Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 300 | correct stringclasses 5
values | annotated_formula stringlengths 7 6.48k | linear_formula stringlengths 8 925 | category stringclasses 6
values |
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what is the area of a square with perimeter 12 p ? | each side is 3 p a = ( 3 p ) ^ 2 = 9 p ^ 2 answer c | ['a ) 16 p ^ 2', 'b ) 4 p', 'c ) 9 p ^ 2', 'd ) p / 16', 'e ) p ^ 2 / 16'] | c | square_area(square_edge_by_perimeter(12)) | square_edge_by_perimeter(n0)|square_area(#0) | geometry |
exactly 10 % of the reporters for a certain wire service cover local politics in country x . if 30 % of the reporters who cover politics for the wire service do not cover local politics in country x , what percent of the reporters for the wire service do not cover politics ? | "you are correct . people who cover local politics are a subset of people who cover politics . 30 % of reporters who cover politics do not cover local politics so 70 % do cover local politics . reporters covering local politics = 70 % of reporters covering politics = 10 % of all reporters reporters covering politics / ... | a ) 20 % , b ) 42 % , c ) 44 % , d ) 85 % , e ) 84 % | d | multiply(subtract(const_1, divide(10, subtract(const_100, 30))), const_100) | subtract(const_100,n1)|divide(n0,#0)|subtract(const_1,#1)|multiply(#2,const_100)| | gain |
sachin is younger than rahul by 7 years . if the ratio of their ages is 5 : 12 , find the age of sachin | explanation : if rahul age is x , then sachin age is x - 7 , so , 12 x - 84 = 5 x 7 x = 84 x = 12 so sachin age is 12 - 7 = 5 answer : e ) 5 | a ) 9 , b ) 24.8 , c ) 2 , d ) 3 , e ) 5 | e | multiply(divide(7, subtract(12, 5)), 5) | subtract(n2,n1)|divide(n0,#0)|multiply(n1,#1) | other |
in a 6 - sided polygon xqrstv ∠ x measures 48 ° less than the average ( arithmetic mean ) of the remaining 5 angles . what does ∠ x equal to , in degrees ? | the sum of interior angle in a polygon is ( n - 2 ) * 180 so for a six sided figure , ( 6 - 2 ) * 180 = 720 let the a . m of the other 5 angles be ' m ' - - > m * 5 + ∠ x = 720 , we know ∠ x = m - 48 - - > m * 5 + m - 48 = 720 , solving we get m = 128 hence ∠ x = 128 - 48 = 80 ans b . | a ) 72 , b ) 80 , c ) 88 , d ) 96 , e ) 104 | b | add(multiply(6, 5), 48) | multiply(n0,n2)|add(n1,#0) | general |
a train crosses a platform of 150 m in 15 sec , same train crosses another platform of length 220 m in 18 sec . then find the length of the train ? | "length of the train be ‘ x ’ ( x + 150 ) / 15 = ( x + 220 ) / 18 5 x + 1100 = 6 x + 900 x = 200 m answer : a" | a ) 200 m , b ) 180 m , c ) 272 m , d ) 210 m , e ) 267 m | a | subtract(multiply(220, divide(15, divide(15, const_3))), multiply(150, divide(18, divide(15, const_3)))) | divide(n1,const_3)|divide(n1,#0)|divide(n3,#0)|multiply(n2,#1)|multiply(n0,#2)|subtract(#3,#4)| | physics |
if ( - 4 ) ^ ( 6 x ) = 4 ^ ( 30 + x ) and x is an integer , what is the value of x ? | "since x is an integer , ( - 4 ) ^ ( 6 x ) is always positive . so , 4 ^ 6 x = 4 ^ ( 30 + x ) 6 x = 30 + x 5 x = 30 x = 6 answer : e" | a ) 5 , b ) 4 , c ) 3 , d ) 2 , e ) 6 | e | divide(30, 4) | divide(n3,n0)| | general |
working alone , a can complete a certain kind of job in 10 hours . a and d , working together at their respective rates , can complete one of these jobs in 5 hours . in how many hours can d , working alone , complete one of these jobs ? | "let total time taken by d to complete the job = d total time taken by a to complete the job = 10 work done by a in an hour 1 / a = 1 / 10 working together a and d can complete the job in 5 hours 1 / a + 1 / d = 1 / 5 = > 1 / d = 1 / 5 - 1 / a = 1 / 5 - 1 / 10 = 1 / 10 = > d = 10 hours answer b" | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 13 | b | inverse(subtract(inverse(5), inverse(10))) | inverse(n1)|inverse(n0)|subtract(#0,#1)|inverse(#2)| | physics |
a train 360 m long is running at a speed of 45 km / hr . in what time will it pass a bridge 160 m long ? | "speed = 45 * 5 / 18 = 25 / 2 m / sec total distance covered = 360 + 160 = 520 m required time = 520 * 2 / 25 = 41.6 sec answer : a" | a ) 41.6 sec , b ) 11 sec , c ) 88 sec , d ) 19 sec , e ) 10 sec | a | divide(360, multiply(subtract(45, 160), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
man can row upstream at 10 kmph and downstream at 20 kmph , and then find the speed of the man in still water ? | us = 10 ds = 20 m = ( 10 + 20 ) / 2 = 15 answer : e | a ) 5 , b ) 20 , c ) 30 , d ) 25 , e ) 15 | e | divide(add(10, 20), const_2) | add(n0,n1)|divide(#0,const_2) | physics |
at what rate percent of simple interest will a sum of money double itself in 44 years ? | "let sum = x . then , s . i . = x . rate = ( 100 * s . i . ) / ( p * t ) = ( 100 * x ) / ( x * 44 ) = 100 / 44 = 2.27 % answer : e" | a ) 6.54 % , b ) 4.54 % , c ) 8 . 2 % , d ) 4.94 % , e ) 2.27 % | e | divide(const_100, 44) | divide(const_100,n0)| | gain |
which is greatest in 16 ( 2 / 3 ) % , 2 / 5 and 0.17 ? | 16 ( 2 / 3 ) % = [ ( 50 / 3 ) * ) 1 / 100 ) ] = 1 / 6 = 0.166 , 2 / 15 = 0.133 . clearly , 0.17 is the greatest . answer is b . | a ) 2 / 3 , b ) 0.17 , c ) 3 / 2 , d ) 0.71 , e ) none of them | b | multiply(0.17, const_1) | multiply(n5,const_1) | gain |
the sum of first n consecutive odd integers is n ^ 2 . what is the sum of all odd integers between 11 and 41 inclusive . | "we ' re dealing with a sequence of consecutive odd integers : 11 to 41 , inclusive . we ' re asked for the sum of this group . 1 ) start with the sum of the smallest and the biggest : 11 + 41 = 52 2 ) now look at the ' next smallest ' and the ' next biggest ' : 13 + 39 = 52 now we have proof that there is no middle te... | a ) 351 , b ) 364 , c ) 416 , d ) 424 , e ) 450 | c | subtract(power(divide(add(41, const_1), 2), 2), power(divide(add(subtract(11, 2), const_1), 2), 2)) | add(n2,const_1)|subtract(n1,n0)|add(#1,const_1)|divide(#0,n0)|divide(#2,n0)|power(#3,n0)|power(#4,n0)|subtract(#5,#6)| | general |
if 25 / 999 = 0.002 , what is the 81 st digit to the right of the decimal point in the decimal equivalent of the fraction ? | "0 . [ u ] 25 [ / u = 0.002002002002 . . . . . every 3 rd digit is 5 and every multiple of 3 will be the digit 2 . since 81 is multiple of 3 , the 81 st digit is 2 . answer e" | a ) 6 , b ) 5 , c ) 4 , d ) 3 , e ) 2 | e | add(const_2, const_3) | add(const_2,const_3)| | general |
if x is an integer and 6.1 × 10 ^ x is less than 620 what is the greatest possible value for x ? | if x = 2 6.1 × 10 ^ 2 = 610 < 620 so , x = 2 answer : a | a ) 2 , b ) 6 , c ) 5 , d ) 4 , e ) 3 | a | floor(divide(log(divide(620, 6.1)), log(10))) | divide(n2,n0)|log(n1)|log(#0)|divide(#2,#1)|floor(#3) | general |
in the game of dubblefud , red balls , blue balls and green balls are each worth 2 , 4 and 5 points respectively . in a certain selection of balls , the product of the point values of the balls is 16000 . if the number of blue balls in this selection equals the number of green balls , how many red balls are in the sele... | this is equivalent to : - 2 x * 4 y * 5 z = 16000 y = z ( given ) 2 x * 4 y * 5 y = 16000 2 x * y ^ 2 = 16000 / 20 2 x * y ^ 2 = 800 now from options given we will figure out which number will divide 800 and gives us a perfect square : - which gives us x = 4 as 2 * 4 * y ^ 2 = 800 y ^ 2 = 100 y = 10 number of red balls... | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | add(add(multiply(divide(2, 2), 2), divide(2, 2)), divide(2, 2)) | divide(n0,n0)|multiply(n0,#0)|add(#0,#1)|add(#2,#0) | general |
if the area of a circle is 49 pi square feet , find its circumference . | "the area is given by pi * r * r . hence pi * r * r = 49 pi r * r = 49 ; hence r = 49 feet the circumference is given by 2 * pi * r = 2 * pi * 7 = 14 pi feet correct answer a" | a ) 14 pi feet , b ) 28 pi feet , c ) 38 pi feet , d ) 48 pi feet , e ) 58 pi feet | a | circumface(sqrt(49)) | sqrt(n0)|circumface(#0)| | geometry |
the chance of rain on any given day in tel - aviv is 50 % . what is the probability q that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ? | "chances of rain on exactly 4 days and not rain on 2 days = ( 1 / 2 ) ^ 4 * ( 1 / 2 ) ^ 2 = 1 / 64 chosing 4 days out of 6 = 6 ! / ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days q = 15 / 64 ans a it is ." | a ) 15 / 64 , b ) 30 / 64 , c ) 1 / 2 , d ) 1 / 4 , e ) 52 / 64 | a | divide(choose(6, 4), inverse(multiply(power(divide(const_1, const_2), 4), power(divide(const_1, const_2), const_2)))) | choose(n2,n1)|divide(const_1,const_2)|power(#1,n1)|power(#1,const_2)|multiply(#2,#3)|inverse(#4)|divide(#0,#5)| | probability |
a man can row 4.5 km / hr in still water . it takes him twice as long to row upstream as to row downstream . what is the rate of the current ? | "speed of boat in still water ( b ) = 4.5 km / hr . speed of boat with stream ( down stream ) , d = b + u speed of boat against stream ( up stream ) , u = b – u it is given upstream time is twice to that of down stream . ⇒ downstream speed is twice to that of upstream . so b + u = 2 ( b – u ) ⇒ u = b / 3 = 1.5 km / hr ... | a ) 1.9 , b ) 1.7 , c ) 1.2 , d ) 1.5 , e ) 1.1 | d | divide(subtract(multiply(4.5, const_2), 4.5), const_3) | multiply(n0,const_2)|subtract(#0,n0)|divide(#1,const_3)| | general |
sides of a rectangular park are in the ratio 4 : 3 and its area is 3750 sq m , the cost of fencing it at 50 ps per meter is ? | "4 x * 3 x = 3750 = > x = 17.7 2 ( 82.3 + 50 ) = 264.6 m 264.6 * 1 / 2 = rs . 132.3 answer : a" | a ) s . 132.3 , b ) s . 129 , c ) s . 125 , d ) s . 120 , e ) s . 121 | a | divide(multiply(50, rectangle_perimeter(sqrt(divide(multiply(3750, 3), 4)), divide(3750, sqrt(divide(multiply(3750, 3), 4))))), 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 |
what is the smallest no . which must be added to 8261955 so as to obtain a sum which is divisible by 11 ? | for divisibility by 11 , the difference of sums of digits at even and odd places must be either zero or divisible by 11 . for 8261955 , difference = ( 8 + 6 + 9 + 5 ) - ( 2 + 1 + 5 ) = 28 - 8 = 20 . the units digit is at odd place . so we add 2 to the number = > 8261955 + 2 = 8261957 now , ( 8 + 6 + 9 + 7 ) - ( 2 + 1 +... | a ) 8 , b ) 11 , c ) 13 , d ) 15 , e ) 16 | b | divide(multiply(8261955, 11), 8261955) | multiply(n0,n1)|divide(#0,n0) | general |
if 3 x + y = 40 , 2 x - y = 20 , for integers of x and y , 2 y ^ 2 = ? | "3 x + y = 40 2 x - y = 20 5 x = 60 x = 12 y = 4 2 y ^ 2 = 2 * 16 = 32 answer is e" | a ) 2 , b ) 4 , c ) 0 , d ) 10 , e ) 32 | e | multiply(3, power(subtract(40, multiply(divide(add(40, 20), add(3, 2)), 3)), 2)) | add(n1,n3)|add(n0,n2)|divide(#0,#1)|multiply(n0,#2)|subtract(n1,#3)|power(#4,n2)|multiply(n0,#5)| | general |
a ’ s speed is 20 / 14 times that of b . if a and b run a race , what part of the length of the race should a give b as a head start , so that the race ends in a dead heat ? | "we have the ratio of a ’ s speed and b ’ s speed . this means , we know how much distance a covers compared with b in the same time . this is what the beginning of the race will look like : ( start ) a _________ b ______________________________ if a covers 20 meters , b covers 14 meters in that time . so if the race i... | a ) 1 / 17 , b ) 3 / 17 , c ) 1 / 10 , d ) 6 / 20 , e ) 3 / 10 | d | divide(subtract(20, 14), 20) | subtract(n0,n1)|divide(#0,n0)| | general |
what is the product of all the possible values of x if x ^ 2 ( x + 2 ) + 7 x ( x + 2 ) + 6 ( x + 2 ) = 0 ? | x ^ 2 ( x + 2 ) + 7 x ( x + 2 ) + 6 ( x + 2 ) = 0 = > ( x + 2 ) ( x ^ 2 + 7 x + 6 ) = 0 = > ( x + 2 ) ( x + 1 ) ( x + 6 ) = 0 = > x = - 1 or - 2 or - 6 product = - 12 answer option - 12 answer : b | a ) – 29 , b ) – 12 , c ) 12 , d ) 29 , e ) 168 | b | multiply(multiply(negate(2), subtract(7, 6)), subtract(7, subtract(7, 6))) | negate(n0)|subtract(n2,n4)|multiply(#0,#1)|subtract(n2,#1)|multiply(#2,#3) | general |
jar x is 1 / 5 full of water . jar y , which has half the capacity of jar x , is 1 / 2 full of water . if the water in jar y is poured into jar x , then jar x will be filled to what fraction of its capacity ? | "let p be the capacity of jar x . the amount of water in jar y is 1 / 2 * p / 2 = p / 4 then the total amount in jar x is p / 5 + p / 4 = 9 / 20 the answer is e ." | a ) 2 / 5 , b ) 3 / 5 , c ) 7 / 10 , d ) 7 / 20 , e ) 9 / 20 | e | add(divide(1, const_2.0), multiply(divide(1, 5), divide(1, 5))) | divide(n0,n3)|multiply(#0,#0)|add(#0,#1)| | general |
a babysitter is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 30 hours per week . for any overtime hours worked in excess of 30 hours per week , the babysitter is paid a rate that is 75 % higher than his regular rate . if last week the babysitter earned $ 760 , how many total hours d... | for 30 hrs = 30 * 16 = 480 excess = 760 - 480 = 280 for extra hours = . 75 ( 16 ) = 12 in addition to $ 16 . thus , 12 + 16 = 28 number of extra hrs = 280 / 28 = 10 total hrs = 30 + 10 = 40 answer a 40 | a ) 40 , b ) 30 , c ) 35 , d ) 25 , e ) 45 | a | add(30, divide(subtract(760, multiply(16, 30)), multiply(16, divide(add(75, const_100), const_100)))) | add(n3,const_100)|multiply(n0,n1)|divide(#0,const_100)|subtract(n4,#1)|multiply(n0,#2)|divide(#3,#4)|add(n1,#5) | gain |
by travelling at 80 kmph , a person reaches his destination on time . he covered two - third the total distance in one - third of the total time . what speed should he maintain for the remaining distance to reach his destination on time ? | "let the time taken to reach the destination be 3 x hours . total distance = 80 * 3 x = 240 x km he covered 2 / 3 * 240 x = 160 x km in 1 / 3 * 3 x = x hours so , the remaining 80 x km , he has to cover in 2 x hours . required speed = 80 x / 2 x = 40 kmph . answer : b" | a ) 25 kmph , b ) 40 kmph , c ) 35 kmph , d ) 46 kmph , e ) 50 kmph | b | divide(subtract(multiply(80, const_3), divide(multiply(multiply(80, const_3), const_2), const_3)), subtract(const_3, const_1)) | multiply(n0,const_3)|subtract(const_3,const_1)|multiply(#0,const_2)|divide(#2,const_3)|subtract(#0,#3)|divide(#4,#1)| | physics |
two employees m and n are paid a total of $ 594 per week by their employer . if m is paid 120 percent of the salary paid to n , how much is n paid per week ? | 1.2 n + n = 594 2.2 n = 594 n = 270 the answer is d . | a ) $ 240 , b ) $ 250 , c ) $ 260 , d ) $ 270 , e ) $ 280 | d | divide(594, add(divide(120, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1) | general |
excluding stoppages , the speed of a bus is 50 kmph and including stoppages , it is 42 kmph . for how many minutes does the bus stop per hour ? | "d 9.6 min due to stoppages , it covers 8 km less . time taken to cover 9 km = ( 8 / 50 x 60 ) min = 9.6 min" | a ) 10 min , b ) 7 min , c ) 8.2 min , d ) 9.6 min , e ) 40 min | d | multiply(const_60, divide(subtract(50, 42), 50)) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_60)| | physics |
tough and tricky questions : functions . let a be a positive integer . let n # a equal n ^ ( 2 a ) if a is odd and n ^ ( 3 a ) if a is even . then ( 2 # 3 ) + ( 2 # 3 ) – ( 3 # 3 ) is equal to | "answer : 3 ^ 6 + 3 ^ 6 - 3 ^ 6 = 3 ^ 6 = 729 ans d" | a ) 64 , b ) 82 , c ) 128 , d ) 729 , e ) 1522 | d | power(2, multiply(const_2.0, const_3.0)) | multiply(n0,n1)|power(n5,#0)| | general |
the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 900 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | "given that the area of the square = 900 sq . units = > side of square = √ 900 = 30 units the radius of the circle = side of the square = 30 units length of the rectangle = 2 / 5 * 30 = 12 units given that breadth = 10 units area of the rectangle = lb = 12 * 10 = 120 sq . units answer : b" | a ) 140 sq . units , b ) 120 sq . units , c ) 187 sq . units , d ) 607 sq . units , e ) 815 sq . units | b | multiply(10, multiply(const_2, divide(sqrt(900), divide(10, const_2)))) | divide(n1,const_2)|sqrt(n0)|divide(#1,#0)|multiply(#2,const_2)|multiply(n1,#3)| | geometry |
a rectangular grass field is 60 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 . 2 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 60 + 55 + 2.5 * 2 ) 2 * 2.5 = > 600 600 * 2 = rs . 1200 answer : b" | a ) 1350 , b ) 1200 , c ) 9676 , d ) 1679 , e ) 2691 | b | multiply(subtract(rectangle_area(add(60, multiply(2.5, 2)), add(55, multiply(2.5, 2))), rectangle_area(60, 55)), 2) | multiply(n2,n3)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)|multiply(n3,#5)| | geometry |
a clock shows the time as 9 a . m . if the minute hand gains 5 minutes every hour , how many minutes will the clock gain by 5 p . m . ? | "there are 8 hours in between 9 a . m . to 5 p . m . 8 * 5 = 40 minutes . answer : d" | a ) 30 min , b ) 35 min , c ) 45 min , d ) 40 min , e ) 55 min | d | multiply(add(const_3, 5), 5) | add(const_3,n2)|multiply(n1,#0)| | physics |
a plant manager must assign 6 new workers to one of five shifts . she needs a first , second , and third shift , and two alternate shifts . each of the shifts will receive 2 new workers . how many different ways can she assign the new workers ? | "whatever : my take selecting team of 2 out of 10 to assign to the shifts = 6 c 2 = 15 ways . now 2 out of 10 means total of 3 group possible . so putting them in shifts = counting methode : first , second , third , alt , alt = 3 * 2 * 1 = 6 here alt and alt are the same : so 6 / 2 = 3 ways . total ways of selecting = ... | a ) 24 , b ) 27 , c ) 33 , d ) 45 , e ) 54 | d | multiply(divide(factorial(divide(6, 2)), const_2), divide(factorial(6), multiply(factorial(subtract(6, 2)), factorial(2)))) | divide(n0,n1)|factorial(n0)|factorial(n1)|subtract(n0,n1)|factorial(#0)|factorial(#3)|divide(#4,const_2)|multiply(#5,#2)|divide(#1,#7)|multiply(#6,#8)| | physics |
the average of first four multiples of 4 is : | "solution average = 4 ( 1 + 2 + 3 + 4 ) / 4 = 40 / 4 = 10 answer c" | a ) 3 , b ) 9 , c ) 10 , d ) 15 , e ) 17 | c | add(4, const_1) | add(n0,const_1)| | 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 28 % 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 43 % , thus he must pay slightly more than $ 57 . only answer choice e fits . answer : e . | a ) $ 92.00 , b ) $ 88.00 , c ) $ 87.04 , d ) $ 80.96 , e ) $ 60.00 | e | multiply(subtract(10, divide(multiply(15, 28), const_100)), const_10) | multiply(n2,n4)|divide(#0,const_100)|subtract(n0,#1)|multiply(#2,const_10) | gain |
what is the sum of the greatest common factor and the lowest common multiple of 45 and 125 ? | "prime factorization of the given numbers 45 = 3 ^ 2 125 = 5 ^ 3 greatest common factor = 5 lowest common multiple = 3 ^ 2 * 5 ^ 2 * 5 = 1125 sum = 5 + 1125 = 1130 answer d" | a ) 1920 , b ) 2201 , c ) 3240 , d ) 1130 , e ) 3840 | d | divide(multiply(45, 125), const_4) | multiply(n0,n1)|divide(#0,const_4)| | general |
the chance of rain on any given day in tel - aviv is 50 % . what is the probability e that it rained in tel - aviv on exactly 4 out of 6 randomly chosen days ? | "chances of rain on exactly 4 days and not rain on 2 days = ( 1 / 2 ) ^ 4 * ( 1 / 2 ) ^ 2 = 1 / 64 chosing 4 days out of 6 = 6 ! / ( 4 ! * 2 ! ) = 15 chances of rain on exactly 4 days out of 6 days e = 15 / 64 ans a it is ." | a ) 15 / 64 , b ) 30 / 64 , c ) 1 / 2 , d ) 1 / 4 , e ) 52 / 64 | a | divide(choose(6, 4), inverse(multiply(power(divide(const_1, const_2), 4), power(divide(const_1, const_2), const_2)))) | choose(n2,n1)|divide(const_1,const_2)|power(#1,n1)|power(#1,const_2)|multiply(#2,#3)|inverse(#4)|divide(#0,#5)| | probability |
you collect pens . suppose you start out with 10 . mike gives you another 22 pens . since her father makes pens , cindy decides to double your pens . since you ' re nice , you give sharon 19 pens . how many pens do you have at the end ? | "solution start with 10 pens . mike gives you 22 pens : 10 + 22 = 39 pens . cindy doubles the number of pens you have : 39 ã — 2 = 78 pens . sharon takes 19 pens from you : 78 - 19 = 59 pens . so you have 59 at the end . correct answer : e" | a ) 39 , b ) 40 , c ) 41 , d ) 42 , e ) 59 | e | subtract(multiply(add(22, 10), const_2), 19) | add(n0,n1)|multiply(#0,const_2)|subtract(#1,n2)| | general |
how many integer values are there for x such that 1 < 3 x + 5 < 17 ? | the inequality given is 1 < 3 x + 5 < 17 it can further reduced to - 4 < 3 x < 12 finally - 4 / 3 < x < 4 so can only take 5 integer values i . e . - 1 , 0,1 , 2,3 correct answer - d | a ) two , b ) three , c ) four , d ) five , e ) six | d | floor(subtract(divide(subtract(17, 5), 3), divide(subtract(1, 5), 3))) | subtract(n3,n2)|subtract(n0,n2)|divide(#0,n1)|divide(#1,n1)|subtract(#2,#3)|floor(#4) | general |
a portion of the 85 % solution of chemicals was replaced with an equal amount of 30 % solution of chemicals . as a result , 40 % solution of chemicals resulted . what part of the original solution was replaced ? | "this is a weighted average question . say x % of the solution was replaced - - > equate the amount of chemicals : 0.85 ( 1 - x ) + 0.3 * x = 0.4 - - > x = 9 / 11 . answer : d ." | a ) 5 / 11 , b ) 6 / 11 , c ) 7 / 11 , d ) 9 / 11 , e ) 13 / 6 | d | divide(subtract(85, 40), subtract(85, 30)) | subtract(n0,n2)|subtract(n0,n1)|divide(#0,#1)| | gain |
if the price of a book is first decreased by 20 % and then increased by 20 % , then the net change in the price will be : | "explanation : let the original price be rs . 100 . decreased by 20 % = 80 then increased 20 % on rs 80 = 80 + 16 = 96 net change in price = 100 - 96 = 4 answer : c" | a ) 10 , b ) 388 , c ) 4 , d ) 29 , e ) 22 | c | subtract(const_100, multiply(multiply(divide(subtract(const_100, divide(multiply(const_100, 20), const_100)), const_100), divide(add(const_100, divide(multiply(const_100, 20), const_100)), const_100)), const_100)) | multiply(n0,const_100)|multiply(n1,const_100)|divide(#0,const_100)|divide(#1,const_100)|add(#3,const_100)|subtract(const_100,#2)|divide(#5,const_100)|divide(#4,const_100)|multiply(#6,#7)|multiply(#8,const_100)|subtract(const_100,#9)| | general |
a basketball is dropped from a height of 30 feet . if it bounces back up to a height that is exactly half of its previous height , and it stops bouncing after hitting the ground for the fourth time , then how many total feet will the ball have traveled after 3 full bounces . | "initial distance = 30 feet first bounce = 15 feet up + 15 feet down = 30 feet second bouche = 7.5 feet up + 7.5 feet down = 15 feet third bounce = 375 feet up and 3.75 feet down = 7.5 feet total distance covered = 30 + 30 + 15 + 7.5 = 82.5 answer is b" | a ) 50 , b ) 82.5 , c ) 60 , d ) 75 , e ) 80 | b | subtract(subtract(subtract(subtract(multiply(add(divide(divide(divide(divide(30, 3), 3), 3), 3), add(add(add(add(add(add(30, divide(30, 3)), divide(30, 3)), divide(divide(30, 3), 3)), divide(divide(30, 3), 3)), divide(divide(divide(30, 3), 3), 3)), divide(divide(divide(30, 3), 3), 3))), 3), divide(divide(divide(30, 3),... | divide(n0,n1)|add(n0,#0)|divide(#0,n1)|add(#1,#0)|divide(#2,n1)|add(#3,#2)|divide(#4,n1)|add(#5,#2)|add(#7,#4)|add(#8,#4)|add(#9,#6)|multiply(n1,#10)|subtract(#11,#4)|subtract(#12,#4)|subtract(#13,#4)|subtract(#14,#4)| | general |
two cyclist start on a circular track from a given point but in opposite direction with speeds of 7 m / s and 8 m / s . if the circumference of the circle is 600 meters , after what time will they meet at the starting point ? | "they meet every 600 / 7 + 8 = 40 sec answer is e" | a ) 20 sec , b ) 15 sec , c ) 30 sec , d ) 50 sec , e ) 40 min | e | divide(600, add(8, 7)) | add(n0,n1)|divide(n2,#0)| | physics |
praveen age after 10 years will be 3 times his age 3 years back . what is the present age of chris ? | chris present age = x after 10 years = x + 10 3 years back = x - 3 x + 10 = 3 ( x - 3 ) x = 20 answer is a | a ) 20 , b ) 25 , c ) 10 , d ) 30 , e ) 15 | a | divide(add(10, multiply(3, 10)), subtract(3, const_1)) | multiply(n0,n1)|subtract(n1,const_1)|add(n0,#0)|divide(#2,#1) | general |
how many multiples of 2 are there between 10 and 52 , 10 and 52 inclusive ? | "2 multiples are . . . 10,12 , 14,16 , 18,20 , 22,24 , 26,28 , 30,32 , 34,36 , 38,40 , 42,44 , 46,48 , 50,52 , . . . , the answer is = 22 answer is e" | a ) 27 , b ) 25 , c ) 24 , d ) 21 , e ) 22 | e | add(divide(subtract(52, 10), 2), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general |
what is the next number 12 13 15 17 111 113 117 119 123 129 ? | 131 . these are the first 10 prime numbers ( 2 , 3 , 5 . . . ) prefixed with a 1 answer : d | a ) 154 , b ) 220 , c ) 129 , d ) 131 , e ) 224 | d | add(const_100, add(reminder(129, const_100), const_2)) | reminder(n9,const_100)|add(#0,const_2)|add(#1,const_100) | general |
david works at a resort from the beginning of march to the end of september . during the month of august this past year , he made 2 times the average ( arithmetic mean ) of his monthly totals in tips for the other months . his total tips for august were what fraction of his total tips for all of the months he worked ? | "first notice the number of months for which he worked - march to sept i . e . 7 months avg of monthly totals in tips for months other than august = x tips in august = 2 x total tips for all months = 6 * x + 2 x = 8 x tips for august / total tips for all months = 2 x / 8 x = 1 / 4 answer : d" | a ) 1 / 6 , b ) 1 / 7 , c ) 1 / 8 , d ) 1 / 4 , e ) 1 / 3 | d | divide(const_4.0, add(subtract(const_10, 2), 2)) | subtract(const_10,const_4.0)|add(n0,#0)|divide(n0,#1)| | general |
if the average of 10 x and 8 y is greater than 70 , and x is twice y , what is the least integer value of x ? | "substitution can be used in the following way : always start with the equation : x = 2 y . it is more straight forward to manage as compared to the inequality . substitute y = x / 2 , not the other way because you need to find the minimum value of x . so you can get rid of y . now go on to the inequality . so 8 y = 8 ... | a ) 12 , b ) 14 , c ) 16 , d ) 10 , e ) 20 | d | multiply(10, const_2) | multiply(n0,const_2)| | 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 35 % water ? | explanation : flower - nectar contains 50 % of non - water part . in honey this non - water part constitutes 65 % ( 100 - 35 ) . therefore 0.5 x amount of flower - nectar = 0.65 x amount of honey = 0.65 x 1 kg therefore amount of flower - nectar needed = ( 0.65 / 0.51 ) kg = 1.3 kgs answer : a | a ) 1.3 kg , b ) 1.5 kg , c ) 1.7 kg , d ) 1.9 kg , e ) none of these | a | divide(divide(subtract(const_100, 35), const_100), divide(50, const_100)) | divide(n1,const_100)|subtract(const_100,n2)|divide(#1,const_100)|divide(#2,#0) | general |
a sum of money is put out at compound interest for 2 years at 20 % . it would fetch rs . 1205 more if the interest were payable half - yearly , then it were pay able yearly . find the sum . | "p ( 11 / 10 ) ^ 4 - p ( 6 / 5 ) ^ 2 = 1205 p = 50000 answer : d" | a ) 22200 , b ) 31897 , c ) 48799 , d ) 50000 , e ) 52782 | d | add(multiply(add(20, const_1), add(20, const_1)), add(20, const_1)) | add(n1,const_1)|multiply(#0,#0)|add(#0,#1)| | general |
cindy has her eye on a sundress but thinks it is too expensive . it goes on sale for 15 % less than the original price . before cindy can buy the dress , however , the store raises the new price by 25 % . if the dress cost $ 61.2 after it went on sale for 15 % off , what is the difference between the original price and... | "0.85 * { original price } = $ 61.2 - - > { original price } = $ 72 . { final price } = $ 61.2 * 1.25 = $ 76.5 . the difference = $ 76.5 - $ 72 = $ 4.5 answer : b ." | a ) $ 0.00 , b ) $ 4.50 , c ) $ 3.40 , d ) $ 5.00 , e ) $ 6.80 | b | subtract(multiply(61.2, divide(add(const_100, 25), const_100)), divide(61.2, divide(subtract(const_100, 15), const_100))) | add(n1,const_100)|subtract(const_100,n0)|divide(#0,const_100)|divide(#1,const_100)|divide(n2,#3)|multiply(n2,#2)|subtract(#5,#4)| | general |
the smallest number which when diminished by 12 , is divisible 12 , 24 , 36 , 48 and 56 is : | "required number = ( l . c . m . of 12 , 24 , 36 , 48,56 ) + 12 = 1008 + 12 = 1020 answer : option c" | a ) 1008 , b ) 1015 , c ) 1020 , d ) 1032 , e ) 1048 | c | add(multiply(multiply(power(const_3, const_2.0), power(const_2.0, const_4)), add(const_3, const_4)), 12) | add(const_3,const_4)|power(const_3,const_2.0)|power(const_2.0,const_4)|multiply(#1,#2)|multiply(#0,#3)|add(#4,n0)| | general |
there are 100 students in 3 sections a , b and c of a class . the average marks of all the 3 sections was 84 . the average of b and c was 87.5 and the average marks of students in a section was ? | numbers of students in section a = x numbers of students in section b and c = ( 100 – x ) x 70 + ( 100 – x ) 87.5 = 84 100 70 x + 87.5 100 – 87.5 x = 8400 8750 – 17.5 x = 8400 17.5 x = 8750 – 8400 = > x = 20 . answer : a | a ) 20 , b ) 99 , c ) 28 , d ) 27 , e ) 12 | a | divide(subtract(multiply(100, 87.5), multiply(100, 84)), subtract(87.5, multiply(add(const_4, const_3), const_10))) | add(const_3,const_4)|multiply(n0,n4)|multiply(n0,n3)|multiply(#0,const_10)|subtract(#1,#2)|subtract(n4,#3)|divide(#4,#5) | general |
a trader bought a car at 5 % discount on its original price . he sold it at a 60 % increase on the price he bought it . what percent of profit did he make on the original price ? | original price = 100 cp = 95 s = 95 * ( 160 / 100 ) = 152 100 - 152 = 52 % answer : c | a ) 17 % , b ) 62 % , c ) 52 % , d ) 19 % , e ) 22 % | c | multiply(subtract(divide(divide(multiply(subtract(const_100, 5), add(const_100, 60)), const_100), const_100), const_1), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|divide(#3,const_100)|subtract(#4,const_1)|multiply(#5,const_100) | gain |
the mean daily profit made by a shopkeeper in a month of 30 days was rs . 350 . if the mean profit for the first fifteen days was rs . 275 , then the mean profit for the last 15 days would be | "average would be : 350 = ( 275 + x ) / 2 on solving , x = 425 . answer : d" | a ) rs . 200 , b ) rs . 350 , c ) rs . 275 , d ) rs . 425 , e ) none of these | d | divide(subtract(multiply(30, 350), multiply(15, 275)), 15) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n3)| | gain |
two trains of length 100 m and 200 m are 100 m apart . they start moving towards each other on parallel tracks , at speeds 18 kmph and 72 kmph . in how much time will the trains cross each other ? | "relative speed = ( 18 + 72 ) * 5 / 18 = 25 mps . the time required = d / s = ( 100 + 100 + 200 ) / 25 = 400 / 25 = 16 sec . answer : b" | a ) 8 , b ) 16 , c ) 11 , d ) 12 , e ) 21 | b | divide(100, multiply(add(18, 72), const_0_2778)) | add(n3,n4)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics |
if x : y = 1 : 3 , find the value of ( 2 x + 3 y ) : ( 5 x – y ) | "explanation : given : x / y = 1 / 3 ( 2 x + 3 y ) / ( 3 x – y ) = ( 2 * 1 + 3 * 3 ) : ( 5 * 1 – 3 ) = 11 : 2 answer : b" | a ) 11 : 3 , b ) 11 : 2 , c ) 3 : 2 , d ) 7 : 2 , e ) 9 : 5 | b | divide(add(power(2, 3), 3), add(add(1, 3), 3)) | add(n0,n1)|power(n2,n3)|add(n3,#1)|add(n3,#0)|divide(#2,#3)| | general |
if x # y is defined to equal x ^ 3 / y for all x and y , then ( - 1 # 2 ) # 3 = | "( - 1 ) ^ 3 / 2 = 1 / 2 ( 1 / 2 ) ^ 3 / 3 = 1 / 24 so e is my answer" | a ) 4 / 3 , b ) 1 / 3 , c ) 1 / 12 , d ) - 1 / 12 , e ) 1 / 24 | e | divide(power(divide(power(negate(1), 3), 3), 3), negate(3)) | negate(n1)|negate(n3)|power(#0,n0)|divide(#2,n0)|power(#3,n0)|divide(#4,#1)| | general |
in a restaurant , the profit is 150 % of the cost . if the cost increases by 12 % but the selling price remains constant , approximately what percentage of the selling price is the profit ? | "explanation : let c . p . = rs . 100 . then , profit = rs . 150 , s . p . = rs . 250 . new c . p . = 112 % of rs . 100 = rs . 112 new s . p . = rs . 250 . profit = rs . ( 250 - 112 ) = rs . 138 . required percentage = ( 138 / 250 * 100 ) % = 55 % appox answer : b" | a ) 30 % , b ) 55 % , c ) 90 % , d ) 100 % , e ) none of these | b | multiply(const_100, divide(subtract(add(multiply(const_100, divide(150, const_100)), const_100), multiply(const_100, add(const_1, divide(12, const_100)))), add(multiply(const_100, divide(150, const_100)), const_100))) | divide(n0,const_100)|divide(n1,const_100)|add(#1,const_1)|multiply(#0,const_100)|add(#3,const_100)|multiply(#2,const_100)|subtract(#4,#5)|divide(#6,#4)|multiply(#7,const_100)| | gain |
a can contains a mixture of liquids a and b is the ratio 7 : 5 . when 9 litres of mixture are drawn off and the can is filled with b , the ratio of a and b becomes 7 : 9 . how many liter e of liquid a was contained by the can initially ? | "as a : b : : 7 : 5 - - - > only option c is a multiple of 7 and hence it is a good place to start . also a : b : : 7 : 5 means that , a = ( 712 ) * total and b = ( 5 / 12 ) * total if a = 21 , b = 15 - - - > remove 9 litres - - - > you remove ( 7 / 12 ) * 9 of a - - - > a remaining = 21 - ( 7 / 12 ) * 9 = 63 / 4 simil... | a ) 10 , b ) 20 , c ) 21 , d ) 25 , e ) 27 | c | multiply(7, divide(multiply(add(7, 9), subtract(9, multiply(divide(5, add(7, 5)), 9))), subtract(multiply(9, 7), multiply(7, 5)))) | add(n0,n2)|add(n0,n1)|multiply(n0,n2)|multiply(n0,n1)|divide(n1,#1)|subtract(#2,#3)|multiply(n2,#4)|subtract(n2,#6)|multiply(#0,#7)|divide(#8,#5)|multiply(n0,#9)| | other |
a can do a piece of work in 14 days . when he had worked for 2 days b joins him . if the complete work was finished in 8 days . in how many days b alone can finish the work ? | "8 / 14 + 6 / x = 1 x = 14 days answer : e" | a ) 18 , b ) 77 , c ) 66 , d ) 55 , e ) 14 | e | subtract(inverse(subtract(multiply(divide(const_1, 8), subtract(const_1, multiply(2, divide(const_1, 14)))), divide(const_1, 14))), add(14, 8)) | add(n0,n2)|divide(const_1,n2)|divide(const_1,n0)|multiply(n1,#2)|subtract(const_1,#3)|multiply(#1,#4)|subtract(#5,#2)|inverse(#6)|subtract(#7,#0)| | physics |
8 chess players take part in a tournament . every player plays twice with each of his opponents . how many games are to be played ? | "though 2 * ( 8 c 2 ) is the correct approcah to do this , but for people like me who find perm , comb n prob a nightmare , an easy approach can be used . the first guy has to play 2 matches with the rest of 7 , so he ' ll play 14 matches . similarly , second guy has to play with the rest of 6 as his 2 games with the f... | a ) 66 , b ) 52 , c ) 58 , d ) 55 , e ) 56 | e | multiply(8, subtract(8, const_1)) | subtract(n0,const_1)|multiply(n0,#0)| | general |
a small college reduced its faculty by approximately 14 percent to 195 professors . what was the original number of faculty members ? | if x is the original number of faculty members , then after 14 % reduction in faculty members number is . 86 x but we are given . 86 x = 195 x = 227 so the original number of faculty members is 227 correct answer - c | a ) 182 , b ) 208 , c ) 227 , d ) 254 , e ) 302 | c | divide(195, divide(subtract(const_100, 14), const_100)) | subtract(const_100,n0)|divide(#0,const_100)|divide(n1,#1) | gain |
one fourth of one third of two fifth of a number is 16 . what will be 40 % of that number | "explanation : ( 1 / 4 ) * ( 1 / 3 ) * ( 2 / 5 ) * x = 16 then x = 16 * 30 = 480 40 % of 480 = 192 answer : option d" | a ) 140 , b ) 150 , c ) 180 , d ) 192 , e ) 250 | d | divide(multiply(divide(16, multiply(multiply(divide(const_1, const_4), divide(const_1, const_3)), divide(const_2, add(const_2, const_3)))), 40), const_100) | add(const_2,const_3)|divide(const_1,const_4)|divide(const_1,const_3)|divide(const_2,#0)|multiply(#1,#2)|multiply(#3,#4)|divide(n0,#5)|multiply(n1,#6)|divide(#7,const_100)| | gain |
hree number are in the ratio of 3 : 4 : 5 and their l . c . m . is 2400 . their h . c . f . is : | let the numbers be 3 x , 4 x and 5 x . then , their l . c . m . = 60 x . so , 60 x = 2400 or x = 40 . the numbers are ( 3 x 40 ) , ( 4 x 40 ) and ( 5 x 40 ) . hence , required h . c . f . = 40 . answer : a | a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | a | divide(2400, multiply(multiply(3, 4), 5)) | multiply(n0,n1)|multiply(n2,#0)|divide(n3,#1) | other |
the length of a rectangle is two - fifths of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 625 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 10 units ? | "given that the area of the square = 625 sq . units = > side of square = √ 625 = 25 units the radius of the circle = side of the square = 25 units length of the rectangle = 2 / 5 * 25 = 10 units given that breadth = 10 units area of the rectangle = lb = 10 * 10 = 100 sq . units answer : c" | a ) 140 , b ) 99 , c ) 100 , d ) 72 , e ) 12 | c | multiply(10, multiply(const_2, divide(sqrt(625), divide(10, const_2)))) | divide(n1,const_2)|sqrt(n0)|divide(#1,#0)|multiply(#2,const_2)|multiply(n1,#3)| | geometry |
if x + y = 250 , x - y = 200 , for integers of x and y , y = ? | "x + y = 250 x - y = 200 2 x = 50 x = 25 y = 225 answer is b" | a ) 200 , b ) 225 , c ) 50 , d ) 115 , e ) 150 | b | divide(add(250, 200), const_2) | add(n0,n1)|divide(#0,const_2)| | general |
if the area of a circle is 36 pi square feet , find its circumference . | "the area is given by pi * r * r . hence pi * r * r = 81 pi r * r = 36 ; hence r = 36 feet the circumference is given by 2 * pi * r = 2 * pi * 6 = 128 pi feet correct answer c" | a ) 65 pi feet , b ) 43 pi feet , c ) 12 pi feet , d ) 18 pi feet , e ) 64 pi feet | c | circumface(sqrt(36)) | sqrt(n0)|circumface(#0)| | geometry |
to fill a tank , 42 buckets of water is required . how many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two - fifths of its present ? | "let the capacity of 1 bucket = x . then , the capacity of tank = 42 x . new capacity of bucket = 2 / 5 x therefore , required number of buckets = ( 42 x ) / ( 2 x / 5 ) = ( 42 x ) x 5 / 2 x = 210 / 2 = 105 answer is d ." | a ) 102 , b ) 103 , c ) 104 , d ) 105 , e ) 106 | d | divide(multiply(42, add(const_4, const_1)), const_2) | add(const_1,const_4)|multiply(n0,#0)|divide(#1,const_2)| | physics |
a factory produces 1 defective bulb out of 10 bulbs a yr . if it produces 870 bulbs a yr , how many defective bulbs are produced ? | 10 out of 1 is defective 20 out of 2 is defective 100 out of 10 is defective and so on 800 out of 80 is defective 70 out of 7 is defective 80 + 7 = 87 answer : d | a ) 84 , b ) 85 , c ) 86 , d ) 87 , e ) 88 | d | divide(870, 10) | divide(n2,n1) | other |
the area of an isosceles trapezoid with sides of length 5 and bases of length 10 and 16 is ? | "isosceles trapezoid indicates that if we draw perpendicular from two vertex of the smaller side , the longer side with side 16 will be divided into 3 parts = 10 , 3 and 3 ( making one square and two right triangles ) for the right triangle , height will be = ( 5 ^ 2 - 3 ^ 2 ) ^ 1 / 2 = 4 and the area of trapezoid = 1 ... | a ) 48 , b ) 40 , c ) 52 , d ) 54 , e ) 58 | c | quadrilateral_area(const_4, 16, 10) | quadrilateral_area(const_4,n2,n1)| | geometry |
find the principal which yields a simple interest of rs . 20 and compound interest of rs . 25 in two years , at the same percent rate per annum ? | "explanation : si in 2 years = rs . 20 , si in 1 year = rs . 10 ci in 2 years = rs . 25 % rate per annum = [ ( ci – si ) / ( si in 1 year ) ] * 100 = [ ( 25 – 20 ) / 20 ] * 100 = 25 % p . a . let the principal be rs . x time = t = 2 years % rate = 25 % p . a . si = ( prt / 100 ) 20 = ( x * 25 * 2 ) / 100 x = rs . 40 an... | a ) s . 40 , b ) s . 48 , c ) s . 49 , d ) s . 20 , e ) s . 60 | a | divide(multiply(subtract(25, 20), const_100), const_2) | subtract(n1,n0)|multiply(#0,const_100)|divide(#1,const_2)| | gain |
a shopkeeper sells 400 metres of cloth for $ 18000 at a loss of rs . 5 per metre . find his cost price for one metre of cloth ? | "c $ 50 sp per metre = 18000 / 400 = $ 45 loss per metre = $ 5 cp per metre = 45 + 5 = $ 50" | a ) $ 20 , b ) $ 43 , c ) $ 50 , d ) $ 70 , e ) $ 40 | c | add(divide(18000, 400), 5) | divide(n1,n0)|add(n2,#0)| | gain |
36 * 36 * 36 = 6 ^ ? | "6 ^ 2 * 6 ^ 2 * 6 ^ 2 = 6 ^ ( 2 + 2 + 2 ) = 6 ^ 6 answer : 6 answer : c" | a ) 3 , b ) 2 , c ) 6 , d ) 1 , e ) 5 | c | multiply(36, 36) | multiply(n0,n1)| | general |
what is the least number which when divided by 5 , 9 , 12 and 18 leaves remainder 4 in each care ? | "explanation : lcm of 5 , 9 , 12 and 18 is 180 required number = 180 + 4 = 184 answer : option d" | a ) 130 , b ) 140 , c ) 136 , d ) 184 , e ) 196 | d | add(lcm(lcm(5, 9), lcm(12, 18)), 4) | lcm(n0,n1)|lcm(n2,n3)|lcm(#0,#1)|add(n4,#2)| | general |
abhinav and bhupathi together have rs . 1210 . if 4 / 15 of abhinav ' s amount is equal to 2 / 5 of bhupathi ' s amount , how much amount does bhupathi have ? | explanation : let a = abhinav , b = bhupathi . 4 / 15 a = 2 / 5 b = > a = ( 2 / 515 / 4 ) b = > a = 3 / 2 b = > a / b = 3 / 2 = > a : b = 3 : 2 b ' s share = rs . ( 12102 / 5 ) = rs . 484 . answer : b | a ) rs 460 , b ) rs 484 , c ) rs 550 , d ) rs 664 , e ) none of these | b | add(divide(1210, 5), divide(1210, 5)) | divide(n0,n4)|add(#0,#0) | general |
difference of two numbers is 1620 . if 7.5 % of the number is 12.5 % of the other number , find the number ? | "let the numbers be x and y . then , 7.5 % of x = 12.5 % of y x = 125 * y / 75 = 5 * y / 3 . now , x - y = 1620 5 * y / 3 – y = 1620 2 * y / 3 = 1620 y = [ ( 1620 * 3 ) / 2 ] = 2430 . one number = 2430 , second number = 5 * y / 3 = 4050 . answer d ." | a ) 2660 , 1000 , b ) 3660 , 2000 , c ) 3000 , 4160 , d ) 2430 , 4050 , e ) 4660 , 3000 | d | multiply(divide(7.5, 12.5), divide(1620, subtract(const_1, divide(7.5, 12.5)))) | divide(n1,n2)|subtract(const_1,#0)|divide(n0,#1)|multiply(#0,#2)| | gain |
a shirt goes on sale for 60 % of its original price . one week later , the sale price is marked down 10 % . the final price is what percent of the original price ? | "just assume original price is 100 . sale price = 60 then it is marked down by 10 % = 60 - 6 = 52 . hence it is 52 % od the original price . hence answer is c ." | a ) 72 % , b ) 70 % , c ) 52 % , d ) 50 % , e ) 28 % | c | subtract(60, multiply(divide(60, const_100), 10)) | divide(n0,const_100)|multiply(n1,#0)|subtract(n0,#1)| | gain |
two trains , one from howrah to patna and the other from patna to howrah , start simultaneously . after they meet , the trains reach their destinations after 9 hours and 16 hours respectively the ratio of their speeds is | "solution let us name the trains as a and b . = ( a ' s speed ) : ( b ' s speed ) = √ b : √ a = √ 16 : √ 9 = 4 : 3 answer b" | a ) 2 : 3 , b ) 4 : 3 , c ) 6 : 7 , d ) 9 : 16 , e ) none | b | divide(sqrt(16), sqrt(9)) | sqrt(n1)|sqrt(n0)|divide(#0,#1)| | physics |
61 small identical cubes are used to form a large cube . how many more cubes are needed to add one top layer of small cube all over the surface of the large cube ? | "61 small cube will make a large cube with 4 cubes in each line i . e . adding one layer will require one cube at each end and hence new cube will have 6 cubes in each line . total number of small cubes in new cube = 6 ^ 3 = 216 extra cube required = 216 - 61 = 155 hence , d is the answer ." | a ) 64 , b ) 128 , c ) 152 , d ) 155 , e ) 256 | d | subtract(volume_cube(add(cube_edge_by_volume(61), const_2)), 61) | cube_edge_by_volume(n0)|add(#0,const_2)|volume_cube(#1)|subtract(#2,n0)| | geometry |
pipe a can fill a tank in 5 hr , pipe b in 10 hr and pipe c in 30 hr . if all the pipes are open in how many hours will tank be filled ? | "part filled by ( a + b + c ) in 1 hour = 1 / 5 + 1 / 10 + 1 / 30 = 1 / 3 all the 3 pipes together will fill the tank in 3 hours answer is c" | a ) 1 hr , b ) 2 hr , c ) 3 hr , d ) 4 hr , e ) 5 hr | c | multiply(add(add(divide(5, 5), divide(5, 10)), divide(5, 30)), 5) | divide(n0,n0)|divide(n0,n1)|divide(n0,n2)|add(#0,#1)|add(#3,#2)|multiply(n0,#4)| | physics |
a pipe takes a hours to fill the tank . but because of a leakage it took 5 times of its original time . find the time taken by the leakage to empty the tank | "pipe a can do a work 60 min . lets leakage time is x ; then 1 / 60 - 1 / x = 1 / 300 x = 75 min answer : e" | a ) 50 min , b ) 60 min , c ) 90 min , d ) 80 min , e ) 75 min | e | multiply(const_10, multiply(const_1, 5)) | multiply(n0,const_1)|multiply(#0,const_10)| | physics |
the average weight of 19 students is 15 kg . by the admission of a new student the average weight is reduced to 14.7 kg . the weight of the new student is ? | "answer weight of new student = total weight of all 20 students - total weight of initial 19 students = ( 20 x 14.7 - 19 x 15 ) kg = 9 kg . correct option : e" | a ) 10.6 kg , b ) 10.8 kg , c ) 11 kg , d ) 14.9 kg , e ) none | e | subtract(multiply(add(19, const_1), 14.7), multiply(19, 15)) | add(n0,const_1)|multiply(n0,n1)|multiply(n2,#0)|subtract(#2,#1)| | general |
sandy walked 20 meters towards south . then sandy turned to her left and walked 20 meters . she then turned to her left and walked 20 meters . she then turned to her right and walked 20 meters . what distance is she from the starting point and in which direction ? | "the net distance is 20 + 20 = 40 meters to the east . the answer is d ." | a ) 20 m east , b ) 40 m north , c ) 30 m west , d ) 40 m east , e ) 20 m west | d | add(20, 20) | add(n0,n3)| | physics |
the following questions are based on the information given . choose the appropriate answer that suits the question based on the information . a cube is cut in two equal parts along a plane parallel to one of its faces . one piece is then colored red on the two larger faces and green on remaining . while the other is co... | 16 = 2 ^ 3 each cube will have 2 ^ 3 cubes painted 1 side . answer : c | ['a ) 32', 'b ) 8', 'c ) 16', 'd ) 24', 'e ) 0'] | c | multiply(divide(32, const_4), const_2) | divide(n0,const_4)|multiply(#0,const_2) | geometry |
the value of a scooter depreciates in such a way that its value of the end of each year is 3 / 4 of its value of the beginning of the same year . if the initial value of the scooter is rs . 40,000 , what is the value at the end of 2 years ? | explanation : 40,000 * ( 3 / 4 ) ^ 2 answer : c | a ) 3277 , b ) 2977 , c ) 22500 , d ) 6077 , e ) 17112 | c | multiply(multiply(4, multiply(const_100, const_100)), power(divide(3, 4), 2)) | divide(n0,n1)|multiply(const_100,const_100)|multiply(n1,#1)|power(#0,n3)|multiply(#2,#3) | general |
a committee is reviewing a total of 20 x black - and - white films and 6 y color films for a festival . if the committee selects y / x % of the black - and - white films and all of the color films , what fraction e of the selected films are in color ? | "it ' s y / xpercentnot y / x . if x = 20 and y = 10 . then : 20 x = 400 black - and - white films ; 6 y = 60 color films . y / x % = 10 / 20 % = 0.5 % of the black - and - white films , so 2 black - and - white films and all 60 color films , thus total of 62 films were selected . color films thus compose e = 60 / 62 =... | a ) 1 / 130 , b ) 1 / 5 , c ) 3 / 13 , d ) 10 / 13 , e ) 30 / 31 | e | divide(6, add(divide(20, const_100), 6)) | divide(n0,const_100)|add(n1,#0)|divide(n1,#1)| | general |
a wooden box of dimensions 8 m x 7 m x 6 m is to carry rectangularboxes of dimensions 4 cm x 7 cm x 6 cm . the maximum number ofboxes that can be carried in the wooden box , is | explanation : number = ( 800 * 700 * 600 ) / 8 * 7 * 6 = 2000000 answer : d | a ) 9800000 , b ) 1000000 , c ) 7500000 , d ) 2000000 , e ) none of these | d | divide(multiply(multiply(multiply(4, const_100), multiply(const_2.0, const_100)), multiply(8, const_100)), multiply(multiply(const_4.0, 7), 7)) | multiply(n3,const_100)|multiply(n1,const_100)|multiply(n3,const_2.0)|multiply(#0,#1)|multiply(n1,#2)|multiply(#3,#0)|divide(#5,#4)| | physics |
a searchlight on top of the watch - tower makes 3 revolutions per minute . what is the probability that a man appearing near the tower will stay in the dark for at least 10 seconds ? | "3 revolutions per minute = 1 revolution every 20 seconds so no matter what anybody appearing at the tower can not stay in the dark for more than 20 seconds . this will be our total number of possibilities i . e the denominator . p ( man in dark for at least 5 seconds ) = 1 - p ( man in dark for max of 10 seconds ) = 1... | a ) 1 / 4 , b ) 1 / 3 , c ) 1 / 2 , d ) 2 / 3 , e ) 3 / 4 | c | subtract(const_1, divide(10, divide(multiply(const_60, const_1), 3))) | multiply(const_1,const_60)|divide(#0,n0)|divide(n1,#1)|subtract(const_1,#2)| | physics |
the time it took car p to travel 750 miles was 2 hours less than the time it took car r to travel the same distance . if car p ’ s average speed was 10 miles per hour greater than that of car r , what was car r ’ s average speed , in miles per hour ? | let speed of car r be = x then speed of car p = x + 10 a / q , ( 750 / x ) - ( 750 / ( x + 10 ) ) = 2 solving for x = 56 miles \ hr . b | a ) 40 , b ) 56 , c ) 60 , d ) 70 , e ) 80 | b | divide(add(negate(10), sqrt(subtract(power(negate(10), const_2), multiply(const_4, negate(divide(multiply(750, 10), const_2)))))), const_2) | multiply(n0,n2)|negate(n2)|divide(#0,const_2)|power(#1,const_2)|negate(#2)|multiply(#4,const_4)|subtract(#3,#5)|sqrt(#6)|add(#1,#7)|divide(#8,const_2) | physics |
if 5 % more is gained by selling an book for $ 350 than by selling it for $ 340 , the cost of the book is : | "c 200 let c . p . be $ x . then , 5 % of x = 350 - 340 = 10 x / 20 = 10 = > x = 200" | a ) 500 , b ) 400 , c ) 200 , d ) 300 , e ) 100 | c | divide(subtract(350, 340), divide(5, const_100)) | divide(n0,const_100)|subtract(n1,n2)|divide(#1,#0)| | gain |
find the number which when added to itself 13 times , gives 112 . | "let the number be x . then , x + 13 x = 112 14 x = 112 x = 8 . answer : b" | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | b | divide(112, subtract(13, const_1)) | subtract(n0,const_1)|divide(n1,#0)| | general |
there are 6 red shoes & 4 green shoes . if two of red shoes are drawn what is the probability of getting red shoes | "taking 2 red shoe the probablity is 6 c 2 from 10 shoes probablity of taking 2 red shoe is 6 c 2 / 10 c 2 = 1 / 3 answer : a" | a ) 1 / 3 , b ) 1 / 4 , c ) 1 / 2 , d ) 1 / 5 , e ) 1 / 6 | a | divide(choose(6, const_2), choose(add(6, 4), const_2)) | add(n0,n1)|choose(n0,const_2)|choose(#0,const_2)|divide(#1,#2)| | probability |
a batsman makes a score of 200 runs in the 17 th inning and thus increases his averages by 10 what is his average after 17 th inning ? | "let the average after 16 th inning = x then total run after 16 th inning = 16 x then total run after 17 th inning = 16 x + 200 then average run after 17 th inning = ( 16 x + 200 ) / 17 ( 16 x + 200 ) / 17 = x + 10 = > 16 x + 200 = 17 x + 170 x = 30 ; average after 17 th inning = 30 + 10 = 40 answer : d" | a ) 39 , b ) 35 , c ) 42 , d ) 40 , e ) 41.5 | d | add(subtract(200, multiply(17, 10)), 10) | multiply(n1,n2)|subtract(n0,#0)|add(n2,#1)| | general |
a certain number when divided by 110 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 110 + 25 = 135 / 15 = 9 ( remainder ) e" | a ) 4 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | e | reminder(25, 15) | reminder(n1,n2)| | general |
if w and y are both odd prime numbers and w < y , how many distinct positive integer factors does 2 wy have ? | since 2 wy prime factors are w ^ 1 * y ^ 1 * 2 ^ 1 , its total number or factors must be ( 1 + 1 ) ( 1 + 1 ) ( 1 + 1 ) = 2 ^ 3 = 8 . thus , i think d would be the correct answer . d | a ) 3 , b ) 4 , c ) 6 , d ) 8 , e ) 12 | d | add(add(const_3, 2), const_3) | add(n0,const_3)|add(#0,const_3) | other |
in goshawk - eurasian nature reserve 30 percent of the birds are hawks , and 40 percent of the non - hawks are paddyfield - warblers . if there are 25 percent as many kingfishers as paddyfield - warblers in the reserve , then what percent of the birds r in the nature reserve are not hawks , paddyfield - warblers , or k... | 1 . we are given the following percentages : 30 ( 70 ) , 40 ( 60 ) , 25 ( 75 ) . there are two threads from here . first starts at 30 % and finishes there . second one starts at 70 , then 40 , and then 25 . we need a value that is divisible by 7 , 2 , and 5 at least once . lets pick a number now , say 700 . so say if n... | a ) 25 % , b ) 35 % , c ) 45 % , d ) 70 % , e ) 80 % | b | add(const_10, divide(add(25, 25), const_2)) | add(n2,n2)|divide(#0,const_2)|add(#1,const_10) | general |
what is the remainder when 8 ^ 1 + 8 ^ 2 + 8 ^ 3 + . . . + 8 ^ 9 is divided by 4 ? | "notice that in the brackets we have the sum of 9 even multiples of 4 , hence the sum in the brackets will be even multiple of 4 ( the sum of 9 even numbers is even ) . which yields remainder of 0 upon division by 4 . answer : e" | a ) 4 , b ) 3 , c ) 2 , d ) 5 , e ) none of the above | e | divide(subtract(power(8, 1), power(8, 1)), 8) | power(n0,n1)|power(n2,n1)|subtract(#0,#1)|divide(#2,n4)| | general |
the sides of a rectangular field are in the ratio 3 : 4 . if the area of the field is 8112 sq . m , the cost of fencing the field @ 25 paise per metre is | "solution let length = ( 3 x ) metres and breadth = ( 4 x ) metres . then , 3 x × 4 x = 7600 ⇔ 12 x 2 = 8112 ⇔ x 2 = 676 ⇔ x = 26 so , length = 78 m and breadth = 104 m . perimeter = [ 2 ( 78 + 104 ) ] m = 364 m . ∴ cost of fencing = rs . ( 0.25 × 364 ) = rs . 91 . answer d" | a ) rs . 55.50 , b ) rs . 67.50 , c ) rs . 86.50 , d ) rs . 91.00 , e ) none of these | d | divide(multiply(rectangle_perimeter(multiply(3, sqrt(divide(8112, multiply(3, 4)))), multiply(4, sqrt(divide(8112, multiply(3, 4))))), 25), const_100) | multiply(n0,n1)|divide(n2,#0)|sqrt(#1)|multiply(n0,#2)|multiply(n1,#2)|rectangle_perimeter(#3,#4)|multiply(n3,#5)|divide(#6,const_100)| | physics |
a sum was put at simple interest at a certain rate for 5 years had it been put at 6 % higher rate , it would have fetched 90 more . find the sum . | "difference in s . i . = p × t / 100 ( r 1 − r 2 ) ⇒ 90 = p × 5 x 6 / 100 ( ∵ r 1 - r 2 = 2 ) ⇒ p = 90 × 100 / 5 x 6 = 300 answer d" | a ) 90 , b ) 100 , c ) 200 , d ) 300 , e ) 400 | d | divide(90, multiply(divide(6, const_100), 5)) | divide(n1,const_100)|multiply(n0,#0)|divide(n2,#1)| | gain |
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