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 |
|---|---|---|---|---|---|---|
a number when divided by 899 gives a remainder 63 . if the same number is divided by 29 , the remainder will be | "sol . number = ( 31 x q ) + 29 . given data is inadequate . answer d" | a ) 11 , b ) 13 , c ) 15 , d ) data inadequate , e ) none | d | subtract(63, multiply(29, const_2)) | multiply(n2,const_2)|subtract(n1,#0)| | general |
the length of a rectangular plot is thrice its breadth . if the area of the rectangular plot is 2700 sq m , then what is the breadth of the rectangular plot ? | "let the breadth of the plot be b m . length of the plot = 3 b m ( 3 b ) ( b ) = 2700 3 b 2 = 2700 b 2 = 900 = 30 ( b > 0 ) b = 30 m . answer : e" | a ) 11 , b ) 17 , c ) 18 , d ) 101 , e ) 30 | e | sqrt(divide(2700, const_3)) | divide(n0,const_3)|sqrt(#0)| | geometry |
two trains a and b start simultaneously in the opposite direction from two points p and q and arrive at their destinations 16 and 9 hours respectively after their meeting each other . at what speed does the second train b travel if the first train travels at 120 km / h | answer : b ) 160 km / h | a ) 334 , b ) 160 , c ) 387 , d ) 278 , e ) 112 | b | divide(multiply(sqrt(16), 120), sqrt(9)) | sqrt(n0)|sqrt(n1)|multiply(n2,#0)|divide(#2,#1) | physics |
a man , a woman and a boy can together complete a piece of work in 3 days . if a man alone can do it in 6 days and a boy alone in 12 days , how long will a woman take to complete the work ? | "explanation : ( 1 man + 1 woman + 1 boy ) β s 1 day β s work = 1 / 3 1 man β s 1 day work = 1 / 6 1 boy β s 1 day β s work = 1 / 1 ( 1 man + 1 boy ) β s 1 day β s work = 1 / 6 + 1 / 12 = 1 / 4 therefore , 1 woman β s 1 day β s work = 1 / 3 β 1 / 4 = 1 / 12 therefore , the woman alone can finish the work in 12 days . a... | a ) 12 days , b ) 10 days , c ) 9 days , d ) 8 days , e ) 11 days | a | inverse(subtract(inverse(3), add(inverse(6), inverse(12)))) | inverse(n0)|inverse(n1)|inverse(n2)|add(#1,#2)|subtract(#0,#3)|inverse(#4)| | physics |
carmen made a sculpture from small pieces of wood . the sculpture is 2 feet 6 inches tall . carmen places her sculpture on a base that is 12 inches tall . how tall are the sculpture andbase together ? | "we know 1 feet = 12 inch then 2 feet = 24 inch 24 + 10 = 34 then 34 + 12 = 44 46 / 12 = 3.83 feet answer : e" | a ) 3.1 feet , b ) 3.2 feet , c ) 3.3 feet , d ) 3.4 feet , e ) 3.83 feet | e | divide(add(add(multiply(add(2, 6), 2), 6), 12), add(2, 6)) | add(n0,n1)|multiply(n0,#0)|add(#1,n1)|add(n2,#2)|divide(#3,#0)| | geometry |
the cost of 3 pens and 5 pencils is rs . 340 . also the cost of one pen and one pencil is in the ratio of 4 : 1 respectively . what is the cost of one dozen pens ? | "explanation : let the cost of one pen is β 4 x β and pencil is β x β 3 x 4 x + 5 x = rs . 340 12 x + 5 x = rs . 340 x = 340 / 17 = 20 : . cost of 1 pen = 4 x = 4 x 20 = 80 : . cost of 12 pens , i . e . ( one dozen ) = 80 x 12 = rs . 960 answer : option a" | a ) rs . 960 , b ) rs . 250 , c ) rs . 300 , d ) rs . 150 , e ) none of these | a | multiply(multiply(3, const_4), divide(340, add(3, 1))) | add(n0,n4)|multiply(n0,const_4)|divide(n2,#0)|multiply(#2,#1)| | other |
the length of a room is 5.5 m and width is 3.75 m . what is the cost of paying the floor by slabs at the rate of $ 400 per sq . metre . | "area = 5.5 Γ 3.75 sq . metre . cost for 1 sq . metre . = $ 400 hence , total cost = 5.5 Γ 3.75 Γ 400 = $ 8250 a" | a ) $ 8250 , b ) $ 8350 , c ) $ 8650 , d ) $ 8450 , e ) $ 8500 | a | multiply(400, multiply(5.5, 3.75)) | multiply(n0,n1)|multiply(n2,#0)| | physics |
the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit . if the temperature f of the coffee t minutes after it was poured can be determined by the formula f = 120 ( 2 ^ - at ) + 60 , where f is in degrees fahrenheit and a is a constant . then the temperature of the coffee 8... | "answer : b the temperature of coffee 10 minutes after it was poured ( 120 f ) will help in solving the constant β a β . 120 = 120 ( 2 ^ 10 a ) + 60 2 ^ - 1 = 2 ^ 10 a a = - 1 / 10 the temperature of coffee 80 minutes after it was poured is : f = 120 ( 2 ^ - 80 / 10 ) + 60 f = 120 * 1 / 256 + 60 f = 15 / 32 + 60 f = 19... | a ) 45 , b ) 56 , c ) 60.4 , d ) 85 , e ) 90 | c | add(multiply(power(2, multiply(divide(60, 10), subtract(const_1, 2))), 120), 60) | divide(n4,n0)|subtract(const_1,n3)|multiply(#0,#1)|power(n3,#2)|multiply(n1,#3)|add(n4,#4)| | general |
the average of 11 results is 20 . the average of first 5 of them is 15 and that of last 5 is 22 . find the 6 th result ? | "6 th result = sum of 11 results - sum of 10 results = 11 * 20 - 5 * 15 - 5 * 22 = 220 - 75 - 110 = 35 answer is a" | a ) 35 , b ) 50 , c ) 100 , d ) 120 , e ) 150 | a | subtract(subtract(multiply(11, 20), multiply(5, 22)), multiply(5, 15)) | multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2)| | general |
1 / 3 + 1 / 2 - 5 / 6 + 1 / 5 + 1 / 4 - 9 / 20 - 5 / 6 = | we need to determine the result of 1 / 3 + 1 / 2 - 5 / 6 + 1 / 5 + 1 / 4 - 9 / 20 let β s add the given fractions in two groups . in the group of the first three fractions , notice that 1 / 3 and 1 / 2 share a common denominator of 6 with 5 / 6 . 1 / 2 + 1 / 3 = 3 / 6 + 2 / 6 = 5 / 6 thus , 5 / 6 β 5 / 6 = 0 looking at... | a ) 0 , b ) 2 / 15 , c ) 2 / 5 , d ) 9 / 20 , e ) 5 / 6 | e | divide(5, 6) | divide(n4,n5) | general |
the product of three consecutive numbers is 210 . then the sum of the smallest two numbers is ? | "product of three numbers = 210 210 = 2 * 3 * 5 * 7 = 5 * 6 * 7 . so , the three numbers are 5 , 6 and 7 . and sum of smallest of these two = 5 + 6 = 11 . answer : option a" | a ) 11 , b ) 15 , c ) 20 , d ) 38 , e ) 56 | a | multiply(power(const_2, 210), factorial(210)) | factorial(n0)|power(const_2,n0)|multiply(#0,#1)| | general |
frank the fencemaker needs to fence in a rectangular yard . he fences in the entire yard , except for one full side of the yard , which equals 40 feet . the yard has an area of 240 square feet . how many feet offence does frank use ? | "area = length x breadth 240 = 40 x breadth so , breadth = 6 units fencing required is - breadth + breadth + length 6 + 6 + 40 = > 52 feet answer must be ( b ) 52" | a ) 14 , b ) 52 , c ) 54 , d ) 180 , e ) 240 | b | add(add(divide(240, 40), divide(240, 40)), 40) | divide(n1,n0)|add(#0,#0)|add(n0,#1)| | geometry |
the total of 334 of 20 paise and 25 paise make a sum of rs . 71 . the no of 20 paise coins is | "explanation : let the number of 20 paise coins be x . then the no of 25 paise coins = ( 334 - x ) . 0.20 * ( x ) + 0.25 ( 334 - x ) = 71 = > x = 250 . . answer : a ) 250" | a ) 250 , b ) 277 , c ) 278 , d ) 200 , e ) 288 | a | divide(subtract(multiply(334, 25), multiply(71, const_100)), subtract(25, 20)) | multiply(n0,n2)|multiply(n3,const_100)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)| | general |
the function f ( n ) is defined as the product of all the consecutive positive integers between 1 and n ^ 2 , inclusive , whereas the function g ( n ) is defined as the product of the squares of all the consecutive positive integers between 1 and n , inclusive . the exponent on 3 in the prime factorization of f ( 3 ) /... | "f ( 3 ) / g ( 3 ) = product ( 1 to 3 ^ 2 ) / 1.2 ^ 2.3 ^ 2 = 1 . 2.3 . 4.5 . 6.7 . 8.9 / 1 . 4.9 = 1 . 2.3 . ( 2 ^ 2 ) . 5 . ( 2.3 ) . 7 . ( 2 ^ 3 ) . 9 / 1 . ( 2 ^ 2 ) . 9 = 1 . ( 2 ^ 7 ) . 3.5 . 7.9 / 1 . ( 2 ^ 2 ) . 9 loof for 2 ^ 7 / 2 ^ 2 = 2 ^ 5 - - - - exponent 1 answer : a" | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | power(2, 2) | power(n1,n1)| | general |
what will be the compound interest on rs . 45000 after 3 years at the rate of 12 % per annum | "explanation : ( 45000 Γ ( 1 + 12 / 100 ) 3 ) = > 45000 Γ 28 / 25 Γ 28 / 25 Γ 28 / 25 = > 63221.76 so compound interest will be 63221.76 - 45000 = rs 18221.76 option a" | a ) rs 18221.76 , b ) rs 18123.30 , c ) rs 18123.40 , d ) rs 18123.50 , e ) none of these | a | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
the guests at a football banquet consumed a total of 337 pounds of food . if no individual guest consumed more than 2 pounds of food , what is the minimum number of guests that could have attended the banquet ? | "to minimize one quantity maximize other . 168 * 2 ( max possible amount of food a guest could consume ) = 336 pounds , so there must be more than 168 guests , next integer is 169 . answer : e ." | a ) 160 , b ) 161 , c ) 162 , d ) 163 , e ) 169 | e | add(floor(divide(337, 2)), const_1) | divide(n0,n1)|floor(#0)|add(#1,const_1)| | general |
a plant manager must assign 10 new workers to one of five shifts . she needs a first , second , and third shift , and two alternate shifts . each of the shifts will receive 2 new workers . how many different ways can she assign the new workers ? | "my take selecting team of 2 out of 10 to assign to the shifts = 10 c 2 = 45 ways . now 2 out of 10 means total of 5 group possible . so putting them in shifts = counting methode : first , second , third , alt , alt = 5 * 4 * 3 * 2 * 1 = 120 here alt and alt are the same : so 120 / 2 = 60 ways . total ways of selecting... | a ) 2430 , b ) 2700 , c ) 3300 , d ) 4860 , e ) 5400 | b | multiply(divide(factorial(divide(10, 2)), const_2), divide(factorial(10), multiply(factorial(subtract(10, 2)), factorial(2)))) | divide(n0,n1)|factorial(n0)|factorial(n1)|subtract(n0,n1)|factorial(#0)|factorial(#3)|divide(#4,const_2)|multiply(#5,#2)|divide(#1,#7)|multiply(#6,#8)| | physics |
there are 5 red shoes & 4 green shoes . if two of red shoes are drawn what is the probability of getting red shoes | taking 2 red shoe the probability is 5 c 2 from 9 shoes probability of taking 2 red shoes is 5 c 2 / 9 c 2 = 5 / 18 answer : c | a ) 1 / 18 , b ) 1 / 14 , c ) 5 / 18 , d ) 1 / 15 , e ) 1 / 16 | c | divide(choose(5, const_2), choose(add(5, 4), const_2)) | add(n0,n1)|choose(n0,const_2)|choose(#0,const_2)|divide(#1,#2) | probability |
to fill a tank , 200 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to 4 - fifths of its present ? | let the capacity of 1 bucket = x . then , the capacity of tank = 200 x . new capacity of bucket = 4 / 5 x therefore , required number of buckets = ( 200 x ) / ( 4 x / 5 ) = ( 200 x ) x 5 / 4 x = 1000 / 4 = 250 answer is e . | a ) 50 , b ) 100 , c ) 150 , d ) 200 , e ) 250 | e | divide(200, divide(4, divide(const_10, const_2))) | divide(const_10,const_2)|divide(n1,#0)|divide(n0,#1) | physics |
what is the probability of getting a sum 9 from two throw of a dice ? | "in two throws of die , n ( s ) = 36 let e = event of getting a sum 9 = { ( 3,6 ) , ( 4,5 ) , ( 5,4 ) , ( 6,3 ) } p ( e ) = 4 / 36 = 1 / 9 answer c 1 / 9" | a ) 1 / 6 , b ) 1 / 8 , c ) 1 / 9 , d ) 1 / 12 , e ) 1 / 13 | c | divide(const_2, choose(add(const_3, const_3), const_3)) | add(const_3,const_3)|choose(#0,const_3)|divide(const_2,#1)| | probability |
john was 31 years old when he married betty . they just celebrated their fifth wedding anniversary , and betty ' s age is now 7 / 9 of john ' s . how old is betty ? | "assume betty ' s age on marriage = x years . john ' s age on marriage = 31 john ' s age after 5 years = 36 years . betty ' s age after 5 years = x + 5 given : x + 5 = 7 / 9 ( 36 ) = 28 therefore betty ' s current age = 28 option c" | a ) 24 , b ) 26 , c ) 28 , d ) 30 , e ) 32 | c | multiply(divide(7, 9), add(31, const_4)) | add(n0,const_4)|divide(n1,n2)|multiply(#0,#1)| | general |
if the volume of two cubes are in the ratio 27 : 1 , the ratio of their edges is : | "explanation : let the edges be a and b of two cubes , then a 3 / b 3 = 27 / 1 = > ( a / b ) 3 = ( 3 / 1 ) 3 a / b = 3 / 1 = > a : b = 3 : 1 option a" | a ) 3 : 1 , b ) 3 : 2 , c ) 3 : 5 , d ) 3 : 7 , e ) none of these | a | cube_edge_by_volume(27) | cube_edge_by_volume(n0)| | geometry |
tom purchased 8 kg of apples at the rate of 70 per kg and 9 kg of mangoes at the rate of 65 per kg . how much amount did he pay to the shopkeeper ? | "cost of 8 kg apples = 70 Γ 8 = 560 . cost of 9 kg of mangoes = 65 Γ 9 = 585 . total cost he has to pay = 560 + 585 = 1145 . b )" | a ) a ) 1040 , b ) b ) 1145 , c ) c ) 1055 , d ) d ) 1060 , e ) e ) 1075 | b | add(multiply(8, 70), multiply(9, 65)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | gain |
a man can row 4 kmph is still water . if the river is running at 2 kmph it takes 90 min to row to a place and back . how far is the place | explanation : speed in still water = 4 kmph speed of the stream = 2 kmph speed upstream = ( 4 - 2 ) = 2 kmph speed downstream = ( 4 + 2 ) = 6 kmph total time = 90 minutes = 90 β 60 hour = 3 β 2 hour let l be the distance . then ( l / 6 ) + ( l / 2 ) = 32 = > l + 3 l = 9 = > 4 l = 9 = > l = 9 β 4 = 2.25 km . answer : op... | a ) 2 km , b ) 4 km , c ) 5 km , d ) 2.25 km , e ) none of these | d | divide(divide(90, const_60), add(inverse(add(4, 2)), inverse(subtract(4, 2)))) | add(n0,n1)|divide(n2,const_60)|subtract(n0,n1)|inverse(#0)|inverse(#2)|add(#3,#4)|divide(#1,#5) | physics |
what is the least number should be added to 929 , so the sum of the number is completely divisible by 30 ? | "( 929 / 30 ) gives remainder 29 29 + 1 = 30 , so we need to add 1 answer : a" | a ) 1 , b ) 2 , c ) 5 , d ) 6 , e ) 8 | a | subtract(multiply(add(floor(divide(929, 30)), const_1), 30), 929) | divide(n0,n1)|floor(#0)|add(#1,const_1)|multiply(n1,#2)|subtract(#3,n0)| | general |
the sale price sarees listed for rs . 495 after successive discount is 15 % and 10 % is ? | "495 * ( 85 / 100 ) * ( 90 / 100 ) = 378 answer : b" | a ) 288 , b ) 378 , c ) 342 , d ) 662 , e ) 262 | b | subtract(subtract(495, divide(multiply(495, 15), const_100)), divide(multiply(subtract(495, divide(multiply(495, 15), const_100)), 10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
how many numbers from 29 to 119 are exactly divisible by 11 ? | "29 / 11 = 2 and 119 / 11 = 10 = = > 10 - 2 = 8 numbers answer : d" | a ) 5 , b ) 7 , c ) 9 , d ) 8 , e ) 12 | d | add(divide(subtract(multiply(floor(divide(119, 11)), 11), multiply(add(floor(divide(29, 11)), const_1), 11)), 11), const_1) | divide(n1,n2)|divide(n0,n2)|floor(#0)|floor(#1)|add(#3,const_1)|multiply(n2,#2)|multiply(n2,#4)|subtract(#5,#6)|divide(#7,n2)|add(#8,const_1)| | general |
the points a ( 0 , 0 ) , b ( 0 , 4 a - 2 ) and c ( 2 a + 1 , 2 a + 6 ) form a triangle . if angle abc = 90 , what is the area of triangle abc ? | 1 / 2 bh = 1 / 2 ( 2 a + 1 ) ( 2 a + 6 ) now 4 a - 2 = 2 a + 6 2 a = 8 . a = 4 therefore , a ( 0,0 ) ; b ( 0,14 ) ; c ( 9,14 ) 1 / 2 * 9 * 14 = 63 answer : c | a ) 58 , b ) 70 , c ) 63 , d ) 65 , e ) 72 | c | divide(multiply(subtract(multiply(divide(add(6, 2), 2), 4), 2), add(multiply(divide(add(6, 2), 2), subtract(4, 2)), const_1)), const_2) | add(n4,n8)|subtract(n3,n4)|divide(#0,n4)|multiply(#2,#1)|multiply(n3,#2)|add(#3,const_1)|subtract(#4,n4)|multiply(#5,#6)|divide(#7,const_2) | geometry |
a fellow borrowed a certain sum of money at 8 % per annum at simple interest and in 8 years the interest amounted to rs . 900 less than the sum lent . what was the sum lent ? | "p - 900 = ( p * 8 * 8 ) / 100 p = 2500 answer : c" | a ) 1050 , b ) 1220 , c ) 2500 , d ) 1060 , e ) 1110 | c | divide(900, subtract(const_1, divide(multiply(8, 8), const_100))) | multiply(n0,n0)|divide(#0,const_100)|subtract(const_1,#1)|divide(n2,#2)| | gain |
aarti can do a piece of work in 3 days . in how many days will she complete 10 time of work of same type ? | "we have the important relation , more work , more time ( days ) a piece of work can be done in 3 days . 10 times of work of same type can be done in 3 x 10 = 30 days answer d" | a ) 6 days , b ) 18 days , c ) 21 days , d ) 30 days , e ) 13 days | d | multiply(const_3, 3) | multiply(n0,const_3)| | physics |
two passenger trains start at the same hour in the day from two different stations and move towards each other at the rate of 29 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 = 29 + 21 = 50 t = 12 d = 50 * 12 = 600 answer : d" | a ) 457 km , b ) 444 km , c ) 547 km , d ) 600 km , e ) 453 km | d | add(multiply(divide(60, subtract(21, 29)), 29), multiply(divide(60, subtract(21, 29)), 21)) | subtract(n1,n0)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | physics |
when sold at a 40 % discount , a sweater nets the merchant a 30 % profit on the wholesale cost at which he initially purchased the item . by what % is the sweater marked up from wholesale at its normal retail price ? | "we should be careful about what are we measuring % on / what is the base . . let the marked up price = 100 . . selling price = 100 - 40 % of 100 = 60 . . profit = 30 % . . therefore the wholesale purchase cost = x . . . . 1.3 x = 60 or x = 46.15 . . . marked price was 100 so . . . so answer is 53.85 % . . d" | a ) 20 % , b ) 40 % , c ) 50 % , d ) 53.85 % , e ) 100 % | d | subtract(const_100, divide(subtract(const_100, 40), add(const_1, divide(30, const_100)))) | divide(n1,const_100)|subtract(const_100,n0)|add(#0,const_1)|divide(#1,#2)|subtract(const_100,#3)| | gain |
what is the greatest positive integer x such that 3 ^ x is a factor of 9 ^ 10 ? | "9 ^ 10 = ( 3 ^ 2 ) ^ 10 = 3 ^ 20 the answer is d . 20" | a ) 5 , b ) 9 , c ) 10 , d ) 20 , e ) 30 | d | multiply(subtract(9, 10), 10) | subtract(n1,n2)|multiply(n2,#0)| | general |
the simple interest and the true discount on a certain sum for a given time and at a given rate are rs . 85 and rs . 78 respectively . the sum is : | sol . sum = s . i . * t . d . / ( s . i ) - ( t . d . ) = 85 * 78 / ( 85 - 78 ) = rs . 947 . answer d | a ) 1360 , b ) 1450 , c ) 1600 , d ) 947 , e ) none | d | divide(multiply(85, 78), subtract(85, 78)) | multiply(n0,n1)|subtract(n0,n1)|divide(#0,#1) | gain |
a cycle is bought for rs . 675 and sold for rs . 1080 , find the gain percent ? | "675 - - - - 180 100 - - - - ? = > 60 % answer : b" | a ) 22 , b ) 60 , c ) 99 , d ) 88 , e ) 11 | b | multiply(divide(subtract(1080, 675), 675), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain |
two workers a and b are engaged to do a work . a working alone takes 8 hours more to complete the job than if both worked together . if b worked alone , he would need 4 1 / 2 hours more to complete the job than they both working together . what time would they take to do the work together ? | let a and be together take x hours to complete the work . then , a alone takes ( x + 8 ) hrs and b alone takes ( x + 9 / 2 ) hrs to complete the work . then , 1 / ( x + 8 ) + 1 / ( x + 9 / 2 ) = 1 / x , 1 / ( x + 8 ) + 2 / ( 2 x + 9 ) = 1 / x x ( 4 x + 25 ) = ( x + 8 ) ( 2 x + 9 ) 2 x ^ 2 = 72 , x ^ 2 = 36 , x = 6 corr... | a ) 4 hours , b ) 5 hours , c ) 6 hours , d ) 7 hours , e ) none of these | c | sqrt(multiply(add(4, divide(1, 2)), 8)) | divide(n2,n3)|add(n1,#0)|multiply(n0,#1)|sqrt(#2) | general |
a rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered . if the area of the field is 80 sq . feet , how many feet of fencing will be required ? | "we have : l = 20 ft and lb = 80 sq . ft . so , b = 4 ft . length of fencing = ( l + 2 b ) = ( 20 + 8 ) ft = 28 ft . answer : a" | a ) 28 , b ) 40 , c ) 68 , d ) 88 , e ) 78 | a | add(multiply(divide(80, 20), const_2), 20) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
a can do a work in 6 days . b can do in 12 days . if both a & b are working together in how many days they can finish the work ? | 1 day work of a = 1 / 6 1 day work of b = 1 / 12 1 day work of a & b = 1 / 6 + 1 / 12 = 1 / 4 a & b finish the work in 4 days answer is a | a ) 4 , b ) 5 , c ) 6 , d ) 10 , e ) 15 | a | divide(const_1, add(divide(const_1, 6), divide(const_1, 12))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2) | physics |
if 2 and 3 are positive integers , then 2 * 3 + 2 is | answer : a | a ) 8 , b ) 10 , c ) 12 , d ) 12 , e ) 16 | a | lcm(2, 3) | lcm(n0,n1)| | general |
a 240 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds . what is the length of the other train ? | "relative speed = ( 120 + 80 ) km / hr = ( 200 x ( 5 / 18 ) ) m / sec = ( 500 / 9 ) m / sec . let the length of the other train be x metres . then , ( x + 240 ) / 9 = 500 / 9 x + 240 = 500 x = 260 . c" | a ) 230 m , b ) 240 m , c ) 260 m , d ) 320 m , e ) 330 m | c | subtract(multiply(multiply(add(120, 80), const_0_2778), 9), 240) | add(n1,n2)|multiply(#0,const_0_2778)|multiply(n3,#1)|subtract(#2,n0)| | physics |
kavi had a stock of 600 bags in his bookshop . he sold 25 on monday , 70 on tuesday , 100 on wednesday , 110 on thursday and 145 on friday . what percentage of the bags were not sold ? | let n be the total number of bags sold . hence n = 25 + 70 + 100 + 110 + 145 = 450 let m be the bags not sold m = 600 - n = 600 - 450 = 150 percentage bags not sold / total number of bags = 150 / 600 = 0.25 = 25 % correct answer b | a ) 10 % , b ) 25 % , c ) 64 % , d ) 42 % , e ) 17 % | b | multiply(divide(subtract(600, add(add(add(add(25, 70), 100), 110), 145)), 600), const_100) | add(n1,n2)|add(n3,#0)|add(n4,#1)|add(n5,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100) | gain |
a man rows his boat 85 km downstream and 45 km upstream , taking 2 1 / 2 hours each time . find the speed of the stream ? | "speed downstream = d / t = 85 / ( 2 1 / 2 ) = 34 kmph speed upstream = d / t = 45 / ( 2 1 / 2 ) = 18 kmph the speed of the stream = ( 34 - 18 ) / 2 = 8 kmph answer : d" | a ) 6 , b ) 7 , c ) 5 , d ) 8 , e ) 9 | d | divide(subtract(divide(85, 2), divide(45, 2)), const_2) | divide(n0,n2)|divide(n1,n2)|subtract(#0,#1)|divide(#2,const_2)| | physics |
if f ( x ) = 1 / x and x is a natural number , what can not be the answer for f ( f ( x ) ) ? | answer a is impossible because the invers of 1 / x is x , and the only way to have something other than a natural number is to input something other than a natural number . with the specification that only natural numbers may be used 1 / 7 is not a possibility for f ( f ( x ) ) | a ) 1 / 7 , b ) 3 , c ) 12 / 3 , d ) 6 / 6 , e ) 1 | a | divide(multiply(1, const_1), subtract(multiply(const_2, const_4), 1)) | multiply(n0,const_1)|multiply(const_2,const_4)|subtract(#1,n0)|divide(#0,#2) | general |
the product of the squares of two positive integers is 900 . how many pairs of positive integers satisfy this condition ? | "ans : e - 4 pairs ( x Λ 2 ) ( y Λ 2 ) = 900 [ square root both sides ] xy = 30 20 = 1 x 30 , 3 x 10 , 6 x 5 , 5 x 6 , 10 x 3 , 30 x 1 , 15 x 2 , 2 x 15 cancel the repeats this leaves us with exactly 4 options . hence , e" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | e | subtract(add(const_2, const_3), const_2) | add(const_2,const_3)|subtract(#0,const_2)| | geometry |
what is the least value of x . so that 23 x 57 is divisible by 3 . | "explanation : the sum of the digits of the number is divisible by 3 , then the number is divisible by 3 . 2 + 3 + x + 5 + 7 = 17 + x least value of x may be 1 therefore 17 + 1 = 18 is divisible by 3 . answer : option c" | a ) 2 , b ) 0 , c ) 1 , d ) 3 , e ) 4 | c | divide(divide(divide(lcm(23, 57), 57), const_4), const_4) | lcm(n0,n1)|divide(#0,n1)|divide(#1,const_4)|divide(#2,const_4)| | general |
six women can do a work in 10 days . ten men can complete the same work in 4 days . what is the ratio between the capacity of a man and a woman ? | "explanation : ( 6 Γ£ β 10 ) women can complete the work in 1 day . Γ’ Λ Β΄ 1 woman ' s 1 day ' s work = 1 / 60 ( 10 Γ£ β 4 ) men can complete the work in 1 day . Γ’ Λ Β΄ 1 man ' s 1 day ' s work = 1 / 40 so , required ratio = 1 / 60 : 1 / 40 = 3 : 2 answer : d" | a ) 1 : 2 , b ) 2 : 1 , c ) 2 : 3 , d ) 3 : 2 , e ) none of these | d | divide(divide(const_1, multiply(10, 4)), divide(const_1, multiply(4, const_10))) | multiply(n0,n1)|multiply(n1,const_10)|divide(const_1,#0)|divide(const_1,#1)|divide(#2,#3)| | physics |
if 28 less than 5 times a certain number is 232 . what is the number ? | 5 x β 28 subtraction is built backwards , multiply the unknown by 5 5 x β 28 = 232 is translates to equals + 28 + 28 add 28 to both sides 5 x = 260 the variable ismultiplied by 5 5 5 divide both sides by 5 x = 52 the number is 52 . correct answer c | a ) 32 , b ) 42 , c ) 52 , d ) 62 , e ) 72 | c | subtract(subtract(subtract(add(multiply(5, 28), 232), multiply(5, 28)), multiply(5, 28)), multiply(const_4, const_10)) | multiply(n0,n1)|multiply(const_10,const_4)|add(n2,#0)|subtract(#2,#0)|subtract(#3,#0)|subtract(#4,#1) | general |
a rope of which a calf is tied is increased from 10 m to 23 m , how much additional grassy ground shall it graze ? | Ο ( 232 β 102 ) = 1348.2 answer : b | a ) 1217 , b ) 1348.2 , c ) 1210 , d ) 1212 , e ) 1312 | b | multiply(subtract(power(23, const_2), power(const_10, const_2)), divide(add(multiply(const_10, const_2), const_2), add(const_4, const_3))) | add(const_3,const_4)|multiply(const_10,const_2)|power(n1,const_2)|power(const_10,const_2)|add(#1,const_2)|subtract(#2,#3)|divide(#4,#0)|multiply(#6,#5) | general |
a boat having a length 3 m and breadth 2 m is floating on a lake . the boat sinks by 1 cm when a man gets into it . the mass of the man is : | explanation : in this type of question , first we will calculate the volume of water displaces then will multiply with the density of water . volume of water displaced = 3 * 2 * 0.01 = 0.06 m cube mass of man = volume of water displaced * density of water = 0.06 * 1000 = 60 kg option b | a ) 50 kg , b ) 60 kg , c ) 70 kg , d ) 80 kg , e ) none of these | b | multiply(multiply(multiply(3, 2), divide(1, const_100)), const_1000) | divide(n2,const_100)|multiply(n0,n1)|multiply(#0,#1)|multiply(#2,const_1000) | physics |
tanks a and b are each in the shape of a right circular cylinder . the interior of tank a has a height of 10 meters and a circumference of 6 meters , and the interior of tank b has a height of 6 meters and a circumference of 10 meters . the capacity of tank a is what percent of the capacity of tank b ? | "the radius of tank a is 6 / ( 2 * pi ) . the capacity of tank a is 10 * pi * 36 / ( 4 * pi ^ 2 ) = 180 / ( 2 * pi ) the radius of tank b is 10 / ( 2 * pi ) . the capacity of tank b is 6 * pi * 100 / ( 4 * pi ^ 2 ) = 300 / ( 2 * pi ) tank a / tank b = 180 / 300 = 6 / 10 = 60 % the answer is a ." | a ) 60 % , b ) 80 % , c ) 100 % , d ) 120 % , e ) 125 % | a | multiply(multiply(power(divide(6, 10), const_2), divide(10, 6)), const_100) | divide(n0,n2)|divide(n1,n3)|power(#1,const_2)|multiply(#0,#2)|multiply(#3,const_100)| | physics |
the tax on a commodity is diminished by 30 % but its consumption is increased by 10 % . find the decrease percent in the revenue derived from it ? | "explanation : 100 * 100 = 10000 70 * 110 = 7700 10000 - - - - - - - 2300 100 - - - - - - - ? = 23 % e )" | a ) 12 % , b ) 14 % , c ) 16 % , d ) 20 % , e ) 23 % | e | subtract(const_100, divide(multiply(add(const_100, 10), subtract(const_100, 30)), const_100)) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(const_100,#3)| | general |
a jogger running at 9 kmph along side a railway track is 270 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction . in how much time will the train pass the jogger ? | "speed of train relative to jogger = ( 45 β 9 ) km / h = 36 km / h = ( 36 Γ 5 β 18 ) m / sec = 10 m / sec distance to be covered = ( 270 + 120 ) m = 390 m . β΄ time taken = ( 390 β 10 ) sec = 39 sec . answer d" | a ) 3.6 sec , b ) 18 sec , c ) 36 sec , d ) 39 sec , e ) none of these | d | multiply(multiply(divide(divide(add(270, 120), const_1000), subtract(45, 9)), const_60), const_60) | add(n1,n2)|subtract(n3,n0)|divide(#0,const_1000)|divide(#2,#1)|multiply(#3,const_60)|multiply(#4,const_60)| | physics |
some consecutive natural numbers , starting with 1 , are written on the board . now , one of the numbers was erased and the average of the remaining numbers is 800 / 39 . find the number which was erased . | we know that average of n consecutive numbes average = n Γ ( n + 1 ) 2 n = ( n + 1 ) 2 n Γ ( n + 1 ) 2 n = ( n + 1 ) 2 if the given n is sufficiently large , the average does not change much even though we exclude one or two numbers from it . so the approximate number of observations is almost double to the average ( r... | a ) 20 , b ) 87 , c ) 266 , d ) 288 , e ) 11 | a | subtract(divide(multiply(add(39, const_1), add(add(39, const_1), const_1)), const_2), 800) | add(n2,const_1)|add(#0,const_1)|multiply(#0,#1)|divide(#2,const_2)|subtract(#3,n1) | general |
find the area of the quadrilateral of one of its diagonals is 10 cm and its off sets 7 cm and 3 cm ? | "1 / 2 * 10 ( 7 + 3 ) = 50 cm 2 answer : a" | a ) 50 cm 2 , b ) 100 cm 2 , c ) 150 cm 2 , d ) 200 cm 2 , e ) 250 cm 2 | a | multiply(multiply(divide(const_1, const_2), add(3, 7)), 10) | add(n1,n2)|divide(const_1,const_2)|multiply(#0,#1)|multiply(n0,#2)| | geometry |
carol and jordan draw rectangles of equal area . if carol ' s rectangle measures 15 inches by 24 inches and jordan ' s rectangle is 8 inches long , how wide is jordan ' s rectangle , in inches ? | "area of first rectangle is 15 * 24 = 360 hence area of second would be 8 x = 360 x x = 45 answer is d" | a ) 30 , b ) 35 , c ) 40 , d ) 45 , e ) 50 | d | divide(rectangle_area(15, 24), 8) | rectangle_area(n0,n1)|divide(#0,n2)| | geometry |
what is the smallest number which when increased by 3 is divisible by 18 , 70 , 25 and 21 ? | "when increased by 3 , the number must include at least 2 * 3 ^ 2 * 5 ^ 2 * 7 = 3150 the answer is c ." | a ) 2327 , b ) 2757 , c ) 3147 , d ) 3587 , e ) 3997 | c | add(lcm(lcm(18, 70), lcm(25, 21)), 3) | lcm(n1,n2)|lcm(n3,n4)|lcm(#0,#1)|add(n0,#2)| | general |
the dimensions of a field are 10 m by 10 m . a pit 10 m long , 5 m wide and 4 m deep is dug in one corner of the field and the earth removed has been evenly spread over the remaining area of the field . what will be the rise in the height of field as a result of this operation ? | "the volume of the earth removed is 10 * 5 * 4 = 200 m ^ 3 . the remaining area of the field is 10 * 10 - 10 * 5 = 50 m ^ 2 . 200 m ^ 3 of the earth evenly spread over the area of 50 m ^ 2 will rise the height by ( height ) = ( volume ) / ( area ) = 200 / 50 = 4 m . answer : c" | a ) 2 m , b ) 3 m , c ) 4 m , d ) 5 m , e ) 1.5 m | c | divide(multiply(10, 10), subtract(rectangle_area(10, 10), rectangle_area(5, 10))) | multiply(n1,n2)|rectangle_area(n0,n1)|rectangle_area(n2,n3)|subtract(#1,#2)|divide(#0,#3)| | other |
in a survey of parents , exactly 7 / 8 of the mothers and 3 / 4 of the fathers held full - time jobs . if 60 percent of the parents surveyed were women , what percent of the parents did not hold full - time jobs ? | "fathers without full - time jobs are 1 / 4 * 2 / 5 = 2 / 20 of all the parents surveyed . mothers without full - time jobs are 1 / 8 * 3 / 5 = 3 / 40 of all the parents surveyed . the percent of parents without full - time jobs is 2 / 20 + 3 / 40 = 7 / 40 = 17.5 % the answer is c ." | a ) 25.5 % , b ) 21.5 % , c ) 17.5 % , d ) 13.5 % , e ) 9.5 % | c | add(subtract(subtract(const_100, 60), multiply(divide(3, 4), subtract(const_100, 60))), subtract(60, multiply(divide(7, 8), 60))) | divide(n2,n3)|divide(n0,n1)|subtract(const_100,n4)|multiply(#0,#2)|multiply(n4,#1)|subtract(#2,#3)|subtract(n4,#4)|add(#5,#6)| | general |
a truck covers a distance of 296 km at a certain speed in 8 hours . how much time would a car take at an average speed which is 18 kmph more than that of the speed of the truck to cover a distance which is 6.5 km more than that travelled by the truck ? | explanation : speed of the truck = distance / time = 296 / 8 = 37 kmph now , speed of car = ( speed of truck + 18 ) kmph = ( 37 + 18 ) = 55 kmph distance travelled by car = 296 + 6.5 = 302.5 km time taken by car = distance / speed = 302.5 / 55 = 5.5 hours . answer β c | a ) 6 hours , b ) 5 hours , c ) 5.5 hours , d ) 8 hours , e ) none | c | divide(add(296, 6.5), add(divide(296, 8), 18)) | add(n0,n3)|divide(n0,n1)|add(n2,#1)|divide(#0,#2) | physics |
a bag contains 12 red jellybeans and 12 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability that all 3 jellybeans removed from the bag are blue ? | "method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total o... | a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10 | e | divide(choose(12, 3), choose(add(12, 12), 3)) | add(n0,n0)|choose(n0,n2)|choose(#0,n2)|divide(#1,#2)| | probability |
a candidate got 10 % of the votes polled and he lost to his rival by 16000 votes . how many votes were cast ? | "10 % - - - - - - - - - - - l 90 % - - - - - - - - - - - w - - - - - - - - - - - - - - - - - - 80 % - - - - - - - - - - 16000 100 % - - - - - - - - - ? = > 20000 answer : b" | a ) 7500 , b ) 20000 , c ) 2775 , d ) 5496 , e ) 6851 | b | divide(16000, subtract(subtract(const_1, divide(10, const_100)), divide(10, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|subtract(#1,#0)|divide(n1,#2)| | gain |
if i walk at 8 km / h , i miss the bus by 14 minutes . if i walk at 9 km / h , i reach 16 minutes before the arrival of the bus . how far i walk to reach the bus stand ? | d = product of speed difference of time / difference of speed d = 8 x 9 / 60 [ 14 Γ’ Λ β ( Γ’ Λ β 16 ) / 9 - 8 ] [ here , Γ’ β¬ β ve sign indicates before the schedule time ] Γ’ β‘ β d = 2.4 km answer c | a ) 3.4 km , b ) 2.9 km , c ) 2.4 km , d ) 2.6 km , e ) 2.8 km | c | multiply(divide(multiply(8, 9), subtract(9, 8)), divide(add(14, 16), const_60)) | add(n1,n3)|multiply(n0,n2)|subtract(n2,n0)|divide(#1,#2)|divide(#0,const_60)|multiply(#3,#4) | physics |
the length of a rectangle is 2 cm more than the width of the rectangle . the perimeter of the rectangle is 20 cm . find the length and the width of the rectangle . | let length l = x , width w = x β 2 and perimeter = p β΄ p = 2 l + 2 w = 2 x + 2 ( x β 2 ) 20 = 2 x + 2 x β 4 4 x = 24 x = 6 l = 6 cm and w = l β 2 = 4 cm answer is c . | ['a ) l = 2 , w = 9', 'b ) l = 5 , w = 8', 'c ) l = 6 , w = 4', 'd ) l = 1 , w = 7', 'e ) l = 2 , w = 3'] | c | add(divide(subtract(divide(20, const_2), 2), const_2), 2) | divide(n1,const_2)|subtract(#0,n0)|divide(#1,const_2)|add(n0,#2) | geometry |
a certain sum becomes 4 times itself at simple interest in 8 years . in how many years does it become 10 times itself ? | let the sum be rs . x , then it becomes rs . 4 x in eight years rs . 3 x is the interest on x for eight years . r = ( 100 * 3 x ) / ( x * 8 ) = 300 / 8 % if the sum becomes ten times itself , then interest is 9 x . the required time period = ( 100 * 9 x ) / ( x * 300 / 8 ) = ( 100 * 9 x * 8 ) / ( x * 300 ) = 24 years .... | a ) 24 , b ) 55 , c ) 77 , d ) 99 , e ) 01 | a | multiply(divide(subtract(10, const_1), subtract(4, const_1)), 8) | subtract(n2,const_1)|subtract(n0,const_1)|divide(#0,#1)|multiply(n1,#2) | general |
product of two natural numbers is 5 . then , the sum of reciprocals of their squares is | "explanation : if the numbers are a , b , then ab = 5 , as 17 is a prime number , so a = 1 , b = 5 . 1 / a 2 + 1 / b 2 = 1 / 1 ( 2 ) + 1 / 5 ( 2 ) = 26 / 25 option b" | a ) 290 / 289 , b ) 26 / 25 , c ) 290 / 90 , d ) 290 / 19 , e ) none of these | b | add(power(divide(const_1, const_1), const_2), power(divide(const_1, 5), const_2)) | divide(const_1,const_1)|divide(const_1,n0)|power(#0,const_2)|power(#1,const_2)|add(#2,#3)| | general |
certain stocks in january were 20 % less than they were in february and 30 % greater than they were in march . what was the percentage decrease in the stocks from february to march ? | "let , stock in february = 100 then , stock in january = 100 - ( 20 / 100 ) * 100 = 80 january = 30 % greater than march = 1.3 * stock in march i . e . stock in march = 90 / 1.3 = 69 approximately % decrease from february to march = ( 69 - 100 ) * 100 / 100 = 31 % approximately answer : option a" | a ) 31 % , b ) 45 % , c ) 52 % , d ) 61 % , e ) 25 % | a | multiply(divide(subtract(divide(const_100, divide(subtract(const_100, 20), const_100)), divide(const_100, divide(add(const_100, 30), const_100))), divide(const_100, divide(subtract(const_100, 20), const_100))), const_100) | add(n1,const_100)|subtract(const_100,n0)|divide(#1,const_100)|divide(#0,const_100)|divide(const_100,#2)|divide(const_100,#3)|subtract(#4,#5)|divide(#6,#4)|multiply(#7,const_100)| | general |
if a ( a + 2 ) = 35 and b ( b + 2 ) = 35 , where a β b , then a + b = | "i . e . if a = 5 then b = - 7 or if a = - 7 then b = 5 but in each case a + b = 5 - 7 = - 2 answer : d" | a ) - 5 , b ) - 6 , c ) - 7 , d ) - 2 , e ) - 10 | d | add(divide(35, const_10), divide(35, divide(35, const_10))) | divide(n1,const_10)|divide(n1,#0)|add(#0,#1)| | general |
if 5 machines can produce 20 units in 10 hours , how long would it take 10 to produce 80 units ? | "5 machines would produce 80 units in 40 hours . increasing the amount of machines by 2 would mean dividing 40 hours by 2 . 40 / 2 = 20 answer : b" | a ) 63 , b ) 20 , c ) 42 , d ) 65 , e ) 84 | b | divide(80, multiply(divide(divide(20, 10), 5), 20)) | divide(n1,n2)|divide(#0,n0)|multiply(n1,#1)|divide(n4,#2)| | physics |
jar a has 26 % more marbles than jar b . what percent of marbles from jar a need to be moved into jar b so that both jars have equal marbles ? | "an easy way to solve this question is by number plugging . assume there are 100 marbles in jar b then in jar a there will be 126 marbles . now , for both jars to have equal marbles we should move 13 marbles from a to b , which is 13 / 126 = ~ 10.3 % of a . answer : d ." | a ) 7.6 % , b ) 8.3 % , c ) 9.6 % , d ) 10.3 % , e ) 11.5 % | d | multiply(divide(divide(26, const_2), add(26, const_100)), const_100) | add(n0,const_100)|divide(n0,const_2)|divide(#1,#0)|multiply(#2,const_100)| | gain |
the average weight of 5 students decreases by 4 kg when one of them weighing 92 kg is replaced by a new student . the weight of the student is | "explanation : let the weight of student be x kg . given , difference in average weight = 4 kg = > ( 92 - x ) / 5 = 4 = > x = 72 answer : d" | a ) 62 kg , b ) 60 kg , c ) 70 kg , d ) 72 kg , e ) none of these | d | subtract(92, multiply(5, 4)) | multiply(n0,n1)|subtract(n2,#0)| | general |
of the final grades received by the students in a certain math course , 1 / 5 are a ' s , 1 / 4 are b ' s , 1 / 2 are c ' s , and the remaining 25 grades are d ' s . what is the number of students in the course ? | "we start by creating a variable for the total number of students in the math course . we can say : t = total number of students in the math course next , we can use variable t in an equation that we translate from the given information . we are given that , of the final grades received by the students in a certain mat... | a ) 80 , b ) 110 , c ) 160 , d ) 500 , e ) 400 | d | divide(25, subtract(1, add(add(divide(1, 5), divide(1, 4)), divide(1, 2)))) | divide(n0,n1)|divide(n0,n3)|divide(n0,n5)|add(#0,#1)|add(#3,#2)|subtract(n0,#4)|divide(n6,#5)| | general |
a 10 - meter long wire is cut into two pieces . if the longer piece is then used to form a perimeter of a square , what is the probability that the area of the square will be more than 4 if the original wire was cut at an arbitrary point ? | "the longer wire will form a square with an area more than 4 if the wire is cut at a point within two meters of either end . the probability of this is 4 / 10 = 2 / 5 . the answer is d ." | a ) 2 / 3 , b ) 3 / 4 , c ) 1 / 5 , d ) 2 / 5 , e ) 3 / 5 | d | divide(square_perimeter(4), 10) | square_perimeter(n1)|divide(#0,n0)| | geometry |
an uneducated retailer marks all his goods at 55 % above the cost price and thinking that he will still make 25 % profit , offers a discount of 25 % on the marked price . what is his actual profit on the sales ? | "sol . let c . p . = rs . 100 . then , marked price = rs . 155 . s . p . = 75 % of rs . 155 = rs . 116.25 . β΄ gain % = 16.25 % . answer e" | a ) 12.50 % , b ) 13.50 % , c ) 14 % , d ) 14.50 % , e ) none | e | multiply(subtract(subtract(add(const_1, divide(55, const_100)), multiply(add(const_1, divide(55, const_100)), divide(25, const_100))), const_1), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|multiply(#2,#1)|subtract(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100)| | gain |
the sum of ages of 5 children born 4 years different each is 70 yrs . what is the age of the elder child ? | "let the ages of children be x , ( x + 4 ) , ( x + 8 ) , ( x + 12 ) and ( x + 16 ) years . then , x + ( x + 4 ) + ( x + 8 ) + ( x + 12 ) + ( x + 16 ) = 70 5 x = 30 x = 6 x + 16 = 6 + 16 = 22 answer : c" | a ) 8 , b ) 9 , c ) 22 , d ) 17 , e ) 18 | c | divide(add(add(add(add(4, const_4), add(4, const_4)), add(const_4, const_4)), 70), 5) | add(n1,const_4)|add(const_4,const_4)|add(#0,#0)|add(#2,#1)|add(n2,#3)|divide(#4,n0)| | general |
the difference between simple and compound interest on rs . 1400 for one year at 10 % per annum reckoned half - yearly is ? | "s . i . = ( 1400 * 10 * 1 ) / 100 = rs . 140 c . i . = [ 1400 * ( 1 + 5 / 100 ) 2 - 1400 ] = rs . 143.5 difference = ( 143.5 - 140 ) = rs . 3.50 answer : b" | a ) 8.25 , b ) 3.5 , c ) 9.0 , d ) 3.15 , e ) 2.0 | b | multiply(subtract(power(add(divide(divide(10, const_2), const_100), const_1), const_2), add(divide(10, const_100), const_1)), 1400) | divide(n1,const_2)|divide(n1,const_100)|add(#1,const_1)|divide(#0,const_100)|add(#3,const_1)|power(#4,const_2)|subtract(#5,#2)|multiply(n0,#6)| | gain |
if a * b * c = ( β ( a + 2 ) ( b + 3 ) ) / ( c + 1 ) , find the value of 6 * 15 * 2 . | "6 * 15 * 2 = ( β ( 6 + 2 ) ( 15 + 3 ) ) / ( 2 + 1 ) = ( β 8 * 18 ) / 3 = ( β 144 ) / 3 = 12 / 3 = 4 . answer is e" | a ) 8 , b ) 5 , c ) 11 , d ) 3 , e ) 4 | e | divide(sqrt(multiply(add(6, 2), add(15, 3))), add(2, 1)) | add(n0,n3)|add(n1,n4)|add(n2,n5)|multiply(#0,#1)|sqrt(#3)|divide(#4,#2)| | general |
the sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 420 m , its area is ? | "5 x + 12 x + 13 x = 420 = > x = 14 a = 70 , b = 168 , c = 182 s = ( 70 + 168 + 182 ) / 2 = 210 answer : d" | a ) 150 , b ) 882 , c ) 277 , d ) 210 , e ) 281 | d | multiply(420, divide(420, add(add(5, 12), 13))) | add(n0,n1)|add(n2,#0)|divide(n3,#1)|multiply(n3,#2)| | geometry |
a train 500 m long , running with a speed of 180 km / hr will pass a tree in ? | speed = 180 * 5 / 18 = 50 m / sec time taken = 500 * 1 / 50 = 10 sec answer : d | a ) 17 sec , b ) 16 sec , c ) 18 sec , d ) 10 sec , e ) 12 sec | d | multiply(divide(500, multiply(180, const_1000)), const_3600) | multiply(n1,const_1000)|divide(n0,#0)|multiply(#1,const_3600) | physics |
33 1 / 3 % of 330 ? | "33 1 / 3 % = 1 / 3 1 / 3 Γ 330 = 110 c )" | a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130 | c | divide(multiply(add(33, divide(1, 3)), 330), const_100) | divide(n1,n2)|add(n0,#0)|multiply(n3,#1)|divide(#2,const_100)| | gain |
the malibu country club needs to drain its pool for refinishing . the hose they use to drain it can remove 60 cubic feet of water per minute . if the pool is 40 feet wide by 150 feet long by 10 feet deep and is currently at 80 % capacity , how long will it take to drain the pool ? | "volume of pool = 40 * 150 * 10 cu . ft , 80 % full = 40 * 150 * 10 * 0.8 cu . ft water is available to drain . draining capacity = 60 cu . ft / min therefore time taken = 40 * 150 * 10 * 0.8 / 60 min = 800 min b" | a ) 1000 , b ) 800 , c ) 600 , d ) 700 , e ) 500 | b | divide(multiply(divide(80, const_100), multiply(multiply(40, 150), 10)), 60) | divide(n4,const_100)|multiply(n1,n2)|multiply(n3,#1)|multiply(#0,#2)|divide(#3,n0)| | gain |
you collect pens . suppose you start out with 7 . mike gives you another 22 pens . since her father makes pens , cindy decides to double your pens . since you ' re nice , you give sharon 19 pens . how many pens do you have at the end ? | "solution start with 7 pens . mike gives you 22 pens : 7 + 22 = 29 pens . cindy doubles the number of pens you have : 29 Γ 2 = 58 pens . sharon takes 19 pens from you : 58 - 19 = 39 pens . so you have 39 at the end . correct answer : a" | a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 43 | a | subtract(multiply(add(22, 7), const_2), 19) | add(n0,n1)|multiply(#0,const_2)|subtract(#1,n2)| | general |
the difference between simple and compound interest on rs . 1300 for one year at 10 % per annum reckoned half - yearly is ? | s . i . = ( 1300 * 10 * 1 ) / 100 = rs . 130 c . i . = [ 1300 * ( 1 + 5 / 100 ) 2 - 1300 ] = rs . 133.25 difference = ( 133.25 - 130 ) = rs . 3.25 answer : b | a ) 8.0 , b ) 3.25 , c ) 9.15 , d ) 3.13 , e ) 2.0 | b | multiply(subtract(power(add(divide(divide(10, const_2), const_100), const_1), const_2), add(divide(10, const_100), const_1)), 1300) | divide(n1,const_2)|divide(n1,const_100)|add(#1,const_1)|divide(#0,const_100)|add(#3,const_1)|power(#4,const_2)|subtract(#5,#2)|multiply(n0,#6) | gain |
the digital sum of a number is the sum of its digits . for how many of the positive integers 24 - 90 inclusive is the digital sum a multiple of 7 ? | "is there other way than just listing ? 25 34 43 52 59 61 68 70 77 86 10 ways . . a" | a ) 10 , b ) 8 , c ) 14 , d ) 16 , e ) 20 | a | subtract(subtract(24, 7), const_2) | subtract(n0,n2)|subtract(#0,const_2)| | general |
ravi and kavi start a business by investing Γ’ β ΒΉ 3000 and Γ’ β ΒΉ 72000 , respectively . find the ratio of their profits at the end of year . | ratio of profit = ratio of investments = 3000 : 72000 = 1 : 24 answer : d | a ) 2 : 24 , b ) 5 : 24 , c ) 7 : 24 , d ) 1 : 24 , e ) 3 : 24 | d | divide(3000, 72000) | divide(n0,n1) | gain |
in august , a cricket team that played 120 matches won 30 % of the games it played . after a continuous winning streak , this team raised its average to 52 % . how many matches did the team win to attain this average ? | "let the no of matches played more = x so , ( 120 + x ) * 52 / 100 = 36 + x by solving we get x = 55 answer : b" | a ) 40 , b ) 55 , c ) 68 , d ) 80 , e ) 98 | b | divide(subtract(multiply(divide(52, const_100), 120), multiply(divide(30, const_100), 120)), subtract(const_1, divide(52, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|multiply(n0,#1)|subtract(const_1,#0)|subtract(#2,#3)|divide(#5,#4)| | general |
jim β s taxi service charges an initial fee of $ 2.05 at the beginning of a trip and an additional charge of $ 0.35 for each 2 / 5 of a mile traveled . what is the total charge for a trip of 3.6 miles ? | let the fixed charge of jim β s taxi service = 2.05 $ and charge per 2 / 5 mile ( . 4 mile ) = . 35 $ total charge for a trip of 3.6 miles = 2.05 + ( 3.6 / . 4 ) * . 35 = 2.05 + 9 * . 35 = 5.2 $ answer a | a ) $ 5.20 , b ) $ 4.45 , c ) $ 4.80 , d ) $ 5.05 , e ) $ 5.40 | a | add(2.05, multiply(0.35, divide(3.6, divide(2, 5)))) | divide(n2,n3)|divide(n4,#0)|multiply(n1,#1)|add(n0,#2) | general |
if a = 105 and a ^ 3 = 21 Γ 25 Γ 45 Γ w , what is the value of w ? | "a = 105 = 3 * 5 * 7 a ^ 3 = 21 Γ 25 Γ 45 Γ w = > a ^ 3 = ( 7 * 3 ) x ( 5 * 5 ) x ( 3 ^ 2 * 5 ) x w = > a ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x w = > ( 3 * 5 * 7 ) ^ 3 = 3 ^ 3 * 5 ^ 3 * 7 x w w = 7 ^ 2 = 49 answer d" | a ) 35 , b ) 42 , c ) 45 , d ) 49 , e ) 54 | d | divide(power(105, 3), multiply(multiply(21, 25), 45)) | multiply(n2,n3)|power(n0,n1)|multiply(n4,#0)|divide(#1,#2)| | general |
a certain car traveled twice as many miles from town a to town b as it did from town b to town c . from town a to town b , the car averaged 40 miles per gallon , and from town b to town c , the car averaged 50 miles per gallon . what is the average miles per gallon that the car achieved on its trip from town a through ... | "step 1 ) took lcm of 40 and 50 . . came as 200 . step 2 ) 200 distance between b to c . . . do 200 / 50 hence 4 gallons used step 3 ) twice distance . . hence 200 * 2 = 400 . . . do as above . . 400 / 40 = 10 gallons used step 4 ) total gallons . . 4 + 10 = 14 gallons step ) total miles = 200 + 400 = 600 miles hence .... | a ) 42.85 , b ) 40 , c ) 35 , d ) 41 , e ) 39 | a | divide(add(multiply(50, const_10), divide(multiply(50, const_10), const_2)), add(divide(multiply(50, const_10), 40), divide(divide(multiply(50, const_10), const_2), 50))) | multiply(n1,const_10)|divide(#0,const_2)|divide(#0,n0)|add(#1,#0)|divide(#1,n1)|add(#2,#4)|divide(#3,#5)| | general |
a fill pipe can fill 2 / 5 of cistern in 30 minutes in how many minutes , it can fill 4 / 5 of the cistern ? | "2 / 5 of the cistern can fill in 30 min 4 / 5 of the cistern can fill in = 30 * 5 / 2 * 4 / 5 = 60 min answer is e" | a ) 48 min , b ) 63 min , c ) 25 min , d ) 30 min , e ) 60 min | e | divide(30, 2) | divide(n2,n0)| | physics |
a mixture contains alcohol and water in the ratio 4 : 3 . if 10 litres of water is added to the mixture , the ratio becomes 4 : 5 . find the quantity of alcohol in the given mixture | "let the quantity of alcohol and water be 4 x litres and 3 x litres respectively 4 x / ( 3 x + 10 ) = 4 / 5 20 x = 4 ( 3 x + 10 ) 8 x = 40 x = 5 quantity of alcohol = ( 4 x 5 ) litres = 20 litres . answer is d ." | a ) 15 litres , b ) 10 litres , c ) 30 litres , d ) 20 litres , e ) 8 litres | d | multiply(10, const_1) | multiply(n2,const_1)| | general |
a car averages 45 mph for the first 4 hours of a trip and averages 75 mph for each additional hour . the average speed for the entire trip was 65 mph . how many hours long is the trip ? | "let the time for which car averages 75 mph = t 65 * ( t + 4 ) = 45 * 4 + 75 t = > 10 t = 80 = > t = 8 total duration of the trip = 8 + 4 = 12 answer a" | a ) 12 , b ) 11 , c ) 10 , d ) 9 , e ) 8 | a | add(divide(subtract(multiply(65, 4), multiply(45, 4)), subtract(75, 65)), 4) | multiply(n3,n1)|multiply(n0,n1)|subtract(n2,n3)|subtract(#0,#1)|divide(#3,#2)|add(n1,#4)| | general |
the sum of the present age of henry and jill is 40 . what is their present ages if 8 years ago henry was twice the age of jill ? | "let the age of jill 8 years ago be x , age of henry be 2 x x + 8 + 2 x + 8 = 40 x = 8 present ages will be 16 and 24 answer : b" | a ) 13 and 27 , b ) 16 and 24 , c ) 18 and 22 , d ) 11 and 29 , e ) none of these | b | subtract(40, divide(add(40, 8), const_3)) | add(n0,n1)|divide(#0,const_3)|subtract(n0,#1)| | general |
the cricket team of 11 members is 25 yrs old & the wicket keeper is 3 yrs older . if the ages ofthese 2 are excluded , the average age of theremaining players is 1 year less than the average age of the whole team . what is the average age of the team ? | let the average age of the whole team be x years . 11 x - ( 25 + 28 ) = 9 ( x - 1 ) = > 11 x - 9 x = 44 = > 2 x = 44 = > x = 22 . so , average age of the team is 22 years . b | a ) 21 , b ) 22 , c ) 23 , d ) 25 , e ) 28 | b | divide(subtract(add(add(25, const_3), 25), subtract(11, 2)), subtract(11, subtract(11, 2))) | add(n1,const_3)|subtract(n0,n3)|add(n1,#0)|subtract(n0,#1)|subtract(#2,#1)|divide(#4,#3) | general |
the speed of a boat in still water is 24 km / hr and the rate of current is 3 km / hr . the distance travelled downstream in 15 minutes is | "explanation : speed downstreams = ( 24 + 3 ) kmph = 27 kmph . distance travelled = ( 27 x 15 / 60 ) km = 6.75 km option d" | a ) 1.6 km , b ) 2 km , c ) 3.6 km , d ) 6.75 km , e ) none of these | d | multiply(divide(15, const_60), add(24, 3)) | add(n0,n1)|divide(n2,const_60)|multiply(#0,#1)| | physics |
louie takes out a 3 - month loan of $ 2000 . the lender charges him 10 % interest per month compounded monthly . the terms of the loan state that louie must repay the loan in 3 equal monthly payments . to the nearest dollar , how much does louie have to pay each month ? | here ' s the calculation for that case , assume monthly payment is x . after 1 st month : ( 2000 ) ( 1.1 ) - x = 2200 - x after 2 nd month : ( 2200 - x ) ( 1.1 ) - x = 2420 - 2.21 x after 3 rd month : ( 2420 - 2.21 x ) ( 1.1 ) - x = 2662 - 3.31 x now , the amount after the last payment in 3 rd month must bring the tota... | a ) a ) 333 , b ) b ) 383 , c ) c ) 402 , d ) d ) 433 , e ) e ) 804 | e | divide(multiply(multiply(2000, add(const_1, divide(const_1, const_10))), add(const_1, divide(const_1, const_10))), 3) | divide(const_1,const_10)|add(#0,const_1)|multiply(n1,#1)|multiply(#1,#2)|divide(#3,n0) | general |
50 ^ 51 ^ 52 / 11 | "we know that 6 ^ 1 = 6 or 6 ^ 2 = 36 so for all power of 6 unit digit is 6 now que is 50 ^ 51 ^ 52 now if we divide 50 / 11 then rem is 6 means 6 ^ 51 = 6 at unit place same for 6 ^ 52 salso give 6 at unit place now finally 50 ^ 51 ^ 52 gives 6 at unit place now 6 / 11 = 6 answer : a" | a ) 6 , b ) 4 , c ) 7 , d ) 3 , e ) 5 | a | divide(power(50, 51), power(50, 11)) | power(n0,n1)|power(n0,n3)|divide(#0,#1)| | general |
some of 50 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 45 % . what fraction of the original paint was replaced ? | "45 % is 20 % - points above 25 % and 5 % - points below 50 % . thus the ratio of 25 % - solution to 50 % - solution is 1 : 4 . 1 / 5 of the original paint was replaced . the answer is b ." | a ) 1 / 30 , b ) 1 / 5 , c ) 2 / 3 , d ) 3 / 4 , e ) 4 / 5 | b | divide(subtract(divide(45, const_100), divide(50, const_100)), subtract(divide(25, const_100), divide(50, const_100))) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#1)|divide(#3,#4)| | gain |
if a certain toy store ' s revenue in november was 3 / 5 of its revenue in december and its revenue in january was 1 / 3 of its revenue in november , then the store ' s revenue in december was how many times the average ( arithmetic mean ) of its revenues in november and january ? | "n = 3 d / 5 j = n / 3 = d / 5 the average of november and january is ( n + j ) / 2 = 4 d / 5 / 2 = 2 d / 5 d is 5 / 2 times the average of november and january . the answer is a ." | a ) 5 / 2 , b ) 7 / 2 , c ) 10 / 3 , d ) 15 / 4 , e ) 25 / 3 | a | divide(1, divide(add(divide(3, 5), multiply(divide(3, 5), divide(1, 3))), const_2)) | divide(n0,n1)|divide(n2,n3)|multiply(#0,#1)|add(#0,#2)|divide(#3,const_2)|divide(n2,#4)| | general |
the cross - section of a cannel is a trapezium in shape . if the cannel is 10 m wide at the top and 6 m wide at the bottom and the area of cross - section is 640 sq m , the depth of cannel is ? | "1 / 2 * d ( 10 + 6 ) = 640 d = 80 answer : d" | a ) 76 , b ) 28 , c ) 27 , d ) 80 , e ) 25 | d | divide(divide(divide(640, divide(add(10, 6), const_2)), 6), const_2) | add(n0,n1)|divide(#0,const_2)|divide(n2,#1)|divide(#2,n1)|divide(#3,const_2)| | physics |
the jogging track in a sports complex is 528 m in circumference . deepak and his wife start from the same point and walk in opposite directions at 4.5 km / hr and 3.75 km / hr respectively . they will meet for the first time in ? | "clearly , the two will meet when they are 528 m apart . to be ( 4.5 + 3.75 ) = 8.25 km apart , they take 1 hour . to be 528 m apart , they take ( 100 / 825 * 528 / 1000 ) hrs = ( 528 / 8250 * 60 ) min = 3.84 min . answer : b" | a ) 5.29 min , b ) 3.84 min , c ) 5.08 min , d ) 9.28 min , e ) 5.988 min | b | multiply(divide(divide(528, const_1000), add(4.5, 3.75)), const_60) | add(n1,n2)|divide(n0,const_1000)|divide(#1,#0)|multiply(#2,const_60)| | general |
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