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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a policeman noticed a criminal from a distance of 240 km . the criminal starts running and the policeman chases him . the criminal and the policeman run at the rate of 8 km and 9 km per hour respectively . what is the distance between them after 3 minutes ? | "explanation : solution : relative speed = ( 9 - 8 ) = 1 km / hr . distance covered in 3 minutes = ( 1 * 3 / 60 ) km = 1 / 20 km = 50 m . . ' . distance between the criminal and policeman = ( 240 - 50 ) m = 190 m . answer : c" | a ) 100 m , b ) 120 m , c ) 190 m , d ) 150 m , e ) none of these | c | subtract(240, multiply(divide(3, const_60), const_1000)) | divide(n3,const_60)|multiply(#0,const_1000)|subtract(n0,#1)| | physics | C |
in the xy - plane , a triangle has vertices ( 0,0 ) , ( 4,0 ) and ( 4,9 ) . if a point ( a , b ) is selected at random from the triangular region , what is the probability that a - b > 0 ? | "the area of the right triangle is ( 1 / 2 ) * 4 * 9 = 18 . only the points ( a , b ) below the line y = x satisfy a - b > 0 . the part of the triangle which is below the line y = x has an area of ( 1 / 2 ) ( 4 ) ( 4 ) = 8 . p ( a - b > 0 ) = 8 / 18 = 2 / 9 the answer is d ." | a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 9 , e ) 4 / 5 | d | divide(const_4, add(0,0, const_10)) | add(n0,const_10)|divide(const_4,#0)| | geometry | D |
a broker invested her own money in the stock market . during the first year , she increased her stock market wealth by 50 percent . in the second year , largely as a result of a slump in the stock market , she suffered a 30 percent decrease in the value of her stock investments . what was the net increase or decrease on her overall stock investment wealth by the end of the second year ? | "assume the broker invested $ 100 . after year 1 ( 50 % increase in wealth ) : $ 100 * 0.5 = $ 50 increase + $ 100 = $ 150 after year 2 ( 30 % decrease in wealth ) : $ 150 * 0.3 = $ 45 ; $ 150 - $ 45 decrease = $ 105 net increase = $ 105 / $ 100 - 1 = 5 % answer is b ." | a ) − 5 % , b ) 5 % , c ) 15 % , d ) 20 % , e ) 80 % | b | multiply(subtract(multiply(add(const_1, divide(50, const_100)), subtract(const_1, divide(30, const_100))), const_1), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain | B |
the sum of two numbers is 528 and their h . c . f is 33 . the number of pairs of numbers satisfying the above conditions is : | "let the required numbers be 33 a and 33 b . then , 33 a + 33 b = 528 = > a + b = 16 . now , co - primes with sum 16 are ( 1 , 15 ) , ( 3 , 13 ) , ( 5 , 11 ) and ( 7 , 9 ) . required numbers are ( 33 * 1 , 33 * 15 ) , ( 33 * 3 , 33 * 13 ) , ( 33 * 5 , 33 * 11 ) , ( 33 * 7 , 33 * 9 ) . the number of such pairs is 4 . answer : a" | a ) 4 , b ) 5 , c ) 6 , d ) 6 , e ) 1 | a | multiply(divide(add(528, 33), add(const_1, const_1)), subtract(divide(add(528, 33), add(const_1, const_1)), 33)) | add(n0,n1)|add(const_1,const_1)|divide(#0,#1)|subtract(#2,n1)|multiply(#2,#3)| | general | A |
a dishonest shopkeeper professes to sell pulses at the cost price , but he uses a false weight of 960 gm . for a kg . his gain is … % . | "his percentage gain is 100 * 40 / 960 as he is gaining 40 units for his purchase of 960 units . so 4.16 % . answer : a" | a ) 4.16 % , b ) 5.36 % , c ) 4.26 % , d ) 6.26 % , e ) 7.26 % | a | multiply(subtract(inverse(divide(960, 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 |
a train running at the speed of 54 km / hr crosses a pole in 9 seconds . what is the length of the train ? | "speed = ( 54 * 5 / 18 ) m / sec = ( 15 ) m / sec length of the train = ( speed x time ) = ( 15 * 9 ) m = 135 m . answer : a" | a ) 135 , b ) 278 , c ) 255 , d ) 150 , e ) 287 | a | multiply(divide(multiply(54, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics | A |
two goods train each 500 m long , are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one . | "explanation : relative speed = = ( 45 + 30 ) km / hr = 75 x 5 / 18 m / sec = 125 / 6 m / sec we have to find the time taken by the slower train to pass the driver of the faster train and not the complete train . so , distance covered = length of the slower train . therefore , distance covered = 500 m . required time = 500 x 6 / 125 = 24 sec . answer is b" | a ) 12 sec , b ) 24 sec , c ) 48 sec , d ) 60 sec , e ) 65 sec | b | multiply(divide(500, divide(multiply(const_1000, add(45, 30)), const_3600)), const_2) | add(n1,n2)|multiply(#0,const_1000)|divide(#1,const_3600)|divide(n0,#2)|multiply(#3,const_2)| | physics | B |
if the average ( arithmetic mean ) of ( 2 a + 16 ) and ( 3 a - 8 ) is 84 , what is the value of a ? | "( ( 2 a + 16 ) + ( 3 a - 8 ) ) / 2 = ( 5 a + 8 ) / 2 = 84 a = 32 the answer is d ." | a ) 25 , b ) 30 , c ) 28 , d ) 32 , e ) 42 | d | divide(subtract(multiply(84, 2), subtract(16, 8)), add(2, 3)) | add(n0,n2)|multiply(n0,n4)|subtract(n1,n3)|subtract(#1,#2)|divide(#3,#0)| | general | D |
find the maximum value of n such that 50 ! is perfectly divisible by 2520 ^ n | 2520 = 2 ^ 3 * 3 ^ 2 * 5 * 7 here , 7 is the highest prime , so find the no . of 7 ' s in 50 ! only no . of 7 ' s in 50 ! = [ 50 / 7 ] + [ 50 / 7 ^ 2 ] = 7 + 1 = 8 answer : b | a ) 7 , b ) 8 , c ) 9 , d ) 6 , e ) 5 | b | add(divide(50, const_10), const_3) | divide(n0,const_10)|add(#0,const_3) | general | B |
if the area of a square with sides of length 7 centimeters is equal to the area of a rectangle with a width of 4 centimeters , what is the length of the rectangle , in centimeters ? | "let length of rectangle = l 7 ^ 2 = l * 4 = > l = 49 / 4 = 13 answer d" | a ) 4 , b ) 8 , c ) 12 , d ) 13 , e ) 18 | d | divide(power(7, const_2), 4) | power(n0,const_2)|divide(#0,n1)| | geometry | D |
if the sales tax reduced from 3 1 / 2 % to 3 1 / 3 % , then what difference does it make to a person who purchases an article with market price of rs . 6600 ? | "required difference = [ 3 ½ % of rs . 6600 ] – [ 3 1 / 3 % of rs . 6600 ] = [ ( 7 / 20 - ( 10 / 3 ) ] % of rs . 6600 = 1 / 6 % of rs . 6600 = rs . [ ( 1 / 6 ) 8 ( 1 / 100 ) * 6600 ] = rs . 11 . answer is a ." | a ) 11 , b ) 13 , c ) 14 , d ) 18 , e ) 10 | a | divide(multiply(subtract(add(divide(1, 2), 3), add(divide(1, 3), 3)), 6600), const_100) | divide(n1,n2)|divide(n1,n0)|add(n0,#0)|add(n0,#1)|subtract(#2,#3)|multiply(n6,#4)|divide(#5,const_100)| | general | A |
the duplicate ratio of 2 : 3 is ? | "2 ^ 2 : 3 ^ 2 = 4 : 9 answer : a" | a ) 4 : 9 , b ) 1 : 4 , c ) 1 : 8 , d ) 1 : 18 , e ) 1 : 13 | a | divide(power(2, const_2), power(3, const_2)) | power(n0,const_2)|power(n1,const_2)|divide(#0,#1)| | other | A |
if the least common addition of two prime numbers x and y is 10 , where x > y , then the value of 2 x + y is | "( x + y ) = 10 and both x an y are prime . the only values of x and y can be 7 and 3 ( x = 7 and y = 3 ) 2 x + y = 2 * 7 + 3 = 17 correct option : d" | a ) 7 , b ) 9 , c ) 14 , d ) 17 , e ) 21 | d | add(add(multiply(10, const_2), divide(10, 2)), add(const_3, const_3)) | add(const_3,const_3)|divide(n0,n1)|multiply(n0,const_2)|add(#1,#2)|add(#3,#0)| | general | D |
a thief is spotted by a policeman from a distance of 100 meters . when the policeman starts the chase , the thief also starts running . if the speed of the thief be 8 km / hr and that of the policeman 10 km / hr , how far the thief will have run before he is overtaken ? | "relative speed of the policeman = ( 10 - 8 ) km / hr = 2 km / hr . time taken by police man to cover ( 100 m / 1000 ) x 1 / 2 hr = 1 / 20 hr . in 1 / 20 hrs , the thief covers a distance of 8 x 1 / 20 km = 2 / 5 km = 400 m answer is c ." | a ) 350 m , b ) 200 m , c ) 400 m , d ) 500 m , e ) none of them | c | divide(multiply(100, 8), subtract(10, 8)) | multiply(n0,n1)|subtract(n2,n1)|divide(#0,#1)| | physics | C |
a truck covers a distance of 296 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 6.5 km more than that travelled by the truck ? | explanation : speed of the truck = distance / time = 296 / 8 = 37 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 37 + 18 ) = 55 kmph distance travelled by car = 296 + 6.5 = 302.5 km time taken by car = distance / speed = 302.5 / 55 = 5.5 hours . answer – c | a ) 6 hours , b ) 5 hours , c ) 5.5 hours , d ) 8 hours , e ) none | c | divide(add(296, 6.5), add(divide(296, 8), 18)) | add(n0,n3)|divide(n0,n1)|add(n2,#1)|divide(#0,#2) | physics | C |
after decreasing 24 % in the price of an article costs rs . 320 . find the actual cost of an article ? | "cp * ( 76 / 100 ) = 320 cp = 4.21 * 100 = > cp = 421 answer : e" | a ) 400 , b ) 520 , c ) 651 , d ) 525 , e ) 421 | e | divide(320, subtract(const_1, divide(24, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain | E |
a can do a piece of work in 40 days ; b can do the same in 20 days . a started alone but left the work after 10 days , then b worked at it for 10 days . c finished the remaining work in 10 days . c alone can do the whole work in ? | "10 / 40 + 10 / 20 + 10 / x = 1 x = 40 days answer : d" | a ) 24 days , b ) 65 days , c ) 86 days , d ) 40 days , e ) 17 days | d | divide(10, subtract(const_1, divide(add(10, 10), 40))) | add(n2,n2)|divide(#0,n0)|subtract(const_1,#1)|divide(n2,#2)| | physics | D |
by approximately what percent is x greater than 2 / 5 if ( 3 / 5 ) ( x ) = 1 ? | "what percent is x greater than 2 / 5 if ( 3 / 5 ) ( x ) = 1 ? = > x = 5 / 3 % change = [ ( 5 / 3 - 2 / 5 ) / ( 2 / 5 ) ] * 100 = 316.5 % = 316 % approx ans , c" | a ) 723 % , b ) 156 % , c ) 316 % , d ) 37 % , e ) 29 % | c | multiply(divide(subtract(5, divide(2, 5)), divide(2, 5)), const_100) | divide(n0,n1)|subtract(n1,#0)|divide(#1,#0)|multiply(#2,const_100)| | general | C |
due to construction , the speed limit along an 9 - mile section of highway is reduced from 45 miles per hour to 35 miles per hour . approximately how many minutes more will it take to travel along this section of highway at the new speed limit than it would have taken at the old speed limit ? | "old time in minutes to cross 9 miles stretch = 9 * 60 / 45 = 9 * 4 / 3 = 12 new time in minutes to cross 9 miles stretch = 9 * 60 / 35 = 9 * 12 / 7 = 15.42 time difference = 3.42 ans : b" | a ) a ) 5.61 , b ) b ) 3.42 , c ) c ) 10 , d ) d ) 15 , e ) e ) 24 | b | max(multiply(subtract(add(45, 9), const_1), subtract(divide(9, 35), divide(9, 45))), const_4) | add(n0,n1)|divide(n0,n2)|divide(n0,n1)|subtract(#0,const_1)|subtract(#1,#2)|multiply(#3,#4)|max(#5,const_4)| | physics | B |
what is the remainder when 43 ^ 88 is divided by 5 ? | "the units digit of the exponents of 3 cycle in a group of 4 : { 3 , 9 , 7 , 1 } 88 has the form 4 k so the units digit of 43 ^ 88 is 1 . the remainder when dividing by 5 is 1 . the answer is b ." | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(divide(5, const_2), multiply(43, 43)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general | B |
if 0.6 : x : : 5 : 8 , then x is equal to : | "( x * 5 ) = ( 0.6 * 8 ) x = 4.8 / 5 x = 0.96 answer = a" | a ) a ) 0.96 , b ) b ) 1.2 , c ) c ) 1.25 , d ) d ) 1.3 , e ) e ) 0.8 | a | divide(multiply(0.6, 8), 5) | multiply(n0,n2)|divide(#0,n1)| | general | A |
1904 x 1904 = ? | "1904 x 1904 = ( 1904 ) 2 = ( 1900 + 4 ) 2 = ( 1900 ) 2 + ( 4 ) 2 + ( 2 x 1900 x 4 ) = 3610000 + 16 + 15200 . = 3625216 . e )" | a ) 3623216 , b ) 3624216 , c ) 3624316 , d ) 3625116 , e ) 3625216 | e | multiply(divide(1904, 1904), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | E |
a train running at a speed of 36 km / h passes an electric pole in 14 seconds . in how many seconds will the whole train pass a 350 - meter long platform ? | "let the length of the train be x meters . when a train crosses an electric pole , the distance covered is its own length x . speed = 36 km / h = 36000 m / 3600 s = 10 m / s x = 14 * 10 = 140 m . the time taken to pass the platform = ( 140 + 350 ) / 10 = 49 seconds the answer is d ." | a ) 43 , b ) 45 , c ) 47 , d ) 49 , e ) 51 | d | divide(add(multiply(multiply(36, const_0_2778), 14), 350), multiply(36, const_0_2778)) | multiply(n0,const_0_2778)|multiply(n1,#0)|add(n2,#1)|divide(#2,#0)| | physics | D |
solve this 6 + 7 = 12 8 + 9 = 16 5 + 6 = 10 7 + 8 = 14 then , 3 + 3 = ? ? | 5 answer : c | a ) 11 , b ) 12 , c ) 5 , d ) 14 , e ) 18 | c | add(3, subtract(3, subtract(7, subtract(12, 6)))) | subtract(n2,n0)|subtract(n1,#0)|subtract(n12,#1)|add(n12,#2) | general | C |
a metallic sheet is of rectangular shape with dimensions 50 m x 36 m . from each of its corners , a square is cut off so as to make an open box . if the length of the square is 8 m , the volume of the box ( in m 3 ) is : | "l = ( 50 - 16 ) m = 34 m , b = ( 36 - 16 ) m = 20 m , h = 8 m . volume of the box = ( 34 x 20 x 8 ) m 3 = 5440 m 3 . answer : option a" | a ) 5440 , b ) 5120 , c ) 6420 , d ) 7500 , e ) 8960 | a | volume_rectangular_prism(subtract(50, multiply(8, const_2)), subtract(36, multiply(8, const_2)), 8) | multiply(n2,const_2)|subtract(n0,#0)|subtract(n1,#0)|volume_rectangular_prism(n2,#1,#2)| | geometry | A |
the average marks of a class of 28 students is 40 and that of another class of 50 students is 60 . find the average marks of all the students ? | "sum of the marks for the class of 28 students = 28 * 40 = 1120 sum of the marks for the class of 50 students = 50 * 60 = 3000 sum of the marks for the class of 78 students = 1120 + 3000 = 4120 average marks of all the students = 4120 / 78 = 52.8 answer : c" | a ) 52.2 , b ) 59.5 , c ) 52.8 , d ) 52.5 , e ) 52.1 | c | divide(add(multiply(28, 40), multiply(50, 60)), add(28, 50)) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|divide(#3,#0)| | general | C |
if x + y = 25 and x 2 y 3 + y 2 x 3 = 25 , what is the value of xy ? | xy = 1 as x + y = 25 x 2 y 3 + y 2 x 3 = 25 x 2 y 2 ( y + x ) = 25 substituting x + y x 2 y 2 = 1 xy = 1 answer : a | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | sqrt(divide(25, 25)) | divide(n0,n0)|sqrt(#0) | general | A |
a cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick . the thickness of the bottom is : | "explanation : let the thickness of the bottom be x cm . then , [ ( 330 - 10 ) × ( 260 - 10 ) × ( 110 - x ) ] = 8000 × 1000 = > 320 × 250 × ( 110 - x ) = 8000 × 1000 = > ( 110 - x ) = 8000 × 1000 / 320 = 100 = > x = 10 cm = 1 dm . answer : b" | a ) 90 cm , b ) 1 dm , c ) 1 m , d ) 1.1 cm , e ) none of these | b | subtract(multiply(multiply(3.3, 2.6), 1.1), divide(8000, const_1000)) | divide(n0,const_1000)|multiply(n1,n2)|multiply(n3,#1)|subtract(#2,#0)| | physics | B |
the length and breadth of a rectangle is increased by 30 % and 45 % respectively . what is the increase in the area ? | "100 * 100 = 10000 130 * 145 = 18850 - - - - - - - - - - - 8850 10000 - - - - 8850 100 - - - - - 88.50 answer e" | a ) 88.25 , b ) 87.5 , c ) 86.0 , d ) 88.0 , e ) 88.5 | e | divide(multiply(subtract(rectangle_area(add(30, const_100), add(45, const_100)), rectangle_area(const_100, const_100)), const_100), rectangle_area(const_100, const_100)) | add(n0,const_100)|add(n1,const_100)|rectangle_area(const_100,const_100)|rectangle_area(#0,#1)|subtract(#3,#2)|multiply(#4,const_100)|divide(#5,#2)| | geometry | E |
mr . hernandez , who was a resident of state x for only 9 months last year , had a taxable income of $ 42,500 for the year . if the state tax rate were 4 percent of the year ’ s taxable income prorated for the proportion of the year during which the taxpayer was a resident , what would be the amount of mr . hernandez ’ s state x tax for last year ? | total tax for the year = 42,500 x 4 % = 1700 as stated annual tax is prorated as per the duration of stay . prorated tax = 1700 ( 9 / 12 ) = 1275 answer b | a ) $ 900 , b ) $ 1275 , c ) $ 1200 , d ) $ 1300 , e ) $ 1060 | b | divide(multiply(multiply(divide(add(multiply(multiply(4, const_100), const_100), multiply(multiply(const_100, const_0_25), const_100)), const_100), const_4), 9), multiply(const_3, const_4)) | multiply(n2,const_100)|multiply(const_0_25,const_100)|multiply(const_3,const_4)|multiply(#0,const_100)|multiply(#1,const_100)|add(#3,#4)|divide(#5,const_100)|multiply(#6,const_4)|multiply(n0,#7)|divide(#8,#2) | gain | B |
simplify : 32 ÷ 8 × 24 + 16 = | "32 ÷ 8 × × 24 + 16 = > 4 × × 24 + 16 = 96 + 16 = 112 option e" | a ) 130 , b ) 124 , c ) 120 , d ) 115 , e ) 112 | e | add(multiply(32, 8), multiply(24, 16)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | general | E |
on dividing 122 by a number , the quotient is 6 and the remainder is 2 . find the divisor ? | "d = ( d - r ) / q = ( 122 - 2 ) / 6 = 122 / 6 = 20 d )" | a ) 14 , b ) 16 , c ) 18 , d ) 20 , e ) 22 | d | floor(divide(122, 6)) | divide(n0,n1)|floor(#0)| | general | D |
what is the hcf of 2 / 3 , 4 / 9 and 6 / 18 | explanation : hcf of fractions = hcf of numerators / lcm of denominators = ( hcf of 2 , 4 , 6 ) / ( lcm of 3 , 9 , 18 ) = 2 / 18 = 1 / 9 answer : option c | a ) 7 / 45 , b ) 2 / 45 , c ) 1 / 9 , d ) 8 / 45 , e ) 9 / 45 | c | divide(2, 18) | divide(n0,n5) | general | C |
find a sum for 1 st 5 prime no . ' s ? | "required sum = ( 2 + 3 + 5 + 7 + 11 ) = 28 note : 1 is not a prime number option b" | a ) 23 , b ) 28 , c ) 30 , d ) 32 , e ) 34 | b | add(add(add(add(add(add(add(add(const_2, const_3), add(const_2, const_3)), add(add(const_2, const_3), const_2)), add(5, const_2)), add(add(5, const_2), const_2)), add(add(add(5, const_2), const_2), const_4)), add(add(add(add(5, const_2), const_2), const_4), const_2)), add(add(add(add(add(5, const_2), const_2), const_4), const_2), const_4)) | add(const_2,const_3)|add(n1,const_2)|add(#0,#0)|add(#0,const_2)|add(#1,const_2)|add(#2,#3)|add(#4,const_4)|add(#5,#1)|add(#6,const_2)|add(#7,#4)|add(#8,const_4)|add(#9,#6)|add(#11,#8)|add(#12,#10)| | general | B |
running at their respective constant rate , machine x takes 2 days longer to produce w widgets than machines y . at these rates , if the two machines together produce 5 w / 4 widgets in 3 days , how many days would it take machine x alone to produce 2 w widgets . | "rx * ( t + 2 ) = w ry * ( t ) = w . also , w / ( t + 2 ) + w / t = ( 5 w / 4 ) * ( 1 / 3 ) or 1 / t + 1 / ( t + 2 ) = 5 / 12 - - > ( 3 + 2 ) / 12 = 1 / 4 + 1 / 6 thus , t = 4 . replacing , rx = w / 6 , w / 6 * t = 2 w or t = 12 days . answer : e" | a ) 4 , b ) 6 , c ) 8 , d ) 10 , e ) 12 | e | multiply(multiply(2, 2), 3) | multiply(n0,n4)|multiply(n3,#0)| | general | E |
two cars are driving toward each other . the first car is traveling at a speed of 90 km / h , which is 50 % faster than the second car ' s speed . if the distance between the cars is 975 km , how many hours will it take until the two cars meet ? | "the speed of the first car is 90 km / h . the speed of the second car is 90 / 1.5 = 60 km / h . the two cars complete a total of 150 km each hour . the time it takes the cars to meet is 975 / 150 = 6.5 hours . the answer is b ." | a ) 6 , b ) 6.5 , c ) 7 , d ) 7.5 , e ) 8 | b | divide(add(multiply(90, const_4), multiply(divide(multiply(const_4, 90), add(const_1, const_4)), const_4)), add(90, divide(multiply(const_4, 90), add(const_1, const_4)))) | add(const_1,const_4)|multiply(n0,const_4)|divide(#1,#0)|add(n0,#2)|multiply(#2,const_4)|add(#1,#4)|divide(#5,#3)| | physics | B |
in a certain company 20 % of the men and 40 % of the women attended the annual company picnic . if 35 % of all the employees are men . what % of all the employee went to the picnic ? | total men in company 35 % means total women in company 65 % ( assume total people in company 100 % ) no of men emplyess attended picnic = 35 x ( 20 / 100 ) = 7 no of women empolyees attened picnin = 65 x ( 40 / 100 ) = 26 total percentage of empolyess attened the picnic = 7 + 26 = 33 % answer : a | a ) 33 % , b ) 34 % , c ) 35 % , d ) 36 % , e ) 37 % | a | multiply(add(multiply(divide(35, const_100), divide(20, const_100)), multiply(divide(subtract(const_100, 35), const_100), divide(40, const_100))), const_100) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(const_100,n2)|divide(#3,const_100)|multiply(#0,#1)|multiply(#4,#2)|add(#5,#6)|multiply(#7,const_100) | gain | A |
two trains of length 250 m and 120 m are 50 m apart . they start moving towards each other on parallel tracks , at speeds 64 kmph and 42 kmph . after how much time will the trains meet ? | "they are moving in opposite directions , relative speed is equal to the sum of their speeds . relative speed = ( 64 + 42 ) * 5 / 18 = 7 * 5 = 29.4 mps . the time required = d / s = 50 / 29.4 = 17 / 10 sec . answer : b" | a ) 15 / 10 , b ) 17 / 10 , c ) 16 / 10 , d ) 18 / 10 , e ) 16 / 10 | b | divide(50, multiply(add(64, 42), const_0_2778)) | add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics | B |
the smallest value of n , for which n + 1 is not a prime number , is | "( 1 + 1 ) = 2 . ( 2 + 1 ) = 3 . ( 3 + 1 ) = 4 . ( 4 + 1 ) = 5 . which is not prime , n = 3 . answer : d" | a ) 5 , b ) 6 , c ) 8 , d ) 3 , e ) 10 | d | add(1, 1) | add(n0,n0)| | general | D |
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 27 , then how old is b ? | "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 = 27 5 x = 25 = > x = 5 hence , b ' s age = 2 x = 10 years . answer : b" | a ) 11 years , b ) 10 years , c ) 18 years , d ) 189 years , e ) 28 years | b | divide(multiply(subtract(27, 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 | B |
a train running at the speed of 60 km / hr crosses a pole in 9 seconds . find the length of the train ? | "speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 9 = 150 meter answer : a" | a ) 150 , b ) 872 , c ) 287 , d ) 288 , e ) 212 | a | multiply(divide(multiply(60, const_1000), const_3600), 9) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics | A |
a searchlight on top of the watchtower makes 2 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 10 seconds ? | "the searchlight completes one revolution every 30 seconds . the probability that the man ' s area will be lit up is 10 / 30 = 1 / 3 . the probability that he will stay in the dark is 1 - 1 / 3 = 2 / 3 the answer is c ." | a ) 5 / 6 , b ) 3 / 4 , c ) 2 / 3 , d ) 1 / 2 , e ) 3 / 5 | c | subtract(const_1, divide(10, multiply(10, const_3))) | multiply(n1,const_3)|divide(n1,#0)|subtract(const_1,#1)| | physics | C |
b completes a work in 3 days . a alone can do it in 10 days . if both work together , the work can be completed in how many days ? | 1 / 3 + 1 / 10 = 13 / 30 30 / 13 = 2.31 days answer : a | a ) 2.31 days , b ) 4.31 days , c ) 5.31 days , d ) 6.31 days , e ) 7.31 days | a | divide(multiply(10, 3), add(3, 10)) | add(n0,n1)|multiply(n0,n1)|divide(#1,#0) | physics | A |
a bucket full of nuts was discovered by the crow living in the basement . the crow eats a sixth of the total number of nuts in 8 hours . how many hours in total will it take the crow to finish a quarter of the nuts ? | "in one hour , the crow eats 1 / 48 of the nuts . ( 1 / 4 ) / ( 1 / 48 ) = 12 hours the answer is b ." | a ) 10 , b ) 12 , c ) 14 , d ) 16 , e ) 18 | b | divide(divide(const_1, const_4), divide(divide(const_1, add(const_2, const_3)), 8)) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,#0)|divide(#2,n0)|divide(#1,#3)| | general | B |
in a survey of 335 employees , 104 of them are uninsured , 54 work part time , and 12.5 percent of employees who are uninsured work part time . if a person is to be randomly selected from those surveyed , what is the probability that the person will neither work part time nor be uninsured ? | "- - - - - - - - - ui - - - - - - - - - - - - - - - - nui - - - - - - - total pt - - - - ( 12.5 / 100 ) * 104 = 13 - - - - - - - - - - - - - 54 npt - - - 104 - 13 - - - - - - - - - - - - - - x - - - - - - - - 281 total - - 104 - - - - - - - - - - - - - - - - - - - - - - - - - - - - 335 we have to find not part time and not uninsured . in other words not part time and insured = x / 335 = ( 281 - 104 + 13 ) / 335 = 38 / 67 answer is a ." | a ) 38 / 67 , b ) 8 / 41 , c ) 9 / 348 , d ) 1 / 8 , e ) 41 / 91 | a | divide(subtract(335, add(subtract(54, divide(multiply(12.5, 104), multiply(const_1, const_100))), 104)), 335) | multiply(n1,n3)|multiply(const_1,const_100)|divide(#0,#1)|subtract(n2,#2)|add(n1,#3)|subtract(n0,#4)|divide(#5,n0)| | physics | A |
a crate measures 4 feet by 8 feet by 12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius , in feet , of the pillar with the largest volume that could still fit in the crate ? | "to fit the cylinder with largest radius inside this cuboid , we should make the base of the crate as wide as possible so we will take the base as 12 feet by 8 feet now since the limiting number in the base is 8 feet ; therefore a cylinder { we can visualise that a cylinder ' s width is its diameter } can only fit inside the crate if it is 8 feet or less . therefore the radius of the cylinder will become diameter 2 = = = > 8 / 2 = 4 diameter answer : d" | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | divide(divide(multiply(multiply(8, 12), 4), 12), 8) | multiply(n1,n2)|multiply(n0,#0)|divide(#1,n2)|divide(#2,n1)| | geometry | D |
the ratio of efficiency of a is to c is 5 : 3 . the ratio of no . of days taken by b is to c is 2 : 3 . a takes 6 days less than c , when a and c completes the work individually . b and c left the work after 2 days . the no of days taken by a to finish the remaining work ? | a : b : c efficiency 10 : 9 : 6 no of days 9 x : 10 x : 15 x given = > 15 x - 9 x = 6 hence , x = 1 number of days taken by a = 9 and daily work done = 10 number of days taken by b = 10 and daily work done = 9 ( total work = 90 ) number of days taken by c = 15 and daily work done = 6 work done by b and c in initial 2 days = 30 rest work = 60 number of days required by a to finish = 60 / 10 = 6 days answer : c | a ) 4.5 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | c | multiply(multiply(divide(6, subtract(5, 3)), 3), subtract(const_1, multiply(add(divide(const_1, multiply(2, divide(multiply(5, divide(6, subtract(5, 3))), 3))), divide(const_1, multiply(5, divide(6, subtract(5, 3))))), 2))) | subtract(n0,n1)|divide(n4,#0)|multiply(n1,#1)|multiply(n0,#1)|divide(#3,n1)|divide(const_1,#3)|multiply(n2,#4)|divide(const_1,#6)|add(#7,#5)|multiply(n2,#8)|subtract(const_1,#9)|multiply(#2,#10) | physics | C |
there were two candidates in an election . winner candidate received 62 % of votes and won the election by 336 votes . find the number of votes casted to the winning candidate ? | "w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 336 62 % - - - - - - - - ? = > 868 answer : d" | a ) 288 , b ) 744 , c ) 788 , d ) 868 , e ) 177 | d | divide(multiply(divide(336, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)|multiply(n0,#3)|divide(#4,const_100)| | gain | D |
guy drives 60 miles to attend a meeting . halfway through , he increases his speed so that his average speed on the second half is 16 miles per hour faster than the average speed on the first half . his average speed for the entire trip is 30 miles per hour . guy drives on average how many miles per hour during the first half of the way ? | let x be the average speed for 1 st half of the distance . then the average speed for 2 nd half of the distance will be x + 16 avg speed = total distance / total time 30 = 60 / { ( 30 / x ) + ( 30 / ( x + 16 ) ) } solving we get x ^ 2 - 14 x - 240 = 0 x = - 10 or 24 x cant be negative hence x = 24 answer : d | a ) 12 , b ) 14 , c ) 16 , d ) 24 , e ) 40 | d | divide(divide(multiply(16, 30), const_2), divide(60, multiply(const_3, const_2))) | multiply(n1,n2)|multiply(const_2,const_3)|divide(#0,const_2)|divide(n0,#1)|divide(#2,#3) | physics | D |
machine a and machine b are each used to manufacture 330 sprockets . it takes machine a 10 hours longer to produce 330 sprockets than machine b . machine b produces 10 % more sprockets per hour than machine a . how many sprockets per hour does machineaproduce ? | "time taken by b = t time taken by a = t + 10 qty produced by a = q qty produced by b = 1.1 q for b : t ( 1.1 q ) = 330 qt = 300 for a : ( t + 10 ) ( q ) = 330 qt + 10 q = 330 300 + 10 q = 330 q = 3 so a can produce 3 / hour . then b can produce = 3 ( 1.1 ) = 3.3 / hour . a" | a ) 3.3 , b ) 6.6 , c ) 60 , d ) 100 , e ) 110 | a | divide(330, divide(multiply(multiply(10, 330), divide(add(const_100, 10), const_100)), subtract(multiply(330, divide(add(const_100, 10), const_100)), 330))) | add(n1,const_100)|multiply(n0,n1)|divide(#0,const_100)|multiply(#2,#1)|multiply(n0,#2)|subtract(#4,n0)|divide(#3,#5)|divide(n0,#6)| | gain | A |
a card is drawn from a pack of 52 cards . the probability of getting a queen of club or a king of heart is : | "here , n ( s ) = 52 . let e = event of getting a queen of club or a king of heart . then , n ( e ) = 2 . p ( e ) = n ( e ) / n ( s ) = 2 / 52 = 1 / 26 answer should be a" | a ) 1 / 26 , b ) 1 / 52 , c ) 2 / 62 , d ) 2 / 26 , e ) 3 / 26 | a | divide(subtract(52, multiply(const_4, const_4)), 52) | multiply(const_4,const_4)|subtract(n0,#0)|divide(#1,n0)| | probability | A |
mysoon collects glass ornaments . 10 more than 1 / 6 of the ornaments in her collection are handmade , and 1 / 2 of the handmade ornaments are antiques . if 1 / 3 of the ornaments in her collection are handmade antiques , how many ornaments are in her collection ? | the number of ornaments = a ten more than 1 / 6 of the ornaments in her collection are handmade = > handmade = 10 + a / 6 1 / 2 of the handmade ornaments are antiques = > handmade ornaments = 1 / 2 * ( 10 + a / 6 ) = 5 + a / 12 1 / 3 of the ornaments in her collection are handmade antiques = > handmade ornaments = a / 3 = > 5 + a / 12 = a / 3 = > a = 200 ans : a | a ) 20 , b ) 60 , c ) 108 , d ) 144 , e ) 180 | a | divide(multiply(10, 3), subtract(2, divide(3, 6))) | divide(n6,n2)|multiply(n0,n6)|subtract(n4,#0)|divide(#1,#2) | general | A |
according to the directions on the can of frozen orange juice concentrate , 1 can of concentrate is to be mixed with 3 cans of water to make orange juice . how many 12 ounces cans of the concentrate are required to prepare 272 6 ounces servings of orange juice ? | "its b . total juice rquired = 272 * 6 = 1632 ounce 12 ounce concentate makes = 12 * 4 = 48 ounce juice total cans required = 1632 / 48 = 34 . answer b" | a ) a ) 25 , b ) b ) 34 , c ) c ) 50 , d ) d ) 67 , e ) e ) 100 | b | divide(divide(multiply(272, 6), 12), const_4) | multiply(n3,n4)|divide(#0,n2)|divide(#1,const_4)| | general | B |
car a runs at the speed of 35 km / hr & reaches its destination in 10 hr . car b runs at the speed of 25 km / h & reaches its destination in 6 h . what is the respective ratio of distances covered by car a & car b ? | "sol . distance travelled by car a = 36 × 10 = 350 km distance travelled by car b = 25 × 6 = 150 km ratio = 350 / 150 = 7 : 3 a" | a ) 7 : 3 , b ) 5 : 7 , c ) 9 : 7 , d ) 3 : 5 , e ) 11 : 6 | a | divide(multiply(35, 10), multiply(25, 6)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,#1)| | physics | A |
the perimeter of a triangle is 48 cm and the in radius of the triangle is 2.5 cm . what is the area of the triangle ? | "area of a triangle = r * s where r is the in radius and s is the semi perimeter of the triangle . area of triangle = 2.5 * 48 / 2 = 60 cm 2 answer : c" | a ) 76 , b ) 88 , c ) 60 , d ) 55 , e ) 35 | c | triangle_area(2.5, 48) | triangle_area(n0,n1)| | geometry | C |
two numbers a and b are such that the sum of 7 % of a and 4 % of b is two - third of the sum of 6 % of a and 8 % of b . find the ratio of a : b . | "explanation : 7 % of a + 4 % of b = 2 / 3 ( 6 % of a + 8 % of b ) 7 a / 100 + 4 b / 100 = 2 / 3 ( 6 a / 100 + 8 b / 100 ) ⇒ 7 a + 4 b = 2 / 3 ( 6 a + 8 b ) ⇒ 21 a + 12 b = 12 a + 16 b ⇒ 9 a = 4 b ⇒ ab = 4 / 9 ⇒ a : b = 4 : 9 answer : option c" | a ) 2 : 1 , b ) 1 : 2 , c ) 4 : 9 , d ) 1 : 1 , e ) 3 : 2 | c | divide(subtract(divide(multiply(divide(8, const_100), 4), const_3), divide(4, const_100)), subtract(divide(7, const_100), divide(multiply(divide(6, const_100), 4), const_3))) | divide(n3,const_100)|divide(n1,const_100)|divide(n0,const_100)|divide(n2,const_100)|multiply(#0,n1)|multiply(#3,n1)|divide(#4,const_3)|divide(#5,const_3)|subtract(#6,#1)|subtract(#2,#7)|divide(#8,#9)| | general | C |
the average of 11 numbers is 22 . average of the first 6 of them is 19 and that of the last 6 is 27 . find the 6 th number ? | sum of all the 11 numbers = 11 * 22 = 242 sum of the first 6 of them = 6 * 19 = 114 sum of the last 6 of them = 6 * 27 = 162 so , the 8 th number = 242 + 114 - 162 = 34 . answer : c | a ) 12 , b ) 25 , c ) 34 , d ) 29 , e ) 27 | c | subtract(multiply(6, 27), subtract(multiply(11, 22), multiply(19, 6))) | multiply(n5,n6)|multiply(n0,n1)|multiply(n2,n3)|subtract(#1,#2)|subtract(#0,#3)| | general | C |
if the area of a circle decreases by 42 % , then the radius of a circle decreases by | "if area of a circle decreased by x % then the radius of a circle decreases by ( 100 − 10 √ 100 − x ) % = ( 100 − 10 √ 100 − 42 ) % = ( 100 − 10 √ 58 ) % = 100 - 76 = 24 % answer c" | a ) 20 % , b ) 18 % , c ) 24 % , d ) 64 % , e ) none of these | c | multiply(subtract(const_1, sqrt(divide(subtract(const_100, 42), const_100))), const_100) | subtract(const_100,n0)|divide(#0,const_100)|sqrt(#1)|subtract(const_1,#2)|multiply(#3,const_100)| | geometry | C |
a rectangular room has the rectangular shaped rug shown as above figure such that the rug ’ s area is 200 square feet and its length is 10 feet longer than its width . if the uniform width between the rug and room is 10 feet , what is the area of the region uncovered by the rug ( shaded region ) , in square feet ? | "rug ' s area = 200 which is ( x ) x ( 10 + x ) = 200 so x = 10 rug maintains a uniform distance of 10 feet so room has dimension 10 + 20 and 20 + 20 i . e . 30 and 40 area of room 30 x 40 = 1200 area covered is 200 so uncovered area is 1200 - 200 = 1000 ( answer c )" | a ) 32 , b ) 1360 , c ) 1000 , d ) 46 , e ) 104 | c | subtract(multiply(add(subtract(divide(200, const_10), const_2), const_10), add(divide(200, const_10), subtract(divide(200, const_10), const_2))), 200) | divide(n0,const_10)|subtract(#0,const_2)|add(#1,const_10)|add(#0,#1)|multiply(#2,#3)|subtract(#4,n0)| | geometry | C |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.10 km and 1.15 km respectively . the time taken by the slower train to cross the faster train in seconds is ? | "relative speed = 60 + 90 = 150 km / hr . = 150 * 5 / 18 = 125 / 3 m / sec . distance covered = 1.10 + 1.15 = 2.25 km = 2250 m . required time = 2250 * 3 / 125 = 54 sec . answer : b" | a ) 65 sec , b ) 54 sec , c ) 48 sec , d ) 33 sec , e ) 12 sec | b | subtract(divide(multiply(1.10, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.15, const_1000), divide(multiply(90, const_1000), const_3600))) | multiply(n2,const_1000)|multiply(n0,const_1000)|multiply(n3,const_1000)|multiply(n1,const_1000)|divide(#1,const_3600)|divide(#3,const_3600)|divide(#0,#4)|divide(#2,#5)|subtract(#6,#7)| | physics | B |
mohit sold an article for $ 21000 . had he offered a discount of 10 % on the selling price , he would have earned a profit of 8 % . what is the cost price of the article ? | "let the cp be $ x . had he offered 10 % discount , profit = 8 % profit = 8 / 100 x and hence his sp = x + 8 / 100 x = $ 1.08 x = 21000 - 10 / 100 ( 21000 ) = 21000 - 2100 = $ 18900 = > 1.08 x = 18900 = > x = 17500 e" | a ) 16000 , b ) 25000 , c ) 15000 , d ) 18000 , e ) 17500 | e | multiply(divide(subtract(21000, divide(multiply(21000, 10), const_100)), add(const_100, 8)), const_100) | add(n2,const_100)|multiply(n0,n1)|divide(#1,const_100)|subtract(n0,#2)|divide(#3,#0)|multiply(#4,const_100)| | gain | E |
a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 50 paisa . if the share of y is rs . 36 , what is the total amount ? | "x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 36 39 - - - ? = > 156 answer : e" | a ) 115 , b ) 116 , c ) 117 , d ) 118 , e ) 156 | e | add(add(multiply(divide(const_100, 45), 36), multiply(divide(50, 45), 36)), 36) | divide(const_100,n0)|divide(n1,n0)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|add(n2,#4)| | general | E |
if p is a prime number greater than 3 , find the remainder when p ^ 2 + 15 is divided by 12 . | "every prime number greater than 3 can be written 6 n + 1 or 6 n - 1 . if p = 6 n + 1 , then p ^ 2 + 15 = 36 n ^ 2 + 12 n + 1 + 15 = 36 n ^ 2 + 12 n + 12 + 4 if p = 6 n - 1 , then p ^ 2 + 15 = 36 n ^ 2 - 12 n + 1 + 15 = 36 n ^ 2 - 12 n + 12 + 4 when divided by 12 , it must leave a remainder of 4 . the answer is b ." | a ) 6 , b ) 4 , c ) 0 , d ) 8 , e ) 7 | b | subtract(add(15, power(add(const_1, const_4), 2)), multiply(12, 3)) | add(const_1,const_4)|multiply(n0,n3)|power(#0,n1)|add(n2,#2)|subtract(#3,#1)| | general | B |
reduce 368 / 598 to the lowest terms | explanation : we can do it easily by in two steps step 1 : we get the hcf of 368 and 598 which is 46 step 2 : divide both by 46 , we will get the answer 8 / 13 answer : option d | a ) 30 / 25 , b ) 28 / 13 , c ) 28 / 29 , d ) 8 / 13 , e ) none of these | d | divide(368, 598) | divide(n0,n1) | general | D |
a sports retailer ordered white and yellow tennis balls in equal number but the dispatch clerk of the tennis ball company erred and dispatched 20 extra yellow balls and making ratio of white balls to yellow balls 8 / 13 . how many tennis balls did the retailer order originally . | "white : yellow = x : ( x + 20 ) = 8 : 13 - - > 13 x = 8 x + 160 - - > x = 32 . the total # of balls originally x + x = 32 + 32 = 64 . answer : a ." | a ) 64 , b ) 130 , c ) 140 , d ) 160 , e ) 120 | a | multiply(divide(multiply(8, 20), subtract(13, 8)), const_2) | multiply(n0,n1)|subtract(n2,n1)|divide(#0,#1)|multiply(#2,const_2)| | general | A |
what number is 75 more than three - fourth of itself ? | "3 / 4 x + 75 = x that means 75 = 1 / 4 x x = 75 * 4 = 300 c is the answer" | a ) 138 , b ) 225 , c ) 300 , d ) 324 , e ) 336 | c | divide(75, subtract(const_1, divide(const_1, const_4))) | divide(const_1,const_4)|subtract(const_1,#0)|divide(n0,#1)| | general | C |
the ratio of expenditure and savings is 3 : 2 . if the income increases by 15 % and the savings increases by 6 % , then by how much percent should his expenditure increases ? | answer : b ) 21 | a ) 22 , b ) 21 , c ) 88 , d ) 76 , e ) 29 | b | divide(add(multiply(subtract(15, 6), 2), multiply(15, 3)), const_3) | multiply(n0,n2)|subtract(n2,n3)|multiply(n1,#1)|add(#2,#0)|divide(#3,const_3) | general | B |
a man can row downstream at 28 kmph and upstream at 10 kmph . find the speed of the man in still water and the speed of stream respectively ? | "let the speed of the man in still water and speed of stream be x kmph and y kmph respectively . given x + y = 28 - - - ( 1 ) and x - y = 10 - - - ( 2 ) from ( 1 ) & ( 2 ) 2 x = 38 = > x = 19 , y = 9 . answer : d" | a ) 2 , 9 , b ) 4 , 9 , c ) 8 , 9 , d ) 19 , 9 , e ) 7 , 9 | d | divide(divide(add(28, 10), const_2), const_2) | add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)| | physics | D |
9 log 9 ( 2 ) = ? | exponential and log functions are inverse of each other . hence aloga ( x ) = x , for all x real and positive . and therefore 9 log 9 ( 2 ) = 2 correct answer b | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(log(multiply(9, 9)), log(const_10)) | log(const_10)|multiply(n0,n0)|log(#1)|divide(#2,#0) | other | B |
a and b can do a work in 10 days and 15 days respectively . a starts the work and b joins him after 5 days . in how many days can they complete the remaining work ? | "work done by a in 5 days = 5 / 10 = 1 / 2 remaining work = 1 / 2 work done by both a and b in one day = 1 / 10 + 1 / 15 = 5 / 30 = 1 / 6 remaining work = 1 / 2 * 6 / 1 = 3 days . answer : e" | a ) 6 days , b ) 2 days , c ) 8 days , d ) 4 days , e ) 3 days | e | subtract(add(inverse(add(inverse(15), inverse(10))), 10), const_3) | inverse(n1)|inverse(n0)|add(#0,#1)|inverse(#2)|add(n0,#3)|subtract(#4,const_3)| | physics | E |
the average age of a family of 6 members is 28 years . if the age of the youngest member is 7 years , what was the average age of the family at the birth of the youngest member ? | "present age of total members = 6 x 28 = 168 7 yrs back their ages were = 6 x 7 = 42 ages at the birth of youngest member = 168 - 42 = 126 therefore , avg age at the birth of youngest member = 126 / 6 = 21 . answer : c" | a ) 15 , b ) 18 , c ) 21 , d ) 12 , e ) 19 | c | subtract(divide(subtract(multiply(6, 28), 7), const_4), 7) | multiply(n0,n1)|subtract(#0,n2)|divide(#1,const_4)|subtract(#2,n2)| | general | C |
if a ã — b = 2 a - 3 b + ab , then 1 ã — 2 + 2 ã — 1 is equal to : | explanation : 1 ã — 2 + 2 ã — 1 = ( 2 ã — 1 - 3 ã — 2 + 1 ã — 2 ) + ( 2 ã — 2 - 3 ã — 1 + 2 ã — 1 ) = ( 2 - 6 + 2 + 4 - 3 + 2 ) = 1 . answer : a | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | add(add(subtract(multiply(2, 1), multiply(3, 2)), multiply(1, 2)), add(subtract(multiply(2, 2), multiply(3, 1)), multiply(2, 1))) | multiply(n0,n2)|multiply(n0,n1)|multiply(n0,n0)|multiply(n1,n2)|subtract(#0,#1)|subtract(#2,#3)|add(#0,#4)|add(#0,#5)|add(#6,#7) | general | A |
the diagonals of a rhombus are 14 cm and 18 cm . find its area ? | "1 / 2 * 14 * 18 = 126 answer : d" | a ) 176 , b ) 190 , c ) 150 , d ) 126 , e ) 292 | d | rhombus_area(14, 18) | rhombus_area(n0,n1)| | geometry | D |
what will be the fraction of 6 % | "explanation : 6 * 1 / 100 = 3 / 50 . option b" | a ) 1 / 20 , b ) 3 / 50 , c ) 1 / 75 , d ) 1 / 25 , e ) none of these | b | divide(circle_area(divide(6, const_2)), const_2) | divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)| | gain | B |
rohan spends 40 % of his salary on food , 20 % on house rent , 10 % on entertainment and 10 % on conveyance . if his savings at the end of a month are rs . 1500 . then his monthly salary is | "sol . saving = [ 100 - ( 40 + 20 + 10 + 10 ] % = 20 % . let the monthly salary be rs . x . then , 20 % of x = 1500 ⇔ 20 / 100 x = 1500 ⇔ x = 1500 × 5 = 7500 . answer b" | a ) rs . 6000 , b ) rs . 7500 , c ) rs . 8000 , d ) rs . 10,000 , e ) none | b | multiply(1500, add(const_4, const_1)) | add(const_1,const_4)|multiply(n4,#0)| | gain | B |
the average ( arithmetic mean ) of all scores on a certain algebra test was 90 . if the average of the 8 male students ’ grades was 85 , and the average of the female students ’ grades was 92 , how many female students took the test ? | "total marks of male = m total marks of female = f number of males = 8 number of females = f given : ( m + f ) / ( 8 + f ) = 90 - - - - - - - - - - - - - 1 also given , m / 8 = 85 thus m = 680 - - - - - - - - - 2 also , f / f = 92 thus f = 92 f - - - - - - - - - 3 put 2 and 3 in 1 : we get ( 680 + 92 f ) / ( 8 + f ) = 90 solving this we get f = 20 ans : e" | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 20 | e | divide(subtract(multiply(90, 8), multiply(85, 8)), subtract(92, 90)) | multiply(n0,n1)|multiply(n1,n2)|subtract(n3,n0)|subtract(#0,#1)|divide(#3,#2)| | general | E |
a train covers a distance of 12 km in 10 min . if it takes 4 sec to pass a telegraph post , then the length of the train is ? | "speed = ( 12 / 10 * 60 ) km / hr = ( 72 * 5 / 18 ) m / sec = 20 m / sec . length of the train = 20 * 4 = 80 m . answer : option d" | a ) 20 , b ) 110 , c ) 120 , d ) 80 , e ) 60 | d | divide(12, subtract(divide(12, 10), 4)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics | D |
a jogger running at 9 km / hr along side a railway track is 200 m ahead of the engine of a 200 m long train running at 45 km / hr in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = 45 - 9 = 36 km / hr . = 36 * 5 / 18 = 10 m / sec . distance to be covered = 200 + 200 = 360 m . time taken = 400 / 10 = 40 sec . answer : option c" | a ) 89 , b ) 20 , c ) 40 , d ) 88 , e ) 34 | c | divide(add(200, 200), multiply(subtract(45, 9), divide(divide(const_10, const_2), divide(subtract(45, 9), const_2)))) | add(n1,n2)|divide(const_10,const_2)|subtract(n3,n0)|divide(#2,const_2)|divide(#1,#3)|multiply(#4,#2)|divide(#0,#5)| | general | C |
a scientist used a unique two - color code to identify each of the test subjects involved in a certain study . if the scientist found that choosing from among 5 colors produced enough color codes to identify all but 6 of the test subjects , how many test subjects were in the study ? ( assume that the order of the colors in the codes does not matter . ) | 5 c 2 = 10 the number of test subjects is 10 + 6 = 16 the answer is c . | a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | c | add(divide(factorial(5), multiply(factorial(subtract(5, const_2)), factorial(const_2))), 6) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)|add(n1,#5) | general | C |
a pharmaceutical company received $ 4 million in royalties on the first $ 50 million in sales of and then $ 12 million in royalties on the next $ 170 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 50 million in sales to the next $ 170 million in sales ? | "( 12 / 170 ) / ( 4 / 50 ) = 15 / 17 = 88 % it means that 12 / 170 represents only 88 % . therefore a decrease of 12 % . answer a" | a ) 12 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 56 % | a | multiply(divide(4, 50), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general | A |
the h . c . f . of two numbers is 12 and their l . c . m . is 600 . if one of the number is 30 , find the other ? | "other number = 12 * 600 / 30 = 240 answer is b" | a ) 100 , b ) 240 , c ) 120 , d ) 200 , e ) 150 | b | multiply(12, 30) | multiply(n0,n2)| | physics | B |
if 13 = 13 w / ( 1 - w ) , then ( 2 w ) 2 = | "13 - 13 w = 13 w 26 w = 13 w = 1 / 2 2 w = 1 2 w * 2 = 1 * 2 = 2 answer : d" | a ) 1 / 4 , b ) 1 / 2 , c ) 1 , d ) 2 , e ) 3 | d | multiply(divide(13, add(13, 13)), 2) | add(n0,n0)|divide(n0,#0)|multiply(n3,#1)| | general | D |
q is as much younger than r as he is older than t . if the sum of the ages of r and t is 50 years , what is definitely the difference between r and q ’ s age ? | "explanation given that : 1 . the difference of age b / w r and q = the difference of age b / w q and t . 2 . sum of age of r and t is 50 i . e . ( r + t ) = 50 . question : r – q = ? . explanation : r – q = q – t ( r + t ) = 2 q now given that , ( r + t ) = 50 so , 50 = 2 q and therefore q = 25 . question is ( r – q ) = ? here we know the value ( age ) of q ( 25 ) , but we don ’ t know the age of r . therefore , ( r - q ) can not be determined . answer d" | a ) 1 year , b ) 2 years , c ) 25 years , d ) data inadequate , e ) none of these | d | divide(50, const_2) | divide(n0,const_2)| | general | D |
a grocer has a sale of rs . 6400 , rs . 7000 , rs . 6800 , rs . 7200 and rs . 6500 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 6500 ? | "total sale for 5 months = rs . ( 6400 + 7000 + 6800 + 7200 + 6500 ) = rs . 33900 required sale = rs . [ ( 6500 x 6 ) - 34009 ] = rs . ( 39000 - 33900 ) = rs . 5100 answer : e" | a ) rs . 4500 , b ) rs . 4700 , c ) rs . 4800 , d ) rs . 5000 , e ) rs . 5100 | e | subtract(multiply(add(5, const_1), 6500), add(add(add(add(6400, 7000), 6800), 7200), 6500)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general | E |
a rectangular floor that measures 15 meters by 18 meters is to be covered with carpet squares that each measure 3 meters by 3 meters . if the carpet squares cost $ 12 apiece , what is the total cost for the number of carpet squares needed to cover the floor ? | "the width of the rectangular floor ( 15 m ) is a multiple of one side of the square ( 3 m ) , and the length of the floor ( 18 m ) is also a multiple of the side of the square . so the number of carpets to cover the floor is ( 15 / 3 ) * ( 18 / 3 ) = 30 . the total cost is 30 * 12 = $ 360 . the answer is , therefore , c ." | a ) $ 200 , b ) $ 240 , c ) $ 360 , d ) $ 960 , e ) $ 1,920 | c | multiply(15, 15) | multiply(n0,n0)| | geometry | C |
two family reunions are happening at the same hotel , the dates reunion and the hall reunion . all 50 guests at the hotel attend at least one of the reunions . if 50 people attend the dates reunion and 60 people attend the hall reunion , how many people attend both reunions ? | no of people in dates reunion = 50 no of people in hall reunion = 60 attending both = x all guests attend at least one . therefore , 50 = 50 + 60 - ( both ) both = 60 answer e | a ) 2 , b ) 5 , c ) 10 , d ) 16 , e ) 60 | e | subtract(add(50, 60), 50) | add(n0,n2)|subtract(#0,n0) | other | E |
it is the new year and mandy has made a resolution to lose weight this year . she plans to exercise and do yoga . for exercise she plans to workout at the gym and ride her bicycle in the ratio of 2 : 3 everyday . she will also do yoga in the ratio , yoga : exercise = 2 : 3 . if she rides her bike for 12 minutes , how much time will she spend doing yoga ? ( rounded to minutes ) | the ratio is 2 : 3 = gym : ride , so ( 12 ) ( 3 / 2 ) = 18 minutes at the gym , and 18 + 12 = 30 minutes exercise , so ( 2 / 3 ) ( 30 ) = 20 minutes yoga . answer : e | a ) 10 min . , b ) 41 min . , c ) 17 min . , d ) 23 min . , e ) 20 min . | e | divide(multiply(12, divide(3, add(2, 3))), multiply(divide(3, add(2, 3)), divide(3, add(2, 3)))) | add(n0,n1)|divide(n1,#0)|multiply(n4,#1)|multiply(#1,#1)|divide(#2,#3) | physics | E |
a train starts full of passengers at the first station it drops 1 / 3 of the passengers and takes 280 more at the second station it drops one half the new total and takes 12 more . on arriving at the third station it is found to have 242 passengers . find the no of passengers in the beginning ? | let no of passengers in the beginning be x after first station no passengers = ( x - x / 3 ) + 280 = 2 x / 3 + 280 after second station no passengers = 1 / 2 ( 2 x / 3 + 280 ) + 12 1 / 2 ( 2 x / 3 + 280 ) + 12 = 242 2 x / 3 + 280 = 2 * 230 = 460 - 280 = 180 2 x / 3 = 180 x = 270 . so the answer is option b ) 270 . | a ) 292 , b ) 270 , c ) 192 , d ) 144 , e ) 168 | b | multiply(multiply(3, const_2), divide(subtract(242, add(divide(280, const_2), 12)), const_2)) | divide(n2,const_2)|multiply(n1,const_2)|add(n3,#0)|subtract(n4,#2)|divide(#3,const_2)|multiply(#4,#1) | general | B |
a 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds . what is the length of the platform ? | "solution speed = ( 300 / 18 ) m / sec = ( 50 / 3 ) m / sec . length of the platform be x metres . then = x + 300 / 39 = 50 / 3 m / sec = 3 ( x + 300 ) = 1950 . = 350 m answer b" | a ) 320 m , b ) 350 m , c ) 650 m , d ) none of these , e ) can not be determined | b | divide(subtract(multiply(divide(300, divide(18, const_3)), 39), multiply(const_3, 300)), const_3) | divide(n2,const_3)|multiply(n0,const_3)|divide(n0,#0)|multiply(n1,#2)|subtract(#3,#1)|divide(#4,const_3)| | physics | B |
tim came second in math . when his mother asked him how much he had scored , he answered that he got the sum of the first 7 even numbers . his mother immediately worked out the answer . how much had he scored in math ? | "c 56 sum = ( n x n ) + n hence , 7 x 7 = 49 + 7 = 56" | a ) 80 , b ) 90 , c ) 56 , d ) 70 , e ) 60 | c | multiply(add(7, const_1), 7) | add(n0,const_1)|multiply(n0,#0)| | physics | C |
a man sold 18 toys for rs . 21000 , gaining thereby the cost price of 3 toy find the cost price of a toy | "let the cost of one toy = x . then , cost of 18 toys = 18 x . gain = 3 x . sp of 18 toys = rs . 21000 . gain = sp â € “ cp 3 x = 21000 â € “ 18 x 21 x = 21000 x = rs . 1000 . answer : option e" | a ) s . 600 , b ) s . 800 , c ) s . 500 , d ) s . 900 , e ) s . 1000 | e | divide(21000, add(18, 3)) | add(n0,n2)|divide(n1,#0)| | gain | E |
on selling 20 balls at rs . 720 , there is a loss equal to the cost price of 5 balls . the cost price of a ball is : | "( c . p . of 20 balls ) - ( s . p . of 20 balls ) = ( c . p . of 5 balls ) c . p . of 15 balls = s . p . of 20 balls = rs . 720 . c . p . of 1 ball = rs . 720 / 15 = rs . 48 . answer : option a" | a ) s . 48 , b ) s . 50 , c ) s . 55 , d ) s . 60 , e ) s . 70 | a | divide(720, subtract(20, 5)) | subtract(n0,n2)|divide(n1,#0)| | gain | A |
a certain bag contains 60 balls — 22 white , 18 green , 2 yellow , 15 red , and 3 purple . if a ball is to be chosen at random , what is the probability that the ball will be neither red nor purple ? | "according to the stem the ball can be white , green or yellow , so the probability is ( white + green + yellow ) / ( total ) = ( 22 + 18 + 2 ) / 60 = 42 / 60 = 0.7 . answer : a ." | a ) 0.7 , b ) 0.15 , c ) 0.54 , d ) 0.85 , e ) 0.91 | a | divide(add(add(22, 18), 2), 60) | add(n1,n2)|add(n3,#0)|divide(#1,n0)| | other | A |
( x ) + 9088 + 1090 - 9156 = 19845 . calculate the value of x | "( x ) + 9088 + 1090 - 9156 = 19845 = x + 9088 + 1090 = 19845 + 9156 = x + 10178 = 29001 = x = 29001 - 10178 = 18823 answer is e" | a ) 11123 , b ) 18153 , c ) 11282 , d ) 18825 , e ) 18823 | e | multiply(subtract(subtract(add(19845, 9156), 1090), 9088), divide(const_60, const_2)) | add(n2,n3)|divide(const_60,const_2)|subtract(#0,n1)|subtract(#2,n0)|multiply(#1,#3)| | general | E |
in a certain company , a third of the workers do not have a retirement plan . 20 % of the workers who do not have a retirement plan are women , and 40 % of the workers who do have a retirement plan are men . if 140 of the workers of that company are men , how many of the workers are women ? | "20 % of the workerswithoutretirement plan are women , this means 80 % of the workerswithoutretirement plan are men right ? makes 100 % = no retirement plan . then 40 % of those wo have a retirement plan are men , which means 60 % of those who have a retirement plan are women . these percentages are always refering to the subgroupsretirement plan = yesandno retirement plan . make the total workers equal x and work from there and it should be fine > x / 3 have no retirement plan at all . i struggled with the question a lot , it was hard for me although usually i have no problems with sets questions like this . i believe it must be a 700 + question . yes , you are right . i read the question incorrectly ! i glazed overnotfor some reason . if 20 % of the workers who do not have a retirement plan are women , 80 % who do not have retirement plan are men . also 40 % who do have retirement plan are men . since 1 / 3 do not have retirement plan and 2 / 3 do have , it becomes a simple weighted average question . cavg = ( 80 * 1 + 40 * 2 ) / 3 = 160 / 3 % are men total = 120 * 300 / 160 = 225 no . of women = 225 - 140 = 85 a" | a ) 85 , b ) 95 , c ) 105 , d ) 120 , e ) 210 | a | multiply(divide(140, add(subtract(divide(const_1, const_3), multiply(divide(const_1, const_3), divide(20, const_100))), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100)))), add(multiply(divide(const_1, const_3), divide(20, const_100)), subtract(subtract(const_1, divide(const_1, const_3)), multiply(subtract(const_1, divide(const_1, const_3)), divide(40, const_100))))) | divide(const_1,const_3)|divide(n0,const_100)|divide(n1,const_100)|multiply(#0,#1)|subtract(const_1,#0)|multiply(#2,#4)|subtract(#0,#3)|add(#5,#6)|subtract(#4,#5)|add(#3,#8)|divide(n2,#7)|multiply(#9,#10)| | gain | A |
if 4 a = 5 b and ab ≠ 0 , what is the ratio of a / 5 to b / 4 ? | a nice fast approach is the first find a pair of numbers that satisfy the given equation : 4 a = 5 b here ' s one pair : a = 5 and b = 4 what is the ratio of a / 5 to b / 4 ? in other words , what is the value of ( a / 5 ) / ( b / 4 ) ? plug in values to get : ( a / 5 ) / ( b / 4 ) = ( 5 / 5 ) / ( 4 / 4 ) = 1 / 1 = 1 c | a ) 125 / 64 , b ) 25 / 16 , c ) 1 , d ) 4 / 5 , e ) 64 / 125 | c | divide(multiply(4, 5), multiply(5, 4)) | multiply(n0,n1)|divide(#0,#0) | general | C |
ravi purchased a refrigerator and a mobile phone for rs . 15000 and rs . 8000 respectively . he sold the refrigerator at a loss of 4 percent and the mobile phone at a profit of 9 percent . overall he make a . | let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 8000 ( 1 + 9 / 100 ) = 8000 + 720 total sp - total cp = r + m - ( 15000 + 8000 ) = - 600 + 720 = rs . 120 as this is positive , an overall profit of rs . 200 was made . answer : d | a ) 228 , b ) 288 , c ) 27 , d ) 120 , e ) 881 | d | subtract(add(multiply(15000, subtract(const_1, divide(4, const_100))), multiply(8000, add(const_1, divide(9, const_100)))), add(15000, 8000)) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|add(#2,const_1)|subtract(const_1,#1)|multiply(n0,#4)|multiply(n1,#3)|add(#5,#6)|subtract(#7,#0) | gain | D |
car z travels 51 miles per gallon of gasoline when driven at a constant rate of 45 miles per hour , but travels 20 percent fewer miles per gallon of gasoline when driven at a constant rate of 60 miles per hour . how many miles does car z travel on 10 gallons of gasoline when driven at a constant rate of 60 miles per hour ? | "the question stem asks us for the distance possible with 10 gallons of fuel at a constant speed of 60 miles per hour . we therefore first calculate the fuel efficiency at that speed . the stem tells us that at 45 miles / hour , the car will run 51 miles / gallon and at 60 miles / hour , that distance decreases by 20 % . we can therefore conclude that the car will travel 40.8 miles / gallon at a constant speed of 60 miles / gallon . with 10 gallons of fuel , the car can therefore travel 40.8 miles / gallon * 10 gallons = 408 miles . answer d ." | a ) 320 , b ) 375.2 , c ) 400 , d ) 408 , e ) 440 | d | multiply(multiply(subtract(const_1, divide(20, const_100)), 51), 10) | divide(n2,const_100)|subtract(const_1,#0)|multiply(n0,#1)|multiply(n4,#2)| | gain | D |
an aeroplane covers a certain distance of 420 kmph in 6 hours . to cover the same distance in 4 2 / 3 hours , it must travel at a speed of ? | "speed of aeroplane = 420 kmph distance travelled in 6 hours = 420 * 6 = 2520 km speed of aeroplane to acver 2520 km in 14 / 3 = 2520 * 3 / 14 = 540 km answer : option ' d '" | a ) 450 km , b ) 480 km , c ) 500 km , d ) 540 km , e ) 590 km | d | divide(multiply(420, 6), divide(add(multiply(3, 4), 2), 3)) | multiply(n0,n1)|multiply(n2,n4)|add(n3,#1)|divide(#2,n4)|divide(#0,#3)| | physics | D |
the ratio of investments of two partners p and q is 7 : 5 and the ratio of their profits is 7 : 10 . if p invested the money for 2 months , find for how much time did q invest the money ? | "7 * 5 : 2 * x = 7 : 10 x = 4 answer : 4 answer is d" | a ) 8 , b ) 10 , c ) 18 , d ) 4 , e ) 12 | d | multiply(multiply(divide(7, 5), divide(10, 7)), 2) | divide(n0,n1)|divide(n3,n2)|multiply(#0,#1)|multiply(n4,#2)| | gain | D |
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