Problem stringlengths 5 628 | Rationale stringlengths 1 2.74k | options stringlengths 37 137 | correct stringclasses 5
values | annotated_formula stringlengths 6 848 | linear_formula stringlengths 7 357 | category stringclasses 6
values |
|---|---|---|---|---|---|---|
calculate the largest 6 digit number which is exactly divisible by 99 ? | b 425 ( 68 ã · 16 ) ã — 100 | a ) 336 , b ) 425 , c ) 275 , d ) 235 , e ) 689 | b | subtract(multiply(multiply(68, const_4), const_2), const_100) | multiply(n0,const_4)|multiply(#0,const_2)|subtract(#1,const_100) | general |
a green grocer received a boxful of tomatoes and on opening the box found that several had gone bad . he then counted them up so that he could make a formal complaint and found that 68 were mouldy , which was 16 per cent of the total contents of the box . how many tomatoes were in the box ? | drop = 12 inches / day 5 days ago = w , means now it ' s equal w - 60 and in 4 days = w - 60 - 48 = w - 108 answer a | a ) w − 108 , b ) w − 56 , c ) w − 14 , d ) w + 14 , e ) w + 126 | a | multiply(12, divide(4, 5)) | divide(n2,n1)|multiply(n0,#0) | gain |
the water level in a reservoir has been dropping at the rate of 12 inches per day . exactly 5 days ago , the water level was at w inches . what will be the water level exactly 4 days from now if the rate at which the level is dropping remains the same ? | sol . apples 250 each carries 25 = 250 / 25 = 10 answer : d | a ) a ) 9 , b ) b ) 5 , c ) c ) 7 , d ) d ) 10 , e ) e ) none of the above | d | divide(250, 25) | divide(n0,n1) | general |
how many boxes do we need if we have to carry 250 apples into boxes that each hold 25 apples ? | "15 km / hr = 15000 m / 3600 s = ( 150 / 36 ) m / s = ( 25 / 6 ) m / s time = 600 / ( 25 / 6 ) = 144 seconds the answer is c ." | a ) 128 , b ) 136 , c ) 144 , d ) 152 , e ) 160 | c | divide(600, multiply(15, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
how many seconds does sandy take to cover a distance of 600 meters , if sandy runs at a speed of 15 km / hr ? | "total profit = 1000 ratio = 600 / 300 = 2 : 1 answer : e" | a ) 3 : 4 , b ) 2 : 3 , c ) 4 : 3 , d ) 1 : 3 , e ) 2 : 1 | e | divide(600, 300) | divide(n0,n1)| | other |
if shares of two persons in profits are rs . 600 and rs . 300 then ratio of their capitals is | "correct sum = ( 36 * 50 + 48 - 23 ) = 1825 . correct mean = 1825 / 50 = 36.5 answer a" | a ) 36.5 , b ) 35 , c ) 34 , d ) 33 , e ) 32.5 | a | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general |
the mean of 50 observations was 36 . it was found later that an observation 48 was wrongly taken as 23 . the corrected new mean is : | "let us suppose there are 100 people . 50 % of them donated $ 25000 ( 500 * 50 ) $ 25000 is 60 % of total amount . so total amount = 25000 * 100 / 60 remaining amount is 40 % of total amount . 40 % of total amount = 25000 * ( 100 / 60 ) * ( 40 / 100 ) = 50000 / 3 this amount has to be divided by 50 ( remaining people a... | a ) $ 200 , b ) $ 333.33 , c ) $ 100.25 , d ) $ 277.78 , e ) $ 377.00 | b | divide(multiply(divide(multiply(divide(50, const_100), 500), divide(60, const_100)), divide(50, const_100)), divide(60, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|divide(#2,#1)|multiply(#3,#0)|divide(#4,#1)| | general |
a school has received 60 % of the amount it needs for a new building by receiving a donation of $ 500 each from people already solicited . people already solicited represent 50 % of the people from whom the school will solicit donations . how much average contribution is requited from the remaining targeted people to c... | "if there is one bacteria colony , then it will reach the limit of its habitat in 20 days . if there are two bacteria colonies , then in order to reach the limit of habitat they would need to double one time less than in case with one colony . thus colonies need to double 18 times . answer : d . similar questions to pr... | a ) 6.33 , b ) 7.5 , c ) 10 , d ) 18 , e ) 19 | d | subtract(19, divide(19, 19)) | divide(n0,n0)|subtract(n0,#0)| | physics |
a certain bacteria colony doubles in size every day for 19 days , at which point it reaches the limit of its habitat and can no longer grow . if two bacteria colonies start growing simultaneously , how many days will it take them to reach the habitat ’ s limit ? | circumference = 2 * pi * r = 2 * pi * 4 / pi = > 8 a | ['a ) 8', 'b ) 4 π', 'c ) 4', 'd ) 6', 'e ) 5'] | a | circumface(divide(4, const_pi)) | divide(n0,const_pi)|circumface(#0) | geometry |
the diameter of a circle is 4 / π . find the circumference of the circle . | "given exp . = 0.3 * 0.3 + ( 0.3 * 0.3 ) = 0.09 + 0.09 = 0.18 answer is c ." | a ) 0.52 , b ) 0.42 , c ) 0.18 , d ) 0.64 , e ) 0.46 | c | add(multiply(0.3, 0.3), multiply(0.3, 0.3)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | general |
simplify : 0.3 * 0.3 + 0.3 * 0.3 | straight a way lets exclude all the even numbers between 260 and 280 . so now the number starts from 261 to 279 ( only odd ) 261 is a divisible of 3 and next odd divisible by 3 will be 261 + 6 = 267 + 6 = 273 + 6 = 279 . also we can eliminate numbers ending with ' 5 ' so in odd , the excluded numbers are 261 , 265,267 ... | a ) none , b ) one , c ) two , d ) three , e ) four | e | subtract(divide(subtract(280, 260), const_4), const_1) | subtract(n1,n0)|divide(#0,const_4)|subtract(#1,const_1) | general |
how many prime numbers exist between 260 and 280 ? | "first the 2 robots work at the rate of 1 + 1 / 2 = 3 / 2 so they complete one robot in 2 / 3 rd of an hour = 40 minutes - ( 1 ) now the 3 robots work together at the rate of 1 + 1 / 2 + 1 / 2 = 4 / 2 = 2 / 1 so they complete one robot in 1 / 2 an hour , i . e 30 minutes - ( 2 ) now the 4 robots work together at the ra... | a ) 70 min , b ) 94 min , c ) 110 min , d ) 131 1 / 7 min , e ) 146 1 / 7 min | e | add(inverse(add(add(inverse(1), inverse(2)), inverse(2))), inverse(add(inverse(1), inverse(2)))) | inverse(n0)|inverse(n1)|add(#0,#1)|add(#2,#1)|inverse(#2)|inverse(#3)|add(#5,#4)| | physics |
one robot builds a robot in 1 hour , another one builds a robot in 2 hours . the 2 robots work together and when a new robot is complete , it joins the others , working at a constant rate of one robot in 2 hours . how much time will it take until there are 8 robots altogether , if the robots build only one robot at a t... | x = - 2.5 prob = 1 / 12 answer - a | a ) 1 / 12 , b ) 1 / 6 , c ) 1 / 4 , d ) 1 / 3 , e ) 1 / 2 | a | divide(1, multiply(6, 2)) | multiply(n1,n14)|divide(n5,#0) | general |
{ - 10 , - 6 , - 5 , - 4 , - 2.5 , - 1 , 0 , 2.5 , 4 , 6 , 7 , 10 } a number is to be selected at random from the set above . what is the probability that the number will be a solution to the equation ( x - 4 ) ( x + 9 ) ( 2 x + 5 ) = 0 ? | "the distance between the nest and the ditch is 300 meters . 15 times mean = a crow leaves its nest , and flies back ( going and coming back ) i . e . 2 times we get total 30 rounds . so the distance is 30 * 300 = 9000 . d = st 9000 / 1.5 = t , i think we can take 9000 meters as 9 km , then only we get t = 6 . ( 1000 m... | a ) 1 , b ) 2 , c ) 4 , d ) 6 , e ) 8 | d | divide(divide(multiply(300, multiply(15, const_2)), const_1000), divide(15, const_10)) | divide(n1,const_10)|multiply(n1,const_2)|multiply(n0,#1)|divide(#2,const_1000)|divide(#3,#0)| | physics |
a crow leaves its nest , and flies back and forth from its nest to a nearby ditch to gather worms . the distance between the nest and the ditch is 300 meters . in one and a half hours , the crow manages to bring worms to its nest 15 times . what is the speed of the crow in kilometers per hour ? | "now given that q is set the consecutive integers between a and b . and q contains 9 multiples of 9 let take a as 36 . then 36 45 54 63 72 81 90 99 108 . . . so b will 108 . now let ' s check the multiples of 4 among this set 108 - 36 / 4 + 1 = > 18 + 1 = > 19 ans option b ." | a ) 18 , b ) 19 , c ) 20 , d ) 21 , e ) 22 | b | subtract(multiply(9, const_2), const_1) | multiply(n1,const_2)|subtract(#0,const_1)| | physics |
a and b are two multiples of 36 , and q is the set of consecutive integers between a and b , inclusive . if q contains 9 multiples of 9 , how many multiples of 4 are there in q ? | "20 c 20 = 1 ( 20 c 2 ) * ( 20 c 20 ) = 20 ! * 1 / 18 ! = 20 * 19 * 18 ! / 18 ! = 20 * 19 * 1 = 380 answer : b" | a ) 400 , b ) 380 , c ) 360 , d ) 350 , e ) 330 | b | multiply(add(divide(18, 20), 20), 20) | divide(n1,n2)|add(n0,#0)|multiply(#1,n2)| | general |
find the value of ( 20 c 18 ) * ( 20 c 20 ) | "this question is a weighted average question with a series of dependent variables . the remaining portion of the class represents 100 % - 30 % - 50 % = 20 % of the class converting the portions of the class population to decimal weights , we find : class average = 0.30 x 95 + 0.50 x 79 + 0.20 x 60 = 80 the class avera... | a ) 76 % , b ) 77 % , c ) 78 % , d ) 79 % , e ) 80 % | e | divide(add(add(multiply(30, 95), multiply(50, 79)), multiply(subtract(const_100, add(30, 50)), 60)), const_100) | add(n0,n2)|multiply(n0,n1)|multiply(n2,n3)|add(#1,#2)|subtract(const_100,#0)|multiply(n4,#4)|add(#3,#5)|divide(#6,const_100)| | gain |
if 30 % of a class averages 95 % on a test , 50 % of the class averages 79 % on the test , and the remainder of the class averages 60 % on the test , what is the overall class average ? ( round final answer to the nearest percent ) . | "the profit on the first kind of vodka = x % ; the profit on the second kind of vodka = y % . when they are mixed in the ratio 1 : 2 ( total of 3 parts ) the average profit is 10 % : ( x + 2 y ) / 3 = 10 . when they are mixed in the ratio 2 : 1 ( total of 3 parts ) the average profit is 20 % : ( 2 x + y ) / 3 = 20 . so... | a ) 20 % , b ) 40 % , c ) 18 % , d ) 23 % , e ) can not be determined | a | add(divide(multiply(10, 4), 6), add(10, 5)) | add(n4,n8)|multiply(n4,n6)|divide(#1,const_2.0)|add(#0,#2)| | general |
two kinds of vodka are mixed in the ratio 2 : 6 and 6 : 3 and they are sold fetching the profit 10 % and 20 % respectively . if the vodkas are mixed in equal ratio and the individual profit percent on them are increased by 4 / 3 and 5 / 3 times respectively , then the mixture will fetch the profit of | if the tens digit of positive integers m , y are 6 , how many values of the tens digit of 2 ( m + y ) can be there ? a . 2 b . 3 c . 4 d . 5 e . 6 - > if m = y = 60 , 2 ( m + y ) = 240 is derived . if m = y = 69 , 2 ( m + y ) = 276 is derived , which makes 4,5 , 6,7 possible for the tens digit . therefore , the answer ... | a ) 2 , b ) 3 , c ) 4 , d ) 5 , e ) 6 | c | subtract(6, 2) | subtract(n0,n1) | physics |
if the tens digit of positive integers m , y are 6 , how many values of the tens digit of 2 ( m + y ) can be there ? | "number of pens = 848 number of pencils = 630 required number of students = h . c . f . of 848 and 630 = 2 answer is c" | a ) 10 , b ) 4 , c ) 2 , d ) 14 , e ) 16 | c | gcd(848, 630) | gcd(n0,n1)| | general |
the maximum number of students among them 848 pens and 630 pencils can be distributed in such a way that each student get the same number of pens and same number of pencils ? | 1 : 2 answer : a | ['a ) 1 : 2', 'b ) 2 : 3', 'c ) 2 : 9', 'd ) 2 : 1', 'e ) 2 : 2'] | a | divide(1, 2) | divide(n0,n1) | geometry |
find the ratio of the curved surfaces of two cylinders of same heights if their radii are in the ratio 1 : 2 ? | "you have 6 digits : 1 , 2 , 3 , 7 , 8 , 9 each digit needs to be used to make two 3 digit numbers . this means that we will use each of the digits only once and in only one of the numbers . the numbers need to be as close to each other as possible . the numbers can not be equal so the greater number needs to be as sma... | a ) 19 , b ) 49 , c ) 58 , d ) 113 , e ) 131 | a | subtract(subtract(const_100, multiply(subtract(8, 1), const_10)), const_1) | subtract(n5,n1)|multiply(#0,const_10)|subtract(const_100,#1)|subtract(#2,const_1)| | general |
n and m are each 3 - digit integers . each of the numbers 1 , 2 , 3 , 7 , 8 and 9 is a digit of either n or m . what is the smallest possible positive difference between n and m ? | "explanation : let the average age of the whole team by x years . 11 x â € “ ( 29 + 32 ) = 9 ( x - 1 ) 11 x â € “ 9 x = 52 2 x = 52 x = 26 . so , average age of the team is 26 years . answer e" | a ) 20 years , b ) 21 years , c ) 22 years , d ) 23 years , e ) 26 years | e | divide(subtract(add(29, add(29, 3)), multiply(3, 3)), const_2) | add(n1,n2)|multiply(n2,n2)|add(n1,#0)|subtract(#2,#1)|divide(#3,const_2)| | general |
the captain of a cricket team of 11 members is 29 years old and the wicket keeper is 3 years older . if the ages of these two are excluded , the average age of the remaining players is one year less than the average age of the whole team . what is the average age of the team ? | "the triangle with sides 31 cm , 29 cm and 15 cm is right angled , where the hypotenuse is 31 cm . area of the triangle = 1 / 2 * 29 * 15 = 217.5 cm 2 answer : e" | a ) 220.75 cm 2 , b ) 258 cm 2 , c ) 225.50 cm 2 , d ) 222.25 cm 2 , e ) 217.5 cm 2 | e | divide(multiply(29, 15), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
if the sides of a triangle are 31 cm , 29 cm and 15 cm , what is its area ? | "let us say the ratio of the quantities of cheaper and dearer varieties = x : y by the rule of allegation , x / y = ( 8.75 - 7.50 ) / ( 7.50 - 6.5 ) = 5 / 4 answer : c" | a ) 5 / 6 , b ) 5 / 9 , c ) 5 / 4 , d ) 5 / 3 , e ) 7 / 6 | c | divide(divide(subtract(8.75, 7.50), subtract(8.75, 6.5)), subtract(const_1, divide(subtract(8.75, 7.50), subtract(8.75, 6.5)))) | subtract(n1,n2)|subtract(n1,n0)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)| | other |
in what ratio should a variety of rice costing rs . 6.5 per kg be mixed with another variety of rice costing rs . 8.75 per kg to obtain a mixture costing rs . 7.50 per kg ? | "drawing two balls of same color from seven green balls can be done in ⁷ c ₂ ways . similarly from eight white balls two can be drawn in ⁸ c ₂ ways . p = ⁷ c ₂ / ¹ ⁵ c ₂ + ⁸ c ₂ / ¹ ⁵ c ₂ = 7 / 15 answer : e" | a ) 7 / 16 , b ) 7 / 12 , c ) 7 / 19 , d ) 7 / 12 , e ) 7 / 15 | e | add(multiply(divide(8, add(7, 8)), divide(subtract(8, const_1), subtract(add(7, 8), const_1))), multiply(divide(7, add(7, 8)), divide(subtract(7, const_1), subtract(add(7, 8), const_1)))) | add(n0,n1)|subtract(n1,const_1)|subtract(n0,const_1)|divide(n1,#0)|divide(n0,#0)|subtract(#0,const_1)|divide(#1,#5)|divide(#2,#5)|multiply(#3,#6)|multiply(#4,#7)|add(#8,#9)| | other |
a bag contains 7 green and 8 white balls . if two balls are drawn simultaneously , the probability that both are of the same colour is | "explanation : manager ' s monthly salary rs . ( 1500 * 21 - 1400 * 20 ) = rs . 3500 . answer : e" | a ) 3600 , b ) 3890 , c ) 88798 , d ) 2789 , e ) 3500 | e | subtract(multiply(add(1400, 100), add(20, const_1)), multiply(1400, 20)) | add(n1,n2)|add(n0,const_1)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general |
the average monthly salary of 20 employees in an organisation is rs . 1400 . if the manager ' s salary is added , then the average salary increases by rs . 100 . what is the manager ' s monthly salary ? | "let 1 man ' s 1 day work = x and 1 woman ' s 1 day work = y . then , 4 x + 6 y = 1 / 8 and 3 x + 7 y = 1 / 10 solving these two equations , we get : x = 11 / 400 and y = 1 / 400 1 woman ' s 1 day work = ( 1 / 400 * 10 ) = 1 / 40 . hence , 10 women will complete the work in 40 days . answer : b" | a ) 21 days , b ) 40 days , c ) 27 days , d ) 18 days , e ) 17 days | b | inverse(multiply(divide(subtract(divide(const_1, 10), multiply(3, divide(subtract(divide(const_1, 8), multiply(divide(6, 7), divide(const_1, 10))), subtract(4, multiply(3, divide(6, 7)))))), 7), 8)) | divide(const_1,n5)|divide(const_1,n2)|divide(n1,n4)|multiply(#2,#0)|multiply(n3,#2)|subtract(#1,#3)|subtract(n0,#4)|divide(#5,#6)|multiply(n3,#7)|subtract(#0,#8)|divide(#9,n4)|multiply(n2,#10)|inverse(#11)| | physics |
4 men and 6 women can complete a work in 8 days , while 3 men and 7 women can complete it in 10 days . in how many days will 10 women complete it ? | "purchase price = 180 selling price = x 180 + 0.4 * x = x 0.6 * x = 180 x = 300 profit = 300 - 180 = 120 answer : d" | a ) $ 40 , b ) $ 60 , c ) $ 80 , d ) $ 120 , e ) $ 100 | d | divide(multiply(subtract(divide(180, subtract(const_1, divide(40, const_100))), 180), const_100), 180) | divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)|subtract(#2,n0)|multiply(#3,const_100)|divide(#4,n0)| | gain |
a furniture dealer purchased a desk for $ 180 and then set the selling price equal to the purchase price plus a markup that was 40 % of the selling price . if the dealer sold the desk at the selling price , what was the amount of the dealer ' s gross profit from the purchase and the sale of the desk ? | "length = speed * time speed = l / t s = 400 / 10 s = 40 m / sec speed = 40 * 18 / 5 ( to convert m / sec in to kmph multiply by 18 / 5 ) speed = 144 kmph answer : b" | a ) 165 kmph , b ) 144 kmph , c ) 172 kmph , d ) 175 kmph , e ) 178 kmph | b | divide(divide(400, const_1000), divide(10, const_3600)) | divide(n0,const_1000)|divide(n1,const_3600)|divide(#0,#1)| | physics |
a train 400 m long can cross an electric pole in 10 sec and then find the speed of the train ? | "ds = 30 us = 14 s = ? s = ( 30 - 14 ) / 2 = 8 kmph answer : e" | a ) 1 kmph , b ) 4 kmph , c ) 5 kmph , d ) 7 kmph , e ) 8 kmph | e | divide(subtract(30, 14), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
a man can row his boat with the stream at 30 km / h and against the stream in 14 km / h . the man ' s rate is ? | explanation : work done by 4 men and 6 women in 1 day = 1 / 8 work done by 3 men and 7 women in 1 day = 1 / 10 let 1 man does m work in 1 day and 1 woman does w work in 1 day . the above equations can be written as 4 m + 6 w = 1 / 8 - - - ( 1 ) 3 m + 7 w = 1 / 10 - - - ( 2 ) solving equation ( 1 ) and ( 2 ) , we get m ... | a ) 50 , b ) 40 , c ) 30 , d ) 20 , e ) 10 | b | inverse(multiply(divide(subtract(divide(const_1, 8), multiply(4, divide(subtract(divide(const_1, 10), multiply(divide(7, 6), divide(const_1, 8))), subtract(3, multiply(4, divide(7, 6)))))), 6), 10)) | divide(const_1,n5)|divide(const_1,n2)|divide(n1,n4)|multiply(#2,#0)|multiply(n3,#2)|subtract(#1,#3)|subtract(n0,#4)|divide(#5,#6)|multiply(n3,#7)|subtract(#0,#8)|divide(#9,n4)|multiply(n2,#10)|inverse(#11) | physics |
3 men and 7 women can complete a work in 10 days . but 4 men and 6 women need 8 days to complete the same work . in how many days will 10 women complete the same work ? | log ( 0.0000134 ) . since there are four zeros between the decimal point and the first significant digit , the characteristic is – 5 . answer : b | a ) 5 , b ) - 5 , c ) 6 , d ) - 6 , e ) 7 | b | floor(divide(log(0.0000134), log(const_10))) | log(n0)|log(const_10)|divide(#0,#1)|floor(#2) | other |
what is the characteristic of the logarithm of 0.0000134 ? | this is equivalent to : - 2 x * 4 y * 5 z = 16000 y / 2 = z ( given ) 2 x * 4 y * 5 y / 2 = 16000 2 x * y ^ 2 = 16000 / 10 2 x * y ^ 2 = 1600 now from options given we will figure out which number will divide 800 and gives us a perfect square : - which gives us x = 2 as 2 * 2 * y ^ 2 = 1600 y ^ 2 = 400 y = 20 number of... | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | b | divide(multiply(multiply(power(2, 4), power(2, const_3)), power(5, const_3)), multiply(power(const_2, multiply(2, const_3)), power(5, const_3))) | multiply(n0,const_3)|power(n0,n1)|power(n0,const_3)|power(n2,const_3)|multiply(#1,#2)|power(const_2,#0)|multiply(#4,#3)|multiply(#5,#3)|divide(#6,#7) | general |
in the game of dubblefud , red chips , blue chips and green chips are each worth 2 , 4 and 5 points respectively . in a certain selection of chips , the product of the point values of the chips is 16000 . if the number of blue chips in this selection doubles the number of green chips , how many red chips are in the sel... | "10 % interest per annum will be 5 % interest half yearly for 3 terms ( 1 1 / 2 years ) so compound interest = 2000 [ 1 + ( 5 / 100 ) ] ^ 3 - 2000 = 2000 [ ( 21 / 20 ) ^ 3 - 1 ] = 2000 ( 9261 - 8000 ) / 8000 = 2 * 1261 / 8 = 315 answer : d" | a ) rs . 473 , b ) rs . 374 , c ) rs . 495 , d ) rs . 315 , e ) none of the above | d | subtract(multiply(2000, power(add(1, divide(divide(10, 2), const_100)), multiply(add(1, divide(1, 2)), 2))), 2000) | divide(n1,n4)|divide(n2,n4)|add(n2,#1)|divide(#0,const_100)|add(#3,n2)|multiply(#2,n4)|power(#4,#5)|multiply(n0,#6)|subtract(#7,n0)| | gain |
compound interest of rs . 2000 at 10 % per annum for 1 1 / 2 years will be ( interest compounded half yearly ) . | "cone curved surface area = ï € rl 22 / 7 ã — 49 ã — 35 = 154 ã — 35 = 5390 m ( power 2 ) answer is b ." | a ) 5160 , b ) 5390 , c ) 6430 , d ) 6720 , e ) 7280 | b | volume_cone(49, 35) | volume_cone(n0,n1)| | geometry |
the radius of a cone is 49 m , slant height is 35 m . find the curved surface area ? | "perimeter = distance covered in 10 min . = ( 12000 / 60 ) x 10 m = 2000 m . let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 2000 or x = 200 . length = 600 m and breadth = 400 m . area = ( 600 x 400 ) m 2 = 240000 m 2 . answer : e" | a ) 153601 , b ) 153600 , c ) 153602 , d ) 153603 , e ) 240000 | e | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 10), const_1000), add(3, 2)), const_2), 2)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 10 minutes , then the area of the park ( in sq . m ) is : | "speed = ( 11 / 10 * 60 ) km / hr = ( 66 * 5 / 18 ) m / sec = 55 / 3 m / sec . length of the train = 55 / 3 * 6 = 110 m . answer : c" | a ) m , b ) m , c ) m , d ) m , e ) m | c | divide(11, subtract(divide(11, 10), 6)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
a train covers a distance of 11 km in 10 min . if it takes 6 sec to pass a telegraph post , then the length of the train is ? | ( 38 g + 20 b ) / ( g + b ) = 30 38 g + 20 b = 30 ( g + b ) 8 g = 10 b b / g = 4 / 5 the answer is d . | a ) 1 / 2 , b ) 2 / 3 , c ) 3 / 4 , d ) 4 / 5 , e ) 5 / 6 | d | divide(30, 38) | divide(n2,n1) | general |
on average , the boys in the class have 20 pencils and the girls have 38 pencils . if the overall class average is 30 pencils , what is the ratio of boys to girls in the class ? | "i find it much easier to understand with real numbers , so choose ( almost ) any numbers to replace m , n and p : in a certain village , m 200 litres of water are required per household per month . at this rate , if there aren 5 households in the village , how long ( in months ) willp 2000 litres of water last ? water... | a ) 9 , b ) 5 , c ) 6 , d ) 2 , e ) 4 | d | divide(2000, multiply(200, 5)) | multiply(n0,n1)|divide(n2,#0)| | gain |
in a certain village , 200 litres of water are required per household per month . at this rate , if there are 5 households in the village , how long ( in months ) will 2000 litres of water last ? | "as two numbers are prime , only two options satisfy ie option c and d . but option c will not make the product of numbers i . e 99 answer : d" | a ) 8 and 12 , b ) 14 and 6 , c ) 19 and 1 , d ) 11 and 9 , e ) 12 and 9 | d | add(99, 20) | add(n0,n1)| | physics |
sum of two numbers prime to each other is 20 and their l . c . m . is 99 . what are the numbers ? | "s = 180 / 20 * 18 / 5 = 32 kmph answer : c" | a ) 37 kmph , b ) 35 kmph , c ) 32 kmph , d ) 38 kmph , e ) 36 kmph | c | multiply(const_3_6, divide(180, 20)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
a train 180 m in length crosses a telegraph post in 20 seconds . the speed of the train is ? | "solution s . i . = rs . ( 956 - 925 ) = rs . 31 rate = ( 100 x 31 / 925 x 3 ) = 124 / 111 % new rate = ( 124 / 111 + 4 ) % = 568 / 111 % new s . i . = rs . ( 925 x 568 / 111 x 3 / 100 ) rs . 142 ∴ new amount = rs . ( 925 + 142 ) = rs . 1067 . answer c" | a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1067 , d ) data inadequate , e ) none of these | c | add(925, divide(multiply(multiply(925, add(divide(multiply(subtract(956, 925), const_100), multiply(925, 3)), 4)), 3), const_100)) | multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7)| | gain |
rs . 925 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 925 become in 3 years ? | "the total number of the people in the room must be a multiple of both 7 and 12 ( in order 3 / 7 and 5 / 12 of the number to be an integer ) , thus the total number of the people must be a multiple of lcm of 7 and 12 , which is 84 . since , the total number of the people in the room is greater than 50 and less than 100... | a ) 21 , b ) 35 , c ) 36 , d ) 60 , e ) 65 | c | divide(multiply(multiply(7, 12), 3), 7) | multiply(n1,n4)|multiply(n0,#0)|divide(#1,n1)| | general |
exactly 3 / 7 of the people in the room are under the age of 21 , and exactly 5 / 12 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 ? | "3 x * 2 x = 2460 = > x = 20.24 2 ( 79.76 + 50 ) = 259.52 m 259.52 * 1 / 2 = rs . 129.76 answer : c" | a ) s . 122 , b ) s . 129 , c ) s . 129.76 , d ) s . 120 , e ) s . 121 | c | divide(multiply(50, rectangle_perimeter(sqrt(divide(multiply(2460, 2), 3)), divide(2460, sqrt(divide(multiply(2460, 2), 3))))), const_100) | multiply(n1,n2)|divide(#0,n0)|sqrt(#1)|divide(n2,#2)|rectangle_perimeter(#3,#2)|multiply(n3,#4)|divide(#5,const_100)| | physics |
sides of a rectangular park are in the ratio 3 : 2 and its area is 2460 sq m , the cost of fencing it at 50 ps per meter is ? | "to solve this problem , all you have to do is take every even number between 24 and 50 and add them together . so we have 25 + 27 + 29 + 31 + 33 + 35 + 37 + 39 + 41 + 43 + 45 + 47 + 49 , which is 481 . final answer : b" | a ) 592 , b ) 481 , c ) 330 , d ) 475 , e ) 483 | b | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 24), add(const_2, const_4)) | add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)| | general |
what is the sum of all the odd numbers between 24 and 50 , inclusive ? | r + b + c + 14 + 15 = 12 * 5 = 60 = > r + b + c = 60 - 29 = 31 r + b + c + 29 = 31 + 29 = 60 average = 60 / 4 = 15 answer d | a ) 12 , b ) 13 , c ) 14 , d ) 15 , e ) 16 | d | divide(add(subtract(multiply(add(const_4, const_1), 12), add(14, 15)), 29), const_4) | add(const_1,const_4)|add(n0,n1)|multiply(n2,#0)|subtract(#2,#1)|add(n3,#3)|divide(#4,const_4) | general |
if the average of r , b , c , 14 and 15 is 12 . what is the average value of r , b , c and 29 | expl : 15 % of a i = 30 % of b = 15 a / 100 = 30 b / 100 = 2 / 1 = 2 : 1 answer : e | a ) 1 : 4 , b ) 4 : 3 , c ) 6 : 7 , d ) 3 : 5 , e ) 2 : 1 | e | divide(divide(30, const_100), divide(15, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(#0,#1) | gain |
if 15 % of a is the same as 30 % of b , then a : b is : | "p ( a and b ) = 1 / 2 * 1 / 5 = 1 / 10 the answer is b ." | a ) 1 / 18 , b ) 1 / 10 , c ) 1 / 5 , d ) 1 / 3 , e ) 1 / 2 | b | multiply(divide(subtract(2, const_1), multiply(subtract(2, const_1), 2)), divide(multiply(subtract(2, const_1), const_2), multiply(subtract(2, const_1), 2))) | subtract(n0,const_1)|multiply(n0,#0)|multiply(#0,const_2)|divide(#0,#1)|divide(#2,#1)|multiply(#3,#4)| | physics |
let a be the event that a randomly selected two digit number is divisible by 2 and let b be the event that a randomly selected two digit number is divisible by 5 . what is p ( a and b ) ? | "required angle = 240 – 24 × ( 11 / 2 ) = 240 – 132 = 108 ° answer d" | a ) 100 ° , b ) 107 ° , c ) 106 ° , d ) 108 ° , e ) none of these | d | subtract(multiply(8, multiply(const_3, const_2)), 2) | multiply(const_2,const_3)|multiply(n1,#0)|subtract(#1,n0)| | geometry |
what is the angle between the 2 hands of the clock at 8 : 24 pm ? | let x per minute be the speed of c and y per minute be the speed of d . after meeting at a point , c travels for 32 mins and d travels for 50 mins . so distance covered by each of them post point of crossing c = 32 x and d = 50 y the distance covered by c and d before they cross each would be distance covered by d and ... | a ) 90 , b ) 80 , c ) 75 , d ) 60 , e ) 65 | a | add(sqrt(multiply(50, 32)), 50) | multiply(n0,n1)|sqrt(#0)|add(n1,#1) | physics |
two friends c and d leave point c and point d simultaneously and travel towards point d and point c on the same route at their respective constant speeds . they meet along the route and immediately proceed to their respective destinations in 32 minutes and 50 minutes respectively . how long will d take to cover the ent... | "a 150 125 % of 120 % of a = 225 125 / 100 * 120 / 100 * a = 225 a = 225 * 2 / 3 = 150 ." | a ) 150 , b ) 120 , c ) 130 , d ) 160 , e ) 210 | a | divide(225, multiply(add(const_1, divide(20, const_100)), add(const_1, divide(25, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|add(#1,const_1)|multiply(#2,#3)|divide(n2,#4)| | gain |
a sells a cricket bat to b at a profit of 20 % . b sells it to c at a profit of 25 % . if c pays $ 225 for it , the cost price of the cricket bat for a is : | "let x be the total worker then 0.5 x = female worker and 0.5 x is male worker then 20 male worker added 05 x / ( 0.5 x + 20 ) = 50 / 100 or 50 x = 50 * ( 0.5 x + 100 ) = 25 x + 5000 or 25 x = 5000 , x = 5000 / 25 = 200 total worker = 200 + 20 = 220 b" | a ) 225 , b ) 220 , c ) 230 , d ) 235 , e ) 240 | b | add(divide(multiply(divide(50, const_100), 20), subtract(divide(const_60.0, const_100), divide(50, const_100))), 20) | divide(n2,const_100)|divide(const_60.0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)| | gain |
the workforce of company x is 50 % female . the company hired 20 additional male workers , and as a result , the percent of female workers dropped to 50 % . how many employees did the company have after hiring the additional male workers ? | "explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 5 ^ 2 = 9 / 25 answer : e" | a ) 3 , b ) 6 , c ) 1 / 9 , d ) 1 / 3 , e ) 9 / 25 | e | divide(multiply(1, power(3, const_2)), power(5, const_2)) | power(n0,const_2)|power(n2,const_2)|multiply(n1,#0)|divide(#2,#1)| | general |
x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 5 will be equal to : | notice that 7 play both baseball and cricket does not mean that out of those 7 , some does not play football too . the same for cricket / football and baseball / football . [ color = # ffff 00 ] { total } = { baseball } + { cricket } + { football } - { hc + ch + hf } + { all three } + { neither } for more checkadvanced... | a ) 10 , b ) 46 , c ) 67 , d ) 68 , e ) 446 | a | add(subtract(5, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18))), add(subtract(7, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18))), subtract(4, subtract(50, add(subtract(add(add(20, 15), 11), add(add(7, 4), 5)), 18))))) | add(n1,n2)|add(n4,n5)|add(n3,#0)|add(n6,#1)|subtract(#2,#3)|add(n7,#4)|subtract(n0,#5)|subtract(n4,#6)|subtract(n5,#6)|subtract(n6,#6)|add(#7,#8)|add(#10,#9) | other |
in a class of 50 students , 20 play baseball , 15 play cricket and 11 play football . 7 play both baseball and cricket , 4 play cricket and football and 5 play baseball and football . if 18 students do not play any of these given sports , how many students play exactly two of these sports ? | "90 * 10 / 60 = 15 kmph answer : a" | a ) 15 , b ) 87 , c ) 99 , d ) 77 , e ) 55 | a | multiply(divide(10, const_60), 90) | divide(n1,const_60)|multiply(n0,#0)| | physics |
the speed of a train is 90 kmph . what is the distance covered by it in 10 minutes ? | "speed in still water = ( 11 + 5 ) / 2 = 8 kmph ans - c" | a ) 2 kmph , b ) 3 kmph , c ) 8 kmph , d ) 9 kmph , e ) 7 kmph | c | stream_speed(11, 5) | stream_speed(n0,n1)| | physics |
in one hour , a boat goes 11 km / hr along the stream and 5 km / hr against the stream . the speed of the boat in still water ( in km / hr ) is : | "sq rt ( 5 x / 3 ) = x = > 5 x / 3 = x ^ 2 = > x = 5 / 3 ans - d" | a ) 9 / 4 , b ) 3 / 2 , c ) 4 / 3 , d ) 5 / 3 , e ) 1 / 2 | d | divide(5, 3) | divide(n0,n1)| | general |
a positive number x is multiplied by 5 , and this product is then divided by 3 . if the positive square root of the result of these two operations equals x , what is the value of x ? | "suppose commodity x will cost 40 paise more than y after z years . then , ( 4.20 + 0.40 z ) - ( 6.30 + 0.15 z ) = 0.40 0.25 z = 0.40 + 2.10 z = 2.50 / 0.25 = 250 / 25 = 10 . therefore , x will cost 40 paise more than y 10 years after 2001 i . e . , 2011 . answer is d ." | a ) 2010 , b ) 2001 , c ) 2012 , d ) 2011 , e ) 2009 | d | add(2001, divide(add(divide(40, const_100), subtract(6.30, 4.20)), subtract(divide(40, const_100), divide(15, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|subtract(n4,n3)|add(#0,#2)|subtract(#0,#1)|divide(#3,#4)|add(n2,#5)| | general |
the price of commodity x increases by 40 paise every year , while the price of commodity y increases by 15 paise every year . if in 2001 , the price of commodity x was rs . 4.20 and that of y was rs . 6.30 , in which year commodity x will cost 40 paise more than the commodity y ? | "a + b = 60 , a = 2 b 2 b + b = 60 = > b = 20 then a = 40 . 5 years , their ages will be 48 and 28 . sum of their ages = 48 + 28 = 76 . answer : d" | a ) 50 , b ) 60 , c ) 70 , d ) 76 , e ) 90 | d | add(add(multiply(divide(60, 8), const_2), 8), add(divide(60, 8), 8)) | divide(n0,n1)|add(#0,n1)|multiply(#0,const_2)|add(#2,n1)|add(#3,#1)| | general |
the sum of the present ages of two persons a and b is 60 . if the age of a is twice that of b , find the sum of their ages 8 years hence ? | "explanation : capital = rs . x , then 5 / 7 x = 61.5 x = 87.86 answer : b ) rs . 87.86" | a ) 22.378 , b ) 87.86 , c ) 246.0 , d ) 78.88 , e ) 127.71 | b | divide(61.50, divide(const_4, 7)) | divide(const_4,n3)|divide(n4,#0)| | gain |
a money lender finds that due to a fall in the annual rate of interest from 8 % to 7 2 / 7 % his yearly income diminishes by rs . 61.50 . his capital is | "let the numbers be x and y . then , xy = 468 and x 2 + y 2 = 289 . ( x + y ) 2 = x 2 + y 2 + 2 xy = 289 + ( 2 x 468 ) = 1225 x + y = 35 . option e" | a ) a ) 23 , b ) b ) 25 , c ) c ) 27 , d ) d ) 31 , e ) e ) 35 | e | sqrt(add(power(sqrt(subtract(289, multiply(const_2, 468))), const_2), multiply(const_4, 468))) | multiply(n0,const_4)|multiply(n0,const_2)|subtract(n1,#1)|sqrt(#2)|power(#3,const_2)|add(#0,#4)|sqrt(#5)| | general |
the product of two numbers is 468 and the sum of their squares is 289 . the sum of the number is ? | "soap : alcohol initial ratio soap : alcohol : water - - > 4 : 20 : 60 initial soap : alcohol = 4 / 20 = 4 : 20 after doubled soap : alcohol = 2 * 4 / 20 = 8 : 20 initial soap : water = 4 / 60 = 4 : 60 after halved soap : water : 1 / 2 * 4 / 60 = 2 / 60 = 2 : 60 after soap : alcohol : water - - > 8 : 20 : 240 - - > 2 :... | a ) 1200 , b ) 1250 , c ) 1300 , d ) 1400 , e ) 1450 | a | divide(divide(divide(divide(divide(volume_rectangular_prism(100, 60, 20), const_3), const_2), 4), 4), 4) | volume_rectangular_prism(n1,n2,n3)|divide(#0,const_3)|divide(#1,const_2)|divide(#2,n0)|divide(#3,n0)|divide(#4,n0)| | geometry |
the ratio , by volume , of soap to alcohol to water in a certain solution is 4 : 20 : 60 . the solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved . if the altered solution will contain 100 cubic centimeters of alcohol , how many cubic centimeters of water... | "since f ( n ) = f ( n - 1 ) - n then : f ( 6 ) = f ( 5 ) - 6 and f ( 5 ) = f ( 4 ) - 5 . as given that f ( 4 ) = 13 then f ( 5 ) = 13 - 5 = 8 - - > substitute the value of f ( 5 ) back into the first equation : f ( 6 ) = f ( 5 ) - 6 = 8 - 6 = 2 . answer : d . questions on funtions to practice :" | a ) - 1 , b ) 0 , c ) 1 , d ) 2 , e ) 4 | d | subtract(subtract(13, add(1, 4)), 6) | add(n0,n1)|subtract(n2,#0)|subtract(#1,n3)| | general |
if n is an integer , f ( n ) = f ( n - 1 ) - n and f ( 4 ) = 13 . what is the value of f ( 6 ) ? | area of semicircle = ½ π r 2 = ½ × 22 ⁄ 7 × 7 × 7 = 77 m 2 answer b | ['a ) 154 sq metres', 'b ) 77 sq metres', 'c ) 308 sq metres', 'd ) 22 sq metres', 'e ) none of these'] | b | divide(circle_area(divide(14, const_2)), const_2) | divide(n0,const_2)|circle_area(#0)|divide(#1,const_2) | geometry |
what will be the area of a semi - circle of 14 metres diameter ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 2200 ( 100 / 110 ) = rs . 2000 answer : b" | a ) 2299 , b ) 2000 , c ) 2670 , d ) 6725 , e ) 2601 | b | divide(2200, add(const_1, divide(10, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
the owner of a furniture shop charges his customer 10 % more than the cost price . if a customer paid rs . 2200 for a computer table , then what was the cost price of the computer table ? | "say the second solution ( which was 1 / 4 th of total ) was x % salt , then 3 / 4 * 0.1 + 1 / 4 * x = 1 * 0.16 - - > x = 0.34 . alternately you can consider total solution to be 100 liters and in this case you ' ll have : 75 * 0.1 + 25 * x = 100 * 0.16 - - > x = 0.34 . answer : b ." | a ) 24 % , b ) 34 % , c ) 22 % , d ) 18 % , e ) 8.5 % | b | multiply(subtract(multiply(divide(16, const_100), const_4), subtract(multiply(divide(10, const_100), const_4), divide(10, const_100))), const_100) | divide(n1,const_100)|divide(n0,const_100)|multiply(#0,const_4)|multiply(#1,const_4)|subtract(#3,#1)|subtract(#2,#4)|multiply(#5,const_100)| | gain |
one fourth of a solution that was 10 % salt by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight . the second solution was what percent salt by weight ? | "the arithmetic mean of a and b = ( a + b ) / 2 = 45 - - a + b = 90 - - 1 similarly for b + c = 170 - - 2 subtracting 1 from 2 we have c - a = 80 ; answer : b" | a ) 25 , b ) 80 , c ) 90 , d ) 140 , e ) it can not be determined from the information given | b | subtract(multiply(85, const_2), multiply(45, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
if the average ( arithmetic mean ) of a and b is 45 and the average of b and c is 85 , what is the value of c â ˆ ’ a ? | the required number of working hours per day x , more pumps , less working hours per day ( indirect ) less days , more working hours per day ( indirect ) pumps 4 : 3 , days 1 : 2 } : : 8 : x therefore 4 * 1 * x = 3 * 2 * 8 , x = ( 3 * 2 * 8 ) / 4 x = 12 correct answer ( d ) | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | d | divide(multiply(multiply(3, 8), 2), 4) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n3) | physics |
3 pumps , working 8 hours a day , can empty a tank in 2 days . how many hours a day must 4 pumps work to empty the tank in 1 day ? | since both peter and david invested the same amount of money at the same rate , they would earn same interest per year . david invested for one year more than peter and hence he got interest amount for one more year . interest earned per year = amount received by david - amount received by peter = 854 - 810 = 44 intere... | a ) 670 , b ) 683 , c ) 698 , d ) 744 , e ) 700 | b | subtract(810, multiply(divide(subtract(854, 810), subtract(divide(4, const_100), divide(3, const_100))), divide(3, const_100))) | divide(n3,const_100)|divide(n1,const_100)|subtract(n2,n0)|subtract(#0,#1)|divide(#2,#3)|multiply(#4,#1)|subtract(n0,#5) | gain |
peter invests a sum of money and gets back an amount of $ 810 in 3 years . david invests an equal amount of money and gets an amount of $ 854 in 4 years . if both amounts were invested at the same rate ( simple interest ) what was the sum of money invested ? | "800 * 9.2 7360.0 gm 7.36 kg answer : e" | a ) 6.6 kg , b ) 6.8 kg , c ) 6.7 kg , d ) 6.9 kg , e ) 7.36 kg | e | divide(multiply(9.2, 800), const_1000) | multiply(n0,n1)|divide(#0,const_1000)| | general |
a envelop weight 9.2 gm , if 800 of these envelop are sent with an advertisement mail . how much wieght ? | "1022 ã · 25 = 40 with remainder = 22 22 + 3 = 25 . hence 3 should be added to 1022 so that the sum will be divisible by 25 answer : option b" | a ) 4 , b ) 3 , c ) 2 , d ) 0 , e ) 5 | b | subtract(25, reminder(1022, 25)) | reminder(n0,n1)|subtract(n1,#0)| | general |
what least number should be added to 1022 , so that the sum is completely divisible by 25 ? | "a hostel had provisions for 250 men for 44 days if 50 men leaves the hostel , remaining men = 250 - 50 = 200 we need to find out how long the food will last for these 200 men . let the required number of days = x days more men , less days ( indirect proportion ) ( men ) 250 : 200 : : x : 44 250 × 44 = 200 x 5 × 44 = 4... | a ) 55 , b ) 40 , c ) 50 , d ) 60 , e ) 65 | a | divide(multiply(250, 44), subtract(250, 50)) | multiply(n0,n1)|subtract(n0,n2)|divide(#0,#1)| | gain |
a hostel had provisions for 250 men for 44 days . if 50 men left the hostel , how long will the food last at the same rate ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 8339 ( 100 / 124 ) = rs . 6725 . answer : c" | a ) rs . 6825 , b ) rs . 6721 , c ) rs . 6725 . , d ) rs . 4298 , e ) rs . 6729 | c | divide(8339, add(const_1, divide(24, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
the owner of a furniture shop charges his customer 24 % more than the cost price . if a customer paid rs . 8339 for a computer table , then what was the cost price of the computer table ? | "63 + 5 * 12 / ( 180 / 3 ) = 63 + 5 * 12 / ( 60 ) = 63 + ( 5 * 12 ) / 60 = 63 + 1 = 64 . answer : d" | a ) 22 , b ) 77 , c ) 29 , d ) 64 , e ) 21 | d | add(63, divide(multiply(5, 12), divide(180, 3))) | divide(n3,n4)|multiply(n1,n2)|divide(#1,#0)|add(n0,#2)| | general |
63 + 5 * 12 / ( 180 / 3 ) = ? | 12 * 1.75 + 0.45 * 12 * 55 = 318 hence - a | a ) 318 $ , b ) 380 $ , c ) 420 $ , d ) 450 $ , e ) 480 $ | a | multiply(multiply(0.45, 55), 12) | multiply(n1,n3)|multiply(n2,#0)| | general |
in a fuel station the service costs $ 1.75 per car , every liter of fuel costs 0.45 $ . assuming that a company owns 12 cars and that every fuel tank contains 55 liters and they are all empty , how much money total will it cost to fuel all cars ? | "since x is an integer , ( - 6 ) ^ 2 x is always positive . so , 6 ^ 2 x = 6 ^ ( 7 + x ) 2 x = 7 + x x = 7 answer : e" | a ) 5 , b ) 4 , c ) 3 , d ) 8 , e ) 7 | e | divide(7, 6) | divide(n3,n0)| | general |
if ( - 6 ) ^ ( 2 x ) = 6 ^ ( 7 + x ) and x is an integer , what is the value of x ? | "given ; 4 duck = 3 hen ; or , duck / hen = 3 / 4 ; let hen ' s 1 leap = 4 meter and ducks 1 leap = 3 meter . then , ratio of speed of hen and duck = 4 * 3 / 3 * 2 = 2 : 1 ' ' answer : 2 : 1 ;" | a ) 2 : 1 , b ) 3 : 4 , c ) 4 : 3 , d ) 1 : 4 , e ) 5 : 6 | a | divide(divide(3, 2), divide(3, 4)) | divide(n0,n1)|divide(n3,n2)|divide(#0,#1)| | other |
a hen leaps 3 leaps for every 2 leaps of a duck , but 4 leaps of the duck are equal to 3 leaps of the hen . what is the ratio of the speed of the hen to that of the duck ? | "let hari ’ s capital be rs . x . then , 3640 * 12 / 7 x = 2 / 3 = > 14 x = 131040 = > x = 9360 . answer : e" | a ) s . 7500 , b ) s . 8000 , c ) s . 8500 , d ) s . 9000 , e ) s . 9360 | e | divide(divide(3640, subtract(const_1, divide(5, const_12))), divide(2, 3)) | divide(n1,const_12)|divide(n2,n3)|subtract(const_1,#0)|divide(n0,#2)|divide(#3,#1)| | other |
praveen starts business with rs . 3640 and after 5 months , hari joins with praveen as his partner . after a year , the profit is divided in the ratio 2 : 3 . what is hari ’ s contribution in the capital ? | "let x be the percentage of dressing p in the new dressing . 0.3 x + 0.1 ( 1 - x ) = 0.20 0.2 x = 0.10 x = 0.5 = 50 % the answer is b ." | a ) 60 % , b ) 50 % , c ) 40 % , d ) 30 % , e ) 20 % | b | divide(subtract(30, 10), subtract(20, 10)) | subtract(n0,n2)|subtract(n4,n2)|divide(#0,#1)| | gain |
salad dressing p is made up of 30 % vinegar and 70 % oil , and salad dressing q contains 10 % vinegar and 90 % oil . if the two dressings are combined to produce a salad dressing that is 20 % vinegar , dressing p comprises what percentage of the new dressing ? | 1 ) i figured there are 101 integers ( 300 - 200 + 1 = 101 ) . since the set begins with an even and ends with an even , there are 51 evens . 2 ) question says integers are not divisible by 2 , leaving all of the odds ( 101 - 51 = 50 integers ) . 3 ) question says integers are not divisible by 5 , removing all the inte... | a ) 3 , b ) 16 , c ) 75 , d ) 24 , e ) 26 | e | subtract(subtract(subtract(add(subtract(300, 200), const_1), add(subtract(divide(300, 2), divide(200, 2)), const_1)), floor(add(subtract(add(subtract(divide(300, 3), divide(200, 3)), const_1), add(add(const_10, 5), 2)), const_1))), subtract(add(subtract(divide(300, 5), divide(200, 5)), const_1), add(const_10, 5))) | add(n4,const_10)|divide(n1,n2)|divide(n0,n2)|divide(n1,n3)|divide(n0,n3)|divide(n1,n4)|divide(n0,n4)|subtract(n1,n0)|add(#7,const_1)|add(n2,#0)|subtract(#1,#2)|subtract(#3,#4)|subtract(#5,#6)|add(#10,const_1)|add(#11,const_1)|add(#12,const_1)|subtract(#8,#13)|subtract(#14,#9)|subtract(#15,#0)|add(#17,const_1)|floor(#19... | other |
how many positive integers q between 200 and 300 ( both inclusive ) are not divisible by 2 , 3 or 5 ? | "solution s . i . = rs . ( 956 - 850 ) = rs . 106 rate = ( 100 x 106 / 850 x 3 ) = 212 / 51 % new rate = ( 212 / 51 + 4 ) % = 416 / 51 % new s . i . = rs . ( 850 x 416 / 51 x 3 / 100 ) rs . 208 . ∴ new amount = rs . ( 850 + 208 ) = rs . 1058 . answer c" | a ) rs . 1020.80 , b ) rs . 1025 , c ) rs . 1058 , d ) data inadequate , e ) none of these | c | add(850, divide(multiply(multiply(850, add(divide(multiply(subtract(956, 850), const_100), multiply(850, 3)), 4)), 3), const_100)) | multiply(n0,n2)|subtract(n1,n0)|multiply(#1,const_100)|divide(#2,#0)|add(n3,#3)|multiply(n0,#4)|multiply(n2,#5)|divide(#6,const_100)|add(n0,#7)| | gain |
rs . 850 becomes rs . 956 in 3 years at a certain rate of simple interest . if the rate of interest is increased by 4 % , what amount will rs . 850 become in 3 years ? | "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 4 gms in 600,000 cm ^ 3 , y = 4 * 600,000 = 2 ,... | a ) 3 , 400,000 , b ) 2 , 800,000 , c ) 55,000 , d ) 28,000 , e ) 280 | a | 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, 4)), 4)), 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 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 4 grams per cubic centimeter , how many grams does a cubic meter of specialized sand combination weigh ? ( 1 meter = 100 centimeters ) | "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 * 10 ) = 1 / 600 of the video per second . the combined rate is 1 / 15 + 1 / 600 = 41 / 600 of the video per second . the time required to load the video is 600 / 41 = 14.63 seconds . the answ... | a ) 13.42 , b ) 13.86 , c ) 14.25 , d ) 14.63 , e ) 14.88 | d | 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 |
it takes ten 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 , rounded to the nearest hundredth ? | since the starting point is given as the $ 4000 scholarship , assume $ 4000 scholarships to be x by the given information , $ 2500 scholarships = 2 x and $ 1250 scholarships = 6 x gievn : total $ 1250 scholarships = $ 75000 6 x * 1250 = 75000 solve for x = 10 option d | a ) 5 , b ) 6 , c ) 9 , d ) 10 , e ) 15 | d | divide(divide(75000, 1250), multiply(const_2, 3)) | divide(n8,n0)|multiply(n5,const_2)|divide(#0,#1) | general |
a certain scholarship committee awarded scholarships in the amounts of $ 1250 , $ 2500 and $ 4000 . the committee awarded twice as many $ 2500 scholarships as $ 4000 and it awarded 3 times as many $ 1250 scholarships as $ 2500 scholarships . if the total of $ 75000 was awarded in $ 1250 scholarships , how many $ 4000 s... | "a dozen eggs cost as much as a pound of rice - - > 12 eggs = 1 pound of rice = 33 cents ; a half - liter of kerosene costs as much as 6 eggs - - > 6 eggs = 1 / 2 liters of kerosene . how many cents does a liter of kerosene cost - - > 1 liter of kerosene = 12 eggs = 12 / 12 * 33 = 33 cents . answer : c ." | a ) 0.33 , b ) 0.44 , c ) 33 , d ) 44 , e ) 55 | c | multiply(divide(divide(6, divide(const_1, const_2)), const_12), multiply(0.33, 100)) | divide(const_1,const_2)|multiply(n1,n2)|divide(n0,#0)|divide(#2,const_12)|multiply(#3,#1)| | general |
in a market , a dozen eggs cost as much as a pound of rice , and a half - liter of kerosene costs as much as 6 eggs . if the cost of each pound of rice is $ 0.33 , then how many cents does a liter of kerosene cost ? [ one dollar has 100 cents . ] | "let there be x pupils in the class . total increase in marks = ( x * 1 / 2 ) = x / 2 x / 2 = ( 73 - 40 ) = > x / 2 = 33 = > x = 66 . answer : c" | a ) 18 , b ) 82 , c ) 66 , d ) 27 , e ) 77 | c | multiply(subtract(73, 40), const_2) | subtract(n0,n1)|multiply(#0,const_2)| | general |
a pupil ' s marks were wrongly entered as 73 instead of 40 . due to the average marks for the class got increased by half . the number of pupils in the class is ? | "let the initial value of baseball card = 100 after first year , value of baseball card = ( 1 - 25 / 100 ) * 100 = 75 after second year , value of baseball card = ( 1 - 10 / 100 ) * 75 = 67.5 total percent decrease of the card ' s value over the two years = ( 100 - 67.5 ) / 100 * 100 % = 31.5 % answer c" | a ) 28 % , b ) 30 % , c ) 32.5 % , d ) 36 % , e ) 72 % | c | subtract(const_100, multiply(multiply(subtract(const_1, divide(10, const_100)), subtract(const_1, divide(25, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | gain |
a baseball card decreased in value 25 % in its first year and 10 % in its second year . what was the total percent decrease of the card ' s value over the two years ? | "total nos of ways in which we can choose n = 96 n ( n + 1 ) ( n + 2 ) will be divisible by 8 ? case 1 : n = odd then n + 2 = odd & n + 1 will be even i . e this needs get divided by 8 , hence is a multiple of 8 so we have 8 . . 96 = 12 multiples to fill the n + 1 pos hence 12 ways case 2 : n is even then n + 2 will be... | a ) 1 / 4 , b ) 3 / 8 , c ) 1 / 2 , d ) 5 / 8 , e ) 3 / 4 | d | divide(add(divide(96, 2), divide(96, 8)), 96) | divide(n1,n3)|divide(n1,n4)|add(#0,#1)|divide(#2,n1)| | general |
if an integer n is to be chosen at random from the integers 1 to 96 , inclusive , what is the probability that n ( n + 1 ) ( n + 2 ) will be divisible by 8 ? | "let car a = car that starts at 9 am car b = car that starts at 9 : 10 am time for which car a travels at speed of 40 m per hour = 1.5 hours distance travelled by car a = 40 * 1.5 = 60 miles since car b catches up car a at 10 : 30 , time = 60 mins = 1 hour speed of car b = 60 / ( 1 ) = 60 miles per hour answer b" | a ) 45 , b ) 60 , c ) 53 , d ) 55 , e ) 50 | b | divide(60, divide(add(multiply(subtract(10, 9), const_60), subtract(40, 10)), const_60)) | subtract(n4,n2)|subtract(n0,n4)|multiply(#0,const_60)|add(#2,#1)|divide(#3,const_60)|divide(n1,#4)| | physics |
a car going at 40 miles per hour set out on an 60 - mile trip at 9 : 00 a . m . exactly 10 minutes later , a second car left from the same place and followed the same route . how fast , in miles per hour , was the second car going if it caught up with the first car at 10 : 30 a . m . ? | "1 = 5,2 = 25,3 = 253,4 = 150,5 = 225 then 150 = ? 150 = 4 check the fourth eqn . answer : c" | a ) 1 , b ) 255 , c ) 4 , d ) 445 , e ) 235 | c | divide(subtract(subtract(225, multiply(multiply(add(const_4, const_2), add(const_4, const_2)), const_10)), 1), const_2) | add(const_2,const_4)|multiply(#0,#0)|multiply(#1,const_10)|subtract(n5,#2)|subtract(#3,n0)|divide(#4,const_2)| | general |
1 = 5,2 = 25,3 = 253,4 = 150,5 = 225 then 150 = ? | "given cash discount - 16 % profit - 25 % items sold - 60 price sold at = list price of 50 assume list price = $ 10 total invoice = $ 500 - 16 % cash discount = $ 420 let cost price of 60 items be x so total cost = 60 * x given the shopkeeper had a profit of 25 % 60 * x * 125 / 100 = 420 or x = $ 7 * 4 / 5 = $ 28 / 5 w... | a ) 50 % , b ) 60 % , c ) 70 % , d ) 75 % , e ) 78 + ( 4 / 7 ) % | e | multiply(subtract(divide(divide(divide(add(const_100, 25), const_100), subtract(const_1, divide(subtract(60, 50), 60))), divide(subtract(const_100, 16), const_100)), const_1), const_100) | add(n1,const_100)|subtract(n2,n3)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,n2)|divide(#2,const_100)|subtract(const_1,#4)|divide(#3,#6)|divide(#7,#5)|subtract(#8,const_1)|multiply(#9,const_100)| | gain |
a dealer offers a cash discount of 16 % and still makes a profit of 25 % when he further allows 60 articles to be sold at the cost price of 50 articles to a particular sticky bargainer . how much percent above the cost price were his articles listed ? | as per question = > n = 8 p + 1 for some integer p hence 3 n = > 24 q + 3 = > remainder = > 3 for some integer q hence b | a ) 1 , b ) 3 , c ) 7 , d ) 5 , e ) 6 | b | multiply(3, 1) | multiply(n1,n2) | general |
if n divided by 8 has a remainder of 1 , what is the remainder when 3 times n is divided by 8 ? | a 3 = 2197 = > a = 13 6 a 2 = 6 * 13 * 13 = 1014 answer : b | a ) 864 , b ) 1014 , c ) 1299 , d ) 1268 , e ) 1191 | b | surface_cube(cube_edge_by_volume(2197)) | cube_edge_by_volume(n0)|surface_cube(#0)| | geometry |
the volume of a cube is 2197 cc . find its surface . | "there are 15 business exec and in each handshake 2 business execs are involved . hence 15 c 2 = 105 also , each of 15 exec will shake hand with every 3 other chairmen for total of 45 handshake . total = 45 + 105 = 150 ans : a" | a ) 150 , b ) 131 , c ) 115 , d ) 90 , e ) 45 | a | add(divide(multiply(15, subtract(15, const_1)), const_2), multiply(15, 3)) | multiply(n0,n1)|subtract(n0,const_1)|multiply(n0,#1)|divide(#2,const_2)|add(#3,#0)| | geometry |
15 business executives and 3 chairmen meet at a conference . if each business executive shakes the hand of every other business executive and every chairman once , and each chairman shakes the hand of each of the business executives but not the other chairmen , how many handshakes would take place ? | "1 2 3 4 5 6 7 8 9 a b c d e f g h i j k l m n o p q r s t u v w x y z sooo . . . mumbo is 43426 . . . answer : a" | a ) 43426 , b ) 14236 , c ) 13436 , d ) 14263 , e ) 15263 | a | divide(79523, add(const_3, const_3)) | add(const_3,const_3)|divide(n0,#0)| | general |
if pintu is coded as 79523 in a certain code language , how would you code mumbo in the same language ? | "a can do the work in 18 / 2 i . e . , 9 days . a and b ' s one day ' s work = 1 / 9 + 1 / 18 = ( 2 + 1 ) / 18 = 1 / 6 so a and b together can do the work in 6 days . answer : d" | a ) 10 , b ) 16 , c ) 18 , d ) 6 , e ) 12 | d | inverse(add(divide(const_1, 18), multiply(divide(const_1, 18), const_2))) | divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2)| | physics |
a is twice as fast as b . if b alone can do a piece of work in 18 days , in what time can a and b together complete the work ? | "1 / 2 * 5.5 * 6 = 16.5 m 2 answer : b" | a ) 11 m 2 , b ) 16.5 m 2 , c ) 18.5 m 2 , d ) 19.5 m 2 , e ) 12 m 2 | b | triangle_area(5.5, 6) | triangle_area(n0,n1)| | geometry |
the area of a triangle is with base 5.5 m and height 6 m ? | "15 * a + 15 * b = x pages in 15 mins printer a will print = 15 / 45 * x pages = 1 / 3 * x pages thus in 15 mins printer printer b will print x - 1 / 3 * x = 2 / 3 * x pages also it is given that printer b prints 3 more pages per min that printer a . in 15 mins printer b will print 45 more pages than printer a thus 2 /... | a ) 125 , b ) 135 , c ) 145 , d ) 155 , e ) 165 | b | multiply(divide(3, subtract(divide(45, 15), const_1)), 45) | divide(n1,n0)|subtract(#0,const_1)|divide(n2,#1)|multiply(#2,n1)| | physics |
working together , printer a and printer b would finish the task in 15 minutes . printer a alone would finish the task in 45 minutes . how many pages does the task contain if printer b prints 3 pages a minute more than printer a ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 8400 ( 100 / 125 ) = rs . 6720 . answer : d" | a ) rs . 5725 , b ) rs . 5275 , c ) rs . 6275 , d ) rs . 6720 , e ) none of these | d | divide(8400, add(const_1, divide(25, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
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