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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find the value of x ? ( 2 / 7 ) ^ - 8 * ( 7 / 2 ) ^ - 2 = ( 2 / 7 ) ^ x | ( 2 / 7 ) ^ - 8 * ( 7 / 2 ) ^ - 2 = ( 2 / 7 ) ^ x now , ( 7 / 2 ) ^ 8 * ( 2 / 7 ) ^ 2 = ( 2 / 7 ) ^ x on solving , ( 7 / 2 ) ^ 6 = ( 2 / 7 ) ^ x ( 2 / 7 ) ^ - 6 = ( 2 / 7 ) ^ x thus , x = - 6 . answer : b | a ) - 5 , b ) - 6 , c ) 4 , d ) 0 , e ) - 4 | b | multiply(7, 7) | multiply(n1,n1) | general |
in a maths test , students were asked to find 5 / 16 of a certain number . one of the students by mistake found 5 / 6 th of that number and his answer was 200 more than the correct answer . find the number . | "explanation : let the number be x . 5 * x / 6 = 5 * x / 16 + 200 25 * x / 48 = 200 x = 384 answer : a" | a ) 384 , b ) 76 , c ) 26 , d ) 28 , e ) 11 | a | divide(multiply(multiply(200, 16), 6), subtract(multiply(5, 16), multiply(5, 6))) | multiply(n1,n4)|multiply(n0,n1)|multiply(n0,n3)|multiply(n3,#0)|subtract(#1,#2)|divide(#3,#4)| | general |
how many seconds does sandy take to cover a distance of 500 meters , if sandy runs at a speed of 18 km / hr ? | "18 km / hr = 18000 m / 3600 s = 5 m / s time = 500 / 5 = 100 seconds the answer is b ." | a ) 80 , b ) 100 , c ) 120 , d ) 140 , e ) 160 | b | divide(500, multiply(18, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics |
8 men can do a piece of work in 12 days . 4 women can do it in 48 days and 10 children can do it in 24 days . in how many days can 10 men , 4 women and 18 children together complete the piece of work ? | "explanation : 1 man β s 1 day β s work = 1 / 8 Γ 12 = 1 / 96 10 men β s 1 day β s work = 1 Γ 10 / 96 = 5 / 48 1 woman β s 1 day β s work = 1 / 192 4 women β s 1 day β s work = 1 / 192 Γ 4 = 1 / 48 1 child β s 1 day β s work = 1 / 240 18 children β s 1 day β s work = 3 / 40 therefore , ( 10 men + 4 women + 18 children ... | a ) 5 days , b ) 15 days , c ) 28 days , d ) 6 days , e ) 7 days | a | inverse(add(multiply(10, inverse(multiply(24, 10))), add(multiply(inverse(multiply(12, 8)), 10), multiply(inverse(multiply(48, 4)), 4)))) | multiply(n0,n1)|multiply(n2,n3)|multiply(n4,n5)|inverse(#0)|inverse(#1)|inverse(#2)|multiply(n4,#3)|multiply(n2,#4)|multiply(n4,#5)|add(#6,#7)|add(#9,#8)|inverse(#10)| | physics |
the circumference of the front wheel of a cart is 30 ft long and that of the back wheel is 35 ft long . what is the distance traveled by the cart , when the front wheel has done five more revolutions than the rear wheel ? | "point to note : both the wheels would have traveled the same distance . now consider , no . of revolutions made by back wheel as x , which implies that the number of revolutions made by the front wheel is ( x + 5 ) . equating the distance traveled by front wheel to back wheel : ( x + 5 ) * 30 = x * 35 . ( formula for ... | a ) 20 ft , b ) 25 ft , c ) 750 ft , d ) 900 ft , e ) 1050 ft | e | multiply(30, add(divide(multiply(30, divide(const_10, const_2)), const_3), divide(const_10, const_2))) | divide(const_10,const_2)|multiply(n0,#0)|divide(#1,const_3)|add(#2,#0)|multiply(n0,#3)| | physics |
a frog can climb up a well at 3 ft per min but due to slipperiness of the well , frog slips down 2 ft before it starts climbing the next minute . if the depth of the well is 57 ft , how much time will the frog take to reach the top ? | explanation : as per given , in 1 min , frog climbs up 3 ft and slips down by 2 ft . so the frog climbs only 1 ft in 1 min so after 54 mins , it would have climbed 54 ft . at the end of 55 mins it climbs up 3 ft to make it 57 ft and come out of the well . once it had reached the destination , it will not slip . so the ... | a ) 29 , b ) 27 , c ) 55 , d ) 17 , e ) 10 | c | divide(57, subtract(3, 2)) | subtract(n0,n1)|divide(n2,#0) | general |
if 50 % of x is 20 less than 5 % of 500 , then x is ? | 50 % of x = x / 2 ; 5 % of 500 = 5 / 100 * 500 = 25 given that , x / 2 = 25 - 20 = > x / 2 = 5 = > x = 10 . answer : e | a ) 50 , b ) 100 , c ) 75 , d ) 25 , e ) 10 | e | divide(subtract(multiply(500, divide(5, const_100)), 20), divide(50, const_100)) | divide(n2,const_100)|divide(n0,const_100)|multiply(n3,#0)|subtract(#2,n1)|divide(#3,#1) | general |
a rectangular room 14 m long , 12 m broad is surrounded by a varandah , 3 m wide . find the area of the varandah ? | area of varandah = ( l + b + 2 p ) 2 p = ( 14 + 12 + 6 ) 6 = 180 m ( power ) 2 answer is b . | ['a ) 170', 'b ) 180', 'c ) 190', 'd ) 210', 'e ) 150'] | b | multiply(add(14, multiply(const_2, divide(3, const_10))), add(12, multiply(const_2, divide(3, const_10)))) | divide(n2,const_10)|multiply(#0,const_2)|add(n0,#1)|add(n1,#1)|multiply(#2,#3) | geometry |
calculate the largest 4 digit number which is exactly divisible by 98 ? | largest 4 digit number is 9999 after doing 9999 Γ· 98 we get remainder 3 hence largest 4 digit number exactly divisible by 98 = 9999 - 3 = 9996 e | a ) 9800 , b ) 9898 , c ) 9702 , d ) 9604 , e ) 9996 | e | multiply(add(const_100, const_2), 98) | add(const_100,const_2)|multiply(n1,#0) | general |
20 6 years ago , a father was 3 times as old as his son . now the father is only twice as old as his son . then the sum of the present ages of the son and the father is : | let the present ages of the father and son be 2 x and x years respectively . then , ( 2 x - 26 ) = 3 ( x - 26 ) x = 52 . required sum = ( 2 x + x ) = 3 x = 156 years . answer is b | a ) 120 , b ) 156 , c ) 108 , d ) 105 , e ) 86 | b | add(multiply(20, 6), multiply(multiply(6, 3), const_2)) | multiply(n0,n1)|multiply(n1,n2)|multiply(#1,const_2)|add(#0,#2) | general |
the compound and the simple interests on a certain sum at the same rate of interest for two years are rs . 11730 and rs . 10200 respectively . find the sum ? | "the simple interest for the first year is 10200 / 2 is rs . 5100 and compound interest for first year also is rs . 5100 . the compound interest for second year on rs . 5100 for one year so rate of the interest = ( 100 * 1530 ) / ( 5100 * 1 ) = 30 % p . a . so p = ( 100 * 10200 ) / ( 30 * 2 ) = rs . 17000 answer : c" | a ) rs . 17029 , b ) rs . 17028 , c ) rs . 17000 , d ) rs . 17008 , e ) rs . 17067 | c | divide(multiply(const_100, 10200), multiply(divide(multiply(subtract(11730, 10200), const_100), divide(10200, const_2)), const_2)) | divide(n1,const_2)|multiply(n1,const_100)|subtract(n0,n1)|multiply(#2,const_100)|divide(#3,#0)|multiply(#4,const_2)|divide(#1,#5)| | gain |
x varies inversely as square of y . given that y = 3 for x = 1 . the value of x for y = 6 will be equal to : | "explanation : solution : given x = k / y ^ 2 , where k is constant . now , y = 3 and x = 1 gives k = 9 . . ' . x = 9 / y ^ 2 = > x = 9 / 6 ^ 2 = 1 / 4 answer : c" | a ) 3 , b ) 6 , c ) 1 / 4 , d ) 1 / 3 , e ) 9 | c | divide(multiply(1, power(3, const_2)), power(6, const_2)) | power(n0,const_2)|power(n2,const_2)|multiply(n1,#0)|divide(#2,#1)| | general |
each child has 3 crayons and 12 apples . if there are 6 children , how many crayons are there in total ? | 3 * 6 = 18 . answer is c . | a ) 22 , b ) 65 , c ) 18 , d ) 36 , e ) 10 | c | multiply(6, 3) | multiply(n0,n2)| | general |
a man can row 6 kmph in still water . when the river is running at 1.2 kmph , it takes him 1 hour to row to a place and black . what is the total distance traveled by the man ? | "m = 6 s = 1.2 ds = 7.2 us = 4.8 x / 7.2 + x / 4.8 = 1 x = 2.88 d = 2.88 * 2 = 5.76 . answer : c" | a ) 5.79 , b ) 5.77 , c ) 5.76 , d ) 5.72 , e ) 5.71 | c | multiply(divide(multiply(add(6, 1.2), subtract(6, 1.2)), add(add(6, 1.2), subtract(6, 1.2))), const_2) | add(n0,n1)|subtract(n0,n1)|add(#0,#1)|multiply(#0,#1)|divide(#3,#2)|multiply(#4,const_2)| | physics |
if the l . c . m of two numbers is 660 and their product is 18480 , find the h . c . f of the numbers . | h . c . f = ( product of the numbers ) / ( their l . c . m ) = 18480 / 660 = 28 . answer : e | a ) 50 , b ) 30 , c ) 125 , d ) 25 , e ) none of these | e | divide(18480, 660) | divide(n1,n0) | physics |
a train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds . its length is : | "explanation : let the length of the train be x meters and its speed be y m / sec . they , x / y = 15 = > y = x / 15 x + 100 / 25 = x / 15 x = 150 m . answer is d" | a ) 100 m , b ) 125 m , c ) 130 m , d ) 150 m , e ) 160 m | d | multiply(100, subtract(const_2, const_1)) | subtract(const_2,const_1)|multiply(n1,#0)| | physics |
show timer statistics to apply for the position of photographer at a local magazine , veronica needs to include 3 or 4 of her pictures in an envelope accompanying her application . if she has pre - selected 5 photos representative of her work , how many choices does she have to provide the photos for the magazine ? | veronica can choose 3 photos in 5 c 3 = 10 ways she can choose 4 photos in 5 c 4 = 5 ways total number of ways = 10 + 5 = 15 answer : d | a ) 5 , b ) 10 , c ) 12 , d ) 15 , e ) 50 | d | add(divide(factorial(5), multiply(factorial(3), factorial(subtract(5, 3)))), divide(factorial(5), multiply(factorial(4), factorial(const_1)))) | factorial(n2)|factorial(n0)|factorial(n1)|factorial(const_1)|subtract(n2,n0)|factorial(#4)|multiply(#2,#3)|divide(#0,#6)|multiply(#1,#5)|divide(#0,#8)|add(#9,#7) | physics |
a train 300 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 300 + x ) / 60 = 65 / 3 x = 1000 m . answer : e" | a ) 298 m , b ) 268 m , c ) 500 m , d ) 267 m , e ) 1000 m | e | divide(300, multiply(subtract(78, 1), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
a consignment of 20 picture tubes contains 5 defectives . two tubes are selected one after the other at random . the probability that both are defective assuming that the first tube is not replaced before drawing the second , would be : | probability of drawing a defective tube in first draw is 5 / 20 probability of drawing a defective tube in second draw ( without replacing first tube ) is 4 / 19 . therefore probability of getting both defective is = ( 5 / 20 ) * ( 4 / 19 ) = 1 / 19 answer : b | a ) 1 / 16 , b ) 1 / 19 , c ) 1 / 4 , d ) 1 / 3 , e ) none of these | b | divide(choose(const_5, const_2), choose(20, const_2)) | choose(const_5,const_2)|choose(n0,const_2)|divide(#0,#1) | other |
two goods trains each 125 m long are running in opposite directions on parallel tracks . their speeds are 45 km / hr and 30 km / hr respectively . find the time taken by the slower train to pass the driver of the faster one ? | "relative speed = 45 + 30 = 75 km / hr . 75 * 5 / 18 = 125 / 6 m / sec . distance covered = 125 + 125 = 250 m . required time = 250 * 6 / 125 = 12 sec . answer : option b" | a ) 40 , b ) 12 , c ) 48 , d ) 51 , e ) 44 | b | add(45, 30) | add(n1,n2)| | physics |
a sum of $ 500 amounts to 600 in 2 years at simple interest . if the interest rate is increased by 10 % it would amount to how much ? | s . i = 600 - 500 = 100 p = $ 500 t = 2 years r = 100 * 100 / 500 * 2 = 10 % new rate = 10 + 10 = 20 % new s . i . = 500 * 20 * 2 / 100 = $ 200 new amount = 500 + 200 = $ 700 answer is b | a ) $ 500 , b ) $ 700 , c ) $ 800 , d ) $ 600 , e ) $ 300 | b | add(600, multiply(500, divide(multiply(2, 10), const_100))) | multiply(n2,n3)|divide(#0,const_100)|multiply(n0,#1)|add(n1,#2) | gain |
if p = 775 Γ 778 Γ 781 , what is the remainder when p is divided by 14 ? | "p 775 / 14 leaves a remainder 5 778 / 14 leaves a remainder 8 781 / 14 leaves a remainder 11 5 * 8 * 11 = 440 so the remainder will be the remainder of 440 / 14 which is 6 ans a" | a ) 6 , b ) 7 , c ) 8 , d ) 9 , e ) 10 | a | divide(multiply(multiply(divide(778, divide(14, add(const_4, const_1))), 781), divide(775, add(const_4, const_1))), multiply(const_100, const_3)) | add(const_1,const_4)|multiply(const_100,const_3)|divide(n0,#0)|divide(n3,#0)|divide(n1,#3)|multiply(n2,#4)|multiply(#2,#5)|divide(#6,#1)| | general |
the present ratio of students to teachers at a certain school is 70 to 1 . if the student enrollment were to increase by 50 students and the number of teachers were to increase by 5 , the ratio of students to teachers would then be 25 to 1 . what is the present number of teachers ? | "we are given that the ratio of students to teacher is 70 to 1 . we can rewrite this using variable multipliers . students : teachers = 70 x : x we are next given that student enrollment increases by 50 and the number of teachers increases by 5 . with this change the new ratio becomes 25 to 1 . we can put all this into... | a ) 2 , b ) 8 , c ) 10 , d ) 12 , e ) 15 | a | divide(add(70, 25), 25) | add(n0,n4)|divide(#0,n4)| | other |
find out the square of a number which when doubled exceeds its one fifth by 9 ? | a let the number be p , then the square will be p ^ 2 according to question : 2 p = ( p / 5 ) + 9 = > 10 p = p + 45 = > p = 5 p ^ 2 = 5 ^ 2 = 25 . answer : b | a ) 16 , b ) 25 , c ) 19 , d ) 26 , e ) 17 | b | power(divide(9, subtract(const_2, divide(const_1, add(const_4, const_1)))), const_2) | add(const_1,const_4)|divide(const_1,#0)|subtract(const_2,#1)|divide(n0,#2)|power(#3,const_2) | general |
a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 50 paisa . if the share of y is rs . 18 , what is the total amount ? | "x : y : z = 100 : 45 : 50 20 : 9 : 10 9 - - - 18 39 - - - ? = > 78 answer : d" | a ) 115 , b ) 116 , c ) 117 , d ) 78 , e ) 119 | d | add(add(multiply(divide(const_100, 45), 18), multiply(divide(50, 45), 18)), 18) | divide(const_100,n0)|divide(n1,n0)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|add(n2,#4)| | general |
the l . c . m . of 2 numbers is 30 . the numbers are in the ratio 2 : 3 . find their sum ? | "let the numbers be 2 x and 3 x l . c . m . = 6 x 6 x = 30 x = 5 the numbers are = 10 and 15 required sum = 10 + 15 = 25 answer is b" | a ) 36 , b ) 25 , c ) 48 , d ) 32 , e ) 56 | b | add(multiply(divide(2, multiply(30, 2)), 2), multiply(divide(2, multiply(30, 2)), 30)) | multiply(n1,n2)|divide(n0,#0)|multiply(n2,#1)|multiply(n1,#1)|add(#2,#3)| | other |
if 15 horses eat 15 bags of gram in 15 days , in how many days will one horse eat one bag of grain ? | horse * days = bags 15 * 15 = 15 and 1 * days = 1 ( 15 * 15 ) / ( 1 * days ) = 15 / 1 days = 15 answer : a | a ) 15 days , b ) 1 / 15 days , c ) 1 day , d ) 30 days , e ) 225 days | a | inverse(inverse(15)) | inverse(n0)|inverse(#0) | physics |
in a certain game , each player scores either 2 points or 5 points . if n players score 2 points and m players score 5 points , and the total number of points scored is 50 , what is the least possible positive t difference between n and m ? | "we have equation 2 n + 5 m = 50 we have factor 2 in first number and we have factor 5 in second number . lcm ( 2 , 5 ) = 10 so we can try some numbers and we should start from 5 because it will be less list than for 2 2 * 5 = 10 and n should be equal 20 4 * 5 = 20 and n should be equal 15 6 * 5 = 30 and n should be eq... | a ) 1 , b ) 3 , c ) 5 , d ) 7 , e ) 9 | b | subtract(5, 2) | subtract(n1,n0)| | general |
if the average of 201 , 202 , 204 , 205 , 206 , 209 , 209 , 210 , 212 and x is 207 , what is the value of x ? | sum of the deviations of the numbers in the set from the mean is always zero 201 , 202 , 204 , 205 , 206 , 209 , 209 , 210 , 212 mean is 207 so the list is - 6 - 5 - 3 - 2 - 1 + 2 + 2 + 3 + 5 . . . this shud total to zero but this is - 5 , hence we need a number that is 5 more than the mean to get a + 5 and make it zer... | a ) 207 , b ) 209 , c ) 211 , d ) 212 , e ) 213 | d | subtract(multiply(207, const_10), add(add(add(add(add(add(add(add(201, 202), 204), 205), 206), 209), 209), 210), 212)) | add(n0,n1)|multiply(n9,const_10)|add(n2,#0)|add(n3,#2)|add(n4,#3)|add(n5,#4)|add(n5,#5)|add(n7,#6)|add(n8,#7)|subtract(#1,#8) | general |
in a group of cows and chickens , the number of legs was 20 more than twice the number of heads . the number of cows was : | "let the number of cows be x and their legs be 4 x . let the number of chicken be y and their legs be 2 x . total number of legs = 4 x + 2 y . total number of heads = x + y . the number of legs was 20 more than twice the number of heads . therefore , 2 Γ ( x + y ) + 20 = 4 x + 2 y . or , 2 x + 2 y + 20 = 4 x + 2 y . or... | a ) 5 , b ) 7 , c ) 10 , d ) 12 , e ) 14 | c | divide(20, subtract(const_4, const_2)) | subtract(const_4,const_2)|divide(n0,#0)| | general |
the average marks in mathematics scored by the pupils of a school at the public examination were 39 . if 4 of these pupils who actually scored 25 , 12 , 15 and 19 marks at the examination had not been sent up , the average marks for the school would have been 44 . find the number of pupils sent up for examination from ... | 39 x = 25 + 12 + 15 + 19 + ( x β 4 ) 44 39 x = 71 + 44 x - 176 5 x = 105 x = 21 answer : b | a ) 20 , b ) 21 , c ) 22 , d ) 23 , e ) 24 | b | divide(subtract(multiply(44, 4), add(add(add(25, 12), 15), 19)), subtract(44, 39)) | add(n2,n3)|multiply(n1,n6)|subtract(n6,n0)|add(n4,#0)|add(n5,#3)|subtract(#1,#4)|divide(#5,#2) | general |
amanda sees a sale for 30 % off all items , she sees a dress on sale that originally cost $ 50 . how much will it cost amanda to buy the dress after the sale amount of 30 % has been take off ? | final number = original number - 30 % ( original number ) = 50 - 30 % ( 50 ) = 50 - 15 = $ 35 . answer b | a ) $ 40 , b ) $ 35 , c ) $ 50 , d ) $ 65 , e ) $ 15 | b | subtract(50, multiply(divide(30, const_100), 50)) | divide(n0,const_100)|multiply(n1,#0)|subtract(n1,#1) | gain |
a cricket bat is sold for $ 900 , making a profit of $ 150 . the profit percentage would be | 150 / ( 900 - 150 ) = 150 / 750 = 0.2 = 20 % . answer : b | a ) 24 % , b ) 20 % , c ) 30 % , d ) 36 % , e ) 40 % | b | multiply(divide(150, subtract(900, 150)), const_100) | subtract(n0,n1)|divide(n1,#0)|multiply(#1,const_100) | gain |
of the 3,600 employees of company x , 1 / 3 are clerical . if the clerical staff were to be reduced by 1 / 4 , what percent of the total number of the remaining employees would then be clerical ? | "welcome , just post the question and the choices let ' s see , the way i did it was 1 / 3 are clerical out of 3600 so 1200 are clerical 1200 reduced by 1 / 4 is 1200 * 1 / 4 so it reduced 300 people , so there is 900 clerical people left but since 300 people left , it also reduced from the total of 3600 so there are 3... | a ) 25 % , b ) 28.1 % , c ) 20 % , d ) 12.5 % , e ) 11.1 % | b | multiply(divide(multiply(divide(1, 3), subtract(1, divide(1, 4))), add(multiply(divide(1, 3), subtract(1, divide(1, 4))), subtract(1, divide(1, 3)))), const_100) | divide(n1,n2)|divide(n3,n4)|subtract(n3,#1)|subtract(n3,#0)|multiply(#0,#2)|add(#4,#3)|divide(#4,#5)|multiply(#6,const_100)| | general |
raja spends 60 % of his monthly income on household items , 10 % of his monthly income on buying cloths , 10 % of his monthly income on medicines and saves the remaining amount which is rs . 5000 . find his monthly income . | "savings 20 % - 5000 expenditure 80 % - 20000 total - 25000 answer : d" | a ) rs . 40000 , b ) rs . 36000 , c ) rs . 50000 , d ) rs . 25000 , e ) none of these | d | divide(5000, subtract(const_1, add(add(divide(60, const_100), divide(10, const_100)), divide(10, const_100)))) | divide(n0,const_100)|divide(n1,const_100)|divide(n2,const_100)|add(#0,#1)|add(#3,#2)|subtract(const_1,#4)|divide(n3,#5)| | gain |
22.085 Γ£ β 0.001 = ? | "22.085 Γ£ β 0.001 = ? or , ? = 0.022085 answer c" | a ) 0.22085 , b ) 2.2085 , c ) 0.022085 , d ) 0.0022085 , e ) none of these | c | multiply(divide(22.085, 0.001), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
0.99999 + 0.11111 = ? | "0.99999 + 0.00001 = 1 0.1111 + 1 = 1.1111 correct answer is e ." | a ) 1 , b ) 1.0001 , c ) 1.0021 , d ) 1.111 , e ) 1.1111 | e | multiply(divide(0.99999, 0.11111), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
a certain company has records stored with a record storage firm in 15 - inch by 12 - inch by 10 - inch boxes . the boxes occupy 1.08 million cubic inches of space . if the company pays $ 0.8 per box per month for the record storage , what is the total amount that the company pays each month for record storage ? | "volume per box : 15 x 12 x 10 = 1,800 total volume : 1 , 080,000 number of boxes : total volume / volume per box = 1 , 080,000 / 1,800 = 600 price per month : number of boxes * price per box = 600 * 0.8 = 480 answer : a" | a ) a . 480 , b ) b . 300 , c ) c . 600 , d ) d . 410 , e ) e . 240 | a | multiply(0.8, divide(1.08, divide(divide(divide(multiply(multiply(15, 12), 10), const_100), const_100), const_100))) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_100)|divide(#2,const_100)|divide(#3,const_100)|divide(n3,#4)|multiply(n4,#5)| | general |
the sector of a circle has radius of 35 cm and central angle 135 o . find its perimeter ? | "perimeter of the sector = length of the arc + 2 ( radius ) = ( 135 / 360 * 2 * 22 / 7 * 35 ) + 2 ( 35 ) = 82.5 + 70 = 152.5 cm answer : c" | a ) 91.5 cm , b ) 92.2 cm , c ) 152.5 cm , d ) 29.2 cm , e ) 98.2 cm | c | multiply(multiply(const_2, divide(multiply(subtract(35, const_3), const_2), add(const_4, const_3))), 35) | add(const_3,const_4)|subtract(n0,const_3)|multiply(#1,const_2)|divide(#2,#0)|multiply(#3,const_2)|multiply(n0,#4)| | physics |
for every even positive integer m , f ( m ) represents the product of all even integers from 2 to m , inclusive . for example , f ( 12 ) = 2 x 4 x 6 x 8 x 10 x 12 . what is the greatest prime factor of f ( 26 ) ? | "f ( 26 ) = 2 * 4 * 6 * 8 * 10 * 12 * 14 * 16 * 18 * 20 * 22 * 24 * 26 the greatest prime factor in this list is 13 . the answer is d ." | a ) 23 , b ) 19 , c ) 17 , d ) 13 , e ) 11 | d | subtract(divide(26, 2), const_1) | divide(n8,n0)|subtract(#0,const_1)| | general |
what is the remainder when 6 ^ 381 is divided by 5 ? | i also agree that the remainder is ' 1 ' ( using the last digit of the powers of 7 ) . could we have the official answer please ? b | a ) 0 , b ) 1 , c ) 2 , d ) 3 , e ) 4 | b | subtract(divide(5, const_2), multiply(6, 6)) | divide(n2,const_2)|multiply(n0,n0)|subtract(#0,#1)| | general |
two trains of equal length are running on parallel lines in the same directions at 46 km / hr . and 36 km / hr . the faster trains pass the slower train in 108 seconds . the length of each train is : | "explanation : the relative speed of train is 46 - 36 = 10 km / hr = ( 10 x 5 ) / 18 = 25 / 9 m / s 10 Γ 518 = 259 m / s in 108 secs the total distance traveled is 108 x 25 / 9 = 300 m . therefore the length of each train is = 300 / 2 = 150 m . answer d" | a ) 82 m , b ) 50 m , c ) 72 m , d ) 150 m , e ) none of these | d | divide(multiply(divide(multiply(subtract(46, 36), const_1000), const_3600), 108), const_2) | subtract(n0,n1)|multiply(#0,const_1000)|divide(#1,const_3600)|multiply(n2,#2)|divide(#3,const_2)| | general |
a team won 40 percent of its first 30 games in a particular season , and 80 percent of its remaining games . if the team won a total of 50 percent of its games that season , what was the total number of games that the team played ? | 50 % is 10 % - points above 40 % and 30 % - points below 80 % . thus the ratio of ` ` the first 30 games ' ' to ` ` remaining games ' ' is 3 : 1 . so the team played a total of 30 + 10 = 40 games . the answer is a . | a ) 40 , b ) 50 , c ) 60 , d ) 70 , e ) 80 | a | divide(multiply(30, divide(40, const_100)), subtract(divide(80, const_100), divide(50, const_100))) | divide(n0,const_100)|divide(n2,const_100)|divide(n3,const_100)|multiply(n1,#0)|subtract(#1,#2)|divide(#3,#4) | gain |
the parameter of a square is equal to the perimeter of a rectangle of length 22 cm and breadth 16 cm . find the circumference of a semicircle whose diameter is equal to the side of the square . ( round off your answer to two decimal places ) | "let the side of the square be a cm . parameter of the rectangle = 2 ( 22 + 16 ) = 76 cm parameter of the square = 76 cm i . e . 4 a = 76 a = 19 diameter of the semicircle = 19 cm circimference of the semicircle = 1 / 2 ( β ) ( 19 ) = 1 / 2 ( 22 / 7 ) ( 19 ) = 418 / 14 = 29.85 cm to two decimal places answer : d" | a ) 77.14 cm , b ) 47.14 cm , c ) 84.92 cm , d ) 29.85 cm , e ) 23.57 cm | d | divide(circumface(divide(square_edge_by_perimeter(rectangle_perimeter(22, 16)), const_2)), const_2) | rectangle_perimeter(n0,n1)|square_edge_by_perimeter(#0)|divide(#1,const_2)|circumface(#2)|divide(#3,const_2)| | geometry |
the ratio of the arithmetic mean of two numbers to one of the numbers is 5 : 9 . what is the ratio of the smaller number to the larger number ? | "for two numbers , the arithmetic mean is the middle of the two numbers . the ratio of the mean to the larger number is 5 : 9 , thus the smaller number must have a ratio of 1 . the ratio of the smaller number to the larger number is 1 : 9 . the answer is e ." | a ) 1 : 5 , b ) 1 : 6 , c ) 1 : 7 , d ) 1 : 8 , e ) 1 : 9 | e | multiply(subtract(divide(5, 9), divide(const_1, const_2)), const_2) | divide(n0,n1)|divide(const_1,const_2)|subtract(#0,#1)|multiply(#2,const_2)| | other |
if | 5 x - 10 | = 100 , then find the sum of the values of x ? | "| 5 x - 10 | = 100 5 x - 10 = 100 or 5 x - 10 = - 100 5 x = 110 or 5 x = - 90 x = 22 or x = - 18 sum = 21 - 18 = 3 answer is c" | a ) 1 , b ) - 2 , c ) 3 , d ) - 3 , e ) 4 | c | subtract(subtract(subtract(100, 10), add(100, 10)), 10) | add(n1,n2)|subtract(n2,n1)|subtract(#1,#0)|subtract(#2,n1)| | general |
a is twice as good a workman as b and together they finish a piece of work in 18 days . in how many days will a alone finish the work ? | "if a takes x days to do a work then b takes 2 x days to do the same work . - - > 1 / x + 1 / 2 x = 1 / 18 - - > 3 / 2 x = 1 / 18 - - > x = 27 days . hence , a alone can finish the work in 27 days . answer : c ." | a ) 31 days , b ) 25 days , c ) 27 days , d ) 29 days , e ) 19 days | c | multiply(add(const_1, const_2), 18) | add(const_1,const_2)|multiply(n0,#0)| | physics |
it takes avery 2.5 hours to build a brick wall while tom can do it in 5 hours . if the two start working together and after an hour avery leaves , how much time will it take tom to complete the wall on his own ? | "avery ' s efficiency is 100 / 2.5 = 40 % tom ' s = 100 / 5 = 20 % they worked together for 1 hour and finished 60 % of the job remaining = 40 % tom will complete 20 % in 60 minutes , 40 % in 120 minutes time taken by tom to finish the remaining on his own = 120 minutes answer : e" | a ) 15 minutes . , b ) 30 minutes . , c ) 1 hour and 30 minutes . , d ) 1 hour and 40 minutes , e ) 2 hours | e | multiply(divide(subtract(const_1, add(divide(const_1, 2.5), divide(const_1, 5))), divide(const_1, 5)), const_60) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|subtract(const_1,#2)|divide(#3,#1)|multiply(#4,const_60)| | physics |
a , b and c invests rs . 4000 , rs . 2000 and rs . 6000 in a business . after one year a removed his money ; b and c continued the business for two more year . if the net profit after 3 years be rs . 3300 , then a ' s share in the profit is ? | 4 * 12 : 2 * 36 : 4 * 36 2 : 3 : 6 2 / 11 * 3300 = 600 answer : d | a ) 6000 , b ) 4000 , c ) 2000 , d ) 600 , e ) 300 | d | multiply(divide(4000, add(4000, add(multiply(2000, 3), multiply(6000, 3)))), 4000) | multiply(n1,n3)|multiply(n2,n3)|add(#0,#1)|add(n0,#2)|divide(n0,#3)|multiply(n0,#4) | gain |
the avg weight of a , b & c is 80 kg . if d joins the group , the avg weight of the group becomes 82 kg . if another man e who weights is 3 kg more than d replaces a , then the avgof b , c , d & e becomes 81 kg . what is the weight of a ? | a + b + c = 3 * 80 = 240 a + b + c + d = 4 * 82 = 328 - - - - ( i ) so , d = 88 & e = 88 + 3 = 91 b + c + d + e = 81 * 4 = 324 - - - ( ii ) from eq . ( i ) & ( ii ) a - e = 328 β 324 = 4 a = e + 4 = 91 + 4 = 95 answer : e | a ) 56 , b ) 65 , c ) 75 , d ) 89 , e ) 95 | e | subtract(multiply(82, const_4), subtract(multiply(81, const_4), add(3, subtract(multiply(82, const_4), multiply(80, const_3))))) | multiply(n1,const_4)|multiply(n3,const_4)|multiply(n0,const_3)|subtract(#0,#2)|add(n2,#3)|subtract(#1,#4)|subtract(#0,#5) | general |
the base of a triangle is 15 cm and height is 12 cm . the height of another triangle of double the area having the base 20 cm is : | explanation : area of triangle , a 1 = 1 / 2 β base β height = 1 / 2 β 15 β 12 = 90 cm 2 area of second triangle = 2 β a 1 = 180 cm 2 1 / 2 β 20 β height = 180 = > height = 18 cm option c | ['a ) 22 cm', 'b ) 20 cm', 'c ) 18 cm', 'd ) 10 cm', 'e ) none of these'] | c | divide(multiply(multiply(const_2, triangle_area(15, 12)), const_2), 20) | triangle_area(n0,n1)|multiply(#0,const_2)|multiply(#1,const_2)|divide(#2,n2) | geometry |
rectangular tile each of size 20 cm by 30 cm must be laid horizontally on a rectangular floor of size 100 cm by 150 cm , such that the tiles do not overlap and they are placed with edges jutting against each other on all edges . a tile can be placed in any orientation so long as its edges are parallel to the edges of f... | "area of tile = 20 * 30 = 600 area of floor = 100 * 150 = 15000 no of tiles = 15000 / 600 = 25 so , the no of tile = 25 answer : c" | a ) 40 , b ) 15 , c ) 25 , d ) 32 , e ) 45 | c | divide(multiply(100, 150), multiply(20, 30)) | multiply(n2,n3)|multiply(n0,n1)|divide(#0,#1)| | geometry |
5000 was divided into two parts such a way that when first part was invested at 3 % and the second at 5 % , the whole annual interest from both the investments is rs . 144 , how much was put at 3 % ? | "( x * 3 * 1 ) / 100 + [ ( 5000 - x ) * 5 * 1 ] / 100 = 144 3 x / 100 + ( 25000 β 5 x ) / 100 = 144 = > x = 10600 answer : a" | a ) 10600 , b ) 11600 , c ) 12600 , d ) 10680 , e ) 20600 | a | divide(subtract(multiply(divide(5, const_100), 5000), 144), subtract(divide(5, const_100), divide(3, const_100))) | divide(n2,const_100)|divide(n1,const_100)|multiply(n0,#0)|subtract(#0,#1)|subtract(#2,n3)|divide(#4,#3)| | gain |
if x is 30 percent more than y and y is 40 percent less than z , then x is what percent of z ? | "z = 100 ; y = 60 so x = 78 x as % of z = 78 / 100 * 100 = > 78 % answer will be ( a )" | a ) 78 % , b ) 250 % , c ) 500 / 3 % , d ) 125 % , e ) 60 % | a | multiply(add(const_1, divide(30, const_100)), subtract(const_100, 40)) | divide(n0,const_100)|subtract(const_100,n1)|add(#0,const_1)|multiply(#2,#1)| | general |
a train running at the speed of 60 km / hr crosses a pole in 12 seconds . find the length of the train . | ": speed = 60 * ( 5 / 18 ) m / sec = 50 / 3 m / sec length of train ( distance ) = speed * time ( 50 / 3 ) * 12 = 200 meter answer : d" | a ) 150 , b ) 278 , c ) 179 , d ) 200 , e ) 191 | d | multiply(divide(multiply(60, const_1000), const_3600), 12) | multiply(n0,const_1000)|divide(#0,const_3600)|multiply(n1,#1)| | physics |
if y > 0 , ( 7 y ) / 20 + ( 3 y ) / 10 is what percent of y ? | "can be reduced to 7 y / 20 + 6 y / 20 = 13 y / 20 = 65 % answer c" | a ) 40 % , b ) 50 % , c ) 65 % , d ) 70 % , e ) 80 % | c | multiply(const_100, add(divide(7, 20), divide(3, 10))) | divide(n1,n2)|divide(n3,n4)|add(#0,#1)|multiply(#2,const_100)| | general |
the value of a machine depreciates at 21 % per annum . if its present value is $ 1 , 50,000 , at what price should it be sold after two years such that a profit of $ 24,000 is made ? | "the value of the machine after two years = 0.79 * 0.79 * 1 , 50,000 = $ 96,000 sp such that a profit of $ 24,000 is made = 93,615 + 24,000 = $ 1 , 17,615 d" | a ) $ 257615 , b ) $ 437615 , c ) $ 127615 , d ) $ 117615 , e ) $ 157615 | d | add(multiply(multiply(subtract(1, divide(21, const_100)), subtract(1, divide(21, const_100))), add(multiply(multiply(const_100, const_100), sqrt(const_100)), multiply(multiply(divide(sqrt(const_100), const_2), const_100), const_100))), multiply(multiply(add(21, const_2), const_100), sqrt(const_100))) | add(n0,const_2)|divide(n0,const_100)|multiply(const_100,const_100)|sqrt(const_100)|divide(#3,const_2)|multiply(#2,#3)|multiply(#0,const_100)|subtract(n1,#1)|multiply(#4,const_100)|multiply(#7,#7)|multiply(#6,#3)|multiply(#8,const_100)|add(#5,#11)|multiply(#12,#9)|add(#13,#10)| | gain |
what is the cp of rs 100 stock at 9 discount , with 1 / 5 % brokerage ? | "explanation : use the formula , cp = 100 Γ’ β¬ β discount + brokerage % cp = 100 - 9 + 1 / 5 91.2 thus the cp is rs 91.2 . answer : c" | a ) 96.9 , b ) 96.3 , c ) 91.2 , d ) 96.7 , e ) 96.21 | c | add(subtract(100, 9), divide(1, 5)) | divide(n2,n3)|subtract(n0,n1)|add(#0,#1)| | gain |
the speeds of three asteroids were compared . asteroids x - 13 and y - 14 were observed for identical durations , while asteroid z - 15 was observed for 2 seconds longer . during its period of observation , asteroid y - 14 traveled three times the distance x - 13 traveled , and therefore y - 14 was found to be faster t... | x 13 : ( t , d , s ) y 14 : ( t , 3 d , s + 8000 mi / hour ) z 15 : ( t + 2 seconds , s , 5 d ) d = ? distance = speed * time x 13 : d = s * t x 14 : 3 d = ( s + 8000 ) * t = = = > 3 d = ts + 8000 t z 15 : 5 d = s * ( t + 2 t ) = = = > 5 d = st + 2 st = = = > 5 d - 2 st = st 3 d = 5 d - 2 st + 8000 t - 2 d = - 2 st + 8... | a ) 500 , b ) 1,600 / 3 , c ) 1,000 , d ) 1,500 , e ) 2,000 | e | multiply(divide(8000, const_2), divide(const_1, const_2)) | divide(n8,const_2)|divide(const_1,const_2)|multiply(#0,#1) | physics |
the average of 15 numbers is 75 . average of the first 8 of them is 68 and that of the last 8.00001 is 77 . find the 8.000001 th number ? | sum of all the 15 numbers = 15 * 75 = 1,125 sum of the first 8 of them = 8 * 68 = 544 sum of the last 8 of them = 8 * 77 = 616 so , the 8 th number = 544 + 616 - 1,125 = 35 . answer : c | a ) 51 , b ) 87 , c ) 35 , d ) 75 , e ) 68 | c | subtract(multiply(8.000001, 77), subtract(multiply(15, 75), multiply(68, 8))) | multiply(n5,n6)|multiply(n0,n1)|multiply(n2,n3)|subtract(#1,#2)|subtract(#0,#3) | general |
a train covers a distance of 12 km in 10 minutes . if it takes 5 seconds to pass a telegraph post , then the length of the train is | "explanation : speed = 12 / 10 x 60 km / hr = 72 x 5 / 18 m / sec = 20 m / sec . length of the train = ( speed x time ) = ( 20 x 5 ) m = 100 m answer : option e" | a ) 110 m , b ) 120 m , c ) 140 m , d ) 160 m , e ) 100 cm | e | divide(12, subtract(divide(12, 10), 5)) | divide(n0,n1)|subtract(#0,n2)|divide(n0,#1)| | physics |
mr yadav spends 60 % of his monthly salary on consumable items and 50 % of the remaining on clothes and transport . he saves the remaining amount . if his savings at the end of the year were 48456 , how much amount per month would he have spent on clothes and transport ? | β΅ amount , he have spent in 1 month on clothes transport = amount spent on saving per month β΅ amount , spent on clothes and transport = 48456 β 12 = 4038 answer a | a ) 4038 , b ) 8076 , c ) 9691.2 , d ) 4845.6 , e ) none of these | a | multiply(divide(divide(48456, divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)), multiply(const_3, const_4)), divide(divide(multiply(subtract(const_100, 60), 50), const_100), const_100)) | multiply(const_3,const_4)|subtract(const_100,n0)|multiply(n1,#1)|divide(#2,const_100)|divide(#3,const_100)|divide(n2,#4)|divide(#5,#0)|multiply(#6,#4) | general |
what is the sum of all the multiples of 10 between 0 and 100 ? | "the multiples of 10 between 0 and 100 are 10 , 20 , 30 , 40 , 50 , 60 , 70 , 80 , 90 and 100 . if these are all added together , the result is 550 . final answer : c" | a ) 500 , b ) 620 , c ) 550 , d ) 340 , e ) 440 | c | add(add(add(add(add(add(const_12, const_2), const_1), add(add(const_12, const_2), add(add(add(add(add(const_2, const_4), const_4), subtract(const_10, const_1)), add(add(const_2, const_4), const_4)), add(const_10, const_2)))), add(add(add(const_12, const_2), const_1), const_1)), 10), 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 field is 7 / 5 its width . if the perimeter of the field is 336 meters , what is the width of the field ? | "let l be the length and w be the width . l = ( 7 / 5 ) w perimeter : 2 l + 2 w = 336 , 2 ( 7 / 5 ) w + 2 w = 336 solve the above equation to find : w = 70 m and l = 98 m . correct answer c ) 70" | a ) 50 , b ) 60 , c ) 70 , d ) 80 , e ) 90 | c | divide(336, add(add(divide(7, 5), divide(7, 5)), const_2)) | divide(n0,n1)|add(#0,#0)|add(#1,const_2)|divide(n2,#2)| | geometry |
difference of two numbers is 1650 . 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 = 1650 5 * y / 3 β y = 1650 2 * y / 3 = 1650 y = [ ( 1650 * 3 ) / 2 ] = 2475 . one number = 2475 , second number = 5 * y / 3 = 4125 . answer e ." | a ) 2660 , 1000 , b ) 3660 , 2000 , c ) 3000 , 4160 , d ) 2490 , 4150 , e ) 2475 , 4125 | e | multiply(divide(7.5, 12.5), divide(1650, subtract(const_1, divide(7.5, 12.5)))) | divide(n1,n2)|subtract(const_1,#0)|divide(n0,#1)|multiply(#0,#2)| | gain |
a certain no . when divided by 140 leaves a remainder 25 , what is the remainder if the same no . be divided by 15 ? | "explanation : 140 + 25 = 165 / 15 = 11 ( remainder ) e" | a ) 2 , b ) 4 , c ) 7 , d ) 8 , e ) 11 | e | subtract(subtract(subtract(140, 25), const_4), const_2) | subtract(n0,n1)|subtract(#0,const_4)|subtract(#1,const_2)| | general |
a room of 6 m 00 cm long and 8 m 00 cm broad is to be paved with square tiles . find the least number of square tiles required to cover the floor . | "explanation : area of the room = 600 * 800 sq cm size of largest square tile = h . c . f of 600 cm and 800 cm = 200 cm area of 1 tile = 200 * 200 sq cm no . of tiles required = ( 600 * 800 ) / ( 200 * 200 ) = 12 answer : a ) 12" | a ) 12 , b ) 200 , c ) 600 , d ) 800 , e ) 20 | a | divide(multiply(add(multiply(6, const_100), 00), add(multiply(8, const_100), 00)), power(divide(add(multiply(6, const_100), 00), power(const_2, const_4)), const_2)) | multiply(n0,const_100)|multiply(n2,const_100)|power(const_2,const_4)|add(n1,#0)|add(n3,#1)|divide(#3,#2)|multiply(#3,#4)|power(#5,const_2)|divide(#6,#7)| | physics |
an inspector rejects 0.07 % of the meters as defective . how many will he examine to reject 2 ? | "let the number of meters to be examined be x then , 0.07 % of x = 2 ( 7 / 100 ) * ( ( 1 / 100 ) * x = 2 x = 2857 answer is c" | a ) a ) 1500 , b ) b ) 2000 , c ) c ) 2857 , d ) d ) 3000 , e ) e ) 3100 | c | divide(multiply(2, const_100), 0.07) | multiply(n1,const_100)|divide(#0,n0)| | gain |
teas worth rs . 126 per kg and rs . 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2 . if the mixture is worth rs 154 per kg , the price of the third variety per kg will be ? | "explanation : since first and second varieties are mixed in equal proportions . so , their average price = rs . ( 126 + 135 ) / 2 . = > rs . 130.50 . so , the mixture is formed by mixing two varieties , one at rs . 130.50 per kg and the other at say , rs . x per kg in the ratio 2 : 2 , i . e . , 1 : 1 . we have to fin... | a ) rs . 147.50 , b ) rs . 785.50 , c ) rs . 176.50 , d ) rs . 258.50 , e ) none of these | c | divide(subtract(multiply(154, add(add(1, 1), 2)), add(126, 135)), 2) | add(n2,n2)|add(n0,n1)|add(n4,#0)|multiply(n5,#2)|subtract(#3,#1)|divide(#4,n4)| | other |
mother , her daughter and her grand child weighs 110 kg . daughter and her daughter ( child ) weighs 60 kg . child is 1 / 5 th of her grand mother . what is the age of the daughter ? | "mother + daughter + child = 110 kg daughter + child = 60 kg mother = 110 - 60 = 50 kg child = 1 / 5 th of mother = ( 1 / 5 ) * 50 = 10 kg so now daughter = 110 - ( mother + child ) = 110 - ( 50 + 10 ) = 50 kg answer : e" | a ) 46 , b ) 47 , c ) 48 , d ) 49 , e ) 50 | e | subtract(60, divide(subtract(110, 60), 5)) | subtract(n0,n1)|divide(#0,n3)|subtract(n1,#1)| | general |
in a simultaneous throw of a pair of dice , find the probability of getting a total more than 10 | "total number of cases = 6 * 6 = 36 favourable cases = [ ( 5,6 ) ( 6,5 ) , ( 6,6 ) ] = 3 so probability = 3 / 36 = 1 / 12 answer is c" | a ) 5 / 13 , b ) 2 / 15 , c ) 1 / 12 , d ) 3 / 17 , e ) 6 / 19 | c | divide(subtract(10, multiply(const_2, const_3)), 10) | multiply(const_2,const_3)|subtract(n0,#0)|divide(#1,n0)| | general |
when a student joe , weighing 42 kg , joins a group of students whose average weight is 30 kg , the average weight goes up by 1 kg . subsequently , if two students , excluding joe , leave the group the average weight comes back to 30 kg . what is the difference between the average weight of the two students who left an... | after two persons leave the group the average remains the same . that means the weight of the two persons = 42 + 30 = 72 so , the average the two persons = 36 that gives the answer 42 - 36 = 6 answer b | a ) 5.5 kg , b ) 6 kg , c ) 30 kg , d ) 36.5 kg , e ) 71 kg | b | subtract(42, divide(subtract(add(multiply(30, subtract(42, add(30, 1))), 42), multiply(subtract(subtract(42, add(30, 1)), const_1), 30)), const_2)) | add(n1,n2)|subtract(n0,#0)|multiply(n1,#1)|subtract(#1,const_1)|add(n0,#2)|multiply(n1,#3)|subtract(#4,#5)|divide(#6,const_2)|subtract(n0,#7) | general |
the sum of two numbers is 44 . 5 times one number is equal to 6 times the other . the smaller of the two numbers is | sol . let the numbers be x and ( 44 - x ) . then , 5 x = 6 ( 44 - x ) β 11 x = 264 β x = 24 . so , the numbers are 24 and 20 . answer e | a ) 10 , b ) 12 , c ) 16 , d ) 22 , e ) 20 | e | multiply(divide(44, add(5, 6)), 5) | add(n1,n2)|divide(n0,#0)|multiply(n1,#1) | general |
youseff lives x blocks from his office . it takes him 1 minute per block to walk to work and 20 seconds per block to ride his bike to work . it is takes him exactly 4 minutes more to walk to work than to ride his bike to work , then x equals ? | "please follow posting guidelines , link is in my signatures . as for your question , x / 60 = blocks / time / block = block ^ 2 / time . this is not what you want . you are given x blocks and 60 seconds per block . thus you need to put it as 60 * x to give you units of seconds as you are equating this to 240 ( which i... | a ) 4 , b ) 6 , c ) 10 , d ) 15 , e ) 20 | b | divide(multiply(4, const_3), const_2) | multiply(n2,const_3)|divide(#0,const_2)| | physics |
a take twice as much time as b or thrice as much time to finish a piece of work . working together , they can finish the work in 5 days . b can do the work alone in ? | suppose a , b and c take x , x / 2 and x / 3 respectively to finish the work . then , ( 1 / x + 2 / x + 3 / x ) = 1 / 5 6 / x = 1 / 5 = > x = 30 so , b takes 15 hours to finish the work . answer : d | a ) 19 , b ) 12 , c ) 11 , d ) 30 , e ) 114 | d | multiply(add(add(const_1, const_2), const_3), 5) | add(const_1,const_2)|add(#0,const_3)|multiply(n0,#1) | physics |
the mass of 1 cubic meter of a substance is 200 kg under certain conditions . what is the volume in cubic centimeters of 1 gram of this substance under these conditions ? ( 1 kg = 1,000 grams and 1 cubic meter = 1 , 000,000 cubic centimeters ) | 200 kg - 1 cubic meter ; 200,000 g - 1 cubic meter ; 200,000 g - 1 , 000,000 cubic centimeters ; 1 g - 1 , 000,000 / 200,000 = 10 / 2 = 5 cubic centimeters . answer : e . | ['a ) 1', 'b ) 2', 'c ) 3', 'd ) 4', 'e ) 5'] | e | divide(multiply(const_1000, const_1000), multiply(200, const_1000)) | multiply(const_1000,const_1000)|multiply(n1,const_1000)|divide(#0,#1) | geometry |
a rectangular field is to be fenced on three sides leaving a side of 8 feet uncovered . if the area of the field is 680 sq . ft , how many feet of fencing will be required ? | "given that length and area , so we can find the breadth . length x breadth = area 8 x breadth = 680 breadth = 85 feet area to be fenced = 2 b + l = 2 ( 85 ) + 8 = 178 feet answer : d ) 178 ft" | a ) 244 ft , b ) 88 ft , c ) 122 ft , d ) 178 ft , e ) 66 ft | d | add(multiply(divide(680, 8), const_2), 8) | divide(n1,n0)|multiply(#0,const_2)|add(n0,#1)| | geometry |
a train 400 m in length crosses a telegraph post in 16 seconds . the speed of the train is ? | "s = 400 / 16 * 18 / 5 = 90 kmph answer : e" | a ) 22 kmph , b ) 77 kmph , c ) 54 kmph , d ) 71 kmph , e ) 90 kmph | e | multiply(const_3_6, divide(400, 16)) | divide(n0,n1)|multiply(#0,const_3_6)| | physics |
ravi can do a piece of work in 50 days while prakash can do it in 75 days . in how many days will they finish it together ? | 1 / 50 + 1 / 75 = 5 / 150 30 / 1 = 30 days answer : a | a ) 30 days , b ) 35 days , c ) 25 days , d ) 27 days , e ) 29 days | a | divide(const_1, add(inverse(50), inverse(75))) | inverse(n0)|inverse(n1)|add(#0,#1)|divide(const_1,#2) | physics |
the marks obtained by polly and sandy are in the ratio 3 : 5 and those obtained by sandy and willy are in the ratio of 5 : 2 . the marks obtained by polly and willy are in the ratio of . . . ? | "polly : sandy = 3 : 5 sandy : willy = 5 : 2 polly : sandy : willy = 3 : 5 : 2 polly : willy = 3 : 2 the answer is b ." | a ) 2 : 1 , b ) 3 : 2 , c ) 4 : 3 , d ) 5 : 4 , e ) 6 : 5 | b | divide(multiply(3, 5), multiply(5, 2)) | multiply(n0,n2)|multiply(n1,n3)|divide(#0,#1)| | other |
add 10 % of 30 and 15 % of 50 . | "10 % of 30 + 15 % of 50 30 * 10 / 100 + 50 * 15 / 100 3 + 7.5 = 10.5 answer b" | a ) 9.5 , b ) 10.5 , c ) 11.5 , d ) 12 , e ) 15 | b | add(divide(multiply(10, 30), const_100), divide(multiply(15, 50), const_100)) | multiply(n0,n1)|multiply(n2,n3)|divide(#0,const_100)|divide(#1,const_100)|add(#2,#3)| | gain |
smita was making a cube with dimensions 5 * 5 * 5 using 1 * 1 * 1 cubes . what is the number of cubes needed to make a hollow cube looking of the same shape ? | cube use to make perfect 5 * 5 * 5 = 125 no need of cube to make hollow = ( 5 - 2 ) * ( 5 - 2 ) * ( 5 - 2 ) = 27 so total cube need to make hollow cube is = 125 - 27 = 98 answer : a | ['a ) 98', 'b ) 104', 'c ) 100', 'd ) 61', 'e ) 51'] | a | subtract(volume_cube(5), volume_cube(const_3)) | volume_cube(n0)|volume_cube(const_3)|subtract(#0,#1) | geometry |
the average of temperatures at noontime from monday to friday is 50 ; the lowest one is 40 , what is the possible maximum range of the temperatures ? | "there are 5 days so the sum of temperature can be 50 * 5 = 250 lowest is 40 . to find the maximum range we can say the temperature was the lowest for 4 of the 5 days so 4 * 40 = 160 . on the fifth day it is 250 - 160 = 90 range is therefore 90 - 40 = 50 answer d" | a ) 20 , b ) 25 , c ) 40 , d ) 50 , e ) 75 | d | subtract(subtract(multiply(50, add(const_2, const_3)), multiply(40, const_4)), 40) | add(const_2,const_3)|multiply(n1,const_4)|multiply(n0,#0)|subtract(#2,#1)|subtract(#3,n1)| | general |
a sun is divided among x , y and z in such a way that for each rupee x gets , y gets 45 paisa and z gets 30 paisa . if the share of y is rs . 36 , what is the total amount ? | "x : y : z = 100 : 45 : 30 20 : 9 : 6 9 - - - 36 35 - - - ? = > 140 answer : b" | a ) 166 , b ) 140 , c ) 178 , d ) 177 , e ) 169 | b | add(add(multiply(divide(const_100, 45), 36), multiply(divide(30, 45), 36)), 36) | divide(const_100,n0)|divide(n1,n0)|multiply(n2,#0)|multiply(n2,#1)|add(#2,#3)|add(n2,#4)| | general |
a retailer sold 2 houses at each $ 10000 . on 1 st house he got 30 % profit and on 2 nd house he loss 10 % then find his net profit or loss ? | a = 30 % b = - 10 % profit % = 30 - 10 + ( - 300 / 100 ) = 17 % answer is b | a ) 15 % , b ) 17 % , c ) 21 % , d ) 19 % , e ) 25 % | b | multiply(subtract(multiply(subtract(const_1, divide(10, const_100)), const_2), divide(10, const_100)), 10) | divide(n5,const_100)|subtract(const_1,#0)|multiply(#1,const_2)|subtract(#2,#0)|multiply(n5,#3) | gain |
the remainder when e = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . + 3 ^ 200 is divided 13 . | e = 1 + 3 + 3 ^ 2 + 3 ^ 3 + . . . . . . . . . . . . . . . . . . + 3 ^ 200 is a geometric progression having common ratio as ' 3 ' and number of terms as ' 201 ' . since sum to n terms in gp = a ( r ^ n - 1 ) / ( r - 1 ) where a = first term and r = common ration hence , 1 * ( 3 ^ 201 - 1 ) / ( 3 - 1 ) rem of ( 3 ^ 201 ... | a ) 12 , b ) 7 , c ) 0 , d ) 5 , e ) 3 | c | multiply(divide(add(200, const_1), const_3), add(add(reminder(1, 13), reminder(3, 13)), reminder(power(3, const_2), 13))) | add(n7,const_1)|power(n1,const_2)|reminder(n0,n8)|reminder(n1,n8)|add(#2,#3)|divide(#0,const_3)|reminder(#1,n8)|add(#4,#6)|multiply(#7,#5) | general |
today joelle opened an interest - bearing savings account and deposited $ 6000 . if the annual interest rate is 5 percent compounded interest , and she neither deposits nor withdraws money for exactly 2 years , how much money will she have in the account ? | interest for 1 st year = 6000 * 5 / 100 = 300 interest for 2 nd year = 6300 * 5 / 100 = 315 total = 6000 + 300 + 315 = 6615 answer : d | a ) $ 6715 , b ) $ 5615 , c ) $ 6415 , d ) $ 6615 , e ) $ 6315 | d | multiply(6000, power(add(const_1, divide(5, const_100)), const_2)) | divide(n1,const_100)|add(#0,const_1)|power(#1,const_2)|multiply(n0,#2) | gain |
a boat can travel with a speed of 24 km / hr in still water . if the speed of the stream is 4 km / hr , find the time taken by the boat to go 196 km downstream | "explanation : speed of the boat in still water = 24 km / hr speed of the stream = 4 km / hr speed downstream = ( 22 + 5 ) = 28 km / hr distance travelled downstream = 196 km time taken = distance / speed = 196 / 28 = 7 hours answer : option e" | a ) 3 hours , b ) 4 hours , c ) 5 hours , d ) 6 hours , e ) 7 hours | e | divide(196, add(24, 4)) | add(n0,n1)|divide(n2,#0)| | physics |
a train 900 m long is running at a speed of 78 km / hr . if it crosses a tunnel in 1 min , then the length of the tunnel is ? | "speed = 78 * 5 / 18 = 65 / 3 m / sec . time = 1 min = 60 sec . let the length of the train be x meters . then , ( 900 + x ) / 60 = 65 / 3 x = 400 m . answer : c" | a ) 200 m , b ) 379 m , c ) 400 m , d ) 589 m , e ) 600 m | c | divide(900, multiply(subtract(78, 1), const_0_2778)) | subtract(n1,n2)|multiply(#0,const_0_2778)|divide(n0,#1)| | physics |
jayant opened a shop investing rs . 30,000 . madhu joined him 2 months later , investing rs . 45,000 . they earned a profit of rs . 54,000 after completion of one year . what will be madhu ' s share of profit ? | "30,000 * 12 = 45,000 * 8 1 : 1 madhu ' s share = 1 / 2 * 54,000 i . e . rs . 27,000 answer : a" | a ) rs . 27,000 , b ) rs . 24,000 , c ) rs . 30,000 , d ) rs . 36,000 , e ) none of these | a | multiply(add(multiply(multiply(multiply(const_4, 2), multiply(add(2, const_3), 2)), const_100), multiply(multiply(add(2, const_3), const_100), const_100)), divide(divide(multiply(add(2, const_3), 2), 2), multiply(const_4, const_3))) | add(n1,const_3)|multiply(n1,const_4)|multiply(const_3,const_4)|multiply(#0,n1)|multiply(#0,const_100)|divide(#3,n1)|multiply(#1,#3)|multiply(#4,const_100)|divide(#5,#2)|multiply(#6,const_100)|add(#9,#7)|multiply(#10,#8)| | gain |
600 men have provisions for 20 days . if 200 more men join them , for how many days will the provisions last now ? | "600 * 20 = 800 * x x = 15 . answer : a" | a ) 15 , b ) 11 , c ) 10 , d ) 8 , e ) 7 | a | divide(multiply(20, 600), add(600, 200)) | add(n0,n2)|multiply(n0,n1)|divide(#1,#0)| | physics |
the hcf of two numbers is 52 and the other two factors of their lcm are 11 and 12 . what is the largest number . | "explanation : hcf of the two numbers = 52 hcf will be always a factor of lcm 52 is factor of lcm other two factors are 11 & 12 then the numbers are ( 52 * 11 ) and ( 52 x 12 ) = 572 and 624 answer : option a" | a ) 624 , b ) 450 , c ) 480 , d ) 504 , e ) 555 | a | multiply(52, 12) | multiply(n0,n2)| | other |
a whale goes on a feeding frenzy that lasts for 9 hours . for the first hour he catches and eats x kilos of plankton . in every hour after the first , it consumes 3 kilos of plankton more than it consumed in the previous hour . if by the end of the frenzy the whale will have consumed a whopping accumulated total 270 ki... | "if you list the amount eaten each hour , you ' ll get an equally spaced list , increasing by 3 each hour . in any equally spaced list , the median equals the mean . here , the mean is 270 / 9 = 30 , so the median is also 30 , and that is the amount eaten in the 5 th hour . we need to add 3 to find the total eaten in t... | a ) 33 , b ) 47 , c ) 50 , d ) 53 , e ) 62 | a | add(divide(270, 9), 3) | divide(n2,n0)|add(n1,#0)| | general |
in a garment industry , 12 men working 8 hours per day complete a piece of work in 10 days . to complete the same work in 8 days , working 5 hours a day , the number of men required is : | "explanation : let the required number of men be x . less days , more men ( indirect proportion ) more working hrs per day , less men ( indirect proportion ) days 8 : 10 working hrs 5 : 8 : : 12 : x = > 8 x 5 x x = 10 x 8 x 12 = > x = 10 x 8 x 12 / ( 8 x 5 ) = > x = 24 answer : a" | a ) 24 , b ) 5 , c ) 6 , d ) 8 , e ) 9 | a | divide(divide(multiply(multiply(12, 8), 10), 5), 8) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,n4)|divide(#2,n1)| | physics |
if the radii of umbra and penumbra cast by an object on a wall are of the ratio 2 : 6 , what is the area of the penumbra ring around the umbra of the latter β s radius is 40 cms ? | et the radius of umbra and penumbra are 2 k and 6 k . then as given radius of umbra = 40 cm so 2 k = 40 k = 20 radius of penumbra = 20 * 6 = 120 area of penumbra ring around the umbra = area of penumbra - area of umbra 22 / 7 * [ ( 120 ) ^ 2 - ( 40 ) ^ 2 ] = 40288.57 cm ^ 2 answer : a | ['a ) 40288.57 cm ^ 2', 'b ) 40388.57 cm ^ 2', 'c ) 40488.57 cm ^ 2', 'd ) 40588.57 cm ^ 2', 'e ) 40688.57 cm ^ 2'] | a | divide(divide(subtract(multiply(const_pi, multiply(multiply(6, 40), multiply(6, 40))), multiply(const_pi, multiply(multiply(40, const_2), multiply(40, const_2)))), 2), 2) | multiply(n1,n2)|multiply(n2,const_2)|multiply(#0,#0)|multiply(#1,#1)|multiply(#2,const_pi)|multiply(#3,const_pi)|subtract(#4,#5)|divide(#6,n0)|divide(#7,n0) | other |
if in a box of dimensions 6 m * 5 m * 4 m smaller boxes of dimensions 60 cm * 50 cm * 40 cm are kept in it , then what will be the maximum number of the small boxes that can be kept in it ? | 6 * 5 * 4 = 60 / 100 * 50 / 100 * 40 / 100 * x 1 = 1 / 10 * 1 / 10 * 1 / 10 * x = > x = 1000 answer : b | a ) 1098 , b ) 1000 , c ) 1628 , d ) 1098 , e ) 1094 | b | divide(multiply(multiply(6, 5), 4), multiply(multiply(divide(60, const_100), divide(50, const_100)), divide(40, const_100))) | divide(n5,const_100)|divide(n3,const_100)|divide(n4,const_100)|multiply(n0,n1)|multiply(n2,#3)|multiply(#1,#2)|multiply(#0,#5)|divide(#4,#6) | physics |
the area of a square field is 64816 sq m . how long will a lady take to cross the field diagonally at the rate of 7.2 km / hr ? | "area of a square field = 64816 sq m let the side of square = a a ^ 2 = 64816 = > a = 254.59 diagonal = ( 2 ) ^ ( 1 / 2 ) * a = 1.414 * 254.59 = 360 speed of lady = 7.2 km / hour = 7200 m / hour = 120 m / min time taken by lady to cross the field diagonally = 360 / 120 = 3 min answer d" | a ) 2 min , b ) 4 min , c ) 5 min , d ) 3 min , e ) 3.5 min | d | divide(64816, multiply(7.2, const_1000)) | multiply(n1,const_1000)|divide(n0,#0)| | geometry |
the ratio 15 : 25 expressed as percent equals to | "explanation : actually it means 15 is what percent of 25 , which can be calculated as , ( 15 / 25 ) * 100 = 15 * 4 = 60 answer : option b" | a ) 70 % , b ) 60 % , c ) 30 % , d ) 75 % , e ) none of above | b | multiply(divide(15, 25), const_100) | divide(n0,n1)|multiply(#0,const_100)| | general |
the ratio between the number of sheep and the number of horses at the stewar farm is 2 to 7 . if each of horse is fed 230 ounces of horse food per day and the farm needs a total 12880 ounces of horse food per day . what is number sheep in the form ? ? | "et no of sheep and horses are 2 k and 7 k no of horses = 12880 / 230 = 56 now 7 k = 56 and k = 8 no of sheep = ( 2 * 8 ) = 16 answer : b" | a ) 18 , b ) 16 , c ) 32 , d ) 56 , e ) 58 | b | multiply(divide(divide(12880, 230), 7), 2) | divide(n3,n2)|divide(#0,n1)|multiply(n0,#1)| | other |
the ratio of the arithmetic mean of two numbers to one of the numbers is 4 : 6 . what is the ratio of the smaller number to the larger number ? | for two numbers , the arithmetic mean is the middle of the two numbers . the ratio of the mean to the larger number is 4 : 6 , thus the smaller number must have a ratio of 2 . the ratio of the smaller number to the larger number is 2 : 6 = 1 : 3 . the answer is c . | a ) 1 : 5 , b ) 1 : 4 , c ) 1 : 3 , d ) 1 : 2 , e ) 2 : 3 | c | multiply(subtract(divide(4, 6), divide(const_1, const_2)), const_2) | divide(n0,n1)|divide(const_1,const_2)|subtract(#0,#1)|multiply(#2,const_2) | other |
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