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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kareem is 3 times as old as his son . after 10 years , the sum of their ages will be 76 years . find their present ages . | sol . let the present age of kareem β s son be x years . then , kareem β s age = 3 x years after 10 years , kareem β s age = 3 x + 10 years and kareem β s son β s age = x + 10 years β΄ ( 3 x + 10 ) + ( x + 10 ) = 76 = > 4 x = 56 = > x = 14 β΄ kareem β s present age = 3 x = 3 Γ 14 = 42 years kareem β s son β s age = x = 1... | a ) 13 , b ) 14 , c ) 15 , d ) 16 , e ) 17 | b | divide(subtract(76, multiply(10, const_2)), add(3, const_1)) | add(n0,const_1)|multiply(n1,const_2)|subtract(n2,#1)|divide(#2,#0) | general |
what is the greatest value of x such that 4 ^ x is a factor of 21 ! ? | pretty simple , really . if m = 6 , then 4 m = 24 , which is 12 x 2 , both of which are included in 21 ! since 6 is the largest number here , its the answer . answer is b | a ) 5 , b ) 6 , c ) 3 , d ) 2 , e ) 4 | b | add(divide(subtract(21, const_1), 4), const_1) | subtract(n1,const_1)|divide(#0,n0)|add(#1,const_1) | other |
if 8 spiders make 8 webs in 8 days , then 1 spider will make 1 web in how many days ? | "let the required number days be x . less spiders , more days ( indirect proportion ) less webs , less days ( direct proportion ) spiders 1 : 8 webs 8 : 1 1 x 8 x x = 8 x 1 x 8 = > x = 8 answer is d" | a ) 7 , b ) 6 , c ) 5 , d ) 8 , e ) 3 | d | multiply(1, 8) | multiply(n0,n3)| | physics |
what is the sum of the integers from - 160 to 162 , inclusive ? | "in an arithmetic progression , the nth term is given by tn = a + ( n - 1 ) d here tn = 162 , a = - 160 , d = 1 hence , 162 = - 160 + ( n - 1 ) or n = 323 sum of n terms can be calculated by sn = n / 2 ( a + l ) a = first term , l = last term , n = no . of terms sn = 323 * ( - 160 + 162 ) / 2 sn = 323 * 2 / 2 = 323 ans... | a ) 325 , b ) 327 , c ) 323 , d ) 330 , e ) 350 | c | divide(multiply(160, 162), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
chloe has $ 27 dollars to go to the fair . she is bringing 2 other friends along . admission to the fair is $ 3 . each ride costs $ 1.70 . how many rides can she and her friends go on and how much money will be left over ? | admission to the fair is $ 3 per person . if she wants her and her friends to go , the cost will be $ 9 . 27 - 9 is 18 . to go on a ride is $ 5.10 . 5.10 x 3 is 15.30 . 18 - 15.30 is 2.70 . they can go on 3 rides and $ 2.70 will be left over . the correct answer is b . | ['a ) they can go on 15 rides and $ 1.50 will be left over .', 'b ) they can go on 3 rides and $ 2.70 will be left over .', 'c ) they can go on 7 rides and $ 0 will be left over .', 'd ) they can go on 6 rides and $ 0 will be left over .', 'e ) they can go on 5 rides and $ 1.50 will be left over .'] | b | subtract(subtract(27, add(multiply(3, 2), 3)), multiply(floor(divide(subtract(27, add(multiply(3, 2), 3)), 1.7)), 1.7)) | multiply(n1,n2)|add(n2,#0)|subtract(n0,#1)|divide(#2,n3)|floor(#3)|multiply(n3,#4)|subtract(#2,#5) | geometry |
if the selling price of 50 articles is equal to the cost price of 30 articles , then the loss or gain percent is : | "let c . p . of each article be re . 1 . then , c . p . of 50 articles = rs . 50 ; s . p . of 50 articles = rs . 30 . loss % = 20 / 50 * 100 = 40 % answer : b" | a ) 10 % , b ) 40 % , c ) 30 % , d ) 25 % , e ) 35 % | b | subtract(const_100.0, 30) | subtract(const_100.0,n1)| | gain |
solution x is 10 percent alcohol by volume , and solution y is 30 percent alcohol by volume . how many milliliters of solution y must be added to 200 milliliters of solution x to create a solution that is 25 percent alcohol by volume ? | we know that x is 10 % , y is 30 % and w . avg = 25 % . what does this mean with respect to w . avg technique ? w . avg is 1 portion away from y and 3 portion away from x so for every 1 portion of x we will have to add 3 portions of y . if x = 200 then y = 600 answer : e | a ) 250 / 3 , b ) 500 / 3 , c ) 400 , d ) 480 , e ) 600 | e | multiply(divide(subtract(25, 10), subtract(30, 25)), 200) | subtract(n3,n0)|subtract(n1,n3)|divide(#0,#1)|multiply(n2,#2) | general |
what will be the compound interest on a sum of rs . 26,000 after 3 years at the rate of 12 % p . a . ? | "amount = [ 26000 * ( 1 + 12 / 100 ) 3 ] = 26000 * 28 / 25 * 28 / 25 * 28 / 25 = rs . 36528.12 c . i . = ( 36528.12 - 26000 ) = rs . 10528.12 answer : e" | a ) s . 10123.77 , b ) s . 10123.21 , c ) s . 10123.20 , d ) s . 10123.28 , e ) s . 10528.12 | e | subtract(multiply(multiply(multiply(const_4, const_100), const_100), power(add(const_1, divide(12, const_100)), 3)), multiply(multiply(const_4, const_100), const_100)) | divide(n2,const_100)|multiply(const_100,const_4)|add(#0,const_1)|multiply(#1,const_100)|power(#2,n1)|multiply(#3,#4)|subtract(#5,#3)| | gain |
the star running back on our football team got most of his total yardage running . the rest was catching passes . he caught passes for 60 yards . his total yardage was 150 yards . the running back for the other team got 200 yards . how many yards did the star running back on our football team get running ? | . the other team is extra information . 150 β 60 = 90 he got 90 yards running . correct answer e | a ) 50 yards , b ) 60 yards , c ) 70 yards , d ) 80 yards , e ) 90 yards | e | subtract(150, 60) | subtract(n1,n0) | general |
in what ratio must rice of rs . 30 per kg be mixed with rice of rs . 56 per kg so that cost of mixture is rs . 35 per kg ? | "( 35 - 56 ) / ( 30 - 35 ) = 6 / 2 = 21 : 5 answer : e" | a ) 28 : 15 , b ) 6 : 7 , c ) 7 : 6 , d ) 5 : 21 , e ) 21 : 5 | e | divide(divide(subtract(35, 56), subtract(30, 56)), subtract(const_1, divide(subtract(35, 56), subtract(30, 56)))) | subtract(n2,n1)|subtract(n0,n1)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)| | other |
an automobile parts supplier charges $ 25 per package of gaskets . when a customer orders more than 10 packages of gaskets , the supplier charges 4 / 5 the price for each package in excess of 10 . during a certain week , the supplier sold 60 packages of gaskets . if 30 percent of the gaskets went to company x , 15 perc... | "$ 25 per packet of gasket in case a customer orders less than 10 in case a customer orders > 10 price per gasket = 25 * 4 / 5 = 20 a certain week the supplier sold 60 gasket 1 . he sold 30 % of the gaskets to x = 18 gaskets = 25 * 10 + 20 * 8 = 250 + 160 = 410 2 . he sold 15 % of the gaskets to y = 9 gaskets = 25 * 9 ... | a ) 1305 , b ) 1375 , c ) 1345 , d ) 1415 , e ) 1455 | c | add(add(multiply(multiply(25, divide(4, 5)), subtract(subtract(subtract(60, multiply(60, divide(30, const_100))), multiply(60, divide(30, const_100))), 10)), multiply(25, 10)), add(multiply(25, multiply(60, divide(30, const_100))), multiply(25, multiply(60, divide(30, const_100))))) | divide(n2,n3)|divide(n6,const_100)|multiply(n0,n1)|multiply(n0,#0)|multiply(n5,#1)|multiply(n0,#4)|subtract(n5,#4)|add(#5,#5)|subtract(#6,#4)|subtract(#8,n1)|multiply(#3,#9)|add(#10,#2)|add(#11,#7)| | general |
two cars start from the opposite places of a main road , 113 km apart . first car runs for 25 km and takes a right turn and then runs 15 km . it then turns left and then runs for another 25 km and then takes the direction back to reach the main road . in the mean time , due to minor break down the other car has run onl... | answer : d ) 28 km | a ) 65 , b ) 38 , c ) 20 , d ) 28 , e ) 21 | d | subtract(subtract(113, 35), add(25, 25)) | add(n1,n1)|subtract(n0,n4)|subtract(#1,#0)| | physics |
a straight line in the xy - plane has a slope of 4 and a y - intercept of 4 . on this line , what is the x - coordinate of the point whose y - coordinate is 800 ? | eq of line = y = mx + c m = 4 , c = 4 y = 4 x + 4 , substitute y by 800 as given in question . 800 = 4 x + 4 , x = 199 . correct option is d | a ) 233 , b ) 299 , c ) 333 , d ) 199 , e ) 339 | d | divide(subtract(800, 4), 4) | subtract(n2,n0)|divide(#0,n0) | general |
a shop owner sells 25 mtr of cloth and gains sp of 10 mtrs . find the gain % ? | "here , selling price of 10 m cloth is obtained as profit . profit of 10 m cloth = ( s . p . of 25 m cloth ) β ( c . p . of 25 m cloth ) selling price of 15 m cloth = selling price of 25 m of cloth let cost of each metre be rs . 100 . therefore , cost price of 15 m cloth = rs . 1500 and s . p . of 15 m cloth = rs . rs ... | a ) 30 % , b ) 40 % , c ) 50 % , d ) 66.67 % , e ) 70 % | d | multiply(divide(10, subtract(25, 10)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100)| | gain |
what number has a 150 : 1 ratio to the number 2 ? | "150 : 1 = x : 2 x = 150 * 2 x = 300 answer : b" | a ) 130 , b ) 300 , c ) 200 , d ) 30 , e ) 75 | b | multiply(2, 150) | multiply(n0,n2)| | other |
on a saturday night , each of the rooms at a certain motel was rented for either $ 40 or $ 60 . if 10 of the rooms that were rented for $ 60 had instead been rented for $ 40 , then the total rent the motel charged for that night would have been reduced by 50 percent . what was the total rent the motel actually charged ... | "let total rent the motel charge for all rooms = x if 10 rooms that were rented for 60 $ had instead been rented for 40 $ , then total difference in prices = 20 $ * 10 = 200 $ total rent the motel charged would have been reduced by 50 % . 5 x = 200 = > x = 400 answer a" | a ) $ 400 , b ) $ 800 , c ) $ 1,000 , d ) $ 1,600 , e ) $ 2,400 | a | divide(multiply(10, subtract(60, 40)), divide(50, const_100)) | divide(n5,const_100)|subtract(n1,n0)|multiply(n2,#1)|divide(#2,#0)| | gain |
a sum fetched a total simple interest of rs . 6016.75 at the rate of 8 p . c . p . a . in 5 years . what is the sum ? | "explanation : principal = rs 100 x 6016.75 / 8 x 5 = rs . 601675 / 40 = rs . 15041.875 answer : option d" | a ) rs . 16241.875 , b ) rs . 15041.85 , c ) rs . 15401.875 , d ) rs . 15041.875 , e ) none of these | d | divide(divide(multiply(6016.75, const_100), 8), 5) | multiply(n0,const_100)|divide(#0,n1)|divide(#1,n2)| | gain |
x starts a business with rs . 45000 . y joins in the business after 2 months with rs . 30000 . what will be the ratio in which they should share the profit at the end of the year ? | "explanation : ratio in which they should share the profit = ratio of the investments multiplied by the time period = 45000 Γ£ β 12 : 30000 Γ£ β 10 = 45 Γ£ β 12 : 30 Γ£ β 10 = 3 Γ£ β 12 : 2 Γ£ β 10 = 9 : 5 answer : option b" | a ) 1 : 2 , b ) 9 : 5 , c ) 1 : 5 , d ) 3 : 1 , e ) 1 : 1 | b | divide(multiply(45000, const_12), multiply(30000, add(const_4, const_3))) | add(const_3,const_4)|multiply(n0,const_12)|multiply(n2,#0)|divide(#1,#2)| | other |
a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 165 m long ? | ": speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 165 = 525 m required time = 525 * 2 / 25 = 42 sec answer : e" | a ) 40 , b ) 99 , c ) 88 , d ) 77 , e ) 42 | e | divide(360, multiply(subtract(45, 165), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
two pipes can fill a tank in 8 minutes and 5 minutes . an outlet pipe can empty the tank in 12 minutes . if all the pipes are opened when the tank is empty , then how many minutes will it take to fill the tank ? | "part of the filled by all the three pipes in one minute = 1 / 8 + 1 / 5 - 1 / 12 = ( 15 + 24 - 10 ) / 120 = 29 / 120 so , the tank becomes full in 14 minutes . answer : e" | a ) 30 minutes , b ) 17 minutes , c ) 15 minutes , d ) 10 minutes , e ) 14 minutes | e | subtract(add(divide(const_1, 8), divide(const_1, 5)), divide(const_1, 12)) | divide(const_1,n0)|divide(const_1,n1)|divide(const_1,n2)|add(#0,#1)|subtract(#3,#2)| | physics |
the owner of a furniture shop charges his customer 15 % more than the cost price . if a customer paid rs . 8325 for a computer table , then what was the cost price of the computer table ? | "cp = sp * ( 100 / ( 100 + profit % ) ) = 8325 ( 100 / 115 ) = rs . 7239.13 . answer : b" | a ) rs . 6725 , b ) rs . 7239.13 , c ) rs . 6908 , d ) rs . 6725 , e ) rs . 6728 | b | divide(8325, add(const_1, divide(15, const_100))) | divide(n0,const_100)|add(#0,const_1)|divide(n1,#1)| | gain |
a boat takes 19 hours for travelling downstream from point a to point b and coming back to a point c midway between a and b . if the velocity of the stream is 4 km / h and the speed of the boat in still water is 14 km / h , what is the distance between a and b ? | "speed of boat for downstream = 14 + 4 = 18 km / hr speed of boat for upstream = 14 β 4 = 10 km / hr distance = x x / 18 + ( x / 2 ) / 10 = 19 x = 180 km answer : b" | a ) 200 km , b ) 180 km , c ) 160 km , d ) 220 km , e ) none of these | b | divide(19, add(divide(const_1, add(14, 4)), divide(const_1, multiply(subtract(14, 4), const_2)))) | add(n1,n2)|subtract(n2,n1)|divide(const_1,#0)|multiply(#1,const_2)|divide(const_1,#3)|add(#2,#4)|divide(n0,#5)| | physics |
if 73 ! has 16 zeroes at the end , how many zeroes will 80 ! have at the end ? | prime factoring : you need a 2 and a 5 to make a 10 ( a ` ` zero ' ' ) , and there are tons of 2 ' s so let ' s skip these and focus on the ( rarer ) 5 s : 80 ! = 1 * 2 * 3 * 4 * 5 * 6 . . . * 78 * 79 * 80 since there are 80 consecutive numbers , there are 16 multiples of 5 in there , but if we ' re prime factoring , w... | a ) 16 , b ) 17 , c ) 18 , d ) 19 , e ) 20 | d | add(divide(80, divide(const_10, const_2)), const_3) | divide(const_10,const_2)|divide(n2,#0)|add(#1,const_3) | other |
in 1990 the budgets for projects q and v were $ 700,000 and $ 780,000 , respectively . in each of the next 10 years , the budget for q was increased by $ 30,000 and the budget for v was decreased by $ 10,000 . in which year was the budget for q equal to the budget for v ? | "let the no of years it takes is x . 700 + 30 x = 780 - 10 x - - > 40 x = 80 and x = 2 . thus , it happens in 1992 . a ." | a ) 1992 , b ) 1993 , c ) 1994 , d ) 1995 , e ) 1996 | a | add(1990, multiply(10, multiply(const_2, const_3))) | multiply(const_2,const_3)|multiply(n3,#0)|add(n0,#1)| | general |
6 bells commence tolling together and toll at the intervals of 24 , 68 , 1012 seconds resp . in 60 minutes how many times they will toll together . | explanation : lcm of 2 - 4 - 6 - 8 - 10 - 12 is 120 seconds , that is 2 minutes . now 60 / 2 = 30 adding one bell at the starting it will 30 + 1 = 31 answer : option d | a ) 15 , b ) 16 , c ) 30 , d ) 31 , e ) none of these | d | divide(multiply(60, const_60), multiply(add(const_10, const_2), const_10)) | add(const_10,const_2)|multiply(n4,const_60)|multiply(#0,const_10)|divide(#1,#2) | physics |
in a full deck of 52 cards , there are 13 spades . a game is played whereby a card from is chosen at random from the deck . if the card is a spade , a winner is declared . if the card is not a spade , it is returned to the deck , the deck is reshuffled and another card is drawn . this process is repeated until a spade ... | favorable case = the spade is picked in the third draw or later unfavorable case = the spade is picked in either first draw or in second draw probability = favorable outcomes / total out comes also probability = 1 - ( unfavorable outcomes / total out comes ) unfavorable case : probability of spade picked in first draw ... | a ) 1 / 8 , b ) 1 / 4 , c ) 1 / 2 , d ) 3 / 4 , e ) 7 / 8 | d | subtract(const_1, divide(13, 52)) | divide(n1,n0)|subtract(const_1,#0) | probability |
of 600 surveyed students , 20 % of those who read book a also read book b and 25 % of those who read book b also read book a . if each student read at least one of the books , what is the difference between the number of students who read only book a and the number of students who read only book b ? | say the number of students who read book a is a and the number of students who read book b is b . given that 20 % of those who read book a also read book b and 25 % of those who read book b also read book a , so the number of students who read both books is 0.2 a = 0.25 b - - > a = 1.25 b . since each student read at l... | a ) 20 , b ) 25 , c ) 30 , d ) 35 , e ) 75 | e | subtract(multiply(multiply(divide(600, subtract(add(divide(divide(25, const_100), divide(20, const_100)), const_1), divide(25, const_100))), divide(divide(25, const_100), divide(20, const_100))), subtract(const_1, divide(20, const_100))), multiply(divide(600, subtract(add(divide(divide(25, const_100), divide(20, const_... | divide(n2,const_100)|divide(n1,const_100)|divide(#0,#1)|subtract(const_1,#1)|subtract(const_1,#0)|add(#2,const_1)|subtract(#5,#0)|divide(n0,#6)|multiply(#7,#2)|multiply(#7,#4)|multiply(#8,#3)|subtract(#10,#9) | gain |
if 4 ( p ' s capital ) = 6 ( q ' s capital ) = 10 ( r ' s capital ) , then out of the total profit of rs 4340 , r will receive | explanation : let p ' s capital = p , q ' s capital = q and r ' s capital = r then 4 p = 6 q = 10 r = > 2 p = 3 q = 5 r = > q = 2 p / 3 r = 2 p / 5 p : q : r = p : 2 p / 3 : 2 p / 5 = 15 : 10 : 6 r ' s share = 4340 * ( 6 / 31 ) = 140 * 6 = 840 . answer : option c | a ) 600 , b ) 700 , c ) 840 , d ) 900 , e ) none of these | c | multiply(4340, divide(6, add(add(add(10, add(4, const_1)), 10), 6))) | add(n0,const_1)|add(n2,#0)|add(n2,#1)|add(n1,#2)|divide(n1,#3)|multiply(n3,#4) | general |
there are 6 red balls and 4 blue balls in a jar . if 3 balls are selected from the jar , what is the probability that all 3 balls selected are red balls ? | "the number of ways of choosing 3 balls from the jar is 10 c 3 = 120 . the number of ways of choosing 3 red balls is 6 c 3 = 20 . p ( 3 red balls ) = 20 / 120 = 1 / 6 . the answer is b ." | a ) 1 / 3 , b ) 1 / 6 , c ) 5 / 12 , d ) 5 / 24 , e ) 7 / 24 | b | divide(choose(4, 3), choose(add(6, 4), 3)) | add(n0,n1)|choose(n1,n2)|choose(#0,n2)|divide(#1,#2)| | probability |
for how many integer values of n will the value of the expression 4 n + 7 be an integer greater than 1 and less than 100 ? | "4 n + 7 > 1 4 n > - 6 n > - ( 3 / 2 ) n > - 1.5 ( n = - 1 , 0 , 1 , 2 3 . . . . . . . . upto infinity ) from second constraint 4 n + 7 < 100 4 n < 93 n < 23 . 25 n = ( - infinity , . . . . . . . - 3 , - 2 , - 1 , 0 , 1 , 2 , . . . . . . . . . upto 23 ) combining the two - 1.5 < n < 23.25 n = 1 to 23 ( 23 integers ) an... | a ) 30 , b ) 28 , c ) 27 , d ) 25 , e ) 26 | d | subtract(floor(divide(subtract(100, 7), 4)), floor(divide(subtract(1, 7), 4))) | subtract(n3,n1)|subtract(n2,n1)|divide(#0,n0)|divide(#1,n0)|floor(#2)|floor(#3)|subtract(#4,#5)| | general |
if the arithmetic mean of p and q is 10 and the arithmetic mean of q and r is 25 , what is the value of r - p ? | "arithmetic mean expression for p and q : ( p + q ) / 2 = 10 ; p + q = 20 - - - - eq 1 arithmetic mean expression for q and r : ( q + r ) / 2 = 20 ; q + r = 50 - - - - eq 2 subtracting eq 1 from eq 2 we get : r - p = 30 hence , the correct answer is c" | a ) 20 , b ) 10 , c ) 30 , d ) 40 , e ) 5 | c | subtract(multiply(25, const_2), multiply(10, const_2)) | multiply(n1,const_2)|multiply(n0,const_2)|subtract(#0,#1)| | general |
if the population of a certain country increases at the rate of one person every 25 seconds , by how many persons does the population increase in 1 hour ? | "answer = 2.4 * 60 = 144 answer is c" | a ) 100 , b ) 120 , c ) 144 , d ) 180 , e ) 160 | c | multiply(divide(const_60, 25), 1) | divide(const_60,n0)|multiply(n1,#0)| | physics |
there were two candidates in an election . winner candidate received 62 % of votes and won the election by 300 votes . find the number of votes casted to the winning candidate ? | "w = 62 % l = 38 % 62 % - 38 % = 24 % 24 % - - - - - - - - 300 62 % - - - - - - - - ? = > 775 answer : e" | a ) 456 , b ) 744 , c ) 912 , d ) 1200 , e ) 775 | e | divide(multiply(divide(300, divide(subtract(62, subtract(const_100, 62)), const_100)), 62), const_100) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)|multiply(n0,#3)|divide(#4,const_100)| | gain |
10 percent of ram ' s monthly salary is equal to 8 percent of shyam ' s monthly salary . shyam ' s monthly salary is twice abhinav ' s monthly salary . if abhinav ' s annual salary is rs . 1.92 lakhs , find ram ' s monthly salary ? | let the monthly salaries of ram and shyam be rs . r and rs . s respectively . 10 / 100 r = 8 / 100 s r = 4 / 5 s monthly salary of abhinav = ( 1.92 lakhs ) / 12 = rs . 0.16 lakhs s = 2 ( 0.16 lakhs ) = 0.32 lakhs r = 4 / 5 ( 0.32 lakhs ) = rs . 25600 answer : c | a ) rs . 18000 , b ) rs . 20000 , c ) rs . 25600 , d ) rs . 32000 , e ) none of these | c | multiply(multiply(divide(multiply(multiply(8, const_100), multiply(divide(1.92, const_12), const_2)), multiply(const_100, 10)), const_100), const_1000) | divide(n2,const_12)|multiply(n1,const_100)|multiply(n0,const_100)|multiply(#0,const_2)|multiply(#1,#3)|divide(#4,#2)|multiply(#5,const_100)|multiply(#6,const_1000) | general |
during a sale of 20 % on everything in a store , a kid is successful in convincing the store manager to give him 20 candies for the discounted price of 10 candies . the store still makes a profit of 12 % on this sale . what is the mark up percentage on each candy ? | "hi akhil , i can share the way i solved it . . let ' s say marked price = m so , there ' s a discount of 20 % on m so , new s . p . = 80 % of m now , the child convinces the owner to sell 20 candies for the price of 14 candies . let ' s say each candy after discount is 1 $ . so , s . p . of 20 candies = 20 $ . the chi... | a ) 100 % , b ) 80 % , c ) 75 % , d ) 66 + 2 / 3 % , e ) 55 % | b | multiply(subtract(divide(add(divide(12, const_100), const_1), multiply(subtract(const_1, divide(20, const_100)), divide(subtract(20, 10), 20))), const_1), const_100) | divide(n3,const_100)|divide(n0,const_100)|subtract(n1,n2)|add(#0,const_1)|divide(#2,n1)|subtract(const_1,#1)|multiply(#4,#5)|divide(#3,#6)|subtract(#7,const_1)|multiply(#8,const_100)| | gain |
if the selling price of 50 articles is equal to the cost price of 25 articles , then the loss or gain percent is : | "c . p . of each article be re . 1 . then , c . p . of 50 articles = rs . 50 ; s . p . of 50 articles = rs . 25 . loss % = 25 / 50 * 100 = 50 % answer e" | a ) 45 % , b ) 23 % , c ) 20 % , d ) 60 % , e ) 50 % | e | subtract(50, 25) | subtract(n0,n1)| | gain |
find the area of trapezium whose parallel sides are 20 cm and 18 cm long , and the distance between them is 15 cm ? | "area of a trapezium = 1 / 2 ( sum of parallel sides ) * ( perpendicular distance between them ) = 1 / 2 ( 20 + 18 ) * ( 15 ) = 285 cm 2 answer : c" | a ) 288 , b ) 276 , c ) 285 , d ) 299 , e ) 261 | c | quadrilateral_area(15, 18, 20) | quadrilateral_area(n2,n1,n0)| | physics |
when the bus started from the first bus stop , the number of male passengers to the number of female passengers was 1 : 3 . at the first stop , 16 passengers got down and 6 more female passengers got in . the ratio of the male to female passengers now became 1 : 2 . what was the total number of passengers in the bus wh... | let , initially the number of males and females in the bus be β x β and β 3 x β respectively . at the first stop , suppose β m β males and β f β females left the bus . therefore at the first stop the number of males is β x - m β and the number of females is β ( 3 x - f ) + 6 β therefore ( x - m ) / [ ( 3 x - f ) + 6 ] ... | a ) 64 , b ) 48 , c ) 54 , d ) 46 , e ) can not be determined | e | subtract(negate(multiply(divide(16, add(3, 1)), const_2)), subtract(6, multiply(divide(16, add(3, 1)), const_3))) | add(n0,n1)|divide(n2,#0)|multiply(#1,const_2)|multiply(#1,const_3)|negate(#2)|subtract(n3,#3)|subtract(#4,#5) | general |
when processing flower - nectar into honey bees ' extract , a considerable amount of water gets reduced . how much flower - nectar must be processed to yield 1 kg of honey , if nectar contains 50 % water , and the honey obtained from this nectar contains 15 % water ? | "explanation : flower - nectar contains 50 % of non - water part . in honey this non - water part constitutes 85 % ( 100 - 15 ) . therefore 0.5 x amount of flower - nectar = 0.85 x amount of honey = 0.85 x 1 kg therefore amount of flower - nectar needed = ( 0.85 / 0.51 ) kg = 1.7 kgs answer : c" | a ) 1.2 kg , b ) 1.5 kg , c ) 1.7 kg , d ) 1.9 kg , e ) none of these | c | divide(divide(subtract(const_100, 15), const_100), divide(50, const_100)) | divide(n1,const_100)|subtract(const_100,n2)|divide(#1,const_100)|divide(#2,#0)| | general |
a batsman scored 90 runs which included 2 boundaries and 7 sixes . what percent of his total score did he make by running between the wickets . | "explanation : number of runs made by running = 90 - ( 2 x 4 + 7 x 6 ) = 90 - ( 50 ) = 40 now , we need to calculate 40 is what percent of 90 . = > 40 / 90 * 100 = 44.44 % option c" | a ) 30 % , b ) 40 % , c ) 44.44 % , d ) 60 % , e ) 80 % | c | multiply(divide(subtract(90, add(multiply(2, 7), multiply(7, 2))), 90), const_100) | multiply(n1,n2)|multiply(n1,n2)|add(#0,#1)|subtract(n0,#2)|divide(#3,n0)|multiply(#4,const_100)| | general |
a bag contains 10 red jellybeans and 10 blue jellybeans . if 3 jellybeans are removed one at a time , at random and are not replaced , what is the probability w that all 3 jellybeans removed from the bag are blue ? | method - 1 10 red jellybeans and 10 blue jellybeans total outcomes = no . of ways to choose 3 jelly bean at random out of a total 20 jellybeans = 20 c 3 = 1140 favourable outcomes = no . of ways to choose 3 jelly bean such that they are all blue out of 10 blue = 10 c 3 = 120 probability = favourable outcomes / total ou... | a ) 9 / 100 , b ) 2 / 19 , c ) 1 / 8 , d ) 3 / 20 , e ) 3 / 10 | b | divide(choose(10, 3), choose(add(10, 10), 3)) | add(n0,n0)|choose(n0,n2)|choose(#0,n2)|divide(#1,#2) | probability |
peter invested a certain sum of money in a simple interest bond whose value grew to $ 300 at the end of 3 years and further to $ 400 at the end of another 5 years . what was the rate of interest in which he invested his sum ? | answer initial amount invested = $ x amount at the end of year 3 = $ 300 amount at the end of year 8 ( another 5 years ) = $ 400 therefore , the interest earned for the 5 year period between the 3 rd year and 8 th year = $ 400 - $ 300 = $ 100 as the simple interest earned for a period of 5 years is $ 100 , interest ear... | a ) 12 % , b ) 12.5 % , c ) 6.67 % , d ) 6.25 % , e ) 8.33 % | e | multiply(divide(divide(subtract(400, 300), 5), subtract(300, multiply(divide(subtract(400, 300), 5), 3))), const_100) | subtract(n2,n0)|divide(#0,n3)|multiply(n1,#1)|subtract(n0,#2)|divide(#1,#3)|multiply(#4,const_100) | gain |
28 buckets of water fill a tank when the capacity of each bucket is 13.5 litres . how many buckets will be required to fill the same tank if the capacity of each bucket is 9 litres ? | "capacity of the tank = 28 Γ£ β 13.5 = 378 litres when the capacity of each bucket = 9 litres , then the required no . of buckets = 378 Γ’ Β β 9 = 42 answer b" | a ) 30 , b ) 42 , c ) 60 , d ) data inadequate , e ) none of these | b | divide(multiply(13.5, 28), 9) | multiply(n0,n1)|divide(#0,n2)| | physics |
the average ( arithmetic mean ) of 4 different positive integers is 16 . if the first of these integers in 3 times the second integer and the second integer is 2 less than the third integer , what is the least possible value of the fourth integer ? | let the second integer be x and the fourth be a . then [ 3 x + x + ( x + 2 ) + a ] / 4 = 16 = > 5 x + 2 + a = 64 = > 5 x + a = 62 = > a = 62 - 5 x from the above equation we can see that a is minimum when x is maximum , provided both are positive the maximum value that x can take in the above equation while still keepi... | a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 1 | d | subtract(subtract(multiply(16, 4), 2), multiply(const_12, add(3, 2))) | add(n2,n3)|multiply(n0,n1)|multiply(#0,const_12)|subtract(#1,n3)|subtract(#3,#2) | general |
36 welders work at a constant rate they complete an order in 3 days . if after the first day , 12 welders start to work on the other project , how many more days the remaining welders will need to complete the rest of the order ? | "1 . we need to find out the time taken by 24 workers after day 1 . 2 . total no . of wokers * total time taken = time taken by 1 worker 3 . time taken by 1 worker = 36 * 3 = 108 days 4 . but on day 1 thirty - six workers had already worked finishing 1 / 3 of the job . so 24 workers have to finish only 2 / 3 of the job... | a ) 3 , b ) 2 , c ) 8 , d ) 4 , e ) 6 | a | divide(divide(subtract(const_1, multiply(divide(const_1, multiply(36, 3)), 36)), subtract(36, 12)), divide(const_1, multiply(36, 3))) | multiply(n0,n1)|subtract(n0,n2)|divide(const_1,#0)|multiply(n0,#2)|subtract(const_1,#3)|divide(#4,#1)|divide(#5,#2)| | physics |
frank the fencemaker needs to fence in a rectangular yard . he fences in the entire yard , except for one full side of the yard , which equals 40 feet . the yard has an area of 200 square feet . how many feet offence does frank use ? | "area = length x breadth 200 = 40 x breadth so , breadth = 5 units fencing required is - breadth + breadth + length 5 + 5 + 40 = > 50 feet answer must be ( c ) 50" | a ) 14 , b ) 47 , c ) 50 , d ) 180 , e ) 240 | c | add(add(divide(200, 40), divide(200, 40)), 40) | divide(n1,n0)|add(#0,#0)|add(n0,#1)| | geometry |
in an examination , a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer . if he attempts all 60 questions and secures 120 marks , the no of questions he attempts correctly is : | "explanation : let the number of correct answers be x . number of incorrect answers = ( 60 β x ) . 4 x β ( 60 β x ) = 120 = > 5 x = 180 = > x = 36 answer : b" | a ) 35 , b ) 36 , c ) 40 , d ) 42 , e ) 44 | b | divide(add(120, 60), add(4, 1)) | add(n2,n3)|add(n0,n1)|divide(#0,#1)| | physics |
a certain car increased its average speed by 4 miles per hour in each successive 5 - minute interval after the first interval . if in the first 5 - minute interval its average speed was 20 miles per hour , how many miles did the car travel in the third 5 - minute interval ? | "in the third time interval the average speed of the car was 20 + 4 + 5 = 29 miles per hour ; in 5 minutes ( 1 / 12 hour ) at that speed car would travel 29 * 1 / 12 = 2.5 miles . answer : d ." | a ) 1.0 , b ) 1.5 , c ) 2.0 , d ) 2.5 , e ) 3.0 | d | multiply(add(add(20, 4), 5), divide(4, const_60)) | add(n0,n3)|divide(n1,const_60)|add(n0,#0)|multiply(#2,#1)| | physics |
an engineer undertakes a project to build a road 10 km long in 15 days and employs 30 men for the purpose . after 5 days , he finds only 2 km of the road has been completed . find the ( approximate ) number of extra men he must employ to finish the work in time . | "30 workers working already let x be the total men required to finish the task in next 10 days 2 km done hence remaining is 8 km also , work has to be completed in next 10 days ( 15 - 10 = 5 ) we know that , proportion of men to distance is direct proportion and , proportion of men to days is inverse proportion hence ,... | a ) 15 , b ) 20 , c ) 25 , d ) 28 , e ) 30 | e | subtract(divide(multiply(multiply(30, subtract(10, 2)), 5), multiply(2, subtract(15, 5))), 30) | subtract(n0,n4)|subtract(n1,n3)|multiply(n2,#0)|multiply(n4,#1)|multiply(n3,#2)|divide(#4,#3)|subtract(#5,n2)| | physics |
chandigarh express of 100 m runs at a speed of 60 km / hr . what will be the time taken to cross a platform of 150 meters long ? | explanation : given : length of train = 100 m , speed of train = 60 km / hr , length of platform = 150 m 1 ) always remember first step is the conversion of units . convert 60 km / hr into m / s by multiplying it with ( 5 / 18 ) speed of the train = 60 x 5 / 18 = 16.66 m / s 2 ) distance covered by the train in passing... | a ) 11.00 sec , b ) 12.50 sec , c ) 15.00 sec , d ) 15.23 sec , e ) 5.23 sec | c | divide(add(100, 150), multiply(60, const_0_2778)) | add(n0,n2)|multiply(n1,const_0_2778)|divide(#0,#1) | physics |
a cube is painted red on all faces . it is then cut into 27 equal smaller cubes . how many p cubes are painted on only 2 faces ? | "1 ) draw a simple cube 2 ) draw 9 squares on each face of the cube ( so that it looks like a rubik ' s cube ) - this is what the cube will look like when it ' s cut into 27 equal smaller cubes . 3 ) remember that the outside of the cube is the part that ' s painted . . . . the mini - cubes with 2 painted sides are all... | a ) 12 , b ) 8 , c ) 6 , d ) 10 , e ) 16 | a | multiply(const_4, power(27, divide(const_1, const_3))) | divide(const_1,const_3)|power(n0,#0)|multiply(#1,const_4)| | geometry |
in a school 50 % of the students are younger than 10 , 1 / 20 are 10 years old and 1 / 10 are older than 10 but younger than 12 , the remaining 70 students are 12 years or older . how many students are 10 years old ? | let us write the fraction for each group of students group a : younger than 10 : 50 % = 50 / 100 = 1 / 2 group b : 10 years old : 1 / 20 group c : older that 10 but younger than 12 : 1 / 10 group d : 12 years or older : 70 students the fraction for group a , b and c together is given by 1 / 2 + 1 / 20 + 1 / 10 = 10 / 2... | a ) 50 , b ) 60 , c ) 10 , d ) 20 , e ) 30 | c | divide(divide(70, subtract(subtract(subtract(const_1, divide(50, const_100)), divide(1, 20)), divide(1, 10))), 20) | divide(n0,const_100)|divide(n2,n3)|divide(n2,n1)|subtract(const_1,#0)|subtract(#3,#1)|subtract(#4,#2)|divide(n9,#5)|divide(#6,n3) | gain |
a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 270 metres long in 20 seconds . the length of the train ( in metres ) is : | "explanation : let the length of train be l m . acc . to question ( 270 + l ) / 20 = l / 8 2160 + 8 l = 20 l l = 2160 / 12 = 180 m answer a" | a ) 180 , b ) 176 , c ) 175 , d ) 96 , e ) none of these | a | multiply(divide(270, subtract(20, 8)), 8) | subtract(n2,n0)|divide(n1,#0)|multiply(n0,#1)| | physics |
if the operation β¬ is defined for all x and y by the equation x β¬ y = 2 * x * y , then 7 β¬ ( 4 β¬ 5 ) = | "working inside out , ( 4 β¬ 5 ) = 2 * 4 * 5 = 40 7 β¬ 40 = 2 * 3 * 40 = 560 hence , answer is e" | a ) 80 , b ) 120 , c ) 160 , d ) 240 , e ) 560 | e | multiply(multiply(2, 7), multiply(multiply(2, 4), 5)) | multiply(n0,n1)|multiply(n0,n2)|multiply(n3,#1)|multiply(#0,#2)| | general |
if a farmer wants to plough a farm field on time , he must plough 90 hectares a day . for technical reasons he ploughed only 85 hectares a day , hence he had to plough 2 more days than he planned and he still has 40 hectares left . what is the area of the farm field and how many days the farmer planned to work initiall... | "let x be the number of days in the initial plan . therefore , the whole field is 90 Γ’ βΉ β¦ x hectares . the farmer had to work for x + 2 days , and he ploughed 85 ( x + 2 ) hectares , leaving 40 hectares unploughed . then we have the equation : 90 x = 85 ( x + 2 ) + 40 5 x = 210 x = 42 so the farmer planned to have the... | a ) 1600 , b ) 2490 , c ) 3780 , d ) 4235 , e ) 6179 | c | multiply(divide(add(40, multiply(2, 85)), subtract(90, 85)), 90) | multiply(n1,n2)|subtract(n0,n1)|add(n3,#0)|divide(#2,#1)|multiply(n0,#3)| | physics |
the workforce of company x is 60 % female . the company hired 30 additional male workers , and as a result , the percent of female workers dropped to 55 % . how many employees did the company have after hiring the additional male workers ? | "let ' s xx be total quantity of employees 0.6 x = females before adding men 0.55 ( x + 30 ) = females after adding men as quantity of women does n ' t change we can make an equation : 0.6 x = 0.55 ( x + 30 ) 0.05 x = 16.5 x = 330 - this is quantity of employees before adding 30 men so after adding it will be 360 answe... | a ) 160 , b ) 220 , c ) 240 , d ) 360 , e ) 420 | d | add(divide(multiply(divide(55, const_100), 30), subtract(divide(60, const_100), divide(55, const_100))), 30) | divide(n2,const_100)|divide(n0,const_100)|multiply(n1,#0)|subtract(#1,#0)|divide(#2,#3)|add(n1,#4)| | gain |
a merchant gets a 5 % discount on each meter of fabric he buys after the first 2,000 meters and a 7 % discount on every meter after the next 1,500 meters . the price , before discount , of one meter of fabric is $ 2 , what is the total amount of money the merchant spends on 4,500 meters of fabric ? | for first 2000 metres he does not get any discount amount = 2 * 2000 = $ 4000 for next 1500 metres he gets 5 % discount amount = 2 * 1500 - ( 1 / 20 ) * 3000 = $ 2850 for the nest 1000 metres he gets 7 % discount amount = 2 * 1000 - 7 * 20 = $ 1860 total amount = $ 4000 + $ 2850 + $ 1860 = $ 8710 answer : b | a ) $ 8,617 , b ) $ 8,710 , c ) $ 8,810 , d ) $ 8,835 , e ) $ 8,915 | b | multiply(multiply(const_2, const_3), const_100) | multiply(const_2,const_3)|multiply(#0,const_100) | gain |
20 litres of mixture contains 20 % alcohol and the rest water . if 3 litres of water be mixed with it , the percentage of alcohol in the new mixture would be ? | "alcohol in the 20 litres of mix . = 20 % of 20 litres = ( 20 * 20 / 100 ) = 4 litres water in it = 20 - 3 = 17 litres new quantity of mix . = 20 + 3 = 23 litres quantity of alcohol in it = 4 litres percentage of alcohol in new mix . = 4 * 100 / 23 = 17.4 % answer is e" | a ) 16.67 % , b ) 23 % , c ) 18.3 % , d ) 19.75 % , e ) 17.4 % | e | multiply(divide(subtract(add(20, 3), add(multiply(divide(subtract(const_100, 20), const_100), 20), 3)), add(20, 3)), const_100) | add(n0,n2)|subtract(const_100,n1)|divide(#1,const_100)|multiply(n0,#2)|add(n2,#3)|subtract(#0,#4)|divide(#5,#0)|multiply(#6,const_100)| | gain |
the heights of a wall is 6 times its width and the length of the wall is 7 times its height . if volume of the wall be 16128 cu . m , its width is | solution let the width of the wall be x metres . then , height = ( 6 x ) m and length = ( 42 x ) m . 42 x Γ x Γ 6 x = 16128 x 3 βΉ = βΊ ( 16128 / 42 Γ 6 ) = 64 βΉ = βΊ x = 4 . answer a | a ) 4 m , b ) 4.5 m , c ) 5 m , d ) 6 m , e ) none | a | divide(sqrt(divide(16128, multiply(multiply(6, 7), 6))), const_2) | multiply(n0,n1)|multiply(n0,#0)|divide(n2,#1)|sqrt(#2)|divide(#3,const_2) | physics |
( 9 ^ 5.6 x 9 ^ 10.3 ) Γ· 9 ^ 2.56256 = 9 ^ ? | 5.6 + 10.3 = 15.9 i . e . . 15.9 - 2.56 = 13.34 so ur answer is 9 ^ 13.34 answer : c | a ) 11.34 , b ) 12.34 , c ) 13.34 , d ) 14.34 , e ) 15.34 | c | subtract(add(5.6, 10.3), 2.56256) | add(n1,n3)|subtract(#0,n5) | general |
3850 * 99 | "explanation : 3850 * ( 100 - 1 ) = 385000 - 3850 = 381150 option e" | a ) 382550 , b ) 384560 , c ) 385690 , d ) 389650 , e ) 381150 | e | multiply(divide(3850, 99), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
a milk man has 20 liters of milk . if he mixes 5 liters of water , which is freely available , in 20 liters of pure milk . if the cost of pure milk is rs . 18 per liter , then the profit of the milkman , when he sells all the mixture at cost price is : | "explanation : when the water is freely available and all the water is sold at the price of the milk , then the water gives the profit on the cost of 20 liters of milk . therefore , profit percentage = \ inline \ frac { 5 } { 20 } \ times 100 = 25 % % [ profit % = \ inline \ frac { profit } { cost \ : price } \ times 1... | a ) 22 , b ) 25 , c ) 27 , d ) 29 , e ) 11 | b | multiply(divide(subtract(multiply(add(add(add(20, 5), const_10), add(divide(5, const_2), const_3)), 18), multiply(20, 18)), multiply(20, 18)), const_100) | add(n1,n2)|divide(n1,const_2)|multiply(n0,n3)|add(#0,const_10)|add(#1,const_3)|add(#3,#4)|multiply(n3,#5)|subtract(#6,#2)|divide(#7,#2)|multiply(#8,const_100)| | gain |
10 + 45 | b | a ) 8 , b ) 55 , c ) 87 , d ) 90 , e ) 2 | b | multiply(divide(10, 45), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
equal amount of water were poured into two empty jars of different capacities , which made one jar 1 / 8 full and other jar 1 / 6 full . if the water in the jar with lesser capacity is then poured into the jar with greater capacity , what fraction of the larger jar will be filled with water ? | "same amount of water made bigger jar 1 / 8 full , then the same amount of water ( stored for a while in smaller jar ) were added to bigger jar , so bigger jar is 1 / 8 + 1 / 8 = 1 / 4 full . answer : e ." | a ) 1 / 7 , b ) 2 / 7 , c ) 1 / 2 , d ) 7 / 12 , e ) 1 / 4 | e | divide(const_2, 8) | divide(const_2,n1)| | general |
a certain bus driver is paid a regular rate of $ 20 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 1000 in total compen... | "for 40 hrs = 40 * 20 = 800 excess = 1000 - 800 = 200 for extra hours = . 75 ( 20 ) = 15 + 20 = 35 number of extra hrs = 200 / 35 = 5.71 = 6 approx . total hrs = 40 + 6 = 46 answer a 46" | a ) 46 , b ) 40 , c ) 44 , d ) 48 , e ) 49 | a | add(40, divide(subtract(1000, multiply(20, 40)), divide(multiply(20, add(const_100, 75)), const_100))) | add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)| | general |
a cistern normally takes 6 hours to be filled by a tap but because of a leak , 2 hours more . in how many hours will the leak empty a full cistern ? | β΅ cistern fill in 6 hours . β΄ in 1 hour , filled part = 1 β 6 th now , due to leakage , filled part in 1 hour = 1 β 8 th part of the cistern emptied , due to leakage in 1 hour = 1 β 6 - 1 β 8 = 1 β 24 th β΄ the leakage will empty the full cistern in 24 hrs . answer b | a ) 20 hours , b ) 24 hours , c ) 26 hours , d ) 18 hours , e ) none of these | b | inverse(subtract(inverse(6), inverse(add(2, 6)))) | add(n0,n1)|inverse(n0)|inverse(#0)|subtract(#1,#2)|inverse(#3) | physics |
what is the greatest integer m for which the number 20 ! / 10 ^ m is an integer ? | "10 ^ m = 2 ^ m * 5 ^ m . let ' s figure out how many 5 ' s are in the prime factorization of 20 ! the multiples of 5 are : 5 , 10 , 15 , 20 . thus 5 ^ 4 will divide 20 ! but 5 ^ 5 will not . clearly 2 ^ 4 will divide 20 ! so m = 4 is the largest possible integer . the answer is b ." | a ) 3 , b ) 4 , c ) 5 , d ) 6 , e ) 7 | b | floor(divide(20, divide(10, const_2))) | divide(n1,const_2)|divide(n0,#0)|floor(#1)| | general |
what is the value of ( p + q ) / ( p - q ) if p / q is 4 ? | "( p + q ) / ( p - q ) = [ ( p / q ) + 1 ] / [ ( p / q ) - 1 ] = ( 4 + 1 ) / ( 4 - 1 ) = 5 / 3 = 5 / 3 answer : a" | a ) 5 / 3 , b ) 2 / 3 , c ) 2 / 6 , d ) 7 / 8 , e ) 8 / 7 | a | divide(multiply(add(add(const_100, const_60), const_1), 4), const_100) | add(const_100,const_60)|add(#0,const_1)|multiply(n0,#1)|divide(#2,const_100)| | general |
doughnuts , cookies and muffins in a pastry shop are in the ratio of 5 : 3 : 1 . if there are 50 doughnuts , the number of muffins in the shop is : . | explanation : let doughnuts = 5 x , cookies = 3 x & muffins = 1 x . now , 5 x = 50 hence x = 10 . number of muffins = 1 x which is 10 . answer : c | a ) 25 , b ) 30 , c ) 10 , d ) 15 , e ) 20 | c | divide(50, 5) | divide(n3,n0) | other |
what least number must be subtracted from 9679 so that the remaining number is divisible by 15 ? | "on dividing 9679 by 15 , we get remainder = 4 . required number be subtracted = 4 answer : d" | a ) 3 , b ) 1 , c ) 16 , d ) 4 , e ) 15 | d | subtract(9679, multiply(floor(divide(9679, 15)), 15)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
the ages of two person differ by 20 years . if 8 years ago , the elder one be 5 times as old as the younger one , their present ages ( in years ) are respectively | let their ages be x and ( x + 20 ) years . then , 5 ( x - 8 ) = ( x + 20 - 8 ) = > 4 x = 52 = > x = 13 their present ages are 33 years and 13 year . answer : a | a ) 33 , 13 , b ) 25 , 5 , c ) 29 , 9 , d ) 50 , 30 , e ) 20,10 | a | add(divide(add(multiply(5, 8), subtract(20, 8)), subtract(5, const_1)), 20) | multiply(n1,n2)|subtract(n0,n1)|subtract(n2,const_1)|add(#0,#1)|divide(#3,#2)|add(n0,#4) | general |
a case of 12 rolls of paper towels sells for $ 9 . the cost of one roll sold individually is $ 1 . what is the percent of savings per roll for the 12 - roll package over the cost of 12 rolls purchased individually ? | "cost of 12 paper towels individually = 1 * 12 = 12 cost of a set of 12 paper towels = 9 cost of one roll = 9 / 12 = 3 / 4 = 0.75 savings per roll = 1 - . 75 = 0.25 % of savings is = . 25 / 1 * 100 = 25 % d is the answer ." | a ) 9 % , b ) 11 % , c ) 15 % , d ) 25 % , e ) 90 % | d | subtract(const_100, multiply(divide(9, 12), const_100)) | divide(n1,n0)|multiply(#0,const_100)|subtract(const_100,#1)| | general |
the difference between compound interest and simple interest on a certain amount of money at 5 % per annum for 2 years is 17 . find the sum : | sol . ( d ) let the sum be 100 . therefore , si = 100 Γ 5 Γ 2100 = 10100 Γ 5 Γ 2100 = 10 and ci = 100 ( 1 + 5100 ) 2 β 100100 ( 1 + 5100 ) 2 β 100 β΄ = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 = 100 Γ 21 Γ 2120 Γ 20 β 100 = 414 difference of ci and si = 41 β 4 - 10 = 1 β 4 if the difference is 1 β 4 , the sum = 100 = > if the d... | a ) 4500 , b ) 7500 , c ) 5000 , d ) 6800 , e ) none of these | d | divide(17, subtract(power(add(const_1, divide(5, const_100)), 2), add(const_1, multiply(2, divide(5, const_100))))) | divide(n0,const_100)|add(#0,const_1)|multiply(n1,#0)|add(#2,const_1)|power(#1,n1)|subtract(#4,#3)|divide(n2,#5) | gain |
a train speeds past a pole in 20 seconds and a platform 100 m long in 25 seconds . its length is ? | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 20 = > y = x / 20 x + 100 / 25 = x / 20 x = 400 m . answer : e" | a ) 188 m , b ) 876 m , c ) 251 m , d ) 150 m , e ) 400 m | e | multiply(100, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
a sum of money at simple interest amounts to rs . 825 in 3 years and to rs . 850 in 4 years . the sum is ? | "s . i . for 1 year = ( 850 - 825 ) = rs . 25 s . i . for 3 years = 25 * 3 = rs . 75 principal = ( 825 - 75 ) = rs . 750 . answer : b" | a ) s . 738 , b ) s . 750 , c ) s . 650 , d ) s . 730 , e ) s . 735 | b | subtract(825, divide(multiply(subtract(850, 825), 3), 4)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain |
the difference between a number and its three - fourth is 100 . what is 25 % of that number ? | "explanation : solution : let the number be x . then , x - 3 / 4 x = 100 = > x / 4 = 100 = > x = 400 . 25 % of x = 25 % of 400 = 25 / 100 * 400 = 100 . answer : b" | a ) 120 , b ) 100 , c ) 125 , d ) 150 , e ) none of these | b | divide(multiply(100, add(const_4, const_1)), add(const_1, const_2)) | add(const_1,const_4)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)| | general |
the average of 1 st 3 of 4 numbers is 16 and of the last 3 are 15 . if the sum of the first and the last number is 11 . what is the last numbers ? | "a + b + c = 48 b + c + d = 45 a + d = 11 a β d = 3 a + d = 11 2 d = 8 d = 4 answer b" | a ) 2 , b ) 4 , c ) 6 , d ) 5 , e ) 7 | b | subtract(subtract(multiply(3, 16), add(subtract(11, 16), 3)), 16) | multiply(n1,n3)|subtract(n6,n3)|add(n1,#1)|subtract(#0,#2)|subtract(#3,n3)| | general |
john left home and drove at the rate of 45 mph for 2 hours . he stopped for lunch then drove for another 3 hours at the rate of 50 mph to reach his destination . how many miles did john drive ? | the total distance d traveled by john is given by d = 45 * 2 + 3 * 50 = 240 miles . answer c | a ) 235 miles . , b ) 245 miles . , c ) 240 miles . , d ) 265 miles . , e ) 275 miles . | c | add(multiply(45, 2), multiply(3, 50)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1) | physics |
what least number must be subtracted from 3830 so that the remaining number is divisible by 15 ? | "on dividing 3830 by 15 , we get remainder = 5 . required number be subtracted = 5 answer : c" | a ) 3 , b ) 1 , c ) 5 , d ) 11 , e ) 15 | c | subtract(3830, multiply(floor(divide(3830, 15)), 15)) | divide(n0,n1)|floor(#0)|multiply(n1,#1)|subtract(n0,#2)| | general |
in a bag containing 3 balls , a white ball was placed and then 1 ball was taken out at random . what isthe probability that theextracted ball would turn on to be white , if all possible hypothesis concerning the color of the balls that initially in the bag were equally possible ? | "since , all possible hypothesis regarding the colour of the balls are equally likely , therefore these could be 3 white balls , initially in the bag . β΄ required probability = 1 / 4 [ 1 + 3 / 4 + 1 / 2 + 1 / 4 ] = 1 / 4 [ ( 4 + 3 + 2 + 1 ) / 4 ] = 5 / 8 b" | a ) 2 / 3 , b ) 5 / 8 , c ) 3 / 5 , d ) 4 / 7 , e ) 5 / 8 | b | divide(add(1, add(divide(1, const_2), add(divide(1, add(1, 3)), divide(3, add(1, 3))))), add(1, 3)) | add(n0,n1)|divide(n1,const_2)|divide(n1,#0)|divide(n0,#0)|add(#2,#3)|add(#4,#1)|add(n1,#5)|divide(#6,#0)| | probability |
find a sum for 1 st 8 prime number ' s ? | "required sum = ( 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 ) = 77 note : 1 is not a prime number option d" | a ) 25 , b ) 28 , c ) 30 , d ) 77 , e ) 36 | d | add(add(add(add(add(add(add(add(const_2, const_3), add(const_2, const_3)), add(add(const_2, const_3), const_2)), add(8, const_2)), add(add(8, const_2), const_2)), add(add(add(8, const_2), const_2), const_4)), add(add(add(add(8, const_2), const_2), const_4), const_2)), add(add(add(add(add(8, const_2), const_2), const_4)... | add(const_2,const_3)|add(n1,const_2)|add(#0,#0)|add(#0,const_2)|add(#1,const_2)|add(#2,#3)|add(#4,const_4)|add(#5,#1)|add(#6,const_2)|add(#7,#4)|add(#8,const_4)|add(#9,#6)|add(#11,#8)|add(#12,#10)| | general |
if 1,000 microns = 1 decimeter , and 1 , 000,000 angstroms = 1 decimeter , how many angstroms equal 1 micron ? | "given that 1,000 microns = 1 decimeter = 1 , 000,000 angstroms so , 1 micron = 1 , 000,000 / 1,000 = 1,000 answer : c" | a ) 100 , b ) 10 , c ) 1,000 , d ) 10,000 , e ) 100,000 | c | multiply(divide(1, multiply(const_100, const_100)), multiply(const_100, const_100)) | multiply(const_100,const_100)|divide(n1,#0)|multiply(#1,#0)| | general |
a flagpole 18 meters high casts a shadow of length 45 meters . if a building under similar conditions casts a shadow of length 50 meters , what is the height of the building ( in meters ) ? | "the height : length ratio will be equal in both cases . 18 / 45 = x / 50 x = 20 the answer is b ." | a ) 16 , b ) 20 , c ) 24 , d ) 28 , e ) 32 | b | multiply(50, divide(18, 45)) | divide(n0,n1)|multiply(n2,#0)| | physics |
if 7 ^ ( x - y ) = 343 and 7 ^ ( x + y ) = 16807 , what is the value of x ? | 7 ^ ( x - y ) = 343 = 7 ^ 3 = > x - y = 3 - - - - - - - - - - - - - - - - - - - - - - - - - - - ( equation 1 ) 7 ^ ( x + y ) = 16807 = 7 ^ 5 = > x + y = 5 - - - - - - - - - - - - - - - - - - - - - - - - - - - ( equation 2 ) ( equation 1 ) + ( equation 2 ) = > 2 x = 3 + 5 = 8 = > x = 8 / 2 = 4 answer is b | a ) 3 , b ) 4 , c ) 2 , d ) 1 , e ) 7 | b | divide(add(subtract(7, const_4), subtract(7, const_2)), const_2) | subtract(n0,const_4)|subtract(n0,const_2)|add(#0,#1)|divide(#2,const_2) | general |
242 students represent x percent of the boys at a school . if the boys at the school make up 50 % of the total school population of x students , what is x ? | "let b be the number of boys in the school . 242 = xb / 100 b = 0.5 x 24200 = 0.5 x ^ 2 x ^ 2 = 48400 x = 220 the answer is e ." | a ) 120 , b ) 150 , c ) 180 , d ) 200 , e ) 220 | e | sqrt(divide(multiply(242, const_100), divide(50, const_100))) | divide(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)|sqrt(#2)| | gain |
a certain number of workers can do a work in 75 days . if there were 10 workers more it could be finished in 10 days less . how many workers are there ? | "number of workers = 10 * ( 75 - 10 ) / 10 = 65 answer is a" | a ) 65 , b ) 30 , c ) 28 , d ) 24 , e ) 32 | a | divide(multiply(subtract(75, 10), 10), subtract(75, subtract(75, 10))) | subtract(n0,n1)|multiply(n1,#0)|subtract(n0,#0)|divide(#1,#2)| | physics |
a man spends 2 / 5 of his salary on house rent , 3 / 10 of his salary on food and 1 / 8 of his salary on conveyance . if he has $ 1400 left with him , find his expenditure on food and conveyance . | remaining part of salary left = 1 - ( 2 / 5 + 3 / 10 + 1 / 8 ) = 1 - ( 33 / 40 ) = 7 / 40 let the monthly salary be $ x then , 7 / 40 of x = $ 1400 x = ( 1400 * 40 / 7 ) = $ 8600 expenditure on food = $ ( 3 / 10 * 800 ) = $ 2400 expenditure on conveyance = $ ( 1 / 8 * 8000 ) = $ 1000 $ 2400 + $ 1000 = $ 3400 so the ans... | a ) $ 8600 , b ) $ 2400 , c ) $ 1000 , d ) $ 3000 , e ) $ 3400 | e | add(multiply(divide(1400, subtract(1, add(add(divide(2, 5), divide(3, 10)), divide(1, 8)))), divide(3, 10)), multiply(divide(1400, subtract(1, add(add(divide(2, 5), divide(3, 10)), divide(1, 8)))), divide(1, 8))) | divide(n0,n1)|divide(n2,n3)|divide(n4,n5)|add(#0,#1)|add(#3,#2)|subtract(n4,#4)|divide(n6,#5)|multiply(#6,#1)|multiply(#6,#2)|add(#7,#8) | general |
some of 15 % - intensity red paint is replaced with 25 % solution of red paint such that the new paint intensity is 30 % . what fraction of the original paint was replaced ? | "let total paint = 1 let amount replaced = x 15 ( 1 - x ) + 25 x = 30 x = 3 / 2 answer : d" | a ) 1 / 5 , b ) 1 / 3 , c ) 1 / 2 , d ) 3 / 2 , e ) 3 / 5 | d | divide(subtract(divide(30, const_100), divide(15, const_100)), subtract(divide(25, const_100), divide(15, const_100))) | divide(n2,const_100)|divide(n0,const_100)|divide(n1,const_100)|subtract(#0,#1)|subtract(#2,#1)|divide(#3,#4)| | gain |
find the compound interest on $ 1200 for 2 years at 20 % p . a . if ci is component yearly ? | a = p ( 1 + r / 100 ) ^ t = 1200 ( 1 + 20 / 100 ) ^ 2 = $ 1728 ci = $ 548 answer is c | a ) $ 120 , b ) $ 150 , c ) $ 548 , d ) $ 250 , e ) $ 300 | c | subtract(multiply(1200, power(add(const_1, divide(20, const_100)), 2)), 1200) | divide(n2,const_100)|add(#0,const_1)|power(#1,n1)|multiply(n0,#2)|subtract(#3,n0) | gain |
it takes avery 2 hours to build a brick wall while tom can do it in 4 hour . if the two start working together and after an hour avery leaves , how much time will it take tom to complete the wall on his own ? | avery takes 2 hours tom takes 4 hours efficiency of avery is 1 / 2 units / hr efficiency of tom is 1 / 4 units / hr combined efficiency of tom and avery is 1 / 2 + 1 / 4 = 3 / 4 units / hr since they worked for 1 hour they completed 3 / 4 units of work and 1 / 4 units of work is left which is to be completed by tom ( s... | a ) 70 , b ) 20 , c ) 50 , d ) 60 , e ) 30 | d | multiply(divide(subtract(const_1, add(divide(const_1, 2), divide(const_1, 4))), divide(const_1, 4)), const_60) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|subtract(const_1,#2)|divide(#3,#1)|multiply(#4,const_60) | physics |
how many of the integers between 25 and 45 are even ? | number start between 25 to 45 is 20 numbers half of them is even . . which is 10 answer : d | a ) 21 , b ) 20 , c ) 11 , d ) 10 , e ) 9 | d | divide(subtract(45, 25), const_2) | subtract(n1,n0)|divide(#0,const_2) | general |
bill made a profit of 10 % by selling a product . if he had purchased that product for 10 % less and sold it at a profit of 30 % , he would have received $ 28 more . what was his original selling price ? | "let p be the original purchase price of the product . bill originally sold the product for 1.1 * p . in the second scenario , the purchase price is 0.9 * p . a 30 % profit means the selling price would be 1.3 * 0.9 * p = 1.17 * p thus , according to the information in the question , 1.17 p - 1.1 p = 28 0.07 = 28 p = 4... | a ) $ 200 , b ) $ 260 , c ) $ 320 , d ) $ 380 , e ) $ 440 | e | multiply(divide(28, subtract(multiply(subtract(const_1, divide(10, const_100)), add(const_1, divide(30, const_100))), add(const_1, divide(10, const_100)))), add(const_1, divide(10, const_100))) | divide(n0,const_100)|divide(n2,const_100)|add(#0,const_1)|add(#1,const_1)|subtract(const_1,#0)|multiply(#3,#4)|subtract(#5,#2)|divide(n3,#6)|multiply(#2,#7)| | general |
if 0.4 of a number is equal to 0.07 of another number , the ratio of the numbers i | "sol . 0.4 a = 0.08 b Γ’ β‘ β a / b = 0.07 / 0.40 = 7 / 40 . Γ’ Λ Β΄ a : b = 7 : 40 . answer b" | a ) 2 : 3 , b ) 7 : 40 , c ) 3 : 20 , d ) 20 : 3 , e ) none | b | divide(multiply(0.07, const_100), multiply(0.4, const_100)) | multiply(n1,const_100)|multiply(n0,const_100)|divide(#0,#1)| | other |
sixty 5 percent of a number is 21 less than 4 fifth of that number . what is the number | let the number be x . then , 4 * x / 5 β ( 65 % of x ) = 21 4 x / 5 β 65 x / 100 = 21 5 x = 2100 x = 140 . answer a 140 | a ) 140 , b ) 165 , c ) 150 , d ) 142 , e ) 174 | a | divide(multiply(21, const_100), multiply(const_3, 5)) | multiply(n1,const_100)|multiply(n0,const_3)|divide(#0,#1) | general |
a wooden cube whose edge length is 7 inches is composed of smaller cubes with edge lengths of one inch . the outside surface of the large cube is painted red and then it is split up into its smaller cubes . if one cube is randomly selected from the small cubes , what is the probability that the cube will have at least ... | "there are a total of 7 * 7 * 7 = 343 cubes . all the exterior cubes will have at least one face painted red . the interior is formed by 5 * 5 * 5 = 125 cubes . the number of cubes with at least one side painted red is 343 - 125 = 218 cubes the probability that a cube has at least one side painted red is 218 / 343 whic... | a ) 59.8 % , b ) 63.6 % , c ) 67.2 % , d ) 70.5 % , e ) 73.6 % | b | multiply(const_100, subtract(const_1, divide(volume_cube(multiply(const_1, const_4)), volume_cube(7)))) | multiply(const_1,const_4)|volume_cube(n0)|volume_cube(#0)|divide(#2,#1)|subtract(const_1,#3)|multiply(#4,const_100)| | geometry |
two numbers a and b are such that the sum of 7 % of a and 4 % of b is two - third of the sum of 6 % of a and 8 % of b . find the ratio of a : b . | "explanation : 7 % of a + 4 % of b = 2 / 3 ( 6 % of a + 8 % of b ) 7 a / 100 + 4 b / 100 = 2 / 3 ( 6 a / 100 + 8 b / 100 ) β 7 a + 4 b = 2 / 3 ( 6 a + 8 b ) β 21 a + 12 b = 12 a + 16 b β 9 a = 4 b β ab = 4 / 9 β a : b = 4 : 9 answer : option c" | a ) 2 : 1 , b ) 1 : 2 , c ) 4 : 9 , d ) 1 : 1 , e ) 3 : 2 | c | divide(subtract(divide(multiply(divide(8, const_100), 4), const_3), divide(4, const_100)), subtract(divide(7, const_100), divide(multiply(divide(6, const_100), 4), const_3))) | divide(n3,const_100)|divide(n1,const_100)|divide(n0,const_100)|divide(n2,const_100)|multiply(#0,n1)|multiply(#3,n1)|divide(#4,const_3)|divide(#5,const_3)|subtract(#6,#1)|subtract(#2,#7)|divide(#8,#9)| | general |
the average weight of a class is x pounds . when a new student weighing 50 pounds joins the class , the average decreases by 1 pound . in a few months the student β s weight increases to 110 pounds and the average weight of the class becomes x + 4 pounds . none of the other students β weights changed . what is the valu... | "when the student weighs 80 pounds the average weight is x - 1 pounds ; when the student weighs 110 pounds the average weight is x + 4 pounds . so , the increase in total weight of 110 - 80 = 30 pounds corresponds to the increase in average weight of ( x + 4 ) - ( x - 1 ) = 5 pounds , which means that there are 30 / 5 ... | a ) 85 , b ) 86 , c ) 88 , d ) 90 , e ) 92 | a | add(add(50, 4), add(4, 1)) | add(n0,n3)|add(n1,n3)|add(#0,#1)| | general |
the speed of a car is 140 km in the first hour and 40 km in the second hour . what is the average speed of the car ? | "s = ( 140 + 40 ) / 2 = 90 kmph answer : d" | a ) 72 kmph , b ) 75 kmph , c ) 30 kmph , d ) 90 kmph , e ) 82 kmph | d | divide(add(140, 40), const_2) | add(n0,n1)|divide(#0,const_2)| | physics |
a train of 50 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 1.5 km long ? | "d = 50 * 60 + 1500 = 3000 m t = 4500 / 60 * 18 / 5 = 270 sec = 4.5 mins answer : d" | a ) 6 , b ) 3 , c ) 4 , d ) 4.5 , e ) 2 | d | add(divide(multiply(add(50, const_1), 60), const_1000), 1.5) | add(n0,const_1)|multiply(n1,#0)|divide(#1,const_1000)|add(n4,#2)| | physics |
how much space , in cubic units , is left vacant when maximum number of 5 x 5 x 5 cubes are fitted in a rectangular box measuring 10 x 15 x 18 ? | "no of cubes that can be accommodated in box = ( 10 * 15 * 18 ) / ( 5 * 5 * 5 ) 15 * 10 in numerator can be perfectly divided by 5 * 5 in denominator . side with length 18 ca n ' t be perfectly divided by 5 and hence is the limiting factor . closet multiple of 5 less that 18 is 15 . so vacant area in cube = = 15 * 10 *... | a ) 50 , b ) 150 , c ) 250 , d ) 350 , e ) 450 | e | multiply(subtract(10, multiply(const_4.0, const_2)), multiply(15, 18)) | multiply(n4,n5)|multiply(const_4.0,const_2)|subtract(n3,#1)|multiply(#0,#2)| | geometry |
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