Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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in a hostel there were 100 students . to accommodate 25 more students the average is decreased by some rupees . but total expenditure increased by rs . 500 . if the total expenditure of the hostel now 8500 , find decrease of average budget ? | "let average is x 100 x + 500 = 8500 x = 80 let decrease = y 125 ( 80 – y ) = 8500 y = 12 answer : c" | a ) 20 , b ) 15 , c ) 12 , d ) 18 , e ) 24 | c | multiply(subtract(divide(add(multiply(add(100, 25), 500), 8500), 25), 500), add(100, 25)) | add(n0,n1)|multiply(n2,#0)|add(n3,#1)|divide(#2,n1)|subtract(#3,n2)|multiply(#0,#4)| | general |
8 men , working 7 hours a day can complete a work in 18 days . how many hours a day must 12 men work to complete the same work in 12 days ? | "the number of hours required to complete the work is 8 * 7 * 18 = 1008 12 × 12 × ( x ) = 1008 x = 7 the answer is b ." | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | divide(multiply(multiply(8, 7), 18), multiply(12, 12)) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|divide(#2,#1)| | physics |
of the four - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? | "of the four - digit positive integers that have no digits equal to zero , how many have two digits that are equal to each other and the remaining digit different from the other two ? a . 24 b . 36 c . 72 d . 144 e . 216 choosing the digit for p - 9 ways ; choosing the digit for q - 8 ways ; choosing the digit for r - ... | a ) 6048 , b ) 3648 , c ) 7248 , d ) 1448 , e ) 2168 | a | multiply(multiply(subtract(const_10, const_1), subtract(const_10, const_2)), const_1) | subtract(const_10,const_1)|subtract(const_10,const_2)|multiply(#0,#1)|multiply(#2,const_1)| | general |
( 5 ) 1.25 × ( 12 ) 0.25 × ( 60 ) 0.75 = ? | explanation : ( 5 ) 1.25 × ( 12 ) 0.25 × ( 60 ) 0.75 = ( 5 ) 1.25 × ( 12 ) 0.25 × ( 12 × 5 ) 0.75 = ( 5 ) 1.25 × ( 12 ) 0.25 × ( 12 ) 0.75 × ( 5 ) 0.75 = ( 5 ) ( 1.25 + 0.75 ) × ( 12 ) ( 0.25 + 0.75 ) = ( 5 ) 2 × ( 12 ) 1 = 25 × 12 = 300 answer : option d | a ) 420 , b ) 260 , c ) 200 , d ) 300 , e ) 400 | d | multiply(power(60, 0.75), multiply(power(5, 1.25), power(12, 0.25))) | power(n0,n1)|power(n2,n3)|power(n4,n5)|multiply(#0,#1)|multiply(#3,#2) | general |
a computer factory produces 4200 computers per month at a constant rate , how many computers are built every 30 minutes assuming that there are 28 days in one month ? | "number of hours in 28 days = 28 * 24 number of 30 mins in 28 days = 28 * 24 * 2 number of computers built every 30 mins = 4200 / ( 28 * 24 * 2 ) = 3.125 answer b" | a ) 2.25 , b ) 3.125 . , c ) 4.5 , d ) 5.225 , e ) 6.25 | b | divide(divide(4200, 28), multiply(subtract(28, const_4), const_2)) | divide(n0,n2)|subtract(n2,const_4)|multiply(#1,const_2)|divide(#0,#2)| | physics |
a number x is multiplied with itself and then added to the product of 4 and x . if the result of these two operations is 12 , what is the value of x ? | "a number x is multiplied with itself - - > x ^ 2 added to the product of 4 and x - - > x ^ 2 + 4 x if the result of these two operations is - 4 - - > x ^ 2 + 4 x = 4 i . e x ^ 2 + 4 x - 12 = 0 is the quadratic equation which needs to be solved . ( x - 2 ) ( x + 6 ) = 0 hence x = 2 , x = - 6 imo c" | a ) - 4 , b ) - 2 , c ) 2 and 6 , d ) 4 , e ) can not be determined . | c | divide(power(12, const_2), 4) | power(n1,const_2)|divide(#0,n0)| | general |
a bowl of fruit contains 14 apples and 20 oranges . how many oranges must be removed so that 70 % of the pieces of fruit in the bowl will be apples ? | "number of apples = 14 number of oranges = 20 let number of oranges that must be removed so that 70 % of pieces of fruit in bowl will be apples = x total number of fruits after x oranges are removed = 14 + ( 20 - x ) = 34 - x 14 / ( 34 - x ) = 7 / 10 = > 20 = 34 - x = > x = 14 answer c" | a ) 3 , b ) 6 , c ) 14 , d ) 17 , e ) 20 | c | divide(subtract(multiply(add(14, 20), divide(70, const_100)), 14), divide(70, const_100)) | add(n0,n1)|divide(n2,const_100)|multiply(#0,#1)|subtract(#2,n0)|divide(#3,#1)| | gain |
the area of a square field is 1225 km 2 . how long will it take for a horse to run around at the speed of 20 km / h ? | explanation area of field = 1225 km 2 . then , each side of field = √ 1225 = 35 km distance covered by the horse = perimeter of square field = 35 × 4 = 140 km ∴ time taken by horse = distances / peed = 140 / 20 = 7 h answer a | a ) 7 h , b ) 10 h , c ) 8 h , d ) 6 h , e ) none of these | a | divide(multiply(const_4, sqrt(1225)), 20) | sqrt(n0)|multiply(#0,const_4)|divide(#1,n2)| | geometry |
xavier starts from p towards q at a speed of 50 kmph and after every 12 mins increases his speed by 10 kmph . if the distance between p and q is 52 km , then how much time does he take to cover the distance ? | first 12 min = 50 * 12 / 60 = 10 km 2 nd 12 min = 60 * 12 / 60 = 12 km 3 rd 12 min = 70 * 12 / 60 = 14 km 4 th 12 min = 80 * 12 / 60 = 16 km total time 12.4 = 48 min c | a ) 40 , b ) 60 , c ) 48 , d ) 70 , e ) 80 | c | add(add(add(12, 12), 12), 12) | add(n1,n1)|add(n1,#0)|add(n1,#1) | physics |
a shopkeeper bought 600 oranges and 400 bananas . he found 15 % of oranges and 5 % of bananas were rotten . find the percentage of fruits in good condition ? | "total number of fruits shopkeeper bought = 600 + 400 = 1000 number of rotten oranges = 15 % of 600 = 15 / 100 × 600 = 9000 / 100 = 90 number of rotten bananas = 5 % of 400 = 20 therefore , total number of rotten fruits = 90 + 20 = 110 therefore number of fruits in good condition = 1000 - 110 = 890 therefore percentage... | a ) 92.5 % , b ) 89.0 % , c ) 85.2 % , d ) 96.8 % , e ) 78.9 % | b | multiply(divide(subtract(add(600, 400), add(multiply(600, divide(15, const_100)), multiply(400, divide(5, const_100)))), add(600, 400)), const_100) | add(n0,n1)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#1)|multiply(n1,#2)|add(#3,#4)|subtract(#0,#5)|divide(#6,#0)|multiply(#7,const_100)| | gain |
if a rectangular room measures 8 meters by 5 meters by 4 meters , what is the volume of the room in cubic centimeters ? ( 1 meter = 100 centimeters ) | b . 160 , 000,000 8 * 100 * 5 * 100 * 4 * 100 = 160 , 000,000 | ['a ) 24,000', 'b ) 16 , 000,000', 'c ) 2 , 400,000', 'd ) 24 , 000,000', 'e ) 240 , 000,000'] | b | multiply(multiply(multiply(multiply(4, 100), divide(1, const_10)), multiply(5, 100)), multiply(8, 100)) | divide(n3,const_10)|multiply(n2,n4)|multiply(n1,n4)|multiply(n0,n4)|multiply(#0,#1)|multiply(#4,#2)|multiply(#5,#3) | geometry |
on a 20 mile course , pat bicycled at an average rate of 30 miles per hour for the first 12 minutes and without a break , ran the rest of the distance at an average rate of 8 miles per hour . how many minutes did pat take to cover the entire course ? | at an average rate of 30 miles per hour in 12 minute ( 1 / 5 hours ) pat covers ( distance ) = ( time ) * ( rate ) = 1 / 5 * 30 = 6 miles , thus she should cover the remaining distance of 20 - 6 = 14 miles at an average rate of 8 miles per hour . to cover 14 miles at an average rate of 8 miles per hour pat needs ( time... | a ) 75 , b ) 105 , c ) 117 , d ) 150 , e ) 162 | c | add(multiply(divide(subtract(20, multiply(30, divide(12, const_60))), 8), const_60), 12) | divide(n2,const_60)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n3)|multiply(#3,const_60)|add(n2,#4) | physics |
the sum of all the integers k such that – 25 < k < 24 is | "- 24 - - - - - - - - - - - - - - - - - - 0 - - - - - - - - - - - - - - - - - 23 values upto + 23 cancels outwe are left with only - 24 - 23 sum of which is - 47 . hence option d . e" | a ) 0 , b ) - 2 , c ) - 25 , d ) - 49 , e ) - 47 | e | add(add(negate(25), const_1), add(add(negate(25), const_1), const_1)) | negate(n0)|add(#0,const_1)|add(#1,const_1)|add(#1,#2)| | general |
sonika deposited rs . 14500 which amounted to rs . 12200 after 3 years at simple interest . had the interest been 3 % more . she would get how much ? | "( 14500 * 3 * 3 ) / 100 = 1305 12200 - - - - - - - - 13505 answer : a" | a ) 13505 , b ) 12004 , c ) 15003 , d ) 14500 , e ) 16400 | a | add(multiply(multiply(add(divide(3, const_100), divide(divide(subtract(12200, 14500), 3), 14500)), 14500), 3), 14500) | divide(n3,const_100)|subtract(n1,n0)|divide(#1,n2)|divide(#2,n0)|add(#0,#3)|multiply(n0,#4)|multiply(n2,#5)|add(n0,#6)| | gain |
the sale price sarees listed for rs . 298 after successive discount is 12 % and 15 % is ? | "explanation : 298 * ( 88 / 100 ) * ( 85 / 100 ) = 223 answer : b" | a ) 321 , b ) 223 , c ) 245 , d ) 265 , e ) 162 | b | subtract(subtract(298, divide(multiply(298, 12), const_100)), divide(multiply(subtract(298, divide(multiply(298, 12), const_100)), 15), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
3 friends a , b , c went for week end party to mcdonald ’ s restaurant and there they measure there weights in some order in 7 rounds . a , b , c , ab , bc , ac , abc . final round measure is 170 kg then find the average weight of all the 7 rounds ? | "average weight = [ ( a + b + c + ( a + b ) + ( b + c ) + ( c + a ) + ( a + b + c ) ] / 7 = 4 ( a + b + c ) / 7 = 4 x 170 / 7 = 97.1 kgs answer : a" | a ) 97.1 kgs , b ) 88.5 kgs , c ) 86.5 kgs , d ) 67.5 kgs , e ) 88.2 kgs | a | divide(multiply(add(const_1, 3), 170), 7) | add(const_1,n0)|multiply(n2,#0)|divide(#1,n1)| | general |
if the area of the equilateral triangle above is 0.75 , what is the area of the adjacent square ? | area of equilateral triangle : a ^ 2 * ( square root of 3 ) / 4 = 0.75 hence , area of square = a ^ 2 = 0.75 * 4 / ( square root of 3 ) = 3 / ( square root of 3 ) = square root of 3 answer : b | ['a ) ( √ 3 ) / 2', 'b ) √ 3', 'c ) √ 6', 'd ) 3', 'e ) 2 √ 3'] | b | divide(multiply(0.75, const_4), sqrt(const_3)) | multiply(n0,const_4)|sqrt(const_3)|divide(#0,#1) | geometry |
a technician makes a round - trip to and from a certain service center by the same route . if the technician completes the drive to the center and then completes 20 percent of the drive from the center , what percent of the round - trip has the technician completed ? | "round trip means 2 trips i . e . to and fro . he has completed one i . e 50 % completed . then he traveled another 20 % of 50 % i . e 10 % . so he completed 50 + 10 = 60 % of total trip a" | a ) 60 , b ) 50 , c ) 40 , d ) 70 , e ) 45 | a | add(divide(const_100, const_2), divide(multiply(20, divide(const_100, const_2)), const_100)) | divide(const_100,const_2)|multiply(n0,#0)|divide(#1,const_100)|add(#0,#2)| | gain |
it has been raining at the rate of 5 centimeters per hour . if the rain filled a cylindrical drum with a depth of 15 centimeters , and area 300 square centimeters , how long did it take to fill the drum completely ? | answer is : a , 3 hours the volume of the drum is irrelevant and only height matters since rain fell all over the city . thus , it takes only 15 / 5 = 3 hours of rain to fill the drum | ['a ) 3 hours', 'b ) 4 hours 15 minutes', 'c ) 6 hours', 'd ) 2 hours 5 minutes', 'e ) 8 hours'] | a | divide(15, 5) | divide(n1,n0) | geometry |
in digging a pond 28 m * 10 m * 5 m the volumes of the soil extracted will be ? | "28 * 10 * 5 = 1400 answer : e" | a ) 3387 , b ) 1000 , c ) 2866 , d ) 2787 , e ) 1400 | e | multiply(multiply(28, 10), 5) | multiply(n0,n1)|multiply(n2,#0)| | general |
suppose you work for a manufacturing plant that pays you $ 12.50 an hour and $ . 16 for each widget you make . how many widgets must you produce in a 40 hour week to earn $ 800 ( before payroll deductions ) ? | "hourly = $ 12.50 / hour totalhours = 40 piecework = $ 0.16 / widget totalpieces = ? ? ? ? total pay = ( hourly * totalhours ) + ( piecework * totalpieces ) so , put it all together : $ 800 = ( $ 12.50 * 40 ) + ( $ 0.16 * total pieces ) totalpieces = ( $ 800 - ( $ 12.50 * 40 ) ) / $ 0.16 = 1875 widgets e . 1875" | a ) 1800 , b ) 1825 , c ) 1850 , d ) 1855 , e ) 1875 | e | divide(subtract(800, multiply(12.50, 40)), 16) | multiply(n0,n2)|subtract(n3,#0)|divide(#1,n1)| | physics |
the diameter of a cylindrical tin is 10 cm and height is 5 cm . find the volume of the cylinder ? | "r = 5 h = 5 π * 5 * 5 * 5 = 125 π cc answer : a" | a ) 125 , b ) 155 , c ) 130 , d ) 120 , e ) 100 | a | divide(volume_cylinder(divide(10, const_2), 5), const_pi) | divide(n0,const_2)|volume_cylinder(#0,n1)|divide(#1,const_pi)| | geometry |
if 8 spiders make 4 webs in 9 days , then how many days are needed for 1 spider to make 1 web ? | "explanation : let , 1 spider make 1 web in x days . more spiders , less days ( indirect proportion ) more webs , more days ( direct proportion ) hence we can write as ( spiders ) 8 : 1 ( webs ) 1 : 9 } : : x : 4 â ‡ ’ 8 ã — 1 ã — 9 = 1 ã — 4 ã — x â ‡ ’ x = 18 answer : option e" | a ) 10 , b ) 20 , c ) 12 , d ) 16 , e ) 18 | e | multiply(1, 8) | multiply(n0,n3)| | physics |
if x ^ 2 + ( 1 / x ^ 2 ) = 9 , x ^ 4 + ( 1 / x ^ 4 ) = ? | "- > x ^ 4 + ( 1 / x ^ 4 ) = ( x ^ 2 ) ^ 2 + ( 1 / x ^ 2 ) ^ 2 = ( x ^ 2 + 1 / x ^ 2 ) ^ 2 - 2 x ^ 2 ( 1 / x ^ 2 ) = 9 ^ 2 - 2 = 79 . thus , the answer is e ." | a ) 10 , b ) 11 , c ) 12 , d ) 14 , e ) 79 | e | subtract(power(2, 2), 2) | power(n0,n0)|subtract(#0,n0)| | general |
( 150 % of 1265 ) ÷ 7 = ? | "explanation : ? = ( 150 x 1265 / 100 ) ÷ 7 = 189750 / 700 = 271 answer : option c" | a ) a ) 125 , b ) b ) 175 , c ) c ) 271 , d ) d ) 375 , e ) e ) 524 | c | divide(multiply(divide(150, const_100), 1265), 7) | divide(n0,const_100)|multiply(n1,#0)|divide(#1,n2)| | general |
two vessels contains equal number of mixtures milk and water in the ratio 7 : 2 and 8 : 1 . both the mixtures are now mixed thoroughly . find the ratio of milk to water in the new mixture so obtained ? | "the ratio of milk and water in the new vessel is = ( 7 / 9 + 8 / 9 ) : ( 2 / 9 + 1 / 9 ) = 15 / 9 : 3 / 9 = 5 : 1 answer is c" | a ) 1 : 3 , b ) 9 : 13 , c ) 5 : 1 , d ) 11 : 3 , e ) 15 : 4 | c | divide(add(multiply(7, divide(add(8, 1), add(7, 2))), 8), add(multiply(2, divide(add(8, 1), add(7, 2))), 1)) | add(n2,n3)|add(n0,n1)|divide(#0,#1)|multiply(n0,#2)|multiply(n1,#2)|add(n2,#3)|add(n3,#4)|divide(#5,#6)| | other |
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 12 kmph and 21 kmph respectively . when they meet , it is found that one train has traveled 60 km more than the other one . the distance between the two stations is ? | "1 h - - - - - 5 ? - - - - - - 60 12 h rs = 12 + 21 = 33 t = 12 d = 33 * 12 = 396 . answer : b" | a ) 477 , b ) 396 , c ) 279 , d ) 276 , e ) 291 | b | add(multiply(divide(60, subtract(21, 12)), 12), multiply(divide(60, subtract(21, 12)), 21)) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | physics |
a part of certain sum of money is invested at 9 % per annum and the rest at 12 % per annum , if the interest earned in each case for the same period is equal , then ratio of the sums invested is ? | "12 : 9 = 4 : 3 answer : c" | a ) 4 : 7 , b ) 4 : 0 , c ) 4 : 3 , d ) 4 : 1 , e ) 4 : 2 | c | multiply(divide(12, const_100), 9) | divide(n1,const_100)|multiply(n0,#0)| | gain |
the hcf and lcm of two numbers m and n are respectively 6 and 210 . if m + n = 58 , then 1 / m + 1 / n is equal to | "answer we have , m x n = 6 x 210 = 1260 â ˆ ´ 1 / m + 1 / n = ( m + n ) / mn = 58 / 1260 = 17 / 620 correct option : e" | a ) 1 / 35 , b ) 3 / 35 , c ) 5 / 37 , d ) 2 / 35 , e ) none | e | divide(58, multiply(6, 210)) | multiply(n0,n1)|divide(n2,#0)| | general |
a 30 kg metal bar made of alloy of tin and silver lost 3 kg of its weight in the water . 10 kg of tin loses 1.375 kg in the water ; 5 kg of silver loses 0.375 kg . what is the ratio of tin to silver in the bar ? | "you can simply use this formula to avoid confusion : w 1 / w 2 = ( a 2 - aavg ) / ( avg - a 1 ) here is how you will find the values of a 1 an a 2 . we have an overall loss ( average loss ) . the average loss is 3 kg when 30 kg alloy is immersed . this is a loss of ( 3 / 30 ) * 100 = 10 % . this is aavg the loss of ti... | a ) 1 / 4 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 2 / 3 | e | divide(divide(subtract(3, multiply(divide(0.375, 5), 30)), subtract(divide(1.375, 10), divide(0.375, 5))), subtract(30, divide(subtract(3, multiply(divide(0.375, 5), 30)), subtract(divide(1.375, 10), divide(0.375, 5))))) | divide(n5,n4)|divide(n3,n2)|multiply(n0,#0)|subtract(#1,#0)|subtract(n1,#2)|divide(#4,#3)|subtract(n0,#5)|divide(#5,#6)| | other |
the diameter of a cylindrical tin is 14 cm and height is 2 cm . find the volume of the cylinder ? | "r = 7 h = 5 π * 7 * 7 * 5 = 98 π cc answer : e" | a ) 33 , b ) 45 , c ) 66 , d ) 77 , e ) 98 | e | divide(volume_cylinder(divide(14, const_2), 2), const_pi) | divide(n0,const_2)|volume_cylinder(#0,n1)|divide(#1,const_pi)| | geometry |
a shopkeeper sold an article for rs 2552.36 . approximately what was his profit percent if the cost price of the article was rs 2400 | explanation : gain % = ( 152.36 * 100 / 2400 ) = 6.34 % = 6 % approx option c | a ) 4 % , b ) 5 % , c ) 6 % , d ) 7 % , e ) 8 % | c | floor(multiply(const_100, divide(subtract(2552.36, 2400), 2400))) | subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)|floor(#2) | gain |
the balance of a trader weighs 10 % less than it should . still the trader marks up his goods to get an overall profit of 30 % . what is the mark up on the cost price ? | "the most natural way to deal with ' weights ' questions is by assuming values . say the trader ' s balance shows 100 gms . it is actually 90 gms because it weighs 10 % less . say , the cost price is $ 90 ( $ 1 / gm ) . since he gets a profit of 30 % , the selling price must be 90 + ( 30 / 100 ) * 90 = $ 117 since the ... | a ) 40 % , b ) 8 % , c ) 25 % , d ) 17 % , e ) 9 % | d | subtract(add(multiply(multiply(const_4, 10), divide(30, const_100)), multiply(const_4, 10)), const_100) | divide(n1,const_100)|multiply(n0,const_4)|multiply(#0,#1)|add(#2,#1)|subtract(#3,const_100)| | gain |
hammers and wrenches are manufactured at a uniform weight per hammer and a uniform weight per wrench . if the total weight of 3 hammers and 4 wrenches is one - third that of 10 hammers and 5 wrenches , then the total weight of one wrench is how many times that of one hammer ? | "x be the weight of a hammer and y be the weight of a wrench . ( 3 x + 4 y ) = 1 / 3 * ( 10 x + 5 y ) 3 ( 3 x + 4 y ) = ( 10 x + 5 y ) 9 x + 12 y = 10 x + 5 y 7 y = x y = x / 7 ans - e" | a ) 1 / 2 , b ) 2 / 3 , c ) 1 , d ) 3 / 2 , e ) 1 / 7 | e | divide(subtract(3, multiply(const_3, const_2)), subtract(multiply(const_3, const_2), 4)) | multiply(const_2,const_3)|subtract(n0,#0)|subtract(#0,n1)|divide(#1,#2)| | general |
if a is an integer greater than 2 but less than 7 and b is an integer greater than 4 but less than 13 , what is the range of a / b ? | "the way to approach this problem is 2 < a < 7 and 4 < b < 13 minimum possible value of a is 3 and maximum is 6 minimum possible value of b is 5 and maximum is 12 range = max a / min b - min a / max b ( highest - lowest ) 6 / 5 - 3 / 12 = 57 / 60 hence b" | a ) 23 / 34 , b ) 57 / 60 , c ) 51 / 67 , d ) 19 / 71 , e ) 75 / 64 | b | subtract(divide(subtract(7, const_1), add(4, const_1)), divide(add(2, const_1), subtract(13, const_1))) | add(n2,const_1)|add(n0,const_1)|subtract(n1,const_1)|subtract(n3,const_1)|divide(#2,#0)|divide(#1,#3)|subtract(#4,#5)| | general |
all factors of a positive integer a are multiplied and the product obtained is a ^ 3 . if a is greater than 1 , how many factors does a have ? | answer = d = 6 number . . . . . . . . . . . . . . . . factors . . . . . . . . . . . . . . . . . . . . . . . . product 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 , ... | a ) 2 , b ) 3 , c ) 5 , d ) 6 , e ) 8 | d | divide(multiply(multiply(multiply(multiply(const_2, const_2), const_2), const_2), 3), multiply(multiply(const_2, const_2), const_2)) | multiply(const_2,const_2)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(n0,#2)|divide(#3,#1) | general |
when 5 + 1 = 56 , 6 + 1 = 67 , 7 + 1 = 78 , then 8 + 1 = ? | "5 + 1 = > 5 x 1 = 5 & 5 + 1 = 6 = > 5 & 6 = > 56 6 + 1 = > 6 x 1 = 6 & 6 + 1 = 7 = > 6 & 7 = > 67 7 + 1 = > 7 x 1 = 7 & 7 + 1 = 8 = > 7 & 8 = > 78 then 8 + 1 = > 8 x 1 = 8 & 8 + 1 = 9 = > 8 & 9 = > 89 answer : c" | a ) 85 , b ) 98 , c ) 89 , d ) 105 , e ) 129 | c | add(multiply(multiply(1, 1), const_10), 5) | multiply(n7,n10)|multiply(#0,const_10)|add(n0,#1)| | general |
express 22 mps in kmph ? | "22 * 18 / 5 = 79.2 kmph answer : a" | a ) 79.2 kmph , b ) 89.2 kmph , c ) 79.6 kmph , d ) 99.2 kmph , e ) 69.2 kmph | a | multiply(divide(22, const_1000), const_3600) | divide(n0,const_1000)|multiply(#0,const_3600)| | physics |
for the past n days , the average ( arithmetic mean ) daily production at a company was 50 units . if today ' s production of 90 units raises the average to 58 units per day , what is the value of n ? | "( average production for n days ) * n = ( total production for n days ) - - > 50 n = ( total production for n days ) ; ( total production for n days ) + 90 = ( average production for n + 1 days ) * ( n + 1 ) - - > 50 n + 90 = 58 * ( n + 1 ) - - > n = 4 . or as 40 extra units increased the average for n + 1 days by 8 u... | a ) 30 , b ) 18 , c ) 10 , d ) 9 , e ) 4 | e | subtract(divide(subtract(90, 50), subtract(58, 50)), const_1) | subtract(n1,n0)|subtract(n2,n0)|divide(#0,#1)|subtract(#2,const_1)| | general |
the manager at a health foods store mixes a unique superfruit juice cocktail that costs $ 1399.45 per litre to make . the cocktail includes mixed fruit juice and a ç ai berry juice , which cost $ 262.85 per litre and $ 3104.35 per litre , respectively . the manager has already opened 32 litres of the mixed fruit juice ... | "262.85 ( 32 ) + 3 , 104.35 x = 1 , 399.45 ( 32 + x ) solve the equation . 262.85 ( 32 ) + 3 , 104.35 x = 1 , 399.45 ( 32 + x ) 8 , 411.2 + 3 , 104.35 x = 44 , 782.4 + 1 , 399.45 x 8 , 411.2 + 1 , 704.9 x = 44 , 782.4 1 , 704.9 x = 36 , 371.2 x ≈ 21.3 answer is b ." | a ) 17 litres , b ) 21.3 litres , c ) 11 litres , d ) 07 litres , e ) 38 litres | b | divide(subtract(multiply(32, 1399.45), multiply(32, 262.85)), subtract(3104.35, 1399.45)) | multiply(n0,n3)|multiply(n1,n3)|subtract(n2,n0)|subtract(#0,#1)|divide(#3,#2)| | general |
the sum of the squares of 3 numbers is 241 , while the sum of their products taken two at a time is 100 . their sum is : | x ^ + y ^ 2 + z ^ 2 = 241 xy + yz + zx = 100 as we know . . ( x + y + z ) ^ 2 = x ^ 2 + y ^ 2 + z ^ 2 + 2 ( xy + yz + zx ) so ( x + y + z ) ^ 2 = 241 + ( 2 * 100 ) ( x + y + z ) ^ 2 = 441 so x + y + z = 21 answer : c | a ) 20 , b ) 11 , c ) 21 , d ) 41 , e ) none of these | c | sqrt(add(241, multiply(100, const_2))) | multiply(n2,const_2)|add(n1,#0)|sqrt(#1) | general |
two goods trains each 720 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 750 + 750 = 1500 m . required time = 1500 * 6 / 125 = 72 sec . answer : e" | a ) 22 sec , b ) 88 sec , c ) 48 sec , d ) 18 sec , e ) 72 sec | e | add(45, 30) | add(n1,n2)| | physics |
if w / x = 1 / 6 and w / y = 1 / 5 , then ( x + y ) / y = | "w / x = 1 / 6 = > x = 6 w and w / y = 1 / 5 = > y = 5 w ( x + y ) / y = ( 6 w + 5 w ) / 5 w = 11 w / 5 w = 11 / 5 correct option : d" | a ) 4 / 5 , b ) 6 / 5 , c ) 7 / 5 , d ) 11 / 5 , e ) 9 / 5 | d | add(divide(divide(1, 1), divide(5, 6)), const_1) | divide(n2,n0)|divide(n3,n1)|divide(#0,#1)|add(#2,const_1)| | general |
35 liters of a mixture is created by mixing liquid p and liquid q in the ratio 4 : 3 . how many liters of liquid q must be added to make the ratio 5 : 7 ? | let x be the amount of liquid q to be added . ( 3 / 7 ) * 35 + x = ( 7 / 12 ) * ( 35 + x ) 1260 + 84 x = 1715 + 49 x 35 x = 455 x = 13 the answer is d . | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | d | subtract(divide(multiply(divide(multiply(add(5, 7), divide(multiply(35, 4), add(4, 3))), 5), 7), add(5, 7)), divide(multiply(35, 3), add(4, 3))) | add(n3,n4)|add(n1,n2)|multiply(n0,n1)|multiply(n0,n2)|divide(#2,#1)|divide(#3,#1)|multiply(#0,#4)|divide(#6,n3)|multiply(n4,#7)|divide(#8,#0)|subtract(#9,#5) | general |
a jar contains a mixture of ab in the ratio 4 : 1 . when 20 l of mixture is replaced with liquid b , ratio becomes 2 : 3 . how many liters of liquid a was present in mixture initially . | "20 litres of mixture that is replaced will contain 16 litres of a and 4 litres of b ( as a : b = 4 : 1 ) let the initial volume of the mixture be 4 k + 1 k = 5 k so by condition , [ 4 k - 16 ] / [ k - 4 + 20 ] = 2 / 3 = > 12 k - 48 = 2 k - 8 + 40 = > 10 k = 80 solve for k which is k = 8 so initial volume of liquid a =... | a ) 12 , b ) 15 , c ) 32 , d ) 20 , e ) 25 | c | multiply(divide(multiply(add(multiply(3, 2), multiply(2, 2)), divide(20, subtract(multiply(3, 2), 1))), add(4, 1)), 4) | add(n0,n1)|multiply(n4,n3)|multiply(n3,n3)|add(#1,#2)|subtract(#1,n1)|divide(n2,#4)|multiply(#3,#5)|divide(#6,#0)|multiply(n0,#7)| | other |
two trains 140 m and 160 m long run at the speed of 60 kmph and 40 kmph in opposite directions in parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 kmph * 5 / 18 = 250 / 9 m / s distance covered in crossing each other = 140 + 160 = 300 m required time = 300 * 9 / 250 = 54 / 5 = 10.8 sec answer is c" | a ) 5.6 sec , b ) 8.9 sec , c ) 10.8 sec , d ) 12.6 sec , e ) 15 sec | c | divide(add(140, 160), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
for a certain exam , a score of 60 was 2 standard deviations below mean and a score of 100 was 3 standard deviations above mean . what was the mean score for the exam ? | "mean - 2 sd = 60 mean + 3 sd = 100 by solving above we get , sd ( absolute value ) = 8 mean = 76 ans . b" | a ) 74 , b ) 76 , c ) 78 , d ) 80 , e ) 82 | b | divide(add(multiply(60, 3), multiply(100, 2)), add(2, 3)) | add(n1,n3)|multiply(n0,n3)|multiply(n1,n2)|add(#1,#2)|divide(#3,#0)| | general |
the sum of the present age of henry and jill is 48 . what is their present ages if 9 years ago henry was twice the age of jill ? | "let the age of jill 9 years ago be x , age of henry be 2 x x + 9 + 2 x + 9 = 48 x = 10 present ages will be 19 and 29 answer : d" | a ) and 27 , b ) and 24 , c ) and 22 , d ) and 29 , e ) of these | d | subtract(48, divide(add(48, 9), const_3)) | add(n0,n1)|divide(#0,const_3)|subtract(n0,#1)| | general |
the average of 15 numbers is calculated as 20 . it is discovered later on that while calculating the average , one number namely 36 was wrongly read as 26 . the correct average is ? | "15 * 20 + 36 – 26 = 300 / 10 = 30 answer : a" | a ) 30 , b ) 26 , c ) 16 , d ) 97 , e ) 12 | a | add(20, divide(subtract(36, 26), 15)) | subtract(n2,n3)|divide(#0,n0)|add(n1,#1)| | general |
what is the area of a square field whose diagonal of length 16 m ? | "d 2 / 2 = ( 16 * 16 ) / 2 = 128 answer : a" | a ) 128 , b ) 289 , c ) 200 , d ) 112 , e ) 178 | a | divide(square_area(16), const_2) | square_area(n0)|divide(#0,const_2)| | geometry |
the avg age of an adult class is 40 years . 12 new students with an avg age of 32 years join the class . therefore decreasingthe avg by 4 year . find what was the original strength of class ? | let original strength = y then , 40 y + 12 x 32 = ( y + 12 ) x 36 ⇒ 40 y + 384 = 36 y + 432 ⇒ 4 y = 48 ∴ y = 12 b | a ) 8 , b ) 12 , c ) 15 , d ) 16 , e ) 19 | b | subtract(subtract(32, 12), subtract(40, 32)) | subtract(n2,n1)|subtract(n0,n2)|subtract(#0,#1) | general |
what distance will be covered by a bus moving at 108 kmph in 30 seconds ? | "108 kmph = 108 * 5 / 18 = 30 mps d = speed * time = 30 * 30 = 900 m . answer : e" | a ) 287 , b ) 600 , c ) 289 , d ) 276 , e ) 900 | e | multiply(multiply(108, const_0_2778), 30) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
vijay sells a cupboard at 12 % below cost price . had he got rs . 1500 more , he would have made a profit of 12 % . what is the cost price of the cupboard ? | "explanation : cost price = 1500 / ( 0.12 + 0.12 ) = 1500 / 0.24 = rs . 6250 answer c" | a ) 7450 , b ) 14900 , c ) 6250 , d ) 6000 , e ) none of these | c | divide(1500, divide(subtract(add(const_100, 12), subtract(const_100, 12)), const_100)) | add(n0,const_100)|subtract(const_100,n0)|subtract(#0,#1)|divide(#2,const_100)|divide(n1,#3)| | gain |
if 20 men can build a wall 112 metres long in 6 days , what length of a similar wall can be built by 30 men in 3 days ? | 20 men is 6 days can build 112 metres 30 men in 3 days can build = 112 * ( 30 / 20 ) x ( 3 / 6 ) = 84 meters answer : b . | a ) 65 mtr . , b ) 84 mtr , c ) 70 mtr . , d ) 78 mtr . , e ) 17 mtr . | b | multiply(112, divide(multiply(30, 3), multiply(20, 6))) | multiply(n3,n4)|multiply(n0,n2)|divide(#0,#1)|multiply(n1,#2) | physics |
if the speed of a man is 57 km per hour , then what is the distance traveled by him in 30 seconds ? | "the distance traveled in 30 sec = 57 * ( 5 / 18 ) * 30 = 475 m answer : c" | a ) 275 m , b ) 360 m , c ) 475 m , d ) 420 m , e ) 440 m | c | multiply(multiply(57, const_0_2778), 30) | multiply(n0,const_0_2778)|multiply(n1,#0)| | physics |
victor gets 90 % marks in examinations . if these are 405 marks , find the maximum marks . | "let the maximum marks be m then 90 % of m = 405 ⇒ 90 / 100 × m = 405 ⇒ m = ( 405 × 100 ) / 90 ⇒ m = 40500 / 90 ⇒ m = 450 therefore , maximum marks in the examinations are 450 answer : d" | a ) 334 , b ) 500 , c ) 376 , d ) 450 , e ) 271 | d | divide(405, divide(90, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
there are 3 fictions and 6 non - fictions . how many cases are there such that 2 fictions and 2 non - fictions are selected from them ? | "number of ways of selecting 2 fiction books = 3 c 2 number of ways of selecting 2 non fiction books = 6 c 2 3 c 2 * 6 c 2 = 3 * 15 = 45 answer : c" | a ) 90 , b ) 120 , c ) 45 , d ) 180 , e ) 200 | c | divide(multiply(multiply(3, const_4), multiply(6, 3)), power(factorial(2), 2)) | factorial(n2)|multiply(n0,const_4)|multiply(n0,n1)|multiply(#1,#2)|power(#0,n2)|divide(#3,#4)| | general |
when positive integer n is divided by positive integer j , the remainder is 16 . if n / j = 134.08 , what is value of j ? | "when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 134.08 here 134 is the quotient . given that remainder = 16 so , 134.08 = 134 + 16 / j so , j = 200 ans e" | a ) 22 , b ) 56 , c ) 78 , d ) 112 , e ) 200 | e | divide(16, subtract(134.08, add(const_100, add(multiply(const_4, const_10), const_2)))) | multiply(const_10,const_4)|add(#0,const_2)|add(#1,const_100)|subtract(n1,#2)|divide(n0,#3)| | general |
find the circumference and area of radius 13 cm . | "area of circle = π r ² = 22 / 7 × 13 × 13 cm ² = 531 cm ² answer : b" | a ) 124 cm ² , b ) 531 cm ² , c ) 354 cm ² , d ) 584 cm ² , e ) 594 cm ² | b | circle_area(13) | circle_area(n0)| | geometry |
( 0.0048 ) ( 3.5 ) / ( 0.05 ) ( 0.1 ) ( 0.004 ) = | ( 0.0048 ) ( 3.5 ) / ( 0.05 ) ( 0.1 ) ( 0.004 ) = 0.0048 * 350 / 5 ( 0.1 ) ( 0.004 ) = 0.048 * 70 / 1 ( 0.004 ) = 48 * 70 / 4 = 12 * 70 = 840 answer : b | a ) 8.4 , b ) 840 , c ) 84.0 , d ) 0.84 , e ) 0.084 | b | divide(multiply(0.0048, 3.5), multiply(multiply(0.05, 0.1), 0.004)) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,#1)|divide(#0,#2) | general |
a batsman in his 17 th innings makes a score of 85 , and thereby increases his average by 3 . what is his average after the 17 th innings ? he had never been ’ not out ’ . | average score before 17 th innings = 85 - 3 × 17 = 34 average score after 17 th innings = > 34 + 3 = 37 answer : b | a ) 47 , b ) 37 , c ) 39 , d ) 43 , e ) 42 | b | add(subtract(85, multiply(3, 17)), 3) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1) | general |
a snooker tournament charges $ 40.00 for vip seats and $ 15.00 for general admission ( “ regular ” seats ) . on a certain night , a total of 300 tickets were sold , for a total cost of $ 7,500 . how many fewer tickets were sold that night for vip seats than for general admission seats ? | "let no of sits in vip enclosure is x then x * 40 + 15 ( 300 - x ) = 7500 or 25 x = 7500 - 4500 , x = 3000 / 25 = 120 vip = 120 general 200 d" | a ) 190 , b ) 180 , c ) 150 , d ) 200 , e ) 300 | d | subtract(300, divide(subtract(add(multiply(add(const_3, const_4), const_1000), multiply(add(const_2, const_3), const_100)), multiply(300, 15.00)), multiply(add(const_2, const_3), add(const_2, const_3)))) | add(const_3,const_4)|add(const_2,const_3)|multiply(n1,n2)|multiply(#0,const_1000)|multiply(#1,const_100)|multiply(#1,#1)|add(#3,#4)|subtract(#6,#2)|divide(#7,#5)|subtract(n2,#8)| | geometry |
a hiker walked for 3 days . she walked 18 miles on the first day , walking 3 miles per hour . on the second day she walked for one less hour but she walked one mile per hour , faster than on the first day . on the third day she walked the same number of hours as on the first day , but at the same speed as on the second... | first day - 18 miles with 3 miles per hours then total - 6 hours for that day second day - 4 miles per hour and 5 hours - 20 miles third day - 4 miles per hour and 6 hours - 24 miles total 18 + 20 + 24 - 62 answer : option e . | a ) 24 , b ) 44 , c ) 58 , d ) 60 , e ) 62 | e | add(add(18, multiply(add(3, const_1), subtract(divide(18, 3), const_1))), multiply(divide(18, 3), add(3, const_1))) | add(n0,const_1)|divide(n1,n0)|multiply(#0,#1)|subtract(#1,const_1)|multiply(#0,#3)|add(n1,#4)|add(#5,#2) | physics |
sum of two numbers is 30 . two times of the first exceeds by 10 from the three times of the other . then the numbers will be ? | "explanation : x + y = 30 2 x ã ¢ â ‚ ¬ â € œ 3 y = 10 x = 16 y = 14 answer : b" | a ) 14 , 16 , b ) 16 , 14 , c ) 18 , 12 , d ) 12 , 18 , e ) 17 , 13 | b | subtract(30, divide(subtract(30, divide(10, const_2)), const_2)) | divide(n1,const_2)|subtract(n0,#0)|divide(#1,const_2)|subtract(n0,#2)| | general |
a runner runs the 40 miles from marathon to athens at a constant speed . halfway through the run she injures her foot , and continues to run at half her previous speed . if the second half takes her 4 hours longer than the first half , how many hours did it take the runner to run the second half ? | "the runner runs the first 20 miles at speed v and the second 20 miles at speed v / 2 . the time t 2 to run the second half must be twice the time t 1 to run the first half . t 2 = 2 * t 1 = t 1 + 4 t 1 = 4 and so t 2 = 8 . the answer is c ." | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | c | divide(40, divide(divide(40, const_2), 4)) | divide(n0,const_2)|divide(#0,n1)|divide(n0,#1)| | physics |
a man can do a job in 15 days . his father takes 20 days and his son finishes it in 25 days . how long will they take to complete the job if they all work together ? | a 6.4 days 1 day work of the three persons = ( 1 / 15 + 1 / 20 + 1 / 25 ) = 47 / 300 so , all three together will complete the work in 300 / 47 = 6.4 days . - | a ) 6.4 days , b ) 4.4 days , c ) 5.4 days , d ) 8.4 days , e ) 2.4 days | a | divide(const_1, add(divide(const_1, 25), add(divide(const_1, 15), divide(const_1, 20)))) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|add(#3,#2)|divide(const_1,#4) | physics |
cricket match is conducted in us . the run rate of a cricket game was only 3.2 in first 10 over . what should be the run rate in the remaining 40 overs to reach the target of 272 runs ? | required run rate = 272 - ( 3.2 x 10 ) = 240 = 6 40 40 a | a ) 6 , b ) 6.25 , c ) 7.25 , d ) 7.5 , e ) 8 | a | divide(subtract(272, multiply(3.2, 10)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2) | gain |
two numbers are less than third number by 34 % and 37 % respectively . how much percent is the second number less than by the first | "let the third number is x . then first number = ( 100 - 34 ) % of x = 66 % of x = 66 x / 100 second number is ( 63 x / 100 ) difference = 66 x / 100 - 63 x / 100 = 3 x / 100 so required percentage is , difference is what percent of first number ( 3 x / 100 * 100 / 63 x * 100 ) % = 22 % answer : e" | a ) 8 % , b ) 10 % , c ) 9 % , d ) 11 % , e ) 22 % | e | subtract(multiply(divide(subtract(37, 34), subtract(const_100, 34)), const_100), const_10) | subtract(n1,n0)|subtract(const_100,n0)|divide(#0,#1)|multiply(#2,const_100)|subtract(#3,const_10)| | gain |
how many liters of a 40 % iodine solution need to be mixed with 35 liters of a 20 % iodine solution to create a 30 % iodine solution ? | "solution 1 : assume the iodine solution to be mixed = x lts . iodine = 0.4 x lts , water = 0.6 x lts . solution 2 : 35 liters of a 20 % iodine solution iodine = 7 lts , water = 28 lts . total iodine = 0.4 x + 7 total water = 0.6 x + 28 the resultant is a 35 % idoine solution . hence ( 0.4 x + 7 ) / ( x + 35 ) = 30 / 1... | a ) 30.5 , b ) 49 , c ) 100 , d ) 105 , e ) 140 | a | add(divide(subtract(multiply(divide(multiply(35, 20), const_100), 30), multiply(divide(multiply(35, 20), const_100), 20)), subtract(multiply(20, divide(40, const_100)), divide(multiply(35, 20), const_100))), 35) | divide(n0,const_100)|multiply(n1,n2)|divide(#1,const_100)|multiply(n2,#0)|multiply(n3,#2)|multiply(n2,#2)|subtract(#3,#2)|subtract(#4,#5)|divide(#7,#6)|add(n1,#8)| | gain |
of the total amount that jill spent on a shopping trip , excluding taxes , she spent 50 percent on clothing , 20 percent on food , and 30 percent on other items . if jill paid a 5 percent tax on the clothing , no tax on the food , and an 10 percent tax on all other items , then the total tax that she paid was what perc... | "let amount spent by jill = 100 clothing = 50 , food = 20 , others = 30 tax on clothing = 2.5 tax on others = 3 percentage = 5.5 / 100 = 5.5 % answer : a" | a ) 5.5 % , b ) 3.6 % , c ) 4.4 % , d ) 5.2 % , e ) 6.0 % | a | multiply(divide(add(multiply(50, divide(5, const_100)), multiply(30, divide(10, const_100))), const_100), const_100) | divide(n3,const_100)|divide(n4,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,const_100)|multiply(#5,const_100)| | general |
in a certain warehouse , 70 percent of the packages weigh less than 75 pounds , and a total of 48 packages weigh less than 25 pounds . if 80 percent of the packages weigh at least 25 pounds , how many of the packages weigh at least 25 pounds but less than 75 pounds ? | if 80 % of the packages weigh at least 25 pounds this means that 20 % of the packages weigh less than 25 pounds let t = total number of packages so , 20 % of t = # of packages that weigh less than 25 pounds 48 packages weigh less than 25 pounds great . so , 20 % of t = 48 rewrite to get : 0.2 t = 48 solve : t = 240 70 ... | a ) 8 , b ) 64 , c ) 120 , d ) 102 , e ) 144 | c | subtract(divide(multiply(multiply(divide(48, subtract(const_100, 80)), const_100), 70), const_100), 48) | subtract(const_100,n4)|divide(n2,#0)|multiply(#1,const_100)|multiply(n0,#2)|divide(#3,const_100)|subtract(#4,n2) | general |
in the standard formulation of a flavored drink the ratio by volume of flavoring to corn syrup to water is 1 : 12 : 30 . in the sport formulation , the ratio of flavoring to corn syrup is 3 times as great as in the standard formulation , and the ratio of flavoring to water is half that of the standard formulation . if ... | standard : fl : corn s : water = 1 : 12 : 30 sport : fl : corn s : water = 3 : 12 : 180 this simplifies to 1 : 4 : 60 if the large bottle has a capacity of x ounces , then 4 x / 65 = 2 . so , x = 32.5 ounces . water = ( 60 / 65 ) * ( 65 / 2 ) = = 30 ounces . ans b | a ) 15 , b ) 30 , c ) 45 , d ) 60 , e ) 90 | b | multiply(multiply(multiply(divide(1, 12), 3), const_2), multiply(30, 2)) | divide(n0,n1)|multiply(n2,n4)|multiply(n3,#0)|multiply(#2,const_2)|multiply(#3,#1) | other |
a machine , working at a constant rate , manufactures 18 pens in 30 minutes . how many pens does it make in 1 hr 45 min ? | change 1 hr 45 min to 105 min . for this , we need to set up a simple proportion of pens per time 18 / 30 = s / 105 the absolutely worst thing you could do at this point in the problem is to cross - multiply . that would be a supremely unstrategic move . instead , cancel before you multiply . for what we can see this p... | a ) 63 , b ) 65 , c ) 62 , d ) 60 , e ) 45 | a | multiply(divide(add(multiply(1, const_60), 45), 30), 18) | multiply(n2,const_60)|add(n3,#0)|divide(#1,n1)|multiply(n0,#2) | physics |
a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 225 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "b 225 * t = 1 km = > t = 1 / 225 km / h v * ( t + 2 / 3600 ) = 1 v ( 1 / 225 + 2 / 3600 ) = 1 = > v = 200 km / h" | a ) 220 , b ) 200 , c ) 210 , d ) 225 , e ) 230 | b | divide(1, divide(add(multiply(const_3600, divide(1, 225)), 2), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics |
if ( - 7 ) ^ ( 8 x ) = 7 ^ ( 12 - 5 x ) and x is an integer , what is the value of x ? | "since x is an integer , ( - 7 ) ^ ( 8 x ) is always positive . so , 7 ^ 8 x = 7 ^ ( 13 - 5 x ) 8 x = 12 - 5 x 12 x = 12 x = 1 answer : c" | a ) 5 , b ) - 4 , c ) 1 , d ) 2 , e ) 3 | c | divide(12, 7) | divide(n3,n0)| | general |
a library branch originally contained 18360 volumes , 30 % of which were fiction novels . 1 / 4 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 = 5,508 transferred to another location = 4590 transferred fiction novels = 1530 non transferred fiction novels = 3,978 percent of the remaining collection was fiction novels = 3978 / ( 18360 - 4590 ) * 100 = > 28.888 . . . % hence answer will be ( c )" | a ) 2.5 % , b ) 17.67 % , c ) 28.8 % , d ) 45.2 % , e ) 73.6 % | c | multiply(divide(multiply(divide(30, const_100), subtract(1, divide(1, 3))), subtract(1, divide(1, 4))), 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 |
( 51 + 52 + 53 + … … … + 100 ) is equal to : | "( 51 + 52 + 53 + … … … + 100 ) = ( 1 + 2 + 3 + … … . + 100 ) - ( 1 + 2 + 3 + 4 + … … + 50 ) = ( 100 * 101 ) / 2 - ( 50 * 51 ) / 2 = ( 5050 - 1275 ) = 3775 . answer : d" | a ) 2525 , b ) 2975 , c ) 3225 , d ) 3775 , e ) 3885 | d | subtract(divide(multiply(100, add(100, const_1)), const_2), divide(multiply(subtract(51, const_1), 51), const_2)) | add(n3,const_1)|subtract(n0,const_1)|multiply(n3,#0)|multiply(n0,#1)|divide(#2,const_2)|divide(#3,const_2)|subtract(#4,#5)| | general |
the mean of 50 observations was 30 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is | "sol . therefore correct sum = ( 30 × 50 + 48 – 23 ) = 1525 . therefore correct mean = 1525 / 50 = 30.5 . answer c" | a ) 35.2 , b ) 36.1 , c ) 30.5 , d ) 39.1 , e ) none | c | divide(add(multiply(30, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
a train running at the speed of 60 km / hr crosses a pole in 42 seconds . what is the length of the train ? | "speed = 60 x 5 / 18 m / sec = 50 / 3 m / sec . length of the train = ( speed x time ) . length of the train = 50 / 3 x 42 m = 150 m . option b" | a ) 120 metres , b ) 700 metres , c ) 324 metres , d ) 828 metres , e ) 600 metres | b | multiply(divide(multiply(60, const_1000), const_3600), 42) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
the average weight of 6 person ' s increases by 3.5 kg when a new person comes in place of one of them weighing 47 kg . what might be the weight of the new person ? | "total weight increased = ( 6 x 3.5 ) kg = 21 kg . weight of new person = ( 47 + 21 ) kg = 68 kg option c" | a ) 60 kg , b ) 75 kg , c ) 68 kg , d ) 85 kg , e ) 90 kg | c | add(multiply(6, 3.5), 47) | multiply(n0,n1)|add(n2,#0)| | general |
ajay can walk 6 km in 1 hour . in how many hours he can walk 70 km ? | "1 hour he walk 6 km he walk 70 km in = 70 / 6 * 1 = 11.6 hours answer is b" | a ) 5 hrs , b ) 11.6 hrs , c ) 15.6 hrs , d ) 20.1 hrs , e ) 30 hrs | b | divide(70, 6) | divide(n2,n0)| | physics |
the speed of a boat in still water in 18 km / hr and the rate of current is 6 km / hr . the distance travelled downstream in 14 minutes is : | "explanation : speed downstream = ( 18 + 6 ) = 24 kmph time = 14 minutes = 14 / 60 hour = 7 / 30 hour distance travelled = time × speed = ( 7 / 30 ) × 26 = 6.06 km answer : option d" | a ) 11.4 km , b ) 10.9 km , c ) 10.4 km , d ) 6.06 km , e ) 12.56 km | d | multiply(add(18, 6), divide(14, const_60)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
tea worth rs . 126 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs . 133 per kg , the price of the third variety per kg will be | "solution since first second varieties are mixed in equal proportions , so their average price = rs . ( 126 + 135 / 2 ) = rs . 130.50 so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to find x . x - 133 / 2... | a ) rs . 169.50 , b ) rs . 1700 , c ) rs . 175.50 , d ) rs . 155.50 , e ) none | d | divide(subtract(multiply(133, add(add(1, 1), 2)), add(126, 126)), 2) | add(n1,n1)|add(n0,n0)|add(n3,#0)|multiply(n4,#2)|subtract(#3,#1)|divide(#4,n3)| | other |
find the simple interest on rs . 71,000 at 16 2 / 3 % per year for 9 months . | "p = rs . 71000 , r = 50 / 3 % p . a and t = 9 / 12 years = 3 / 4 years . simple interest = ( p * r * t ) / 100 = rs . ( 71,000 * ( 50 / 3 ) * ( 3 / 4 ) * ( 1 / 100 ) ) = rs . 8875 answer is c ." | a ) 7500 , b ) 6500 , c ) 8875 , d ) 9500 , e ) none of them | c | multiply(multiply(multiply(add(multiply(multiply(multiply(2, 3), const_100), const_100), multiply(multiply(multiply(3, 3), const_100), multiply(add(3, 2), 2))), divide(add(multiply(16, 3), 2), 3)), divide(multiply(3, 3), multiply(2, multiply(2, 3)))), divide(const_1, const_100)) | add(n2,n3)|divide(const_1,const_100)|multiply(n3,n3)|multiply(n2,n3)|multiply(n1,n3)|add(n2,#4)|multiply(n2,#3)|multiply(#3,const_100)|multiply(#2,const_100)|multiply(#0,n2)|divide(#2,#6)|divide(#5,n3)|multiply(#7,const_100)|multiply(#8,#9)|add(#12,#13)|multiply(#14,#11)|multiply(#10,#15)|multiply(#1,#16)| | gain |
john makes $ 65 a week from his job . he earns a raise andnow makes $ 72 a week . what is the % increase ? | "increase = ( 7 / 65 ) * 100 = ( 7 / 65 ) * 100 = 10.76 % . b" | a ) 16 % , b ) 10.76 % , c ) 10.69 % , d ) 10.98 % , e ) 10 % | b | multiply(divide(subtract(72, 65), 65), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
if 20 men take 15 days to to complete a job , in how many days can 20 men finish that work ? | ans . 15 days | a ) 15 , b ) 16 , c ) 17 , d ) 18 , e ) 19 | a | divide(multiply(20, 15), 20) | multiply(n0,n1)|divide(#0,n2)| | physics |
a can do a piece of work 40 days . b can do work in 60 days . in how many days they will complete the work together ? | "lcm = 120 , ratio = 40 : 60 = 2 : 3 no of days = 120 / ( 2 + 3 ) = 90 / 5 = 24 days answer : e" | a ) 15 days , b ) 16 days , c ) 19 days , d ) 17 days , e ) 24 days | e | divide(const_1, add(divide(const_1, 40), divide(const_1, 60))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
a profit of rs . 800 is divided between x and y in the ratio of 1 / 2 : 1 / 3 . what is the difference between their profit shares ? | "a profit of rs . 800 is divided between x and y in the ratio of 1 / 2 : 1 / 3 or 3 : 2 . so profits are 480 and 320 . difference in profit share = 480 - 320 = 160 answer : b" | a ) s . 260 , b ) s . 160 , c ) s . 360 , d ) s . 50 , e ) s . 90 | b | subtract(divide(divide(800, add(divide(1, 2), divide(1, 3))), 2), divide(divide(800, add(divide(1, 2), divide(1, 3))), 3)) | divide(n1,n2)|divide(n1,n4)|add(#0,#1)|divide(n0,#2)|divide(#3,n2)|divide(#3,n4)|subtract(#4,#5)| | general |
two trains are moving in opposite directions at 60 km / hr and 90 km / hr . their lengths are 1.50 km and 1.0 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 = 1.50 + 1.0 = 2.5 km = 2500 m . required time = 2500 * 3 / 125 = 60 sec . answer : d" | a ) 48 , b ) 9 , c ) 7 , d ) 60 , e ) 15 | d | subtract(divide(multiply(1.50, const_1000), divide(multiply(60, const_1000), const_3600)), divide(multiply(1.0, 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 |
what is the smallest integer that is multiple of 3 , 5,9 | "it is the lcm of 3 , 5 and 9 which is 45 . the answer is b ." | a ) a ) 70 , b ) b ) 45 , c ) c ) 200 , d ) d ) 280 , e ) e ) 140 | b | add(const_3, const_4) | add(const_3,const_4)| | general |
a book is bought for $ 60 and sold for $ 78 . what is the profit in percentage ? | "78 / 60 = 1.3 the answer is c ." | a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) 50 | c | multiply(divide(subtract(78, 60), 60), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
a is a positive integer and multiple of 2 ; p = 4 ^ a , what is the remainder when p is divided by 10 ? | it is essential to recognize that the remainder when an integer is divided by 10 is simply the units digit of that integer . to help see this , consider the following examples : 4 / 10 is 0 with a remainder of 4 14 / 10 is 1 with a remainder of 4 5 / 10 is 0 with a remainder of 5 105 / 10 is 10 with a remainder of 5 it... | a ) 10 , b ) 6 , c ) 4 , d ) 0 , e ) it can not be determined | b | reminder(power(const_4, const_4), 10) | power(const_4,const_4)|reminder(#0,n2) | general |
product of two natural numbers is 7 . then , the sum of reciprocals of their squares is | "explanation : if the numbers are a , b , then ab = 7 , as 17 is a prime number , so a = 1 , b = 7 . 1 / a 2 + 1 / b 2 = 1 / 1 ( 2 ) + 1 / 7 ( 2 ) = 50 / 49 option a" | a ) 50 / 49 , b ) 1 / 289 , c ) 290 / 90 , d ) 290 / 19 , e ) none of these | a | add(power(divide(const_1, const_1), const_2), power(divide(const_1, 7), const_2)) | divide(const_1,const_1)|divide(const_1,n0)|power(#0,const_2)|power(#1,const_2)|add(#2,#3)| | general |
in a fuel station the service costs $ 1.15 per car , every liter of fuel costs 0.4 $ . assuming that you own 2 sports cars and 2 executive cars and all fuel tanks are empty . how much will it cost to fuel all cars together if a sports car tank is 32 liters and an executive car tank is 75 % bigger ? | "total cars = 4 1.15 * 4 = 4.6 - > service cost fuel cost in sports car = 2 * 32 * 0.4 = 25.6 fuel cost in executive car = 25.6 * 7 / 4 = 44.8 total fuel cost = 25.6 + 44.8 = 70.4 cost to fuel car = 70.4 + 4.6 = 75 answer : b" | a ) 37.5 $ , b ) 75 $ , c ) 87.5 $ , d ) 94.5 $ , e ) 98.4 $ | b | add(multiply(multiply(add(32, divide(multiply(32, 75), const_100)), 2), 0.4), multiply(multiply(32, 2), 0.4)) | multiply(n4,n5)|multiply(n2,n4)|divide(#0,const_100)|multiply(n1,#1)|add(n4,#2)|multiply(n3,#4)|multiply(n1,#5)|add(#6,#3)| | general |
a , band c can do a piece of work in 11 days , 20 days and 67 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 / 20 = 31 / 220 ( a + c ) ' s 1 day ' s work = 1 / 11 + 1 / 67 = 78 / 737 work done in 2 day ' s = 31 / 220 + 78 / 737 = 19 / 77 19 / 77 th work done in 2 days work done = 77 / 19 * 2 = 8.1 days answer : b" | a ) 7.1 days , b ) 8.1 days , c ) 9.1 days , d ) 10 days , e ) 11 days | b | divide(67, divide(add(add(divide(67, 11), divide(67, 20)), add(divide(67, 11), divide(67, 67))), 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 telephone company needs to create a set of 3 - digit area codes . the company is entitled to use only digits 6 , 4 and 3 , which can be repeated . if the product of the digits in the area code must be even , how many different codes can be created ? | "total # of codes possible is 3 * 3 * 3 = 27 . oit of those 27 codes only the product of 333 and will be odd , the remaining 26 will have either 2 or 4 in them , which ensures that their product will be even . therefore the number of codes where the product of the digits is even = ( total ) - ( restriction ) = 27 - 1 =... | a ) 20 , b ) 22 , c ) 26 , d ) 24 , e ) 30 | c | subtract(power(3, 3), const_1) | power(n0,n0)|subtract(#0,const_1)| | general |
in a rectangular axis system , what is the area of a parallelogram with the coordinates : ( 3,5 ) , ( 9,5 ) , ( 2,7 ) , ( 8,7 ) ? | "delta x will give us the dimension of one side of the parallelogram = 9 - 3 = 6 unit delta y will give us the dimension of the other side of parallelogram = 7 - 5 = 2 unit area of parallelogram = 6 * 2 = 12 answer is d" | a ) 21 . , b ) 28 . , c ) 35 . , d ) 12 . , e ) 52 . | d | add(const_3, const_2) | add(const_2,const_3)| | geometry |
find the c . i . on a sum of rs . 10000 for 6 months at 25 % per annum , interest being compounded quarterly ? | c . i . = 2000 ( 21 / 20 ) ^ 2 - 1800 = 1289 answer : c | a ) 10290 , b ) 5290 , c ) 1289 , d ) 1290 , e ) 2290 | c | subtract(multiply(10000, multiply(add(const_1, divide(const_0_25, const_4)), add(const_1, divide(const_0_25, const_4)))), 10000) | divide(const_0_25,const_4)|add(#0,const_1)|multiply(#1,#1)|multiply(n0,#2)|subtract(#3,n0) | gain |
you collect pens . suppose you start out with 5 . mike gives you another 20 pens . since her father makes pens , cindy decides to double your pens . since you ' re nice , you give sharon 10 pens . how many pens do you have at the end ? | "solution start with 5 pens . mike gives you 20 pens : 5 + 20 = 25 pens . cindy doubles the number of pens you have : 25 ã — 2 = 50 pens . sharon takes 10 pens from you : 50 - 10 = 40 pens . so you have 40 at the end . correct answer : b" | a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 43 | b | subtract(multiply(add(20, 5), const_2), 10) | add(n0,n1)|multiply(#0,const_2)|subtract(#1,n2)| | general |
a sum of money at simple interest amounts to rs . 1560 in 4 years and to rs . 1590 in 6 years . the sum is : | "s . i . for 2 years = rs . ( 1590 - 1560 ) = rs . 30 . s . i . for 4 years = rs . ( 30 x 2 ) = rs . 60 . principal = rs . ( 1560 - 60 ) = rs . 1500 . answer : option c" | a ) rs . 1400 , b ) rs . 1450 , c ) rs . 1500 , d ) rs . 1550 , e ) rs . 1525 | c | subtract(1560, divide(multiply(subtract(1590, 1560), 4), 6)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain |
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