Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
|---|---|---|---|---|---|---|---|
the mean of 50 observations was 36 . it was found later that an observation 28 was wrongly taken as 23 . the corrected new mean is : | "explanation : correct sum = ( 36 * 50 + 28 - 23 ) = 1825 . correct mean = = 1805 / 50 = 36.1 answer : c ) 36.1" | a ) 36.0 , b ) 36.5 , c ) 36.1 , d ) 36.2 , e ) 36.8 | c | divide(add(multiply(36, 50), subtract(subtract(50, const_2), 23)), 50) | multiply(n0,n1)|subtract(n0,const_2)|subtract(#1,n3)|add(#0,#2)|divide(#3,n0)| | general | C |
a store reported total sales of $ 385 million for february of this year . if the total sales for the same month last year was $ 340 million , approximately what was the percent increase in sales ? | "new value – old value ) / old value x 100 we are given : february sales this year = 385 million february sales last year = 340 million we need to determine the percent increase between sales from last year to sales this year . thus , the new value = 385 million and the old value = 340 million . let ’ s plug them into our percent change formula . ( new value – old value ) / old value x 100 [ ( 385 – 340 ) / 340 ] x 100 65 / 340 x 100 ≈ 13 % . the answer is e ." | a ) 2 % , b ) 17 % , c ) 20 % , d ) 65 % , e ) 13 % | e | multiply(divide(subtract(385, 340), 340), const_100) | subtract(n0,n1)|divide(#0,n1)|multiply(#1,const_100)| | general | E |
if the price of a tv is first decreased by 20 % and then increased by 30 % , then the net change in the price will be : | explanation : solution : let the original price be rs . 100 . new final price = 130 % of ( 80 % of 100 ) = rs . 130 / 100 * 80 / 100 * 100 = rs . 104 . . ' . increase = 4 % answer : a | a ) 4 % increase , b ) 10 % increase , c ) 10 % decrease , d ) 6 % increase , e ) none of these | a | subtract(divide(multiply(subtract(const_100, 20), add(const_100, 30)), const_100), const_100) | add(n1,const_100)|subtract(const_100,n0)|multiply(#0,#1)|divide(#2,const_100)|subtract(#3,const_100) | general | A |
a man sold 18 toys for rs . 27300 , gaining thereby the cost price of 3 toy find the cost price of a toy | "let the cost of one toy = x . then , cost of 18 toys = 18 x . gain = 3 x . sp of 18 toys = rs . 27300 . gain = sp â € “ cp 3 x = 27300 â € “ 18 x 21 x = 27300 x = rs . 1300 . answer : option e" | a ) s . 600 , b ) s . 800 , c ) s . 500 , d ) s . 900 , e ) s . 1300 | e | divide(27300, add(18, 3)) | add(n0,n2)|divide(n1,#0)| | gain | E |
if 40 % of a number is equal to two - thirds of another number , what is the ratio of the first number to the second ? | "suppose the first number is x and the second numbery . therefore , 40 % of x = 2 ⁄ 3 of y ∴ x ⁄ y = 2 ⁄ 3 × 100 ⁄ 40 = 5 ⁄ 3 answer d" | a ) 7 : 3 , b ) 3 : 7 , c ) 2 : 5 , d ) 5 : 3 , e ) none of these | d | divide(divide(const_1, const_4), divide(40, const_100)) | divide(const_1,const_4)|divide(n0,const_100)|divide(#0,#1)| | general | D |
a fruit seller sells mangoes at the rate of rs . 13 per kg and thereby loses 25 % . at what price per kg , he should have sold them to make a profit of 15 % ? | "solution 75 : 13 = 115 : x x = ( 13 ã — 115 / 75 ) = rs . 19.93 hence , s . p per kg = rs . 19.93 answer b" | a ) rs . 11.81 , b ) rs . 19.93 , c ) rs . 12.25 , d ) rs . 12.31 , e ) none | b | divide(multiply(13, add(const_100, 15)), subtract(const_100, 25)) | add(n2,const_100)|subtract(const_100,n1)|multiply(n0,#0)|divide(#2,#1)| | gain | B |
a man sold 18 toys for rs . 16800 , gaining thereby the cost price of 3 toy find the cost price of a toy | "let the cost of one toy = x . then , cost of 18 toys = 18 x . gain = 3 x . sp of 18 toys = rs . 16800 . gain = sp – cp 3 x = 16800 – 18 x 21 x = 16800 x = rs . 800 . answer : b" | a ) rs . 600 , b ) rs . 800 , c ) rs . 500 , d ) rs . 900 , e ) rs . 970 | b | divide(16800, add(18, 3)) | add(n0,n2)|divide(n1,#0)| | gain | B |
of the diplomats attending a summit conference , 22 speak french , 32 do not speak russian , and 20 % of the diplomats speak neither french nor russian . if 10 % of the diplomats speak both languages , then how many diplomats attended the conference ? | "{ total } = { french } + { russian } - { both } + { neither } { total } = 22 + ( { total } - 32 ) - ( 0.1 * { total } ) + 0.2 * { total } solving gives { total } = 100 . answer : c ." | a ) 72 , b ) 96 , c ) 100 , d ) 120 , e ) 150 | c | divide(subtract(32, 22), subtract(divide(20, const_100), divide(10, const_100))) | divide(n2,const_100)|divide(n3,const_100)|subtract(n1,n0)|subtract(#0,#1)|divide(#2,#3)| | other | C |
there are 8 students . 4 of them are men and 4 of them are women . if 4 students are selected from the 8 students . what is the probability that the number of men is equal to that of women ? | men = women means 2 men 2 women total such combinations = 4 c 2 * 4 c 2 = 4 ! / 2 ! . 2 ! * 4 ! / 2 ! . 2 ! = 6 * 6 total combinations = 8 c 4 = 8 * 7 * 6 * 5 / 4 * 3 * 2 * 1 = 70 so probability = 36 / 70 = 18 / 35 hence a | a ) 18 / 35 , b ) 16 / 35 , c ) 14 / 35 , d ) 13 / 35 , e ) 12 / 35 | a | divide(multiply(choose(4, divide(4, const_2)), choose(4, divide(4, const_2))), choose(8, 4)) | choose(n0,n1)|divide(n1,const_2)|choose(n1,#1)|multiply(#2,#2)|divide(#3,#0) | probability | A |
in a certain candy store , 22 % of the customers are caught sampling the candy and are charged a small fine , but 12 % of the customers who sample the candy are not caught . what is the total percent of all customers who sample candy ? | "since 12 % of the customers who sample the candy are not caught , then 88 % of the customers who sample the candy are caught : { % of customers who sample candy } * 0.88 = 0.22 ; { % of customers who sample candy } = 0.25 . answer : d ." | a ) 22 % , b ) 23 % , c ) 24 % , d ) 25 % , e ) 34 % | d | divide(22, divide(subtract(const_100, 12), const_100)) | subtract(const_100,n1)|divide(#0,const_100)|divide(n0,#1)| | gain | D |
each of the integers from 0 to 8 , inclusive , is written on a separate slip of blank paper and the 10 slips are dropped into hat . if the slips are then drawn one at a time without replacement , how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10 ? | you should consider the worst case scenario : if you pick numbers 0 , 1 , 2 , 3 , 4 and 5 then no two numbers out of these 6 add up to 10 . now , the next , 7 th number whatever it ' ll be ( 6,7 , or 8 ) will guarantee that two number will add up to 10 . so , 7 slips must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10 answer : c | a ) 8 , b ) 9 , c ) 7 , d ) 6 , e ) 5 | c | add(const_1, subtract(8, const_2)) | subtract(n1,const_2)|add(#0,const_1) | gain | C |
a , b and c started a business with capitals of rs . 8000 , rs . 10000 and rs . 12000 respectively . at the end of the year , the profit share of b is rs . 3000 . the difference between the profit shares of a and c is ? | "explanation : ratio of investments of a , b and c is 8000 : 10000 : 12000 = 4 : 5 : 6 and also given that , profit share of b is rs . 3000 = > 5 parts out of 15 parts is rs . 3000 now , required difference is 6 - 4 = 2 parts required difference = 2 / 5 ( 3000 ) = rs . 1200 answer : e" | a ) 621 , b ) 276 , c ) 236 , d ) 600 , e ) 1200 | e | multiply(subtract(divide(12000, 10000), divide(8000, 10000)), 3000) | divide(n2,n1)|divide(n0,n1)|subtract(#0,#1)|multiply(n3,#2)| | gain | E |
in what time will a railway train 30 m long moving at the rate of 36 kmph pass a telegraph post on its way ? | "t = 30 / 36 * 18 / 5 = 3 sec answer : c" | a ) 5 sec , b ) 4 sec , c ) 3 sec , d ) 6 sec , e ) 1 sec | c | divide(30, multiply(36, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | C |
a number when divided by 342 gives a remainder 38 , what remainder will be obtained by dividing the same number 21 ? | "explanation : 342 + 38 = 380 / 21 = 2 ( remainder ) answer : d" | a ) 7 , b ) 6 , c ) 9 , d ) 2 , e ) 3 | d | subtract(38, multiply(21, const_2)) | multiply(n2,const_2)|subtract(n1,#0)| | general | D |
if p = 125 × 243 × 16 / 405 , how many digits are in p ? | p = 125 * 243 * 16 / 405 p = 5 ^ 3 * 3 * 9 ^ 2 * 4 ^ 2 / ( 5 * 9 ^ 2 ) p = 5 ^ 2 * 3 * 4 ^ 2 p = 20 ^ 2 * 3 = 1200 answer d | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | d | divide(multiply(multiply(divide(243, divide(405, add(const_4, const_1))), 16), divide(125, 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 | D |
if x and y are the two digits of the number 653 xy such that this number is divisible by 80 , then x + y = ? | "80 = 2 x 5 x 8 since 653 xy is divisible by 2 and 5 both , so y = 0 . now , 653 x is divisible by 8 , so 13 x should be divisible by 8 . this happens when x = 6 . x + y = ( 6 + 0 ) = 6 . answer : a" | a ) 6 , b ) 4 , c ) 8 , d ) 8 , e ) none of these | a | divide(reminder(multiply(653, const_100), 80), const_10) | multiply(n0,const_100)|reminder(#0,n1)|divide(#1,const_10)| | general | A |
two positive integers differ by 4 , and sum of their reciprocals is 4 . then one of the numbers is | "algebraic approach : let n be the smaller integer = > 1 / n + 1 / ( n + 4 ) = 4 or ( ( n + 4 ) + n ) / n ( n + 4 ) = 4 or ( n ^ 2 + 4 n ) * 4 = 2 n + 4 or n = 2 as n can not be - negative solve for n = > n = 2 . hence , c" | a ) a ) 3 , b ) b ) 1 , c ) 2 , d ) d ) 21 , e ) e ) 28 | c | add(4, 4) | add(n1,n1)| | general | C |
in an election between two candidates , one got 55 % of the total valid votes , 20 % of the votes were invalid . if the total number of votes was 7500 , what was the number of valid votes that the other candidate got ? | "explanation : total number of votes = 7500 given that 20 % of percentage votes were invalid = > valid votes = 80 % total valid votes = 7500 × ( 80 / 100 ) 1 st candidate got 55 % of the total valid votes . hence the 2 nd candidate should have got 45 % of the total valid votes = > valid votes that 2 nd candidate got = total valid votes × ( 45 / 100 ) = 7500 × ( 80 / 100 ) × ( 45 / 100 ) = 75 × 45 × 45 = 75 × 4 × 9 = 300 × 9 = 2700 answer : option b" | a ) 2800 , b ) 2700 , c ) 2100 , d ) 2500 , e ) none of these | b | multiply(multiply(subtract(const_1, divide(20, const_100)), subtract(const_1, divide(55, const_100))), 7500) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|subtract(const_1,#1)|multiply(#2,#3)|multiply(n2,#4)| | gain | B |
suresh started a business , investing rs . 18000 . after 3 months and 4 months respectively , rohan and sudhir joined him with capitals of 12000 and 9000 . at the end of the year the total profit was rs . 3960 . what is the difference between rohan ’ s and sudhir ’ s share in the profit ? | "suresh : rohan : sudhir ratio of their investments = 18000 × 12 : 12000 × 9 : 9000 × 8 = 6 : 3 : 2 the difference between rohan ’ s and sudhir ’ s share = 1 share : . i . e . = rs . 3960 × 1 / 11 = rs . 360 . d" | a ) s . 345 , b ) s . 350 , c ) s . 352 , d ) s . 360 , e ) s . 368 | d | multiply(subtract(multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))), divide(3960, add(add(18000, multiply(12000, divide(subtract(multiply(3, 4), 3), multiply(3, 4)))), multiply(9000, divide(subtract(multiply(3, 4), 4), multiply(3, 4)))))) | multiply(n1,n2)|subtract(#0,n1)|subtract(#0,n2)|divide(#1,#0)|divide(#2,#0)|multiply(n3,#3)|multiply(n4,#4)|add(n0,#5)|subtract(#5,#6)|add(#7,#6)|divide(n5,#9)|multiply(#10,#8)| | gain | D |
the number of diagonals of a polygon of n sides is given by the formula c = n ( n - 3 ) / 2 . if a polygon has twice as many diagonals as sides , how many sides does it have ? | c = n ( n - 3 ) c = 2 * n 2 n = n ( n - 3 ) = > 2 = n - 3 = > n = 5 answer b | a ) 3 , b ) 5 , c ) 6 , d ) 7 , e ) 8 | b | add(3, 2) | add(n0,n1) | general | B |
in a group of 22 boys , 13 play basketball and 15 play football . 3 of the boys play neither . how many play both ? | 18 boys play something . if b play both , then 13 + 15 - b = 18 answer is a | a ) 18 , b ) 20 , c ) 21 , d ) 22 , e ) 25 | a | subtract(add(13, 15), add(subtract(subtract(22, 3), 15), subtract(subtract(22, 3), 13))) | add(n1,n2)|subtract(n0,n3)|subtract(#1,n2)|subtract(#1,n1)|add(#2,#3)|subtract(#0,#4) | other | A |
the contents of a certain box consist of 48 apples and 30 kiwis . how many kiwis must be added to the box so that exactly 30 % of the pieces of fruit in the box will be apples ? | "apple = ( apple + kiwi + x ) * 0.3 48 = ( 30 + 48 + x ) * 0.3 x = 82 answer : c" | a ) 90 , b ) 72 , c ) 82 , d ) 54 , e ) 65 | c | subtract(add(48, 30), divide(48, divide(30, const_100))) | add(n0,n1)|divide(n2,const_100)|divide(n0,#1)|subtract(#0,#2)| | general | C |
the length of a rectangle is increased by 20 % and its breadth is decreased by 20 % . what is the effect on its area ? | "100 * 100 = 10000 120 * 80 = 9600 answer : b" | a ) 1288 , b ) 9600 , c ) 1000 , d ) 10000 , e ) 2887 | b | multiply(add(20, const_100), subtract(const_100, 20)) | add(n0,const_100)|subtract(const_100,n1)|multiply(#0,#1)| | geometry | B |
a bakery has 6 employees . it pays annual salaries of $ 15,000 to each of 2 employees , $ 17,000 to 1 employee , and $ 18,000 to each of the remaining 3 employees . which number is closest to the average ( arithmetic mean ) annual salary of these employees ? | "average = 2 ( 15000 ) + ( 17000 ) + 3 ( 18000 ) / 6 = $ 16,833 the answer is c ." | a ) $ 16,200 , b ) $ 16,500 , c ) $ 16,800 , d ) $ 17,000 , e ) $ 17,400 | c | divide(add(add(multiply(add(add(add(const_10, const_10), 1), 1), 3), multiply(subtract(add(const_10, const_10), 1), 2)), add(add(const_10, const_10), 1)), 6) | add(const_10,const_10)|add(#0,n4)|subtract(#0,n4)|add(#1,n4)|multiply(#2,n2)|multiply(#3,n6)|add(#5,#4)|add(#6,#1)|divide(#7,n0)| | general | C |
a certain sum is invested at simple interest at 15 % p . a . for two years instead of investing at 12 % p . a . for the same time period . therefore the interest received is more by rs . 720 . find the sum ? | "let the sum be rs . x . ( x * 15 * 2 ) / 100 - ( x * 12 * 2 ) / 100 = 720 = > 30 x / 100 - 24 x / 100 = 720 = > 6 x / 100 = 720 = > x = 12000 . answer : d" | a ) s . 7000 , b ) s . 9000 , c ) s . 14000 , d ) s . 12000 , e ) s . 27000 | d | divide(720, divide(multiply(subtract(15, 12), const_2), const_100)) | subtract(n0,n1)|multiply(#0,const_2)|divide(#1,const_100)|divide(n2,#2)| | gain | D |
( 208 % of 1265 ) ÷ 6 = ? | "explanation : ? = ( 208 x 1265 / 100 ) ÷ 6 = 263120 / 600 = 439 answer : option b" | a ) a ) 125 , b ) b ) 439 , c ) c ) 225 , d ) d ) 375 , e ) e ) 524 | b | divide(multiply(divide(208, const_100), 1265), 6) | divide(n0,const_100)|multiply(n1,#0)|divide(#1,n2)| | general | B |
two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 21 seconds . the ratio of their speeds is ? | "let the speeds of the two trains be x m / sec and y m / sec respectively . then , length of the first train = 27 x meters , and length of the second train = 17 y meters . ( 27 x + 17 y ) / ( x + y ) = 21 = = > 27 x + 17 y = 21 x + 21 y = = > 6 x = 4 y = = > x / y = 2 / 3 . answer : c" | a ) 3 / 6 , b ) 3 / 2 , c ) 2 / 3 , d ) 3 / 5 , e ) 5 / 2 | c | divide(subtract(27, 21), subtract(21, 17)) | subtract(n0,n2)|subtract(n2,n1)|divide(#0,#1)| | physics | C |
in a class of 40 students , 10 did not opt for math . 15 did not opt for science and 2 did not opt for either . how many students of the class opted for both math and science ? | 40 students total 10 did not opt for math 15 did not opt for science 2 did not opt for either total of 30 students in math and 13 did not opt for sci but did for math 30 - 13 = 7 7 students of the class opted for both math and science answer : c . 7 | a ) 13 , b ) 15 , c ) 7 , d ) 17 , e ) 18 | c | subtract(subtract(40, 10), subtract(subtract(40, 15), 2)) | subtract(n0,n1)|subtract(n0,n2)|subtract(#1,n3)|subtract(#0,#2) | other | C |
80 is what percent of 160 ? | 80 / 160 × 100 = 50 % answer : d | a ) 35 % , b ) 40 % , c ) 45 % , d ) 50 % , e ) 55 % | d | multiply(divide(80, 160), const_100) | divide(n0,n1)|multiply(#0,const_100)| | gain | D |
in one hour , a boat goes 11 km along the stream and 5 km against the stream . the sped of the boat in still water ( in km / hr ) is : | "solution speed in still water = 1 / 2 ( 11 + 5 ) km / hr = 8 kmph . answer c" | a ) 3 , b ) 5 , c ) 8 , d ) 9 , e ) 10 | c | divide(add(11, 5), const_2) | add(n0,n1)|divide(#0,const_2)| | gain | C |
in the coordinate plane , points ( x , 5 ) and ( 7 , y ) are on line k . if line k passes through the origin and has slope 5 / 7 , then x + y = | "line k passes through the origin and has slope 5 / 7 means that its equation is y = 5 / 7 * x . thus : ( x , 5 ) = ( 7,5 ) and ( 7 , y ) = ( 7,5 ) - - > x + y = 7 + 5 = 12 . answer : e" | a ) 13 , b ) 14 , c ) 9 , d ) 10 , e ) 12 | e | multiply(multiply(7, 7), divide(5, 7)) | divide(n0,n3)|multiply(n1,n3)|multiply(#0,#1)| | general | E |
income and expenditure of a person are in the ratio 10 : 8 . if the income of the person is rs . 10000 , then find his savings ? | "let the income and the expenditure of the person be rs . 10 x and rs . 8 x respectively . income , 10 x = 10000 = > x = 1000 savings = income - expenditure = 10 x - 8 x = 2 x = 2 ( 1000 ) so , savings = rs . 2000 . answer : c" | a ) 2003 , b ) 2002 , c ) 2000 , d ) 2001 , e ) 2004 | c | subtract(10000, multiply(divide(8, 10), 10000)) | divide(n1,n0)|multiply(n2,#0)|subtract(n2,#1)| | other | C |
ifaequals the sum of the even integers from 2 to 40 , inclusive , andbequals the sum of the odd integers from 1 to 39 , inclusive , what is the value of a - b ? | this is a solution from beatthegmat : even numbers : ( 40 - 2 ) / 2 + 1 = 20 even integers . ( 40 + 2 ) / 2 = 21 is the average of the even set . sum = avg * ( # of elements ) = 21 * 20 = 420 = a odd numbers : ( 39 - 1 ) / 2 + 1 = 20 odd integers . ( 39 + 1 ) / 2 = 20 is the average of the odd set . sum = avg * ( # of elements ) = 20 * 20 = 400 = b a - b = 420 - 400 = 20 . answer : d | a ) 1 , b ) 10 , c ) 19 , d ) 20 , e ) 21 | d | subtract(multiply(divide(40, const_2), add(divide(40, const_2), 1)), multiply(divide(add(39, 1), const_2), add(divide(subtract(39, 1), 2), 1))) | add(n2,n3)|divide(n1,const_2)|subtract(n3,n2)|add(n2,#1)|divide(#2,n0)|divide(#0,const_2)|add(n2,#4)|multiply(#3,#1)|multiply(#6,#5)|subtract(#7,#8) | general | D |
a , b , and c are integers and a < b < c . s is the set of all integers from a to b , inclusive . q is the set of all integers from b to c , inclusive . the median of set s is ( 3 / 4 ) * b . the median of set q is ( 7.5 / 8 ) * c . if r is the set of all integers from a to c , inclusive , what fraction of c is the median of set r ? | "the answer isc : 11 / 16 . the key to this problem is remembering that the median for a consecutive set of numbers is equivalent to its mean . for example , the mean and median of a set consisting of x , x + 1 , x + 2 , . . . , y will always be ( x + y ) / 2 . for set s , consisting of numbers ( a , a + 1 , . . . , b ) , the median is given to be 3 / 4 * b : ( a + b ) / 2 = ( 3 / 4 ) * b a = b / 2 for set q , consisting of numbers ( b , b + 1 , . . . , c ) , the median is given to be 7.5 / 8 * c : ( b + c ) / 2 = ( 7.5 / 8 ) * c b = ( 7 / 8 ) * c for set r , consisting of numbers ( a , a + 1 , . . . c ) , the median needs to be found : a = b / 2 = ( 7 / 8 * c ) / 2 = ( 7 / 16 ) * c median = ( a + c ) / 2 = ( 7 / 16 * c + c ) / 2 = ( 23 / 16 ) * c / 2 = ( 23 / 32 ) * c ( answer d )" | a ) 3 / 8 , b ) 1 / 2 , c ) 11 / 16 , d ) 23 / 32 , e ) 3 / 4 | d | divide(add(const_1, divide(subtract(multiply(divide(7.5, 8), const_2), const_1), divide(const_1, subtract(multiply(divide(3, 4), const_2), const_1)))), const_2) | divide(n2,n3)|divide(n0,n1)|multiply(#0,const_2)|multiply(#1,const_2)|subtract(#2,const_1)|subtract(#3,const_1)|divide(const_1,#5)|divide(#4,#6)|add(#7,const_1)|divide(#8,const_2)| | general | D |
tanks c and b are each in the shape of a right circular cylinder . the interior of tank c has a height of 10 meters and a circumference of 8 meters , and the interior of tank b has a height of 8 meters and a circumference of 10 meters . the capacity of tank c is what percent of the capacity of tank b ? | b . for c , r = 8 / 2 pi . its capacity = ( 4 pi ) ^ 2 * 10 = 160 pi for b , r = 10 / pi . its capacity = ( 5 pi ) ^ 2 * 8 = 200 pi c / b = 160 pi / 200 pi = 0.8 | ['a ) 75 %', 'b ) 80 %', 'c ) 100 %', 'd ) 120 %', 'e ) 125 %'] | b | multiply(divide(volume_cylinder(divide(8, const_2), 10), volume_cylinder(divide(10, const_2), 8)), const_100) | divide(n1,const_2)|divide(n0,const_2)|volume_cylinder(#0,n0)|volume_cylinder(#1,n1)|divide(#2,#3)|multiply(#4,const_100) | physics | B |
without stoppages , a train travels certain distance with an average speed of 100 km / h , and with stoppages , it covers the same distance with an average speed of 95 km / h . how many minutes per hour the train stops ? | due to stoppages , it covers 5 km less . time taken to cover 5 km = 5 â „ 100 h = 1 â „ 20 h = 1 â „ 20 ã — 60 min = 3 min answer c | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | subtract(const_60, multiply(divide(95, 100), const_60)) | divide(n1,n0)|multiply(#0,const_60)|subtract(const_60,#1) | physics | C |
a 1200 m long train crosses a tree in 120 sec , how much time will i take to pass a platform 500 m long ? | "l = s * t s = 1200 / 120 s = 10 m / sec . total length ( d ) = 1700 m t = d / s t = 1700 / 10 t = 170 sec answer : b" | a ) 266 sec , b ) 170 sec , c ) 776 sec , d ) 166 sec , e ) 997 sec | b | divide(add(1200, 500), divide(1200, 120)) | add(n0,n2)|divide(n0,n1)|divide(#0,#1)| | physics | B |
after decreasing 24 % in the price of an article costs rs . 532 . find the actual cost of an article ? | "cp * ( 76 / 100 ) = 532 cp = 7 * 100 = > cp = 700 answer : c" | a ) 118 , b ) 677 , c ) 700 , d ) 2688 , e ) 1991 | c | divide(532, subtract(const_1, divide(24, const_100))) | divide(n0,const_100)|subtract(const_1,#0)|divide(n1,#1)| | gain | C |
a certain car can travel 40 minutes on a gallon of gasoline at 50 miles per hour . if the car had started with a full tank and had 8 gallons of gasoline left in its tank at the end , then what percent of the tank was used to travel 80 miles at 60 mph ? | "let , tank capacity = t gallon used fuel = ( t - 8 ) gallons distance travelled ( @ 50 miles / hr ) = 80 miles distance travelled in 1 gallon = distance travelled in 40 mins ( @ 50 miles / hr ) = ( 50 / 50 ) * 40 = 40 miles fuel used to travel 80 miles = ( 80 / 40 ) = 2 gallon i . e . used fuel = ( t - 8 ) = 2 gallon i . e . t = 10 gallons i . e . used fuel = ( 2 / 10 ) * 100 = 20 % b" | a ) 15 % , b ) 20 % , c ) 25 % , d ) 30 % , e ) 40 % | b | divide(divide(multiply(40, 50), multiply(80, 60)), add(divide(multiply(40, 50), multiply(80, 60)), 8)) | multiply(n0,n1)|multiply(n3,n4)|divide(#0,#1)|add(n2,#2)|divide(#2,#3)| | physics | B |
ab + cd = jjj , where ab and cd are two - digit numbers and jjj is a 3 digit number ; a , b , c , and d are distinct positive integers . in the addition problem above , what is the value of c ? | ab and cd are two digit integers , their sum can give us only one three digit integer of a kind of jjj it ' s 111 . so , a = 1 . 1 b + cd = 111 now , c can not be less than 9 , because no to digit integer with first digit 1 ( mean that it ' s < 20 ) can be added to two digit integer less than 90 to have the sum 111 ( if cd < 90 meaning c < 9 cd + 1 b < 111 ) - - > c = 9 answer : d . | a ) 1 , b ) 3 , c ) 7 , d ) 9 , e ) can not be determined | d | multiply(3, 3) | multiply(n0,n0) | general | D |
a certain protective additive increases from 45 days to 60 days the time between required maintenance checks on an industrial vehicle . by what percent is the time between maintenance checks increased by using the additive ? | "general formula for percent increase or decrease , ( percent change ) : percent = change / original ∗ 100 so , the time between maintenance checks increased by 60 − 45 / 45 ∗ 100 = 33 1 / 3 answer : b ." | a ) 25 % , b ) 33 1 / 3 % , c ) 50 % , d ) 66 2 / 3 % , e ) 75 % | b | multiply(divide(subtract(60, 45), 45), const_100) | subtract(n1,n0)|divide(#0,n0)|multiply(#1,const_100)| | general | B |
if a rectangular billboard has an area of 130 square feet and a perimeter of 46 feet , what is the length of each of the longer sides ? | "this question can be solved algebraically or by testing the answers . we ' re told that a rectangle has an area of 130 and a perimeter of 46 . we ' re asked for the length of one of the longer sides of the rectangle . since the answers are all integers , and the area is 130 , the shorter side will almost certainly be closer to 10 ( since 10 x 10 = 100 , but we ' re not dealing with a square ) . answer b ( 7 ) does not divide evenly into 130 , if . . . the shorter side = 10 . . . the area = 130 / 10 = 13 = the longer side perimeter = 10 + 10 + 13 + 13 = 46 d" | a ) 4 , b ) 7 , c ) 8 , d ) 13 , e ) 26 | d | divide(subtract(divide(46, const_2), sqrt(subtract(power(divide(46, const_2), const_2), multiply(const_4, 130)))), const_2) | divide(n1,const_2)|multiply(n0,const_4)|power(#0,const_2)|subtract(#2,#1)|sqrt(#3)|subtract(#0,#4)|divide(#5,const_2)| | geometry | D |
the average of runs of a cricket player of 10 innings was 37 . how many runs must he make in his next innings so as to increase his average of runs by 4 ? | "explanation : average = total runs / no . of innings = 37 so , total = average x no . of innings = 37 x 10 = 370 . now increase in avg = 4 runs . so , new avg = 37 + 4 = 41 runs total runs = new avg x new no . of innings = 41 x 11 = 451 runs made in the 11 th inning = 451 - 370 = 81 answer : b" | a ) 76 , b ) 81 , c ) 85 , d ) 87 , e ) 89 | b | subtract(multiply(add(10, const_1), add(4, 37)), multiply(10, 37)) | add(n0,const_1)|add(n1,n2)|multiply(n0,n1)|multiply(#0,#1)|subtract(#3,#2)| | general | B |
find the principle on a certain sum of money at 5 % per annum for 2 2 / 5 years if the amount being rs . 1344 ? | "1344 = p [ 1 + ( 5 * 12 / 5 ) / 100 ] p = 1200 answer : c" | a ) 1000 , b ) 2217 , c ) 1200 , d ) 2777 , e ) 2991 | c | divide(1344, add(divide(multiply(divide(add(multiply(2, 5), 2), 5), 5), const_100), const_1)) | multiply(n1,n3)|add(n1,#0)|divide(#1,n3)|multiply(n0,#2)|divide(#3,const_100)|add(#4,const_1)|divide(n4,#5)| | general | C |
a salesperson received a commission of 3 percent of the sale price for each of the first 100 machines that she sold and 4 percent of the sale price for each machine that she sold after the first 100 . if the sale price of each machine was $ 10,000 and the salesperson received a $ 40,000 commission , how many machines did she sell ? | "first 100 machines = 3 % commission = 0.03 * 100 * 10000 = 30000 commission from sale of next machines = 40000 - 30000 = 10000 so 25 more machines . . total = 125 machines imo e . . ." | a ) 90 , b ) 103 , c ) 105 , d ) 115 , e ) 125 | e | add(100, divide(subtract(multiply(multiply(multiply(add(4, 3), multiply(3, const_2)), 100), multiply(add(4, const_1), const_2)), multiply(multiply(multiply(100, 100), divide(3, 100)), 100)), multiply(multiply(100, 100), divide(4, 100)))) | add(n0,n2)|add(const_1,n2)|divide(n0,n1)|divide(n2,n1)|multiply(const_2,n0)|multiply(n1,n1)|multiply(#0,#4)|multiply(#1,const_2)|multiply(#2,#5)|multiply(#3,#5)|multiply(#6,n1)|multiply(n1,#8)|multiply(#10,#7)|subtract(#12,#11)|divide(#13,#9)|add(n1,#14)| | gain | E |
if each year the population of the country grows by 30 % , how many years will elapse before the population of the country doubles ? | "till year 2000 , population is 100 . year 2001 : population becomes 130 . . . . . . . . . . . . . 1 year elapsed year 2002 : population becomes 169 . . . . . . . . . . . . . 2 year elapsed year 2004 : population > 200 . . . . . . . . . . . . . . . . . . 3 year elapsed answer : d" | a ) 2 , b ) 4 , c ) 5 , d ) 3 , e ) 7 | d | add(const_4, const_4) | add(const_4,const_4)| | gain | D |
an employee ’ s annual salary was increased $ 25,000 . if her new annual salary now equals $ 90,000 , what was the percent increase ? | new annual salary = $ 90,000 salary increase = $ 25,000 . original salary = $ 90,000 - $ 25,000 . = $ 75,000 % increase = ( $ 25,000 / $ 65,000 ) * 100 = 38.46 % hence d . | a ) 15 % , b ) 162 ⁄ 3 % , c ) 20 % , d ) 38.46 % , e ) 24 % | d | multiply(subtract(divide(multiply(subtract(const_100, const_10), const_1000), subtract(multiply(subtract(const_100, const_10), const_1000), multiply(multiply(const_0_25, const_100), const_1000))), const_1), const_100) | multiply(const_0_25,const_100)|subtract(const_100,const_10)|multiply(#1,const_1000)|multiply(#0,const_1000)|subtract(#2,#3)|divide(#2,#4)|subtract(#5,const_1)|multiply(#6,const_100) | general | D |
on a partly cloudy day , derek decides to walk back from work . when it is sunny , he walks at a speed of s miles / hr ( s is an integer ) and when it gets cloudy , he increases his speed to ( s + 1 ) miles / hr . if his average speed for the entire distance is 2.8 miles / hr , what fraction q of the total distance did he cover while the sun was shining on him ? | if s is an integer and we know that the average speed is 2.8 , s must be = 2 . that meanss + 1 = 3 . this implies that the ratio of time for s = 2 is 1 / 4 of the total time . the formula for distance / rate is d = rt . . . so the distance travelled when s = 2 is 2 t . the distance travelled for s + 1 = 3 is 3 * 4 t or 12 t . therefore , total distance covered while the sun was shining over him is q = 2 / 14 = 1 / 7 . answer : e | a ) 1 / 4 , b ) 4 / 5 , c ) 1 / 5 , d ) 1 / 6 , e ) 1 / 7 | e | subtract(divide(lcm(const_2, const_3), 2.8), const_2) | lcm(const_2,const_3)|divide(#0,n1)|subtract(#1,const_2)| | general | E |
what profit percent is made by selling an article at a certain price , if by selling at 2 / 3 rd of that price , there would be a loss of 18 % ? | sp 2 = 2 / 3 sp 1 cp = 100 sp 2 = 82 2 / 3 sp 1 = 82 sp 1 = 123 100 - - - 23 = > 23 % answer : e | a ) 20 % , b ) 26 % , c ) 42 % , d ) 27 % , e ) 23 % | e | subtract(divide(subtract(const_100, 18), divide(2, 3)), const_100) | divide(n0,n1)|subtract(const_100,n2)|divide(#1,#0)|subtract(#2,const_100) | gain | E |
the product of a and b is equal to 11 more than twice the sum of a and b . if b = 7 , what is the value of b - a ? | "given : ab = 2 ( a + b ) + 11 and b = 7 hence 7 a = 2 a + 14 + 11 5 a = 25 , a = 5 ( a potential trap . do not get excited here . we need to find b - a , not a ) b - a = 7 - 5 = 2 correct option : a" | a ) 2 , b ) 5 , c ) 7 , d ) 24 , e ) 35 | a | subtract(7, divide(add(multiply(7, const_2), 11), subtract(7, const_2))) | multiply(n1,const_2)|subtract(n1,const_2)|add(n0,#0)|divide(#2,#1)|subtract(n1,#3)| | general | A |
the speed at which a woman can row a boat in still water is 30 kmph . if he rows downstream , where the speed of current is 6 kmph , what time will he take to cover 100 metres ? | "speed of the boat downstream = 30 + 6 = 36 kmph = 36 * 5 / 18 = 10 m / s hence time taken to cover 100 m = 100 / 10 = 10 seconds . answer : c" | a ) 8 , b ) 9 , c ) 10 , d ) 11 , e ) 12 | c | divide(100, multiply(add(30, 6), const_0_2778)) | add(n0,n1)|multiply(#0,const_0_2778)|divide(n2,#1)| | physics | C |
when 15 is divided by integer u , the remainder is 4 . for how many values of u is this be true ? | when 15 is divided by u , the remainder is 4 i . e . 4 mangoes left over after grouping , so u must be greater than 4 . it also means that 11 is completely divisible by u . factors of 11 are 1 and 11 . out of these , u could be 11 . answer ( a ) | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | a | divide(add(subtract(15, 4), 4), 15) | subtract(n0,n1)|add(n1,#0)|divide(#1,n0) | general | A |
the area of a triangle will be when a = 5 m , b = 7 m , c = 8 m , a , b , c being lengths of respective sides ? | s = ( 5 + 7 + 8 ) / 2 = e answer : e | a ) 3 , b ) 6 , c ) 4 , d ) 9 , e ) 10 | e | divide(add(add(5, 7), 8), 7) | add(n0,n1)|add(n2,#0)|divide(#1,n1)| | geometry | E |
if both 5 ^ 3 and 3 ^ 4 are factors of n x ( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) x ( 10 ^ 2 ) , what is the smallest possible positive value of n ? | "( 2 ^ 5 ) x ( 12 ^ 2 ) x ( 7 ^ 3 ) x ( 10 ) has two appearances of 3 ( in 12 ^ 2 ) and two appearances of 5 ( in 10 ^ 2 ) . thus n must include at least 3 ^ 2 * 5 = 45 the answer is b ." | a ) 15 , b ) 45 , c ) 75 , d ) 125 , e ) 150 | b | add(add(add(add(add(multiply(multiply(5, 2), 3), multiply(multiply(5, 2), 3)), multiply(multiply(5, 2), 3)), 2), const_4), const_4) | multiply(n0,n7)|multiply(n1,#0)|add(#1,#1)|add(#2,#1)|add(n7,#3)|add(#4,const_4)|add(#5,const_4)| | other | B |
the average of first five prime numbers greater than 17 is ? | "19 + 23 + 29 + 31 + 37 = 139 / 5 = 27.8 answer : e" | a ) 32.2 , b ) 32.98 , c ) 32.3 , d ) 32.8 , e ) 27.8 | e | add(17, const_1) | add(n0,const_1)| | general | E |
one half of a two digit number exceeds its one third by 6 . what is the sum of the digits of the number ? | "explanation : x / 2 – x / 3 = 6 = > x = 6 3 + 6 = 9 b )" | a ) 7 , b ) 9 , c ) 11 , d ) 12 , e ) 15 | b | add(reminder(multiply(6, const_4), const_10), const_1) | multiply(n0,const_4)|reminder(#0,const_10)|add(#1,const_1)| | general | B |
a car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 60 kilometers per hour . at what speed , in kilometers per hour , is the car traveling ? | "d 75 * t = 1 km = > t = 1 / 60 km / h v * ( t + 2 / 3600 ) = 1 v ( 1 / 60 + 2 / 3600 ) = 1 = > v = 58 km / h" | a ) 59 , b ) 57 , c ) 59.5 , d ) 58 , e ) 55.5 | d | divide(1, divide(add(multiply(const_3600, divide(1, 60)), 2), const_3600)) | divide(n1,n3)|multiply(#0,const_3600)|add(n0,#1)|divide(#2,const_3600)|divide(n1,#3)| | physics | D |
two employees a and b are paid a total of rs . 570 per week by their employer . if a is paid 150 percent of the sum paid to b , how much is b paid per week ? | "let the amount paid to a per week = x and the amount paid to b per week = y then x + y = 570 but x = 150 % of y = 150 y / 100 = 15 y / 10 ∴ 15 y / 10 + y = 570 ⇒ y [ 15 / 10 + 1 ] = 570 ⇒ 25 y / 10 = 570 ⇒ 25 y = 5700 ⇒ y = 5700 / 25 = rs . 228 d )" | a ) s . 130 , b ) s . 140 , c ) s . 150 , d ) s . 228 , e ) s . 282 | d | divide(570, add(divide(150, const_100), const_1)) | divide(n1,const_100)|add(#0,const_1)|divide(n0,#1)| | general | D |
find the value of ( √ 1.21 ) / ( √ 0.81 ) + ( √ 1.44 ) / ( √ 0.49 ) is | "( √ 1.21 ) / ( √ 0.81 ) + ( √ 1.44 ) / ( √ 0.49 ) 11 / 9 + 12 / 7 = > ( 77 + 108 ) / 63 = > 185 / 63 answer is e ." | a ) 195 / 63 , b ) 145 / 63 , c ) 155 / 63 , d ) 125 / 63 , e ) 185 / 63 | e | add(divide(sqrt(1.21), sqrt(0.81)), divide(sqrt(1.44), sqrt(0.49))) | sqrt(n0)|sqrt(n1)|sqrt(n2)|sqrt(n3)|divide(#0,#1)|divide(#2,#3)|add(#4,#5)| | general | E |
a certain class of students is being divided into teams . the class can either be divided into 10 teams with an equal number of players on each team or 24 teams with an equal number of players on each team . what is the lowest possible number of students in the class ? | "let total no of students in the class be n so , we are told that n is divisible by both 10 24 so , lets find the least common multiple of 10 24 , ie 120 so our answer is ( c ) 120" | a ) 6 , b ) 36 , c ) 120 , d ) 60 , e ) 72 | c | lcm(10, 24) | lcm(n0,n1)| | general | C |
two trains of equal length , running with the speeds of 60 and 40 kmph , take 60 seconds to cross each other while they are running in the same direction . what time will they take to cross each other if they are running in opposite directions ? | "rs = 60 - 40 = 20 * 5 / 18 = 100 / 18 t = 50 d = 60 * 100 / 18 = 1000 / 3 rs = 60 + 40 = 100 * 5 / 18 t = 1000 / 3 * 18 / 500 = 12 sec answer : c" | a ) 10 sec , b ) 16 sec , c ) 12 sec , d ) 67 sec , e ) 13 sec | c | multiply(multiply(multiply(const_0_2778, subtract(60, 40)), 60), inverse(multiply(const_0_2778, add(60, 40)))) | add(n0,n1)|subtract(n0,n1)|multiply(#0,const_0_2778)|multiply(#1,const_0_2778)|inverse(#2)|multiply(n2,#3)|multiply(#4,#5)| | physics | C |
the ratio of radius of a circle and the side of a square is 2 : 3 . find the ratio of their areas : | "radius / side = 2 / 3 â ‡ ’ area of circle / area of square = 4 / 9 answer : b" | a ) 2 : 1 , b ) 4 : 9 , c ) 8 : 77 , d ) 4 : 121 , e ) none | b | power(divide(2, 3), 2) | divide(n0,n1)|power(#0,n0)| | geometry | B |
there is a 20 % chance sandy will visit china this year , while there is a 70 % chance that she will visit malaysia this year . what is the probability that sandy will visit either china or malaysia this year , but not both ? | "p ( china and not malaysia ) = 0.2 * 0.3 = 0.06 p ( malaysia and not china ) = 0.7 * 0.8 = 0.56 total probability = 0.06 + 0.56 = 0.62 = 62 % the answer is d ." | a ) 44 % , b ) 48 % , c ) 54 % , d ) 62 % , e ) 70 % | d | divide(add(20, 70), multiply(multiply(const_5, const_5), const_4)) | add(n0,n1)|multiply(const_5,const_5)|multiply(#1,const_4)|divide(#0,#2)| | probability | D |
3 men and 8 women complete a task in same time as 6 men and 2 women do . how much fraction of work will be finished in same time if 2 men and 2 women will do that task . | "3 m + 8 w = 6 m + 2 w 3 m = 6 w 1 m = 2 w therefore 3 m + 8 w = 14 w 2 m + 2 w = 6 w answer is 6 / 14 = 3 / 7 answer : a" | a ) 3 / 7 , b ) 3 / 10 , c ) 3 / 18 , d ) 3 / 16 , e ) 3 / 11 | a | divide(add(multiply(divide(subtract(8, 2), subtract(6, 3)), 2), 2), add(multiply(6, divide(subtract(8, 2), subtract(6, 3))), 2)) | subtract(n1,n3)|subtract(n2,n0)|divide(#0,#1)|multiply(n4,#2)|multiply(n2,#2)|add(n5,#3)|add(n3,#4)|divide(#5,#6)| | physics | A |
find the average of all the numbers between 6 and 36 which are divisible by 5 . | "solution average = ( 10 + 15 + 20 + 25 + 30 + 35 ) / 6 ) = 135 / 6 = 22.5 answer e" | a ) 18 , b ) 20 , c ) 24 , d ) 30 , e ) 22.5 | e | divide(add(add(6, const_4), subtract(36, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general | E |
the monthly rent of a shop of dimension 20 feet × 15 feet is rs . 3600 . what is the annual rent per square foot of the shop ? | "sol . monthly rent per square feet = 3600 / ( 20 * 15 ) = 12 & annual rent per square feet = 12 * 12 = 144 answer : b" | a ) 43 , b ) 144 , c ) 68 , d ) 87 , e ) 92 | b | multiply(add(const_10, const_2), divide(3600, rectangle_area(20, 15))) | add(const_10,const_2)|rectangle_area(n0,n1)|divide(n2,#1)|multiply(#0,#2)| | geometry | B |
on a certain day , orangeade was made by mixing a certain amount of orange juice with an equal amount of water . on the next day , orangeade was made by mixing the same amount of orange juice with twice the amount of water . on both days , all the orangeade that was made was sold . if the revenue from selling the orangeade was the same for both days and if the orangeade was sold at $ 0.60 per glass on the first day , what was the price w per glass on the second day ? | "on the first day 1 unit of orange juice and 1 unit of water was used to make 2 units of orangeade ; on the second day 1 unit of orange juice and 2 units of water was used to make 3 units of orangeade ; so , the ratio of the amount of orangeade made on the first day to the amount of orangeade made on the second day is 2 to 3 . naturally the ratio of the # of glasses of orangeade made on the first day to the # of glasses of orangeade made on the second day is 2 to 3 . we are told thatthe revenue from selling the orangeade was the same for both daysso the revenue from 2 glasses on the first day equals to the revenue from 3 glasses on the second day . say the price of the glass of the orangeade on the second day was $ x then 2 * 0.6 = 3 * x - - > x = $ 0.4 . answer : d ." | a ) $ 015 , b ) $ 0.20 , c ) $ 0.30 , d ) $ 0.40 , e ) $ 0.45 | d | divide(multiply(add(const_1, const_1), 0.60), add(const_1, const_2)) | add(const_1,const_1)|add(const_1,const_2)|multiply(n0,#0)|divide(#2,#1)| | general | D |
a train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively . what is the length of the train ? | explanation : let x is the length of the train in meter and v is its speed in kmph x / 9 = ( v - 2 ) ( 10 / 36 ) - - - ( 1 ) x / 10 = ( v - 4 ) ( 10 / 36 ) - - - ( 2 ) dividing equation 1 with equation 2 10 / 9 = ( v - 2 ) / ( v - 4 ) = > 10 v - 40 = 9 v - 18 = > v = 22 substituting in equation 1 , x / 9 = 200 / 36 = > x = 9 × 200 / 36 = 50 m answer : option c | a ) 62 m , b ) 54 m , c ) 50 m , d ) 55 m , e ) 56 m | c | multiply(9, subtract(subtract(multiply(divide(multiply(4, const_1000), const_3600), 10), multiply(divide(multiply(2, const_1000), const_3600), 9)), divide(multiply(2, const_1000), const_3600))) | multiply(n1,const_1000)|multiply(n0,const_1000)|divide(#0,const_3600)|divide(#1,const_3600)|multiply(n3,#2)|multiply(n2,#3)|subtract(#4,#5)|subtract(#6,#3)|multiply(n2,#7) | physics | C |
sonika bought a v . c . r . at the list price of 18,800 . if the rate of sales tax was 8 % , find the amount she had to pay for purchasing the v . c . r . | "sol . list price of v . c . r . = 18,800 rate of sales tax = 8 % ∴ sales tax = 8 % of 18,800 = 8 ⁄ 100 × 18800 = 1504 so , total amount which sonika had to pay for purchasing the v . c . r . = 18,800 + 1504 = 20,304 . answer a" | a ) 20,304 , b ) 19,780 , c ) 19,680 , d ) 19,380 , e ) none of these | a | floor(divide(divide(multiply(add(multiply(multiply(add(multiply(add(const_1, const_4), const_2), 8), const_100), multiply(add(const_1, const_4), const_2)), multiply(add(const_1, const_4), const_100)), add(const_100, 8)), const_100), multiply(multiply(add(const_1, const_4), const_100), const_2))) | add(const_1,const_4)|add(n1,const_100)|multiply(#0,const_2)|multiply(#0,const_100)|add(n1,#2)|multiply(#3,const_2)|multiply(#4,const_100)|multiply(#6,#2)|add(#7,#3)|multiply(#8,#1)|divide(#9,const_100)|divide(#10,#5)|floor(#11)| | gain | A |
johnny travels a total of one hour to and from school . on the way there he jogs at 5 miles per hour and on the return trip he gets picked up by the bus and returns home at 21 miles per hour . how far is it to the school ? | "answer : c ) 6.5 miles . average speed for round trip = 2 * a * b / ( a + b ) , where a , b are speeds so , average speed was = 2 * 5 * 21 / ( 5 + 21 ) = 6.5 m / hr the distance between schoolhome should be half of that . ie . 6.5 miles answer c" | a ) 2 miles , b ) 4 miles , c ) 6.5 miles , d ) 8 miles , e ) 10 miles | c | multiply(divide(const_1, add(divide(const_1, 5), divide(const_1, 21))), const_1_6) | divide(const_1,n0)|divide(const_1,n1)|add(#0,#1)|divide(const_1,#2)|multiply(#3,const_1_6)| | physics | C |
a group of hikers is planning a trip that will take them up a mountain using one route and back down using another route . they plan to travel down the mountain at a rate of one and a half times the rate they will use on the way up , but the time each route will take is the same . if they will go up the mountain at a rate of 3 miles per day and it will take them two days , how many miles long is the route down the mountain ? | "on the way down , the rate is 1.5 * 3 = 4.5 miles per day . the distance of the route down the mountain is 2 * 4.5 = 9 miles . the answer is c ." | a ) 7 , b ) 8 , c ) 9 , d ) 10 , e ) 11 | c | multiply(multiply(3, const_2), divide(const_3, const_2)) | divide(const_3,const_2)|multiply(n0,const_2)|multiply(#0,#1)| | physics | C |
a tank can supply water to a village for 20 days . if a leak at the bottom of the tank drains out 10 liters per day , the supply lasts for 15 days only . for how many days will the supply last if the leak drains out 20 liters per day ? | "losing 10 liters per day results in a loss of 150 liters in 15 days . so , those 150 liters were for 5 days , making daily consumption of the village 30 liters per day . thus the capacity of the tank is 30 * 20 = 600 liters . losing 20 liters plus 30 liters gives 50 liters per day . at this rate the supply will last 600 / 50 = 12 days . the answer is d ." | a ) 9 , b ) 10 , c ) 11 , d ) 12 , e ) 14 | d | divide(multiply(divide(multiply(10, 15), subtract(20, 15)), 20), add(20, divide(multiply(10, 15), subtract(20, 15)))) | multiply(n1,n2)|subtract(n0,n2)|divide(#0,#1)|add(n3,#2)|multiply(n0,#2)|divide(#4,#3)| | physics | D |
the sum and the product of two numbers are 24 and 23 respectively , the difference of the number is ? | explanation : x + y = 24 xy = 23 ( x - y ) 2 = ( x + y ) 2 - 4 xy ( x - y ) 2 = 576 - 92 = > ( x - y ) = 22 answer : e | a ) 1 , b ) 6 , c ) 7 , d ) 8 , e ) 22 | e | sqrt(subtract(subtract(power(24, const_2), multiply(23, const_2)), multiply(23, const_2))) | multiply(n1,const_2)|power(n0,const_2)|subtract(#1,#0)|subtract(#2,#0)|sqrt(#3) | general | E |
10 is subtracted from 50 % of a number , the result is 25 . find the number ? | "( 50 / 100 ) * x â € “ 10 = 25 5 x = 350 x = 70 answer : b" | a ) 75 , b ) 70 , c ) 35 , d ) 170 , e ) 50 | b | divide(add(10, 25), divide(50, const_100)) | add(n0,n2)|divide(n1,const_100)|divide(#0,#1)| | gain | B |
a train of 24 carriages , each of 60 meters length , when an engine also of 60 meters length is running at a speed of 60 kmph . in what time will the train cross a bridge 1.5 km long ? | "d = 25 * 60 + 1500 = 3000 m t = 3000 / 60 * 18 / 5 = 180 sec = 3 mins answer : b" | a ) 6 , b ) 3 , c ) 4 , d ) 9 , e ) 2 | b | add(divide(multiply(add(24, const_1), 60), const_1000), 1.5) | add(n0,const_1)|multiply(n1,#0)|divide(#1,const_1000)|add(n4,#2)| | physics | B |
when jessica withdrew $ 400 from her bank account , her account balance decreased by 2 / 5 . if she deposits an amount equal to 1 / 4 of the remaining balance , what will be the final balance in her bank account ? | as per the question 400 = 2 a / 5 thus - a which is the total amount = 1000 the amount thus left = 600 she then deposited 1 / 4 of 600 = 150 total amount in her account = 750 answer e | a ) 300 , b ) 375 , c ) 400 , d ) 500 , e ) 750 | e | multiply(subtract(divide(400, subtract(const_1, divide(const_3, 5))), 400), add(1, divide(1, 4))) | divide(n3,n4)|divide(const_3,n2)|add(n3,#0)|subtract(const_1,#1)|divide(n0,#3)|subtract(#4,n0)|multiply(#2,#5) | general | E |
sam invested rs . 15000 @ 10 % per annum for one year . if the interest is compounded half - yearly , then the amount received by sam at the end of the year will be ? | "p = rs . 15000 ; r = 10 % p . a . = 5 % per half - year ; t = 1 year = 2 half - year amount = [ 15000 * ( 1 + 5 / 100 ) 2 ] = ( 15000 * 21 / 20 * 21 / 20 ) = rs . 16537.50 answer : c" | a ) rs . 16,500 , b ) rs . 16 , 525.50 , c ) rs . 16 , 537.50 , d ) rs . 18,150 , e ) rs . 19,150 | c | multiply(power(add(divide(divide(10, const_2), const_100), const_1), const_2), 15000) | divide(n1,const_2)|divide(#0,const_100)|add(#1,const_1)|power(#2,const_2)|multiply(n0,#3)| | gain | C |
the maximum number of students among them 890 pens and 630 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : | "explanation : required number of students = h . c . f of 890 and 630 = 10 . answer : a" | a ) 10 , b ) 910 , c ) 1001 , d ) 1911 , e ) none of these | a | gcd(890, 630) | gcd(n0,n1)| | general | A |
( 17 ) 5.25 x ( 17 ) ? = 178 | "solution let ( 17 ) 5.25 * ( 17 ) x = 178 . then , ( 17 ) 5.25 + x = ( 17 ) 8 . ∴ 5.25 + x = 8 ⇔ x = ( 8 - 5.25 ) ⇔ x = 2.75 answer b" | a ) 2.29 , b ) 2.75 , c ) 4.25 , d ) 4.5 , e ) none of these | b | multiply(17, 17) | multiply(n0,n2)| | general | B |
the area of a parallelogram is 162 sq m and its altitude is twice the corresponding base . then the length of the base is ? | "2 x * x = 162 = > x = 9 answer : b" | a ) 8 , b ) 9 , c ) 7 , d ) 62 , e ) 2 | b | sqrt(divide(162, const_2)) | divide(n0,const_2)|sqrt(#0)| | geometry | B |
one - third of 1200 is what percent of 236 ? | "answer let one - third of 1200 is n % of 236 . ∵ 1200 / 3 = ( n x 236 ) / 100 ∴ n = ( 400 x 100 ) / 236 = 169.49 correct option : a" | a ) 169.49 , b ) 30 , c ) 120.32 , d ) 125.32 , e ) 15.152 | a | multiply(const_100, divide(divide(1200, const_3), 236)) | divide(n0,const_3)|divide(#0,n1)|multiply(#1,const_100)| | gain | A |
nicky and cristina are running a race . since cristina is faster than nicky , she gives him a 48 meter head start . if cristina runs at a pace of 5 meters per second and nicky runs at a pace of only 3 meters per second , how many seconds will nicky have run before cristina catches up to him ? | "used pluging in method say t is the time for cristina to catch up with nicky , the equation will be as under : for nicky = n = 3 * t + 48 for cristina = c = 5 * t @ t = 24 , n = 120 c = 120 right answer ans : c" | a ) 15 seconds , b ) 18 seconds , c ) 24 seconds , d ) 30 seconds , e ) 45 seconds | c | divide(48, subtract(5, 3)) | subtract(n1,n2)|divide(n0,#0)| | physics | C |
15 lts are taken of from a container full of liquid a and replaced with liquid b . again 15 more lts of the mixture is taken and replaced with liquid b . after this process , if the container contains liquid a and b in the ratio 9 : 16 , what is the capacity of the container w ? | "if you have a 37.5 liter capacity , you start with 37.5 l of a and 0 l of b . 1 st replacement after the first replacement you have 37.5 - 15 = 22.5 l of a and 15 l of b . the key is figuring out how many liters of a and b , respectively , are contained in the next 15 liters of mixture to be removed . the current ratio of a to total mixture is 22.5 / 37.5 ; expressed as a fraction this becomes ( 45 / 2 ) / ( 75 / 2 ) , or 45 / 2 * 2 / 75 . canceling the 2 s and factoring out a 5 leaves the ratio as 9 / 15 . note , no need to reduce further as we ' re trying to figure out the amount of a and b in 15 l of solution . 9 / 15 of a means there must be 6 / 15 of b . multiply each respective ratio by 15 to get 9 l of a and 6 l of b in the next 15 l removal . final replacement the next 15 l removal means 9 liters of a and 6 liters of b are removed and replaced with 15 liters of b . 22.5 - 9 = 13.5 liters of a . 15 liters of b - 6 liters + 15 more liters = 24 liters of b . test to the see if the final ratio = 9 / 16 ; 13.5 / 24 = ( 27 / 2 ) * ( 1 / 24 ) = 9 / 16 . choice c is correct ." | a ) a : 45 , b ) b : 25 , c ) c : 37.5 , d ) d : 36 , e ) e : 42 | c | divide(15, subtract(const_1, sqrt(divide(9, add(9, 16))))) | add(n2,n3)|divide(n2,#0)|sqrt(#1)|subtract(const_1,#2)|divide(n0,#3)| | general | C |
a work which could be finished in 10 days was finished 3 days earlier after 10 more men joined . the number of men employed was ? | "x - - - - - - - 10 ( x + 10 ) - - - - 6 x * 10 = ( x + 10 ) 6 x = 15 \ answer : a" | a ) 15 , b ) 20 , c ) 88 , d ) 71 , e ) 11 | a | divide(multiply(multiply(3, const_2), 10), subtract(10, multiply(3, const_2))) | multiply(n1,const_2)|multiply(n2,#0)|subtract(n0,#0)|divide(#1,#2)| | physics | A |
the tax on a commodity is diminished by 15 % and its consumption increased by 10 % . the effect on revenue is ? | "100 * 100 = 10000 85 * 110 = 9350 - - - - - - - - - - - 10000 - - - - - - - - - - - 650 100 - - - - - - - - - - - ? = > 6.5 % decrease answer : b" | a ) 8.5 , b ) 6.5 , c ) 5.5 , d ) 5.6 , e ) 4.4 | b | subtract(const_100, multiply(multiply(add(const_1, divide(10, const_100)), subtract(const_1, divide(15, const_100))), const_100)) | divide(n1,const_100)|divide(n0,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|multiply(#4,const_100)|subtract(const_100,#5)| | general | B |
if x / y = 3 / z , then 4 x ^ 2 = | "this question is most easily solved by isolating y in the equation and substituting into the expression 4 x ² : x / y = 3 / z x = 3 y / z if we substitute 3 y / z into the expression for x , we get : 4 ( 3 y / z ) ² = 4 ( 9 y ² / z ² ) = 36 y ² / z ² . the correct answer is choice ( d ) ." | a ) y / z , b ) xy , c ) y ² / z ² , d ) 36 y ² / z ² , e ) 15 y ² / z ² | d | divide(add(3, 4), subtract(3, 4)) | add(n0,n1)|subtract(n0,n1)|divide(#0,#1)| | general | D |
a work as fast as b . if b can complete a work in 12 days independently , the number of days in which a and b can together finish the work in ? | ratio of rates of working of a and b = 2 : 1 ratio of times taken = 1 : 2 a ' s 1 day work = 1 / 6 b ' s 1 day work = 1 / 12 a + b 1 day work = 1 / 6 + 1 / 12 = 3 / 12 = 1 / 4 a and b can finish the work in 4 days answer is c | a ) 2 days , b ) 3 days , c ) 4 days , d ) 5 days , e ) 6 days | c | inverse(add(inverse(12), multiply(const_2, inverse(12)))) | inverse(n0)|multiply(#0,const_2)|add(#0,#1)|inverse(#2) | physics | C |
after paying a 10 percent tax on all income over $ 3000 , a person had a net income of $ 12000 . what was the income before taxes ? | let x be the income over 3000 then ( x - x / 10 ) + 3000 = 12000 = > x = 10000 therefore income before taxes = 10000 + 3000 = 13000 option ( b ) | a ) $ 13300 , b ) $ 13000 , c ) $ 12900 , d ) $ 10000 , e ) $ 9000 | b | divide(subtract(12000, multiply(divide(10, const_100), 3000)), subtract(const_1, divide(10, const_100))) | divide(n0,const_100)|multiply(n1,#0)|subtract(const_1,#0)|subtract(n2,#1)|divide(#3,#2) | gain | B |
a and b can do a piece of work in 60 days and 60 days respectively . they work together for 10 days and b leaves . in how many days the whole work is completed ? | "explanation : ( a + b ) ’ s 10 days work = 10 [ 1 / 60 + 1 / 60 ] = 10 [ 1 + 1 / 60 ] = 1 / 3 a complete remaining work in 1 / 3 * 60 = 20 total work = 10 + 20 = 30 days answer : option a" | a ) 30 days , b ) 35 days , c ) 40 days , d ) 45 days , e ) 50 days | a | divide(subtract(10, add(divide(10, 60), divide(10, 60))), divide(10, 60)) | divide(n2,n1)|divide(n2,n0)|add(#0,#1)|subtract(n2,#2)|divide(#3,#0)| | physics | A |
an industrial loom weaves 0.128 metres of cloth every second . approximately , how many seconds will it take for the loom to weave 26 metre of cloth ? | "explanation : let the time required by x seconds . then , more cloth means more time ( direct proportion ) so , 0.128 : 1 : : 26 : x = > x = { \ color { blue } \ frac { 26 \ times 1 } { 0.128 } } = > x = 203.13 so time will be approx 203 seconds answer : e" | a ) 175 seconds , b ) 195 seconds , c ) 155 seconds , d ) 115 seconds , e ) 203 seconds | e | divide(26, 0.128) | divide(n1,n0)| | physics | E |
how many pieces of 0.35 meteres can be cut from a rod 12.5 meteres long | "explanation : we need so simple divide 42.5 / 0.85 , = ( 1250 / 35 ) = 35 option c" | a ) 30 , b ) 40 , c ) 35 , d ) 60 , e ) 70 | c | divide(12.5, 0.35) | divide(n1,n0)| | physics | C |
a two - digit number is such that the product of the digits is 8 . when 18 is added to the number , then the digits are reversed . the number is ? | let the ten ' s and unit digit be x and 8 respectively . x then , 10 x + 8 + 18 = 10 x 8 + x x x 10 x 2 + 8 + 18 x = 80 + x 2 9 x 2 + 18 x - 72 = 0 x 2 + 2 x - 8 = 0 ( x + 4 ) ( x - 2 ) = 0 x = 2 option b | a ) 1 , b ) 2 , c ) 6 , d ) 8 , e ) 9 | b | divide(18, subtract(const_10, const_1)) | subtract(const_10,const_1)|divide(n1,#0) | general | B |
a train running at a speed of 36 kmph crosses an electric pole in 12 seconds . in how much time will it cross a 320 m long platform ? | "a 44 min let the length of the train be x m . when a train crosses an electric pole , the distance covered is its own length . so , x = 12 * 36 * 5 / 18 m = 120 m . time taken to cross the platform = ( 120 + 320 ) / 36 * 5 / 18 = 44 min ." | a ) 44 min , b ) 55 min , c ) 47 min , d ) 67 min , e ) 45 min | a | divide(add(320, multiply(multiply(const_0_2778, 36), 12)), multiply(const_0_2778, 36)) | multiply(n0,const_0_2778)|multiply(n1,#0)|add(n2,#1)|divide(#2,#0)| | physics | A |
in a certain school , 25 % of students are below 8 years of age . the number of students above 8 years of age is 2 / 3 of the number of students of 8 years of age which is 36 . what is the total number of students in the school ? | "explanation : let the number of students be x . then , number of students above 8 years of age = ( 100 - 25 ) % of x = 75 % of x . 75 % of x = 36 + 2 / 3 of 36 75 / 100 x = 60 x = 80 . answer : option b" | a ) 72 , b ) 80 , c ) 120 , d ) 150 , e ) 100 | b | divide(add(36, multiply(36, divide(2, 3))), subtract(const_1, divide(25, const_100))) | divide(n3,n4)|divide(n0,const_100)|multiply(n6,#0)|subtract(const_1,#1)|add(n6,#2)|divide(#4,#3)| | general | B |
the sum of two consecutive integers is 41 . find the numbers . | "n + ( n + 1 ) = 41 2 n + 1 = 41 2 n = 40 n = 20 answer : d" | a ) 17 , 18 , b ) 7 , 8 , c ) 5 , 6 , d ) 20,21 , e ) 8 , 9 | d | add(add(power(add(add(divide(subtract(subtract(41, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(41, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(41, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(41, const_10), const_2), const_4), const_2), const_2))) | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics | D |
each month , after jill pays for rent , utilities , food , and other necessary expenses , she has one fifth of her net monthly salary left as discretionary income . of this discretionary income , she puts 30 % into a vacation fund , 20 % into savings , and spends 35 % on eating out and socializing . this leaves her with $ 102 dollar , which she typically uses for gifts and charitable causes . what is jill ’ s net monthly salary ? | "let x be the monthly salary 15 % of 1 / 5 * x = 102 x = 3400 answer a" | a ) $ 3400 , b ) $ 3200 , c ) $ 6000 , d ) $ 6400 , e ) $ 9600 | a | multiply(multiply(const_100, add(const_3, const_2)), divide(102, subtract(const_100, add(add(30, 20), 35)))) | add(n0,n1)|add(const_2,const_3)|add(n2,#0)|multiply(#1,const_100)|subtract(const_100,#2)|divide(n3,#4)|multiply(#5,#3)| | gain | A |
the average weight of 8 persons increases by 1.5 kg . if a person weighing 65 kg is replaced by a new person , what could be the weight of the new person ? | "total weight increases = 8 × 1.5 = 12 kg so the weight of new person = 65 + 12 = 77 kg answer b" | a ) 76 kg , b ) 77 kg , c ) 76.5 kg , d ) data inadequate , e ) none of these | b | add(65, multiply(8, 1.5)) | multiply(n0,n1)|add(n2,#0)| | general | B |
three walls have wallpaper covering a combined area of 300 square meters . by overlapping the wallpaper to cover a wall with an area of 180 square meters , the area that is covered by exactly two layers of wallpaper is 38 square meters . what is the area that is covered with three layers of wallpaper ? | 300 - 180 = 120 sq m of the wallpaper overlaps ( in either two layers or three layers ) if 36 sq m has two layers , 120 - 38 = 82 sq m of the wallpaper overlaps in three layers . 82 sq m makes two extra layers hence the area over which it makes two extra layers is 41 sq m . answer ( e ) . | ['a ) 5 square meters', 'b ) 36 square meters', 'c ) 42 square meters', 'd ) 83.3 square meters', 'e ) 41 square meters'] | e | divide(subtract(subtract(300, 180), 38), const_2) | subtract(n0,n1)|subtract(#0,n2)|divide(#1,const_2) | geometry | E |
sonika bought a v . c . r . at the list price of 18,700 . if the rate of sales tax was 8 % , find the amount she had to pay for purchasing the v . c . r . | "sol . list price of v . c . r . = 18,700 rate of sales tax = 8 % ∴ sales tax = 8 % of 18,700 = 8 ⁄ 100 × 18700 = 1496 so , total amount which sonika had to pay for purchasing the v . c . r . = 18,700 + 1496 = 20,196 . answer a" | a ) 20,196 , b ) 19,780 , c ) 19,680 , d ) 19,380 , e ) none of these | a | floor(divide(divide(multiply(add(multiply(multiply(add(multiply(add(const_1, const_4), const_2), 8), const_100), multiply(add(const_1, const_4), const_2)), multiply(add(const_1, const_4), const_100)), add(const_100, 8)), const_100), multiply(multiply(add(const_1, const_4), const_100), const_2))) | add(const_1,const_4)|add(n1,const_100)|multiply(#0,const_2)|multiply(#0,const_100)|add(n1,#2)|multiply(#3,const_2)|multiply(#4,const_100)|multiply(#6,#2)|add(#7,#3)|multiply(#8,#1)|divide(#9,const_100)|divide(#10,#5)|floor(#11)| | gain | A |
if w is the set of all the integers between 59 and 99 , inclusive , that are either multiples of 3 or multiples of 2 or multiples of both , then w contains how many numbers ? | multiples of 2 from 59 to 99 = multiples of 2 from 1 to 99 - multiples of 2 from 1 to 58 = [ 99 / 2 ] - [ 58 / 2 ] = 49 - 29 = 20 multiples of 3 from 59 to 99 = multiples of 3 from 1 to 99 - multiples of 3 from 1 to 58 = [ 99 / 3 ] - [ 58 / 3 ] = 33 - 169 = 14 multiples of 2 and 3 bothi . e . 6 from 59 to 99 = multiples of 6 from 1 to 99 - multiples of 6 from 1 to 58 = [ 99 / 6 ] - [ 58 / 6 ] = 16 - 9 = 7 these 8 numbers have been counted twice in both the above calculation while calculating multiples of 2 and 3 i . e . total numbers in w = 20 + 14 - 7 = 27 answer option a | a ) 27 , b ) 32 , c ) 33 , d ) 34 , e ) 35 | a | subtract(add(floor(divide(subtract(99, 59), 3)), divide(subtract(99, 59), 2)), floor(divide(subtract(99, 59), multiply(const_2, const_3)))) | multiply(const_2,const_3)|subtract(n1,n0)|divide(#1,n3)|divide(#1,n2)|divide(#1,#0)|floor(#3)|floor(#4)|add(#2,#5)|subtract(#7,#6) | other | A |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.