Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
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liam is pulled over for speeding just as he is arriving at work . he explains to the police officer that he could not afford to be late today , and has arrived at work only four minutes before he is to start . the officer explains that if liam had driven 5 mph slower for his whole commute , he would have arrived at wor... | "let t be the number of hours he would need to reach office on time . when he is driving with over speed , he reached office 4 min earlier ! so the equation for this is s ( t - 4 / 60 ) = 60 where s is the speed and 30 is the distance . if he decreases his speed by 5 mph then he would have reached his office on time : ... | a ) 50 mph , b ) 45 mph , c ) 48 mph , d ) 52 mph , e ) 70 mph | e | add(60, multiply(5, const_4)) | multiply(n0,const_4)|add(n1,#0)| | physics |
a bus covered a distance of 250 km , partly at an average speed of 40 kmph and partly at 60 kmph . if the total time taken is 5.4 hours , then the distance covered at 40 kmph is | let the partial distance covered at 40 kmph be x let the another partial distance covered at 60 kmph be ( 250 - x ) thus , x / 40 - ( 250 - x ) / 60 = 5.4 or , x / 40 + ( 250 - x ) / 60 = 5.4 or , ( 3 x + 500 - 2 x ) / / 120 = 5.4 or 500 + x = 648 x = 148 answer : e | a ) 130 km , b ) 120 km , c ) 100 km , d ) 140 km , e ) 148 | e | multiply(divide(subtract(multiply(60, 5.4), 250), negate(subtract(40, 60))), 40) | multiply(n2,n3)|subtract(n1,n2)|negate(#1)|subtract(#0,n0)|divide(#3,#2)|multiply(n1,#4) | physics |
a merchant has selected two items to be placed on sale , one of which currently sells for 35 percent less than the other . if he wishes to raise the price of the cheaper item so that the two items are equally priced , by what percentage must he raise the price of the less expensive item ? | "expensive item = $ 100 ; cheap item = $ 65 ; we must increase $ 65 to $ 100 , so by $ 35 , which is approximately 50 % increase : ( 100 - 65 ) / 65 = ~ 0.53 . answer : c ." | a ) 30 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | c | multiply(divide(divide(35, const_100), subtract(const_1, divide(35, const_100))), const_100) | divide(n0,const_100)|subtract(const_1,#0)|divide(#0,#1)|multiply(#2,const_100)| | general |
the population of a bacteria colony doubles every day . if it was started 11 days ago with 2 bacteria and each bacteria lives for 12 days , how large is the colony today ? | "2 ^ 11 ( 2 ) = 2 ^ 12 = 4096 the answer is e ." | a ) 512 , b ) 768 , c ) 1024 , d ) 2048 , e ) 4096 | e | subtract(power(2, add(11, const_1)), const_1) | add(n0,const_1)|power(n1,#0)|subtract(#1,const_1)| | physics |
the unit digit in the product 7858 * 1086 * 4582 * 9783 is ? | "unit digit in the given product = unit digit in 8 * 6 * 2 * 3 = 8 answer is c" | a ) 0 , b ) 1 , c ) 8 , d ) 3 , e ) 4 | c | subtract(multiply(multiply(multiply(7858, 1086), 4582), 9783), subtract(multiply(multiply(multiply(7858, 1086), 4582), 9783), add(const_4, const_4))) | add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4)| | general |
a train speeds past a pole in 20 seconds and passes through a tunnel 500 m long in 40 seconds . its length is : | sol . let the length of the train be x metres and its speed be y m / sec . then , x / y = 20 β y = x / 20 β΄ ( x + 500 ) / 40 = x / 20 β x = 500 m . answer c | a ) 200 m , b ) 400 m , c ) 500 m , d ) 700 m , e ) none of the above | c | multiply(500, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
yesterday ' s closing prices of 1,980 different stocks listed on a certain stock exchange were all different from today ' s closing prices . the number of stocks that closed at a higher price today than yesterday was 20 percent greater than the number that closed at a lower price . how many of the stocks closed at a hi... | lets consider the below - the number of stocks that closed at a higher price = h the number of stocks that closed at a lower price = l we understand from first statement - > h + l = 1980 - - - - ( 1 ) we understand from second statement - > h = ( 120 / 100 ) l = > h = 1.2 l - - - - ( 2 ) solve eq ( 1 ) ( 2 ) to get h =... | a ) 484 , b ) 1080 , c ) 1,100 , d ) 1,320 , e ) 1,694 | b | multiply(divide(subtract(subtract(multiply(20, const_100), const_10), const_10), add(add(const_1, divide(20, const_100)), const_1)), add(const_1, divide(20, const_100))) | divide(n1,const_100)|multiply(n1,const_100)|add(#0,const_1)|subtract(#1,const_10)|add(#2,const_1)|subtract(#3,const_10)|divide(#5,#4)|multiply(#2,#6) | gain |
20 - 8 percent of the programmers in a startup company weigh 200 pounds or more . 20 - 5 percent of the programmers that are under 200 pounds in that same company weigh 100 pounds or less . what percent of the programmers in the startup company weigh between 100 and 200 pounds ? | initially 72 % and 28 % split 80 % is further divided as 25 % and 75 % q is asking about that 75 % let total be ' 100 ' then that 75 % is ( 3 / 4 ) β 72 so , the required % is [ ( 3 / 4 ) β 72 / 100 ] β 100 = 54 % answer : d | a ) 15 % , b ) 20 % , c ) 25 % , d ) 54 % , e ) 75 % | d | multiply(subtract(const_100, add(20, 8)), divide(subtract(const_100, add(20, 5)), const_100)) | add(n0,n4)|add(n0,n1)|subtract(const_100,#0)|subtract(const_100,#1)|divide(#2,const_100)|multiply(#4,#3) | general |
the racing magic takes 90 seconds to circle the racing track once . the charging bull makes 40 rounds of the track in an hour . if they left the starting point together , how many minutes will it take for them to meet at the starting point for the second time ? | "time taken by racing magic to make one circle = 90 seconds time taken bycharging bullto make one circle = 60 mins / 40 = 1.5 mins = 90 seconds lcm of 90 and 90 seconds = 90 seconds time taken for them to meet at the starting point for the second time = 2 * 90 sec = 180 seconds = 3 mins answer a" | a ) 3 , b ) 6 , c ) 9 , d ) 12 , e ) 15 | a | multiply(divide(multiply(multiply(90, divide(90, 40)), const_2), 90), const_2) | divide(n0,n1)|multiply(#0,n0)|multiply(#1,const_2)|divide(#2,n0)|multiply(#3,const_2)| | physics |
a lent rs . 5000 to b for 2 years and rs . 3000 to c for 4 years on simple interest at the same rate of interest and received rs . 3300 in all from both of them as interest . the rate of interest per annum is ? | "let the rate be r % p . a . then , ( 5000 * r * 2 ) / 100 + ( 3000 * r * 4 ) / 100 = 3300 100 r + 120 r = 3300 r = 15 % answer : e" | a ) 16 % , b ) 12 % , c ) 74 % , d ) 10 % , e ) 15 % | e | multiply(divide(3300, add(multiply(5000, 2), multiply(3000, 4))), const_100) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)|divide(n4,#2)|multiply(#3,const_100)| | gain |
in the first 10 overs of a cricket game , the run rate was only 3.2 . what should be the run rate in the remaining 40 overs to reach the target of 282 runs ? | "runs scored in the first 10 overs = 10 Γ 3.2 = 32 total runs = 282 remaining runs to be scored = 282 - 32 = 250 remaining overs = 40 run rate needed = 25040 = 6.25 answer : a" | a ) 6.25 , b ) 7.25 , c ) 7.75 , d ) 7.35 , e ) 8.25 | a | divide(subtract(282, multiply(10, 3.2)), 40) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,n2)| | gain |
what is the unit β s digit of 7 ^ 6 | "7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 1 ( last digit ) 7 ^ 5 = 7 ( last digit ) and the cycle repeats after every 4 powers therefore , last digit of 7 ^ 6 = 9 answer e" | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | e | reminder(power(7, const_2), const_10) | power(n0,const_2)|reminder(#0,const_10)| | general |
a can finish a work in 4 days and b can do same work in half the time taken by a . then working together , what part of same work they can finish in a day ? | explanation : please note in this question , we need to answer part of work for a day rather than complete work . it was worth mentioning here because many do mistake at this point in hurry to solve the question so lets solve now , a ' s 1 day work = 1 / 4 b ' s 1 day work = 1 / 2 [ because b take half the time than a ... | a ) 1 / 4 , b ) 1 / 6 , c ) 1 / 7 , d ) 3 / 4 , e ) none of these | d | add(divide(const_1, 4), multiply(divide(const_1, 4), const_2)) | divide(const_1,n0)|multiply(#0,const_2)|add(#0,#1) | physics |
two pipes a and b can separately fill a tank in 16 and 10 minutes respectively . a third pipe c can drain off 60 liters of water per minute . if all the pipes are opened , the tank can be filled in 10 minutes . what is the capacity of the tank ? | "1 / 16 + 1 / 10 - 1 / x = 1 / 10 x = 12 16 * 60 = 960 answer : b" | a ) 590 liters , b ) 960 liters , c ) 740 liters , d ) 560 liters , e ) 580 liters | b | multiply(multiply(inverse(subtract(add(add(divide(const_1, 16), divide(const_1, 10)), divide(const_1, 60)), divide(const_1, 10))), const_3), 10) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|divide(const_1,n3)|add(#0,#1)|add(#4,#2)|subtract(#5,#3)|inverse(#6)|multiply(#7,const_3)|multiply(n1,#8)| | physics |
a train speeds past a pole in 15 sec and a platform 110 m long in 25 sec , its length is ? | "let the length of the train be x m and its speed be y m / sec . then , x / y = 15 = > y = x / 15 ( x + 110 ) / 25 = x / 15 = > x = 165 m . answer : option d" | a ) 100 , b ) 150 , c ) 160 , d ) 165 , e ) 180 | d | multiply(110, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
if ( t - 8 ) is a factor of t ^ 2 - kt - 47 , then k = | "t ^ 2 - kt - 48 = ( t - 8 ) ( t + m ) where m is any positive integer . if 48 / 8 = 6 , then we know as a matter of fact that : m = + 6 and thus k = 8 - 6 = 6 t ^ 2 - kt - m = ( t - a ) ( t + m ) where a > m t ^ 2 + kt - m = ( t - a ) ( t + m ) where a < m t ^ 2 - kt + m = ( t - a ) ( t - m ) t ^ 2 + kt + m = ( t + a ... | a ) 16 , b ) 12 , c ) 2 , d ) 6 , e ) 14 | d | add(const_10, 2) | add(n1,const_10)| | general |
during a special promotion , a certain filling station is offering a 10 percent discount on gas purchased after the first 10 gallons . if kim purchased 15 gallons of gas , and isabella purchased 25 gallons of gas , then isabella β s total per - gallon discount is what percent of kim β s total per - gallon discount ? | "kim purchased 15 gallons of gas . she paid for 10 + 0.9 * 5 = 14.5 gallons , so the overall discount she got was 0.5 / 15 = 3.33 % . isabella purchased 25 gallons of gas . she paid for 10 + 0.9 * 15 = 23.5 gallons , so the overall discount she got was 1.5 / 25 = 6 % . 6 is 6 / 3.33 * 100 = 180 % of 3.33 . answer : e .... | a ) 80 % , b ) 100 % , c ) 116.7 % , d ) 120 % , e ) 180 % | e | multiply(divide(multiply(subtract(const_1, divide(add(divide(multiply(add(10, subtract(10, const_1)), subtract(const_100, 10)), const_100), 10), 25)), const_100), multiply(subtract(const_1, divide(add(10, divide(multiply(add(10, const_4), subtract(const_100, 10)), const_100)), 15)), const_100)), const_100) | add(n0,const_4)|subtract(n0,const_1)|subtract(const_100,n0)|add(n0,#1)|multiply(#0,#2)|divide(#4,const_100)|multiply(#3,#2)|add(n1,#5)|divide(#6,const_100)|add(n1,#8)|divide(#7,n2)|divide(#9,n3)|subtract(const_1,#10)|multiply(#12,const_100)|subtract(const_1,#11)|multiply(#14,const_100)|divide(#15,#13)|multiply(#16,cons... | gain |
a man can row downstream at the rate of 16 km / hr and upstream at 6 km / hr . find man ' s rate in still water and the rate of current ? | "explanation : rate of still water = 1 / 2 ( 16 + 6 ) = 11 km / hr rate of current = 1 / 2 ( 16 - 6 ) = 5 km / hr answer : option b" | a ) 9,6 , b ) 11,5 , c ) 9,3 , d ) 6,6 , e ) none of these | b | divide(subtract(16, 6), const_2) | subtract(n0,n1)|divide(#0,const_2)| | gain |
a rectangular lawn of dimensions 100 m * 60 m has two roads each 10 m wide running in the middle of the lawn , one parallel to the length and the other parallel to the breadth . what is the cost of traveling the two roads at rs . 3 per sq m ? | "explanation : area = ( l + b β d ) d ( 100 + 60 β 10 ) 10 = > 1500 m 2 1500 * 3 = rs . 4500 answer : option d" | a ) a ) 3820 , b ) b ) 930 , c ) c ) 9309 , d ) d ) 4500 , e ) e ) 8302 | d | multiply(multiply(subtract(add(100, 60), 10), 10), 3) | add(n0,n1)|subtract(#0,n2)|multiply(n2,#1)|multiply(n3,#2)| | geometry |
find avrg speed if a man travels at speed of 18 km / hr up and 24 km / hr dawn at an altitude of 300 m . | "avg speed = 2 * x * y / ( x + y ) = 2 * 18 * 24 / ( 18 + 24 ) = 20.5 answer : d" | a ) 25.8 , b ) 26.8 , c ) 27.8 , d ) 20.5 , e ) 21.8 | d | divide(multiply(multiply(const_2, 18), 24), add(18, 24)) | add(n0,n1)|multiply(n0,const_2)|multiply(n1,#1)|divide(#2,#0)| | physics |
nitin ranks 27 th in a class of 89 students . what is rank from the last ? | "explanation : number students behind the nitin in rank = ( 89 - 27 ) = 62 nitin is 63 th from the last answer : a ) 63" | a ) 63 , b ) 65 , c ) 67 , d ) 87 , e ) 89 | a | subtract(89, 27) | subtract(n1,n0)| | other |
set a consists of the integers from 3 to 12 , inclusive , while set b consists of the integers from 6 to 20 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "a = { 3,4 , 5,6 , 7,8 , 9,10 , 11,12 } b = { 6 , 7,8 , 9,10 , 11,12 . . . } thus we see that there are 7 distinct integers that are common to both . b is the correct answer ." | a ) 5 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | b | add(6, 3) | add(n0,n2)| | other |
what is the units digit of the expression 14 ^ 7 β 20 ^ 4 ? | i think answer on this one should be a too . since we know that 14 ^ 7 > 20 ^ 4 , as will said one should always check if the number is positive . | a ) 8 , b ) 3 , c ) 4 , d ) 6 , e ) 0 | a | add(add(const_4, const_3), const_2) | add(const_3,const_4)|add(#0,const_2)| | general |
the telephone company wants to add an area code composed of 2 letters to every phone number . in order to do so , the company chose a special sign language containing 424 different signs . if the company used 422 of the signs fully and two remained unused , how many additional area codes can be created if the company u... | "# of 2 - letter codes possible from 424 different signs = 424 * 424 . # of 2 - letter codes possible from 422 different signs = 422 * 422 . the difference = 424 ^ 2 - 422 ^ 2 = ( 424 - 422 ) ( 424 + 422 ) = 1692 . answer : c ." | a ) 246 , b ) 248 , c ) 1692 , d ) 15,128 , e ) 30,256 | c | multiply(multiply(422, subtract(424, 422)), subtract(424, 422)) | subtract(n1,n2)|multiply(n2,#0)|multiply(#1,#0)| | general |
thin rectangular slab of potato was cut into two pieces for an osmosis lab . one piece is 50 mm greater than the other . if the original uncut slab is 600 mm , what is the length of the other piece of the potato after it is cut . | piece one as p 1 , and piece two as p 2 . p 2 + 50 = p 1 as eq . a p 1 + p 2 = 600 as eq . b substitute eq . a into eq . b , resulting in ( p 2 + 50 ) + p 2 = 600 mm . group like terms , p 2 + p 2 = 600 - 50 ( 2 * p 2 / 2 ) = 550 / 2 solves to p 2 = 275 . ansewr is b . | ['a ) 550', 'b ) 275', 'c ) 50', 'd ) 300', 'e ) 250'] | b | divide(subtract(600, 50), const_2) | subtract(n1,n0)|divide(#0,const_2) | geometry |
a alone can complete a work in 16 days and b alone can do in 12 days . star Ι΅ ng with a , they work on alternate days . the total work will be completed in | explana Ι΅ on : a ' s 1 day work = 1 / 16 b ' s 1 day work = 1 / 12 as they are working on alternate day ' s so their 2 days work = ( 1 / 16 ) + ( 1 / 12 ) = 7 / 48 [ here is a small technique , total work done will be 1 , right , then mul Ι΅ ply numerator Ι΅ ll denominator , as 7 * 6 = 42 , 7 * 7 = 49 , as 7 * 7 is more ... | a ) 1314 , b ) 1312 , c ) 1334 , d ) 1344 , e ) none of these | c | multiply(const_100, multiply(inverse(add(inverse(12), inverse(16))), const_2)) | inverse(n1)|inverse(n0)|add(#0,#1)|inverse(#2)|multiply(#3,const_2)|multiply(#4,const_100) | physics |
a cistern which could be filled in 6 hours takes one hour more to be filled owing to a leak in its bottom . if the cistern is full in what time will the leak empty it ? | "1 / 6 - 1 / x = 1 / 10 = > 15 hrs answer : c" | a ) 76 hrs , b ) 99 hrs , c ) 15 hrs , d ) 90 hrs , e ) 11 hrs | c | inverse(subtract(divide(const_1, 6), divide(const_1, const_10))) | divide(const_1,n0)|divide(const_1,const_10)|subtract(#0,#1)|inverse(#2)| | physics |
george is 8 years more than christopher and ford is 2 years younger than christopher . the sum of their ages is 60 . find the ages of ford . | "christopher age = x george age , y = x + 8 - - - - - - - - - - > ( 1 ) ford age , z = x - 2 - - - - - - - - - - - - > ( 2 ) sum of their ages , x + y + z = 60 - - - - > ( 3 ) substitute z and y values in equation ( 3 ) therefore , x + ( x + 8 ) + ( x - 2 ) = 60 = > 3 x + 8 - 2 = 60 = > 3 x = 60 - 6 = > 3 x = 54 = > x ... | a ) 14 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | c | divide(subtract(60, subtract(8, 2)), const_3) | subtract(n0,n1)|subtract(n2,#0)|divide(#1,const_3)| | general |
on sunday , bill ran 4 more miles than he ran on saturday . julia did not run on saturday , but she ran twice the number of miles on sunday that bill ran on sunday . if bill and julia ran a total of 28 miles on saturday and sunday , how many miles did bill run on sunday ? | "let bill run x on saturday , so he will run x + 4 on sunday . . julia will run 2 * ( x + 4 ) on sunday . . totai = x + x + 4 + 2 x + 8 = 28 . . 4 x + 12 = 28 . . x = 4 . . ans = x + 4 = 4 + 4 = 8 aswer d" | a ) 5 , b ) 6 , c ) 7 , d ) 8 , e ) 9 | d | add(divide(subtract(28, add(4, multiply(const_2, 4))), 4), 4) | multiply(n0,const_2)|add(n0,#0)|subtract(n1,#1)|divide(#2,n0)|add(n0,#3)| | general |
bhanu spends 30 % of his income on petrol on scooter 14 % of the remaining on house rent and the balance on food . if he spends rs . 300 on petrol then what is the expenditure on house rent ? | "given 30 % ( income ) = 300 β β income = 1000 after having spent rs . 300 on petrol , he left with rs . 700 . his spending on house rent = 14 % ( 700 ) = rs . 98 answer : a" | a ) 98 , b ) 140 , c ) 62 , d ) 797 , e ) 123 | a | multiply(subtract(divide(300, divide(30, const_100)), 300), divide(14, const_100)) | divide(n1,const_100)|divide(n0,const_100)|divide(n2,#1)|subtract(#2,n2)|multiply(#0,#3)| | gain |
a man covered a certain distance at some speed . if he had moved 3 kmph faster , he would have taken 40 minutes less . if he had moved 2 kmph slower , he would have taken 40 minutes more . what is the distance in km ? | explanation : speed = 2 v 1 v 2 / v 1 β v 2 = 2 Γ 3 Γ 23 β 2 = 12 km / h = hr distance = vt 1 ( 1 + v / v 1 ) = 12 Γ 40 / 60 ( 1 + 12 / 3 ) = 40 km answer is c | a ) 36 , b ) 38 , c ) 40 , d ) 42 , e ) 45 | c | multiply(divide(add(multiply(2, divide(add(multiply(2, divide(40, const_60)), multiply(3, divide(40, const_60))), subtract(3, 2))), multiply(2, divide(40, const_60))), divide(40, const_60)), divide(add(multiply(2, divide(40, const_60)), multiply(3, divide(40, const_60))), subtract(3, 2))) | divide(n1,const_60)|subtract(n0,n2)|multiply(n2,#0)|multiply(n0,#0)|add(#2,#3)|divide(#4,#1)|multiply(n2,#5)|add(#6,#2)|divide(#7,#0)|multiply(#8,#5) | physics |
if x * y = xy β 2 ( x + y ) for all integers x and y , then 2 * ( β 3 ) = | "the quickest ( rather only ) way is to apply the given formula : 2 * ( - 3 ) = 2 * ( - 3 ) - 2 ( 2 + ( - 3 ) ) = - 6 + 2 = - 4 option ( c )" | a ) β 16 , b ) β 11 , c ) β 4 , d ) 4 , e ) 16 | c | add(negate(multiply(add(negate(3), 2), 2)), multiply(negate(3), 2)) | negate(n2)|add(n1,#0)|multiply(n1,#0)|multiply(#1,n0)|negate(#3)|add(#2,#4)| | general |
a man is 18 years older than his son . in eight years , his age will be twice the age of his son . the present age of this son is | "explanation : let ' s son age is x , then father age is x + 18 . = > 2 ( x + 8 ) = ( x + 18 + 8 ) = > 2 x + 16 = x + 26 = > x = 10 years answer : option b" | a ) 5 years , b ) 10 years , c ) 15 years , d ) 20 years , e ) 25 years | b | divide(subtract(18, subtract(multiply(const_2, const_2), const_2)), subtract(const_2, const_1)) | multiply(const_2,const_2)|subtract(const_2,const_1)|subtract(#0,const_2)|subtract(n0,#2)|divide(#3,#1)| | general |
operation # is defined as : a # b = 4 a ^ 2 + 4 b ^ 2 + 8 ab for all non - negative integers . what is the value of ( a + b ) + 5 , when a # b = 100 ? | "official solution : ( b ) we know that a # b = 100 and a # b = 4 a Β² + 4 b Β² + 8 ab . so 4 a Β² + 4 b Β² + 8 ab = 100 we can see that 4 a Β² + 4 b Β² + 8 ab is a well - known formula for ( 2 a + 2 b ) Β² . therefore ( 2 a + 2 b ) Β² = 100 . ( 2 a + 2 b ) is non - negative number , since both a and b are non - negative numbe... | a ) 5 , b ) 8 , c ) 10 , d ) 13 , e ) 17 | c | add(sqrt(divide(100, 4)), const_4.0) | divide(n6,n0)|sqrt(#0)|add(n5,#1)| | general |
a student got 72 % in math and 84 % in history . to get an overall average of 75 % , how much should the student get in the third subject ? | "72 + 84 + x = 3 * 75 x = 69 the answer is c ." | a ) 65 % , b ) 67 % , c ) 69 % , d ) 71 % , e ) 73 % | c | subtract(multiply(const_3, 75), add(72, 84)) | add(n0,n1)|multiply(n2,const_3)|subtract(#1,#0)| | general |
if f ( x ) = ax ^ 2 β 6 x + ax + 1 , then f ( b ) β f ( - b ) will equal : | f ( x ) = ax ^ 2 β 6 x + ax + 1 f ( b ) = ab ^ 2 β 6 b + ab + 1 f ( - b ) = ab ^ 2 + 6 b - ab + 1 f ( b ) - f ( - b ) = ab ^ 2 β 6 b + ab + 1 - ab ^ 2 - 6 b + ab - 1 = - 12 b + 2 ab answer b | a ) 0 , b ) - 12 b + 2 ab , c ) 2 ab ^ 4 β 8 b ^ 2 β 6 , d ) - 2 ab ^ 4 + 8 b ^ 2 + 6 , e ) 2 ab ^ 4 β 8 b ^ 2 + 2 ab β 6 | b | multiply(1, const_10) | multiply(n2,const_10) | general |
a 100 meter long train crosses a man standing on the platform in 5 sec . what is the speed of the train ? | "s = 100 / 5 * 18 / 5 = 72 kmph answer : a" | a ) 72 kmph , b ) 108 kmph , c ) 176 kmph , d ) 134 kmph , e ) 161 kmph | a | multiply(divide(100, 5), const_3_6) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
what is the sum r of all 3 digit numbers that leave a remainder of ' 2 ' when divided by 3 ? | the series is = > 101 + . . . 998 number of terms = > 300 sum = > 300 / 2 * [ 101 + 998 ] hence sum r = > 164850 i . e . option b | a ) 897 , b ) 164850 , c ) 164749 , d ) 149700 , e ) 156720 | b | divide(multiply(add(add(multiply(add(floor(divide(subtract(divide(power(const_10, 3), const_10), 2), 3)), const_1), 3), 2), add(multiply(floor(divide(subtract(power(const_10, 3), add(multiply(add(floor(divide(subtract(divide(power(const_10, 3), const_10), 2), 3)), const_1), 3), 2)), 3)), 3), add(multiply(add(floor(divi... | power(const_10,n0)|divide(#0,const_10)|subtract(#1,n1)|divide(#2,n0)|floor(#3)|add(#4,const_1)|multiply(n0,#5)|add(n1,#6)|subtract(#0,#7)|divide(#8,n0)|floor(#9)|multiply(n0,#10)|add(#7,#11)|add(#7,#12)|multiply(#13,#10)|divide(#14,const_2) | general |
find the cost of carpeting a room 13 m long and 9 m broad with a carpet 75 cm wide at the rate of rs . 12 per square metre . | explanation : area of the carpet = area of the room = ( 13 Γ 9 ) m 2 = 117 m 2 . length of the carpet = ( area / width ) = 117 Γ ( 4 / 3 ) m = 156 m . therefore cost of carpeting = ( 156 Γ 12 ) = rs . 1872 answer : option c | a ) 1824 , b ) 1854 , c ) 1872 , d ) 1888 , e ) 1999 | c | multiply(multiply(13, 12), 12) | multiply(n0,n3)|multiply(n3,#0) | physics |
each week a restaurant serving mexican food uses the same volume of chili paste , which comes in either 25 - ounce cans or 15 - ounce cans of chili paste . if the restaurant must order 30 more of the smaller cans than the larger cans to fulfill its weekly needs , then how many larger cans are required to fulfill its we... | let x be the number of 25 ounce cans . therefore ( x + 30 ) is the number of 15 ounce cans . total volume is same , therefore 25 x = 15 ( x + 30 ) 10 x = 450 x = 45 ans - e | a ) 60 , b ) 70 , c ) 80 , d ) 100 , e ) 45 | e | add(15, 30) | add(n1,n2) | general |
a boat running downstream covers a distance of 22 km in 4 hours while for covering the same distance upstream , it takes 5 hours . what is the speed of the boat in still water ? | "speed downstream = 22 / 4 = 5.5 kmph speed upstream = 22 / 5 = 4.4 kmph speed of the boat in still water = ( 5.5 + 4.4 ) / 2 = 4.95 kmph answer : e" | a ) 2.75 , b ) 5.95 , c ) 7.95 , d ) 3.25 , e ) 4.95 | e | divide(add(divide(22, 4), divide(22, 5)), 4) | divide(n0,n1)|divide(n0,n2)|add(#0,#1)|divide(#2,n1)| | physics |
express a speed of 54 kmph in meters per second ? | "54 * 5 / 18 = 15 mps answer : c" | a ) 10 mps , b ) 76 mps , c ) 15 mps , d ) 97 mps , e ) 16 mps | c | multiply(const_0_2778, 54) | multiply(n0,const_0_2778)| | physics |
the cash difference between the selling prices of an article at a profit of 8 % and 6 % is rs 3 . the ratio of two selling prices is | "explanation : let the cost price of article is rs . x required ratio = ( 108 % of x ) / ( 106 % of x ) = 108 / 106 = 54 / 53 = 54 : 53 . answer : d" | a ) 51 : 52 , b ) 52 : 53 , c ) 53 : 54 , d ) 54 : 53 , e ) none of these | d | divide(add(const_100, 8), add(const_100, 6)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)| | gain |
the tax on a commodity is diminished by 16 % and its consumption increased by 15 % . the effect on revenue is ? | 100 * 100 = 10000 84 * 115 = 9660 - - - - - - - - - - - 10000 - - - - - - - - - - - 340 100 - - - - - - - - - - - ? = > 3.4 % decrease answer : c | a ) 2 % , b ) 3.8 % , c ) 3.4 % , d ) 3.6 % , e ) 1 % | c | subtract(const_100, multiply(multiply(add(const_1, divide(15, const_100)), subtract(const_1, divide(16, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5) | general |
the maximum number of student amoung them 2010 pens and 1050 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | "solution required number of student = h . c . f of 2010 and 1050 = 30 . answer c" | a ) 90 , b ) 60 , c ) 30 , d ) 1911 , e ) none of these | c | gcd(2010, 1050) | gcd(n0,n1)| | general |
two quarts containing 2 β 3 water and 1 β 3 formula are mixed with 3 quarts containing 3 β 8 water and 5 β 8 formula . approximately what percent of the combined 5 - quart mixture is water ? | since we are asking of water in the end , lets concentrate on water only . . a solution has 2 quarts with fraction of water = 2 / 3 , almost 67 % . . b solution has 3 quarts with fraction of water = 3 / 8 almost 37 % . . so the ratio a : b = 2 : 3 . . . the average will therefore be closer to b % and will be 2 / ( 2 + ... | a ) 40 % , b ) 45 % , c ) 50 % , d ) 55 % , e ) 60 % | c | multiply(divide(add(multiply(divide(2, 3), 2), multiply(divide(3, 8), 3)), 5), const_100) | divide(n0,n1)|divide(n1,n6)|multiply(n0,#0)|multiply(n1,#1)|add(#2,#3)|divide(#4,n7)|multiply(#5,const_100) | general |
two numbers are in the ratio 3 : 4 . if their l . c . m . is 48 . what is sum of the numbers ? | "explanation : let the numbers be 3 x and 4 x lcm of 3 x and 4 x = 12 x ( since lcm of 3 and 4 is 12 . hence lcm of 3 x and 4 x is 12 x ) given that lcm of 3 x and 4 x is 48 = > 12 x = 48 = > x = 48 / 12 = 4 sum of the numbers = 3 x + 4 x = 7 x = 7 x 4 = 28 answer : option a" | a ) 28 , b ) 56 , c ) 76 , d ) 84 , e ) 85 | a | add(multiply(multiply(3, const_4.0), const_100), multiply(4, 48)) | multiply(n0,const_4.0)|multiply(n1,n2)|multiply(#0,const_100)|add(#2,#1)| | other |
a cricketer scored 134 runs which included 12 boundaries and 2 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 134 - ( 12 x 4 + 2 x 6 ) = 134 - ( 60 ) = 74 now , we need to calculate 72 is what percent of 134 . = > 74 / 134 * 100 = 55.22 % answer : a" | a ) 55.22 % , b ) 54.54 % , c ) 63 % , d ) 70 % , e ) none of these | a | multiply(divide(subtract(134, add(multiply(12, const_4), multiply(2, multiply(2, const_3)))), 134), const_100) | multiply(n1,const_4)|multiply(n2,const_3)|multiply(n2,#1)|add(#0,#2)|subtract(n0,#3)|divide(#4,n0)|multiply(#5,const_100)| | general |
a worker makes a toy in every 2 h . if he works for 80 h , then how many toys will he make ? | "answer let number of toys be n . more hours , more toys ( direct proportion ) 2 : 80 : : 1 : n β n = 80 / 2 = 40 toys correct option : a" | a ) 40 , b ) 54 , c ) 45 , d ) 39 , e ) none | a | divide(80, 2) | divide(n1,n0)| | physics |
bhavan travelled for 20 hours . he covered the first half of the distance at 50 kmph and remaining half of the distance at 40 kmph . find the distance travelled by bhavan ? | let the distance travelled be x km . total time = ( x / 2 ) / 50 + ( x / 2 ) / 40 = 20 = > x / 100 + x / 80 = 20 = > ( 4 x + 5 x ) / 400 = 20 = > x = 888.89 km answer : a | a ) 888.89 , b ) 788.89 , c ) 899.89 , d ) 688.89 , e ) 899.99 | a | add(multiply(divide(20, const_2), 50), multiply(divide(20, const_2), 40)) | divide(n0,const_2)|multiply(n1,#0)|multiply(n2,#0)|add(#1,#2) | physics |
what is the tens digit of 7 ^ 957 ? | "7 ^ 1 = 7 7 ^ 2 = 49 7 ^ 3 = 343 7 ^ 4 = 2401 7 ^ 5 = 16807 7 ^ 6 = 117649 we should see this as pattern recognition . we have a cycle of 4 . ( we can multiply the last 2 digits only as we care about ten ' s digit ) 0 , 4 , 4 , 0 . 957 = 4 * 239 + 1 the ten ' s digit will be 0 . answer d" | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | d | floor(divide(reminder(power(7, reminder(957, add(const_4, const_1))), const_100), const_10)) | add(const_1,const_4)|reminder(n1,#0)|power(n0,#1)|reminder(#2,const_100)|divide(#3,const_10)|floor(#4)| | general |
? x 40 = 173 x 240 | "let y x 40 = 173 x 240 then y = ( 173 x 240 ) / 40 = 1038 answer : c" | a ) 545 , b ) 685 , c ) 1038 , d ) 495 , e ) 534 | c | divide(multiply(173, 240), 40) | multiply(n1,n2)|divide(#0,n0)| | general |
rose made a part payment of $ 300 toward buying a brand new car , which represents 5 % of the total cost of the car , how much remains to be paid ? | explanation : let ' s start with what the total price of the car would be . if 5 % is equal to $ 300 then 100 % equals $ x . we just have to multiply $ 300 by 20 to get total amount = $ 6000 . out of this amount we then need to deduct the amount already paid which was $ 300 so we have $ 6000 - $ 300 = $ 5700 answer : o... | a ) $ 5700 , b ) $ 5500 , c ) $ 5800 , d ) $ 5200 , e ) $ 5880 | a | subtract(divide(multiply(300, const_100), 5), 300) | multiply(n0,const_100)|divide(#0,n1)|subtract(#1,n0) | general |
a pump can fill a tank with water in 2 hours . because of a leak , it took 2 x 1 / 3 hours to fill the tank . the leak can drain all the water of the tank in | "work done by the leak in 1 hour = 1 / 2 - 3 / 7 = 1 / 14 . leak will empty the tank in 14 hrs . answer : d" | a ) 5 hours , b ) 7 hours , c ) 8 hours , d ) 14 hours , e ) 18 hours | d | inverse(subtract(divide(1, 2), inverse(divide(add(multiply(2, 3), 1), 3)))) | divide(n2,n0)|multiply(n0,n3)|add(n2,#1)|divide(#2,n3)|inverse(#3)|subtract(#0,#4)|inverse(#5)| | physics |
in a rectangular coordinate system , what is the area of a rhombus whose vertices have the coordinates ( 0 , 3.5 ) , ( 12 , 0 ) , ( 0 , - 3.5 ) , ( - 12 , 0 ) ? | "area of rhombus = 1 / 2 * d 1 * d 2 length of 1 st diagonal = 12 + 12 = 24 length of 2 nd diagonal = 3.5 + 3.5 = 7 area = 1 / 2 * 24 * 7 = 84 a is the answer" | a ) 84 , b ) 88 , c ) 112 , d ) 116 , e ) 120 | a | rhombus_area(multiply(12, const_2), multiply(3.5, const_2)) | multiply(n2,const_2)|multiply(n1,const_2)|rhombus_area(#0,#1)| | geometry |
rs . 6000 is lent out in two parts . one part is lent at 7 % p . a simple interest and the other is lent at 10 % p . a simple interest . the total interest at the end of one year was rs . 480 . find the ratio of the amounts lent at the lower rate and higher rate of interest ? | "let the amount lent at 7 % be rs . x amount lent at 10 % is rs . ( 6000 - x ) total interest for one year on the two sums lent = 7 / 100 x + 10 / 100 ( 6000 - x ) = 600 - 3 x / 100 = > 600 - 3 / 100 x = 480 = > x = 5000 amount lent at 10 % = 2000 required ratio = 2000 : 1000 = 2 : 1 answer : a" | a ) 2 : 1 , b ) 5 : 6 , c ) 5 : 2 , d ) 5 : 8 , e ) 5 : 2 | a | divide(divide(subtract(multiply(480, const_100), multiply(6000, 7)), subtract(10, 7)), divide(subtract(multiply(480, const_100), multiply(6000, 7)), subtract(10, 7))) | multiply(n3,const_100)|multiply(n0,n1)|subtract(n2,n1)|subtract(#0,#1)|divide(#3,#2)|divide(#4,#4)| | gain |
the banker ' s gain on a bill due due 1 year hence at 12 % per annum is rs . 7.2 . the true discount is | "solution t . d = [ b . g x 100 / r x t ] = rs . ( 7.2 x 100 / 12 x 1 ) = rs . 60 . answer d" | a ) rs . 72 , b ) rs . 36 , c ) rs . 54 , d ) rs . 60 , e ) none | d | divide(7.2, divide(12, const_100)) | divide(n1,const_100)|divide(n2,#0)| | gain |
n is the greatest number which divides 1305 , 4665 and 6905 and gives the same remainder in each case . what is the sum of the digits in n ? | if the remainder is same in each case and remainder is not given , hcf of the differences of the numbers is the required greatest number 6905 - 1305 = 5600 6905 - 4665 = 2240 4665 - 1305 = 3360 hence , the greatest number which divides 1305 , 4665 and 6905 and gives the same remainder , n = hcf of 5600 , 2240 , 3360 = ... | a ) 23 , b ) 28 , c ) 4 , d ) 1 , e ) 25 | c | add(add(divide(subtract(divide(subtract(divide(gcd(gcd(subtract(4665, 1305), subtract(6905, 4665)), subtract(6905, 1305)), const_10), reminder(divide(gcd(gcd(subtract(4665, 1305), subtract(6905, 4665)), subtract(6905, 1305)), const_10), const_10)), const_10), reminder(divide(subtract(divide(gcd(gcd(subtract(4665, 1305)... | subtract(n1,n0)|subtract(n2,n1)|subtract(n2,n0)|gcd(#0,#1)|gcd(#3,#2)|divide(#4,const_10)|reminder(#5,const_10)|subtract(#5,#6)|divide(#7,const_10)|reminder(#8,const_10)|subtract(#8,#9)|divide(#10,const_10)|add(#11,#9)|add(#12,#6) | general |
john purchased a grinder and a mobile for rs . 15000 & rs . 8000 respectively . he sold the grinder at a loss of 2 % and the mobile phone at a profit of 10 % . overall how much he make a profit . | "let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 2 / 100 ) = 15000 - 300 m = 8000 ( 1 + 10 / 100 ) = 8000 + 800 total sp - total cp = r + m - ( 15000 + 8000 ) = - 300 + 800 = rs . 500 as this is positive , an overall profit of rs . 200 was made . e" | a ) s . 90 , b ) s . 120 , c ) s . 200 , d ) s . 250 , e ) s . 500 | e | subtract(multiply(10, divide(8000, const_100)), multiply(2, divide(15000, const_100))) | divide(n1,const_100)|divide(n0,const_100)|multiply(n3,#0)|multiply(n2,#1)|subtract(#2,#3)| | gain |
mid intended to type a 7 - digit number , but the two 3 ' s he meant to type did not appear . what appeared instead was the 5 - digit number 52115 . how many different 7 - digit numbers could mid have meant to type ? | should be 21 . mid intended to type a seven - digit number there are two possibilities for placing 2 3 s . case 1 : two 3 s were missed consecutively . i . e . he typed 33 and it came blank on screen . - 5 - 2 - 1 - 1 - 5 - in this arrangement we can fit 33 in 6 ways . ( six dashes , each dash represent one possible pl... | a ) 10 , b ) 16 , c ) 21 , d ) 24 , e ) 27 | c | divide(factorial(7), multiply(factorial(5), factorial(subtract(7, 5)))) | factorial(n0)|factorial(n2)|subtract(n0,n2)|factorial(#2)|multiply(#1,#3)|divide(#0,#4) | general |
in a clothing store , there are 6 different colored neckties ( red , orange , yellow , green , blue , and indigo ) and 4 different colored shirts ( red , orange , yellow , green , blue , and indigo ) that must be packed into boxes for gifts . if each box can only fit one necktie and one shirt , what is the probability ... | 4 ties and 4 shirts . . . red tie can take any of 4 shirts . . orange can take any of the remaining 3 shirts yellow any of remaining 2 . . and so on till last indigo chooses the 1 remaining . . total ways = 4 * 3 * 2 * 1 = 24 out of this 720 , only 1 way will have same colour tie and shirt . . prob = 1 / 720 b | a ) 719 / 720 , b ) 1 / 24 , c ) 2 / 233 , d ) 3 / 543 , e ) 1 / 720 | b | divide(const_1, factorial(4)) | factorial(n1)|divide(const_1,#0) | general |
a , b , c subscribe rs . 50000 for a business , a subscribes rs . 4000 more than b and b rs . 5000 more than c . out of a total profit of rs . 70000 , a receives : | explanation : let c = x . then , b = x + 5000 and a = x + 5000 + 4000 = x + 9000 . so , x + x + 5000 + x + 9000 = 50000 < = > 3 x = 36000 < = > x = 12000 . a : b : c = 21000 : 17000 : 12000 = 21 : 17 : 12 . a ' s share = rs . ( 70000 * 21 / 50 ) = rs . 29,400 . answer : c ) | a ) 14700 , b ) 24703 , c ) 29400 , d ) 33777 , e ) 34778 | c | divide(multiply(add(divide(subtract(50000, add(5000, add(4000, 5000))), const_3), add(4000, 5000)), 70000), 50000) | add(n1,n2)|add(n2,#0)|subtract(n0,#1)|divide(#2,const_3)|add(#0,#3)|multiply(n3,#4)|divide(#5,n0) | general |
a canoe has two oars , left and right . each oar either works or breaks . the failure or non - failure of each oar is independent of the failure or non - failure of the other . you can still row the canoe with one oar . the probability that the left oar works is 3 / 5 . the probability that the right oar works is also ... | "simply look at the question from the other side . what is the probability that you can β t row the canoe ? this would be 2 / 5 x 2 / 5 = 4 / 25 . using the idea that the probability q of something happening is 1 β the probability that it doesn β t happen , you can use the following equation to reach the right answer :... | a ) 9 / 25 , b ) 10 / 25 , c ) 6 / 10 , d ) 2 / 3 , e ) 21 / 25 | e | add(multiply(divide(3, 5), divide(3, 5)), add(multiply(divide(3, 5), divide(const_2, 5)), multiply(divide(3, 5), divide(const_2, 5)))) | divide(n0,n1)|divide(const_2,n1)|multiply(#0,#1)|multiply(#0,#0)|add(#2,#2)|add(#4,#3)| | general |
john purchased a grinder and a mobile for rs . 15000 & rs . 8000 respectively . he sold the grinder at a loss of 4 % and the mobile phone at a profit of 10 % . overall how much he make aprofit . | "let the sp of the refrigerator and the mobile phone be rs . r and rs . m respectively . r = 15000 ( 1 - 4 / 100 ) = 15000 - 600 m = 8000 ( 1 + 10 / 100 ) = 8000 + 800 total sp - total cp = r + m - ( 15000 + 8000 ) = - 600 + 800 = rs . 200 as this is positive , an overall profit of rs . 200 was made . c" | a ) rs . 90 , b ) rs . 120 , c ) rs . 200 , d ) rs . 250 , e ) rs . 290 | c | subtract(multiply(10, divide(8000, const_100)), multiply(4, divide(15000, const_100))) | divide(n1,const_100)|divide(n0,const_100)|multiply(n3,#0)|multiply(n2,#1)|subtract(#2,#3)| | gain |
0.03 x 0.5 = ? | "explanation : 3 x 5 = 15 . sum of decimal places = 3 0.02 x 0.5 = 0.015 answer - c" | a ) 0.0001 , b ) 0.001 , c ) 0.015 , d ) 0.1 , e ) none of these | c | multiply(divide(0.03, 0.5), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
how much interest can a person get on rs . 7200 at 17.5 % p . a . simple interest for a period of two years and 6 months ? | i = ( 7200 * 2.5 * 17.5 ) / 100 = ( 7200 * 5 * 35 ) / ( 100 * 2 * 2 ) = rs . 3150 answer : d | a ) rs . 4150 , b ) rs . 3850 , c ) rs . 3770 , d ) rs . 3150 , e ) none of these | d | multiply(7200, divide(multiply(add(const_2, divide(6, const_12)), 17.5), const_100)) | divide(n2,const_12)|add(#0,const_2)|multiply(n1,#1)|divide(#2,const_100)|multiply(n0,#3) | gain |
what is the units digit of 13 ^ 4 * 17 ^ 2 * 29 ^ 3 ? | "3 ^ 1 = 3 3 ^ 2 = 9 3 ^ 3 = 7 3 ^ 4 = 1 , it is first digit 7 ^ 2 = 9 , it is second digit 9 ^ 3 = 9 , it is third digit 1 * 9 * 9 = 1 answer : e" | a ) 9 , b ) 7 , c ) 5 , d ) 3 , e ) 1 | e | divide(add(multiply(factorial(13), factorial(4)), multiply(factorial(13), factorial(2))), 13) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general |
donovan took a math test and got 35 correct and 13 incorrect answers . what was the percentage of correct answers ? ( round to the nearest hundredth ) | 35 correct our of 48 total ( 35 + 13 ) 35 / 48 correct answer c | a ) 57.78 % , b ) 67.78 % , c ) 72.92 % , d ) 47.78 % , e ) 97.78 % | c | multiply(divide(35, add(35, 13)), const_100) | add(n0,n1)|divide(n0,#0)|multiply(#1,const_100) | gain |
the slant height of a cone is 14 cm and radius of the base is 12 cm , find the curved surface of the cone . | "Ο * 12 * 14 = 528 answer : d" | a ) 150 , b ) 170 , c ) 48 , d ) 528 , e ) 600 | d | multiply(multiply(const_pi, 12), 14) | multiply(n1,const_pi)|multiply(n0,#0)| | geometry |
triangle xyz is an isosceles right triangle . if side xy is longer than side yz , and the area of the triangle is 49 , what is the measure of side xy ? | ans c . . 14 . . xy being larger means it is the hyp . . area = ( 1 / 2 ) * ( yz ) ^ 2 = 4 or yz = 7 * \ sqrt { 2 } . . therefore hyp = xy = 14 | ['a ) 4', 'b ) 4 β 2', 'c ) 14', 'd ) 8 β 2', 'e ) can not be determined from the information provided'] | c | sqrt(add(power(multiply(sqrt(49), sqrt(const_2)), const_2), power(multiply(sqrt(49), sqrt(const_2)), const_2))) | sqrt(n0)|sqrt(const_2)|multiply(#0,#1)|power(#2,const_2)|add(#3,#3)|sqrt(#4) | geometry |
if the price of sugar rises from rs . 6 per kg to rs . 7.50 per kg , a person , to have no increase in his expenditure on sugar , will have to reduce his consumption of sugar by | "solution let the original consumption = 100 kg and new consumption = x kg . so , 100 x 6 = x Γ 7.50 x = 80 kg . reduction in consumption = 20 % . answer b" | a ) 10 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) none | b | multiply(subtract(const_1, divide(multiply(const_1, 6), 7.50)), const_100) | multiply(n0,const_1)|divide(#0,n1)|subtract(const_1,#1)|multiply(#2,const_100)| | general |
in a lottery there are 10 prizes and 25 blanks . a lottery is drawn at random . what is the probability of getting a blank ? | total draws = prizes + blanks = 10 + 25 = 35 probability of getting a blank = 25 / 35 = 5 / 7 correct option is c | a ) 2 / 3 , b ) 6 / 7 , c ) 5 / 7 , d ) 8 / 7 , e ) 4 / 3 | c | divide(25, add(10, 25)) | add(n0,n1)|divide(n1,#0) | probability |
what is the maximum number of pieces that a circular pizza can be cut into with 6 linear cuts ? | 1 cut makes 2 pieces a second cut can make 4 pieces by cutting through 2 of the pieces a third cut can make 7 pieces by cutting through 3 of the pieces a fourth cut can make 11 pieces by cutting through 4 of the pieces a fifth cut can make 16 pieces by cutting through 5 of the pieces a sixth cut can make 22 pieces by c... | ['a ) 12', 'b ) 13', 'c ) 16', 'd ) 20', 'e ) 22'] | e | multiply(6, const_3) | multiply(n0,const_3) | geometry |
walking at the rate of 4 kmph a man cover certain distance in 2 hr 45 min . running at a speed of 16.5 kmph the man will cover the same distance in . | distance = speed * time 4 * 11 / 4 = 11 km new speed = 16.5 kmph therefore time = d / s = 11 / 16.5 = 40 min answer : c | a ) 12 min , b ) 25 min , c ) 40 min , d ) 48 min , e ) 52 min | c | multiply(divide(multiply(4, add(2, divide(45, const_60))), 16.5), const_60) | divide(n2,const_60)|add(n1,#0)|multiply(n0,#1)|divide(#2,n3)|multiply(#3,const_60) | physics |
how many positive factors do 120 and 96 have in common ? | "the number of common factors will be same as number of factors of the highest common factor ( hcf ) hcf of 120 and 96 is 24 number of factors of 24 = 8 answer : a" | a ) 8 , b ) 12 , c ) 16 , d ) 18 , e ) 24 | a | divide(subtract(96, const_10), const_10) | subtract(n1,const_10)|divide(#0,const_10)| | other |
if p and q are positive integers each greater than 1 , and 15 ( p + 1 ) = 29 ( q + 1 ) , what is the least possible value of p + q ? | 17 ( p + 1 ) = 29 ( q + 1 ) - - > ( p + 1 ) / ( q + 1 ) = 29 / 17 - - > the least positive value of p + 1 is 29 , so the least value of p is 28 and the least positive value of q + 1 is 15 , so the least value of q is 14 - - > the least value of p + q is 28 + 14 = 42 . answer : b . | a ) 36 , b ) 42 , c ) 44 , d ) 46 , e ) none | b | subtract(add(29, 15), const_2) | add(n1,n3)|subtract(#0,const_2) | general |
how many numbers from 10 to 50 are exactly divisible by 3 ? | "12 , 15 , 18 , 21 , 24 , 27 , 30 , 33 , 36 , 39 , 42 , 45,48 . 13 numbers . 10 / 3 = 3 and 50 / 3 = 16 = = > 16 - 3 = 13 . therefore 13 digits a )" | a ) 13 , b ) 15 , c ) 16 , d ) 17 , e ) 18 | a | add(divide(subtract(multiply(floor(divide(50, 3)), 3), multiply(add(floor(divide(10, 3)), const_1), 3)), 3), 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 percentage profit earned by selling an article for rs . 1820 is equal to the percentage loss incurred by selling the same article for rs . 1280 . at what price should the article be sold to make 25 % profit ? | "c . p . be rs . x . then , ( 1820 - x ) / x * 100 = ( x - 1280 ) / x * 100 1820 - x = x - 1280 2 x = 3100 = > x = 1550 required s . p . = 125 % of rs . 1550 = 125 / 100 * 1550 = rs . 1937.5 . answer e" | a ) 3000 , b ) 1230 , c ) 2000 , d ) 5600 , e ) 1937.5 | e | multiply(divide(add(const_100, 25), const_100), divide(add(1820, 1280), const_2)) | add(n2,const_100)|add(n0,n1)|divide(#0,const_100)|divide(#1,const_2)|multiply(#2,#3)| | gain |
if the remainder is 3 when positive integer n is divided by 18 , what is the remainder when n is divided by 6 ? | assume x is quotient here , n = 18 x + 3 - - - - - - - - - - ( 1 ) and n = 6 x + ? we can also write first term as n = ( 18 x + 6 ) = 6 ( 3 x + 1 ) + 3 ie 6 ( 3 x + 1 ) + 3 ie the first term is perfectly divisible by 6 . so , the remainder left is 3 . so , answer ( a ) is right choice . | a ) 3 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | a | divide(18, 6) | divide(n1,n2) | general |
how many 4 digit numbers are there , if it is known that the first digit is even , the second is odd , the third is prime , the fourth ( units digit ) is divisible by 3 , and the digit 7 can be used only once ? | "4 options for the first digit : 2 , 4 , 6 , 8 ; 5 options for the second digit : 1 , 3 , 5 , 7 , 9 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 3 , 6 , 9 . four digit # possible without the restriction ( about the digit 2 ) : 4 * 5 * 4 * 4 = 320 numbers with seven 7 - s , 7 x ... | a ) 20 , b ) 150 , c ) 304 , d ) 310 , e ) 320 | c | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), 4)) | add(n0,const_1)|multiply(n0,n0)|multiply(#0,#0)|multiply(#2,#1)|multiply(#2,n0)|subtract(#3,#4)| | physics |
two kinds of vodka are mixed in the ratio 1 : 2 and 2 : 1 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 | "mixture 1 and 2 having individual profit % of 10 and 20 respectively . individual profits of vodka a for mixture 1 = 1 / 3 * 10 = 3.33 individual profits of vodka b for mixture 1 = 2 / 3 * 10 = 6.77 individual profits for a and b are increased 4 / 3 and 5 / 3 times . means profits of mixtures 1 and 2 will increase too... | a ) 18 % , b ) 20 % , c ) 21 % , d ) 23 % , e ) can not be determined | b | add(divide(multiply(10, 4), 2), add(10, 5)) | add(n4,n8)|multiply(n4,n6)|divide(#1,n1)|add(#0,#2)| | general |
farm tax is levied on the 40 % of the cultivated land . the tax department collected total $ 3840 through the farm tax from the village of mr . willam . mr . willam paid only $ 480 as farm tax . the percentage of total land of mr . willam over the total taxable land of the village is : | "only trick n this question is to ignore 40 % information as farm tax is levied uniformly in the village and that includes mr william ' s land . what percentage of tax mr william paid ? this will be equal to the percentage of total cultivated land he holds over the total cultivated land in the village . that leads to (... | a ) 31.25 % , b ) 25 % , c ) 12.5 % , d ) 0.2083 % , e ) none | a | divide(multiply(multiply(divide(480, 3840), const_100), const_100), 40) | divide(n2,n1)|multiply(#0,const_100)|multiply(#1,const_100)|divide(#2,n0)| | general |
the area of a circular field is 13.86 hectares . find the cost of fencing it at the rate of rs . 4.50 per metre . | "explanation : area = ( 13.86 x 10000 ) sq . m = 138600 sq . m circumference = cost of fencing = rs . ( 1320 x 4.50 ) = rs . 5940 . answer : d ) 5940" | a ) 2399 , b ) 3888 , c ) 2999 , d ) 5940 , e ) 2888 | d | multiply(circumface(multiply(sqrt(divide(13.86, const_pi)), const_100)), 4.50) | divide(n0,const_pi)|sqrt(#0)|multiply(#1,const_100)|circumface(#2)|multiply(#3,n1)| | geometry |
a mixture contains alcohol and water in the ratio 4 : 3 . if 6 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 + 6 ) = 4 / 5 20 x = 4 ( 3 x + 6 ) 8 x = 24 x = 3 quantity of alcohol = ( 4 x 3 ) litres = 12 litres . answer is d" | a ) 15 litres , b ) 10 litres , c ) 30 litres , d ) 12 litres , e ) 8 litres | d | multiply(6, const_1) | multiply(n2,const_1)| | general |
if the sides of a triangle are 26 cm , 25 cm and 10 cm , what is its area ? | "the triangle with sides 26 cm , 25 cm and 10 cm is right angled , where the hypotenuse is 26 cm . area of the triangle = 1 / 2 * 25 * 10 = 125 cm 2 answer : option e" | a ) 120 , b ) 110 , c ) 130 , d ) 140 , e ) 125 | e | divide(multiply(25, 10), const_2) | multiply(n1,n2)|divide(#0,const_2)| | geometry |
( 7.5 Γ 7.5 + 37.5 + 2.5 Γ 2.5 ) is equal to : | solution given expression = ( 7.5 Γ 7.5 + 2 Γ 7.5 Γ 2.5 + 2.5 Γ 2.5 ) = ( a 2 + 2 ab + b 2 ) = ( a + b ) 2 = ( 7.5 + 2.5 ) 2 = 102 = 100 . answer d | a ) 30 , b ) 60 , c ) 80 , d ) 100 , e ) none of these | d | add(add(multiply(7.5, 7.5), multiply(2.5, 2.5)), 37.5) | multiply(n0,n0)|multiply(n3,n3)|add(#0,#1)|add(n2,#2) | general |
a theater charges $ 12 for seats in the orchestra and $ 8 for seats in the balcony . on a certain night , a total of 350 tickets were sold for a total cost of $ 3,320 . how many more tickets were sold that night for seats in the balcony than for seats in the orchestra ? | "orchestra seats - a balcony seats - b a + b = 350 and 12 a + 8 b = 3320 solving equations simultaneously ( multiply equation 1 with 8 and subtract from second equation ) 4 a = 3320 - 8 * 350 = 3320 - 2800 = 520 i . e . a = 130 and b = 350 - 130 = 220 more seats in balcony than orchestra = b - a = 220 - 130 = 90 answer... | a ) 90 , b ) 110 , c ) 120 , d ) 130 , e ) 220 | a | subtract(divide(subtract(multiply(12, 350), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtract(12, 8)), subtract(350, divide(subtract(multiply(12, 350), add(add(multiply(const_3, const_1000), multiply(const_3, const_100)), multiply(const_2, const_10))), subtrac... | multiply(n0,n2)|multiply(const_1000,const_3)|multiply(const_100,const_3)|multiply(const_10,const_2)|subtract(n0,n1)|add(#1,#2)|add(#5,#3)|subtract(#0,#6)|divide(#7,#4)|subtract(n2,#8)|subtract(#8,#9)| | general |
if a - b = 3 and a 2 + b 2 = 29 , find the value of ab . | "2 ab = ( a 2 + b 2 ) - ( a - b ) 2 = 29 - 9 = 20 ab = 10 . answer : a" | a ) 10 , b ) 12 , c ) 13 , d ) 15 , e ) 18 | a | divide(subtract(29, power(3, 2)), 2) | power(n0,n1)|subtract(n3,#0)|divide(#1,n1)| | general |
each child has 4 pencils and 13 skittles . if there are 8 children , how many pencils are there in total ? | 4 * 8 = 32 . answer is e . | a ) 16 , b ) 12 , c ) 18 , d ) 22 , e ) 32 | e | multiply(4, 8) | multiply(n0,n2)| | general |
sandy had $ 140 left after spending 30 % of the money she took for shopping . how much money did sandy take along with her ? | let the money sandy took for shopping be x . 0.7 x = 140 x = 200 the answer is b . | a ) $ 180 , b ) $ 200 , c ) $ 220 , d ) $ 240 , e ) $ 260 | b | divide(140, divide(subtract(const_100, 30), const_100)) | subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1) | gain |
three pipes , a , b , & c are attached to a tank . a & b can fill it in 20 & 30 minutes respectively while c can empty it in 15 minutes . if a , b & c are kept open successively for 1 minute each , how soon will the tank be filled ? | "in three minute 1 / 20 + 1 / 30 - 1 / 15 = 1 / 60 part is filled 3 min - - - - - - - - 1 / 60 parts x min - - - - - - - - - 1 part ( full ) x = 180 min = 3 hours answer : c" | a ) 2 hours , b ) 4 hours , c ) 3 hours , d ) 5 hours , e ) 6 hours | c | add(add(1, 1), 1) | add(n3,n3)|add(n3,#0)| | physics |
a total of $ 20,000 was invested in two certificates of deposit at simple annual interest rates of 6 percent and 8 percent , respectively . if the total interest on the two certificates was $ 1,440 at the end of one year , what fractional part of the $ 20.000 was invested at the higher rate ? | "let ` ` $ x ' ' be invested with the higher rate of 8 % thus , ` ` 20000 - x ' ' must have been invested with lower rate of 6 % simple interest = principal * rate * time for the amount invested at 8 % principal = x rate = 0.08 time = 1 simple interest = x * 0.08 * 1 = 0.08 x for the amount invested at 6 % principal = ... | a ) 3 / 8 , b ) 2 / 5 , c ) 1 / 2 , d ) 3 / 5 , e ) 3 / 4 | d | divide(subtract(add(divide(divide(add(add(multiply(const_100, multiply(const_2, add(const_1, const_4))), multiply(const_4, const_100)), multiply(const_4, multiply(const_2, add(const_1, const_4)))), 20.000), 8), 8), 6), 20.000) | add(const_1,const_4)|multiply(const_100,const_4)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#2,const_4)|add(#3,#1)|add(#5,#4)|divide(#6,n4)|divide(#7,n2)|add(n2,#8)|subtract(#9,n1)|divide(#10,n4)| | gain |
the average of 5 quantities is 6 . the average of 3 of them is 4 . what is the average of remaining 2 numbers ? | answer : a | a ) 9 , b ) 10 , c ) 8 , d ) 9.5 , e ) none of these | a | divide(subtract(multiply(5, 6), multiply(3, 4)), 2) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n4)| | general |
having received his weekly allowance , john spent 3 / 5 of his allowance at the arcade . the next day he spent one third of his remaining allowance at the toy store , and then spent his last $ 0.40 at the candy store . what is john β s weekly allowance ? | "x = 3 x / 5 + 1 / 3 * 2 x / 5 + 40 4 x / 15 = 40 x = 150 = $ 1.50 the answer is a ." | a ) $ 1.50 , b ) $ 2.00 , c ) $ 2.50 , d ) $ 3.00 , e ) $ 3.50 | a | divide(0.40, subtract(const_1, add(divide(3, 5), multiply(divide(const_1, 3), subtract(const_1, divide(3, 5)))))) | divide(n0,n1)|divide(const_1,n0)|subtract(const_1,#0)|multiply(#1,#2)|add(#0,#3)|subtract(const_1,#4)|divide(n2,#5)| | general |
how much space , in cubic units , is left vacant when maximum number of 2 x 2 x 2 cubes are fitted in a rectangular box measuring 8 x 10 x 9 ? | "no of cubes that can be accommodated in box = ( 8 * 10 * 9 ) / ( 2 * 2 * 2 ) 8 * 10 in numerator can be perfectly divided by 2 * 2 in denominator . side with length 9 ca n ' t be perfectly divided by 2 and hence is the limiting factor . closet multiple of 2 less that 9 is 8 . so vacant area in cube = = 10 * 8 ( 9 - 8 ... | a ) 60 , b ) 65 , c ) 70 , d ) 75 , e ) 80 | e | multiply(subtract(8, multiply(2, const_2)), multiply(10, 9)) | multiply(n4,n5)|multiply(n0,const_2)|subtract(n3,#1)|multiply(#0,#2)| | geometry |
how many 4 digit numbers are there , if it is known that the first digit is even , the second is odd , the third is prime , the fourth ( units digit ) is divisible by 3 , and the digit 3 can be used only once ? | "4 options for the first digit : 2 , 4 , 6 , 8 ; 5 options for the second digit : 1 , 3 , 5 , 7 , 9 ; 4 options for the third digit : 2 , 3 , 5 , 7 ; 4 options for the fourth digit : 0 , 3 , 6 , 9 . four digit # possible without the restriction ( about the digit 6 ) : 4 * 5 * 4 * 4 = 320 numbers with three 3 - s , 3 x ... | a ) 20 , b ) 150 , c ) 225 , d ) 250 , e ) 316 | e | subtract(multiply(multiply(add(4, const_1), add(4, const_1)), multiply(4, 4)), multiply(multiply(add(4, const_1), add(4, const_1)), 4)) | add(n0,const_1)|multiply(n0,n0)|multiply(#0,#0)|multiply(#2,#1)|multiply(#2,n0)|subtract(#3,#4)| | physics |
in an examination , there were 2,000 candidates , out of which 900 candidates were girls and rest were boys . if 32 % of the boys and 32 % of the girls passed , then the total percentage of failed candidates is ? | "girls = 900 , boys = 1100 passed = ( 32 % of 1100 ) + ( 32 % of 900 ) = 352 + 288 = 640 failed = 2000 - 640 = 1360 failed % = [ ( 1360 / 2000 ) x 100 ] % = 68 % . answer : c" | a ) 35.67 % , b ) 64.75 % , c ) 68.00 % , d ) 69.57 % , e ) none of these | c | multiply(divide(subtract(subtract(multiply(const_2, multiply(const_100, const_10)), multiply(divide(32, const_100), subtract(multiply(const_2, multiply(const_100, const_10)), 900))), multiply(divide(32, const_100), 900)), multiply(const_2, multiply(const_100, const_10))), const_100) | divide(n2,const_100)|divide(n3,const_100)|multiply(const_10,const_100)|multiply(#2,const_2)|multiply(n1,#1)|subtract(#3,n1)|multiply(#0,#5)|subtract(#3,#6)|subtract(#7,#4)|divide(#8,#3)|multiply(#9,const_100)| | general |
find ( 7 x + 5 y ) / ( x - 2 y ) if x / 2 y = 3 / 2 ? | "x / 2 y = 3 / 2 = > x = 6 y / 2 = 3 y = > ( 7 x + 5 y ) / ( x - 2 y ) = ( ( 7 * ( 3 y ) ) + 5 y ) / ( 3 y - 2 y ) = > 26 y / y = 26 answer : c" | a ) 24 , b ) 25 , c ) 26 , d ) 27 , e ) 29 | c | divide(add(multiply(multiply(divide(3, 2), 2), 7), 5), subtract(multiply(divide(3, 2), 2), 2)) | divide(n4,n2)|multiply(n2,#0)|multiply(n0,#1)|subtract(#1,n2)|add(n1,#2)|divide(#4,#3)| | general |
when a student weighing 45 kgs left a class , the average weight of the remaining 59 students increased by 200 g . what is the average weight of the remaining 59 students | explanation : let the average weight of the 59 students be a . so the total weight of the 59 of them will be 59 * a . the questions states that when the weight of this student who left is added , the total weight of the class = 59 a + 45 when this student is also included , the average weight decreases by 0.2 kgs 59 a ... | a ) 55 , b ) 56 , c ) 57 , d ) 58 , e ) 59 | c | add(45, multiply(add(59, const_1), divide(200, const_1000))) | add(n1,const_1)|divide(n2,const_1000)|multiply(#0,#1)|add(n0,#2) | general |
mike drives his new corvette from san francisco to las vegas , a journey of 640 miles . he drives the first half of the trip at an average rate of 80 miles per hour , but has to slow down for the second half of his journey . if the second half of the trip takes him 200 percent longer than the first half , what is his a... | veritas prepofficial solution correct answer : c using the formula : time = distance / rate , we find that mike takes 4 hours to cover the first 320 miles of his trip . since the 2 nd 320 miles take 200 % longer than the first , it takes mike 8 hours longer , or 12 hours . ( note : 200 % longer than the first half is n... | a ) 26.7 , b ) j = 30.0 , c ) j = 40.0 , d ) j = 53.3 , e ) 60.0 | c | divide(640, add(add(multiply(divide(divide(640, const_2), 80), const_2), divide(divide(640, const_2), 80)), divide(divide(640, const_2), 80))) | divide(n0,const_2)|divide(#0,n1)|multiply(#1,const_2)|add(#1,#2)|add(#3,#1)|divide(n0,#4) | physics |
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