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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a is 1.5 times as fast as b . a alone can do the work in 30 days . if a and b work on alternative days , beginning with a , in how many days will the work be completed ? | "a can finish 1 work in 30 days b can finish 1 / 1.5 work in 30 days - since a is 1.5 faster than b this means b can finish 1 work in 30 * 1.5 days = 45 days now using the awesome gmat formula when two machines work together they can finish the job in = ab / ( a + b ) = 45 * 30 / ( 45 + 30 ) = 20 * 30 / 50 = 18 days si... | a ) 23 , b ) 22 , c ) 21 , d ) 36 , e ) 25 | d | divide(const_1, add(divide(const_1, 30), divide(divide(const_1, 30), 1.5))) | divide(const_1,n1)|divide(#0,n0)|add(#0,#1)|divide(const_1,#2)| | physics |
5355 x 51 = ? | "5355 x 51 = 5355 x ( 50 + 1 ) = 5355 x 50 + 5355 x 1 = 267750 + 5355 = 273105 a" | a ) 273105 , b ) 273243 , c ) 273247 , d ) 273250 , e ) 273258 | a | multiply(divide(5355, 51), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
he ratio between the sale price and the cost price of an article is 3 : 1 . what is the ratio between the profit and the cost price of that article ? | "let c . p . = rs . x and s . p . = rs . 3 x . then , gain = rs . 2 x required ratio = 2 x : x = 2 : 1 . answer : a" | a ) 2 : 1 , b ) 2 : 9 , c ) 2 : 4 , d ) 2 : 1 , e ) 2 : 5 | a | inverse(subtract(const_1, divide(1, 3))) | divide(n1,n0)|subtract(const_1,#0)|inverse(#1)| | other |
p and q started a business investing rs . 85,000 and rs . 45,000 respectively . in what ratio the profit earned after 2 years be divided between p and q respectively ? | "p : q = 85000 : 45000 = 17 : 9 . answer : d" | a ) 17 : 6 , b ) 17 : 0 , c ) 17 : 4 , d ) 17 : 9 , e ) 17 : 3 | d | divide(add(multiply(add(add(2, const_3), const_3), multiply(add(2, const_3), 2)), add(2, const_3)), add(multiply(const_3, multiply(add(2, const_3), 2)), add(2, const_3))) | add(n2,const_3)|add(#0,const_3)|multiply(n2,#0)|multiply(#1,#2)|multiply(#2,const_3)|add(#0,#3)|add(#0,#4)|divide(#5,#6)| | gain |
there are a total of 100 jars on the shelf . small jars hold 3 liters and large jars hold 5 liters . the jars can hold a total of 376 liters . how many small jars are on the shelf ? | let s be the number of small jars and let l be the number of large jars . s + l = 100 . l = 100 - s . 3 s + 5 l = 376 . 3 s + 5 ( 100 - s ) = 376 . - 2 s + 500 = 376 . 2 s = 124 . s = 62 . the answer is e . | a ) 49 , b ) 53 , c ) 56 , d ) 59 , e ) 62 | e | subtract(100, divide(subtract(376, multiply(3, 100)), const_2)) | multiply(n0,n1)|subtract(n3,#0)|divide(#1,const_2)|subtract(n0,#2) | general |
trapezoid jklm in the x - y plane has coordinates j = ( – 2 , – 5 ) , k = ( – 2 , 1 ) , l = ( 6 , 7 ) , and m = ( 6 , – 5 ) . what is its perimeter ? | "jk = 6 lm = 11 kl = using distance formula 10 jm = using distance formula 8 sum of all is 35 b" | a ) 34 , b ) 35 , c ) 38 , d ) 40 , e ) ( f ) 42 | b | add(add(add(const_10, add(1, 5)), add(2, 6)), add(7, 5)) | add(n1,n3)|add(n0,n4)|add(n1,n5)|add(#0,const_10)|add(#3,#1)|add(#4,#2)| | physics |
find large no . from below question the difference of two numbers is 1365 . on dividing the larger number by the smaller , we get 6 as quotient and the 10 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 10 5 x = 1355 x = 271 large number = 271 + 1365 = 1636 d" | a ) 1235 , b ) 1456 , c ) 1567 , d ) 1636 , e ) 1635 | d | add(multiply(divide(subtract(1365, 10), subtract(6, const_1)), 6), 10) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|add(n2,#3)| | general |
1 , 0.2 , 0.08 , 0.48 , 0.0384 , ____ | "1 , 0.2 , 0.08 , 0.48 , 0.0384 , . . . . . 1 * 0.2 = 0.2 0.2 * 0.4 = 0.08 0.08 * 0.6 = 0.048 0.048 * 0.8 = 0.0384 so 0.0384 * 1 = 0.0384 answer : d" | a ) 0.0584 , b ) 0.0484 , c ) 0.0434 , d ) 0.0384 , e ) none | d | subtract(negate(0.48), multiply(subtract(0.2, 0.08), divide(subtract(0.2, 0.08), subtract(1, 0.2)))) | negate(n3)|subtract(n1,n2)|subtract(n0,n1)|divide(#1,#2)|multiply(#3,#1)|subtract(#0,#4)| | general |
a certain number when divided by 45 leaves a remainder 31 , what is the remainder if the same number be divided by 15 ? | "explanation : 45 + 31 = 76 / 15 = 1 ( remainder ) answer : d" | a ) 2 , b ) 7 , c ) 6 , d ) 1 , e ) 9 | d | reminder(31, 15) | reminder(n1,n2)| | general |
a person is saving money to purchase a car . he earns â £ 4000 per month and saves â £ 500 to buy the car . if he needs â £ 45000 to buy his dream car . how much money he will earn , before he saves enough money to buy the car ? | he is saving â £ 500 per month . he needs â £ 45000 and it takes â £ 45000 / â £ 500 = 90 months to save the money . in 90 months , he earns 90 * â £ 4000 = â £ 36000 . answer : c | a ) a - 352000 , b ) b - 356000 , c ) c - 360000 , d ) d - 348000 , e ) e - 340000 | c | multiply(divide(45000, 500), 4000) | divide(n2,n1)|multiply(n0,#0) | general |
by investing rs . 1800 in 9 % stock , syam earns rs . 120 . the stock is then quoted at | "explanation : assume that face value = rs . 100 . dividend per share = rs . 9 ( as it is a 9 % stock ) by investing rs . 1800 , he earns rs . 120 investment needed to earn rs . 9 = 1800 × 9 / 120 = rs . 135 ie , stock is then quoted ( then market value ) = rs . 135 answer : option a" | a ) rs . 135 , b ) rs . 96 , c ) rs . 85 , d ) rs . 122 , e ) rs . 142 | a | multiply(multiply(divide(1800, 120), divide(9, const_100)), const_100) | divide(n0,n2)|divide(n1,const_100)|multiply(#0,#1)|multiply(#2,const_100)| | gain |
the rowing athletes in a rowing conference voted for coach of the year . each rower who voted chose exactly 3 coaches to vote for among the 36 coaches in the conference . if each of the 36 coaches received exactly 5 votes ( a 36 way tie ) , how many rowers voted for coach of the year ? | there were 36 * 5 = 180 total votes made . if each voter picked 3 coaches , there were 180 / 3 = 60 voters . a | a ) 60 , b ) 70 , c ) 75 , d ) 84 , e ) 90 | a | divide(multiply(36, 5), 3) | multiply(n1,n3)|divide(#0,n0) | general |
i . x + 2 y + 3 z = 2 ii . x + y - z = 0 iii . 2 x + 2 y - z = 1 what is the value of y in the system above ? | 1 . from the given system of equations i can make equation ii . to be x + y = z 2 . now put that in iii . : 2 x + 2 y - ( x + y ) = 1 = 2 x + 2 y - x - y = 1 = x + y = 1 that gives usz = 1 ! 3 . put z = 1 in i . and solve : x + 2 y + 3 = 2 ; calculate minus 3 = x + 2 y = - 1 ; calculate minus x = 2 y = - 1 - x ; 4 . no... | a ) - 2 , b ) - 1 , c ) 0 , d ) 1 , e ) 2 | a | subtract(2, add(3, 1)) | add(n1,n6)|subtract(n0,#0) | general |
if grapes are 90 % water and raisins are 15 % water , then how many kilograms did a quantity of raisins , which currently weighs 12 kilograms , weigh when all the raisins were grapes ? ( assume that the only difference between their raisin - weight and their grape - weight is water that evaporated during their transfor... | "let x be the original weight of the grapes . the weight of the grape pulp was 0.1 x . since the grape pulp is 85 % of the raisins , 0.1 x = 0.85 ( 12 kg ) . then x = 8.5 * 12 = 102 kg . the answer is d ." | a ) 96 , b ) 98 , c ) 100 , d ) 102 , e ) 104 | d | divide(multiply(divide(subtract(const_100, 15), const_100), 12), divide(subtract(const_100, 90), const_100)) | subtract(const_100,n1)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|multiply(n2,#2)|divide(#4,#3)| | general |
salesperson a ' s compensation for any week is $ 210 plus 6 percent of the portion of a ' s total sales above $ 1,000 for that week . salesperson b ' s compensation for any week is 8 percent of b ' s total sales for that week . for what amount of total weekly sales would both salespeople earn the same compensation ? | "210 + 0.06 ( x - 1000 ) = 0.08 x 0.02 x = 150 x = $ 7,500 the answer is c ." | a ) $ 3500 , b ) $ 5500 , c ) $ 7500 , d ) $ 9500 , e ) $ 11,500 | c | divide(add(divide(subtract(210, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100))), divide(subtract(210, multiply(divide(6, const_100), 1,000)), subtract(divide(8, const_100), divide(6, const_100)))), 1,000) | divide(n1,const_100)|divide(n3,const_100)|multiply(#0,n2)|subtract(#1,#0)|subtract(n0,#2)|divide(#4,#3)|add(#5,#5)|divide(#6,n2)| | general |
a car travels uphill at 30 km / hr and downhill at 50 km / hr . it goes 100 km uphill and 50 km downhill . find the average speed of the car ? | "avg speed = total distance / total time . total distance traveled = 100 + 50 = 150 km ; time taken for uphill journey = 100 / 30 = 10 / 3 ; time taken for down hill journey = 50 / 50 = 5 / 5 ; avg speed = 150 / ( 10 / 3 + 5 / 5 ) = 35 kmph answer : d" | a ) 32 kmph , b ) 33 kmph , c ) 34 kmph , d ) 35 kmph , e ) 36 kmph | d | divide(add(100, 50), add(divide(100, 30), divide(50, 50))) | add(n2,n3)|divide(n2,n0)|divide(n3,n1)|add(#1,#2)|divide(#0,#3)| | general |
if [ [ x ] ] = x ^ 2 + 2 x + 4 , what is the value of [ [ 7 ] ] ? | "these functions questions might look intimidating , but they just test your knowledge about how well you can substitute values [ [ x ] ] = x ^ 2 + 2 x + 4 [ [ 7 ] ] = 7 ^ 2 + 2 * 7 + 4 = 67 . option d" | a ) 3 , b ) 9 , c ) 15 , d ) 67 , e ) 25 | d | add(add(power(2, 2), multiply(2, 2)), 4) | multiply(n0,n0)|power(n0,n0)|add(#0,#1)|add(n2,#2)| | general |
a fruit seller had some oranges . he sells 40 % oranges and still has 540 oranges . how many oranges he had originally ? | "60 % of oranges = 540 100 % of oranges = ( 540 × 100 ) / 6 = 900 total oranges = 900 answer : b" | a ) 700 , b ) 900 , c ) 720 , d ) 730 , e ) 740 | b | add(540, multiply(540, divide(40, const_100))) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1)| | gain |
if | 5 x - 15 | = 100 , then find the sum of the values of x ? | "| 5 x - 15 | = 100 5 x - 15 = 100 or 5 x - 15 = - 100 5 x = 115 or 5 x = - 85 x = 23 or x = - 17 sum = 23 - 17 = 6 answer is c" | a ) 1 , b ) - 2 , c ) 6 , d ) - 3 , e ) 4 | c | subtract(subtract(subtract(100, 15), add(100, 15)), 15) | add(n1,n2)|subtract(n2,n1)|subtract(#1,#0)|subtract(#2,n1)| | general |
the speed of a boat in still water is 25 kmph . if it can travel 10 km upstream in 1 hr , what time it would take to travel the same distance downstream ? | speed of boat in still water = 25 km / hr speed upstream = 10 ⁄ 1 = 10 km / hr speed of the stream = ( 25 - 10 ) = 15 km / hr speed downstream = ( 25 + 15 ) = 40 km / hr time taken to travel 10 km downstream = 10 / 40 hours = ( 10 × 60 ) / 40 = 15 minutes answer is c | a ) 26 minutes , b ) 23 minutes , c ) 15 minutes , d ) 19 minutes , e ) 28 minutes | c | subtract(25, 10) | subtract(n0,n1) | physics |
a train 125 m long passes a man , running at 10 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is ? | "speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 10 ) km / hr . x - 10 = 45 = = > x = 55 km / hr answer : c" | a ) 36 , b ) 50 , c ) 55 , d ) 26 , e ) 29 | c | divide(divide(subtract(125, multiply(multiply(10, const_0_2778), 10)), 10), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics |
on an order of 5 dozen boxes of a consumer product , a retailer receives an extra dozen free . this is equivalent to allowing him a discount of : | "clearly , the retailer gets 1 dozen out of 6 dozens free . equivalent discount = 1 / 6 * 100 = 16 2 / 3 % . answer : c" | a ) 15 % , b ) 16 1 / 6 % , c ) 16 2 / 3 % , d ) 20 % , e ) 30 % | c | subtract(const_100, multiply(divide(5, const_4), const_100)) | divide(n0,const_4)|multiply(#0,const_100)|subtract(const_100,#1)| | general |
seller selling an apple for rs . 17 , a seller loses 1 / 6 th of what it costs him . the cp of the apple is ? | "sp = 17 loss = cp 18 loss = cp − sp = cp − 17 ⇒ cp 18 = cp − 17 ⇒ 17 cp 18 = 17 ⇒ cp 18 = 1 ⇒ cp = 18 c" | a ) 10 , b ) 12 , c ) 18 , d ) 19 , e ) 20 | c | add(17, 1) | add(n0,n1)| | general |
there are two positive numbers in the ratio 7 : 11 . if the larger number exceeds the smaller by 16 , then find the smaller number ? | let the two positive numbers be 7 x and 11 x respectively . 11 x - 7 x = 16 4 x = 16 = > x = 4 = > smaller number = 7 x = 28 . answer : e | a ) 25 , b ) 26 , c ) 30 , d ) 24 , e ) 28 | e | divide(multiply(16, 7), const_4) | multiply(n0,n2)|divide(#0,const_4) | other |
what is rate of interest if principal . amount be 400 , simple interest 120 and time 2 year . | "s . i = ( p * r * t ) / 100 120 = 800 r / 100 r = 120 / 8 = 15 % answer e" | a ) 10 , b ) 12.5 , c ) 25 , d ) 12 , e ) 15 | e | multiply(divide(120, multiply(400, 2)), const_100) | multiply(n0,n2)|divide(n1,#0)|multiply(#1,const_100)| | gain |
a fair coin is tossed 15 times . what is the probability of getting more heads than tails in 15 tosses ? | "on each toss , the probability of getting a head is 1 / 2 and the probability of getting a tail is 1 / 2 . there is no way to get the same number of heads and tails on an odd number of tosses . there will either be more heads or more tails . then there must be more heads on half of the possible outcomes and more tails... | a ) 1 / 2 , b ) 63 / 128 , c ) 4 / 7 , d ) 61 / 256 , e ) 63 / 64 | a | divide(add(add(add(choose(15, const_2), choose(15, const_3)), choose(15, const_4)), choose(15, 15)), power(const_2, 15)) | choose(n0,const_2)|choose(n0,const_3)|choose(n0,const_4)|choose(n0,n0)|power(const_2,n0)|add(#0,#1)|add(#5,#2)|add(#6,#3)|divide(#7,#4)| | probability |
a bowler can take max . 3 wickets in a over . if he bowls 6 overs in an innings , how many maximum wickets can he take ? | 10 because after 10 wickets , the innings is complete . answer : e | a ) 8 , b ) 9 , c ) 11 , d ) 7 , e ) 10 | e | subtract(subtract(multiply(6, 3), 6), const_2) | multiply(n0,n1)|subtract(#0,n1)|subtract(#1,const_2) | general |
if the personal income tax rate is lowered from 42 % to 32 % , what is the differential savings for a tax payer having an annual income before tax to the tune of $ 42400 ? | "saving = ( 42 - 32 ) % of 42400 = 4240 . answer : c" | a ) $ 3500 , b ) $ 5000 , c ) $ 4240 , d ) $ 7000 , e ) $ 10000 | c | multiply(divide(42400, const_100), subtract(42, 32)) | divide(n2,const_100)|subtract(n0,n1)|multiply(#0,#1)| | gain |
( 5568 / 87 ) 1 / 3 + ( 72 x 2 ) 1 / 2 = ( ? ) 1 / 2 ? | answer ? ) 1 / 2 = ( 5568 / 87 ) 1 / 3 + ( 72 x 2 ) 1 / 2 = ( 64 ) 1 / 3 + ( 144 ) 1 / 2 ∴ ? = ( 4 + 12 ) 2 = 256 correct option : a | a ) 256 , b ) 4 , c ) √ 2 , d ) 16 , e ) none | a | power(add(power(divide(5568, 87), divide(1, 3)), power(multiply(72, 2), divide(1, 2))), 2) | divide(n0,n1)|divide(n2,n3)|divide(n2,n5)|multiply(n4,n5)|power(#0,#1)|power(#3,#2)|add(#4,#5)|power(#6,n5) | general |
at the end of the first quarter , the share price of a certain mutual fund was 20 percent higher than it was at the beginning of the year . at the end of the second quarter , the share price was 50 percent higher than it was at the beginning of the year . what was the percent increase in the share price from the end of... | say price at the beginning of year = 100 end of 1 st quarter = 100 + 20 = 120 end of 2 nd quarter = 100 + 50 = 150 percentage increase between 1 st & 2 nd quarter = 150 − 120 / 120 ∗ 100 = 25 answer = b | a ) 20 % , b ) 25 % , c ) 30 % , d ) 33 % , e ) 40 % | b | multiply(subtract(divide(add(const_100, 50), add(const_100, 20)), const_1), const_100) | add(n1,const_100)|add(n0,const_100)|divide(#0,#1)|subtract(#2,const_1)|multiply(#3,const_100) | gain |
find the 25 % of rs . 800 . | "explanation : 25 % of 800 = > 25 / 100 * 800 = rs . 200 answer : b" | a ) s . 50 , b ) s . 200 , c ) s . 100 , d ) s . 80 , e ) s . 60 | b | multiply(800, divide(25, const_100)) | divide(n0,const_100)|multiply(n1,#0)| | gain |
what is the 50 th digit to the right of the decimal point in the decimal form of 48 / 88 ? | "we have to use some long division . this long division allows us to get 48 / 88 in decimal form , which is 0.545454 … where “ 54 ” is repeating . we can see that the 1 st , 3 rd , 5 th digit to the right of the decimal point is a 5 and that the 2 nd , 4 th , 6 th digit to the right of the decimal point is a 4 . in oth... | a ) 6 , b ) 7 , c ) 9 , d ) 11 , e ) 4 | e | divide(48, 88) | divide(n1,n2)| | general |
a crate measures 5 feet by 8 feet by 12 feet on the inside . a stone pillar in the shape of a right circular cylinder must fit into the crate for shipping so that it rests upright when the crate sits on at least one of its six sides . what is the radius , in feet , of the pillar with the largest volume that could still... | "we can find the radius of all the three cases of cylinders . the only crux to find the answer faster is that : voulme is pi * r ^ 2 * h . the volume is a function of r ^ 2 . so r has to be the highest to find the largest volume . so r = 5 for the surface 8 * 12 face . volume = 125 pi answer c" | a ) 2 , b ) 4 , c ) 5 , d ) 8 , e ) 12 | c | divide(divide(multiply(multiply(8, 12), 5), 12), 8) | multiply(n1,n2)|multiply(n0,#0)|divide(#1,n2)|divide(#2,n1)| | geometry |
a pharmaceutical company received $ 4 million in royalties on the first $ 50 million in sales of and then $ 12 million in royalties on the next $ 170 million in sales . by approximately what percentage did the ratio of royalties to sales decrease from the first $ 50 million in sales to the next $ 170 million in sales ? | "( 12 / 170 ) / ( 4 / 50 ) = 15 / 17 = 88 % it means that 12 / 170 represents only 88 % . therefore a decrease of 12 % . answer a" | a ) 12 % , b ) 15 % , c ) 45 % , d ) 52 % , e ) 56 % | a | multiply(divide(4, 50), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
what is the minimum number of square tiles required to tile a floor of length 5 metres 78 cm and width 3 metres 74 cm ? | length = width = = > square = 5 m 78 cm and 3 m 74 5 m 78 cm = 578 cm and 3 m 74 cm = 374 cm hcf of 578 and 374 = 34 square is 34 = 578 * 374 / 34 * 34 = = 17 * 11 = 187 answer a | a ) 187 , b ) 180 , c ) 190 , d ) 195 , e ) 197 | a | multiply(divide(add(multiply(const_100, 5), 78), add(multiply(multiply(const_4, const_4), const_2), const_2)), divide(add(multiply(const_100, 3), 74), add(multiply(multiply(const_4, const_4), const_2), const_2))) | multiply(n0,const_100)|multiply(const_4,const_4)|multiply(n2,const_100)|add(n1,#0)|add(n3,#2)|multiply(#1,const_2)|add(#5,const_2)|divide(#3,#6)|divide(#4,#6)|multiply(#7,#8) | physics |
three rugs have a combined area of 204 square meters . by overlapping the rugs to cover floor area of 140 square meters , the area that is covered by exactly two layers of rug is 24 square meters . what is the area that is covered with three layers of rug ? | "total = rug 1 + rug 2 + rug 3 - { overlap of exactly 2 rugs } - 2 * { overlap of exactly 3 rugs } 140 = 204 - 24 - 2 * { overlap of exactly 2 rugs } - - > { overlap of exactly 3 rugs } = 20 . answer : b ." | a ) 18 square meters , b ) 20 square meters , c ) 24 square meters , d ) 28 square meters , e ) 30 square meters | b | divide(subtract(subtract(204, 140), 24), const_2) | subtract(n0,n1)|subtract(#0,n2)|divide(#1,const_2)| | geometry |
there is a train and car . the ratio between the speed of a train & a car is 24 : 19 respectively . also , a bus covered a distance of 320 km in 5 hours . the speed of the bus is 2 / 3 rd of the speed of the train . how many kilometers will the car cover in 4 hours ? | "the speed of the bus is 320 / 5 = 64 km / hr the speed of the train is ( 64 * 3 ) / 2 = 96 km / hr the speed of the car is 96 / 24 * 19 = 76 km / hr the distance covered by the car in 4 hours is 76 × 4 = 304 km the answer is c ." | a ) 276 , b ) 290 , c ) 304 , d ) 318 , e ) 332 | c | multiply(divide(multiply(divide(multiply(divide(320, 5), 5), 2), 19), 24), 4) | divide(n2,n3)|multiply(n3,#0)|divide(#1,n4)|multiply(n1,#2)|divide(#3,n0)|multiply(n6,#4)| | physics |
in a group of 100 cars , 37 cars do not have air conditioning . if at least 41 cars have racing stripes , what is the greatest number of cars that could have air conditioning but not racing stripes ? | "lets assume ac = 63 ( includesonly ac carsandcars with ac and racing stripes ) lets assume rs ( racing stripes ) > = 41 ( includescars with ac and racing stripesandonly racing stripes ) . now since we want to maximize ( only ac ) we have to see to it thatcars with ac and racing stripesis minimal ( assume 0 ) but since... | a ) 45 , b ) 47 , c ) 59 , d ) 51 , e ) 53 | c | subtract(100, 41) | subtract(n0,n2)| | other |
a man goes downstream at 12 kmph , and upstream 8 kmph . the speed of the stream is | "speed of the stream = 1 / 2 ( 12 - 8 ) kmph = 2 kmph . correct option a" | a ) 2 kmph , b ) 4 kmph , c ) 16 kmph , d ) 2.5 kmph , e ) 26 kmph | a | divide(subtract(12, 8), const_2) | subtract(n0,n1)|divide(#0,const_2)| | physics |
micheal and adam can do together a piece of work in 20 days . after they have worked together for 18 days micheal stops and adam completes the remaining work in 10 days . in how many days micheal complete the work separately . | "rate of both = 1 / 20 together they do = 1 / 20 * 18 = 9 / 10 left work = 1 - 9 / 10 = 1 / 10 adam completes 1 / 10 work in 10 day so he took 10 * 10 = 100 days to complete the left work alone . thus the rate of adam is 1 / 100 rate of micheal = 1 / 20 - 1 / 100 = 1 / 25 thus micheal takes 25 days to complete the whol... | a ) 25 days , b ) 100 days , c ) 120 days , d ) 110 days , e ) 90 days | a | inverse(subtract(inverse(20), inverse(multiply(inverse(subtract(const_1, multiply(inverse(20), 18))), 10)))) | inverse(n0)|multiply(n1,#0)|subtract(const_1,#1)|inverse(#2)|multiply(n2,#3)|inverse(#4)|subtract(#0,#5)|inverse(#6)| | physics |
sides of a rectangular park are in the ratio 3 : 2 and its area is 3750 sq m , the cost of fencing it at 30 ps per meter is ? | "3 x * 2 x = 3750 = > x = 25 2 ( 75 + 30 ) = 210 m 210 * 1 / 2 = rs . 105 answer : b" | a ) s . 122 , b ) s . 105 , c ) s . 125 , d ) s . 120 , e ) s . 121 | b | divide(multiply(30, rectangle_perimeter(sqrt(divide(multiply(3750, 2), 3)), divide(3750, sqrt(divide(multiply(3750, 2), 3))))), const_100) | multiply(n1,n2)|divide(#0,n0)|sqrt(#1)|divide(n2,#2)|rectangle_perimeter(#3,#2)|multiply(n3,#4)|divide(#5,const_100)| | physics |
how many 7 in between 1 to 80 ? | "7 , 17,27 , 37,47 , 57,67 , 70,71 , 72,73 , 74,75 , 76,77 ( two 7 ' s ) , 78,79 18 7 ' s between 1 to 80 answer : a" | a ) 18 , b ) 20 , c ) 22 , d ) 23 , e ) 24 | a | divide(80, multiply(7, 1)) | multiply(n0,n1)|divide(n2,#0)| | general |
a batsman in his 12 th innings makes a score of 115 and thereby increases his average by 3 runs . what is his average after the 12 th innings if he had never been ‘ not out ’ ? | "let ‘ x ’ be the average score after 12 th innings ⇒ 12 x = 11 × ( x – 4 ) + 115 ∴ x = 82 answer d" | a ) 42 , b ) 43 , c ) 44 , d ) 82 , e ) 46 | d | add(subtract(115, multiply(12, 3)), 3) | multiply(n0,n2)|subtract(n1,#0)|add(n2,#1)| | general |
the ratio 3 : 5 expressed as a percent equals | "solution 3 : 5 = 3 / 5 = ( 3 / 5 x 100 ) % . = 60 % . answer a" | a ) 60 % , b ) 40 % , c ) 80 % , d ) 125 % , e ) none | a | multiply(divide(3, 5), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
when positive integer n is divided by positive integer j , the remainder is 28 . if n / j = 142.07 , what is value of j ? | when a number is divided by another number , we can represent it as : dividend = quotient * divisor + remainder so , dividend / divisor = quotient + remainder / divisor given that n / j = 142.07 here 142 is the quotient . given that remainder = 28 so , 142.07 = 142 + 28 / j so , j = 400 answer - c | a ) 300 , b ) 375 , c ) 400 , d ) 460 , e ) 500 | c | divide(28, subtract(142.07, add(const_100, add(multiply(const_4, const_10), const_2)))) | multiply(const_10,const_4)|add(#0,const_2)|add(#1,const_100)|subtract(n1,#2)|divide(n0,#3) | general |
find the annual dividend received by nishita from 1200 preferred shares and 3000 common shares both of par value rs . 50 each if the dividend paid on preferred shares is 10 % and semi - annual dividend of 3 ½ % is declared on common shares . | total number of preferred shares = 1200 face value = rs . 50 dividend paid on preferred shares is 10 % dividend per share = 50 × 10 / 100 = rs . 5 total dividend = 1200 × 5 = 6000 total number of common shares = 3000 face value = rs . 50 semi - annual dividend of 3 ½ % is declared on common shares . semi - annual divid... | a ) 16000 , b ) 16500 , c ) 17000 , d ) 17500 , e ) 18000 | b | add(add(divide(multiply(multiply(1200, 50), 10), const_100), divide(multiply(multiply(3000, 50), add(divide(const_1, const_2), 3)), const_100)), divide(multiply(multiply(3000, 50), add(divide(const_1, const_2), 3)), const_100)) | divide(const_1,const_2)|multiply(n0,n2)|multiply(n1,n2)|add(n4,#0)|multiply(n3,#1)|divide(#4,const_100)|multiply(#3,#2)|divide(#6,const_100)|add(#5,#7)|add(#8,#7) | general |
length of a rectangular plot is 32 mtr more than its breadth . if the cost of fencin gthe plot at 26.50 per meter is rs . 5300 , what is the length of the plot in mtr ? | "let breadth = x metres . then , length = ( x + 32 ) metres . perimeter = 5300 m = 200 m . 26.50 2 [ ( x + 32 ) + x ] = 200 2 x + 32 = 100 2 x = 68 x = 34 . hence , length = x + 32 = 66 m b" | a ) 56 m , b ) 66 m , c ) 76 m , d ) 86 m , e ) 96 m | b | divide(add(divide(5300, 26.50), multiply(const_2, 32)), const_4) | divide(n2,n1)|multiply(n0,const_2)|add(#0,#1)|divide(#2,const_4)| | physics |
the distance from city a to city b is 120 miles . while driving from city a to city b , bob drives at a constant speed of 40 miles per hour . alice leaves city a 30 minutes after bob . what is the minimum constant speed in miles per hour that alice must exceed in order to arrive in city b before bob ? | the time it takes bob to drive to city b is 120 / 40 = 3 hours . alice needs to take less than 2.5 hours for the trip . alice needs to exceed a constant speed of 120 / 2.5 = 48 miles per hour . the answer is b . | a ) 45 , b ) 48 , c ) 50 , d ) 52 , e ) 54 | b | divide(120, subtract(divide(120, 40), divide(30, const_60))) | divide(n0,n1)|divide(n2,const_60)|subtract(#0,#1)|divide(n0,#2) | physics |
solve : 12.05 * 5.4 + 0.6 | = 12.05 * ( 5.4 / 0.6 ) = ( 12.05 * 9 ) = 108.45 answer is a . | a ) 108.45 , b ) 110.45 , c ) 106.45 , d ) 109.45 , e ) none of them | a | multiply(12.05, divide(5.4, 0.6)) | divide(n1,n2)|multiply(n0,#0) | general |
an investment yields an interest payment of $ 225 each month . if the simple annual interest rate is 9 % , what is the amount of the investment ? | "let the principal amount = p simple annual interest = 9 % simple monthly interest = ( 9 / 12 ) = ( 3 / 4 ) % ( 3 / 4 ) * ( p / 100 ) = 225 = > p = ( 225 * 4 * 10 ^ 2 ) / 3 = 75 * 4 * 10 ^ 2 = 300 * 10 ^ 2 = 30000 answer a" | a ) $ 30,000 , b ) $ 30,400 , c ) $ 31,300 , d ) $ 32,500 , e ) $ 35,100 | a | multiply(divide(225, divide(9, multiply(const_3, const_4))), const_100) | multiply(const_3,const_4)|divide(n1,#0)|divide(n0,#1)|multiply(#2,const_100)| | gain |
list i : { y , 2 , 4 , 7 , 10 , 11 } list ii : { 3 , 3 , 4 , 6 , 7 , 10 } if the median q of list i is equal to the sum of the median of list ii and the mode of list ii , then y equals | mode of list ii = 3 median q of list ii = 4 + 6 / 2 = 5 sum of mode + mean = 3 + 5 = 8 now to make 8 as the median we need to find a value of y such that if the no . of terms in list 1 are odd then y = 8 else if even then 7 + y / 2 = 8 here its even so 7 + y / 2 = 8 from this y = 9 ( d ) | a ) 5 , b ) 7 , c ) 8 , d ) q = 9 , e ) 10 | d | add(6, 3) | add(n5,n8) | general |
in a college the ratio of the numbers of boys to the girls is 8 : 5 . if there are 135 girls , the total number of students in the college is ? | "let the number of boys and girls be 8 x and 5 x then , 5 x = 135 x = 27 total number of students = 13 x = 13 * 27 = 351 answer is b" | a ) 562 , b ) 351 , c ) 452 , d ) 416 , e ) 512 | b | add(multiply(divide(8, 5), 135), 135) | divide(n0,n1)|multiply(n2,#0)|add(n2,#1)| | other |
an agent , gets a commission of 2.5 % on the sales of cloth . if on a certain day , he gets rs . 18 as commission , the cloth sold through him on that day is worth | "explanation : let the total sale be rs . x . then , 2.5 % . of x = 18 < = > ( 25 / 10 * 1 / 100 * x ) = 18 < = > x = 720 . answer : d" | a ) 333 , b ) 500 , c ) 887 , d ) 720 , e ) 132 | d | divide(18, divide(2.5, const_100)) | divide(n0,const_100)|divide(n1,#0)| | gain |
if 45 - [ 28 - { 37 - ( 15 - * ) } ] = 58 , then * is equal to : | "45 - [ 28 - { 37 - ( 15 - * ) } ] = 58 = > 45 - [ 28 - { 37 - 15 + * } ] = 58 45 - [ 28 - 37 + 15 - * ] = 58 = > 45 [ 43 - 37 - * ] = 58 45 - [ 6 - * ] = 58 = > 45 - 6 + * = 58 39 + * = 58 = > * = 58 - 39 = 19 answer : c" | a ) - 29 , b ) - 19 , c ) 19 , d ) 29 , e ) 39 | c | subtract(58, subtract(45, add(subtract(28, 37), 15))) | subtract(n1,n2)|add(n3,#0)|subtract(n0,#1)|subtract(n4,#2)| | general |
if the wheel is 14 cm then the number of revolutions to cover a distance of 1056 cm is ? | "2 * 22 / 7 * 14 * x = 1056 = > x = 12 answer : d" | a ) 17 , b ) 19 , c ) 17 , d ) 12 , e ) 91 | d | divide(1056, multiply(multiply(const_2, divide(add(add(multiply(const_3, const_100), multiply(const_1, const_10)), const_4), const_100)), 14)) | multiply(const_100,const_3)|multiply(const_1,const_10)|add(#0,#1)|add(#2,const_4)|divide(#3,const_100)|multiply(#4,const_2)|multiply(n0,#5)|divide(n1,#6)| | physics |
a man walked a certain distance south and then the same distance plus 7 km due west . he is now 13 km from his starting point . what are the distances south and west that he walked ? | if a man walked distance ( in km ) ' x ' towards south & then ' x + 7 ' towards west . distance from starting point = hypotenuse of a right triangle with sides x & ( x + 7 ) so 13 ^ 2 = x ^ 2 + ( x + 7 ) ^ 2 x ^ 2 + 7 x - 60 = 0 or ( x + 12 ) ( x - 5 ) = 0 , so x = 5 distance towards west = x + 7 = 5 + 7 = 12 km . answ... | a ) 512 , b ) 612 , c ) 712 , d ) 513 , e ) 613 | a | subtract(multiply(multiply(multiply(13, 7), const_3), const_2), multiply(const_10, const_4)) | multiply(n0,n1)|multiply(const_10,const_4)|multiply(#0,const_3)|multiply(#2,const_2)|subtract(#3,#1) | physics |
out of 40 applicants to a law school , 15 majored in political science , 20 had a grade point average higher than 3.0 , and 10 did not major in political science and had a gpa equal to or lower than 3.0 . how many r applicants majored in political science and had a gpa higher than 3.0 ? | "total applicants = 40 political science = 15 and non political science = 40 - 15 = 25 gpa > 3.0 = 20 and gpa < = 3.0 = 20 10 non political science students had gpa < = 3.0 - - > 15 non political science students had gpa > 3.0 gpa > 3.0 in political science = total - ( gpa > 3.0 in non political science ) r = 20 - 15 =... | a ) 5 , b ) 10 , c ) 15 , d ) 25 , e ) 35 | a | subtract(20, subtract(40, add(10, 15))) | add(n1,n4)|subtract(n0,#0)|subtract(n2,#1)| | general |
a grocer is storing soap boxes in cartons that measure 25 inches by 42 inches by 60 inches . if the measurement of each soap box is 7 inches by 12 inches by 5 inches , then what is the maximum number of soap boxes that can be placed in each carton ? | "however the process of dividing the volume of box by the volume of a soap seems flawed but it does work in this case due to the numbers dimensions of the box = 25 * 42 * 60 dimensions of the soap = 5 * 12 * 7 placing the 7 inch side along 42 inch side we get 6 soaps in a line and in a similar way 5 along 25 and 6 alon... | a ) 210 , b ) 150 , c ) 280 , d ) 300 , e ) 420 | b | divide(multiply(multiply(25, 42), 60), multiply(multiply(7, 12), 5)) | multiply(n0,n1)|multiply(n3,n4)|multiply(n2,#0)|multiply(n5,#1)|divide(#2,#3)| | general |
if 25 % of x is 15 less than 15 % of 1600 , then x is ? | "25 % of x = x / 4 ; 15 % of 1600 = 15 / 100 * 1600 = 240 given that , x / 4 = 240 - 15 = > x / 4 = 225 = > x = 900 . answer : c" | a ) 872 , b ) 738 , c ) 900 , d ) 840 , e ) 83 | c | divide(subtract(multiply(1600, divide(15, const_100)), 15), divide(25, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n3,#0)|subtract(#2,n1)|divide(#3,#1)| | general |
30 ladies and 60 gentlemen are present at a party . there are 23 couples among them . if a lady and a gentleman is selected at random , what is the probability that they will be a couple ? | in how many ways we can select a woman and a man from 30 women and 60 men ? in 30 * 60 = 1800 ways . we have a total of 23 couples so , the probability of selecting a couple is 23 / 1800 = 23 / 1800 . ans - e | a ) 1 / 200 , b ) 1 / 100 , c ) 1 / 50 , d ) 1 / 40 , e ) 23 / 1800 | e | divide(23, multiply(30, 60)) | multiply(n0,n1)|divide(n2,#0) | probability |
in a class , 30 % of the students speaks truth , 20 % speaks lie and 10 % speaks both . if a student is selected at random , what is the probability that he has speak truth or lie ? | d ) 2 / 5 | a ) 1 / 4 , b ) 2 / 3 , c ) 3 / 5 , d ) 2 / 5 , e ) 2 / 7 | d | subtract(const_1, add(divide(10, multiply(multiply(const_5, const_5), const_4)), add(divide(30, multiply(multiply(const_5, const_5), const_4)), divide(20, multiply(multiply(const_5, const_5), const_4))))) | multiply(const_5,const_5)|multiply(#0,const_4)|divide(n0,#1)|divide(n1,#1)|divide(n2,#1)|add(#2,#3)|add(#5,#4)|subtract(const_1,#6) | other |
the milk level in a rectangular box measuring 50 feet by 25 feet is to be lowered by 6 inches . how many gallons of milk must be removed ? ( 1 cu ft = 7.5 gallons ) | "6 inches = 1 / 2 feet ( there are 12 inches in a foot . ) , so 50 * 25 * 1 / 2 = 625 feet ^ 3 of milk must be removed , which equals to 625 * 7.5 = 4687.5 gallons . answer : d ." | a ) 100 , b ) 250 , c ) 750 , d ) 4687.5 , e ) 5635.5 | d | multiply(multiply(multiply(50, 25), divide(1, const_2)), 7.5) | divide(n3,const_2)|multiply(n0,n1)|multiply(#0,#1)|multiply(n4,#2)| | general |
in the new budget , the price of kerosene oil rose by 25 % . by how much percent must a person reduce his consumption so that his expenditure on it does not increase ? | reduction in consumption = [ ( ( r / ( 100 + r ) ) * 100 ] % = [ ( 25 / 125 ) * 100 ] % = 20 % . answer is b . | a ) 10 , b ) 20 , c ) 30 , d ) 40 , e ) none of them | b | subtract(subtract(25, const_4), const_1) | subtract(n0,const_4)|subtract(#0,const_1) | general |
the length of minute hand of a clock is 5.2 cm . what is the area covered by this in 10 minutes | "area of circle is pi * r ^ 2 but in 10 minutes area covered is ( 10 / 60 ) * 360 = 60 degree so formula is pi * r ^ 2 * ( angle / 360 ) = 3.14 * ( 5.2 ^ 2 ) * ( 60 / 360 ) = 14.15 cm ^ 2 answer : b" | a ) 15.27 , b ) 14.15 , c ) 17.27 , d ) 18.27 , e ) 19.27 | b | multiply(divide(add(multiply(const_2, 10), const_2), add(const_3, const_4)), multiply(multiply(5.2, 5.2), divide(multiply(const_1, const_60), multiply(const_100, const_3_6)))) | add(const_3,const_4)|multiply(n1,const_2)|multiply(const_1,const_60)|multiply(const_100,const_3_6)|multiply(n0,n0)|add(#1,const_2)|divide(#2,#3)|divide(#5,#0)|multiply(#6,#4)|multiply(#7,#8)| | physics |
the lcm of two numbers is 2310 and hcf is 26 . if one of the numbers is 210 . then what is the other number ? | "first number * second number = lcm * hcf other number = 2310 * 26 / 210 = 11 * 26 = 286 answer : c" | a ) 715 , b ) 825 , c ) 286 , d ) 582 , e ) 465 | c | divide(multiply(2310, 26), 210) | multiply(n0,n1)|divide(#0,n2)| | physics |
find the compound interest accrued on an amount of rs . 14,800 at 13.5 % p . a at the end of two years . ( round off your answer to nearest integer ) | ci = 14800 { [ 1 + 13.5 / 100 ] 2 - 1 } = 14800 { [ 1 + 27 / 200 ] 2 - 1 = 14800 { 2 + 27 / 200 } { 27 / 200 } = ( 74 ) [ 2 + 27 / 200 ] ( 27 ) = 1998 [ 2 + 27 / 200 ] = 3996 + 269.73 = rs . 4266 answer : d | a ) 3337 , b ) 2782 , c ) 2788 , d ) 4266 , e ) 2321 | d | multiply(add(multiply(multiply(add(const_3, const_4), const_2), multiply(const_100, multiply(add(const_2, const_3), const_2))), multiply(subtract(multiply(add(const_2, const_3), const_2), const_2), const_100)), subtract(power(add(const_1, divide(13.5, const_100)), const_2), const_1)) | add(const_3,const_4)|add(const_2,const_3)|divide(n1,const_100)|add(#2,const_1)|multiply(#0,const_2)|multiply(#1,const_2)|multiply(#5,const_100)|power(#3,const_2)|subtract(#5,const_2)|multiply(#4,#6)|multiply(#8,const_100)|subtract(#7,const_1)|add(#9,#10)|multiply(#12,#11) | gain |
if all of the telephone extensions in a certain company must be even numbers , and if each of the extensions uses all 4 of the digits 1 , 2 , 3 , and 8 , what is the greatest number of 4 - digit extensions that the company can have ? | since the phone number must be even , the unit ' s digit can be either 2 or 8 . when the unit ' s digit is 2 - - > number of possibilities is 3 ! = 6 when the unit ' s digit is 8 - - > number of possibilities is 3 ! = 6 largest number of extensions = 6 + 6 = 12 answer : c | a ) 4 , b ) 6 , c ) 12 , d ) 16 , e ) 24 | c | multiply(4, const_3) | multiply(n0,const_3) | general |
a rectangular grass field is 75 m * 55 m , it has a path of 2.5 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 7 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 2.5 * 2 ) 2 * 2.5 = > 675 675 * 7 = rs . 4725 answer : b" | a ) 4350 , b ) 4725 , c ) 4328 , d ) 4329 , e ) 4829 | b | multiply(subtract(rectangle_area(add(75, multiply(2.5, const_2)), add(55, multiply(2.5, 7))), rectangle_area(75, 55)), 7) | multiply(n2,const_2)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)|multiply(n3,#5)| | geometry |
of 40 applicants for a job , 24 had at least 4 years ' experience , 15 had degrees , and 4 had less than 4 years ' experience and did not have a degree . how many of the applicants had at least 4 years ' experience and a degree ? | "set a : people with more than 4 years exp set b : people with degree aub = total - ( less than 4 exp and no degree ) aub = 40 - 4 = 36 aub = a + b - aib aib = 15 + 24 - 36 = 3 answer b" | a ) 14 , b ) 3 , c ) 9 , d ) 7 , e ) 5 | b | add(subtract(add(24, 15), subtract(40, 4)), subtract(15, 24)) | add(n1,n3)|subtract(n0,n4)|subtract(n3,n1)|subtract(#0,#1)|add(#3,#2)| | general |
what is the sum of all the multiples of 3 between 30 and 80 ? | "you first have to know all the multiples of 3 between 30 and 80 . they are 3 , 6,9 , 12,15 , 18,21 , 24,27 , 30,33 , 36,39 , 42,45 , 48,51 , 54,57 , 60,63 , 66,69 , 72,75 , and 78 . if you add all these numbers together , you get 1053 . final answer : c" | a ) 2103 , b ) 2106 , c ) 1053 , d ) 1252 , e ) 1535 | c | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 3), add(const_2, const_4)) | add(const_12,const_2)|add(const_2,const_4)|add(const_10,const_2)|subtract(const_10,const_1)|add(#0,const_1)|add(#1,const_4)|add(#5,#3)|add(#4,const_1)|add(#6,#5)|add(#8,#2)|add(#0,#9)|add(#4,#10)|add(#11,#7)|add(n0,#12)|add(#13,#1)| | general |
the first half of the 20 th century , the population of a particular country increased by 200 percent . in the second half of the century , the population increased by 300 percent . what was the percent increase for the 20 th century as a whole ? | say initially population was 100 . what is 200 % of 100 ? it is 200 / 100 * 100 = 200 . an increase of 200 % means the new population became 100 + 200 = 300 what is 300 % of 300 ? it is 300 / 100 * 300 = 900 an increase of 300 % means the new population now is 300 + 900 = 1200 so from 100 , the population increased to ... | a ) 500 % , b ) 600 % , c ) 800 % , d ) 1100 % , e ) 1200 % | d | subtract(add(multiply(multiply(const_100, divide(300, const_100)), divide(300, const_100)), multiply(const_100, divide(300, const_100))), const_100) | divide(n2,const_100)|multiply(#0,const_100)|multiply(#0,#1)|add(#2,#1)|subtract(#3,const_100) | gain |
at a certain restaurant , the average ( arithmetic mean ) number of customers served for the past x days was 60 . if the restaurant serves 120 customers today , raising the average to 70 customers per day , what is the value of x ? | "withoutusing the formula , we can see that today the restaurant served 50 customers above the average . the total amount above the average must equal total amount below the average . this additional 50 customers must offset the “ deficit ” below the average of 70 created on the x days the restaurant served only 60 cus... | a ) 2 , b ) 5 , c ) 9 , d ) 15 , e ) 30 | b | subtract(divide(subtract(120, 70), subtract(70, 60)), divide(subtract(120, const_100), const_100)) | subtract(n1,n2)|subtract(n2,n0)|subtract(n1,const_100)|divide(#0,#1)|divide(#2,const_100)|subtract(#3,#4)| | general |
100 students appeared in 2 tests . 60 students passed 1 st test . 40 students passed in the 2 nd test . 20 students passed in both 1 and 2 tests . what is the probability of the students who failed in both tests ? | 20 student passed both two test 40 student passed only 1 st test 20 student passed only 2 nd test so 100 - ( 20 + 40 + 20 ) = 20 student failed in both sub so ans is 20 % answer : c | a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | c | subtract(const_100, subtract(add(60, 40), 20)) | add(n2,n4)|subtract(#0,n6)|subtract(const_100,#1) | other |
how many 3 - digit even numbers are possible such that if one of the digits is 5 , the next / succeeding digit to it should be 6 | 560 , 562 , 564 , 566 , and 568 , so total 5 . hence option a . | a ) 5 , b ) 305 , c ) 365 , d ) 405 , e ) 495 | a | add(add(3, 6), 6) | add(n0,n2)|add(n2,#0)| | general |
there has been successive increases of 30 % and then 20 % in the price of gas from the previous month . by what percentage should a driver reduce gas consumption so that the expenditure does not change ? | "let p be the original price per unit of gas . let x be the original gas consumption . let y be the reduced gas consumption . y * 1.2 * 1.3 * p = x * p y = x / ( 1.2 * 1.3 ) which is about 0.64 x which is a decrease of about 36 % . the answer is e ." | a ) 20 % , b ) 24 % , c ) 28 % , d ) 32 % , e ) 36 % | e | multiply(subtract(const_1, divide(const_100, add(add(const_100, 30), divide(multiply(add(const_100, 30), 20), const_100)))), const_100) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|add(#0,#2)|divide(const_100,#3)|subtract(const_1,#4)|multiply(#5,const_100)| | general |
a train speeds past a pole in 50 seconds and a platform 500 m long in 100 seconds . its length is : | "let the length of the train be x meters and its speed be y m / sec . they , x / y = 50 = > y = x / 50 x + 500 / 100 = x / 50 x = 500 m . answer : option e" | a ) 550 m . , b ) 300 m . , c ) 600 m . , d ) 400 m . , e ) 500 m . | e | multiply(500, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
if m is an integer such that ( - 2 ) ^ 2 m = 2 ^ ( 24 - m ) then m = ? | "2 m = 24 - m 3 m = 24 m = 8 the answer is b ." | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | divide(24, add(2, const_1)) | add(n0,const_1)|divide(n3,#0)| | general |
a seller of used cars has 16 cars to sell and each of his clients selected 2 cars that he liked most . if each car was selected exactly thrice , how many clients visited the garage ? | ifno caris selected more than once then the number of clients = 16 / 2 = 8 but since every car is being selected three times so no . of clients must be thrice as well = 8 * 3 = 24 answer : option e | a ) 8 , b ) 10 , c ) 12 , d ) 14 , e ) 24 | e | multiply(divide(16, 2), const_3) | divide(n0,n1)|multiply(#0,const_3) | general |
two numbers are in the ratio 3 : 5 . if 9 be subtracted from each , they are in the ratio of 9 : 17 . the first number is ? | "( 3 x - 9 ) : ( 5 x - 9 ) = 9 : 17 x = 12 = > 3 x = 36 answer : a" | a ) 36 , b ) 76 , c ) 98 , d ) 27 , e ) 24 | a | add(multiply(3, divide(9, multiply(3, 5))), multiply(5, divide(9, multiply(3, 5)))) | multiply(n0,n1)|divide(n2,#0)|multiply(n0,#1)|multiply(n1,#1)|add(#2,#3)| | other |
the smallest fraction , which each of 6 / 7 , 5 / 14 , 10 / 21 will divide exactly is ? | "required fraction = l . c . m of 6 / 7 , 5 / 14 , 10 / 21 = ( l . c . m of 6 , 5 , 10 ) / ( h . c . f of 7 , 14 , 21 ) = 30 / 7 answer : a" | a ) 30 / 7 , b ) 30 / 9 , c ) 30 / 2 , d ) 30 / 3 , e ) 30 / 6 | a | divide(multiply(6, 5), 7) | multiply(n0,n2)|divide(#0,n1)| | general |
a solution contains 8 parts of water for every 7 parts of lemonade syrup . how many parts of the solution should be removed and replaced with water so that the solution will now contain 40 % lemonade syrup ? | "let the total solution is 150 l with 80 l water & 70 l syrup . to make 40 % syrup solution , the result solution must have 90 l syrup and 60 l syrup . therefore we are taking 10 l of syrup from initial solution and replacing with water . using urinary method : 70 l syrup in 150 l solution 10 l syrup in 21.4 l solution... | a ) 1.5 , b ) 1.75 , c ) 2.14 , d ) 2.34 , e ) 2.64 | c | multiply(divide(subtract(divide(7, add(8, 7)), divide(const_2, add(const_2, const_3))), divide(7, add(8, 7))), add(8, 7)) | add(n0,n1)|add(const_2,const_3)|divide(n1,#0)|divide(const_2,#1)|subtract(#2,#3)|divide(#4,#2)|multiply(#0,#5)| | gain |
the length of a rectangular floor is more than its breadth by 200 % . if rs . 640 is required to paint the floor at the rate of rs . 5 per sq m , then what would be the length of the floor ? | "let the length and the breadth of the floor be l m and b m respectively . l = b + 200 % of b = l + 3 b = 3 b area of the floor = 640 / 5 = 128 sq m l b = 128 i . e . , l * l / 3 = 128 l 2 = 384 = > l = 19.6 m . answer : a" | a ) 19.6 m . , b ) 20.0 m . , c ) 19.3 m . , d ) 18.5 m . , e ) 18.9 m . | a | multiply(sqrt(divide(divide(640, 5), const_3)), const_3) | divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)| | gain |
a reduction of 15 % in the price of wheat enables a house wife to obtain 3 kgs more for rs . 500 , what is the reduced price for kg ? | explanation : 500 * ( 15 / 100 ) = 75 - - - - 3 ? - - - - 1 = > rs . 25 answer : d | a ) rs . 10 , b ) rs . 15 , c ) rs . 20 , d ) rs . 25 , e ) rs . 30 | d | divide(subtract(500, multiply(divide(subtract(const_100, 15), const_100), 500)), 3) | subtract(const_100,n0)|divide(#0,const_100)|multiply(n2,#1)|subtract(n2,#2)|divide(#3,n1) | gain |
in a certain corporation , there are 300 male employees and 150 female employees . it is known that 20 % of the male employees have advanced degrees and 40 % of the females have advanced degrees . if one of the 450 employees is chosen at random , what is the probability this employee has an advanced degree or is female... | "p ( female ) = 150 / 450 = 1 / 3 p ( male with advanced degree ) = 0.2 * 300 / 450 = 60 / 450 = 2 / 15 the sum of the probabilities is 7 / 15 the answer is a ." | a ) 7 / 15 , b ) 3 / 10 , c ) 3 / 5 , d ) 2 / 5 , e ) 3 / 4 | a | add(divide(multiply(subtract(const_1, divide(40, multiply(20, 20))), 150), 450), divide(add(multiply(divide(20, multiply(20, 20)), 300), multiply(divide(40, multiply(20, 20)), 150)), 450)) | multiply(n2,n2)|divide(n3,#0)|divide(n2,#0)|multiply(n0,#2)|multiply(n1,#1)|subtract(const_1,#1)|add(#3,#4)|multiply(n1,#5)|divide(#7,n4)|divide(#6,n4)|add(#8,#9)| | other |
the ration of the father ’ s age to his son ’ s age is 7 : 3 . the product of their ages is 756 . the ratio of their ages after 6 years will be : | solution let the present ages of the father and son be 7 x and 3 x years respectively . then , 7 x 3 x = 756 ⇔ 21 x 2 = 756 ⇔ x 2 = 36 ⇔ x = 6 . ∴ required ratio = ( 7 x + 6 ) : ( 3 x + 6 ) = 48 : 24 = 2 : 1 . answer b | a ) 5 : 2 , b ) 2 : 1 , c ) 11 : 7 , d ) 13 : 9 , e ) none of these | b | divide(add(multiply(7, sqrt(divide(756, multiply(7, 3)))), 6), add(multiply(3, sqrt(divide(756, multiply(7, 3)))), 6)) | multiply(n0,n1)|divide(n2,#0)|sqrt(#1)|multiply(n0,#2)|multiply(n1,#2)|add(n3,#3)|add(n3,#4)|divide(#5,#6) | general |
at a certain university , 68 % of the professors are women , and 70 % of the professors are tenured . if 90 % of the professors are women , tenured , or both , then what percent of the men are tenured ? | total women = 68 % total men = 40 % total tenured = 70 % ( both men and women ) therefore , women tenured + women professors + men tenured = 90 % men tenured = 22 % but question wants to know the percent of men that are tenured 22 % / 40 % = 55 % c | a ) 25 , b ) 37.5 , c ) 55 , d ) 62.5 , e ) 75 | c | add(subtract(const_100, 68), subtract(90, 68)) | subtract(const_100,n0)|subtract(n2,n0)|add(#0,#1) | gain |
how many times are the hands of a clock at right angles in a day ? | in 12 hours , they are at right angles 22 times . = = > in 24 hours , they are at right angles 44 times . answer is b . | a ) 42 , b ) 44 , c ) 49 , d ) 41 , e ) 47 | b | multiply(multiply(subtract(const_12, const_1), const_2), const_2) | subtract(const_12,const_1)|multiply(#0,const_2)|multiply(#1,const_2) | physics |
a cyclist traveled for two days . on the second day the cyclist traveled 4 hours longer and at an average speed 10 mile per hour slower than she traveled on the first day . if during the two days she traveled a total of 240 miles and spent a total of 12 hours traveling , what was her average speed on the second day ? | solution : d = 280 mi t = 12 hrs đ â y 1 time = t 1 d â y 2 time = t 2 t 2 - t 1 = 4 hrs - - - - - ( i ) t 1 + t 2 = 12 hrs - - - - - ( ii ) adding i and ii , t 2 = 8 hrs and t 1 = 4 hrs d à y 1 rate = r 1 d â y 2 rate = r 2 r 1 - r 2 = 10 mph í . ẹ . r 1 = 10 + r 2 280 = 8 r 2 + 4 r 1 í . ẹ . 280 = 8 r 2 + 4 ( 10 + r ... | a ) 5 mph , b ) 10 mph , c ) 20 mph , d ) 30 mph , e ) 40 mph | b | divide(subtract(240, multiply(divide(subtract(12, 4), const_2), 10)), add(divide(subtract(12, 4), const_2), add(divide(subtract(12, 4), const_2), 4))) | subtract(n3,n0)|divide(#0,const_2)|add(n0,#1)|multiply(n1,#1)|add(#2,#1)|subtract(n2,#3)|divide(#5,#4) | physics |
a is twice as good as workman as b and together they finish a piece of work in 18 days . in how many days will b alone finish the work . | explanation : as per question , a do twice the work as done by b . so a : b = 2 : 1 also ( a + b ) one day work = 1 / 18 to get days in which b will finish the work , lets calculate work done by b in 1 day = = ( 118 ∗ 13 ) = 154 [ please note we multiplied by 1 / 3 as per b share and total of ra ɵ o is 1 / 3 ] so b wil... | a ) 27 days , b ) 54 days , c ) 56 days , d ) 68 days , e ) none of these | b | multiply(18, const_3) | multiply(n0,const_3) | physics |
in the xy - coordinate system , what is the slope of the line that goes through the origin and is equidistant from the two points p = ( 4 , 6 ) and q = ( 6 , 2 ) ? | first , get the middle coordinate between ( 46 ) and ( 62 ) . x = 4 + ( 6 - 4 ) / 2 = 5 y = 2 + ( 6 - 2 ) / 2 = 4 second , get the slope of ( 54 ) and ( 00 ) . m = 4 - 0 / 5 - 0 = 4 / 5 = 0.8 answer : a | a ) 0.8 , b ) 1.0 , c ) 1.2 , d ) 1.4 , e ) 1.6 | a | divide(divide(add(6, 2), const_2), divide(add(4, 6), const_2)) | add(n1,n3)|add(n0,n1)|divide(#0,const_2)|divide(#1,const_2)|divide(#2,#3) | general |
an item is being sold for $ 10 each . however , if a customer will “ buy at least 3 ” they have a promo discount of 15 % . also , if a customer will “ buy at least 10 ” items they will deduct an additional 8 % to their “ buy at least 3 ” promo price . if sam buys 10 pcs of that item how much should he pay ? | "without any discount sam should pay 10 * 10 = $ 100 . now , the overall discount would be slightly less than 23 % , thus he must pay slightly more than $ 77 . answer : e ." | a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 80.00 | e | multiply(subtract(10, divide(multiply(15, 8), const_100)), 10) | multiply(n2,n4)|divide(#0,const_100)|subtract(n0,#1)|multiply(#2,n0)| | gain |
the perimeter of a rectangle is 60 mete ( 132 ) 7 × ( 132 ) ? = ( 132 ) 11.5 . | "7 + x = 11.5 x = 11.5 - 7 x = 4.5 answer : d" | a ) 3 , b ) 3.5 , c ) 4 , d ) 4.5 , e ) 5 | d | multiply(multiply(power(divide(60, multiply(add(132, 7), 7)), 7), 132), 7) | add(n1,n2)|multiply(#0,n2)|divide(n0,#1)|power(#2,n2)|multiply(n1,#3)|multiply(n2,#4)| | geometry |
the youngest of 4 children has siblings who are 3 , 5 , and 8 years older than she is . if the average ( arithmetic mean ) age of the 4 siblings is 21 , what is the age of the youngest sibling ? | "total age of the 4 sibling is 21 x 4 = 84 years . . we already have the total age of all the children is 4 y + 16 so , 4 y + 16 = 84 or , 4 y = 68 or , y = 17 so , age of the youngest child is 17 years . answer : a" | a ) 17 , b ) 18 , c ) 19 , d ) 21 , e ) 22 | a | divide(subtract(multiply(4, 21), add(add(4, 5), 8)), 4) | add(n0,n2)|multiply(n0,n5)|add(n3,#0)|subtract(#1,#2)|divide(#3,n0)| | general |
a can do a piece of work in 15 days and b can do it in 15 days and c can do it 20 days . they started the work together and a leaves after 2 days and b leaves after 4 days from the beginning . how long will work lost ? | "2 / 15 + 4 / 15 + x / 20 = 1 x = 12 answer : c" | a ) 15 , b ) 10 , c ) 12 , d ) 11 , e ) 13 | c | add(divide(subtract(const_1, add(multiply(subtract(4, 2), add(inverse(15), inverse(20))), multiply(add(inverse(20), add(inverse(15), inverse(15))), 2))), inverse(15)), 4) | inverse(n1)|inverse(n2)|inverse(n0)|subtract(n4,n3)|add(#0,#1)|add(#2,#0)|add(#5,#1)|multiply(#4,#3)|multiply(n3,#6)|add(#7,#8)|subtract(const_1,#9)|divide(#10,#0)|add(n4,#11)| | physics |
the speed of a boat in still water is 36 kmph . what is the speed of the stream if the boat can cover 80 km downstream or 40 km upstream in the same time ? | "x = the speed of the stream ( 36 + x ) / ( 36 - x ) = 2 / 1 36 + x = 72 - 2 x 3 x = 36 x = 12 km / hour if the speed of the stream is 12 km / hour , then the ' downstream ' speed of the boat is 36 + 12 = 48 km / hour and the ' upstream ' speed of the boat is 36 - 12 = 24 km / hour . in that way , if the boat traveled ... | a ) 10 kmph , b ) 14 kmph , c ) 12 kmph , d ) 16 kmph , e ) 15 kmph | c | divide(36, add(const_1, const_2)) | add(const_1,const_2)|divide(n0,#0)| | physics |
a man buys a cycle for rs . 1800 and sells it at a loss of 10 % . what is the selling price of the cycle ? | "s . p . = 90 % of rs . 1800 = 90 / 100 x 1800 = rs . 1620 answer : b" | a ) 1410 , b ) 1620 , c ) 1430 , d ) 1440 , e ) 1540 | b | divide(multiply(subtract(const_100, 10), 1800), const_100) | subtract(const_100,n1)|multiply(n0,#0)|divide(#1,const_100)| | gain |
if n divided by 11 has a remainder of 1 , what is the remainder when 5 times n is divided by 11 ? | "as per question = > n = 11 p + 1 for some integer p hence 5 n = > 55 q + 5 = > remainder = > 5 for some integer q hence e" | a ) 1 , b ) 2 , c ) 3 , d ) 9 , e ) 5 | e | multiply(5, 1) | multiply(n1,n2)| | general |
an artist wishes to paint a circular region on a square poster that is 3.4 feet on a side . if the area of the circular region is to be 1 / 2 the area of the poster , what must be the radius of the circular region in feet ? | area of the poster is 3.4 x 3.4 = 11.56 1 / 2 the area = 5.78 pi * r ^ 2 = 5.78 r ^ 2 = 5.78 / pi r = sqrt ( 5.78 / pi ) answer ( e ) | ['a ) 1 / pi', 'b ) sqrt ( 2 / pi )', 'c ) 1', 'd ) 2 / sqrt ( pi )', 'e ) sqrt ( 5.78 / pi )'] | e | sqrt(divide(divide(square_area(3.4), 2), const_pi)) | square_area(n0)|divide(#0,n2)|divide(#1,const_pi)|sqrt(#2) | geometry |
the value of ( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.5 * 7.57 ) is close to | "( 34.31 * 0.473 * 1.567 ) / ( 0.0673 * 23.5 * 7.57 ) = 25.4303 / 11.972 = 2.12 answer : d" | a ) 2 , b ) 1.15 , c ) 2.05 , d ) 2.12 , e ) 2.35 | d | divide(divide(multiply(multiply(34.31, 0.473), 1.567), multiply(multiply(7.57, 23.5), 0.0673)), const_10) | multiply(n0,n1)|multiply(n4,n5)|multiply(n2,#0)|multiply(n3,#1)|divide(#2,#3)|divide(#4,const_10)| | general |
a is two years older than b who is twice as old as c . if the total of the ages of a , b and c be 47 , then how old is b ? | "let c ' s age be x years . then , b ' s age = 2 x years . a ' s age = ( 2 x + 2 ) years . ( 2 x + 2 ) + 2 x + x = 47 5 x = 45 = > x = 9 hence , b ' s age = 2 x = 18 years . answer : c" | a ) 17 years , b ) 19 years , c ) 18 years , d ) 10 years , e ) 12 years | c | divide(multiply(subtract(47, const_2), const_2), add(const_4, const_1)) | add(const_1,const_4)|subtract(n0,const_2)|multiply(#1,const_2)|divide(#2,#0)| | general |
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