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
values |
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how many different values of positive integer x , for which | x + 7 | < x , are there ? | "answer a i opted to put the random value option . i used 0 , 7 , - 7 and the the extreme of 30 and - 30 . . i was able to solve it in 1 : 09 b" | a ) 0 , b ) 2 , c ) 3 , d ) 8 , e ) 16 | b | add(7, 7) | add(n0,n0)| | general |
you buy a piece of land with an area of Γ’ Λ Ε‘ 100 , how long is one side of the land plot ? | "try filling the numbers into the answer y x y = find the closest to 100 . answer c" | a ) 28 , b ) 29 , c ) 10 , d ) 31 , e ) 32 | c | sqrt(100) | sqrt(n0)| | geometry |
find the cost of fencing around a circular field of diameter 14 m at the rate of rs . 2.50 a meter ? | "2 * 22 / 7 * 7 = 44 44 * 2.5 = rs . 110 answer : b" | a ) 288 , b ) 110 , c ) 772 , d ) 592 , e ) 261 | b | multiply(circumface(divide(14, const_2)), 2.50) | divide(n0,const_2)|circumface(#0)|multiply(n1,#1)| | physics |
a farmer used 1,034 acres of land for beans , wheat , and corn in the ratio of 5 : 2 : 4 , respectively . how many y acres were used for corn ? | "consider 5 x acres of land used for bean consider 2 x acres of land used for wheat consider 4 x acres of land used for corn total given is 1034 acres 11 x = 1034 x = 94 land used for corn y = 4 * 94 = 376 correct option - c" | a ) 188 , b ) 258 , c ) 376 , d ) 470 , e ) 517 | c | multiply(divide(add(multiply(const_1000, const_1), add(multiply(const_10, const_3), 4)), add(add(5, 2), 4)), 4) | add(n1,n2)|multiply(const_10,const_3)|multiply(const_1,const_1000)|add(n3,#1)|add(n3,#0)|add(#3,#2)|divide(#5,#4)|multiply(n3,#6)| | other |
10 play kabadi , 15 play kho kho only , 5 play both gmaes . then how many in total ? | "10 play kabadi = > n ( a ) = 10 , 5 play both gmaes . = > n ( anb ) = 5 15 play kho kho only , = > n ( b ) = n ( b only ) + n ( anb ) = 15 + 5 = 20 total = > n ( aub ) = n ( a ) + n ( b ) - n ( anb ) = 10 + 20 - 5 = 25 answer : a" | a ) 25 , b ) 35 , c ) 38 , d ) 40 , e ) 45 | a | subtract(add(10, add(15, 5)), 5) | add(n1,n2)|add(n0,#0)|subtract(#1,n2)| | general |
a tailor trims 7 feet from opposite edges of a square piece of cloth , and 5 feet from the other two edges . if 45 square feet of cloth remain , what was the length of a side of the original piece of cloth ? | "let the original side of the square be x . ( x - 14 ) * ( x - 10 ) = 45 = 5 * 9 x = 19 the answer is d ." | a ) 13 , b ) 15 , c ) 17 , d ) 19 , e ) 21 | d | divide(add(add(multiply(const_2, 7), multiply(const_2, const_3.0)), sqrt(add(multiply(7, subtract(45, multiply(multiply(const_2, 5), multiply(const_2, 7)))), power(add(multiply(const_2, 7), multiply(const_2, 5)), const_2)))), const_2) | multiply(n0,const_2)|multiply(const_3.0,const_2)|add(#0,#1)|multiply(#1,#0)|power(#2,const_2)|subtract(n2,#3)|multiply(#5,n0)|add(#6,#4)|sqrt(#7)|add(#2,#8)|divide(#9,const_2)| | geometry |
$ 500 will become $ 1000 in 5 years find the rate of interest ? | si = simple interest = a - p = 1000 - 500 = $ 500 r = 100 si / pt = 100 * 500 / 500 * 5 = 20 % answer is b | a ) 10 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 50 % | b | divide(subtract(1000, 500), multiply(500, divide(5, const_100))) | divide(n2,const_100)|subtract(n1,n0)|multiply(n0,#0)|divide(#1,#2) | gain |
( ( - 1.9 ) ( 0.6 ) β ( 2.6 ) ( 1.2 ) ) / 3.0 = ? | "dove straight into calculation ( ( - 1.9 ) ( 0.6 ) β ( 2.6 ) ( 1.2 ) ) / 3.0 = - 1.42 answer d" | a ) - 0.71 , b ) 1.0 , c ) 1.07 , d ) - 1.42 , e ) 2.71 | d | divide(subtract(negate(multiply(1.9, 0.6)), multiply(2.6, 1.2)), 3.0) | multiply(n0,n1)|multiply(n2,n3)|negate(#0)|subtract(#2,#1)|divide(#3,n4)| | general |
the number 523 rbc is divisible by 7 , 89 . then what is the value of r * b * c | lcm of 7 , 8 and 9 is 504 , thus 523 rbc must be divisible by 504 . 523 rbc = 523000 + rbc 523000 divided by 504 gives a remainder of 352 . hence , 352 + rbc = k * 504 . k = 1 rbc = 152 - - > r * b * c = 10 k = 2 rbc = 656 - - > r * b * c = 180 as rbc is three digit number k can not be more than 2 . two answers ? well ... | a ) 504 , b ) 532 , c ) 210 , d ) 180 , e ) 280 | d | multiply(multiply(multiply(const_3, const_2), add(const_2, const_3)), multiply(const_3, const_2)) | add(const_2,const_3)|multiply(const_2,const_3)|multiply(#0,#1)|multiply(#2,#1) | general |
a man buys rs . 20 shares paying 9 % dividend . the man wants to have an interest of 12 % on his money . what is the market value of each share ? | "explanation : face value of each share = rs . 20 dividend per share = 9 % of 20 = 9 Γ 20 / 100 = 9 / 5 he needs to have an interest of 12 % on his money ie , money paid for a share 9 Γ 12 / 100 = 9 / 5 money paid for a share = 9 / 5 Γ 100 / 12 = 15 ie , market value of the share = rs . 15 answer : option c" | a ) rs . 12 , b ) rs . 18 , c ) rs . 15 , d ) rs . 21 , e ) rs . 25 | c | multiply(divide(const_100, 12), multiply(divide(9, const_100), 20)) | divide(const_100,n2)|divide(n1,const_100)|multiply(n0,#1)|multiply(#0,#2)| | gain |
what is the smallest no . which must be added to 27452 so as to obtain a sum which is divisible by 9 ? | "for 27452 , 2 + 7 + 4 + 5 + 2 = 20 . 7 must be added to 27452 to make it divisible by 9 . now , 2 + 7 + 4 + 5 + 9 = 27 = > 27 is a multiple of 9 and hence 27452 is also divisible by 9 a" | a ) 7 , b ) 2 , c ) 9 , d ) 4 , e ) 6 | a | divide(multiply(27452, 9), 27452) | multiply(n0,n1)|divide(#0,n0)| | general |
find the average of first 4 multiples of 5 ? | "average = ( 5 + 10 + 15 + 20 ) / 4 = 12.5 answer is c" | a ) 10 , b ) 15 , c ) 12.5 , d ) 13 , e ) 21 | c | divide(add(add(add(4, const_1), add(add(4, const_1), const_2)), add(subtract(5, 4), subtract(5, const_2))), 4) | add(n0,const_1)|subtract(n1,n0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,n0)| | general |
the sale price sarees listed for rs . 400 after successive discount is 10 % and 8 % is ? | "400 * ( 90 / 100 ) * ( 92 / 100 ) = 331.2 answer : c" | a ) 338 , b ) 277 , c ) 331.2 , d ) 882 , e ) 212 | c | subtract(subtract(400, divide(multiply(400, 10), const_100)), divide(multiply(subtract(400, divide(multiply(400, 10), const_100)), 8), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
the h . c . f of two numbers is 11 and their l . c . m is 7700 . if one of the numbers is 275 , then the other is | "explanation : other number = \ inline \ fn _ jvn \ left ( \ frac { 11 \ times 7700 } { 275 } \ right ) = 308 answer : c ) 308" | a ) 344 , b ) 387 , c ) 308 , d ) 299 , e ) 882 | c | multiply(11, 275) | multiply(n0,n2)| | physics |
what is the smallest positive integer x , such that 2160 x is a perfect cube ? | "take out the factors of 2160 that will come 6 ^ 3 * 10 . for perfect cube you need every no . raise to the power 3 . for 2160 to be a perfect cube , you need two 2 and two 5 that means 100 a is the answer ." | a ) 100 , b ) 6 , c ) 8 , d ) 12 , e ) 18 | a | add(const_3, const_4) | add(const_3,const_4)| | geometry |
a batch of cookies was divided amomg 6 tins : 2 / 3 of all the cookies were placed in either the blue or the green tin , and the rest were placed in the red tin . if 1 / 4 of all the cookies were placed in the blue tin , what fraction of the cookies that were placed in the other tins were placed in the green tin | this will help reduce the number of variables you have to deal with : g + b = 2 / 3 r = 1 / 6 b = 1 / 4 we can solve for g which is 5 / 12 what fraction ( let it equal x ) of the cookies that were placed in the other tins were placed in the green tin ? so . . x * ( g + r ) = g x * ( 5 / 12 + 1 / 6 ) = 5 / 12 x = 5 / 7 ... | a ) 15 / 2 , b ) 9 / 4 , c ) 5 / 9 , d ) 5 / 7 , e ) 9 / 7 | d | add(subtract(const_1, divide(2, 3)), subtract(divide(2, 3), divide(1, 4))) | divide(n1,n2)|divide(n3,n4)|subtract(const_1,#0)|subtract(#0,#1)|add(#2,#3) | general |
in a certain game , a large bag is filled with blue , green , purple and red chips worth 1 , 5 , x and 11 points each , respectively . the purple chips are worth more than the green chips , but less than the red chips . a certain number of chips are then selected from the bag . if the product of the point values of the... | "28160 = 1 * 5 * 8 ^ 3 * 11 the factors of 8 must come from the purple point value , so there are 3 purple chips . the answer is c ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(5, const_2) | subtract(n1,const_2)| | general |
two numbers have a h . c . f of 12 and a product of two numbers is 2460 . find the l . c . m of the two numbers ? | "l . c . m of two numbers is given by ( product of the two numbers ) / ( h . c . f of the two numbers ) = 2460 / 12 = 205 . answer : c" | a ) 140 , b ) 150 , c ) 205 , d ) 170 , e ) 180 | c | divide(2460, 12) | divide(n1,n0)| | physics |
if a - b = 9 and a ^ 2 + b ^ 2 = 281 , find the value of ab | "2 ab = ( a ^ 2 + b ^ 2 ) - ( a - b ) ^ 2 = 281 - 81 = 200 = > ab = 100 answer : a" | a ) 100 , b ) 12 , c ) 15 , d ) 18 , e ) 19 | a | multiply(multiply(add(9, divide(subtract(sqrt(281), 9), 2)), divide(subtract(sqrt(281), 9), 2)), 2) | sqrt(n3)|subtract(#0,n0)|divide(#1,n1)|add(n0,#2)|multiply(#3,#2)|multiply(n1,#4)| | general |
if x and y are sets of integers , x # y denotes the set of integers that belong to set x or set y , but not both . if x consists of 16 integers , y consists of 18 integers , and 6 of the integers are in both x and y , then x # y consists of how many integers ? | "the number of integers that belong to set x only is 16 - 6 = 10 ; the number of integers that belong to set y only is 18 - 6 = 12 ; the number of integers that belong to set x or set y , but not both is 10 + 12 = 22 . answer : c" | a ) 6 , b ) 16 , c ) 22 , d ) 30 , e ) 174 | c | add(subtract(18, 6), subtract(16, 6)) | subtract(n1,n2)|subtract(n0,n2)|add(#0,#1)| | other |
a train is moving at 6 / 7 of its usual speed . the train is 15 minutes too late . what is the usual time ( in hours ) for the train to complete the journey ? | "new time = d / ( 6 v / 7 ) = 7 / 6 * usual time 15 minutes represents 1 / 6 of the usual time . the usual time is 1.5 hours . the answer is b ." | a ) 1 , b ) 1.5 , c ) 2 , d ) 2.5 , e ) 3 | b | divide(multiply(multiply(15, divide(6, 7)), inverse(subtract(const_1, divide(6, 7)))), const_60) | divide(n0,n1)|multiply(n2,#0)|subtract(const_1,#0)|inverse(#2)|multiply(#3,#1)|divide(#4,const_60)| | physics |
exactly 3 / 5 of the people in the room are under the age of 21 , and exactly 5 / 13 of the people in the room are over the age of 65 . if the total number of the people in the room is greater than 50 and less than 100 , how many people in the room are under the age of 21 ? | "the total number of the people in the room must be a multiple of both 5 and 14 ( in order 3 / 5 and 5 / 14 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 5 and 14 , which is 70 . since , the total number of the people in the room is greater than 50 and less than 100... | a ) 21 , b ) 35 , c ) 39 , d ) 60 , e ) 42 | e | divide(multiply(multiply(5, 13), 3), 5) | multiply(n1,n4)|multiply(n0,#0)|divide(#1,n1)| | general |
there are 2 sections a and b in a class , consisting of 60 and 70 students respectively . if the average weight of section a is 60 kg and that of section b is 80 kg , find the average of the whole class ? | "total weight of 60 + 70 students = 60 * 60 + 70 * 80 = 3600 + 5600 average weight of the class is = 9200 / 130 = 70.76 kg answer is c" | a ) 50.78 kg , b ) 49.32 kg , c ) 70.76 kg , d ) 69.15 kg , e ) 70.89 kg | c | divide(add(multiply(60, 60), multiply(70, 80)), add(60, 70)) | add(n1,n2)|multiply(n1,n3)|multiply(n2,n4)|add(#1,#2)|divide(#3,#0)| | general |
a furniture store owner decided to drop the price of her recliners by 20 % to spur business . by the end of the week she had sold 80 % more recliners . what is the percentage increase of the gross ? | "say a recliner is actually worth $ 100 if she sells 100 recliners then she earns $ 10000 after the discount of 20 % , she will earn $ 80 per recliner and she sells 80 % more ie . , 180 recliners hence her sales tields 180 * 80 = $ 14400 increase in sales = 14400 - 10000 = $ 4400 so % increase = 4400 * 100 / 10000 = 44... | a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 44 % | e | subtract(multiply(multiply(subtract(const_1, divide(20, const_100)), add(const_1, divide(80, const_100))), const_100), const_100) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(#5,const_100)| | general |
in a family 19 people eat only vegetarian , 9 people eat only non veg . , 12 people eat both veg and non veg . . how many people eat veg in the family ? | "total people eat veg = only veg + both veg and non veg total = 19 + 12 = 31 answer = b" | a ) 20 , b ) 31 , c ) 9 , d ) 31 , e ) 21 | b | add(19, 12) | add(n0,n2)| | other |
two friends plan to walk along a 36 - km trail , starting at opposite ends of the trail at the same time . if friend p ' s rate is 25 % faster than friend q ' s , how many kilometers will friend p have walked when they pass each other ? | if q complete x kilometers , then p completes 1.25 x kilometers . x + 1.25 x = 36 2.25 x = 36 x = 16 then p will have have walked 1.25 * 16 = 20 km . the answer is b . | a ) 19 , b ) 20 , c ) 21 , d ) 22 , e ) 23 | b | multiply(divide(36, add(add(divide(25, const_100), const_1), const_1)), add(divide(25, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|divide(n0,#2)|multiply(#1,#3) | gain |
for any integer k greater than 1 , the symbol k * denotes the product of all integers between 1 and k , inclusive . if k * is a multiple of 3,675 , what is the least possible value of k ? | "we basically need 5,5 , 3,7 and 7 in the number . 14 ! , has 3 , 5 , 7 , 10 , 14 now 10 = 2 x 5 14 = 2 x 7 cross check : k ! = 3675 * m ( m is a positive integer ) k ! = 3 * 5 * 5 * 7 * 7 14 ! has all the numbers above . . therefore b . . ! answer : b" | a ) 12 , b ) 14 , c ) 15 , d ) 21 , e ) 25 | b | divide(divide(divide(3,675, const_3), const_3), add(1, const_4)) | add(n0,const_4)|divide(n2,const_3)|divide(#1,const_3)|divide(#2,#0)| | general |
l . c . m of two numbers is 192 and there h . c . f is 16 . if one of them is 48 . find the other | explanation : product of two numbers = product of their l . c . m and h . c . f 48 x a = 192 x 16 a = ( 192 x 16 ) / 48 = 64 answer : option b | a ) 32 , b ) 64 , c ) 48 , d ) 68 , e ) 78 | b | divide(multiply(192, 16), 48) | multiply(n0,n1)|divide(#0,n2) | physics |
at 3.40 , the hour hand and the minute hand of a clock form an angle of | sol : use formula ΞΈ = β£ β£ β£ 30 h β 112 m β£ β£ β£ ΞΈ = | 30 h β 112 m | angle = 30 Γ 3 β 11 / 2 Γ 40 = 90 β 220 = 130 Β° answer : e | a ) 378 , b ) 277 , c ) 297 , d ) 267 , e ) 130 | e | subtract(multiply(const_10, multiply(const_3, power(const_2, const_3))), multiply(subtract(subtract(multiply(3.4, add(const_4, const_1)), add(const_4, const_1)), const_1), const_10)) | add(const_1,const_4)|power(const_2,const_3)|multiply(#1,const_3)|multiply(n0,#0)|multiply(#2,const_10)|subtract(#3,#0)|subtract(#5,const_1)|multiply(#6,const_10)|subtract(#4,#7) | physics |
x and y started a business with capitals rs . 20000 and rs . 25000 . after few months z joined them with a capital of rs . 30000 . if the share of z in the annual profit of rs . 50000 is rs . 14000 , then after how many months from the beginning did z join ? | investments of x , y and z respectively are rs . 20000 , rs . 25000 and rs . 30000 let investment period of z be x months . ratio of annual investments of x , y and z is ( 20000 * 12 ) : ( 25000 * 12 ) : ( 30000 * x ) = 240 : 300 : 30 x = 8 : 10 : x the share of z in the annual profit of rs . 50000 is rs . 14000 . = > ... | a ) 3 , b ) 4 , c ) 6 , d ) 9 , e ) 5 | e | subtract(const_12, divide(divide(multiply(divide(14000, 50000), multiply(add(20000, 25000), const_12)), subtract(const_1, divide(14000, 50000))), 30000)) | add(n0,n1)|divide(n4,n3)|multiply(#0,const_12)|subtract(const_1,#1)|multiply(#1,#2)|divide(#4,#3)|divide(#5,n2)|subtract(const_12,#6) | gain |
if 18 men take 15 days to to complete a job , in how many days can 24 men finish that work ? | ans . 10 days | a ) 10 , b ) 11 , c ) 12 , d ) 13 , e ) 14 | a | divide(multiply(18, 15), 24) | multiply(n0,n1)|divide(#0,n2)| | physics |
a boat goes 100 km downstream in 10 hours , and 75 m upstream in 15 hours . the speed of the stream is ? | 100 - - - 10 ds = 10 ? - - - - 1 75 - - - - 15 us = 5 ? - - - - - 1 s = ( 10 - 5 ) / 2 = 2 21 / 2 kmph answer : d | a ) 22 1 / 9 kmph , b ) 22 8 / 2 kmph , c ) 22 1 / 8 kmph , d ) 22 1 / 2 kmph , e ) 22 1 / 4 kmph | d | divide(subtract(divide(100, 10), divide(75, 15)), const_2) | divide(n0,n1)|divide(n2,n3)|subtract(#0,#1)|divide(#2,const_2) | physics |
if k is an integer and 121 < k ^ 2 < 225 , then k can have at most how many values ? | given k is an integer 121 < k ^ 2 < 225 11 < | k | < 15 k = - 12 , - 13 , - 1412 , 1314 answer : d | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 8 | d | multiply(subtract(subtract(sqrt(225), sqrt(121)), const_1), const_2) | sqrt(n2)|sqrt(n0)|subtract(#0,#1)|subtract(#2,const_1)|multiply(#3,const_2) | general |
a specialized type of sand consists of 40 % mineral x by volume and 60 % mineral y by volume . if mineral x weighs 2.5 grams per cubic centimeter and mineral y weighs 3 grams per cubic centimeter , how many grams does a cubic meter of specialized sand combination weigh ? ( 1 meter = 100 centimeters ) | "let the volume be 1 m ^ 3 = 1 m * 1 m * 1 m = 100 cm * 100 cm * 100 cm = 1 , 000,000 cm ^ 3 by volume 40 % is x = 400,000 cm ^ 3 60 % is y = 600,000 cm ^ 3 by weight , in 1 cm ^ 3 , x is 2.5 gms in 400,000 cm ^ 3 , x = 2.5 * 400,000 = 1 , 000,000 grams in 1 cm ^ 3 , y is 3 gms in 600,000 cm ^ 3 , y = 3 * 600,000 = 1 ,... | a ) 5 , 500,000 , b ) 2 , 800,000 , c ) 55,000 , d ) 28,000 , e ) 280 | b | subtract(add(multiply(multiply(divide(volume_cube(100), const_10), 2.5), 2.5), multiply(multiply(divide(volume_cube(100), const_10), multiply(const_2, 3)), 3)), volume_cube(100)) | multiply(const_2,n3)|volume_cube(n5)|divide(#1,const_10)|multiply(#2,n2)|multiply(#2,#0)|multiply(#3,n2)|multiply(#4,n3)|add(#5,#6)|subtract(#7,#1)| | geometry |
a sum of rs . 2040 has been divided among a , b and c such that a gets of what b gets and b gets of what c gets . b β s share is : | "explanation let c β s share = rs . x then , b β s share = rs . x / 4 , a β s share = rs . ( 2 / 3 x x / 4 ) = rs . x / 6 = x / 6 + x / 4 + x = 2040 = > 17 x / 12 = 2040 = > 2040 x 12 / 17 = rs . 1440 hence , b β s share = rs . ( 1440 / 4 ) = rs . 360 . answer d" | a ) rs . 120 , b ) rs . 160 , c ) rs . 240 , d ) rs . 360 , e ) none | d | subtract(subtract(multiply(divide(2040, const_10), const_2), const_12), const_12) | divide(n0,const_10)|multiply(#0,const_2)|subtract(#1,const_12)|subtract(#2,const_12)| | general |
an error 17 % in excess is made while measuring the side of a square . now what is the percentage of error in the calculated area of the square ? | "percentage error in calculated area = ( 17 + 17 + ( 17 Γ£ β 17 ) / 100 ) % = 36.89 % answer : b" | a ) 6.64 % , b ) 36.89 % , c ) 15.64 % , d ) 26.64 % , e ) 10.64 % | b | divide(multiply(subtract(square_area(add(const_100, 17)), square_area(const_100)), const_100), square_area(const_100)) | add(n0,const_100)|square_area(const_100)|square_area(#0)|subtract(#2,#1)|multiply(#3,const_100)|divide(#4,#1)| | gain |
a boat can travel with a speed of 40 km / hr in still water . if the speed of the stream is 5 km / hr , find the time taken by the boat to go 45 km downstream . | "speed downstream = ( 40 + 5 ) km / hr = 45 km / hr . time taken to travel 45 km downstream = 45 / 45 hrs = 1 hrs . answer : a" | a ) 1 hr , b ) 2 hrs , c ) 3 hrs , d ) 4 hrs , e ) 5 hrs | a | divide(45, add(40, 5)) | add(n0,n1)|divide(n2,#0)| | physics |
3 candidates in an election and received 1036 , 4636 and 11628 votes respectively . what % of the total votes did the winning candidate gotin that election ? | total number of votes polled = ( 1036 + 4636 + 11628 ) = 17300 so , required percentage = 11628 / 17300 * 100 = 67.2 % e | a ) 40 % , b ) 55 % , c ) 57 % , d ) 60 % , e ) 67.2 % | e | multiply(divide(11628, add(add(1036, 4636), 11628)), const_100) | add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100) | gain |
find the amount on rs . 5000 in 2 years , the rate of interest being 2 % per first year and 3 % for the second year ? | "5000 * 102 / 100 * 103 / 100 = > 5253 answer : e" | a ) 3377 , b ) 2678 , c ) 5460 , d ) 1976 , e ) 5253 | e | divide(multiply(divide(multiply(5000, add(const_100, 2)), const_100), add(const_100, 3)), const_100) | add(n3,const_100)|add(n2,const_100)|multiply(n0,#1)|divide(#2,const_100)|multiply(#0,#3)|divide(#4,const_100)| | gain |
in plutarch enterprises , 60 % of the employees are marketers , 30 % are engineers , and the rest are managers . marketers make an average salary of $ 50,000 a year , and engineers make an average of $ 80,000 . what is the average salary for managers if the average for all employees is also $ 80,000 ? | "for sake of ease , let ' s say there are 10 employees : 6 marketers , 3 engineers , and 1 manager . average company salary * number of employees = total company salary > > > $ 80,000 * 10 = $ 800,000 subtract the combined salaries for the marketers ( 6 * $ 50,000 ) and the engineers ( 3 * $ 80,000 ) > > > $ 800,000 - ... | a ) $ 80,000 , b ) $ 130,000 , c ) $ 260,000 , d ) $ 290,000 , e ) $ 320,000 | c | divide(subtract(subtract(multiply(multiply(add(60, 30), const_1000), const_100), multiply(30, multiply(add(60, 30), const_1000))), multiply(multiply(multiply(divide(30, const_2), 30), const_1000), 60)), 30) | add(n0,n1)|divide(n1,const_2)|multiply(#0,const_1000)|multiply(#1,n1)|multiply(#2,const_100)|multiply(#2,n1)|multiply(#3,const_1000)|multiply(n0,#6)|subtract(#4,#5)|subtract(#8,#7)|divide(#9,n1)| | general |
find the missing figures : 0.25 % of ? = 0.04 | "let 0.25 % of x = 0.04 . then , 0.25 * x / 100 = 0.04 x = [ ( 0.04 * 100 ) / 0.25 ] = 16 . answer is a ." | a ) 16 , b ) 20 , c ) 17 , d ) 12 , e ) 18 | a | divide(0.04, divide(0.25, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
how many 6 digit number contain number 4 ? | "total 6 digit no . = 9 * 10 * 10 * 10 * 10 * 10 = 900000 not containing 4 = 8 * 9 * 9 * 9 * 9 * 9 = 472392 total 6 digit number contain 4 = 900000 - 472392 = 427608 answer : d" | a ) 317608 , b ) 327608 , c ) 337608 , d ) 427608 , e ) 357608 | d | add(subtract(subtract(const_1000, const_10), multiply(multiply(const_10, multiply(6, 6)), multiply(const_4, const_2))), const_10) | multiply(n0,n0)|multiply(const_2,const_4)|subtract(const_1000,const_10)|multiply(#0,const_10)|multiply(#3,#1)|subtract(#2,#4)|add(#5,const_10)| | general |
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 262 runs ? | "required run rate = 262 - ( 3.2 x 10 ) = 230 = 5.75 40 40 c" | a ) 6 , b ) 6.25 , c ) 5.75 , d ) 7.5 , e ) 8 | c | divide(subtract(262, multiply(3.2, 10)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
the mean of 40 observations was 36 . it was found later that an observation 34 was wrongly taken as 20 . the corrected new mean is | "explanation : correct sum = ( 36 * 40 + 34 - 20 ) = 1454 correct mean = = 1454 / 40 = 36.35 answer : b" | a ) 76.55 , b ) 36.35 , c ) 46.15 , d ) 16.05 , e ) 20 | b | divide(add(multiply(36, 40), subtract(subtract(40, const_2), 20)), 40) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
a small pool filled only with water will require an additional 300 gallons of water in order to be filled to 80 % of its capacity . if pumping in these additional 300 gallons of water will increase the amount of water in the pool by 40 % , what is the total capacity of the pool in gallons ? | "300 gallons of water increases capacity by 40 % that means 40 % is 300 gallons , so 100 % would be = 300 * 100 / 40 = 750 gallons now 750 + 300 gallons is 80 % capacity of tank . so 100 % capacity would be = 1050 * 100 / 80 = 1312.5 b is the answer" | a ) 1000 , b ) 1312.5 , c ) 1400 , d ) 1600 , e ) 1625 | b | divide(add(divide(multiply(300, const_100), 40), 300), divide(80, const_100)) | divide(n1,const_100)|multiply(n0,const_100)|divide(#1,n3)|add(n0,#2)|divide(#3,#0)| | general |
the speed of a car is 10 m / s . what is the its speed in kmph . | explanation : m / sec to km / hr conversion : x m / sec = [ x x ( 18 / 5 ) ] km / hr 10 x ( 18 / 5 ) 36 kmph answer : option d | a ) 25 kmph , b ) 30 kmph , c ) 50 kmph , d ) 36 kmph , e ) 37 kmph | d | multiply(10, const_3_6) | multiply(n0,const_3_6) | physics |
total dinning bill of 5 people was $ 139.00 and 10 % tip divided the bill evenly ? what is the bill amount each person shared . | "dinner bill of 5 person = 139 + 10 % tip so , 10 % of 139 = ( 139 * 10 ) / 100 = 13.9 so , the actual total amount = 139 + 13.9 = $ 152.9 so per head bill = 152.9 / 5 = $ 30.58 answer : c" | a ) 32.84 , b ) 22.84 , c ) 30.58 , d ) 24.84 , e ) 30.84 | c | divide(multiply(139.00, add(divide(const_1, 10), const_1)), 5) | divide(const_1,n2)|add(#0,const_1)|multiply(n1,#1)|divide(#2,n0)| | general |
at a monthly meeting , 3 / 5 of the attendees were males and 4 / 5 of the male attendees arrived on time . if 5 / 6 of the female attendees arrived on time , what fraction of the attendees at the monthly meeting did not arrive on time ? | "males who did not arrive on time are 1 / 5 * 3 / 5 = 3 / 25 of the attendees . females who did not arrive on time are 1 / 6 * 2 / 5 = 1 / 15 of the attendees . the fraction of all attendees who did not arrive on time is 3 / 25 + 1 / 15 = 14 / 75 the answer is c ." | a ) 6 / 25 , b ) 11 / 50 , c ) 14 / 75 , d ) 23 / 100 , e ) 31 / 150 | c | add(multiply(subtract(const_1, divide(5, 6)), subtract(const_1, divide(3, 5))), multiply(subtract(const_1, divide(4, 5)), divide(3, 5))) | divide(n4,n5)|divide(n0,n1)|divide(n2,n3)|subtract(const_1,#0)|subtract(const_1,#1)|subtract(const_1,#2)|multiply(#3,#4)|multiply(#1,#5)|add(#6,#7)| | general |
the sum of 7 consecutive integers is 980 . how many of them are prime ? | the middle number is 140 and the seven numbers are 137 , 138 , 139 , Β· Β· Β· , 143 . now 138 , 140 , and 142 are even , 141 is divisible by 3 , and 143 - by 11 . the remaining numbers 137 and 139 are prime ( it is easy to check that they are not divisible by 23 , 57 and 11 ) . correct answer c | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | c | subtract(subtract(const_10, add(add(const_1, const_2), const_3)), const_2) | add(const_1,const_2)|add(#0,const_3)|subtract(const_10,#1)|subtract(#2,const_2) | physics |
if n = 8 ^ 8 β 7 , what is the units digit of n ? | "8 ^ 8 - 8 = 8 ( 8 ^ 7 - 1 ) = = > 8 ( 2 ^ 21 - 1 ) last digit of 2 ^ 21 is 2 based on what explanation livestronger is saying . 2 ^ 24 - 1 yields 2 - 1 = 1 as the unit digit . now on multiply this with 7 , we get unit digit as 7 answer : d" | a ) 0 , b ) 1 , c ) 2 , d ) 7 , e ) 4 | d | divide(log(7), log(power(8, 8))) | log(n2)|power(n0,n1)|log(#1)|divide(#0,#2)| | general |
how many prime numbers between 1 and 100 are factors of 210 ? | factor of 210 = 2 * 3 * 5 * 7 - - - 4 prime numbers b | a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | b | multiply(const_4, const_1) | multiply(const_1,const_4) | other |
in 1998 the profits of company n were 10 percent of revenues . in 1999 , the revenues of company n fell by 20 percent , but profits were 12 percent of revenues . the profits in 1999 were what percent of the profits in 1998 ? | "0,096 r = x / 100 * 0.1 r answer c" | a ) 80 % , b ) 105 % , c ) 96 % , d ) 124.2 % , e ) 138 % | c | multiply(divide(multiply(subtract(const_1, divide(20, const_100)), divide(12, const_100)), divide(10, const_100)), const_100) | divide(n4,const_100)|divide(n3,const_100)|divide(n1,const_100)|subtract(const_1,#1)|multiply(#0,#3)|divide(#4,#2)|multiply(#5,const_100)| | gain |
when positive integer x is divided by 7 , the quotient is y and the remainder is 3 . when 2 x is divided by 6 , the quotient is 3 y and the remainder is 2 . what is the value of 11 y β x ? | "( 1 ) x = 7 y + 3 ( 2 ) 2 x = 18 y + 2 ( 2 ) - ( 1 ) : x = 11 y - 1 11 y - x = 1 the answer is d ." | a ) 4 , b ) 3 , c ) 2 , d ) 1 , e ) 0 | d | subtract(multiply(11, divide(subtract(multiply(2, 3), 2), subtract(multiply(6, 3), multiply(2, 7)))), add(multiply(7, divide(subtract(multiply(2, 3), 2), subtract(multiply(6, 3), multiply(2, 7)))), 3)) | multiply(n1,n2)|multiply(n3,n4)|multiply(n0,n2)|subtract(#0,n5)|subtract(#1,#2)|divide(#3,#4)|multiply(n6,#5)|multiply(n0,#5)|add(n1,#7)|subtract(#6,#8)| | general |
n is a positive integer . when n + 1 is divided by 9 , the remainder is 1 . what is the remainder when n is divided by 9 ? | "n + 1 = 9 a + 1 i . e . n + 1 = 10 , 19 , 28 , 37 , . . . etc . i . e . n = 9 , 18 , 27 , 36 , . . . etc . when n is divided by 9 remainder is always 0 answer : a" | a ) 0 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | a | divide(9, add(1, 9)) | add(n2,n3)|divide(n1,#0)| | general |
how many positive integers less than 23 are prime numbers , odd multiples of 5 , or the sum of a positive multiple of 2 and a positive multiple of 4 ? | "8 prime numbers less than 28 : { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 } 2 odd multiples of 5 : { 5 , 15 } 9 numbers which are the sum of a positive multiple of 2 and a positive multiple of 4 : { 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 } notice , that 5 is in two sets , thus total # of integers satisfying the given condi... | a ) 17 , b ) 25 , c ) 24 , d ) 22 , e ) 20 | a | subtract(subtract(subtract(23, 2), const_1), const_1) | subtract(n0,n2)|subtract(#0,const_1)|subtract(#1,const_1)| | general |
we need to carve out 125 identical cubes from a cube . what is the minimum number of cuts needed ? | total no . of cubes = n ^ 3 here n ^ 3 = 125 which makes n = 5 , also minimum no . of cuts required = 3 ( n - 1 ) hence , 3 ( 5 - 1 ) = 12 cuts . answer : e | ['a ) 8', 'b ) 9', 'c ) 10', 'd ) 11', 'e ) 12'] | e | multiply(const_3, subtract(power(125, const_0_33), const_1)) | power(n0,const_0_33)|subtract(#0,const_1)|multiply(#1,const_3) | geometry |
if a : b : : 3 : 7 , then what is ( 5 a + 6 b ) : ( a - 2 b ) ? | a / b = 3 / 7 dividing numerator & denominator of ' ( 5 a + 6 b ) / ( a - 2 b ) ' by b , [ 5 ( a / b ) + 6 ] / [ ( a / b ) - 2 ] = [ 5 * ( 3 / 7 ) + 6 ] / [ ( 3 / 7 ) - 2 ] = - 57 / 11 answer : b | a ) 57 : 11 , b ) - 57 : 11 , c ) 11 : 10 , d ) - 11 : 10 , e ) - 1 : 10 | b | divide(add(multiply(5, 3), multiply(7, 6)), subtract(3, multiply(2, 7))) | multiply(n0,n2)|multiply(n1,n3)|multiply(n1,n4)|add(#0,#1)|subtract(n0,#2)|divide(#3,#4) | general |
a block of wood has dimensions 10 cm x 10 cm x 50 cm . the block is painted red and then cut evenly at the 25 cm mark , parallel to the sides , to form two rectangular solids of equal volume . what percentage of the surface area of each of the new solids is not painted red ? | "the area of each half is 100 + 4 ( 250 ) + 100 = 1200 the area that is not painted is 100 . the fraction that is not painted is 100 / 1200 = 1 / 12 = 8.3 % the answer is b ." | a ) 5.5 % , b ) 8.3 % , c ) 11.6 % , d ) 14.2 % , e ) 17.5 % | b | multiply(divide(const_100, add(add(multiply(multiply(const_4, const_100), const_4), const_100), const_100)), const_100) | multiply(const_100,const_4)|multiply(#0,const_4)|add(#1,const_100)|add(#2,const_100)|divide(const_100,#3)|multiply(#4,const_100)| | geometry |
a mobile battery in 1 hour charges to 20 percent . how much time ( in minute ) will it require more to charge to 30 percent . | 1 hr = 20 percent . thus 15 min = 5 percent . now to charge 30 percent 90 min . answer : b | a ) 145 , b ) 90 , c ) 175 , d ) 160 , e ) 130 | b | multiply(divide(30, 20), const_60) | divide(n2,n1)|multiply(#0,const_60)| | physics |
how many 4 digit numbers are there , if it is known that the first digit is even , the second is odd , the third is prime , the fourth ( units digit ) is divisible by 3 , and the digit 5 can be used only once ? | "4 options for the first digit : 2 , 4 , 6 , 8 ; 5 options for the second digit : 1 , 3 , 5 , 7 , 9 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 3 , 6 , 9 . four digit # possible without the restriction ( about the digit 2 ) : 4 * 5 * 4 * 4 = 320 numbers with five 5 - s , 5 x 5... | a ) 20 , b ) 150 , c ) 225 , d ) 300 , e ) 304 | e | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), 4)) | add(n0,const_1)|multiply(n0,n0)|multiply(#0,#0)|multiply(#2,#1)|multiply(#2,n0)|subtract(#3,#4)| | physics |
two brothers ram and ravi appeared for an exam . the probability of selection of ram is 6 / 7 and that of ravi is 1 / 5 . find the probability that both of them are selected . | "let a be the event that ram is selected and b is the event that ravi is selected . p ( a ) = 6 / 7 p ( b ) = 1 / 5 let c be the event that both are selected . p ( c ) = p ( a ) x p ( b ) as a and b are independent events : = 6 / 7 x 1 / 5 = 6 / 35 answer : c" | a ) 2 / 35 , b ) 2 / 3 , c ) 6 / 35 , d ) 5 / 7 , e ) 7 / 5 | c | multiply(divide(6, 7), divide(1, 5)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)| | general |
a volunteer organization is recruiting new members . in the fall they manage to increase their number by 6 % . by the spring however membership falls by 19 % . what is the total change in percentage from fall to spring ? | "( 100 % + 6 % ) * ( 100 % - 19 % ) = 1.06 * . 81 = 0.8586 . 1 - 0.8586 = 14.14 % lost = - 14.14 % the answer is c the organization has lost 14.14 % of its total volunteers from fall to spring ." | a ) - 16.16 % , b ) - 15.15 % , c ) - 14.14 % , d ) - 13.13 % , e ) - 12.12 % | c | subtract(const_100, multiply(multiply(add(const_1, divide(6, const_100)), subtract(const_1, divide(19, const_100))), const_100)) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | general |
one hour after yolanda started walking from x to y , a distance of 31 miles , bob started walking along the same road from y to x . if yolanda Γ’ s walking rate was 1 miles per hour and bob Γ’ s was 2 miles per hour , how many miles had bob walked when they met ? | "let t be the number of hours that bob had walked when he met yolanda . then , when they met , bob had walked 4 t miles and yolanda had walked ( t + 1 ) miles . these distances must sum to 31 miles , so 2 t + ( t + 1 ) = 31 , which may be solved for t as follows 2 t + ( t + 1 ) = 31 2 t + t + 1 = 31 3 t = 30 t = 10 ( h... | a ) 19 , b ) 20 , c ) 22 , d ) 21 , e ) 19.5 | b | multiply(divide(subtract(31, 1), add(1, 2)), 2) | add(n1,n2)|subtract(n0,n1)|divide(#1,#0)|multiply(n2,#2)| | physics |
the length of a rectangular plot is 20 metres more than its breadth . if the cost of fencing the plot @ rs . 26.50 per metre is rs . 5565 , what is the length of the plot in metres ? | "let length of plot = l meters , then breadth = l - 20 meters and perimeter = 2 [ l + l - 20 ] = [ 4 l - 40 ] meters [ 4 l - 40 ] * 26.50 = 5565 [ 4 l - 40 ] = 5565 / 26.50 = 210 4 l = 250 l = 250 / 4 = 62.5 meters . answer : d" | a ) 333 , b ) 200 , c ) 288 , d ) 210 , e ) 1999 | d | subtract(divide(divide(5565, 26.50), const_2), multiply(const_2, 20)) | divide(n2,n1)|multiply(n0,const_2)|divide(#0,const_2)|subtract(#2,#1)| | physics |
if the perimeter of Ξ΄ acd is 9 + 3 β 3 , what is the perimeter of equilateral triangle Ξ΄ abc ? | the altitude of an equilateral triangle is side * β 3 / 2 . as perimeter of triangle acd is 9 + 3 β 3 , ac + cd + ad = ( side + side / 2 + side * β 3 / 2 ) = 9 + 3 β 3 or side = 6 . perimeter of equilateral triangle , abc is 3 ( side ) or 18 . answer : ( optionc ) | ['a ) 9', 'b ) 18 β 3 β 3', 'c ) 18', 'd ) 18 + 3 β 3', 'e ) 27'] | c | multiply(3, divide(multiply(multiply(3, const_2), subtract(const_3, sqrt(const_3))), subtract(const_3, sqrt(const_3)))) | multiply(n1,const_2)|sqrt(const_3)|subtract(const_3,#1)|multiply(#0,#2)|divide(#3,#2)|multiply(n1,#4) | geometry |
if 8 boys meet at a reunion and each boy shakes hands exactly once with each of the others , then what is the total number of handshakes | "n ( n - 1 ) / 2 = 8 * 7 / 2 = 28 answer : b" | a ) 41 , b ) 28 , c ) 43 , d ) 44 , e ) 45 | b | divide(factorial(8), multiply(factorial(subtract(8, const_2)), factorial(const_2))) | factorial(n0)|factorial(const_2)|subtract(n0,const_2)|factorial(#2)|multiply(#3,#1)|divide(#0,#4)| | general |
a man ' s basic pay for a 40 hour week is rs . 20 . overtime is paid for at 25 % above the basic rate . in a certain week he worked overtime and his total wage was rs . 25 . he therefore worked for a total of : | explanation : basic rate per hour = rs . ( 20 / 40 ) = rs . 1 / 2 overtime per hour = 125 % of rs . 1 / 2 = 125 / 100 Γ 1 / 2 = rs . 5 / 8 suppose he worked x hours overtime . then , 20 + 5 / 8 x = 25 or 5 / 8 x = 5 x = 5 Γ 8 / 5 = 8 hours so he worked in all for ( 40 + 8 ) hours = 48 hours . correct option : c | a ) 45 hours , b ) 47 hours , c ) 48 hours , d ) 50 hours , e ) none | c | add(divide(subtract(25, 20), multiply(divide(20, 40), divide(add(const_100, 25), const_100))), 40) | add(n2,const_100)|divide(n1,n0)|subtract(n2,n1)|divide(#0,const_100)|multiply(#1,#3)|divide(#2,#4)|add(n0,#5) | gain |
a can complete a work in 15 days and b can do the same work in 7 days . if a after doing 3 days , leaves the work , find in how many days b will do the remaining work ? | "the required answer = ( 15 - 3 ) * 7 / 15 = 84 / 15 = 5 1 / 2 days answer is b" | a ) 2 days , b ) 5 1 / 2 days , c ) 6 1 / 2 days , d ) 7 1 / 2 days , e ) 10 days | b | add(multiply(15, 3), divide(15, 3)) | divide(n0,n2)|multiply(n0,n2)|add(#0,#1)| | physics |
how many positive integers less than 60 have a reminder 01 when divided by 3 ? | "1 also gives the remainder of 1 when divided by 3 . so , there are total of 20 numbers . answer : e ." | a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 20 | e | divide(factorial(subtract(add(const_4, 01), const_1)), multiply(factorial(01), factorial(subtract(const_4, const_1)))) | add(n1,const_4)|factorial(n1)|subtract(const_4,const_1)|factorial(#2)|subtract(#0,const_1)|factorial(#4)|multiply(#1,#3)|divide(#5,#6)| | general |
it takes six minutes to load a certain video on a cellphone , and fifteen seconds to load that same video on a laptop . if the two devices were connected so that they operated in concert at their respective rates , how many seconds would it take them to load the video ? | "the laptop can load the video at a rate of 1 / 15 of the video per second . the phone can load the video at a rate of 1 / ( 60 * 6 ) = 1 / 360 of the video per second . the combined rate is 1 / 15 + 1 / 360 = 25 / 360 of the video per second . the time required to load the video is 360 / 25 = 14.4 seconds . the answer... | a ) 13.8 , b ) 14.1 , c ) 14.4 , d ) 14.6 , e ) 14.8 | c | subtract(inverse(add(inverse(multiply(add(add(const_2, const_3), const_4), const_60)), inverse(add(multiply(const_3, const_4), const_3)))), divide(subtract(multiply(multiply(const_4, const_4), const_3), const_2), multiply(const_100, const_100))) | add(const_2,const_3)|multiply(const_3,const_4)|multiply(const_4,const_4)|multiply(const_100,const_100)|add(#0,const_4)|add(#1,const_3)|multiply(#2,const_3)|inverse(#5)|multiply(#4,const_60)|subtract(#6,const_2)|divide(#9,#3)|inverse(#8)|add(#11,#7)|inverse(#12)|subtract(#13,#10)| | physics |
a computer store offers employees a 10 % discount off the retail price . if the store purchased a computer from the manufacturer for $ 1200 dollars and marked up the price 10 % to the final retail price , how much would an employee save if he purchased the computer at the employee discount ( 10 % off retail price ) as ... | "cost price = 1200 profit = 10 % = 10 % of 1200 = 120 selling price = cp + profit sp = 1320 a discount of 10 % to employees means 10 % off on 1320 so 10 % of 1320 = 132 ans b" | a ) 122 , b ) 132 , c ) 142 , d ) 152 , e ) 162 | b | divide(add(divide(multiply(1200, 10), const_100), 1200), multiply(divide(1200, const_100), const_2)) | divide(n1,const_100)|multiply(n0,n1)|divide(#1,const_100)|multiply(#0,const_2)|add(n1,#2)|divide(#4,#3)| | gain |
for all real numbers v , an operation is defined by the equation v * = v - v / 3 . if ( v * ) * = 12 , then v = | ( v * ) * = ( v - v / 3 ) - ( v - v / 3 ) / 3 12 = 2 v / 3 - 2 v / 9 = 4 v / 9 v = 27 the answer is e . | a ) 15 , b ) 18 , c ) 21 , d ) 24 , e ) 27 | e | divide(divide(12, subtract(const_1, divide(const_1, 3))), subtract(const_1, divide(const_1, 3))) | divide(const_1,n0)|subtract(const_1,#0)|divide(n1,#1)|divide(#2,#1) | general |
the radius of a wheel is 22.4 cm . what is the distance covered by the wheel in making 500 resolutions . | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 500 resolutions . = 500 * 2 * 22 / 7 * 22.4 = 70400 cm = 704 m answer : b" | a ) 287 m , b ) 704 m , c ) 168 m , d ) 278 m , e ) 107 m | b | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 22.4), const_2), 500), const_100) | add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)| | physics |
20 percent of the women in a college class are science majors , and the non - science majors make up 60 % of the class . what percentage of the men are science majors if 40 % of the class are men ? | science majors make up 0.4 of the class . 60 % of the class are women and 0.2 * 0.6 = 0.12 of the class are female science majors . then 0.28 of the class are male science majors . 0.4 x = 0.28 x = 0.7 = 70 % the answer is d . | a ) 40 % , b ) 50 % , c ) 60 % , d ) 70 % , e ) 80 % | d | multiply(divide(subtract(40, subtract(60, multiply(divide(subtract(const_100, 20), const_100), subtract(const_100, 40)))), 40), const_100) | subtract(const_100,n0)|subtract(const_100,n2)|divide(#0,const_100)|multiply(#2,#1)|subtract(n1,#3)|subtract(n2,#4)|divide(#5,n2)|multiply(#6,const_100) | gain |
a grocer has a sale of rs . 8435 , rs . 8927 , rs . 8855 , rs . 9230 and rs . 8562 for 5 consecutive months . how much sale must he have in the sixth month so that he gets an average sale of rs . 8500 ? | "explanation : total sale for 5 months = rs . ( 8435 + 8927 + 8855 + 9230 + 8562 ) = rs . 44009 . required sale = rs . [ ( 8500 x 6 ) Γ’ β¬ β 44009 ] = rs . ( 51000 Γ’ β¬ β 44009 ) = rs . 6991 . answer e" | a ) s . 1991 , b ) s . 2991 , c ) s . 3991 , d ) s . 4991 , e ) s . 6991 | e | subtract(multiply(add(5, const_1), 8500), add(add(add(add(8435, 8927), 8855), 9230), 8562)) | add(n5,const_1)|add(n0,n1)|add(n2,#1)|multiply(n6,#0)|add(n3,#2)|add(n4,#4)|subtract(#3,#5)| | general |
two stations a and b are 110 km apart on a straight line . one train starts from a at 5 a . m . and travels towards b at 20 kmph . another train starts from b at 8 a . m . and travels towards a at a speed of 25 kmph . at what time will they meet ? | "suppose they meet x hours after 5 a . m . distance covered by a in x hours = 20 x km . distance covered by b in ( x - 1 ) hours = 25 ( x - 1 ) km . therefore 20 x + 25 ( x - 1 ) = 110 45 x = 135 x = 3 . so , they meet at 8 a . m . answer : c" | a ) 11 , b ) 10 , c ) 8 , d ) 12 , e ) 15 | c | add(8, divide(subtract(110, 20), add(25, 20))) | add(n2,n4)|subtract(n0,n2)|divide(#1,#0)|add(n3,#2)| | physics |
a circular ground whose diameter is 40 metres , has a garden of area 1100 m ^ 2 around it . what is the wide of the path of the garden ? | "req . area = Γ― β¬ [ ( 20 ) 2 Γ’ β¬ β ( r ) 2 ] = 22 Γ’ Β β 7 Γ£ β ( 400 - r ^ 2 ) [ since a 2 - b 2 = ( a + b ) ( a - b ) ] ie ) 22 / 7 ( 400 - r ^ 2 ) = 1100 , ie ) r ^ 2 = 50 , r = 7.07 m answer b" | a ) 8.07 , b ) 7.07 , c ) 6.07 , d ) 7.0 , e ) 8.5 | b | subtract(circle_area(add(divide(40, 1100), 1100)), circle_area(divide(40, 1100))) | divide(n0,n1)|add(n1,#0)|circle_area(#0)|circle_area(#1)|subtract(#3,#2)| | physics |
in a manufacturing plant , it takes 36 machines 4 hours of continuous work to fill 4 standard orders . at this rate , how many hours of continuous work by 72 machines are required to fill 12 standard orders ? | "the choices give away the answer . . 36 machines take 4 hours to fill 4 standard orders . . in next eq we aredoubling the machines from 36 to 72 , but thework is not doubling ( only 1 1 / 2 times ) , = 4 * 48 / 72 * 12 / 4 = 6 ans b" | a ) 3 , b ) 6 , c ) 8 , d ) 9 , e ) 12 | b | divide(divide(multiply(multiply(36, 12), 4), 72), 4) | multiply(n0,n4)|multiply(#0,n1)|divide(#1,n3)|divide(#2,n2)| | physics |
a man gains 20 % by selling an article for a certain price . if he sells it at double the price , the percentage of profit will be . | "explanation : let the c . p . = x , then s . p . = ( 120 / 100 ) x = 6 x / 5 new s . p . = 2 ( 6 x / 5 ) = 12 x / 5 profit = 12 x / 5 - x = 7 x / 5 profit % = ( profit / c . p . ) * 100 = > ( 7 x / 5 ) * ( 1 / x ) * 100 = 140 % option b" | a ) 130 % , b ) 140 % , c ) 150 % , d ) 160 % , e ) 170 % | b | add(multiply(subtract(multiply(add(const_1, divide(20, const_100)), const_2), const_1), const_100), const_100) | divide(n0,const_100)|add(#0,const_1)|multiply(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)|add(#4,const_100)| | gain |
if the cost price is 90 % of the selling price , then what is the profit percent ? | "let s . p . = $ 100 c . p . = $ 90 profit = $ 10 profit % = 10 / 90 * 100 = 25 / 6 = 11 % approximately answer is b" | a ) 5 % , b ) 11 % , c ) 13 % , d ) 21 % , e ) 19 % | b | multiply(divide(subtract(const_100, 90), 90), const_100) | subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
marginal cost is the cost of increasing the quantity produced ( or purchased ) by one unit . if the fixed cost for n products is $ 10,000 and the marginal cost is $ 200 , and the total cost is $ 40,000 , what is the value of n ? | "total cost for n products = fixed cost for n products + n * marginal cost - - > $ 40,000 = $ 10,000 + n * $ 200 - - > n = 150 . answer : e ." | a ) 30 , b ) 50 , c ) 60 , d ) 80 , e ) 150 | e | divide(200, const_10) | divide(n1,const_10)| | general |
7 does not occur in 1000 . so we have to count the number of times it appears between 1 and 999 . any number between 1 and 999 can be expressed in the form of xyz where 0 < x , y , z < 9 . | 1 . the numbers in which 7 occurs only once . e . g 7 , 17 , 78 , 217 , 743 etc this means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits ( i . e 0 to 9 with the exception of 7 ) you have 1 * 9 * 9 = 81 such numbers . however , 7 could appear as the first or the second or the... | a ) 200 , b ) 300 , c ) 400 , d ) 500 , e ) 470 | b | add(add(multiply(const_3, multiply(9, 9)), multiply(multiply(const_3, 9), const_2)), const_3) | multiply(n7,n7)|multiply(n7,const_3)|multiply(#0,const_3)|multiply(#1,const_2)|add(#2,#3)|add(#4,const_3) | general |
a person can swim in still water at 4 km / h . if the speed of water 2 km / h , how many hours will the man take to swim back against the current for 16 km ? | "m = 4 s = 2 us = 4 - 2 = 2 d = 16 t = 16 / 2 = 8 answer : c" | a ) 3 , b ) 6 , c ) 8 , d ) 9 , e ) 7 | c | divide(16, subtract(4, 2)) | subtract(n0,n1)|divide(n2,#0)| | physics |
in 1982 and 1983 , company b β s operating expenses were $ 12.0 million and $ 15.0 million , respectively , and its revenues were $ 15.6 million and $ 18.8 million , respectively . what was the percent increase in company b β s profit ( revenues minus operating expenses ) from 1982 to 1983 ? | profit in 1982 = 15.6 - 12 = 3.6 million $ profit in 1983 = 18.8 - 15 = 3.8 million $ percentage increase in profit = ( 3.8 - 3.6 ) / 3.6 * 100 % = 5 5 / 9 % answer b | a ) 3 % , b ) 5 5 / 9 % , c ) 25 % , d ) 33 1 / 3 % , e ) 60 % | b | subtract(const_100, divide(multiply(subtract(15, 12), const_100), subtract(18.8, 15.6))) | subtract(n3,n2)|subtract(n5,n4)|multiply(#0,const_100)|divide(#2,#1)|subtract(const_100,#3) | general |
peter can cover a certain distance in 1 hr . 24 min . by covering two - third of the distance at 4 kmph and the rest at 5 kmph . find the total distance . | time = distance / speed let total distance travelled be x in 84 / 60 hrs 2 / 3 rd of x travelled in 4 km / hr 1 / 3 rd of distance travelled in 5 km / hr 2 x / ( 3 * 4 ) + x / ( 3 * 5 ) = 84 / 60 x = 6 km answer : c | a ) 4 km , b ) 5 km , c ) 6 km , d ) 7 km , e ) 8 km | c | divide(add(1, divide(24, subtract(const_100, multiply(const_4, const_10)))), add(divide(divide(1, const_3), 5), divide(divide(const_2, const_3), const_4))) | divide(n0,const_3)|divide(const_2,const_3)|multiply(const_10,const_4)|divide(#0,n3)|divide(#1,const_4)|subtract(const_100,#2)|add(#3,#4)|divide(n1,#5)|add(n0,#7)|divide(#8,#6) | physics |
if the area of a circle decreases by 20 % , 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 β 10 β 100 β x ) % = ( 100 β 10 β 100 β 20 ) % = ( 100 β 10 β 80 ) % = 100 - 89 = 11 % answer d | ['a ) 20 %', 'b ) 18 %', 'c ) 36 %', 'd ) 11 %', 'e ) none of these'] | d | multiply(subtract(const_1, sqrt(divide(subtract(const_100, 20), const_100))), const_100) | subtract(const_100,n0)|divide(#0,const_100)|sqrt(#1)|subtract(const_1,#2)|multiply(#3,const_100) | geometry |
if n is a natural number , then ( 6 n ^ 2 + 6 n ) is always divisible by ? | "( 6 n ^ 2 + 6 n ) = 6 n ( n + 1 ) , which is always divisible by 6 and 12 both , since n ( n + 1 ) is always even . correct option : b" | a ) 6 only , b ) 6 and 12 , c ) 12 only , d ) by 18 only , e ) none of these | b | add(multiply(6, const_100), multiply(2, 6)) | multiply(n0,const_100)|multiply(n0,n1)|add(#0,#1)| | general |
two trains 140 m and 180 m long run at the speed of 60 km / hr and 40 km / hr respectively in opposite directions on parallel tracks . the time which they take to cross each other is ? | "relative speed = 60 + 40 = 100 km / hr . = 100 * 5 / 18 = 250 / 9 m / sec . distance covered in crossing each other = 140 + 180 = 320 m . required time = 320 * 9 / 250 = 11.52 sec . answer : a" | a ) 11.52 sec , b ) 10.1 sec , c ) 10.6 sec , d ) 10.8 sec , e ) 10.2 sec | a | divide(add(140, 180), multiply(add(60, 40), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
the average of first seven multiples of 8 is : | "explanation : ( 8 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) / 7 = 8 x 28 / 7 = 32 answer : d" | a ) 9 , b ) 16 , c ) 15 , d ) 32 , e ) 10 | d | add(8, const_1) | add(n0,const_1)| | general |
a box contains 100 balls , numbered from 1 to 100 . if 3 balls are selected at random and with replacement from the box . if the 3 numbers on the balls selected contain two odd and one even . what is the probability j that the first ball picked up is odd numbered ? | "answer - d selecting the balls either even or odd is having probability 50 / 100 = 1 / 2 we have already selected 3 balls with 2 odd numbers and 1 even number . so we have 3 combinations ooe , oeo , eoo . we have 3 outcomes and 2 are favourable as in 2 cases 1 st number is odd . so probability j is 2 / 3 . d" | a ) 0 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 1 | d | divide(const_2, 3) | divide(const_2,n3)| | probability |
the ratio between the perimeter and the width of a rectangle is 5 : 1 . if the area of the rectangle is 54 square centimeters , what is the length of the rectangle in centimeters ? | "perimeter = 2 ( w + l ) = 5 w 3 w = 2 l w = 2 l / 3 wl = 54 2 l ^ 2 / 3 = 54 l ^ 2 = 81 l = 9 cm the answer is d ." | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | d | divide(54, const_10) | divide(n2,const_10)| | geometry |
carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 5 inches by 24 inches and jordan ' s rectangle is 3 inches long , how wide is jordan ' s rectangle , in inches ? | "area of carol ' s rectangle = 24 * 5 = 120 let width of jordan ' s rectangle = w since , the areas are equal 3 w = 120 = > w = 40 answer d" | a ) 25 , b ) 43 , c ) 42 , d ) 40 , e ) 18 | d | divide(rectangle_area(5, 24), 3) | rectangle_area(n0,n1)|divide(#0,n2)| | geometry |
dacid obtained 76 , 65 , 82 , 67 and 85 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "average = ( 76 + 65 + 82 + 67 + 85 ) / 5 = 375 / 5 = 75 . answer : e" | a ) 98 , b ) 78 , c ) 76 , d ) 87 , e ) 75 | e | divide(add(add(add(add(76, 65), 82), 67), 85), divide(const_10, const_2)) | add(n0,n1)|divide(const_10,const_2)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general |
16 ltr of water is added with 24 ltr of a solution containing 90 % of alcohol in the water . the % of alcohol in the new mixture is ? | "we have a 24 litre solution containing 90 % of alcohol in the water . = > quantity of alcohol in the solution = 24 Γ£ β 90 / 100 now 16 litre of water is added to the solution . = > total quantity of the new solution = 24 + 16 = 40 percentage of alcohol in the new solution = 24 Γ£ β 90 / 100 40 Γ£ β 100 = 24 Γ£ β 9010040 ... | a ) 48 % , b ) 52 % , c ) 54 % , d ) 60 % , e ) 70 % | c | multiply(const_100, divide(multiply(divide(90, const_100), 24), add(16, 24))) | add(n0,n1)|divide(n2,const_100)|multiply(n1,#1)|divide(#2,#0)|multiply(#3,const_100)| | general |
a bag contains 3 white marbles and 3 black marbles . if each of 3 girls and 3 boys randomly selects and keeps a marble , what is the probability that all of the girls select the same colored marble ? | first , total ways to select for all boys and girls , i . e 6 ! / ( 3 ! * 3 ! ) = 6 * 5 * 4 * 3 * 2 * 1 / 3 * 2 * 1 * 3 * 2 * 1 = 20 then there are one two way girls can have all same colors , either white or black . the number of ways in which 3 girls can select 3 white balls = 3 c 3 = 1 the number of ways in which 3 ... | a ) 1 / 35 , b ) 1 / 10 , c ) 1 / 15 , d ) 1 / 20 , e ) 1 / 25 | b | divide(const_2, choose(add(3, 3), 3)) | add(n0,n0)|choose(#0,n0)|divide(const_2,#1) | probability |
a cube has four of its faces painted half red and half white . the other faces are completely painted white . what is the ratio between the red painted areas and the white painted areas of the cube ? | "let x be the area of each face of the cube . the area painted red is 4 ( x / 2 ) = 2 x the area painted white is 4 ( x / 2 ) + 2 x = 4 x the ratio of red to white is 2 x : 4 x which is 1 : 2 . the answer is c ." | a ) 1 : 4 , b ) 1 : 3 , c ) 1 : 2 , d ) 2 : 5 , e ) 2 : 11 | c | divide(multiply(multiply(add(const_1, const_4), divide(const_1, const_2)), const_2), multiply(add(multiply(add(const_1, const_4), divide(const_1, const_2)), const_1), const_2)) | add(const_1,const_4)|divide(const_1,const_2)|multiply(#0,#1)|add(#2,const_1)|multiply(#2,const_2)|multiply(#3,const_2)|divide(#4,#5)| | geometry |
if a book is sold at 5 % profit instead of 5 % loss , it would have brought rs 13 more . find out the cost price of the book | "let c . p . of the book be rs . β x β given , 1.05 x - 0.95 x = 13 = > 0.1 x = 13 = 13 / 0.1 = rs 130 answer : c" | a ) 75 , b ) 72 , c ) 130 , d ) 70 , e ) 80 | c | divide(multiply(const_100, divide(5, const_2)), 5) | divide(n0,const_2)|multiply(#0,const_100)|divide(#1,n0)| | gain |
if 4 and 8 are factors of 60 n , what is the minimum value of n ? | 60 n / 4 * 8 should be integer = > 2 * 2 * 3 * 5 * n / 2 * 2 * 2 * 2 * 2 = 3 * 5 * n / 8 must be an integer for this to be true n must multiple of 8 , thus min of n = 8 hence c | a ) 2 , b ) 3 , c ) 8 , d ) 14 , e ) 56 | c | lcm(4, 8) | lcm(n0,n1) | other |
if 3 ^ 8 x 3 ^ 7 = 3 ^ n what is the value of n ? | "3 ^ 8 * 3 ^ 7 = 3 ^ n or 3 ^ 8 + 7 = 3 ^ n n = 15 e" | a ) 20 , b ) 18 , c ) 17 , d ) 16 , e ) 15 | e | divide(log(multiply(power(3, 7), power(3, 8))), log(3)) | log(n4)|power(n2,n3)|power(n0,n1)|multiply(#1,#2)|log(#3)|divide(#4,#0)| | general |
in karthik ' s opinion , his weight is greater than 55 kg but less than 62 kg . his brother does not agree with karthik and he thinks that karthik ' s weight is greater than 50 kg but less than 60 kg . his father ' s view is that his weight can not be greater than 58 kg . if all of them are correct in their estimation ... | explanation : solution : assume karthik ' s weight be x kg . according to karthik , 55 < x < 62 according to karthik ; s brother , 50 < x < 60 . according to karthik ' s mother , x < 58 . the values satisfying all the above conditions are 56 and 57 . . ' . required average = ( 56 + 57 ) / 2 = 56.5 . answer : b | a ) 54.5 , b ) 56.5 , c ) 59.2 , d ) 61 , e ) 62 | b | divide(add(55, 58), const_2) | add(n0,n4)|divide(#0,const_2) | general |
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