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the average salary per head of the entire staff of an office including the officers and clerks is rs . 90 . the average salary of officers is rs . 500 and that of the clerks is rs . 84 . if the number of officers is 2 , find the number of officers in the office ? | "500 84 \ / 90 / \ 6 410 3 : 205 3 - > 12 205 - > ? 820 answer : c" | a ) 49 , b ) 820 , c ) 9 , d ) 1840 , e ) 100 | b | divide(90, add(84, add(const_3, const_3))) | add(const_3,const_3)|add(n2,#0)|divide(n0,#1)| | general |
what is the least integer greater than – 4 + 0.5 ? | "this question is just about doing careful arithmetic and remembering what makes a numberbiggerorsmallercompared to another number . first , let ' s take care of the arithmetic : ( - 4 ) + ( 0.5 ) = - 3.5 on a number line , since we ' re adding + . 5 to a number , the total moves to the right ( so we ' re moving from - 4 to - 3.5 ) . next , the question asks for the least integer that is greater than - 3.5 again , we can use a number line . numbers become greater as you move to the right . the first integer to the right of - 3.5 is - 3 . final answer : a" | a ) $ 32500 , b ) 3 , c ) 900 , d ) – 3 , e ) 48 | d | add(0.5, negate(4)) | negate(n0)|add(n1,#0)| | general |
if â € œ * â € is called â € œ + â € , â € œ / â € is called â € œ * â € , â € œ - â € is called â € œ / â € , â € œ + â € is called â € œ - â € . 240 / 80 â € “ 60 * 40 - 10 = ? | "explanation : given : 240 / 80 â € “ 60 * 40 - 10 = ? substituting the coded symbols for mathematical operations , we get , 240 * 80 / 60 + 40 / 10 = ? 240 * 1.33 + 4 = ? 319.2 + 4 = 323.2 answer : c" | a ) 101 kg , b ) 135 deg , c ) 10.6 , d ) 4.2 , e ) 323.2 | e | add(multiply(divide(60, 40), divide(240, 80)), 10) | divide(n2,n3)|divide(n0,n1)|multiply(#0,#1)|add(n4,#2)| | general |
60 boys can complete a work in 24 days . how many men need to complete twice the work in 20 days | "one man can complete the work in 24 * 60 = 1440 days = one time work to complete the work twice it will be completed in let m be the no . of worker assign for this therefore the eqn becomes m * 20 = 2 * 1440 m = 144 workers answer : a" | a ) 130 , b ) 144 , c ) 16 , d ) $ 20.00 , e ) 43.1 | b | divide(multiply(60, multiply(24, const_2)), 20) | multiply(n1,const_2)|multiply(n0,#0)|divide(#1,n2)| | physics |
jill has 21 gallons of water stored in quart , half - gallon , and one gallon jars . she has equal numbers of each size jar holding the liquid . what is the total number of water filled jars ? | let the number of each size of jar = wthen 1 / 4 w + 1 / 2 w + w = 21 1 3 / 4 w = 21 w = 12 the total number of jars = 3 w = 36 answer : d | a ) 9.9 % , b ) 170 , c ) 25 % , d ) 36 , e ) 36 min | d | multiply(divide(21, add(const_1, add(const_0_25, divide(const_1, const_2)))), const_3) | divide(const_1,const_2)|add(#0,const_0_25)|add(#1,const_1)|divide(n0,#2)|multiply(#3,const_3)| | general |
jacob is 12 years old . he is 3 times as old as his brother . how old will jacob be when he is twice as old ? | "j = 12 ; j = 3 b ; b = 12 / 3 = 4 ; twice as old so b = 4 ( now ) + ( 4 ) = 8 ; jacob is 12 + 4 = 16 answer : d" | a ) $ 19500 , b ) 26250 , c ) 16 , d ) 2.5 sec , e ) 270 m | c | multiply(12, 3) | multiply(n0,n1)| | general |
if two integers x , y ( x > y ) are selected from - 10 to 1 ( inclusive ) , how many possible cases are there ? | "if two integers x , y ( x > y ) are selected from - 10 to 9 ( inclusive ) , how many possible cases are there ? a . 150 b . 180 c . 190 d . 210 e . 240 - - > 12 c 2 = 12 * 11 / 2 = 66 . therefore , the answer is b ." | a ) 6 hours , b ) 24 , c ) 41 , d ) 66 , e ) 4 | d | add(add(add(add(add(add(add(1, 10), add(1, const_2)), add(1, const_1)), 1), 10), const_2), const_1) | add(n0,n1)|add(n1,const_2)|add(n1,const_1)|add(#0,#1)|add(#3,#2)|add(#4,n1)|add(#5,n0)|add(#6,const_2)|add(#7,const_1)| | probability |
a trader sells 40 metres of cloth for rs . 8200 at a profit of rs . 20 per metre of cloth . how much profit will the trder earn on 40 metres of cloth ? | "explanation : sp of 1 metre cloth = 8200 / 40 = rs . 205 . cp of 1 metre cloth = rs . 205 – 20 = rs . 185 cp on 40 metres = 185 x 40 = rs . 7400 profit earned on 40 metres cloth = rs . 8200 – rs . 7400 = rs . 800 . answer : option a" | a ) 40 % , b ) 64 , c ) rs . 800 , d ) 0.3 % , e ) $ 11000 | c | multiply(20, 40) | multiply(n0,n2)| | gain |
what least value must be given to * so that the number 451 * 603 is exactly divisible by 9 ? | sum of digits = ( 4 + 5 + 1 + x + 6 + 0 + 3 ) = 19 + x divisible by 9 clearly x = 9 answer e 9 | a ) 9 , b ) 60 , c ) 57 % , d ) 120 , e ) 2 | a | log(451) | log(n0) | general |
24 machines can do a work in 10 days . how many machines are needed to complete the work in 40 days ? | "required number of machines = 24 * 10 / 40 = 6 answer is b" | a ) 2000 , b ) 6 , c ) 6.6 kmph , d ) 157 , e ) 0 | b | divide(multiply(24, 10), 40) | multiply(n0,n1)|divide(#0,n2)| | physics |
in a company of 170 employees , 110 are females . a total of 80 employees have advanced degrees and the rest have a college degree only . if 25 employees are males with college degree only , how many employees are females with advanced degrees ? | the number of males is 170 - 110 = 60 . the number of males with advanced degrees is 60 - 25 = 35 . the number of females with advanced degrees is 80 - 35 = 45 . the answer is b . | a ) 2,180 , b ) 192 , c ) 45 , d ) 50 days , e ) 4327 | c | subtract(80, subtract(subtract(170, 110), 25)) | subtract(n0,n1)|subtract(#0,n3)|subtract(n2,#1) | geometry |
at a particular graduation party with 300 guests , 70 % of the guests brought gifts , and 40 % of the female guests brought gifts . if 36 males did not bring gifts to the party , how many females did bring gifts ? | "the correct method total = 300 . . 70 % of 300 = 210 got gifts . . 90 did not get gift , out of which 48 are males , so remaining 90 - 36 = 54 are females . . but 40 % females brought gift , so 60 % did not get it . . so 60 % = 54 , 100 % = 54 * 100 / 60 = 90 . . ans 40 % of 90 = 36 b" | a ) 10,000 , b ) 6 days , c ) 2 , d ) 36 , e ) 20 | d | divide(multiply(divide(multiply(subtract(subtract(300, divide(multiply(70, 300), const_100)), 36), const_100), subtract(const_100, 40)), 40), const_100) | multiply(n0,n1)|subtract(const_100,n2)|divide(#0,const_100)|subtract(n0,#2)|subtract(#3,n3)|multiply(#4,const_100)|divide(#5,#1)|multiply(n2,#6)|divide(#7,const_100)| | gain |
if 8 a = 9 b and ab ≠ 0 , what is the ratio of a / 9 to b / 8 ? | "if ab ≠ 0 then a and b has two integer sets of pair if a = 9 then b = 8 and if a = - 9 then b = - 8 also in fraction if a = 1 / 8 then b = 1 / 9 any of the pair we check the ratio 8 a / 9 b = 1 answer : c" | a ) 2000 , b ) 75 , c ) 5994 , d ) 1 , e ) 245 | d | divide(multiply(8, 9), multiply(9, 8)) | multiply(n0,n1)|divide(#0,#0)| | general |
in a theater , the first row has 17 seats and each row has 3 more seats than previous row . if the last row has 44 seats , what is the total number of seats in the theater ? | "the number of seats in the theater is 17 + ( 17 + 3 ) + . . . + ( 17 + 27 ) = 10 ( 17 ) + 3 ( 1 + 2 + . . . + 9 ) = 10 ( 17 ) + 3 ( 9 ) ( 10 ) / 2 = 170 + 135 = 305 the answer is e ." | a ) 5 , b ) 27.2 % . , c ) 305 , d ) 40 , e ) 596', ' | c | multiply(divide(add(17, 44), const_2), divide(add(subtract(44, 17), 3), 3)) | add(n0,n2)|subtract(n2,n0)|add(n1,#1)|divide(#0,const_2)|divide(#2,n1)|multiply(#3,#4)| | general |
the list price of an article is rs . 69 . a customer pays rs . 56.16 for it . he was given two successive discounts , one of them being 10 % . the other discount is ? | "69 * ( 90 / 100 ) * ( ( 100 - x ) / 100 ) = 56.16 x = 9.56 % answer : d" | a ) 315 , b ) 10 % , c ) 8 : 5 , d ) 9.56 % , e ) 1 : 2 | d | multiply(divide(subtract(subtract(69, multiply(69, divide(10, const_100))), 56.16), subtract(69, multiply(69, divide(10, const_100)))), const_100) | divide(n2,const_100)|multiply(n0,#0)|subtract(n0,#1)|subtract(#2,n1)|divide(#3,#2)|multiply(#4,const_100)| | gain |
how many seconds will a 200 m long train take to cross a man walking with a speed of 3 km / hr in the direction of the moving train if the speed of the train is 63 km / hr ? | "speed of train relative to man = 63 - 3 = 60 km / hr . = 60 * 5 / 18 = 50 / 3 m / sec . time taken to pass the man = 200 * 3 / 50 = 12 sec . answer : c" | a ) 125 , b ) s . 300 , c ) 28 / 55 , d ) 12 sec , e ) 94 | d | divide(200, multiply(subtract(63, 3), const_0_2778)) | subtract(n2,n1)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
the radius of a cone is 4 m , height 5 m . find the curved surface area ? | "cone curved surface area = π rl = 22 / 7 × 4 × 5 = 440 / 7 = 62 6 / 7 slant height ( l ) = √ r ( power 2 ) + h ( power 2 ) = √ 16 + 9 = √ 25 = 5" | a ) 192 , b ) 62 / 69 , c ) 5 , d ) 1764713 , e ) 8 .', ' | c | volume_cone(4, 5) | volume_cone(n0,n1)| | geometry |
a constructor estimates that 8 people can paint mr khans house in 3 days . if he uses 4 people instead of 8 , how long will they take to complete the job ? | "explanation : use formula for a work members ã — days = constant 8 ã — 3 = 4 ã — a a = 6 so answer is 6 days answer : a" | a ) 12', ' , b ) 7.94 , c ) 132 , d ) 6 , e ) 180 km | d | divide(const_1, multiply(divide(const_1, multiply(const_4.0, 8)), 3)) | multiply(n0,n1)|divide(const_1,#0)|multiply(n2,#1)|divide(const_1,#2)| | physics |
the radius of a semi circle is 70 cm then its perimeter is ? | "diameter = 140 cm 1 / 2 * 22 / 7 * 140 + 140 = 360 answer : c" | a ) 360 , b ) rs . 440 , c ) 15552 , d ) $ 555.55 , e ) 4.8 | a | add(divide(circumface(70), const_2), multiply(70, const_2)) | circumface(n0)|multiply(n0,const_2)|divide(#0,const_2)|add(#2,#1)| | physics |
a cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.3 m and its walls are 5 cm thick . the thickness of the bottom is : | "explanation : let the thickness of the bottom be x cm . then , [ ( 330 - 10 ) × ( 260 - 10 ) × ( 130 - x ) ] = 8000 × 1000 = > 320 × 250 × ( 130 - x ) = 8000 × 1000 = > ( 130 - x ) = 8000 × 1000 / 320 = 100 = > x = 30 cm = 3 dm . answer : b" | a ) 3 dm , b ) $ 4000 , c ) 220 , d ) 120 , e ) 8 | a | subtract(multiply(multiply(3.3, 2.6), 1.3), divide(8000, const_1000)) | divide(n0,const_1000)|multiply(n1,n2)|multiply(n3,#1)|subtract(#2,#0)| | physics |
operation # is defined as adding a randomly selected two digit multiple of 12 to a randomly selected two digit prime number and reducing the result by half . if operation # is repeated 10 times , what is the probability that it will yield at least two integers ? | "any multiple of 12 is even . any two - digit prime number is odd . ( even + odd ) / 2 is not an integer . thus # does not yield an integer at all . therefore p = 0 . answer : a ." | a ) 3 , 400,000 , b ) 0 % , c ) 16.41 , d ) 265 miles . , e ) 10 | b | divide(add(12, add(10, const_3)), const_2) | add(n1,const_3)|add(n0,#0)|divide(#1,const_2)| | general |
running at their respective constant rate , machine x takes 2 days longer to produce w widgets than machines y . at these rates , if the two machines together produce 5 w / 4 widgets in 3 days , how many days would it take machine x alone to produce 6 w widgets . | i am getting 12 . e . hope havent done any calculation errors . . approach . . let y = no . of days taken by y to do w widgets . then x will take y + 2 days . 1 / ( y + 2 ) + 1 / y = 5 / 12 ( 5 / 12 is because ( 5 / 4 ) w widgets are done in 3 days . so , x widgets will be done in 12 / 5 days or 5 / 12 th of a widget in a day ) solving , we have y = 4 = > x takes 6 days to doing x widgets . so , he will take 36 days to doing 6 w widgets . answer : b | a ) 14.63 , b ) 45 , c ) 40 , d ) 100 , e ) 36 | e | multiply(multiply(6, 2), 3) | multiply(n0,n4)|multiply(n3,#0) | general |
the cost price of a radio is rs . 1500 and it was sold for rs . 1230 , find the loss % ? | "1500 - - - - 270 100 - - - - ? = > 18 % answer : a" | a ) 18 % , b ) 5 : 6 , c ) 7600 , d ) - 4 , e ) 9 days | a | multiply(divide(subtract(1500, 1230), 1500), const_100) | subtract(n0,n1)|divide(#0,n0)|multiply(#1,const_100)| | gain |
among the two clocks , clock a gains 20 seconds per minute . if clock a and b are set at 2 0 ' clock , when clock a is at 7 : 20 , what does clock b show ? | clock a gains 20 seconds per minute , 1200 seconds per hour or 20 minutes per hour . the two clocks show 2 : 00 at 2 0 ' clock at 3 : 00 - clock b is at 3 : 00 clock a is at 3 : 20 ( 1 hour + gains 20 minutes ) at 4 : 00 - clock b is at 4 : 00 clock a is at 4 : 40 ( 2 hours + gains 40 minutes ) in 4 hours the clock a gains 4 * 20 = 80 minutes or 1 hour 20 minutes if clock a is at 7 : 20 the clock b is at 6 : 00 answer is b | a ) 24 , b ) 27 sec , c ) 1 , d ) 6 : 00 , e ) 1 / 12 | d | divide(add(add(multiply(subtract(7, 2), const_60), 20), divide(multiply(add(multiply(subtract(7, 2), const_60), 20), 20), const_60)), const_60) | subtract(n3,n1)|multiply(#0,const_60)|add(n0,#1)|multiply(n0,#2)|divide(#3,const_60)|add(#2,#4)|divide(#5,const_60) | physics |
the sale price sarees listed for rs . 500 after successive discount is 10 % and 5 % is ? | "500 * ( 90 / 100 ) * ( 95 / 100 ) = 427.5 answer : a" | a ) 427.5 , b ) 5.45 % , c ) 40 , d ) 12 , e ) 32 | a | subtract(subtract(500, divide(multiply(500, 10), const_100)), divide(multiply(subtract(500, divide(multiply(500, 10), const_100)), 5), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)|multiply(n2,#2)|divide(#3,const_100)|subtract(#2,#4)| | gain |
the number of timeshare condos available at sunset beach is 2 / 5 the number of timeshare condos available at playa del mar . if the total number of timeshare condos available at the two beaches combined is 350 , what is the difference between the number of condos available at sunset beach and the number of condos available at playa del mar ? | let x be the number of timeshare condos available at playa del mar . then number of timeshare condos available at sunset beach = 3 / 5 x we know , x + 2 / 5 x = 350 hence , x = 250 . so , number of timeshare condos available at playa del mar = 250 the difference between the number of condos available at sunset beach and the number of condos available at playa del mar = x - 2 / 5 x = 3 / 5 x = 3 / 5 ( 250 ) = 150 the correct answer is d | a ) 150 , b ) 60 , c ) 70 , d ) 113 , e ) 427.5 | a | add(divide(multiply(350, 2), 5), multiply(2, 5)) | multiply(n0,n2)|multiply(n0,n1)|divide(#0,n1)|add(#2,#1) | general |
a can do a piece of work in 8 days and b alone can do it in 12 days . how much time will both take to finish the work ? | "this question can be solved by different methods . we need to conserve time in exams so solving this problem using equations is the good idea . time taken to finish the job = xy / ( x + y ) = 8 x 12 / ( 8 + 12 ) = 96 / 20 = 4.8 days answer : a" | a ) 6 : 40 , b ) 4.8 , c ) 0.358 , d ) 9 , e ) 2160 | b | divide(const_1, add(divide(const_1, 8), divide(const_1, 12))) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)| | physics |
the measurement of a rectangular box with lid is 25 cmx 6 cmx 18 cm . find the volume of the largest sphere that can be inscribed in the box ( in terms of π cm 3 ) . ( hint : the lowest measure of rectangular box represents the diameter of the largest sphere ) | d = 6 , r = 2 ; volume of the largest sphere = 4 / 3 π r 3 = 4 / 3 * π * 3 * 3 * 3 = 36 π cm 3 answer : c | a ) 60 , b ) $ 330 , c ) 20 , d ) 343 cc , e ) 36', ' | e | multiply(divide(const_4, const_3), power(3, const_3)) | divide(const_4,const_3)|power(n3,const_3)|multiply(#0,#1) | geometry |
a and b go around a circular track of length 150 m on a cycle at speeds of 36 kmph and 54 kmph . after how much time will they meet for the first time at the starting point ? | "time taken to meet for the first time at the starting point = lcm { length of the track / speed of a , length of the track / speed of b } = lcm { 150 / ( 36 * 5 / 18 ) , 150 / ( 54 * 5 / 18 ) } = lcm ( 15 , 10 ) = 30 sec . answer : a" | a ) 240 , b ) 30 sec , c ) $ 28.44 , d ) 27 , e ) 3 | b | divide(150, subtract(multiply(54, const_0_2778), multiply(36, const_0_2778))) | multiply(n2,const_0_2778)|multiply(n1,const_0_2778)|subtract(#0,#1)|divide(n0,#2)| | physics |
maxwell leaves his home and walks toward brad ' s house at the same time that brad leaves his home and runs toward maxwell ' s house . if the distance between their homes is 50 kilometers , maxwell ' s walking speed is 4 km / h , and brad ' s running speed is 6 km / h , what is the distance traveled by brad ? | "time taken = total distance / relative speed total distance = 50 kms relative speed ( opposite side ) ( as they are moving towards each other speed would be added ) = 6 + 4 = 10 kms / hr time taken = 50 / 10 = 5 hrs distance traveled by brad = brad ' s speed * time taken = 6 * 5 = 30 kms . . . answer - e" | a ) 25 days , b ) 30 , c ) 1 , d ) 48 , e ) 64 | b | multiply(4, divide(50, add(4, 6))) | add(n1,n2)|divide(n0,#0)|multiply(n1,#1)| | physics |
two trains of length 150 m and 280 m are running towards each other on parallel lines at 42 kmph and 30 kmph respectively . in what time will they be clear of each other from the moment they meet ? | "relative speed = ( 42 + 30 ) * 5 / 18 = 4 * 5 = 20 mps . distance covered in passing each other = 150 + 280 = 430 m . the time required = d / s = 430 / 20 = 21.5 sec . answer : d" | a ) - 14 , b ) 0 , c ) 21.5 sec , d ) 6 cm', ' , e ) 84 | c | divide(add(150, 280), multiply(add(42, 30), const_0_2778)) | add(n0,n1)|add(n2,n3)|multiply(#1,const_0_2778)|divide(#0,#2)| | physics |
source : knewton a cyclist ' s speed varies , depending on the terrain , between 6.0 miles per hour and 14.0 miles per hour , inclusive . what is the maximum distance , in miles , that the cyclist could travel in 5 hours ? | we are told that : generallya cyclist ' s speed varies , depending on the terrain , between 6.0 miles per hour and 14.0 miles per hour , inclusive . is it possible the cyclist to travel with maximum speed for some time ? why not , if there is right terrain for that . so , if there is long enough terrain for the maximum speed of 14 mph then the maximum distance , in miles , that the cyclist could travel in 5 hours would be 5 * 14 = 70 miles . answer : c . | a ) 1 / 2 , b ) 71.4 % , c ) 70 , d ) 242 , e ) 34 min | c | multiply(14, 5) | multiply(n1,n2) | physics |
elena purchased brand x pens for $ 4.00 apiece and brand y for $ 2.40 apiece . if elena purchased a total of 12 of these pens for $ 42.00 , how many brand x pens did she purchase ? | "4 x + 2.8 y = 42 - - > multiply by 2.5 ( to get the integers ) - - > 10 x + 7 y = 105 - - > only one positive integers solutions x = 5 and y = 5 ( how to solve : 7 y must have the last digit of 5 in order the last digit of the sum to be 5 ) . answer : b ." | a ) 11 , b ) 14.3 % , c ) 1215 , d ) $ 130 , e ) 5 | e | subtract(multiply(4.00, 12), 42.00) | multiply(n0,n2)|subtract(#0,n3)| | general |
the length of a rectangle is two - seventh of the radius of a circle . the radius of the circle is equal to the side of the square , whose area is 5929 sq . units . what is the area ( in sq . units ) of the rectangle if the rectangle if the breadth is 25 units ? | given that the area of the square = 5929 sq . units = > side of square = â ˆ š 5929 = 77 units the radius of the circle = side of the square = 77 units length of the rectangle = 2 / 7 * 77 = 22 units given that breadth = 25 units area of the rectangle = lb = 22 * 25 = 550 sq . units answer : d | a ) b . 1.2 , b ) t = 32 , c ) 34 , d ) 52 , e ) 550 sq . units', ' | e | rectangle_area(25, multiply(sqrt(5929), divide(const_2, add(const_3, const_4)))) | add(const_3,const_4)|sqrt(n0)|divide(const_2,#0)|multiply(#2,#1)|rectangle_area(n1,#3) | geometry |
if 20 men take 15 days to to complete a job , in how many days can 25 men finish that work ? | ans . 12 days | a ) 625 , b ) 38 kg , c ) 12 , d ) $ 250 , e ) 9 : 2 | c | divide(multiply(20, 15), 25) | multiply(n0,n1)|divide(#0,n2)| | physics |
carl is facing very difficult financial times and can only pay the interest on a $ 30,000 loan he has taken . the bank charges him a quarterly compound rate of 5 % . what is the approximate interest he pays annually ? | "usually , you are given the annual rate of interest and it is mentioned that it is annual rate . the bank charges him a quarterly compounded annual rate of 20 % . here you find per quarter rate as ( 20 / 4 ) % = 5 % i have actually never seen a question with quarter rate given but since this question did not mentionannual rate of interestand since the options did not make sense with 5 % annual rate of interest , it is apparent that the intent was a 5 % quarterly rate . so the bank charges 5 % every quarter and compounds it in the next quarter . had it been a simple quarterly rate , we would have just found 4 * 5 % of 30,000 = $ 6000 as our answer . but since , the interest is compounded , it will be a bit more than $ 6000 . option ( d ) looks correct ." | a ) 38 sec , b ) 1400 , c ) 5 , d ) $ 6000 , e ) $ 325 | d | subtract(multiply(multiply(const_100, const_100), power(add(const_1, divide(5, const_100)), const_4)), multiply(const_100, const_100)) | divide(n1,const_100)|multiply(const_100,const_100)|add(#0,const_1)|power(#2,const_4)|multiply(#1,#3)|subtract(#4,#1)| | gain |
on a certain transatlantic crossing , 20 percent of a ship ’ s passengers held round - trip tickets and also took their cars abroad the ship . if 50 percent of the passengers with round - trip tickets did not take their cars abroad the ship , what percent of the ship ’ s passengers held round - trip tickets ? | "let t be the total number of passengers . let x be the number of people with round trip tickets . 0.2 t had round trip tickets and took their cars . 0.5 x had round trip tickets and took their cars . 0.5 x = 0.2 t x = 0.4 t the answer is b ." | a ) 62 , b ) 64 , c ) 5 / 3 , d ) 425 , e ) 40 % | e | divide(20, subtract(const_1, divide(50, const_100))) | divide(n1,const_100)|subtract(const_1,#0)|divide(n0,#1)| | gain |
the pinedale bus line travels at an average speed of 60 km / h , and has stops every 5 minutes along its route . yahya wants to go from his house to the pinedale mall , which is 7 stops away . how far away , in kilometers , is pinedale mall away from yahya ' s house ? | "number of stops in an hour : 60 / 5 = 12 distance between stops : 60 / 12 = 5 km distance between yahya ' s house and pinedale mall : 5 x 7 = 35 km imo , correct answer is ` ` b . ' '" | a ) 23 , b ) 25300 , c ) 35 km , d ) 30 days , e ) 2 | c | multiply(60, divide(multiply(5, 7), 60)) | multiply(n1,n2)|divide(#0,n0)|multiply(n0,#1)| | physics |
if p represents the product of the first 18 positive integers , then p is not a multiple of | "answer is e . since prime factor of 95 is 5 x 19 . 19 is a prime number and not part of the first 18 positive integers . a ) 9 x 11 b ) 12 x 7 c ) 12 x 6 d ) 13 x 5 e ) 3 x 19" | a ) 153600 m 2', ' , b ) 95 , c ) 31.67 % , d ) 35 , e ) 3024 | b | multiply(add(18, const_3), const_3) | add(n0,const_3)|multiply(#0,const_3)| | general |
the fuel indicator in a car shows 1 / 5 th of the fuel tank as full . when 22 more liters of fuel are poured in to the tank , the indicator rests at the 3 / 4 of the full mark . find the capacity of the tank . | x / 5 + 22 = 3 x / 4 = > x = 40 litres answer : d | a ) 120 , b ) 40 litres , c ) 33 % , d ) 50 , e ) 39 % | b | divide(22, subtract(divide(3, 4), divide(1, 5))) | divide(n3,n4)|divide(n0,n1)|subtract(#0,#1)|divide(n2,#2) | general |
a salesman sold twice as much pears in the afternoon than in the morning . if he sold $ 450 kilograms of pears that day , how many kilograms did he sell in the afternoon ? | 3 x = 450 x = 150 therefore , the salesman sold 150 kg in the morning and 2 ⋅ 150 = 300 kg in the afternoon . so answer is c . | a ) 108 ° , b ) $ 25 , c ) 1 / 32 , d ) 300 , e ) 548 | d | multiply(divide(450, const_3), const_2) | divide(n0,const_3)|multiply(#0,const_2) | other |
a and b together can complete work in 10 days . a alone starts working and leaves it after working for 6 days completing only half of the work . in how many days it can be completed if the remaining job is undertaken by b ? | "explanation : ( a + b ) one day work = 1 / 10 now a does half of the work in 6 days so a can complete the whole work in 12 days a ’ s one day work = 1 / 12 b ’ s one day work = 1 / 10 - 1 / 12 = 2 / 120 = 1 / 60 b alone can complete the work in 60 days so half of the work in 30 days answer : option e" | a ) 7.5 sec , b ) 30 , c ) 40 , d ) 11 % , e ) 120 liters | b | multiply(10, divide(6, 10)) | divide(n1,n0)|multiply(n0,#0)| | physics |
set a consists of the integers from 4 to 16 , inclusive , while set b consists of the integers from 6 to 20 , inclusive . how many distinct integers do belong to the both sets at the same time ? | "a = { 4 , 5,6 , 7,8 , 9,10 , 11,12 , . . . 16 } b = { 6 , 7,8 , 9,10 , 11,12 . . . } thus we see that there are 11 distinct integers that are common to both . e is the correct answer ." | a ) 11 , b ) 40 , c ) 5625 , d ) 7.9 % , e ) 45 | a | add(6, 4) | add(n0,n2)| | other |
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 30 as remainder | "let the smaller number be x . then larger number = ( x + 1365 ) . x + 1365 = 6 x + 30 5 x = 1335 x = 267 large number = 267 + 1365 = 1632 d" | a ) 1632 , b ) 625 , c ) 5 , d ) 11 , e ) 9 : 16 | a | add(multiply(divide(subtract(1365, 30), subtract(6, const_1)), 6), 30) | subtract(n0,n2)|subtract(n1,const_1)|divide(#0,#1)|multiply(n1,#2)|add(n2,#3)| | general |
the sum of the ages of 5 children born at the intervals of 3 years each is 50 years . what is the age of the youngest child ? | "let the ages of the children be x , ( x + 3 ) , ( x + 6 ) , ( x + 9 ) and ( x + 12 ) years . then , x + ( x + 3 ) + ( x + 6 ) + ( x + 9 ) + ( x + 12 ) = 50 5 x = 20 = > x = 4 . age of youngest child = x = 4 years . answer : a" | a ) 2 / 3 , b ) 6 , c ) 4 years , d ) 112 , e ) 40 | c | divide(subtract(divide(50, divide(5, 3)), multiply(subtract(5, const_1), 3)), 3) | divide(n0,n1)|subtract(n0,const_1)|divide(n2,#0)|multiply(n1,#1)|subtract(#2,#3)|divide(#4,n1)| | general |
paul sells encyclopedias door - to - door . he earns $ 150 on every paycheck , regardless of how many sets he sells . in addition , he earns commission as follows : commission sales 10 % $ 0.00 - $ 10 , 000.00 5 % $ 10 , 000.01 - - - > he does not earn double commission . that is , if his sales are $ 12,000 , he earns 10 % on the first $ 10,000 and 5 % on the remaining $ 2,000 . his largest paycheck of the year was $ 1,320 . what were his sales for that pay period ? | his pay check was $ 1320 . out of this , $ 150 was his fixed salary so the total commission he earned was $ 1320 - $ 150 = $ 1170 he earns 10 % on the sales of first $ 10,000 which gives a commission of $ 1000 . he earns 5 % on every subsequent dollar . since he earns another $ 170 , he must have had sales of another 170 * ( 100 / 5 ) = 3400 so his total sales must have been $ 10,000 + $ 3400 = $ 13,400 a | a ) 280 , b ) 13,400 , c ) 1612.5 , d ) 2 sqrt ( 6 , e ) 10 min | b | divide(add(multiply(150, divide(const_100, 5)), multiply(const_100, const_100)), multiply(10, const_100)) | divide(const_100,n5)|multiply(const_100,const_100)|multiply(n1,const_100)|multiply(n0,#0)|add(#3,#1)|divide(#4,#2) | general |
a rectangle with width 8 and diagonal 30 . find the area ? | then the area is : 8 ' ' x 30 ' ' = 240 square inches , or 240 square units hence a | a ) 400 , b ) 200 cm 2' , c ) 240 square units', ' , d ) 4 / 7 , e ) 9 / 5 | c | rectangle_area(sqrt(subtract(power(30, const_2), power(8, const_2))), 8) | power(n1,const_2)|power(n0,const_2)|subtract(#0,#1)|sqrt(#2)|rectangle_area(n0,#3) | geometry |
train a leaves the station traveling at 30 miles per hour . two hours later train в leaves the same station traveling in the same direction at 35 miles per hour . how many miles from the station was train a overtaken by train b ? | "after two hours , train a is ahead by 60 miles . train b can catch up at a rate of 5 miles per hour . the time to catch up is 60 / 5 = 12 hours . in 12 hours , train a travels another 30 * 12 = 360 miles for a total of 420 miles . the answer is a ." | a ) 48 - 6 π', ' , b ) $ 28.44 , c ) 7200 , d ) 0.7 , e ) 420 | e | multiply(divide(multiply(30, const_2), subtract(35, 30)), 35) | multiply(n0,const_2)|subtract(n1,n0)|divide(#0,#1)|multiply(n1,#2)| | physics |
a box contains 25 electric bulbs , out of which 4 are defective . two bulbs are chosen at random from this box . the probability that at least one of these is defective is | "solution p ( none is defective ) = 21 c 2 / 25 c 2 = 7 / 10 p ( at least one is defective ) = ( 1 - 7 / 10 ) = 3 / 10 . answer a" | a ) 15 , b ) 3 / 10 , c ) 10 , d ) 15552 , e ) 720 | b | subtract(const_1, divide(choose(subtract(25, 4), const_2), choose(25, const_2))) | choose(n0,const_2)|subtract(n0,n1)|choose(#1,const_2)|divide(#2,#0)|subtract(const_1,#3)| | probability |
how many prime numbers are between 28 / 3 and 86 / 6 ? | "28 / 3 = 9 . xxx 86 / 6 = 14 . xxx so we need to find prime numbers between 9 ( exclusive ) - 13 ( inclusive ) there are 2 prime numbers 1113 hence answer will be ( b ) 2 b" | a ) 0.075 days , b ) 2 , c ) 63 , d ) 40 sec , e ) 20 | b | floor(const_2) | floor(const_2)| | general |
in a school of 800 students , 44 % wear blue shirts , 28 % wear red shirts , 10 % wear green shirts , and the remaining students wear other colors . how many students wear other colors ( not blue , not red , not green ) ? | "44 + 28 + 10 = 82 % 100 – 82 = 18 % 800 * 18 / 100 = 144 the answer is c ." | a ) 1000 , b ) 300 m , c ) 1534 , d ) 756 , e ) 144 | e | subtract(800, add(add(multiply(divide(44, const_100), 800), multiply(divide(28, const_100), 800)), multiply(divide(10, const_100), 800))) | divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|add(#3,#4)|add(#6,#5)|subtract(n0,#7)| | gain |
a man whose speed is 6 kmph in still water rows to a certain upstream point and back to the starting point in a river which flows at 1.5 kmph , find his average speed for the total journey ? | "m = 6.0 s = 1.5 ds = 7.5 us = 4.5 as = ( 2 * 7.5 * 4.5 ) / 12 = 5.6 answer : b" | a ) 1717.85 , b ) 6 cm', ' , c ) 11 , d ) 5.6 , e ) 5.5 % | d | divide(add(6, subtract(6, 1.5)), const_2) | subtract(n0,n1)|add(n0,#0)|divide(#1,const_2)| | general |
in what ratio must rice of rs . 25 per kg be mixed with rice of rs . 12 per kg so that cost of mixture is rs . 20 per kg ? | "( 20 - 12 ) / ( 25 - 20 = 8 / 5 = 8 : 5 answer : a" | a ) 12 , b ) 50 , c ) 8 : 5 , d ) 21 , e ) 24 | c | divide(divide(subtract(20, 12), subtract(25, 12)), subtract(const_1, divide(subtract(20, 12), subtract(25, 12)))) | subtract(n2,n1)|subtract(n0,n1)|divide(#0,#1)|subtract(const_1,#2)|divide(#2,#3)| | other |
10 men and 15 women together can complete a work in 4 days . it takes 100 days for one man alone to complete the same work . how many days will be required for one woman alone to complete the same work ? | "1 man ' s 1 day work = 1 / 100 ( 10 men + 15 women ) ' s 1 day work = 1 / 4 15 women ' s 1 day work = ( 1 / 4 - 10 / 100 ) = 3 / 20 1 woman ' s 1 day work = 1 / 100 1 woman alone can complete the work in 100 days . answer : a" | a ) 100 days , b ) 48 , c ) 500 cm 2 , d ) 5.48 days , e ) 70 | a | multiply(divide(multiply(const_1, 100), subtract(multiply(const_1, 100), multiply(10, 4))), multiply(15, 4)) | multiply(n3,const_1)|multiply(n0,n2)|multiply(n1,n2)|subtract(#0,#1)|divide(#0,#3)|multiply(#4,#2)| | physics |
if john makes a contribution to a charity fund at school , the average contribution size will increase by 50 % reaching $ 75 per person . if there were 4 other contributions made before john ' s , what is the size of his donation ? | cavg = average contribution before john cavg * 1.5 = 75 , therefore the average cont is $ 50 before john . if he needs to increase the average contribution by $ 25 , he must put in $ 25 for each of the 4 people . so $ 100 . but , he also has to put in the average for himself ( the fiveth person ) , so add $ 75 . so $ 175 is your answer . answer b | a ) 540 % , b ) $ 175 , c ) 600 , d ) 16 , e ) 12,526 | b | add(subtract(multiply(add(4, const_1), 75), multiply(add(4, const_1), 50)), 50) | add(n2,const_1)|multiply(n1,#0)|multiply(n0,#0)|subtract(#1,#2)|add(n0,#3) | general |
if 0.5 : x : : 5 : 8 , then x is equal to : | "( x * 5 ) = ( 0.5 * 8 ) x = 4 / 5 x = 0.8 answer = e" | a ) 0.8 , b ) $ 3.75 , c ) 6 , d ) 405 meter , e ) 1 | a | divide(multiply(0.5, 8), 5) | multiply(n0,n2)|divide(#0,n1)| | general |
the average weight of 8 people increases by 2.5 kg when a new person comes in place of one of them weighing 75 kg . what is the weight of the new person ? | "the total weight increase = ( 8 x 2.5 ) kg = 20 kg weight of new person = ( 75 + 20 ) kg = 95 kg the answer is c ." | a ) 14.3 % , b ) 3 kmph , c ) 129.5 , d ) 95 kg , e ) 10 | d | add(multiply(2.5, 8), 75) | multiply(n0,n1)|add(n2,#0)| | general |
the radius of a wheel is 12.6 cm . what is the distance covered by the wheel in making 200 resolutions ? | "in one resolution , the distance covered by the wheel is its own circumference . distance covered in 200 resolutions . = 200 * 2 * 22 / 7 * 12.6 = 15825.6 cm = 158.256 m answer : d" | a ) 5 , b ) 158.256 m , c ) 34 min , d ) 2 , e ) 5 / 26 | b | divide(multiply(multiply(multiply(divide(add(multiply(add(const_3, const_4), const_3), const_1), add(const_3, const_4)), 12.6), const_2), 200), const_100) | add(const_3,const_4)|multiply(#0,const_3)|add(#1,const_1)|divide(#2,#0)|multiply(n0,#3)|multiply(#4,const_2)|multiply(n1,#5)|divide(#6,const_100)| | physics |
simplify : 250 x 250 - 100 x 100 | "( 250 ) ^ 2 - ( 100 ) ^ 2 = ( 250 + 100 ) ( 250 - 100 ) = 350 x 150 = 52500 . answer is c ." | a ) 2068 , b ) 661.25 , c ) 52500 , d ) 1200 , e ) 6 | c | add(multiply(250, 250), multiply(100, 100)) | multiply(n0,n1)|multiply(n2,n3)|add(#0,#1)| | general |
on my sister ' s birthday , she was 143 cm in height , having grown 10 % since the year before . how tall was she the previous year ? | "let the previous year ' s height be x . 1.1 x = 143 x = 130 the answer is c ." | a ) 24', ' , b ) 35 , c ) 300 , d ) 3 / 50 , e ) 130 cm | e | subtract(143, divide(multiply(143, 10), const_100)) | multiply(n0,n1)|divide(#0,const_100)|subtract(n0,#1)| | physics |
if the selling price of 50 articles is equal to the cost price of 35 articles , then the loss or gain percent is : | "c . p . of each article be re . 1 . then , c . p . of 50 articles = rs . 50 ; s . p . of 50 articles = rs . 35 . loss % = 15 / 50 * 100 = 30 % answer c" | a ) 30 % , b ) 4 , c ) 5.5 mph , d ) 285 , e ) 2 | a | subtract(50, 35) | subtract(n0,n1)| | gain |
the length of each side of an equilateral triangle having an area of 4 â ˆ š 3 cm 2 is ? | explanation : â ˆ š 3 / 4 a 2 = 4 â ˆ š 3 - > a = 4 answer is d | a ) 27,550 , b ) 6 : 24 , c ) 50 , d ) 8100 , e ) 4 cm', ' | e | sqrt(divide(multiply(4, sqrt(3)), multiply(multiply(divide(const_1, 2), divide(const_1, 2)), sqrt(3)))) | divide(const_1,n2)|sqrt(n1)|multiply(n0,#1)|multiply(#0,#0)|multiply(#3,#1)|divide(#2,#4)|sqrt(#5) | geometry |
if the sides of a rectangle are increased by 25 % , what is the percentage increase in the area ? | "if sides are a and b , after increase sides would be 1.25 a and 1.25 b . percentage increase in area = ( 1.25 a * 1.25 b - ab ) * 100 / ab = 56.25 % answer : b" | a ) 38 sec , b ) 10780 , c ) 56.25 $ , d ) 71 , e ) 96 | c | multiply(subtract(power(add(divide(25, const_100), const_1), const_2), const_1), const_100) | divide(n0,const_100)|add(#0,const_1)|power(#1,const_2)|subtract(#2,const_1)|multiply(#3,const_100)| | geometry |
the charge for a single room at hotel p is 30 percent less than the charge for a single room at hotel r and 10 percent less than the charge for a single room at hotel g . the charge for a single room at hotel r is what percent greater than the charge for a single room at hotel g ? | "let rate in r = 100 x then p = 70 x g = 100 y p = 90 y thus 70 x = 90 y or x = 1.28 y ans r = 128 y so increase = 28 % answer : a" | a ) 24 days , b ) 11 , c ) 120 , d ) 18 days , e ) 28 % | e | multiply(divide(subtract(const_100, multiply(divide(subtract(const_100, 30), subtract(const_100, 10)), const_100)), multiply(divide(subtract(const_100, 30), subtract(const_100, 10)), const_100)), const_100) | subtract(const_100,n0)|subtract(const_100,n1)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)|divide(#4,#3)|multiply(#5,const_100)| | gain |
a trader marked the selling price of an article at 11 % above the cost price . at the time of selling , he allows certain discount and suffers a loss of 1 % . he allowed a discount of : | "sol . let c . p . = rs . 100 . then , marked price = rs . 110 , s . p . = rs . 99 . ∴ discount % = [ 11 / 111 * 100 ] % = 9.9 % answer c" | a ) 360 , b ) 200 , c ) 9.9 % , d ) 66 min . , e ) 9 | c | multiply(const_100, divide(add(multiply(add(const_2, const_3), const_2), 1), add(const_100, 11))) | add(const_2,const_3)|add(n0,const_100)|multiply(#0,const_2)|add(#2,n1)|divide(#3,#1)|multiply(#4,const_100)| | gain |
bottle r contains 250 capsules and costs $ 5.25 . bottle t contains 130 capsules and costs $ 2.99 . what is the difference between the cost per capsule for bottle r and the cost per capsule for bottle t ? | cost per capsule in r is 5.25 / 250 = 0.525 / 25 = 0.021 cost per capsule in t is 2.99 / 130 = 0.023 the difference is 0.002 the answer is e | a ) 38 , b ) 11 am , c ) 1 / 6 , d ) $ 0.002 , e ) 14 : 00 | d | subtract(divide(2.99, 130), divide(5.25, 250)) | divide(n3,n2)|divide(n1,n0)|subtract(#0,#1) | general |
sushil got thrice as many marks in english as in science . his total marks in english , science and maths are 162 . if the ratio of his marks in english and maths is 4 : 5 , find his marks in science ? | "s : e = 1 : 3 e : m = 4 : 5 - - - - - - - - - - - - s : e : m = 4 : 12 : 15 4 / 31 * 162 = 20.9 answer : b" | a ) 360 , b ) 13 % , c ) $ 330 , d ) 20.9 , e ) 9 / 49 | d | add(add(add(multiply(4, const_3), 4), multiply(multiply(4, const_3), add(multiply(4, const_3), 4))), 4) | multiply(n1,const_3)|add(n1,#0)|multiply(#1,#0)|add(#1,#2)|add(n1,#3)| | general |
solve the equation for x : 21 ( x + y ) + 5 = 21 ( - x + y ) + 47 | "e 1 21 ( x + y ) + 5 = 21 ( - x + y ) + 47 21 x + 21 y + 5 = - 21 x + 21 y + 47 21 x + 5 = - 21 x + 47 42 x = 42 = > x = 1" | a ) 17 , b ) 10 , c ) 1 , d ) 4 , e ) 120 | c | divide(add(5, 47), add(21, 21)) | add(n1,n3)|add(n0,n0)|divide(#0,#1)| | general |
area of four walls of a room is 99 m 2 . the length and breadth of the room are 7.5 m and 3.5 m respectively . the height of the room is : | 2 ( 7.5 + 3.5 ) × h = 99 2 ( 11 ) h = 99 22 h = 99 h = 99 / 22 = 9 / 2 = 4.5 m answer is d . | a ) 4 % , b ) 10.6 , c ) 20 , d ) 4.5 m', ' , e ) 5 % | d | divide(99, add(multiply(7.5, const_2), multiply(3.5, const_2))) | multiply(n2,const_2)|multiply(n3,const_2)|add(#0,#1)|divide(n0,#2) | physics |
x is a positive integer less than 300 . when x is divided by 7 , the remainder is 1 ; when x is divided by 3 , the remainder is 2 . how many x are there ? | "the nubmer which when divided by 7 leaves remainder 1 should be of the form 7 k + 1 this number when divided by 3 leaves remainder 2 . so , ( 7 k + 1 ) - 2 should be divisible by 3 or 7 k - 1 should be divisible by 3 . we now put the values of k starting from 0 to find first number divisible by 3 we find 1 st number at k = 1 thus smallest number will be 7 ( 1 ) + 1 = 8 now , next number will be = 8 + lcm of 37 i . e 29 now we will find number of all such values less than 500 by using the formula for last term of an a . p 8 + ( n - 1 ) 21 = 300 n = 22.13 or n = 22 answer : - b" | a ) 22 , b ) 13.9 , c ) 71 % , d ) 1540 , e ) none | a | subtract(add(multiply(reminder(7, 300), 3), reminder(3, 300)), reminder(1, 300)) | reminder(n1,n0)|reminder(n3,n0)|reminder(n2,n0)|multiply(n3,#0)|add(#3,#1)|subtract(#4,#2)| | general |
how many different ways can 2 students be seated in a row of 4 desks , so that there is always at least one empty desk between the students ? | "total cases : 12 ( student one has 4 options and student two has three options , 4 x 3 = 12 ) non - favourable cases : 6 ( when two students sit together . students in desk 1 and desk 2 , in desk 2 and desk 3 , in desk 3 and desk 4 ) for each of these cases there are two possibilities because the positions can be interchanged . hence 2 x 3 = 6 . so favourable cases : 12 - 6 = 6 . answer : d" | a ) 32 square meters , b ) 77 , c ) 1000 , d ) 44 , e ) 6 | e | permutation(subtract(4, const_1), 2) | subtract(n1,const_1)|permutation(#0,n0)| | probability |
if each side of a square is increased by 25 % , find the percentage change in its area ? | "let each side of the square be a , then area = a x a new side = 125 a / 100 = 5 a / 4 new area = ( 5 a x 5 a ) / ( 4 x 4 ) = ( 25 a ² / 16 ) increased area = = ( 25 a ² / 16 ) - a ² increase % = [ ( 9 a ² / 16 ) x ( 1 / a ² ) x 100 ] % = 56.25 % answer : b" | a ) 6 , b ) 5 , c ) 56.25 % , d ) 15 sec , e ) 72 days | c | square_area(add(const_1, divide(25, const_100))) | divide(n0,const_100)|add(#0,const_1)|square_area(#1)| | gain |
a dishonest dealer professes to sell goods at the cost price but uses a weight of 720 grams per kg , what is his percent ? | "explanation : 720 - - - 280 100 - - - ? = > 38.9 % answer : c" | a ) 36 , b ) 38.9 % , c ) 62 / 69 , d ) 25 % , e ) 100 % | b | subtract(multiply(divide(const_100, 720), multiply(const_100, multiply(add(const_3, const_2), const_2))), const_100) | add(const_2,const_3)|divide(const_100,n0)|multiply(#0,const_2)|multiply(#2,const_100)|multiply(#1,#3)|subtract(#4,const_100)| | gain |
if the mean of numbers 28 , x , 42 , 78 , 82 and 104 is 62 , then what is the mean of 128 , 255 , 511 , 1023 and x ? | "the mean of numbers 28 , x , 42 , 78 and 104 is 62 : 28 + x + 42 + 78 + 82 + 104 = 62 * 6 - - > x = 38 ; so , the mean of 128 , 255 , 511 , 1023 and x is ( 128 + 255 + 511 + 1023 + 38 ) / 5 = 391 . answer : c ." | a ) 4 % , b ) 391 , c ) 54.2 % , d ) 80 / 243 , e ) 0.2 | b | divide(add(add(add(add(subtract(multiply(104, add(const_4, const_1)), add(add(add(28, 42), 78), 82)), 62), 128), 255), 511), add(const_4, const_1)) | add(const_1,const_4)|add(n0,n1)|add(n2,#1)|multiply(n4,#0)|add(n3,#2)|subtract(#3,#4)|add(n5,#5)|add(n6,#6)|add(n7,#7)|add(n8,#8)|divide(#9,#0)| | general |
machine a produces 100 parts twice as fast as machine b does . machine b produces 100 parts in 80 minutes . if each machine produces parts at a constant rate , how many parts does machine a produce in 20 minutes ? | machine b produces 100 part in 80 minutes . machine a produces 100 parts twice as fast as b , so machine a produces 100 parts in 80 / 2 = 40 minutes . now , machine a produces 100 parts in 40 minutes which is 100 / 40 = 10 / 4 parts / minute . 10 / 4 parts x a total of 20 minutes = 10 / 4 * 20 = 50 e | a ) s . 352 , b ) 50 , c ) 184 , d ) 3 , 400,000 , e ) 270 | b | multiply(multiply(divide(100, 80), const_2), 20) | divide(n0,n2)|multiply(#0,const_2)|multiply(n3,#1) | gain |
if x and y are integers such that | y + 3 | ≤ 3 and 2 y – 3 x + 6 = 0 , what is the least possible value t of the product xy ? | "how to deal with inequalities involving absolute values ? first example shows us the so callednumber case in this case we have | y + 3 | ≤ 3 which is generalized | something | ≤ some number . first we solve as if there were no absolute value brackets : y + 3 ≤ 3 y ≤ 0 so y is 0 or negative second scenario - remove the absolute value brackets . put a negative sign around the other side of the inequality , andflip the sign : y + 3 > = - 3 y > = - 6 therefore we have a possible range for y : - 6 = < y < = 0 ok , so far so good , we ' re half way through . what about x ? here ' s the formula : 2 y – 3 x + 6 = 0 , rewrite it as 2 y + 6 = 3 x . you can say that 2 y + 6 is a multiple of 3 ( = 3 x ) . so all values which must be integer must also satisfy this constraint . i ' m just saying that , so it ' s easier to evaluate all the possible numbers ( - 6 , - 3 , 0 ) . if you plug in y = 0 , x will be 2 and xy = 0 as the lowest possible value t . hence , answer choice c is the one to go ." | a ) 8 , b ) 5 cm , c ) 3 : 4 , d ) 49 , e ) 0 | e | multiply(divide(add(6, multiply(negate(add(3, 3)), 2)), 3), negate(add(3, 3))) | add(n0,n0)|negate(#0)|multiply(n2,#1)|add(n4,#2)|divide(#3,n0)|multiply(#4,#1)| | general |
a certain characteristic in a large population has a distribution that is symmetric about the mean m . if 66 percent of the distribution lies within one standard deviation d of the mean , what percent of the distribution is less than m + d ? | "this is easiest to solve with a bell - curve histogram . m here is equal to µ in the gaussian normal distribution and thus m = 50 % of the total population . so , if 66 % is one st . dev , then on either side of m we have 66 / 2 = 33 % . so , 33 % are to the right and left of m ( = 50 % ) . in other words , our value m + d = 50 + 33 = 83 % goingfrom the mean m , to the right of the distributionin the bell shaped histogram . . this means that 83 % of the values are below m + d . like i said , doing it on a bell - curve histogram is much easier to fullygethow this works , or you could apply gmat percentile jargon / theory to it c" | a ) 5994 , b ) 3000000 , c ) 83 % , d ) 15 % , e ) 40 seconds | c | subtract(const_100, divide(subtract(const_100, 66), const_2)) | subtract(const_100,n0)|divide(#0,const_2)|subtract(const_100,#1)| | general |
in a class of 54 students , 12 enrolled for both english and german . 22 enrolled for german . if the students of the class enrolled for at least one of the two subjects , then how many students enrolled for only english and not german ? | "total = english + german - both + neither - - > 54 = english + 22 - 12 + 0 - - > english = 44 - - > only english = english - both = 44 - 12 = 32 . answer : e ." | a ) 4 % , b ) 3 / 7 , c ) 9 , d ) 600 cm 2 , e ) 32 | e | subtract(subtract(add(54, 12), 22), 12) | add(n0,n1)|subtract(#0,n2)|subtract(#1,n1)| | other |
usc invited each south carolina high school to send up to 39 students to watch a football game . a section which has 199 seats in each row is reserved for those students . what is the least number of rows needed to guarantee that if 2006 students show up , then all students from the same high school can be seated in the same row ? | the answer is 12 rows . if 59 schools send 34 students each , then we can sit at most 5 groups of students in the same row , so we will need 12 rows . next , 12 rows are sufficient . assume that this is not the case . suppose the groups of students are seated like this : first the largest group , then the second largest group , then the third largest group , etc . suppose we run out of space - there are not enough seats in any row to seat together the next group . suppose the first group that can not be seated together is the kth group and it consists of n students . then k 61 since any row fits at least 5 groups . also , n 2006 / k 2006 / 61 < 33 ( all groups already seated are no smaller than the kth group ) . so , n 32 . since there is not enough space in any of the 12 rows to seat the kth group , then there must be at least 168 students seated in each of the 12 rows . now , 12 × 168 = 2016 > 2006 a contradiction . so , 12 rows are sufficient . correct answer b | a ) 4 , b ) 16 , c ) 36 % , d ) 12 , e ) 21 years | d | add(divide(2006, 199), const_2) | divide(n2,n1)|add(#0,const_2) | physics |
a rectangular photograph is surrounded by a border that is 1 inch wide on each side . the total area of the photograph and the border is m square inches . if the border had been 3 inches wide on each side , the total area would have been ( m + 64 ) square inches . what is the perimeter of the photograph , in inches ? | let x and y be the width and length of the photograph . ( x + 2 ) ( y + 2 ) = m and so ( 1 ) xy + 2 x + 2 y + 4 = m ( x + 6 ) ( y + 6 ) = m and so ( 2 ) xy + 6 x + 6 y + 36 = m + 64 let ' s subtract equation ( 1 ) from equation ( 2 ) . 4 x + 4 y + 32 = 64 2 x + 2 y = 16 , which is the perimeter of the photograph . the answer is a . | a ) 82 , b ) 501 , c ) 10 , d ) 16', ' , e ) 288,889 | d | divide(subtract(64, subtract(power(multiply(3, const_2), const_2), power(multiply(1, const_2), const_2))), const_2) | multiply(n1,const_2)|multiply(n0,const_2)|power(#0,const_2)|power(#1,const_2)|subtract(#2,#3)|subtract(n2,#4)|divide(#5,const_2) | geometry |
a “ palindromic integer ” is an integer that remains the same when its digits are reversed . so , for example , 43334 and 516615 are both examples of palindromic integers . how many 6 - digit palindromic integers are both even and greater than 300,000 ? | "the first digit and last digit are the same so the 3 possibilities are 4 , 6 , or 8 . the second and third digits can be any digit from 0 to 9 . the total number of palindromic integers is 3 * 10 * 10 = 300 the answer is c ." | a ) 10749 , b ) 2 / 9 , c ) 76 , d ) 300 , e ) 7.5 sec | d | multiply(multiply(subtract(6, const_4), const_10), const_10) | subtract(n2,const_4)|multiply(#0,const_10)|multiply(#1,const_10)| | general |
the unit digit in the product 4556 * 3432 * 4581 * 2784 is ? | unit digit in the given product = unit digit in 6 * 2 * 1 * 4 = 8 answer is a | a ) 68 kmph , b ) $ 6 , c ) $ 840 , d ) 3 / 4 , e ) 8 | e | subtract(multiply(multiply(multiply(4556, 3432), 4581), 2784), subtract(multiply(multiply(multiply(4556, 3432), 4581), 2784), add(const_4, const_4))) | add(const_4,const_4)|multiply(n0,n1)|multiply(n2,#1)|multiply(n3,#2)|subtract(#3,#0)|subtract(#3,#4) | general |
for any positive number x , the function [ x ] denotes the greatest integer less than or equal to x . for example , [ 1 ] = 1 , [ 1.367 ] = 1 and [ 1.996 ] = 1 . if k is a positive integer such that k ^ 2 is divisible by 45 and 80 , what is the units digit of k ^ 3 / 4000 ? | "k = [ lcm of 80 and 45 ] * ( any integer ) however minimum value of k is sq . rt of 3 ^ 2 * 4 ^ 2 * 5 ^ 2 = 60 * any integer for value of k ( 60 ) * any integer unit value will be always zero . c" | a ) 4749 , b ) 0 , c ) 552 , d ) 1575 , e ) 540 % | b | divide(multiply(multiply(multiply(3, 2), multiply(3, 2)), multiply(3, 2)), const_4) | multiply(n6,n9)|multiply(#0,#0)|multiply(#1,#0)|divide(#2,const_4)| | general |
a train passes a man standing on a platform in 8 seconds and also crosses the platform which is 276 metres long in 20 seconds . the length of the train ( in metres ) is : | "explanation : let the length of train be l m . acc . to question ( 276 + l ) / 20 = l / 8 2208 + 8 l = 20 l l = 2208 / 12 = 184 m answer a" | a ) $ 400 , b ) 3 , c ) $ 0.40 , d ) 184 , e ) 5 : 11 | d | multiply(divide(276, subtract(20, 8)), 8) | subtract(n2,n0)|divide(n1,#0)|multiply(n0,#1)| | physics |
to be considered for “ movie of the year , ” a film must appear in at least 1 / 4 of the top - 10 - movies lists submitted by the cinematic academy ’ s 760 members . what is the smallest number of top - 10 lists a film can appear on and still be considered for “ movie of the year ” ? | "total movies submitted are 760 . as per question we need to take 1 / 4 of 760 to be considered for top 10 movies = 190 approximate the value we 190 . imo option b is the correct answer . . ." | a ) 190 , b ) 1800 , c ) rs . 7700 , d ) 50 % , e ) 1.745 % | a | divide(760, 4) | divide(n3,n1)| | general |
each child has 4 crayons and 14 apples . if there are 9 children , how many crayons are there in total ? | 4 * 9 = 36 . answer is d . | a ) 16 , b ) 18 m , c ) 40 minutes , d ) 510 , e ) 36 | e | multiply(9, 4) | multiply(n0,n2) | general |
a man cycles round the boundary of a rectangular park at the rate of 12 kmph and completes one full round in 8 minutes . if the ratio between the length and breadth of the park be 3 : 2 , then its area is : | perimeter = distance covered in 8 min = ( 12000 / 60 * 8 ) m = 1600 m let , length = 3 x meters and breadth = 2 x meters then , 2 ( 3 x + 2 x ) = 1600 or x = 160 therefore , length = 480 m and breadth = 320 m therefore , area = ( 480 * 320 ) m 2 = 153600 m 2 answer : c | a ) 300 , b ) 64 % , c ) 153600 m 2', ' , d ) 6825 , e ) 54 : 53 | c | multiply(multiply(divide(multiply(8, divide(multiply(12, const_1000), multiply(multiply(const_3, const_2), const_10))), const_10), 2), multiply(divide(multiply(8, divide(multiply(12, const_1000), multiply(multiply(const_3, const_2), const_10))), const_10), 3)) | multiply(n0,const_1000)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(#0,#2)|multiply(n1,#3)|divide(#4,const_10)|multiply(n3,#5)|multiply(n2,#5)|multiply(#6,#7) | physics |
find the value of m 12519 x 9999 = m ? | "12519 x 9999 = 12519 x ( 10000 - 1 ) = 12519 x 10000 - 12519 x 1 = 125190000 - 12519 = 125177481 a" | a ) 60 , b ) $ 96 , c ) . 2 , d ) 125177481 , e ) 12.36 % | d | multiply(subtract(9999, const_4), 12519) | subtract(n1,const_4)|multiply(#0,n0)| | general |
a is twice as good a workman as b and they took 9 days together to do the work b alone can do it in . | "wc = 2 : 1 2 x + x = 1 / 9 x = 1 / 27 = > 27 days answer : e" | a ) 27 days , b ) 16 , c ) 32 square meters , d ) 3 : 8 , e ) 1500 | a | multiply(divide(multiply(9, add(const_2, const_1)), const_2), const_2) | add(const_1,const_2)|multiply(n0,#0)|divide(#1,const_2)|multiply(#2,const_2)| | physics |
find the value of ( 875 233 / 899 ) × 899 | "( 875 233 / 899 ) × 899 ( 786625 + 233 ) / 899 × 899 786858 / 899 × 899 786858 c" | a ) 7 2 / 3 , b ) 9 : 5 , c ) 786858 , d ) 52 , e ) $ 900 | c | multiply(add(divide(233, 899), 875), 899) | divide(n1,n2)|add(n0,#0)|multiply(#1,n2)| | general |
a certain store sold pens for $ 0.35 each and pencils for $ 0.25 each . if a customer purchased both pens and pencils from the store for a total of $ 2.00 , what total number of pens and pencils did the customer purchase ? | "answer : aalgebraically , the question looks like this : 2.5 = 0.35 x + 0.25 ythere are six possible numbers of pens that fit that requirement : 1 pen : $ 0.35 2 pens : $ 0.70 3 pens : $ 1.05 4 pens : $ 1.40 5 pens : $ 1.75 and 1 pencils for $ 0.25 each . that ' s 5 pens and 1 pencils for a total of 6 pens and pencils . choice ( a ) is correct ." | a ) 5 / 6 , b ) 6 , c ) 62 , d ) 8200 , e ) 60 | b | multiply(divide(2.00, add(0.35, 0.25)), const_2) | add(n0,n1)|divide(n2,#0)|multiply(#1,const_2)| | other |
the length of the bridge , which a train 130 metres long and travelling at 108 km / hr can cross in 30 seconds , is : | "speed = [ 108 x 5 / 18 ] m / sec = 30 m / sec time = 10 sec let the length of bridge be x metres . then , ( 130 + x ) / 30 = 10 = > 130 + x = 300 = > x = 170 m . answer : d" | a ) 4.0 , b ) 170 m , c ) 90 ° , d ) 6 , e ) 58.65 ft | b | subtract(multiply(divide(multiply(108, speed(const_1000, const_1)), speed(const_3600, const_1)), 30), 130) | speed(const_1000,const_1)|speed(const_3600,const_1)|multiply(n1,#0)|divide(#2,#1)|multiply(n2,#3)|subtract(#4,n0)| | physics |
what is the smallest number which , when increased by 9 , is divisible by 7 , 8 , and 24 ? | "lcm ( 7 , 8,24 ) = 24 x 7 = 168 so the least divisible number is 168 , and the number we are looking for is 168 - 9 = 159 . the answer is c ." | a ) 10 , b ) 159 , c ) 36 kmph , d ) 859 , e ) 24 % | b | subtract(lcm(24, 8), 9) | lcm(n2,n3)|subtract(#0,n0)| | general |
what is the sum of all the prime numbers greater than 50 but less than 70 ? | "required sum = ( 53 + 59 + 61 + 67 ) = 240 note : 1 is not a prime number answer d" | a ) 500 , b ) 62 , c ) 32 , d ) 240 , e ) 60 m | d | 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)), 50), 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 length of a rectangular floor is more than its breadth by 200 % . if rs . 150 is required to paint the floor at the rate of rs . 2 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 + 2 b = 3 b area of the floor = 150 / 2 = 75 sq m l b = 75 i . e . , l * l / 3 = 75 l 2 = 225 = > l = 15 answer : b" | a ) 9 , b ) 15 , c ) 18 : 13 , d ) 110 , e ) 4.55 | b | multiply(sqrt(divide(divide(150, 2), const_3)), const_3) | divide(n1,n2)|divide(#0,const_3)|sqrt(#1)|multiply(#2,const_3)| | gain |
a car traveled from san diego to san francisco at an average speed of 66 miles per hour . if the journey back took twice as long , what was the average speed of the trip ? | "let the time taken be = x one way distance = 66 x total distance traveled = 2 * 66 x = 132 x total time taken = x + 2 x = 3 x average speed = 132 x / 3 x = 44 answer : e" | a ) 44 . , b ) 11 , c ) 1386 , d ) 113 , e ) 2 | a | inverse(add(inverse(66), divide(inverse(66), const_2))) | inverse(n0)|divide(#0,const_2)|add(#1,#0)|inverse(#2)| | physics |
a cube of side 5.5 meter length is cut into small cubes of side 110 cm each . how many such small cubes can be obtained ? | "along one edge , the number of small cubes that can be cut = 550 / 110 = 5 along each edge 5 cubes can be cut . ( along length , breadth and height ) . total number of small cubes that can be cut = 5 * 5 * 5 = 125 answer : a" | a ) s . 4400 , b ) 11 , c ) 17 hr , d ) 35 km , e ) 125 | e | divide(power(power(5.5, const_2), const_3), power(5.5, const_3)) | power(n0,const_2)|power(n0,const_3)|power(#0,const_3)|divide(#2,#1)| | physics |
rates for having a manuscript typed at a certain typing service are $ 5 per page for the first time a page is typed and $ 2 per page each time a page is revised . if a certain manuscript has 100 pages , of which 40 were revised only once , 10 were revised twice , and the rest required no revisions , what was the total cost of having the manuscript typed ? | "for 100 - 40 - 10 = 50 pages only cost is 5 $ per page for the first time page is typed - 50 * 5 = 250 $ ; for 40 pages the cost is : first time 5 $ + 2 $ of the first revision - 40 * ( 5 + 2 ) = 280 $ ; for 10 pages the cost is : first time 5 $ + 2 $ of the first revision + 2 $ of the second revision - 10 ( 5 + 2 + 2 ) = 90 $ ; total : 250 + 280 + 90 = 620 $ . answer : b ." | a ) 34 % , b ) $ 620 , c ) 28 , d ) 37.5 , e ) 274 | b | add(add(multiply(100, 5), multiply(40, 2)), multiply(multiply(10, 2), const_2)) | multiply(n0,n2)|multiply(n1,n3)|multiply(n1,n4)|add(#0,#1)|multiply(#2,const_2)|add(#3,#4)| | general |
some persons can do a piece of work in 32 days . two times the number of these people will do half of that work in ? | 32 / ( 2 * 2 ) = 8 days answer : e | a ) 10,800 , b ) 8 , c ) 5 cm , d ) 26.66 % , e ) 7 | b | multiply(multiply(32, divide(const_1, const_2)), divide(const_1, const_2)) | divide(const_1,const_2)|multiply(n0,#0)|multiply(#0,#1) | physics |
what is the units digit of 33 ^ 2 * 17 ^ 3 * 39 ^ 2 ? | "the units digit of 33 ^ 2 is the units digit of 3 * 3 = 9 which is 9 . the units digit of 17 ^ 3 is the units digit of 7 * 7 * 7 = 343 which is 3 . the units digit of 39 ^ 2 is the units digit of 9 * 9 = 81 which is 1 . the units digit of 9 * 3 * 1 = 27 is 7 . the answer is d ." | a ) rs . 240 , b ) 450 , c ) 67 % , d ) 1200 , e ) 7 | e | divide(add(multiply(factorial(33), factorial(2)), multiply(factorial(33), factorial(3))), 33) | factorial(n0)|factorial(n1)|factorial(n3)|multiply(#0,#1)|multiply(#0,#2)|add(#3,#4)|divide(#5,n0)| | general |