weight . to make a delicious and healthy mixture that is 4 % sugar , what should be the ratio of cereal a to cereal b , by weight ? | "2 % is 2 % - points below 4 % and 11 % is 7 % - points above 4 % . the ratio of a : b should be 2 : 7 . the answer is c ." | a ) 2 : 5 , b ) 1 : 3 , c ) 2 : 7 , d ) 3 : 4 , e ) 1 : 5 | c | divide(subtract(4, 2), subtract(11, 4)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | general | C |
a vessel of capacity 45 litres is fully filled with pure milk . nine litres of milk is removed from the vessel and replaced with water . nine litres of the solution thus formed is removed and replaced with water . find the quantity of pure milk in the final milk solution ? | "explanation : let the initial quantity of milk in vessel be t litres . let us say y litres of the mixture is taken out and replaced by water for n times , alternatively . quantity of milk finally in the vessel is then given by [ ( t - y ) / t ] ^ n * t for the given problem , t = 45 , y = 9 and n = 2 . hence , quantity of milk finally in the vessel = [ ( 45 - 9 ) / 45 ] ^ 2 ( 45 ) = 28.8 litres . answer : option a" | a ) 28.8 , b ) 72.9 , c ) 38.3 , d ) 78.3 , e ) 79.3 | a | subtract(subtract(45, multiply(const_3, const_3)), multiply(divide(subtract(45, multiply(const_3, const_3)), 45), multiply(const_3, const_3))) | multiply(const_3,const_3)|subtract(n0,#0)|divide(#1,n0)|multiply(#2,#0)|subtract(#1,#3)| | physics | A |
the average temperature for monday , tuesday , wednesday and thursday was 48 degrees and for tuesday , wednesday , thursday and friday was 46 degrees . if the temperature on monday was 44 degrees . find the temperature on friday ? | "m + tu + w + th = 4 * 48 = 192 tu + w + th + f = 4 * 46 = 184 m = 44 tu + w + th = 192 - 44 = 148 f = 184 – 148 = 36 answer : c" | a ) 65 degrees , b ) 73 degrees , c ) 36 degrees , d ) 34 degrees , e ) 74 degrees | c | subtract(44, subtract(multiply(48, const_4), multiply(46, const_4))) | multiply(n0,const_4)|multiply(n1,const_4)|subtract(#0,#1)|subtract(n2,#2)| | general | C |
in a mixed college 160 students are there in one class . out of this 160 students 5 / 8 students are girls . how many boys are there ? | "total number of students : 160 total girls : 160 * 5 / 8 = 100 total boys : 160 - 100 = 60 answer is b" | a ) a ) 40 , b ) b ) 60 , c ) c ) 80 , d ) d ) 120 , e ) e ) 140 | b | multiply(divide(160, 8), const_3) | divide(n0,n3)|multiply(#0,const_3)| | general | B |
how many positive integers less than 7,000 are there in which the sum of the digits equals 5 ? | "basically , the question asks how many 4 digit numbers ( including those in the form 0 xxx , 00 xx , and 000 x ) have digits which add up to 5 . think about the question this way : we know that there is a total of 5 to be spread among the 4 digits , we just have to determine the number of ways it can be spread . let x represent a sum of 1 , and | represent a seperator between two digits . as a result , we will have 5 x ' s ( digits add up to the 5 ) , and 3 | ' s ( 3 digit seperators ) . so , for example : xx | x | x | x = 2111 | | xxx | xx = 0032 etc . there are 8 c 3 ways to determine where to place the separators . hence , the answer is 8 c 3 = 56 . c" | a ) 52 , b ) 54 , c ) 56 , d ) 58 , e ) 60 | c | divide(factorial(subtract(add(const_4, 5), const_1)), multiply(factorial(5), 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 | C |
triangle atriangle b are similar triangles with areas 1536 units square and 1734 units square respectively . the ratio of there corresponding height would be | "let x be the height of triangle a and y be the height of triangle of b . since triangles are similar , ratio of area of a and b is in the ratio of x ^ 2 / y ^ 2 therefore , ( x ^ 2 / y ^ 2 ) = 1536 / 1734 ( x ^ 2 / y ^ 2 ) = ( 16 * 16 * 6 ) / ( 17 * 17 * 7 ) ( x ^ 2 / y ^ 2 ) = 16 ^ 2 / 17 ^ 2 x / y = 16 / 17 ans = d" | a ) 9 : 10 , b ) 17 : 19 , c ) 23 : 27 , d ) 16 : 17 , e ) 15 : 23 | d | sqrt(divide(1536, 1734)) | divide(n0,n1)|sqrt(#0)| | geometry | D |
a train speeds past a pole in 23 seconds and a platform 125 m long in 33 seconds . its length is ? | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 23 = > y = x / 23 x + 125 / 33 = x / 23 x = 287.5 m . answer : b" | a ) 275 , b ) 287.5 , c ) 288 , d ) 233.5 , e ) 245.6 | b | multiply(125, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics | B |
in a company with 48 employees , some part - time and some full - time , exactly ( 1 / 3 ) of the part - time employees and ( 1 / 4 ) of the full - time employees take the subway to work . what is the greatest possible number of employees who take the subway to work ? | "let part time emp = x let full time emp = y then , 48 = x + y . . . . . . . . . ( 1 ) ( 1 / 3 ) x + ( 1 / 4 ) y = no . of ppl taking the subway 4 x + 3 y / 12 = no . of ppl taking the subway using 1 x / 12 + 3 * 48 / 12 = no . of ppl taking the subway so , the minimum value of x has to be 12 . hence to maximize the no . put 36 for x 36 / 12 + 12 / 4 = 15 answer : d" | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | add(divide(multiply(48, 3), multiply(3, 4)), divide(subtract(48, multiply(3, 4)), divide(multiply(48, 3), multiply(3, 4)))) | multiply(n0,n2)|multiply(n2,n4)|divide(#0,#1)|subtract(n0,#1)|divide(#3,#2)|add(#2,#4)| | general | D |
a part - time employee whose hourly wage was increased by 50 percent decided to reduce the number of hours worked per week so that the employee ' s total weekly income would remain unchanged . by what percent should the number of hours worked be reduced ? | "let original hourly wage be x and let the no of hours worked be y total wage will be = x * y after the increment the wage will be = 1.5 x now we need to find number of hours worked so that x * y = 1.5 x * z i . e z = 1 / 1.5 y = 2 / 3 y % decrease = ( y - 2 / 3 y ) / y * 100 = 100 / 3 = 33.33 % . thus my answer is e ." | a ) 9 , b ) 13 , c ) 17 , d ) 20 , e ) 33.33 | e | multiply(divide(const_1, multiply(divide(add(50, const_100), const_100), 50)), const_100) | add(n0,const_100)|divide(#0,const_100)|multiply(#1,n0)|divide(const_1,#2)|multiply(#3,const_100)| | general | E |
a 300 m long train crosses a platform in 33 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 33 = 50 / 3 = > x = 250 m . answer : b" | a ) 286 m , b ) 250 m , c ) 277 m , d ) 278 m , e ) 267 m | b | subtract(multiply(speed(300, 18), 33), 300) | speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics | B |
the area of a rectangular plot is 23 times its breadth . if the difference between the length and the breadth is 10 metres , what is its breadth ? | "l × b = 23 × b ∴ l = 23 m and l – b = 10 ∴ b = 23 – 10 = 13 m answer a" | a ) 13 metres , b ) 5 metres , c ) 7.5 metres , d ) data inadequate , e ) none of these | a | subtract(23, 10) | subtract(n0,n1)| | geometry | A |
a , band c can do a piece of work in 11 days , 5 days and 55 days respectively , working alone . how soon can the work be done if a is assisted by band c on alternate days ? | ( a + b ) ' s 1 day ' s work = 1 / 11 + 1 / 5 = 16 / 55 ( a + c ) ' s 1 day ' s work = 1 / 11 + 1 / 55 = 6 / 55 work done in 2 day ' s = 16 / 55 + 6 / 55 = 2 / 5 2 / 5 th work done in 2 days work done = 5 / 2 * 2 = 5 days answer : a | a ) 5 days , b ) 8 days , c ) 9 days , d ) 10 days , e ) 11 days | a | divide(55, divide(add(add(divide(55, 11), divide(55, 5)), add(divide(55, 11), divide(55, 55))), const_2)) | divide(n2,n0)|divide(n2,n1)|divide(n2,n2)|add(#0,#1)|add(#0,#2)|add(#3,#4)|divide(#5,const_2)|divide(n2,#6) | physics | A |
if a is the smallest positive integer such that 5880 multiplied by a is the square of an integer , then ya must be | 5880 = 2 * 2 * 7 * 7 * 5 * 6 , so we need one 5 and one 6 to make it a square of a number . so 5 * 6 = 30 ans : a | a ) 30 , b ) 15 , c ) 12 , d ) 23 , e ) 17 | a | divide(5880, multiply(power(add(const_3, const_4), const_2), power(const_2, const_2))) | add(const_3,const_4)|power(const_2,const_2)|power(#0,const_2)|multiply(#2,#1)|divide(n0,#3) | geometry | A |
a present value of a machine is $ 900 . its value depletion rate is 10 % per annum then find the machine value after 2 years ? | "p = $ 900 r = 10 % t = 2 years machine value after 2 years = p [ ( 1 - r / 100 ) ^ t ] = 900 * 9 / 10 * 9 / 10 = $ 729 answer is e" | a ) $ 900 , b ) $ 810 , c ) $ 915 , d ) $ 715 , e ) $ 729 | e | multiply(900, power(subtract(const_1, divide(10, const_100)), 2)) | divide(n1,const_100)|subtract(const_1,#0)|power(#1,n2)|multiply(n0,#2)| | gain | E |
in x game of billiards , x can give y 20 points in 60 and he can give z 30 points in 60 . how many points can y give z in x game of 100 ? | "c 25 x scores 60 while y score 40 and z scores 30 . the number of points that z scores when y scores 100 = ( 100 * 30 ) / 40 = 25 * 3 = 75 . in x game of 100 points , y gives ( 100 - 75 ) = 25 points to c ." | a ) 30 , b ) 20 , c ) 25 , d ) 40 , e ) 50 | c | subtract(multiply(subtract(60, 20), const_3), multiply(subtract(60, 30), const_3)) | subtract(n1,n0)|subtract(n1,n2)|multiply(#0,const_3)|multiply(#1,const_3)|subtract(#2,#3)| | general | C |
a can complete the job in 3 hours and b can complete the same job in 3 hours . a works for 1 hour and then b joins and both complete the job . what fraction of the job did b complete | 1 / 6 = a | a ) 1 / 6 , b ) 3 / 10 , c ) 1 / 2 , d ) 5 / 6 , e ) 8 / 9 | a | divide(subtract(subtract(3, 1), 3), multiply(3, 3)) | multiply(n0,n1)|subtract(n0,n2)|subtract(#1,n1)|divide(#2,#0)| | physics | A |
find the area of trapezium whose parallel sides are 18 cm and 12 cm long , and the distance between them is 14 cm ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 18 + 12 ) * ( 14 ) = 224 cm 2 answer : c" | a ) 288 cm 2 , b ) 277 cm 2 , c ) 224 cm 2 , d ) 226 cm 2 , e ) 227 cm 2 | c | quadrilateral_area(14, 12, 18) | quadrilateral_area(n2,n1,n0)| | physics | C |
one men and three women working 7 hours a day finish a work in 5 days . four men and four women working 3 hours a day complete the work in 7 days . the number of days in which only 7 men working 4 hours a day will finish the work is ? | "1 m + 3 w - - - - - 35 h 4 m + 4 w - - - - - - - 21 h 7 m - - - - - - - ? d 35 m + 105 w = 84 m + 84 m 21 w = 49 m 4 * 35 = 7 * x = > x = 20 hours 20 / 4 = 5 days answer : b" | a ) 4 days , b ) 5 days , c ) 6 days , d ) 7 days , e ) 8 days | b | divide(const_1, multiply(multiply(4, 7), divide(subtract(divide(3, multiply(3, 7)), divide(4, multiply(5, 7))), 4))) | multiply(n0,n2)|multiply(n0,n1)|multiply(n0,n5)|divide(n2,#0)|divide(n5,#1)|subtract(#3,#4)|divide(#5,n5)|multiply(#6,#2)|divide(const_1,#7)| | physics | B |
what is the least common multiple of 152 and 190 ? | "yes there is a shorter way 152 = 2 * 2 * 2 * 19 190 = 2 * 5 * 19 i think everyone knows how to do this . then choose 2 * 2 * 2 and choose 5 and choose 19 2 * 2 * 2 * 5 * 19 = 760 answer is c" | a ) 304 , b ) 14440 , c ) 760 , d ) 28880 , e ) 1520 | c | multiply(add(152, const_1), const_2) | add(n0,const_1)|multiply(#0,const_2)| | general | C |
a rectangular tiled patio is composed of 160 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 160 tiles , and it will still be rectangular . how many rows are in the tile patio before the change in layout ? | "suppose there are c columns and there are r rows original situation so , number of tiles = c * r = 160 also . reach column has r tiles and each row has c tiles new situation number of tiles in each column is r - 2 and number of tiles in each row is c + 4 so , number of rows = r - 2 and number of columns is c + 4 so , number of tiles = ( r - 2 ) * ( c + 4 ) = 160 comparing both of them we get c * r = ( r - 2 ) * ( c + 4 ) = > 4 r - 2 c = 8 c = 2 r - 4 putting it in c * r = 160 ( 2 r - 4 ) * r = 160 2 r ^ 2 - 4 r - 160 = 0 r can not be negative so r = 10 and c = 16 so , answer will be c" | a ) 5 , b ) 6 , c ) 10 , d ) 13 , e ) 28 | c | divide(160, divide(add(negate(4), sqrt(add(power(4, 2), multiply(4, multiply(160, 2))))), 2)) | multiply(n0,n1)|negate(n2)|power(n2,n1)|multiply(n2,#0)|add(#3,#2)|sqrt(#4)|add(#1,#5)|divide(#6,n1)|divide(n0,#7)| | geometry | C |
find the 12 th term of an arithmetic progression whose first term is 2 and the common difference is 8 . | "n th term of a . p = a + ( n - 1 ) * d = 2 + ( 12 - 1 ) * 8 , = 2 + 88 = 90 . answer : d" | a ) 45 , b ) 38 , c ) 44 , d ) 90 , e ) 96 | d | add(multiply(subtract(12, const_1), 8), 2) | subtract(n0,const_1)|multiply(n2,#0)|add(n1,#1)| | general | D |
how many multiples of 2 are there between 1 and 54 , exclusive ? | "26 multiples of 2 between 1 and 54 exclusive . from 2 * 1 upto 2 * 26 , ( 1,2 , 3,4 , . . . , 26 ) . hence , 26 multiples ! correct option is d" | a ) 21 , b ) 22 , c ) 24 , d ) 26 , e ) 28 | d | add(divide(subtract(54, 1), 2), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general | D |
a man can swim in still water at 9 km / h , but takes twice as long to swim upstream than downstream . the speed of the stream is ? | "m = 9 s = x ds = 9 + x us = 9 - x 9 + x = ( 9 - x ) 2 9 + x = 18 - 2 x 3 x = 9 x = 3 answer : c" | a ) 1.7 , b ) 1.9 , c ) 3 , d ) 1.5 , e ) 1.2 | c | divide(9, const_3) | divide(n0,const_3)| | general | C |
a thief goes away with a santro car at a speed of 40 kmph . the theft has been discovered after half an hour and the owner sets off in a bike at 50 kmph when will the owner over take the thief from the start ? | "d 20 hours | - - - - - - - - - - - 20 - - - - - - - - - - - - - - - - - - - - | 50 40 d = 20 rs = 50 – 40 = 10 t = 20 / 10 = 2 hours" | a ) 22 hours , b ) 21 hours , c ) 23 hours , d ) 20 hours , e ) 28 hours | d | subtract(divide(multiply(divide(const_1, const_2), 40), subtract(50, 40)), divide(const_1, const_2)) | divide(const_1,const_2)|subtract(n1,n0)|multiply(n0,#0)|divide(#2,#1)|subtract(#3,#0)| | physics | D |
in what ratio mental a at rs . 68 per kg be mixed with another metal at rs . 96 per kg so that cost of alloy ( mixture ) is rs . 72 per kg ? | "( 96 - 72 ) / ( 72 - 68 ) = 24 / 4 = 6 / 1 answer : a" | a ) 6 : 1 , b ) 4 : 7 , c ) 3 : 7 , d ) 9 : 5 , e ) 9 : 8 | a | divide(divide(subtract(96, 72), subtract(96, 68)), subtract(const_1, divide(subtract(96, 72), subtract(96, 68)))) | subtract(n1,n2)|subtract(n1,n0)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)| | other | A |
the h . c . f . of two numbers is 59 and the other two factors of their l . c . m . are 13 and 16 . the larger of the two numbers is : | : explanation : clearly , the numbers are ( 59 x 13 ) and ( 59 x 16 ) . { \ color { blue } \ therefore } larger number = ( 59 x 16 ) = 944 . answer : b ) 944 | a ) 678 , b ) 944 , c ) 456 , d ) 323 , e ) 899 | b | multiply(59, 16) | multiply(n0,n2) | other | B |
how many bricks each measuring 21 cm x 10 cm x 8 cm , will be needed to build a wall 9 m x 5 m x 18.5 m | "explanation : no . of bricks = volume of the wall / volume of 1 brick = ( 900 x 500 x 18.5 ) / ( 21 x 10 x 8 ) = 4955 answer : a" | a ) 4955 , b ) 4899 , c ) 4650 , d ) 7200 , e ) none of these | a | divide(multiply(multiply(9, 5), 18.5), divide(divide(multiply(multiply(21, 10), 8), const_100), const_100)) | multiply(n3,n4)|multiply(n0,n1)|multiply(n5,#0)|multiply(n2,#1)|divide(#3,const_100)|divide(#4,const_100)|divide(#2,#5)| | physics | A |
a certain sum is invested at simple interest at 18 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 240 . find the sum ? | "let the sum be rs . x . ( x * 18 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 240 = > 36 x / 100 - 24 x / 100 = 240 = > 12 x / 100 = 240 = > x = 2000 . answer : e" | a ) 7000 , b ) 7029 , c ) 2778 , d ) 2800 , e ) 2000 | e | divide(240, divide(multiply(subtract(18, 12), const_2), const_100)) | subtract(n0,n1)|multiply(#0,const_2)|divide(#1,const_100)|divide(n2,#2)| | gain | E |
a machine , working at a constant rate , manufactures 72 staplers in 28 minutes . how many staplers does it make in 1 hr 52 min ? | "change 1 hr 52 min to 112 min . for this , we need to set up a simple proportion of staplers per time 72 / 28 = s / 112 . 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 28 in the two denominators . 72 / 28 * 28 = s / 112 * 28 72 / 1 = s / 4 s = 4 * 72 s = 288 the machine would be 288 staplers in 1 hr 52 min . answer : d" | a ) 268 , b ) 262 , c ) 300 , d ) 288 , e ) 250 | d | multiply(divide(add(multiply(1, const_60), 52), 28), 72) | multiply(n2,const_60)|add(n3,#0)|divide(#1,n1)|multiply(n0,#2)| | physics | D |
the length of rectangle is thrice its breadth and its perimeter is 96 m , find the area of the rectangle ? | "2 ( 3 x + x ) = 96 l = 36 b = 12 lb = 36 * 12 = 432 answer : a" | a ) 432 , b ) 288 , c ) 376 , d ) 397 , e ) 592 | a | multiply(multiply(divide(96, add(multiply(const_3, const_2), multiply(const_1, const_2))), const_3), divide(96, add(multiply(const_3, const_2), multiply(const_1, const_2)))) | multiply(const_2,const_3)|multiply(const_1,const_2)|add(#0,#1)|divide(n0,#2)|multiply(#3,const_3)|multiply(#3,#4)| | geometry | A |
mrs . evans gave a test to her freshmen economics class , which has 25 students enrolled and 22 of them answered question 1 correctly . if 20 answered question 2 correctly and 3 did not take the test then how many answered both questions correctly ? | "total number of enrolled students = 25 number of students who did not take test = 3 hence , number of students who took test = 25 - 3 = 22 number of students who answered q 2 correctly = 20 , therefore , all students who took test answered q 2 correctly . so , number of students who answered q 1 correctly , also answered q 2 correctly = 20 . number of students who answered both q 1 & q 2 correctly = 20 . answer : b" | a ) 30 , b ) 20 , c ) 23 , d ) 17 , e ) 19 | b | subtract(add(add(22, 20), 3), 25) | add(n1,n3)|add(n5,#0)|subtract(#1,n0)| | other | B |
7 does not occur in 1000 . so we have to count the number of times it appears between 1 and 999 . any number between 1 and 999 can be expressed in the form of xyz where 0 < x , y , z < 9 . | 1 . the numbers in which 7 occurs only once . e . g 7 , 17 , 78 , 217 , 743 etc this means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits ( i . e 0 to 9 with the exception of 7 ) you have 1 * 9 * 9 = 81 such numbers . however , 7 could appear as the first or the second or the third digit . therefore , there will be 3 * 81 = 243 numbers ( 1 - digit , 2 - digits and 3 - digits ) in which 7 will appear only once . in each of these numbers , 7 is written once . therefore , 243 times . 2 . the numbers in which 7 will appear twice . e . g 772 or 377 or 747 or 77 in these numbers , one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7 ) . there will be 9 such numbers . however , this digit which is not 7 can appear in the first or second or the third place . so there are 3 * 9 = 27 such numbers . in each of these 27 numbers , the digit 7 is written twice . therefore , 7 is written 54 times . 3 . the number in which 7 appears thrice - 777 - 1 number . 7 is written thrice in it . therefore , the total number of times the digit 7 is written between 1 and 999 is 243 + 54 + 3 = 300 option b | a ) 200 , b ) 300 , c ) 400 , d ) 500 , e ) 470 | b | add(add(multiply(const_3, multiply(9, 9)), multiply(multiply(const_3, 9), const_2)), const_3) | multiply(n7,n7)|multiply(n7,const_3)|multiply(#0,const_3)|multiply(#1,const_2)|add(#2,#3)|add(#4,const_3) | general | B |
a library branch originally contained 18365 volumes , 30 % of which were fiction novels . 1 / 3 of the volumes were transferred to another location and 1 / 3 of the volumes transferred were fiction novels . what percent of the remaining collection was fiction novels ? | "fiction novels = 5,509 transferred to another location = 6,121 transferred fiction novels = 2,040 non transferred fiction novels = 3,469 percent of the remaining collection was fiction novels = 3469 / ( 18365 - 6121 ) * 100 = > 28.332 . . . % hence answer will be ( c )" | a ) 2.5 % , b ) 17.67 % , c ) 28.3 % , d ) 45.2 % , e ) 73.6 % | c | multiply(divide(multiply(divide(30, const_100), subtract(1, divide(1, 3))), subtract(1, divide(1, 3))), const_100) | divide(n1,const_100)|divide(n2,n5)|divide(n2,n3)|subtract(n2,#1)|subtract(n2,#2)|multiply(#0,#3)|divide(#5,#4)|multiply(#6,const_100)| | gain | C |
the interest on a certain deposit at 4.5 % p . a . is rs . 202.50 in one year . how much will the additional interest in one year be on the same deposit at 5 % p . a . ? | "solution s . i . = rs . 202.50 , r = 4.5 % , t = 1 year . principal = rs . ( 100 x 202.50 / 4.5 x 1 ) = rs . 4500 . now , p = rs . 4500 , r = 5 % , t = 1 year . s . i . = rs . ( 4500 x 5 x 1 / 1000 = rs . 225 . ∴ difference in interest = rs . ( 225 - 202.50 ) = rs . 22.50 . answer b" | a ) rs . 20.25 , b ) rs . 22.50 , c ) rs . 25 , d ) rs . 42.75 , e ) none | b | subtract(divide(multiply(divide(202.50, divide(4.5, const_100)), 5), const_100), 202.50) | divide(n0,const_100)|divide(n1,#0)|multiply(n2,#1)|divide(#2,const_100)|subtract(#3,n1)| | gain | B |
a is twice as good a workman as b and they took 8 days together to do the work b alone can do it in ? | "wc = 2 : 1 2 x + x = 1 / 8 x = 1 / 24 = > 24 days answer : e" | a ) 25 days , b ) 88 days , c ) 21 days , d ) 11 days , e ) 24 days | e | multiply(divide(multiply(8, add(const_2, const_1)), const_2), const_2) | add(const_1,const_2)|multiply(n0,#0)|divide(#1,const_2)|multiply(#2,const_2)| | physics | E |
a man spends 1 / 3 rd of his salary on food . he spends 1 / 4 th on rent and 1 / 5 th on cloths . if he is left with 1760 , then who much salary does he earn | let total salary be x a / q = > x = ( x / 3 ) + ( x / 4 ) + ( x / 5 ) + 1760 = > x - 47 x / 60 = 1760 = > 13 x / 60 = 1760 = > x = 1760 * 60 / 13 = > x = 8123 ( approax ) answer : b | a ) 8223 , b ) 8123 , c ) 8323 , d ) 8423 , e ) 8523 | b | divide(1760, subtract(const_1, add(add(divide(1, 3), divide(1, 4)), divide(1, 5)))) | divide(n0,n1)|divide(n0,n3)|divide(n0,n5)|add(#0,#1)|add(#3,#2)|subtract(const_1,#4)|divide(n6,#5) | general | B |
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 42 , the how old is b ? | "explanation : let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 42 ⇒ 5 x = 40 ⇒ x = 8 . hence , b ' s age = 2 x = 16 years . answer : a" | a ) 16 , b ) 9 , c ) 8 , d ) 11 , e ) 10 | a | divide(multiply(subtract(42, const_2), const_2), add(const_4, const_1)) | add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)| | general | A |
huey ' s hip pizza sells two sizes of square pizzas : a small pizza that measures 8 inches on a side and costs $ 10 , and a large pizza that measures 12 inches on a side and costs $ 20 . if two friends go to huey ' s with $ 30 apiece , how many more square inches of pizza can they buy if they pool their money than if they each purchase pizza alone ? | "in the first case each can buy one pizza of $ 10 and one pizza of $ 20 . in square inches that would be ( 8 * 8 = 64 ) for the small pizza and ( 12 * 12 = 144 ) for the large pizza . in total sq inches that would be ( 64 + 144 ) * 2 = 416 sq inches . in the second case if they pool their money together they can buy 3 large pizzas . in terms of square inches that would be 3 * 144 = 432 sq inches . hence , the difference is 16 square inches more ( 432 - 416 ) . the correct answer is c" | a ) 5 square inches , b ) 10 square inches , c ) 16 square inches , d ) 25 square inches , e ) 350 square inches | c | subtract(multiply(power(12, const_2), const_3), add(add(power(8, const_2), power(12, const_2)), add(power(8, const_2), power(12, const_2)))) | power(n2,const_2)|power(n0,const_2)|add(#1,#0)|multiply(#0,const_3)|add(#2,#2)|subtract(#3,#4)| | geometry | C |
if 1 = 6 , 2 = 12 , 3 = 18 , 4 = 24 , 5 = 30 , then 6 = ? | solution : 1 as stated 1 = 6 = > 6 = 1 answer c | a ) 5 , b ) 3 , c ) 1 , d ) 7 , e ) 9 | c | subtract(2, 1) | subtract(n2,n0) | general | C |
if a no . when divided by 44 , gives 432 as quotient and 0 as remainder . what will be the remainder when dividing the same no . by 39 | "p ã · 44 = 432 = > p = 432 * 44 = 19008 p / 39 = 19008 / 39 = 487 , remainder = 15 c" | a ) 11 , b ) 13 , c ) 15 , d ) 17 , e ) 19 | c | reminder(multiply(432, 44), 39) | multiply(n0,n1)|reminder(#0,n3)| | general | C |
at an upscale fast - food restaurant , shin can buy 3 burgers , 7 shakes , and one cola for $ 120 . at the same place it would cost $ 162.50 for 4 burgers , 10 shakes , and one cola . how much would it cost for a meal of one burger , one shake , and one cola ? | "let ' s suppose that the price of a burger is bb , of a shake - ss and that of a cola is cc . we can then construct these equations : 3 b + 7 s + c = 120 4 b + 10 s + c = 162.5 subtracting the first equation from the second gives us b + 3 s = 42.5 now if we subtract the new equation two times from first or 3 times from second we will get b + s + c = 35 . in any case , there is no necessity to know each item ' s price , just the sum . answer : b" | a ) $ 21 , b ) $ 35 , c ) $ 31 , d ) $ 41 , e ) it can not be determined | b | subtract(add(120, subtract(162.50, 120)), multiply(subtract(162.50, 120), 3)) | subtract(n3,n2)|add(n2,#0)|multiply(#0,n0)|subtract(#1,#2)| | general | B |
if y > 0 , ( 8 y ) / 20 + ( 3 y ) / 10 is what percent of y ? | "can be reduced to 4 y / 10 + 3 y / 10 = 7 y / 10 = 70 % answer d" | a ) 40 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 80 % | d | multiply(const_100, add(divide(8, 20), divide(3, 10))) | divide(n1,n2)|divide(n3,n4)|add(#0,#1)|multiply(#2,const_100)| | general | D |
a rectangular tiled patio is composed of 30 square tiles . the rectangular patio will be rearranged so that there will be 2 fewer columns of tiles and 4 more rows of tiles . after the change in layout , the patio will still have 30 tiles , and it will still be rectangular . how many rows are in the tile patio before the change in layout ? | "suppose there are c columns and there are r rows original situation so , number of tiles = c * r = 30 also . reach column has r tiles and each row has c tiles new situation number of tiles in each column is r - 2 and number of tiles in each row is c + 4 so , number of rows = r - 2 and number of columns is c + 4 so , number of tiles = ( r - 2 ) * ( c + 4 ) = 30 comparing both of them we get c * r = ( r - 2 ) * ( c + 4 ) = > 4 r - 2 c = 8 c = 2 r - 4 putting it in c * r = 30 ( 2 r - 4 ) * r = 30 2 r ^ 2 - 4 r - 30 = 0 r can not be negative so r = 5 and c = 6 so , answer will be a" | a ) 5 , b ) 6 , c ) 10 , d ) 13 , e ) 28 | a | divide(30, divide(add(negate(4), sqrt(add(power(4, 2), multiply(4, multiply(30, 2))))), 2)) | multiply(n0,n1)|negate(n2)|power(n2,n1)|multiply(n2,#0)|add(#3,#2)|sqrt(#4)|add(#1,#5)|divide(#6,n1)|divide(n0,#7)| | geometry | A |
a group of 4 investment bankers and 5 clients recently frequented the bonbon ribs restaurant . the total bill for the meal , including 20 % gratuity , came to $ 756 . on average , how much did the meal of each individual cost before gratuity ? | 4 ibs and 5 clients - so total 9 people the bill $ 756 includes 20 % gratuity . . . so the actual cost of dinner was $ 630 now , the cost per person will be $ 630 / 9 which is $ 70 option b | a ) $ 160 , b ) $ 70 , c ) $ 90 , d ) $ 80 , e ) $ 55 | b | divide(divide(756, divide(add(20, const_100), const_100)), add(4, 5)) | add(n2,const_100)|add(n0,n1)|divide(#0,const_100)|divide(n3,#2)|divide(#3,#1) | general | B |
if there are only 2 wheelers and 4 wheelers parked in a school located at the heart of the city , find the number of 4 wheelers parked there if the total number of wheels is 58 ? | "four wheeler = 14 * 4 = 56 ( max ) 2 wheel = 1 so no of 4 wheeler = 14 answer : d" | a ) 11 , b ) 12 , c ) 13 , d ) 14 , e ) 15 | d | divide(subtract(58, 2), 4) | subtract(n3,n0)|divide(#0,n1)| | general | D |
20 machines can do a work in 5 days . how many machines are needed to complete the work in 10 days ? | "required number of machines = 20 * 5 / 10 = 10 answer is a" | a ) 10 , b ) 15 , c ) 8 , d ) 12 , e ) 20 | a | divide(multiply(20, 5), 10) | multiply(n0,n1)|divide(#0,n2)| | physics | A |
john traveled 80 % of the way from yellow - town to green - fields by train at an average speed of 80 miles per hour . the rest of the way john traveled by car at an average speed of v miles per hour . if the average speed for the entire trip was 45 miles per hour , what is v in miles per hour ? | "hibunuel the question seems incorrect . it should not be 80 % at the speed of 80 . however if it ' s 20 % at the speed of 80 , answer comes out 55 . the question is correct . here ' s the explanation : let distance be d . we can find the total timeequate it , which comes as : 0.8 d / 80 + 0.2 d / v = d / 40 = > v = 40 ( option b ) ." | a ) 30 , b ) 40 , c ) 50 , d ) 55 , e ) 70 | b | multiply(subtract(const_100, 80), subtract(divide(const_100, 45), divide(80, 80))) | divide(const_100,n2)|divide(n0,n0)|subtract(const_100,n0)|subtract(#0,#1)|multiply(#2,#3)| | physics | B |
a sum of 25 was paid for a work which a can do in 32 days , b in 20 days , b and c in 12 days and d in 24 days . how much did c receive if all the 4 work together ? | a ' s one day ' s work = 1 ⁄ 32 b ' s one day ' s work = 1 ⁄ 20 ( b + c ) ' s one day ' s work = 1 ⁄ 12 ∴ c ' s one day ' s work = 1 ⁄ 12 - 1 ⁄ 20 = 1 ⁄ 30 d ' s one day ' s work = 1 ⁄ 24 ∴ ( a + b + c + d ) ' s one day ' s work 1 / 32 + 1 / 20 + 1 / 30 + 1 / 24 = 75 + 120 + 80 + 100 / 2400 = 375 ⁄ 2400 = 15 ⁄ 96 = 5 ⁄ 32 ∴ out of 5 ⁄ 32 of work done 1 ⁄ 30 of the work is done by c . ⇒ out of 25 paid for the work , c will receive 1 / 30 / 5 / 32 × 25 , i . e , 1 ⁄ 30 × 32 ⁄ 5 × 25 , i . e , 16 ⁄ 3 answer b | a ) 14 ⁄ 3 , b ) 16 ⁄ 3 , c ) 15 ⁄ 3 , d ) 17 ⁄ 3 , e ) none of these | b | multiply(divide(lcm(lcm(32, 20), 24), divide(const_1, subtract(divide(const_1, 12), divide(const_1, 20)))), divide(25, add(add(add(divide(lcm(lcm(32, 20), 24), 32), divide(lcm(lcm(32, 20), 24), 20)), divide(lcm(lcm(32, 20), 24), divide(const_1, subtract(divide(const_1, 12), divide(const_1, 20))))), divide(lcm(lcm(32, 20), 24), 24)))) | divide(const_1,n3)|divide(const_1,n2)|lcm(n1,n2)|lcm(n4,#2)|subtract(#0,#1)|divide(const_1,#4)|divide(#3,n1)|divide(#3,n2)|divide(#3,n4)|add(#6,#7)|divide(#3,#5)|add(#9,#10)|add(#11,#8)|divide(n0,#12)|multiply(#10,#13) | physics | B |
a semicircle has a radius of 11 . what is the approximate perimeter of the semicircle ? | the perimeter of a circle is 2 * pi * r . the perimeter of a semicircle is 2 * pi * r / 2 + 2 r = pi * r + 2 r the perimeter is pi * 11 + 2 * 11 which is about 57 . the answer is c . | ['a ) 45', 'b ) 51', 'c ) 57', 'd ) 63', 'e ) 69'] | c | add(divide(circumface(11), const_2), multiply(const_2, 11)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1) | geometry | C |
zoey won the lottery and got $ 7 , 348340 . she wants to split it among herself and 5 friends evenly . how much money must she add if she wants to split it evenly ? | zoey and her 5 friends is 6 people in all . $ 7 , 348340 must be divisible by 6 if she wants to split it evenly . the money is divisible by 6 if it ' s divisible by 2 and 3 . 7 + 3 + 4 + 8 + 3 + 4 + 0 = 29 . 2 and 3 cant go into 29 . adding 1 dollar makes it 30 . 2 and 3 can go into 30 so 6 can also . the answer is d . | a ) $ 3 , b ) $ 2 , c ) $ 9 , d ) $ 1 , e ) $ 4 | d | subtract(reminder(add(multiply(7, multiply(const_1000, const_1000)), 348340), add(5, const_1)), const_1) | add(n2,const_1)|multiply(const_1000,const_1000)|multiply(n0,#1)|add(n1,#2)|reminder(#3,#0)|subtract(#4,const_1) | general | D |
if the wheel is 14 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 14 * x = 1056 = > x = 12 answer : d" | a ) 17 , b ) 19 , c ) 17 , d ) 12 , e ) 91 | d | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14)) | multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)| | physics | D |
a man sold 20 articles for $ 60 and gained 20 % . how many articles should he sell for $ 50 to incur a loss 20 % ? | production cost per article : $ 60 * ( 100 % - 20 % ) / 20 = $ 2.40 required production costs for a loss of 20 % : $ 50 * ( 100 % + 20 % ) = $ 60 number of articles to be sold for $ 60 to incur a 20 % loss : $ 60 / $ 2.40 = 25 thus , solution a is correct . | a ) 25 , b ) 36 , c ) 40 , d ) 50 , e ) 48 | a | divide(original_price_before_loss(20, 50), divide(original_price_before_gain(20, 60), 20)) | original_price_before_gain(n0,n1)|original_price_before_loss(n0,n3)|divide(#0,n0)|divide(#1,#2) | gain | A |
find the value of 1 / ( 3 + 1 / ( 3 + 1 / ( 3 - 1 / 3 ) ) ) ? | "1 / [ 3 + ( 1 / ( 3 + 1 / ( 3 - 1 / 3 ) ) ) ] = > 1 / [ 3 + 1 / ( 3 + 1 / ( 8 / 3 ) ) ] = > 1 / [ 3 + 1 / ( 3 + 3 / 8 ) ] = > 1 / [ 3 + 8 / 27 ] = > 1 / ( 89 / 27 ) = > 27 / 89 c )" | a ) 13 / 89 , b ) 15 / 87 , c ) 27 / 89 , d ) 27 / 87 , e ) 89 / 27 | c | divide(1, add(3, divide(1, add(3, divide(1, subtract(3, divide(1, 3))))))) | divide(n0,n1)|subtract(n1,#0)|divide(n0,#1)|add(n1,#2)|divide(n0,#3)|add(n1,#4)|divide(n0,#5)| | general | C |
a no . when divided by the sum of 555 and 445 gives 2 times their difference as quotient & 30 as remainder . find the no . is ? | "( 555 + 445 ) * 2 * 110 + 30 = 220000 + 30 = 220030 e" | a ) 122443 , b ) 154546 , c ) 165454 , d ) 186545 , e ) 220030 | e | add(multiply(multiply(add(555, 445), 2), subtract(555, 445)), 30) | add(n0,n1)|subtract(n0,n1)|multiply(n2,#0)|multiply(#2,#1)|add(n3,#3)| | general | E |
steve traveled the first 2 hours of his journey at 50 mph and the remaining 3 hours of his journey at 80 mph . what is his average speed for the entire journey ? | "distance traveled in 2 hours = 2 * 50 = 100 m distance traveled in 3 hours = 3 * 80 = 240 m total distance covered = 240 + 100 = 340 m total time = 2 + 3 = 5 h hence avg speed = total distance covered / total time taken = 340 / 5 = 68 mph answer : a" | a ) 68 mph , b ) 56.67 mph , c ) 53.33 mph , d ) 64 mph , e ) 66.67 mph | a | add(divide(add(multiply(80, 3), multiply(50, 2)), add(3, 2)), subtract(divide(const_100, 3), const_0_33)) | add(n0,n2)|divide(const_100,n2)|multiply(n2,n3)|multiply(n0,n1)|add(#2,#3)|subtract(#1,const_0_33)|divide(#4,#0)|add(#6,#5)| | physics | A |
if a fraction is multiplied by itself and then divided by the reciprocal of the same fraction , the result is 18 26 / 27 . find the fraction . | answer if the required fraction be p according to the question ( p x p ) / ( 1 / p ) = 1826 / 27 ⇒ p 3 = 512 / 27 ∴ p = 8 / 3 = 2 2 / 3 correct option : c | a ) 8 / 27 , b ) 1 1 / 3 , c ) 2 2 / 3 , d ) 3 / 3 , e ) none of these | c | add(subtract(multiply(power(add(divide(26, 27), 18), divide(const_1, const_3)), const_3), power(add(divide(26, 27), 18), divide(const_1, const_3))), const_2) | divide(n1,n2)|divide(const_1,const_3)|add(n0,#0)|power(#2,#1)|multiply(#3,const_3)|subtract(#4,#3)|add(#5,const_2) | general | C |
how many paying stones , each measuring 2 1 / 2 m * 2 m are required to pave a rectangular court yard 70 m long and 16 1 / 2 m board ? | "70 * 33 / 2 = 5 / 2 * 2 * x = > x = 231 answer : e" | a ) 99 , b ) 18 , c ) 16 , d ) 10 , e ) 231 | e | divide(multiply(70, add(16, divide(1, 2))), multiply(add(2, divide(1, 2)), 2)) | divide(n1,n0)|add(n5,#0)|add(n0,#0)|multiply(n4,#1)|multiply(n0,#2)|divide(#3,#4)| | general | E |
a train 825 m long passes a man , running at 5 km / hr in the same direction in which the train is going , in 90 seconds . the speed of the train is : | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 5 ) km / hr . x - 5 = 45 = = > x = 50 km / hr . answer : d" | a ) 33 , b ) 27 , c ) 27 , d ) 50 , e ) 81 | d | divide(divide(subtract(825, multiply(multiply(5, const_0_2778), 5)), 5), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics | D |
the sale price sarees listed for rs . 350 after successive discount is 20 % and 5 % is ? | "350 * ( 80 / 100 ) * ( 95 / 100 ) = 266 answer : a" | a ) 266 , b ) 278 , c ) 342 , d ) 787 , e ) 191 | a | subtract(subtract(350, divide(multiply(350, 20), const_100)), divide(multiply(subtract(350, divide(multiply(350, 20), const_100)), 5), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain | A |
find the product of the local value and absolute value of 6 in 564823 | "explanation : place value = local value face value = absolute value the place value of 6 in 564823 is 6 x 10000 = 60000 the face value of 6 in 564823 is nothing but 6 . = > 60000 x 6 = 360,000 answer : option e" | a ) 8000 , b ) 16000 , c ) 12000 , d ) 18000 , e ) 360000 | e | multiply(multiply(6, const_1000), 6) | multiply(n0,const_1000)|multiply(n0,#0)| | general | E |
when a train travels at a speed of 60 kmph , it reaches the destination on time . when the same train travels at a speed of 50 kmph , it reaches its destination 15 min late . what is the length of journey ? | "let x be the time reached with the speed 60 km / h 50 km / h - - - - > x + 15 distance is equal so 60 ( km / h ) × xhr = 50 ( km / h ) × ( x + 15 ) hr so 60 x = 50 x + 750 so the would be in km and x = 75 so 75 km answer : a" | a ) 75 km , b ) 50 km , c ) 60 km , d ) 85 km , e ) 95 km | a | multiply(const_60, divide(multiply(divide(15, const_60), 50), 50)) | divide(n2,const_60)|multiply(n1,#0)|divide(#1,n1)|multiply(#2,const_60)| | physics | A |
in a company the manager wants to give some gifts to all of the workers . in each block there are about 200 workers are there . the total amount for giving the gifts for all the workers is $ 6000 . the worth of the gift is 2 $ . how many blocks are there in the company ? | "each employee will get a gift worth of = $ 2 total employees = 6000 / 2 = 3000 total blocks = 3000 / 200 = 15 correct option is e" | a ) 10 , b ) 12 , c ) 18 , d ) 20 , e ) 15 | e | divide(divide(6000, 2), 200) | divide(n1,n2)|divide(#0,n0)| | general | E |
a and b undertake to do a piece of work for rs . 600 . a alone can do it in 6 days while b alone can do it in 8 days . with the help of c , they finish it in 3 days . ! find the share of each . | c ' s 1 day ' s work = 1 / 3 - ( 1 / 6 + 1 / 8 ) = 24 a : b : c = ratio of their 1 day ' s work = 1 / 6 : 1 / 8 : 1 / 24 = 4 : 3 : 1 . a ’ s share = rs . ( 600 * 4 / 8 ) = rs . 300 , b ' s share = rs . ( 600 * 3 / 8 ) = rs . 225 . c ' s share = rs . [ 600 - ( 300 + 225 » ) = rs . 75 . answer is c | a ) 25 , b ) 50 , c ) 75 , d ) 80 , e ) none of them | c | multiply(divide(const_1, add(add(const_4, const_3), const_1)), 600) | add(const_3,const_4)|add(#0,const_1)|divide(const_1,#1)|multiply(n0,#2) | physics | C |
a boat running up stram takes 6 hours to cover a certain distance , while it takes 7 hours to cover the same distance running down stream . what is the ratio between the speed of the boat and the speed of water current respectively ? | "explanation : let speed of boat is x km / h and speed stream is y km / hr 6 ( x + y ) = 7 ( x - y ) 6 x + 6 y = 7 x - 7 y 13 y = x 13 y = x x / y = 13 / 1 13 : 1 answer : option d" | a ) 2 : 3 , b ) 5 : 6 , c ) 4 : 5 , d ) 13 : 1 , e ) 8 : 1 | d | subtract(7, 6) | subtract(n1,n0)| | physics | D |
the cash difference between the selling prices of an article at a profit of 10 % and 6 % is rs . 3 . the ratio of the two selling prices is ? | "let c . p . of the article be rs . x . then , required ratio = 110 % of x / 106 % of x = 110 / 106 = 55 / 53 = 55 : 53 answer : b" | a ) 52 : 56 , b ) 55 : 53 , c ) 52 : 50 , d ) 22 : 56 , e ) 52 : 51 | b | divide(add(const_100, 10), add(const_100, 6)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)| | gain | B |
n and m are each 3 - digit integers . each of the numbers 2 , 3,4 , 6 , 7 , and 8 is a digit of either n or m . what is the smallest possible positive difference between n and m ? | you have 6 digits : 2 , 3 , 4 , 6 , 7 , 8 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as small as possible and the smaller number needs to be as large as possible to be close to each other . the first digit ( hundreds digit ) of both numbers should be consecutive integers now let ' s think about the next digit ( the tens digit ) . to minimize the difference between the numbers , the tens digit of the greater number should be as small as possible and the tens digit of the smaller number should be as large as possible . so let ' s not use 2 and 8 in the hundreds places and reserve them for the tens places . now what are the options ? try and make a pair with ( 3 * * and 4 * * ) . make the 3 * * number as large as possible and make the 4 * * number as small as possible . 387 and 426 ( difference is 39 ) or try and make a pair with ( 6 * * and 7 * * ) . make the 6 * * number as large as possible and make the 7 * * number as small as possible . we get 684 and 723 ( difference is 39 ) a | a ) 39 , b ) 49 , c ) 58 , d ) 113 , e ) 131 | a | subtract(subtract(const_100, multiply(subtract(8, 2), const_10)), const_1) | subtract(n5,n1)|multiply(#0,const_10)|subtract(const_100,#1)|subtract(#2,const_1) | general | A |
a train passes a station platform in 46 sec and a man standing on the platform in 16 sec . if the speed of the train is 54 km / hr . what is the length of the platform ? | "speed = 54 * 5 / 18 = 15 m / sec . length of the train = 15 * 16 = 240 m . let the length of the platform be x m . then , ( x + 240 ) / 46 = 15 = > x = 450 m . answer : d" | a ) 227 , b ) 240 , c ) 460 , d ) 450 , e ) 455 | d | multiply(16, multiply(54, const_0_2778)) | multiply(n2,const_0_2778)|multiply(n1,#0)| | physics | D |
in what time will a train 95 m long cross an electric pole , it its speed be 214 km / hr ? | "speed = 214 * 5 / 18 = 59 m / sec time taken = 95 / 59 = 1.6 sec . answer : a" | a ) 1.6 sec , b ) 2.9 sec , c ) 2.7 sec , d ) 8.7 sec , e ) 8.5 sec | a | divide(95, multiply(214, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | A |
find the value of ( 28 + 48 / 69 ) × 69 | "= ( 28 + 48 / 69 ) × 69 = ( 1932 + 48 ) / 69 × 69 = 1980 / 69 × 69 = 1980 answer is a ." | a ) 1980 , b ) 1982 , c ) 1930 , d ) 1988 , e ) 1680 | a | multiply(add(divide(48, 69), 28), 69) | divide(n1,n2)|add(n0,#0)|multiply(#1,n2)| | general | A |
express 15 mps in kmph ? | "15 * 18 / 5 = 54 kmph answer : e" | a ) 22 , b ) 88 , c ) 90 , d ) 21 , e ) 54 | e | multiply(divide(15, const_1000), const_3600) | divide(n0,const_1000)|multiply(#0,const_3600)| | physics | E |
a shopkeeper sold an article offering a discount of 5 % and earned a profit of 21.6 % . what would have been the percentage of profit earned if no discount was offered ? | "let c . p . be rs . 100 . then , s . p . = rs . 121.60 let marked price be rs . x . then , 95 / 100 x = 121.60 x = 12160 / 95 = rs . 128 now , s . p . = rs . 128 , c . p . = rs . 100 profit % = 28 % . answer : b" | a ) 60 % , b ) 28 % , c ) 30 % , d ) 56 % , e ) 73 % | b | subtract(divide(multiply(add(const_100, 21.6), const_100), subtract(const_100, 5)), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,const_100)|divide(#2,#1)|subtract(#3,const_100)| | gain | B |
cole drove from home to work at an average speed of 75 kmh . he then returned home at an average speed of 105 kmh . if the round trip took a total of 4 hours , how many minutes did it take cole to drive to work ? | first round distance travelled ( say ) = d speed = 75 k / h time taken , t 2 = d / 75 hr second round distance traveled = d ( same distance ) speed = 105 k / h time taken , t 2 = d / 105 hr total time taken = 4 hrs therefore , 4 = d / 75 + d / 105 lcm of 75 and 105 = 525 4 = d / 75 + d / 105 = > 4 = 7 d / 525 + 5 d / 525 = > d = 525 / 3 km therefore , t 1 = d / 75 = > t 1 = 525 / ( 3 x 75 ) = > t 1 = ( 7 x 60 ) / 3 - - in minutes = > t 1 = 140 minutes . e | a ) 66 , b ) 70 , c ) 95 , d ) 112 , e ) 140 | e | multiply(divide(multiply(105, 4), add(75, 105)), const_60) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60) | physics | E |
which no . need to add to 859622 to get a no . exactly divisible by 456 ? | dividend = quotient * divisor + reminder 859622 / 456 gives quotient = 1885 and reminder = 62 . so , the next number divisible by 456 is 456 places infront of 456 * 1885 which means 456 – 62 = 394 should be added to 859622 . e | a ) 546577 , b ) 674645 , c ) 566578 , d ) 465766 , e ) 859622 | e | subtract(859622, reminder(859622, 456)) | reminder(n0,n1)|subtract(n0,#0) | general | E |
what is the unit ’ s digit of 7 ^ 7 | "7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 1 ( last digit ) 7 ^ 5 = 7 ( last digit ) and the cycle repeats after every 4 powers therefore , last digit of 7 ^ 7 = 3 answer c" | a ) 9 , b ) 5 , c ) 3 , d ) 7 , e ) 1 | c | reminder(power(7, const_2), const_10) | power(n0,const_2)|reminder(#0,const_10)| | general | C |
what quantity of water should taken out to concentrate 18 liters of 40 % acidic liquid to 60 % acidic liquid ? | "required answer is = 18 ( 60 - 40 ) / 60 = 6 liters answer is e" | a ) 5 liters , b ) 10 liters , c ) 15 liters , d ) 8 liters , e ) 6 liters | e | subtract(18, divide(multiply(18, 40), 60)) | multiply(n0,n1)|divide(#0,n2)|subtract(n0,#1)| | gain | E |
15 litres of mixture contains 20 % alcohol and the rest water . if 3 litres of water be mixed with it , the percentage of alcohol in the new mixture would be ? | "alcohol in the 15 litres of mix . = 20 % of 15 litres = ( 20 * 15 / 100 ) = 3 litres water in it = 15 - 3 = 12 litres new quantity of mix . = 15 + 3 = 18 litres quantity of alcohol in it = 3 litres percentage of alcohol in new mix . = 3 * 100 / 18 = 50 / 3 = 16.67 % answer is a" | a ) 16.67 % , b ) 23 % , c ) 18.3 % , d ) 19.75 % , e ) 21.23 % | a | multiply(divide(subtract(add(15, 3), add(multiply(divide(subtract(const_100, 20), const_100), 15), 3)), add(15, 3)), const_100) | add(n0,n2)|subtract(const_100,n1)|divide(#1,const_100)|multiply(n0,#2)|add(n2,#3)|subtract(#0,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain | A |
if 20 men can build a water fountain 56 metres long in 42 days , what length of a similar water fountain can be built by 35 men in 3 days ? | "explanation : let the required length be x metres more men , more length built ( direct proportion ) less days , less length built ( direct proportion ) men 20 : 35 days 42 : 3 : : 56 : x therefore ( 20 x 42 x x ) = ( 35 x 3 x 56 ) x = ( 35 x 3 x 56 ) / 840 = 7 hence , the required length is 7 m . answer : c" | a ) 3 m , b ) 4 m , c ) 7 m , d ) 9 m , e ) 10 m | c | multiply(divide(56, multiply(20, 42)), multiply(35, 3)) | multiply(n0,n2)|multiply(n3,n4)|divide(n1,#0)|multiply(#2,#1)| | physics | C |
a certain lab experiments with white and brown mice only . in one experiment , 4 / 9 of the mice are brown . if there are 20 brown mice in the experiment , how many mice in total are in the experiment ? | "let total number of mice = m number of brown mice = 4 / 9 m number of white mice = 5 / 9 m = 20 = > m = 36 answer c" | a ) 32 , b ) 26 , c ) 36 , d ) 28 , e ) 24 | c | subtract(divide(20, divide(4, 9)), 20) | divide(n0,n1)|divide(n2,#0)|subtract(#1,n2)| | general | C |
the length of a rectangle is reduced by 15 % . by what % would the width have to be increased to maintain the original area ? | sol . required change = ( 15 * 100 ) / ( 100 - 15 ) = 17.6 % c | ['a ) 10 %', 'b ) 20 %', 'c ) 17.6 %', 'd ) 30 %', 'e ) 35 %'] | c | multiply(divide(subtract(const_1, divide(subtract(const_100, 15), const_100)), divide(subtract(const_100, 15), const_100)), const_100) | subtract(const_100,n0)|divide(#0,const_100)|subtract(const_1,#1)|divide(#2,#1)|multiply(#3,const_100) | geometry | C |
mike earns $ 17.5 per hour and phil earns $ 10.5 per hour . approximately how much less , as a percentage , does phil earn than mike per hour ? | "what % less of 17.5 is 10.5 let it be x % less , then = 17.5 ( 1 - x / 100 ) = 10.5 1 - x / 100 = 10.5 / 17.5 x = 200 / 5 x = 40 % ans e" | a ) 25 % , b ) 32.5 % , c ) 37 % , d ) 37.5 % , e ) 40 % | e | multiply(divide(10.5, 17.5), const_100) | divide(n1,n0)|multiply(#0,const_100)| | general | E |
when n is divided by 24 , the remainder is 6 . what is the remainder when 3 n is divided by 8 ? | "let n = 6 ( leaves a remainder of 6 when divided by 24 ) 3 n = 3 ( 6 ) = 18 , which leaves a remainder of 2 when divided by 8 . answer e" | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 2 | e | subtract(6, reminder(3, 8)) | reminder(n2,n3)|subtract(n1,#0)| | general | E |
let f ( x ) = x ^ 2 + bx + c . if f ( 5 ) = 0 and f ( - 3 ) = 0 , then b + c = | "f ( x ) = x ^ 2 + bx + c . if f ( 5 ) = 0 and f ( - 3 ) = 0 , then b + c = f ( 5 ) = 0 = 25 + 5 b + c - - - taking 25 to the other side - > 5 b + c = - 25 f ( - 3 ) = 0 = 9 - 3 b + c - - - taking - 3 b + c to the other side - > 3 b - c = 9 when we add these 2 equations , we get 8 b = - 16 - - - > b = - 2 and while substituting b = - 2 we get c = - 15 . b + c = - 17 - - - answer b" | a ) 18 , b ) - 17 , c ) - 15 , d ) - 21 , e ) - 24 | b | negate(divide(subtract(power(3, 2), 5), add(3, 5))) | add(n3,n1)|power(n3,n0)|subtract(#1,n1)|divide(#2,#0)|negate(#3)| | general | B |
susan made a block with small cubes of 8 cubic cm volume to make a block , 3 small cubes long , 9 small cubes wide and 5 small cubes deep . she realizes that she has used more small cubes than she really needed . she realized that she could have glued a fewer number of cubes together to lock like a block with same dimensions , if it were made hollow . what is the minimum number of cubes that she needs to make the block ? | the total volume ( in terms of number of cubes ) of the solid = 3 * 9 * 5 = 135 the total volume ( in terms of number of cubes ) of the hollow = ( 3 - 2 ) * ( 9 - 2 ) * ( 5 - 2 ) = 21 so number of cubes required = 135 - 21 = 114 answer : b | ['a ) 113', 'b ) 114', 'c ) 115', 'd ) 116', 'e ) 117'] | b | add(multiply(multiply(const_1, 9), 5), add(subtract(multiply(multiply(const_1, 9), 5), multiply(add(const_4, const_3), const_3)), multiply(multiply(const_1, 9), 5))) | add(const_3,const_4)|multiply(n2,const_1)|multiply(n3,#1)|multiply(#0,const_3)|subtract(#2,#3)|add(#2,#4)|add(#5,#2) | geometry | B |
during a sale , the price of a pair of shoes is marked down 12 % from the regular price . after the sale ends , the price goes back to the original price . what is the percent of increase to the nearest percent from the sale price back to the regular price for the shoes ? | "assume the price = 100 price during sale = 88 price after sale = 100 percent increase = 12 / 88 * 100 = 14 % approx . correct option : d" | a ) 9 % , b ) 10 % , c ) 11 % , d ) 14 % , e ) 90 % | d | divide(multiply(12, const_100), subtract(const_100, 12)) | multiply(n0,const_100)|subtract(const_100,n0)|divide(#0,#1)| | gain | D |
in a division , divident is 690 , divisior is 36 and quotient is 19 . find the remainder . | "explanation : 690 = 36 x 19 + r 690 = 684 + r r = 690 - 684 = 6 answer : option e" | a ) a ) 4 , b ) b ) 3 , c ) c ) 2 , d ) d ) 5 , e ) e ) 6 | e | reminder(690, 36) | reminder(n0,n1)| | general | E |
if ( t - 8 ) is a factor of t ^ 2 - kt - 40 , then k = | "t ^ 2 - kt - 48 = ( t - 8 ) ( t + m ) where m is any positive integer . if 48 / 8 = 6 , then we know as a matter of fact that : m = + 6 and thus k = 8 - 6 = 1 t ^ 2 - kt - m = ( t - a ) ( t + m ) where a > m t ^ 2 + kt - m = ( t - a ) ( t + m ) where a < m t ^ 2 - kt + m = ( t - a ) ( t - m ) t ^ 2 + kt + m = ( t + a ) ( t + m ) b" | a ) 16 , b ) 1 , c ) 2 , d ) 6 , e ) 14 | b | add(const_10, 2) | add(n1,const_10)| | general | B |
what is the sum of the numbers between 1 and 10 , inclusive ? | "sol . add all the numbers between 1 and 10 . final answer = e , 55 ." | a ) 85 , b ) 95 , c ) 100 , d ) 105 , e ) 55 | e | divide(multiply(1, 10), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general | E |
a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 9 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter m of liquid a was contained by the can initially ? | "as a : b : : 7 : 5 - - - > only option c is a multiple of 7 and hence it is a good place to start . also a : b : : 7 : 5 means that , a = ( 712 ) * total and b = ( 5 / 12 ) * total if a = 21 , b = 15 - - - > remove 9 litres - - - > you remove ( 7 / 12 ) * 9 of a - - - > a remaining = 21 - ( 7 / 12 ) * 9 = 63 / 4 similarly , for b , you remove ( 5 / 12 ) * 9 - - - > b remaining = 15 - ( 5 / 12 ) * 9 = 45 / 4 and then add 9 more litres of b - - - > 9 + 45 / 4 = 81 / 4 thus a / b ( final ratio ) = ( 45 / 4 ) / ( 81 / 4 ) = 7 : 9 , the same as the final ratio mentioned in the question . hence c is the correct answer . a / b = 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) , where 7 x and 5 x are initial quantities of a and b respectively . thus , 7 / 9 = ( 7 x - ( 7 / 12 ) * 9 ) / ( 5 x - ( 5 / 12 ) * 9 + 9 ) - - - > giving you x = 3 . thus a ( original ) m = 7 * 3 = 21 . c" | a ) 10 , b ) 20 , c ) 21 , d ) 25 , e ) 27 | c | multiply(7, divide(multiply(add(7, 9), subtract(9, multiply(divide(5, add(7, 5)), 9))), subtract(multiply(9, 7), multiply(7, 5)))) | add(n0,n2)|add(n0,n1)|multiply(n0,n2)|multiply(n0,n1)|divide(n1,#1)|subtract(#2,#3)|multiply(n2,#4)|subtract(n2,#6)|multiply(#0,#7)|divide(#8,#5)|multiply(n0,#9)| | physics | C |
mixture contains alcohol and water in the ratio 4 : 3 . if 7 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture . | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 28 x = 4 ( 3 x + 5 ) 16 x = 20 x = 1.25 quantity of alcohol = ( 4 x 1.25 ) litres = 5 litres . answer : c" | a ) 10 , b ) 99 , c ) 5 , d ) 22 , e ) 29 | c | multiply(divide(multiply(4, 5), subtract(multiply(7, 4), multiply(3, 4))), 4) | multiply(n0,n4)|multiply(n0,n2)|multiply(n0,n1)|subtract(#1,#2)|divide(#0,#3)|multiply(n0,#4)| | general | C |
calculate the time it will take for a full tank of water to become completely empty due to a leak given that the tank could be filled in 10 hours , but due to the leak in its bottom it takes 11 hours to be filled ? | part filled without leak in 1 hour = 1 / 10 part filled with leak in 1 hour = 1 / 11 work done by leak in 1 hour = 1 / 10 â ˆ ’ 1 / 11 = 110 hours answer : e | a ) 90 hours , b ) 80 hours , c ) 50 hours , d ) 120 hours , e ) 110 hours | e | divide(multiply(10, 11), subtract(divide(multiply(10, 11), 10), divide(multiply(10, 11), 11))) | multiply(n0,n1)|divide(#0,n0)|divide(#0,n1)|subtract(#1,#2)|divide(#0,#3) | physics | E |
the average of 30 results is 20 and the average of other 20 results is 30 . what is the average of all the results ? | "answer sum of 50 result = sum of 30 result + sum of 20 result . = 30 x 20 + 20 x 30 = 1200 correct option : a" | a ) 24 , b ) 25 , c ) 48 , d ) 50 , e ) none | a | divide(add(multiply(30, 20), multiply(20, 30)), add(30, 20)) | add(n0,n1)|multiply(n0,n1)|add(#1,#1)|divide(#2,#0)| | general | A |
a and b go around a circular track of length 400 m on a cycle at speeds of 36 kmph and 36 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 400 / ( 36 * 5 / 18 ) , 400 / ( 36 * 5 / 18 ) } = lcm ( 40 , 40 ) = 40 sec . answer : a" | a ) 40 sec , b ) 198 sec , c ) 178 sec , d ) 665 sec , e ) 276 sec | a | divide(400, subtract(multiply(36, const_0_2778), multiply(36, const_0_2778))) | multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)| | physics | A |
a picnic attracts 240 persons . there are 80 more men than women , and 80 more adults than children . how many men are at this picnic ? | "adult + children = 240 let , children = y then , adult = y + 80 i . e . y + ( y + 80 ) = 240 i . e . y = 80 i . e . adult = 80 + 80 = 160 adults include only men and women i . e . men + women = 160 let women , w = x then men , m = x + 80 i . e . x + ( x + 20 ) = 2 x + 80 = 160 i . e . x = 40 i . e . men , m = 40 + 80 = 120 answer : option d" | a ) 240 , b ) 75 , c ) 110 , d ) 120 , e ) 200 | d | add(divide(subtract(add(divide(subtract(240, 80), const_2), 80), 80), const_2), 80) | subtract(n0,n1)|divide(#0,const_2)|add(n1,#1)|subtract(#2,n1)|divide(#3,const_2)|add(n1,#4)| | general | D |
two same glasses are respectively 1 / 4 th 1 / 5 th full of milk . they are then filled with water and the contents mixed in a tumbler . the ratio of milk and water in the tumbler is ? | 1 / 4 : 3 / 4 = ( 1 : 3 ) 5 = 5 : 15 1 / 5 : 4 / 5 = ( 1 : 4 ) 4 = 4 : 16 - - - - - - 9 : 31 answer : b | a ) 9 : 39 , b ) 9 : 31 , c ) 9 : 36 , d ) 9 : 32 , e ) 9 : 34 | b | divide(add(5, 4), add(multiply(const_3, const_10), 1)) | add(n1,n3)|multiply(const_10,const_3)|add(n0,#1)|divide(#0,#2) | general | B |
a bag contains 3 white marbles and 3 black marbles . if each of 3 girls and 3 boys randomly selects and keeps a marble , what is the probability that all of the girls select the same colored marble ? | first , total ways to select for all boys and girls , i . e 6 ! / ( 3 ! * 3 ! ) = 6 * 5 * 4 * 3 * 2 * 1 / 3 * 2 * 1 * 3 * 2 * 1 = 20 then there are one two way girls can have all same colors , either white or black . the number of ways in which 3 girls can select 3 white balls = 3 c 3 = 1 the number of ways in which 3 girls can select 3 black balls = 3 c 3 = 1 therefore , total favorable outcomes / total outcomes = 2 / 20 = 1 / 10 b | a ) 1 / 35 , b ) 1 / 10 , c ) 1 / 15 , d ) 1 / 20 , e ) 1 / 25 | b | divide(const_2, choose(add(3, 3), 3)) | add(n0,n0)|choose(#0,n0)|divide(const_2,#1) | probability | B |
convert the 11 / 36 m / s into kilometers per hour ? | "11 / 36 m / s = 11 / 36 * 18 / 5 = 11 / 10 = 1.1 kmph . answer : b" | a ) 2.9 kmph , b ) 1.1 kmph , c ) 1.3 kmph , d ) 1.2 kmph , e ) 5.7 kmph | b | multiply(const_3_6, divide(11, 36)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics | B |
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